Thermodynamics

THERMODYNAMICS

For

MECHANICAL ENGINEERING

THERMODYNAMICS

SYLLABUS Thermodynamic systems and processes; properties of pure substances, behaviour of ideal and real gases; zeroth and first laws of thermodynamics, calculation of work and heat in various processes; second law of thermodynamics; thermodynamic property charts and tables, availability and irreversibility; thermodynamic relations. Power Engineering: Air and gas compressors; vapour and gas power cycles, concepts of regeneration and reheat. I.C. Engines: Air-standard Otto, Diesel and dual cycles. Refrigeration and air-conditioning: Vapour and gas refrigeration and heat pump cycles; properties of moist air, psychrometric chart, basic psychrometric processes.

ANALYSIS OF GATE PAPERS

Exam Year 1 Mark Ques. 2 Mark Ques. Total 2003 5 6 17 2004 3 6 15 2005 2 7 16 2006 - 7 14 2007 2 6 14 2008 2 9 20 2009 3 4 11 2010 2 4 10 2011 2 7 16 2012 1 2 5 2013 1 4 9 2014 Set-1 2 4 10 2014 Set-2 2 2 6 2014 Set-3 1 3 7 2014 Set-4 - 4 8

2015 Set-1 4 3 10

2015 Set-2 3 4 11

2015 Set-3 2 3 8 2016 Set-1 3 3 9 2016 Set-2 3 2 7 2016 Set-3 1 2 5 2017 Set-1 1 4 9 2017 Set-2 2 4 10 2018 Set-1 2 3 8 2018 Set-2 2 4 10

CONTENTS

Topics Page No

1. BASIC CONCEPTS OF THERMODYNAMICS

1.1 Thermodynamic 01 1.2 Microscopic Approach and Macroscopic Approach 01 1.3 System 01 1.4 Property of System 02 1.5 State 03 1.6 Thermodynamic Equilibrium 04 1.7 Zeroth Law of Thermodynamics 04 1.8 Ideal gas Equation and Process 06 1.9 Work Transfer 08 1.10 Heat Transfer 10 1.11 Example 11

2. FIRST LAW OF THERMODYNAMICS

2.1 First Law of Thermodynamics for a Cycle 13 2.2 Application of First Law of Thermodynamics 13 2.3 Meyer’s Formula 14 2.4 Heat Transfer in Different Process 15 2.5 Free Expansion 15 2.6 First Law of Thermodynamics for Flow Process 16 2.7 First Law of Thermodynamics for Variable Flow Process 17 2.8 Examples 18

3. SECOND LAW OF THERMODYNAMICS - ENTROPY

3.1 Thermal Reservoir 23 3.2 Equivalence of Kelvin Plank and Clausius Statement 25 3.3 Carnot Cycle 25 3.4 Refrigeration & Heat Pump working on Reversed Carnot Cycle 26 3.5 Clausius Theorem 26 3.6 Entropy 28 3.7 Combined First Law and Second Law of Thermodynamics 28 3.8 Entropy change for an ideal gas 29 3.9 Entropy change for finite body 29 3.10 Slope of isobaric process & isochoric process on T-S Diagram 29 3.11 Example 29

4. AVAILABLE ENERGY & THERMODYNAMIC RELATIONS

4.1 Available Energy 34

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 4.2 Unavailable Energy 35 4.3 Loss in Available Energy 35 4.4 Available Function 35 4.5 Irreversibility 36 4.6 Exact Differential Equations 36 4.7 Maxwell’s Equations 36 4.8 T – DS Equations 36 4.9 Energy Equation 38 4.10 Enthalpy Equation 38 4.11 Difference in Heat Capacities 38 4.12 Joule Kelvin Coefficient 39 4.13 Examples 39

5. PROPERTIES OF PURE SUBSTANCE & GAS MIXTURE

5.1 Pure substance 44 5.2 P – V diagram of Pure substance 44 5.3 Triple Point 45 5.4 Gibbs phase rule 45 5.5 Phase Change of Pure substance 45 5.6 T – S diagram of pure substance 46 5.7 Properties of Pure Substance 47 5.8 Specific volume, Enthalpy and Entropy of different Phases 48 5.9 Equation of State 49 5.10 Properties of Gas mixture 51 5.11 Examples 51

6. REFRIGERATION

6.1 Introduction 56 6.2 Air Standard Refrigeration Cycle 56 6.3 Refrigerator working on reversed brayton Cycle 58 6.4 Vapour Compression Refrigeration Cycle 60 6.5 Effect of Parameters on COP of Vapour Compression Refrigeration cycle 62 6.6 Vapour Absorption Refrigeration Systems 63 6.7 Refrigerant 65 6.8 Designation of refrigerants 66 6.9 Examples 66

7. PSYCHOMETRY

7.1 Properties of Moist Air 72 7.2 Psychometric Chart 73 7.3 Significance, Different Lines on Psychometric Chart 74 7.4 Different Process on Psychometric Chart 78 7.5 Examples 79

8.1 Stirling cycle 84 8.2 Ericson cycle 84 8.3 Gas Power Plant 84 8.4 Steam Power Plant 88 8.5 Nozzle and Diffuser 90 8.6 Compressor 92 8.7 Steam turbine 93 8.8 Examples 93

9. INTERNAL COMBUSTION ENGINE

9.1 Heat Engine 101 9.2 Applications of IC Engines 101 9.3 Classifications of IC Engines 101 9.4 Basic components of IC Engine 101 9.5 Terms used in internal combustion engine 102 9.6 Difference between four stroke and two stroke Engine 103 9.7 Performance Parameters 103 9.8 Air standard cycle and efficiency 103 9.9 Comparison of Otto, Diesel, dual Cycle 106 9.10 Examples 107

10. GATE QUESTIONS 112

1.1 THERMODYNAMICS 1.3 SYSTEM

Thermodynamics is the science of energy A quantity of matter or region in space transfer and its effect on the physical upon which attention is concentrated is properties of the substances. known as thermodynamic System. The Some students have difficulties with system and the boundary are specified by thermodynamics because of global nature the analyst, these are not specified in a of its applicability. Most students are used problem statement. to courses that focus on a few specific topics like statics, dynamics, fluid flows etc. 1.3.1 SURROUNDING all deals with the limited range of topics. Thermodynamics, on the other hands, deals Anything external to the system is called as with many issues that are inherent in every surrounding or Environment. The engineering system. A thermodynamic combination of the system and surrounding analysis can span from analyzing a makes universe. It means in the universe if complete power plant to analyzing the anything is specified by the analyst as a smallest component in power plant. system the things except that system are We begin by introducing some basic considered as the surrounding. thermodynamic terms and definitions. Some of these terms are already in our everyday vocabulary as a result of the broad result of thermodynamics concepts in non engineering concepts (for example cooling process of tea in the open environment is a thermodynamic process).

1.2 MICROSCOPIC APPROACH AND Fig.1.1system, surrounding and boundary MACROSCOPIC APPROACH 1.3.2 BOUNDARY In microscopic approach, a certain quantity of matter is considered with The separation line which separates the considering the event occurring at system from surrounding is called as molecular level. It is called as statistical Boundary. Thermodynamics. In Macroscopic  Boundary may be fixed or rigid, may be approach, a certain quantity of matter is real or imaginary. considered without considering the event occurring at molecular level. In this, 1.3.3 UNIVERSE average behaviour of molecules is considered. It is called as Dynamic The combination of system and surrounding Thermodynamic. is called as Universe. At the higher altitude or where the density Universe = system + Surrounding of the system is low, the microscopic approach is used for checking the 1.3.4 TYPE OF SYSTEM behaviour of the system. There are three types of thermodynamic systems.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1) Closed System: fluid id transmitted from the outlet valve (for example the process of compression in The system in which only energy can compressor),the mass with energy leaves transfer without transferring the mass from the system it means in this system across the system boundary is called as mass and energy both crosses the closed system. An example of closed system boundary so this is considered as the Open is mass of gas in the piston cylinder System. without valve. Let us consider the piston cylinder 3) Isolated System: arrangement as shown in fig 1.2. When heat is supplied to system, gas inside the The system in which, neither energy nor cylinder expands and thermal energy is mass can transfer across the system converted into mechanical work. But mass boundary is called as isolated system. of the gas will be the same after the process Example: Universe, Insulated thermal flask. in cylinder it means there is no mass which Let consider a proper insulated thermal crosses the boundary of the system so it is flask and hot water is filled in the thermal considered as closed system. flask after passing of the time mass and temperature are same which shows neither energy nor mass crosses the thermal boundary of the flask, so it is considered as an isolated system.

fig.:1.2 closed system

2) Open System:

The system in which energy and the mass Fig. 1.4 isolated system can transfer across the system boundary is called as open system. Most of engineering 1.4 PROPERTY OF SYSTEM devices are open systems. Examples: Turbine, Pump, Boiler, condenser Etc. A property of the system is a characteristic of system that depends upon the state of the system .as long as state is fixed, property of system is fixed. It does not depend how the state is reached. So properties depend upon end points only. These are point functions and exact differential. Examples: Pressure, temperature, entropy etc. Fig. 1.3 open system There are two types of properties: Let us consider the piston cylinder arrangement with inlet and outlet valve. (1) Extensive property: When mass of a fluid enters in control These properties depend upon the mass of volume of system, the mass and energy of the system. If the system mass is changed the system change and then this mass of then these types of properties will change.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission For example, volume, internal energy,  Because in a cycle initial and the final enthalpy, entropy etc. point are same. So for the thermodynamic cycle, a thermodynamic 2) Intensive Property: property remains same. These properties do not depend upon the  Difference of properties for reversible mass of the system. For example, specific and irreversible process is same. volume, specific internal energy, specific enthalpy, specific entropy, pressure, Temperature etc. Most of the extensive properties can be converted in to intensive properties by dividing the extensive property by system mass or number of moles in the system. Intensive properties created in this way are called as specific property. It means all the specific properties are intensive property. 1.5.1 PROCESS Let us we an example: we know that volume of the system is considered as the A process is said to be done by the system extensive property and its unit is m3. when it change its state. Process may be V But Specific volume (v) = . The specific non flow process in mass does not transfer m volume is given by m3/kg. It is given as or it may be flow process in which mass intensive property. and energy both will transfer. Type of Process: A Process may be reversible 1.5 STATE process or irreversible process.

It is the condition of a system at the instant 1.5.1.1 REVERSIBLE PROCESS of time. Each property has its single value The process is said to be reversible process at every state. As long state is fixed the if when it is reversed in direction it follows property at that state is fixed. It means that the same path as former path without the property does not depend the leaving any effect on system and parameter other than the state or the surrounding. Reversible process is most specific point. efficient process. Let us consider a state 1 on P- V diagram as

shown in fig 1.5. the properties at point 1 1.5.1.2 IRREVERSIBLE PROCESS are P1 , V1 . The state change from 1 to 2 and the properties at state 2 are P2, V2. then we A process which does not follow the follow the same points but change the path condition of the reversible process is called between the state 1 and state 2 .but the as the irreversible process. The friction is properties remains same. It means main reason for the process being the properties do not depends upon the path. it irreversible process. only depends upon the point, So result of this discussion is 1.5.1.3 QUASISTATIC PROCESS  All properties are the point function.  Differentiation of all the properties is Quasi means ‘almost’ static means exact differential (dT, dP etc.) ‘slowness’. When process is carried out in  Properties do not depend upon the past very slow manner, it is called as Quasistatic history. These only depends the present process. Infinite slowness is characteristics condition of the system of a Quasistatic process. Friction less

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission quasistatic process is considered as A system is said to be in thermodynamic reversible process. equilibrium when it follows: i) Mechanical Equilibrium ii) Thermal Equilibrium iii)Chemical Equilibrium

i) Mechanical Equilibrium:

If there is no unbalanced force within the system and also between the system and surrounding, the system is said to be in mechanical equilibrium.

ii) Thermal Equilibrium:

Fig.1.6 quasistatic process if temperature of the system is same in all the parts of system, the system is said to be Let us consider the expansion of a gas from in Thermal equilibrium. initial state 1 to final state 2 as shown in fig 1.6. In first case the weight W is release iii) Chemical Equilibrium: from the system and the gas expands quickly from state 1 to 2. But when weight if there is chemical reaction or transfer of W is divided into number of parts and then matter from one part of system to another, allow to release from the gas in second the system is said to be in Chemical case. the gas passes through the number of equilibrium. equilibrium state and then reach the final state 2. So second process is very slow 1.7 ZEROTH LAW OF THERMODYNAMICS process it is known as quasistatic process. The P-V diagram for quasistatic process is It states that when body A is in thermal shown in fig 1.7. equilibrium with body B is in thermal equilibrium with body C , then body A,B and C will be in thermal equilibrium. It is the basic of temperature measurement.

T T A C

Fig. 1.7 TB

1.5.2 CYCLE fig. 1.8 Zeroth law of thermodynamics

A thermodynamic cycle is defined as a If TA =TB & TB = TC, then TA = TB = TC. series of state changes such that final state is identical with initial state. for a cycle 1.7.1 APPLICATION OF ZEROTH LAW difference of all the properties is zero. The main application of Zeroth law of 1.6 THERMODYNAMIC EQUILIBRIUM thermodynamics is to measure the

A system is said to be in thermodynamic temperature. First thermometric property equilibrium when there is no change in the is measure which change with respect to macroscopic property of the system. change of temperature.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Let us consider Mercury in glass temperature resistance change .so thermometer (generally which is used to thermometric Property in resistance measure the temperature of human body) thermometer is resistance. So to find the as shown in fig 1.9. temperature of any state first resistance of The length of the Hg change with respected that state is found and then temperature is to temperature. So the thermometric measured. property in this type of thermometer is length.

Fig. 1.9 Hg in glass thermometer Fig. 1.10 Resistance Thermometer It is more important to find the change of length so that we can calculate the (b) Thermocouple temperature. So first length of the Hg which is change is found then temperature is This works on See beck Effect. When two calculated. In next topic we explain the dissimilar metal at different temperature is process of measurement of the joint with each other, an electro magnetive temperature. force is generated the induced e.m.f. depends upon the temperature difference 1.7.2 TEMPERATURE MEASUREMENT of two ends of the dissimilar metal. So thermometric property in this In the thermometer, Zeroth law of thermocouple is induced e.m.f. thermometer is used to measure the temperature of the body. Thermometric Property: The property which changes with change in temperature is called as thermometric property.

1.7.2.1 THERMOMETRIC PRINCIPLE

The thermometric property which change with the change of temperature is found Fig. 1.11 thermocouple first and with help of this thermometric c) Constant Volume gas thermometer

property temperature of that specified When the volume in constant volume gas state is calculated. thermometer is constant, the pressure

1.7.2.2 TYPE OF THERMOMETER changes with respect to temperature. In this type of thermometers, pressure is There are five types of thermometers. considered as the thermometric property.

a) Resistance thermometer d) Constant pressure gas thermometer

It works on Wheatstone bridge circuit. As When the pressure in constant Pressure temperature hang with respect to change of gas thermometer is constant, the volume

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission changes with respect to temperature. in Ti = a.Xi ------(1) this type of thermometers, volume is and T = a.X ------(2) considered as the thermometric property. Then, X e) Mercury –in- glass thermometer T 372.15 ------(3) Xi In this type of thermometers, length change with change of the temperature so in mercury in glass thermometer, length is 1.7.3.3 TEMPERATURE SCALES considered as the thermometric property. There are following temperature scales 1.7.3 METHOD FOR TEMPERATURE such as Rankin, Celsius, Kelvin and MEASUREMENT Fahrenheit. There are two common absolute temperature scales Rankin (R), 1.7.3.1 METHOD FOR TEMPERATURE and Kelvin scale (K). They are related as MEASUREMENT: BEFORE 1954 follows: 9 Before 1954 temperature measurement T(R)T(K) method, the measurement of temperature 5 is based on two reference point freezing point and boiling point. Let temperature at freezing point is T1 and thermometric property at that temperature is X1 and Temperature at boiling point is T2 and thermometric property at that temperature is X2. Let thermometric Property at temperature T is X .then,

TX11  ------(1) TX and Temperature at boiling point is T2 and thermometric property at that temperature is X2. Let thermometric Property at temperature T is X.then,

TX22  ------(2) TX From equation (1) and (2) Fig. 1.12 different temperature scale TTXX 1 2 1 2 TX The relationship between the Celsius scale TT and Fahrenheit scale can be given as T 12 .X ------(3) XX 009 12 T( F) T( C) 32 ------(2) 5 1.7.3.2 METHOD FOR TEMPERATURE The relationship between the Celsius scale MEASUREMENT: AFTER 1954 and Kelvin scale can be given as 0 The measurement of temperature is based T(K) T( C) 273.15 ------(3) on single reference point triple point of water. Let temperature at triple point is Ti 1.8 IDEAL GAS EQUATION AND PROCESS and thermometric property at that temperature is Xi. Let thermometric The ideal gas law is the equation of Property at temperature T is X. then state of a hypothetical ideal gas. It is a good

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas law is often introduced in its common form: PV nRT Where P is the absolute pressure of the gas, V is the volume of the gas, n is number

of moles of gas, R̅ is universal gas constant, & T is the absolute temperature of the gas. 3) Isothermal Process: For air R̅ = 8.314 KJ/mol.K An is other mal process is a change of n (in moles) is equal to the mass m (in gm) a system, in which the temperature remains divided by the molar mass M ( in gm/ constant. It is also called as hyperbolic mole) process. For Isothermal process, T = m n  Constant or PV = Constant. M For the isothermal process, By replacing n with m / M, we get: PVPV1 1 2 2 m PV RT M Defining the specific gas constant R as the ratio R̅ /M PV mRT Where R is specific Gas Constant. For air R =0.287 KJ/Kg.K

Different process: (4) Adiabatic Process 1) Constant volume process An adiabatic process is a process that An isochoric process, also called a constant- occurs without the transfer of heat or volume process, or an isometric process, is matter between a system and its a thermodynamic process during which surroundings. Such processes are usually the volume of the system undergoing such followed or preceded by events that do a process remains constant. involve heat transfer. For isocoric process, V = Constant For adiabatic process, PVγ = Constant. PP12 P V γ = P V γ  1 1 2 2 γ1 TT12 γ TP22 OR   TP11

2) Constant pressure process An isobaric process is a Thermodynamic

process in which the pressure stays constant. For isobaric process, P = Constant. Adiabatic Process γ1 VV12  TV22 OR,  TT12  TV11

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 5) Polytropic Process Slop of isothermal and adiabatic process A polytropic process is a thermodynamic on P-V diagram process that obeys the relation:  For isothermal process pCn PV = C Where n is polytropic index. Taking log in both the sides logP + logV = log C P1V1n = P2V2n n1 Differentiating the equation n TP22 dP dV OR,  0  PV TP11 dP dV  PV dP P  dV V

Polytropic Process n1 TV22 OR ,   TV11

 For adiabatic process Representation of different process on PVγ = C P-V diagram: Taking log in both the sides PVk  C log P + γ logV = log C If k=0, P= C, process is isobaric process. Differentiating the equation dP dV If k = ∞, V =C, process is isocoric process. γ0 If k=1 PV=C, process is isothermal process. PV If k=n PVn =C, process is polytropic process. dP dV γ γ If k=γ PV =C, process is adiabatic process. PV Different processes on P-V diagram are Slopof adiabaticprocess shown in fig.  γ Slopof isothermalprocess  The slop of adiabatic process is higher than the slop of isothermal process.

WORK AND HEAT TRANSFER

1.9 WORK TRANSFER

Work is said to be done by a system if the Sole effect on things external to system can

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission be reduced to rising of a Weight. The different so work done for different path is Weight may not be raised, but the net effect different. So work transfer is path function. external to system would be the raising of a It is inexact differential. Weight. Let consider the Battery and Motor. 1.9.1 WORK TRANSFER FOR CLOSED SYSTEM IN VARIOUS REVERSIBLE PROCESSES

A) Constant Pressure Process In the isobaric process pressure remains constant. So work transfer 2 W p.dv  P V  V The nature of work changes when 1 2  2 1  boundary between system and 1 Surrounding changes. So work done is boundary phenomenon. It is transient form of energy. Wout

Surrounding System Win B) Constant Volume Process In the isochoric process volume remains constant. → When Work is done on the system it is So work transfer in this process is zero. considered as negative. V= constant → When work is done by the system, it is 2 Work transfer W p.dv 0 considered as positive. 12  Let Consider a piston cylinder arrangement 1 (i.e. closed).

C) Isothermal Process

Work done by the system In isothermal process, T= C or

dW = F. dl = P.A.ds = P.dV P.V PV1 1  P 2 V 2  C 2 PVPV W P.dV ------(1) 1 1 2 2 12  P  ---- (1) 1 VV It is valid for closed system undergoing a 22dv W p.dv  P V  reversible process. 1 2 1 1 V So work done by closed system in 11  reversible process is area under p v V2  P11 V In  diagram about V axis. V1 If Path A and Path B are two paths but VV22    initial and final point is same. Area is W1 2 P 1 V 1 In  m.RT In   VV11   

Heat is defined as a form of energy that is transferred across the boundary by virtue of a temperature difference. The temperature difference is the potential and heat transfer is the flux. Heat flow into the system is taken as positive and heat flow out of system is taken as negative. The D) Adiabatic Process: process in which that does not cross the boundary is adiabatic process. Heat In adiabatic Process, transfer is a boundary phenomenon. Heat γ γ γ PV PV1 1  P 2 V 2  C transfer is transient form of energy. Heat γ γ transfer is Path function. PV11  PV P 22 For the m mass of substance and ∆T VVγγ temperature difference, Heat transfer 22PV γ W pdv 12 Q α m.∆T 12 γ 11V Q m.C. T 1- 1-γ Where m= mass of substance, γ VV21 WPV1 2 1 1  1 ΔT = temperature difference  C = Specific heat and its unit is KJ/Kg.K. P Vγ .V 1- P V γ .V 1-  2 2 2 1 1 1 If m = 1Kg, 1 γ ΔT = 1 0 C, then C = Q P V P V MR(T T ) W 1 1 2 2 1 2 1.10.1 SPECIFIC HEAT 12  11  

The specific heat of a substance is defined as the amount of heat required to raise the unit mass of substance through a unit rise of temperature. For the gases, in constant volume process it is taken as C and in constant pressure V Adiabatic Process Process it is taken as Cp.

(E) Polytrophic Process Examples

P1 V 1 P 2 V 2 MR(T1 T 2 ) 3 W12  Q.1 If a gas of Volume 6000 cm and at a n 1 n 1 pressure of 100 KPa is compressed Where n = Polytrophic index in reversible Process according to PV2 = C until volume becomes 2000 cm3, determine the final pressure and work done ? Solution: Given V1 = 6000 cm3 P1 = 100 KPa, V2 = 2000 cm3

P2 =? , W?12  22 PVPV1 1 2 2 Polytropic Process 100 ×(6000)2 = P2 ×(2000)2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 100 60002 Work done in A-B process = Area P  2 20002 under AB 5 P 900KPa WA-B = 50 × 10 × (0.4 – 0.2) J 2 = 106 J = +1 MJ For the polytrophic process, nn PVPVBBBC 50×0.41.3=PB0.81.3 PB = 8.25 bar = 20.3 × 105 Pa Work done in B-C process = Area Work Transfer under BC on P-V diagram 2 PVBBBPVC W p.dv ----- (i) WBC  12  n 1 1 55 In the process PV2  PVPVC22 50 10  0.4   20.3  10  0.8 11 22  C n1 6 P  2 ----- (ii) V WBC 1.25  10 J 1.25MJ Substituting the Value a/p in So the total work done

equation (1) W  WAB WBC   1  1.25 MJ 2 C W dv = 2.25 MJ 12  2 1 V Q.3 The limiting value of the ratio of the 11  C   pressure of gas at the steam point VV21 and at the triple point of water 22 when the gas is kept at constant PVPV2 2 1 1    VV volume is found to be 1.36605. What 21 is the ideal gas temperature of the 12  PVW121 P V2 steam point? =100×(6000)×10-6-900×(2000)×10-6 Solution: W 1.2K J P 12 Given : 1.36605 It means 1.2 KJ work is done on the Pt system. P T = 273.16 =273.16 × 1.36605 Pt Q.2 Determine the total work done by a gas system, which follows the = 373.15 K 0 expansion Process as shown in T = 100 C Figure. Q.4 A platinum wire is used as a resistance thermometer. The wire resistance was found to be 10 ohm and 16 ohm at ice point and steam point respectively, and 30 ohm at sulphur boiling point of 444.6°C. Find the resistance of the wire at 500°C, if the resistance varies with Solution: temperature by the relation. Given P = P = 50 bar =50 × 105 Pa, A B R = R (1 +α t + βt2) V =0.2 m3, V = 0.4 m3, V =0.8 m3, 0 A B C Solution: WA-C = ?

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Given: ti=0°C, Ri = 10 ohm, ts =100°C, ∴ P1V1 = P2V2 = C R = 16 ohm m 1.5 s V 1   1.29 m3 For sulphur, tb=444.6°C, Rb=30 ohm, 1 1 1.16 t = 500°C, R=? Work done 10 = R0 (1+0×α+β×02) VP    16 = R0 (1+100×α + β×1002) 21 W1 2 P 1 V 1 In  P1 V 1 In   2 30 = R0 (1 +444.6 ×α + β ×444.6 ) VP12    Solving the equation, we get 0.1 R= 10 (1 +6.45 ×10-3t+ 4.48×10-6t2) 0.1 1.29In21.5 MJ 0.7 Substituting the value of t = 5000C

R= 10 (1 +6.45 ×10-3×500-4.48× Q.7 680 kg of fish at 5°C are to be frozen 10-6×5002) and stored at – 12°C. The specific R = 31.05 ohm. heat of fish Above freezing point is

3.182 KJ/Kg.K, and below freezing Q.5 The piston of an oil engine, of area point is 1.717 KJ/Kg K. The freezing 0.0045 m2, moves downwards 75 point is – 2°C, and the latent heat of mm, drawing. In 0.00028 m3 of fresh fusion is 234.5 KJ/Kg. How much air from the atmosphere. The heat must be removed to cool the pressure in the cylinder is uniform fish, and what percent of this is during the process at 80 KPa, while latent heat? the atmospheric pressure is 101.325 Solution: KPa, the difference being due to the Given: m=680 kg, T =50C, T = -120C, flow resistance in the induction pipe 1 2 C = 3.182 KJ/Kg.K, and the inlet valve. Estimate the p1 C = 1.717 KJ /Kg.K, T = -20C LH = displacement work done by the air p2 f 234.5 KJ/Kg, finally in the cylinder. Heat to be removed above freezing Solution: point Given: A=0.0045 m2, length of stroke = 680 × 3.182 × {5 – (-2)} kJ = 75 mm = 0.075 m = 15.146 MJ Volume of piston stroke = 0.0045 × Heat to be removed latent heat 0.075 m3 = 680 × 234.5 kJ = 0.0003375 m3 = 159.460 MJ ∴ V -V = 0.0003375 m3 2 1 Heat to be removed below freezing As pressure is constant P = 80 KPa point So work done = P. (V -V ) 2 1 = 680 × 1.717 × {– 2 – (– 12)} kJ = 80 ×0.0003375 KJ = 11.676 MJ = 0.027 KJ ∴ Total Heat = 186.2816 MJ = 27 J Percentage of Latent heat = (Latent

heat/total heat) × 100 Q.6 A mass of 1.5 kg of air is compressed = (159.460/186.2816) × 100 in a quasi-static process from 0.1 = 85.6 % MPa to 0.7 MPa for which PV =

constant. The initial density of air is 1.16 kg/m3. Find the work done by the piston to compress the air. Solution: Given: m=1.5 kg, P1=0.1 MPa, P2 =0.7 MPa, = 1.16 kg/m3, W=? For quasi-static process, PV = C

2.1 FIRST LAW OF THERMODYNAMIC 2.2.2 ENERGY IS A PROPERTY OF FOR A CYCLE SYSTEM

Heat and work are different forms of From the equation (III), We get energy. Both form of energy can be ------(III) Conserved. As the result of Joule’s Rearranging the equation (III)

experiment, for a closed system undergoing (δQ)BC (δW)B  (δQ)C  (δw) a Cycle (δQ) (δW) (δQ) (δW)  BC   (ΣQ) cycle =J (ΣW) cycle J is Joule’s equivalent. This is first law of For path B and C, the difference of heat thermodynamics for a closed system transfer and work transfer is constant. So undergoing a cycle. Both Heat and work are this different must be a point function and measured in the desired unit of energy property of the system. This property is Joule (J). So constant of proportionality ‘J’ is known as energy. unity. Q  W  dE ------(IV) This is known as first law of 2.2 APPLICATION OF FIRST LAW OF thermodynamic for closed system THERMODYNAMICS undergoing a process. It is valid for the reversible and irreversible process. 2.2.1 HEAT TRANSFER IS A PATH Total Energy E = Kinetic Energy + Potential FUNCTION Energy + Internal Energy. In absence of motion and gravity, kinetic & Let consider a cycle 1-A-2-B-1 and apply potential energy is Zero. In that case E = U first law of thermodynamics for a cycle δQ ̶ δW = dU ----- (V) (ΣQ) cycle =J (ΣW) cycle

(dQ)ABAB (dQ)  (δw)  (δw) ---- (I) Let consider another cycle 1-A-2-B-1 and apply the first law of Thermodynamics again

(dQ)ACAC (dQ)  (δw)  (δw) ------(II) Solving the equation (I) or equation (II) (dQ) (dQ)  (δw)  (δw) ------(III) BCBC Where, Difference (δw)BC (δw) is not Zero  U = Internal energy because work transfer is a Path function as  Unit of internal energy = KJ it was proved with help of chapter 1, so  Internal energy is an extensive

(dQ)BC (dQ) is not equal to Zero. So heat property. transfer for the different path has the  This is first Law of thermodynamics for different value. It depends upon the path. reversible and irreversible process. So results of this discussion are: But we know that δW=P.dV for a  Heat is a path function. reversible process, so  The differentiation of the heat transfer Q  P.dV  dU ---- (VI) is inexact differential (δQ).  This is first Law of thermodynamics for  It is boundary phenomenon. a closed System undergoing a reversible  it is transient form of energy. Process.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2.2.3 ENERGY OF AN ISOLATED SYSTEM H = P.V+ U IS CONSTANT Enthalpy is an extensive property and its unit is KJ. According to I Law of thermodynamics Enthalpy of an ideal gas is given by δQ = δw + dE H = m.Cp.T For isolated System δQ = 0 δw = 0 dE = 0 2.2.5.1 SPECIFIC ENTHALPY

EE12  Energy of an isolated System is always It is amount of enthalpy per unit mass. constant. Specific enthalpy is an intensive property.  Universe is an isolated system so Its unit is KJ/Kg. energy of the Universe remains Constant. 2.2.6 INTERNAL ENERGY  It is known as law of conservation of energy. All fluids store energy. The store of energy within any fluid can be increased or 2.2.4 PERPETUAL MOTION MACHINE decreased as the result of various (PMM-I) IS NOT POSSIBLE processes carried out on or by the fluid. The energy stored within the fluid which Perpetual machine is the machine which results from the internal motion of its atom delivers the work without taking any input. and molecules is called as Internal Energy. There is no such machine which would The Internal energy of any ideal gas is continuously supply mechanical work given: without some other form of energy U = m.Cv.T dissipating. Such a machine is called as Internal Energy is an extensive property. PMM-I. The efficiency of PMM-1 is given by It is a point function. WW Unit of internal energy is KJ. n  PMM I QO 2.2.6.1 SPECIFIC INTERNAL ENERGY n  PMM I It is amount of internal energy per unit It means efficiency of PMM-1 is Infinite. mass. Specific internal energy is an But according to first law of intensive property. It is denoted by u. Its thermodynamics unit is KJ/Kg. (ΣQ) = (ΣW) cycle cycle 2.3 MEYER’S FORMULA

We know that Enthalpy can be given as H = U + PV For an ideal gas, H = m.Cp.T, U = m.Cv.T and PV = mRT If it is possible it will violate of first law of Substituting the values in equation thermodynamics but it is law of nature. So H = m.Cv.T + mRT  PMM-1 is not possible. m.Cp.T = mT (CV+R)  The machine having the infinite CP R efficiency is not possible. CP C  ,C  V = R CV p 1

2.2.5 ENTHALPY This formula is known as Meyer’s formula. We know that Enthalpy of a Substance is defined as

According to first law of thermodynamics 2.4 HEAT TRANSFER IN DIFFERENT δQ= δW + dU PROCESS mr(T T ) mC dT 12 1) Constant Volume Process: V n1 According to first law of thermodynamics mR(T T ) mR(T T )   1 2  1 2 δQ = P.dV + dU  1 n 1 δQ = O + dU 11 δQ = dU = mCVdT mR(T12  T )   n 1   1 mR(T T ) n  12 n 1   1 n Q   work done 1 2) Constant Pressure Process:  According to first law of thermodynamics δQ = P.dV + dU We know that for isobaric Process dH = P.dV + dU δQ = dH = m.Cp.dT

2.5 FREE EXPANSION

3) Isothermal Process: According to first law of thermodynamics δQ = P.dV + dU For isothermal Process dU = 0 δQ = P.dV = δW V δQ = δW = P V l 2 1 1 n V The expansion of gas against vacuum is 1 known as free Expansion. There is no work done in free Expansion.

2 2  W0although  P.dv 0 4) Adiabatic Process: 1 1

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 Steady state energy flow Since work transfer =  P.dv is considered [Internal Energy + Kinetic energy + 1 Potential energy + Heat] for reversible Process but free expansion is 1-1 = [Internal Energy + Kinetic Energy + highly irreversible Process. Vacuum does Potential Energy + Work] not offer any resistance, so there is no work 2-2 1122 transfer involved in free expansion. U1  mC1  mgZ 1   Q  U2  mC2  mgZ 2   w Total 22 2.6 FIRST LAW OF THERMODYNAMICS 1 2 U1  mC1  mgZ 1   Q  U2  FOR FLOW PROCESS 2 1 mC2  mgZ   w  P V  P V Steady Flow Process 2 2 2 cv 1 1 2 2 1 U P V  mC2  mgZ   Q Steady flow means the rate of flow of mass 1 1 1 2 1 1 and energy across the control volume are 1 2 constant. At steady state of a System, any U2  P 2 V 2  mC2  mgZ 2   w cv 2 thermodynamic property will have a fixed 1 2 value at the same state and does not change H1  mC1  mgZ 1   Q with respect to time. 2 1 H  mC2  mgZ   w Or Flow Work 2 2 2 2 cv 2 C1 Q The Flow work is the work done by Fluid of H11  gZ   2m Mass m either at inlet or at the outlet of 1 2 wcv control Volume. Flow work at inlet is taken H2  C 2  gZ 2  as (-P1V1) and at outlet of control volume is 2m (+P1V1).where P1, V1 is pressure and It is known as steady Flow energy Equation volume at the inlet of control volume and (SFEE) P2,V2 is the pressure and volume at the Where outlet of control volume. h1,h2 = Specific enthalpy at inlet and outlet KJ/Kg so total work WWPVPVtotal CV  2 2  1 1 C1,C2 = Velocity at inlet and outlet m/Sec. 2.6.1 STEADY FLOW ENERGY EQUATION wcv = Work done by System KJ δQ = Heat transfer KJ By Appling the energy balance equation at Z1, Z2 = Elevation above on arbitrary datum section 1-1 and section 2-2 , m (Energy at Section 1-1)= (Energy at Section 2-2) 2.6.2 APPLICATION OF S.F.E.E.

1) Nozzle

For a turbine which is well insulated, and

ZZ12 SFEE equation: CC22QW h112  gZ   h   gZ  21 moo2 22 m QW Here 0 mmoo

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission ZZ12 CC22 hh12   1222 22 C2 C 1  2(h 1  h 2 ) If C1 is very less as Compared to C2, then

C2  2(h1 h 2 ) m/s.

0 Wcv  m (h1  h 2 )KJ But in Case of turbine h1 is greater than h2 and in Case of Compressor h2 is greater than h1.

4) Throttling Process

When the fluid flows through a constricted passage, like an opened valve or an orifice it is called as throttling Process. Throttling 2) Turbine process is irreversible process. In throttling Process, For a turbine which is well insulated, the QW 0 flow Velocities are often small, so the K.E mm can be neglected and And change of P.E. P.K.E. is very small and SFEE equation: ignored, then 22 CC22QW CC12QW12 h  gZ   h   gZ  h1   gZ1   h2   gZ2  1 21 moo2 22 m 2 m 2 m

Wcv hh12 hh12 mo So Enthalpy of fluid before throttling and 0 W  m (h12  h )KJ after throttling remains Constant.

3) Compressor 2.7 FIRST LAW OF THERMODYNAMICS FOR VARIABLE FLOW PROCESS For a turbine which is well insulated, the

flow Velocities are often small, so the K.E Variable Flow can be neglected and =

SFEE equation: If there is a change of property at the same state, Process is called as Variable flow or unsteady Flow Process. In the unsteady flow then there is an accumulation of mass and energy.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The mass Accumulated in control volume is then removed and 105 kJ of heat given by flow to the surroundings as the fluid

dm dm12 dm goes to state 3. The following data  dt dt dt are observed for the fluid at states 1, cv 2 and 3. Applying the law of conservation of energy State v (m3) t (°C) Rate of energy accumulation = Rate of 1 0.003 20 energy inlet Rate of energy outlet 2 0.3 370 dE  d  02 1  3 0.06 20 cv m1 h 1  m 1 c 1  m 1 gZ 1   Q dt  dt  2  For the fluid system, calculate E2

d1and E3, if E1 = 0 m h  m c2  m Z   W Solution: dt2 2 2 2 2 2 2  From first law of thermodynamics δQ = dE +PdV

Q1-2 = (E2 – E1) +

= (E2 – E1) + 0.1× 103 (0.3- 0.003) Because Process 1-2 is insulated so Q1-2 = 0 and E1 = 0 So E2 = - 29.7 KJ

Q2-3 = (E3 – E2) + 3 -105=(E3+29.7)+0.1× 10 (0.06- 0.3) Unsteady flow E3 = -110.7 KJ Q.2 A domestic refrigerator is loaded dE   0 1 0 2 0 δQ cv=mh+1 1 mc+mg2+ 1 1 1 1 - with food and the door closed. dt   2 δt  During a certain period the machine 0 1 0 2 0 δw consumes 1 kWh of energy and the me h 2 + m e c 2 +m e g2 2 + 2 δt internal energy of the system drops Neglecting the Kinetic energy and potential by 5000 kJ. Find the net heat energy transfer for the system.

dU 0 0 0 0 Solution: cv m1 h 1  m e h 2   Q   W dt According to first law of Where thermodynamics δQ ̶ δW = dE m0  Mass Flow rate at inlet Kg/Sec l δQ ̶ δW = dE 0 me  Mass Flow tare at outlet Kg/Sec δW = -1× 3600 = - 3600 KJ. Q0 Heat transfer Rate KJ/Sec dE = -5000 KJ δQ + 3600 = -5000 W0 Work done KJ/Sec δQ = - 8600 KJ = -8.6 MJ h ,h Specific Enthalpy at Inlet or Outlet. 12 Q.3 The properties of a certain fluid are Examples related as follows: u = 196 + 0.718t Q.1 A slow chemical reaction takes place pv = 0.287 (t + 273) in a fluid at the constant pressure of Where u is the specific internal 0.1 MPa. The fluid is surrounded by energy (kJ/kg), t is in °C, p is a perfect heat insulator during the pressure (KN/m2), and v is specific reaction which begins at state 1 and volume (m3/kg). For this fluid, find ends at state 2. The insulation is Cv and Cp.

dh Q dE  W12   320  2932.5 we know that Cp =  dT P Q = 2612.5KJ d u pu  2612.5 KJ heat is supplied to the   system. dT P d 196 0.718t  0.287 t  273  Q.5 The heat capacity at constant   pressure of a certain system is a dT P function of temperature only and 0 0.718dt  0.287(dt  0) may be expressed as   41.87 0 dT P Cp  2.093 J / C 0.718dt 0.287dt t 100   Where t is the temperature of the dT P system in °C. The system is heated 1.005dt while it is maintained at a pressure   dT P of 1 atmosphere until its volume but T = t + 273 so dT = dt increases from 2000 cm3 to 2400 Cp =1.005KJ/Kg.K cm3 and its temperature increases du from 0°C to 100°C. again Cv   a) Find the magnitude of heat dT V interaction. d(196 0.718t) b) How much does the internal   dT V energy of the system increase? 0.718dt Solution:   If T is temperature in K, then t = T- dT V 273 and t + 100 = T – 173 but T = t + 273 so dT = dt 373 41.87 CV =0.718KJ/Kg.K Q12   2.093 dT  T 173 273  373 Q.4 A mass of 8 kg gas expands within a 418.7 flexible container so that the P-V 2.093T ln T 173 relationship is PV1.2 = constant. The 273 200 initial pressure is 1000 KPa and the = 2.093(373 - 273) + ln initial volume is 1 m3. The final 100 pressure is 5 KPa. If specific internal = 209.3 + 41.87ln2 = 238.32 J energy of the gas decreases by 40 2 Q E – E PdV KJ/kg, find the heat transfer in 1 2 2 1   1 magnitude and direction. E2 – E1 = Q1-2 –P.(V2-V1) Solution: =238.32-101.325(0.0024 0.0020)×1000J For the polytrophic process = 197.79 J n n P1V1 = P2V2 1000 ×11.2 = 5 ×V21.2 Q.6 a gas undergoes a thermodynamic V2 = 82.7 m3 cycle consisting of three processes PVPV beginning at an initial state where work done W  1 1 2 2 12 n1 P1 = 1 bar, V1 = 1.5 m3 and U1 = 512 1000 1  5  82.7 kJ. The processes are as follows:  1.2 1 i) Process 1–2: Compression with PV = constant to P = 2 bar, U = W 2932.5KJ 2 2 12 690 kJ

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission ii) Process 2–3: W23 = 0, Q23 = –150 Q.8 A turbine operates under steady kJ, and flow conditions, receiving steam at iii) Process 3–1: W31 = +50 kJ. the following state: Pressure 1.2 Neglecting KE and PE changes, MPa, temperature 188°C, enthalpy determine the heat interactions 2785 kJ/kg, velocity 33.3 m/s and Q12 and Q31. elevation 3 m. The steam leaves the Solution: turbine at the following state: Process (1-2): process is isothermal Pressure 20 KPa, enthalpy 2512 process. for isothermal process, kJ/kg, velocity 100 m/s, and

Q1-2 = (U2-U1) elevation 0 m. Heat is lost to the surroundings at the rate of 0.29 = (690 – 512) + 1 105 1.5 KJ/s. If the rate of steam flow through the turbine is 0.42 kg/s, = (690 – 512) + 1 102 1.5 what is the power output of the = 178 – 103.972 turbine in kW? = 74.03 KJ Solution: As W2-3 is ZERO so it is constant P1 = 1.2 MPa t1 = 188°Ch1 = 2785 volume process. As W31 is positive kJ/kg C1 = 33.3 m/s Z1= 3 m P2 = so expansion is done. 20 kPa For the process (2-3) h2 = 2512 kJ/kg C2 = 100 m/s Q2-3 = W2-3 + (U3-U2) dQ/dt = – 0.29 kJ/s dW/dt =? –150 = 0 + (U3 - 690) U3 = 540 KJ Process (3-1) Q3-1=W3-1+ (U1-U3) = 50+(512-540) = 50 -28 = 22 KJ

Q.7 A blower handles 1 kg/s of air at 20°C and consumes a power of 15 kW. The inlet and outlet velocities of By SFEE air are 100 m/s and 150 m/s C2 gZ Q m(h 11 )   respectively. Find the exit air 1 2000 1000 dt temperature, assuming adiabatic C2 gZ W conditions. Take Cp of air is 1.005 m(h 22 )  2 2000 1000 dt KJ/Kg-K. 33.32 9.8 3 Solution: 0.42(2785  )  0.29  2000 1000 From S.F.E.E. c2 1002 9.8 0 W mh m1  mgZ   Q / dt 0.42(2512  )  112 2000 1000 dt 1 W mh  mC2  mgZ   W / dt 1169.655 1057.14 2 2 2 2 dt W 112.515kW 1×h1 + 1× (100)2/2000 + 1×9.8×Z dt + 0 = 1×h2 + 1× (200)2/2000 + 1×9.8×Z + 15 Q.9 A nozzle is a device for increasing h2 – h1 = 8.75 the velocity of a steadily flowing Cp (T2-T1) = 8.75 stream. At the inlet to a certain T2 = 20 + 8.7 nozzle, the enthalpy of the fluid = 28.7 0C passing is 3000 kJ/kg and the

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission velocity is 60 m/s. At the discharge compressor at a temperature of end, the enthalpy is 2762 KJ/kg. The 16°C, a pressure of 100 kPa, and an nozzle is horizontal and there is enthalpy of 391.2kJ/ kg. The gas negligible heat loss from it. leaves the compressor at a a) Find the velocity at exists from temperature of 245°C, a pressure of the nozzle. 0.6 MPa, and an enthalpy of b) If the inlet area is 0.1 m2 and the 534.5kJ/ kg. There is no heat specific volume at inlet is 0.187 transfer to or from the gas as it m3/Kg, find the mass flow rate. flows through the compressor. c) If the specific volume at the a) Evaluate the external work done nozzle exit is 0.498 m3/Kg, find per unit mass of gas assuming the exit area of the nozzle. the gas velocities at entry and Solution: exit to be negligible. Velocity at exit from S.F.E.E. b) Evaluate the external work done C2 gZ Q per unit mass of gas when the m(h 11 )   1 1000 1000 dt gas velocity at entry is 80 m/s 2 and that at exit is 160 m/s. C22 gZ W m(h2  )  Solution : 2000 1000 dt t1 = 16°C, P1 = 100 kPa, h1 = 391.2 kJ/kg,t2=245°C P2=0.6 MPa=600 kPa h2 = 534.5 kJ/kg For V1 and V2 negligible and Z1= Z2 So SFEE

m(h12 0)  0  m(h  0)   W

W  1  (h12  h )  (391.2  5345)kJ / kg = –143.3 kJ/kg h1 = 3000 kJ/kg C1 = 60 m/s h2 = C1 = 80 m/s; C2 = 160 m/s 2762 kJ/kg Then SFEE WQ 2  0 Z12  Z c1 dt dt m(h1   0)  0  22 1000 cc    2 11 c h12   0   h    0 2 1000 1000 m(h2   0)   W     1000 2 2 2 60 c2 80 (3000 )  0  (2762  )  0 1000 1000 1 391.2   0 1000 C22 = 602/ 2000 +3000 - 2762 2 C = 692.532 m/ s 160 2 1  534.5   0   W AC 0.1 60 1000 Mass flow rate (m) 11  v1 0.187 δW= (–143.3 – 9.6) kJ/kg = 32.1 Kg/sec = –152.9 kJ/kg AC Mass flow rate (m) = 22 Q.11 Air flows steadily at the rate of 0.4 v2 kg/s through an air compressor, A 692.532 entering at 6 m/s with a pressure 32.1 2 0.498 of 1 bar and a specific volume of 3 A2 = 8.02 m2 0.85 m /kg, and leaving at 4.5 m/s with a pressure of 6.9 bar and a Q.10 A gas flows steadily through a rotary specific volume of 0.16 m3/kg. The compressor. The gas enters the internal energy of the air leaving is

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 88 KJ/kg greater than that of the air m.V 0.4 0.16 A 2   0.01422 m2 entering. Cooling water in a jacket 2 V2 4.5 surrounding the cylinder absorbs heat from the air at the rate of 59 W. Calculate the power required to drive the compressor and the inlet and outlet cross-sectional areas. Solution : m1 = 0.4 kg/s,C1 = 6 m/s P1 = 1 bar = 100 KPa v1 = 0.85 m3 /kg m2 = 0.4 kg/s C2 = 4.5 m/s P2 = 6.9 bar = 690 kPa v2 = 0.16 m3 /kg u2 = u1 + 88 kJ/kg

by SFEE c2 Q m(h1  0)  1 1000 dt c2 W m(h 2  0)  2 2000 dt c2 Q m(u P v 1 )  1 1 1 1000 dt

c2 W m(u  P v 2 )  2 2 2 2000 dt W  m dt cc22 Q u u  P v  P v 12 )  1 2 1 1 2 2 2000 dt 0.4(  88  85  110.4  0.0076)  0.059  45.357  0.059 = – 45.416 kJ AC Mass flow ratem   11 V1

m.v1 0.4 0.85 2 A1    0.0567m c61 AC Mass flow rate m   22 V2

The first law of thermodynamics states that A heat engine that violates the certain energy is balanced when system Kelvin‐Planck statement of the second law undergoes a thermodynamic process. But it cannot be built. does not give any idea whether a particular process is in feasible condition or not. It 3.1.2 CLAUSIUS STATEMENT OF SECOND does not predict the direction of process. It LAW is second law of thermodynamics which predict whether process is in feasible or It is impossible to construct a device that not with the concept of entropy. operates in a cycle and produces transfer of heat from a lower‐temperature body to 3.1 THERMAL RESERVOIR higher‐temperature body without any external work. If such a device is possible A thermal energy reservoir is defined as a its C.O.P. will be infinite. So Refrigerator large body of infinite heat capacity which is having COP infinite is not possible. capable absorbing or rejecting the heat without change of temperature. Source: The thermal energy reservoir from which heat can be absorbed without change of temperature is known as Source. Sink: The thermal energy reservoir in which heat can be rejected without change of temperature is known as Sink. There are two statements of second law of thermodynamics A refrigerator that violates the Clausius 3.1.1 KELVIN – PLANK STATEMENT OF statement of the second law cannot be SECOND LAW built.

It is impossible for any device that operates 3.1.3 HEAT ENGINE on a cycle to receive heat from a single reservoir and produce a net amount of Heat engines convert heat to work. Heat work. In other words, no heat engine can engine is a thermodynamic system have a thermal efficiency of 100%. The operating in a thermodynamic cycle to machine having 100% efficiency is known which heat is transfer and it is converted in as PMM-II. So PMM-II is not possible. to work.

Fig.: Heat Engine 3.1.4 THERMAL EFFICIENCY

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Thermal efficiency is the index of low temperature body and delivers heat to performance of a heat engine. It is the high temperature body. To accomplish this fraction of the heat input that is converted energy transfer Heat pumps receive the to the network. external energy in the form of work or heat. Work output Wout = Q1- Q2 Efficiency W QQQ   1 2 1  2 ------(3.1) QQQ1 1 1

3.1.5 REFRIGERATOR AND HEAT PUMP

The transfer of heat from a low temperature region to a high‐temperature one requires special devices called refrigerators. Refrigerators are cyclic

devices, and the working fluids used in the Fig: Heat Pump cycles are called refrigerant. 3.1.8 CO-EFFICIENT OF PERFORMANCE

The index of performance of a heat pump is given in the term of co-efficient of performance (COP).It is defined as the ratio of desired effect to given input. For the Heat pump desired effect = Heat supplied to high temperature body (Q1) Work done on the refrigerator W = Q1 - Q2 Fig: Refrigerator So Desired Effect Q (COP) 1 3.1.6 CO-EFFICIENT OF PERFORMANCE R workinput W Q The index of performance of a refrigerator (COP)  1 ------(3.3) HP QQ is given in the term of co-efficient of 12 performance (COP).It is defined as the ratio Subtracting the equation (3.2) from of desired effect to given input. equation (3.3) QQ For the refrigerator COP COP 121  HP  R Desired effect = Heat rejected from low QQ12 temperature body (Q2) (COP)HP  (COP)R 1 ------(3.4) Work done on the refrigerator W = Q -Q 1 2 From the equation (3.1) and equation (3.3) So 1 Desised Effect Q (COP)  ------(3.5) COP    2 HP  R workinputW Refrigerators and heat pumps are Q2 (COP)R  essentially the same devices; they differ in QQ12 their Objectives only. Refrigerator is to maintain the refrigerated space at a low 3.1.7 HEAT PUMP temperature. On the other hand, a heat pump absorbs heat from a low‐ Heat pump is a thermodynamic system temperature source and supplies the heat operating in a cycle that removes heat from to a warmer medium.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 3.2 EQUIVALENCE OF KELVIN PLANK surrounding and as a result of expansion AND CLAUSIUS STATEMENT the gas temperature reduces from TH to TL.

The two statements of the second law are equivalent. In other words, any device violates the Kelvin‐Planck statement also violates the Clausius statement and vice versa.

Process (3‐4): Reversible isothermal compression: The gas is allowed to exchange heat with a sink at temperature TL as the gas is being slowly compressed. The heat is transferred from the system to the surroundings such that the gas temperature remains constant at TL. Fig:-The violation of the Kelvin‐Planck statement leads to violation of Clausius. Process (4‐1): Reversible adiabatic compression: The gas 3.3 CARNOT CYCLE temperature is increasing from TL to TH as a result of reversible adiabatic compression. The efficiency of a heat‐engine cycle greatly Carnot cycle is the most efficient cycle depends on how the individual processes operating between two specified that make up the cycle are executed. The temperature limits. The efficiency of all net work (or efficiency) can be maximized reversible heat engines operating between by using reversible processes. The best the two same reservoirs are the same. known reversible cycle is the Carnot cycle. The thermal efficiency of a heat engine Consider a gas in a cylinder‐piston (closed (reversible or irreversible) is: system). The Carnot cycle has four W Q   1  L processes: th QQ HH For the Carnot cycle, it can be shown: Process (1‐2) Reversible isothermal TL expansion: th 1  ------(3.6) The gas expands slowly, doing work on the TH surroundings. The heat transfers in the 3.3.1 CARNOT THEOREM Reversible process from the heat source at

T to the gas. H It states that all heat engines operating

between same temperature limit none has Process (2‐3) Reversible adiabatic a higher efficiency than a reversible engine. expansion: The efficiency of an irreversible (real) cycle The cylinder‐piston is now insulated is always less than the efficiency of the (adiabatic) and gas continues to expand Carnot cycle operating between the same reversibly. So, the gas is doing work on the two reservoirs.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  If ηth ≤ ηth Carnot, it may be reversible or Entropy of a system is defined as the irreversible heat engine measure of degree of molecular disorder or  If ηth ˃ ηth Carnot, Heat engine is not random existing in the system. If higher is possible. the randomness higher will be entropy. When heat is supplied to the system, 3.4 REFRIGERATION AND HEAT PUMP randomness of the system increases, so the WORKING ON REVERSED CARNOT CYCLE entropy of the system increases. Reversed effect can be measured when heat is A refrigerator or heat pump that operates removed from the system. on the reverse Carnot cycle is called a Carnot Refrigerator, or a Carnot heat pump. 3.6.2 CLAUSIUS INEQUALITY LAW FOR The Coefficient of performance of any REVERSIBLE CYCLE refrigerator or heat pump (reversible or irreversible) is given by: Consider the cycle shown below composed QQof two reversible processes A and B. Apply COP LH, COP  R  HP the Clausius inequality for this cycle. QQHLQQH L COP of all reversible refrigerators or heat pumps can be given by: TT COP LH and COP   RH  P THLTTTH L If (COP)R˂ (COP)R,Rev The refrigerator is irreversible Refrigerator If (COP) = (COP) The refrigerator is R R,Rev reversible Refrigerator Apply the Clausius inequality for the cycle If (COP)R˃ (COP)R,Rev Refrigerator is not made of two internally reversible possible. processes: 3.5 CLAUSIUS THEOREM Qnet  0  T it states that for a cycle (reversible or int rev 22QQ    irreversible cycle) the cyclic integral of net  net δQ     is equal to zero or less than Zero. It 11TT int rev  int rev T   means along path A along path B Q Since the quantity (Qnet/T) rev is  0 ------(3.8) independent of the path and must be a  T  property, this property is known as δQ entropy S. if∮˂ 0, the cycle is irreversible cycle. T The entropy change during a process is δQ related to the heat transfer and the if∮ 0 the cycle is reversible cycle. temperature of the system. The entropy is T given the symbol S (kJ/K), and the specific δQ if∮ 0 the cycle is not possible. entropy is s (kJ/kgK). T The entropy change during a reversible process is defined as 3.6 ENTROPY Q dS   3.6.1 PHYSICAL SIGNIFICANCE OF T Rev ENTROPY 2 Q SS ------(3.9) 21 1 T Rev

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 3.6.3 CLAUSIUS INEQUALITY LAW FOR the net heat transfer is positive so the IRREVERSIBLE CYCLE entropy change will be positive. It means entropy of the system will increase. Consider the cycle 1-A-2-B-1, shown below, where process A is arbitrary that is, it can ii) Heat is rejected from the system be either reversible or irreversible, and process B is internally reversible. if heat is rejected from the system, for reversible process the net heat transfer is negative so the entropy change will be negative. It means entropy of the system will decrease.

iii) No heat Transfer or adiabatic Process

A cycle composed of reversible and If there is no heat transfer between the irreversible processes. system and surrounding the net heat

Qnet transfer will be zero. It means entropy of   0 the system will be constant. T int rev 22 QQnet  net       0 TT    11int rev int rev dS = 0 along path A along path B S2 = S1 So reversible adiabatic process is known as The integral along the reversible path, isentropic process. process B, is the entropy change S1 –S2. Therefore 3.6.5 ENTROPY CHANGE FOR 2 Qnet IRREVERSIBLE PROCESS S12 S 0  T 1 2 2 Q kJ Q net  net SSSsys  2  1   SS21  TK  T 1  1 Here, the inequality is to remind us that the In general the entropy change during a entropy change of a system during an process is defined as irreversible process is always greater than Q dS  net ------(3.10)  δQ T , called the entropy transfer. That is,  = holds for the internally reversible T process some entropy is generated or created during an irreversible process, and this  >holds for the irreversible process. generation is due entirely to the presence

of irreversibility. The entropy generated 3.6.4 ENTROPY CHANGE OF SYSTEM IN during a process is called entropy REVERSIBLE PROCESS generation and is denoted as S . We can gen remove the inequality by noting the i) Heat is added to the system following if heat is added to the system, for a 2 Q kJ reversible process, net  SSSSsys  2  1   gen  dQ  TK ds  1 T

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 Sgen is always a positive quantity or zero. Q TdS (3.11) Its value depends upon the process and net  thus it is not a property. S is zero for an 1 gen This is the second law of thermodynamics internally reversible process. And its value for a reversible process. increases with the irreversibility. So entropy generation is a path function. Entropy Change of an Isolated System:- For the isolated system the heat transfer is zero. δQ So, dS  for the isolated system, δQ = 0 T

In the above figure, the heat transfer in an internally reversible process is shown as the area under the process curve plotted on the T-S diagram.

3.7 COMBINED FIRST LAW AND SECOND LAW OF THERMODYNAMICS

According to first law of thermodynamics So δQ = P.dV + dU ------(3.12) dS≥0 for the reversible process The total entropy change of an isolated According to first law of thermodynamics system during a process always increases δQ = T.dS ------(3.13) or, in the limiting case of a reversible for the reversible process process, remains constant. Universe is an From the equation (3.12) and (3.13) isolated system so entropy of universe T.dS = P.dV + dU ------(3.14) during a process always increases or, in the it is valid for reversible as well as limiting case of a reversible process, irreversible process. remains constant. Equation (3.14) can be written as The increase in entropy principle can be T.dS = dH – VdP ------(3.15) summarized as follows:   0 Irreversibleprocesses 3.8 ENTROPY CHANGE FOR AN IDEAL  GAS Sgen  S total   0 Re versibleprocesses  0 Impossibleprocesses  From the Equation (3.14) entropy change is Heat Transfer as the Area under a T-S given: Curve: P.dV dU For the reversible process, the equation for dS  TT dS implies that For an ideal gas, Q dS  net P m.R T dU m.C dT and  V TV Q TdS net then The heat transfer in a reversible process is m.R.dV m.CVdT the differential area under the process dS  VT curve plotted on the T-S diagram.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Integrating the equation from state 1 to From the Equation (3.14) state 2 T.dS = P.dV + dU 2 2 m.R.dV 2 m.C dT For the constant volume process, dV = 0 dS V    So T.dS = dU = m.CV dT 1 1 VT1 dT T VT22       (3.20) S2 S 1  m.R In  m.Cv .In        (3.16)  dSv m.Cv VT11 From the Equation (3.1) From the equation (3.15) T.dS = dH - V.dP dH VdP For the constant pressure process, dP = 0 dS  So TT T.dS = dH = m.CP dT For an ideal gas,dH = m.CP dT and dT T V m.R       (3.21)  dS m.C TP p p m.R.dP m.C dT From the equation 3.20 and 3.21 it is Then dS P PT cleared that slop of isobaric process is 2 2 2 lesser than slop of isocoric process on T – S m.R.dP m.C dT dS P diagram.  PT  1 1 1 PT22 Examples S2 S 1  m.R In  m.Cp .In        (3.17) PT11 Q.1 An ideal gas cycle is represented by a rectangle on a P-V diagram. If P1 Solving the equation (3.16) and (3.17), We and P2 are the lower and higher get pressures and V1 and V2, the smaller VP and larger volume, respectively. 22 S2 S 1  m.C p In  m.Cv .In        (3.18) VP11 Then a) Calculate the work done per cycle. 3.9 ENTROPY CHANGE FOR FINITE BODY b) Indicate which parts of the cycle involve heat flow in the gas For the finite body entropy change may be c) Show that 1 written as  2 2 2 δQ m.c.dT PV21 dS      PPVV2 1 2 1 1 1TT 1 If heat capacities are constant. 2 S21 S  m.C.In      (3.19) 1

3.10 SLOP OF ISOBARIC PROCESS AND ISOCHORIC PROCESS ON T-S DIAGRAM

Solution: a) w=area of the cycle

(P2  P 1 )(V 2  V 1 ) b) Process a-b and b-c Heat absorbed by 1 Mole of gas in one cycle

Isobaric and isocoric process

P1 Now, TTab and P2 V 1 RT b P2

VPV2 2 1 Tc T b ;T b VR1

PV12    Q CV T b  1    CPP T   1   PV21     PVPPVV2 1 2 1  2 1  CCPP    RPV 21    (a) Maximum efficiencies of the Ans….c) heat engine cycle

T2 313 nmax  1   1   1  0.358  0.642 W PPVV2 1  2 1     T1 873 Q  PVPPVV2 1 2 1  2 1  = 64.2 % CCPV    RP V W  21   Again 1 = 0.642 Q RPPVV2 1  2 1  1  W 0.642  2000  1284KJ CVPPCPVVV 1 2 1   P 2  2  1  1 CC Maximum cop of the refrigerator  PV VP cycle CC12 T VPVVPP (cop)  3 2 1 2 1 max TT γ1 23  V γP 253 12 4.22 VVPP 313 253 2 1 2 1 Q Also cop4 4.22 W Q.2 A reversible heat engine operates 2 between two reservoirs at Since ww12 w360KJ temperatures of 600℃ and 40℃. w21 w w 1284  36 0 924KJ The engine drives a reversible Q4 = 4.22 × 924 = 3899 KJ refrigerator which operates between reservoirs at temperatures  Q2  Q 1  W 1  2000 –1284  716 KJ 0 at 40℃ and -20℃. The heat transfer Heat rejection to t w  40 C reservoir to the heat engine is 2000 KJ and the QQ 716  4823  5539 KJ net work output of the combined 23 b) Efficiency of the actual heat engine refrigerator plant is 360 KJ. engine Cycle (a)Evaluate the heat transfer to the N = 0.4 ×η = 0.4 × 0.642 refrigerant and the net heat transfer max W 0.4  0.642  2000  513.6 KJ to the reservoir at 400C (b) 1

Reconsider (i) given that the W2 513.6 – 360 153.6 KJ efficiency of the heat engine and the Cop of the actual refrigerator as do. cop of the refrigerator are each 40% COP = 0.4 × 4.22 = 1.69 of the maximum possible values. Therefore

Q4 153.6  1.69  259.6 KJ

Q3 259.6  153.6  413. 2KJ

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission QQ2 1 W 1  2000 – 513.6 1486.4 K J Since S0  0 Heat rejected the 40 C reservoir 98 103 W =Q Q  413.2  1486.4  1899.6 KJ 630   0 23 100 W Q.3 A System has a heat capacity at 980  630  0 100 constant volume C = AT2 Where V W A=0.042J/K3. The system is  350 originally at 200 K , and a thermal 100 reservoir at 100 K is available. What W (max) = 35000 J = 35 KJ is the maximum amount of work that can be recovered as the system Q.4 A body of constant heat capacity CP is cooled down to the temp of the and initial temperature T1 is placed reservoir? in contact with a heat reservoir at Solution temperature T2 and comes to Heat removed from the system thermal equilibrium with it. If TT21 , Calculate the entropy change of the universe and show that it is always positive. Solution: T2 dT T S body  m m l 2 CP  TTCP n T1 1

TT21  S reservoi mr CP T2 TT T T 100K 21 22 S universe=mCP  In 1 2 Q C dT 0.042Td T  TT12  V T11T 200K TT1 Putting 1- 12=n, =  0.042[T3 / 3] (100K) (200K) T21 T 1-n 0.042 T J / K3 (100 3 200 3 )K 3 Since 1 >1, n is positive 3 T2 3  98  10 J Suniverse n2 x 3 n 4 100K 100K ln1  ln  1  n   n     dT 2 dT S System  C  0.042T mCP 2 3 4   V 200K TT200K (ds) > 0 It means entropy of universe is always positive. 0.042  2 Q.5 A hypothetical device is supplied J / K3 100 2 200 2 K 2   630J / K with 2 kg/s of air at 4 bar, 300 K.   Two separate streams of air leave QW 98X103  W the device, as shown in figure below. S res= 1 Tres 100 Each stream is at ambient pressures S  Working flued in HE = 0 of 1bar, and the mass flow rate is the same for both streams. One of the = system = ros exits streams is said to be at 330 K 98X103  W while the other is at 270 K. The  6 30 100 ambient temperature is 300 K.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Determine whether such a device is Where C = Specific heat of fluid. possible. Solution: Solution: Heat given by part-1 The entropy generation rate for the = heat taken by part-2

control Volume mc T12 T   mc  T  T 

TT12 T   2 Change of entropy

ds (dS)1 (dS) 2 TTdT dT mc  mc TT TT12 TT    mcln  mcln   TT12    2   2    Sgen  me S e m i S i TT mcln  mcln   T .T T .T  m2 S 2 m  3 S 3 m  1 S 1 1 2  1 2 

 m S m  S  m   m  S TT12 2 2 3 3 2 3  1 But T  m S  S   m  S  S  2 2 2 1 3 3 1 Substituting the value of T in the TP22 Expression Now S2 S 1  C P In  R In TP11 TT   12 330 1 2 1.005 In 0.28 7In 0.494KJ / kg.K dS 2mcIn  300 4  T12 .T TP33  S3 S 1  C P In  RIn  TP11 270 1 1.005 In  0.287In 300 4  0.294 KJ / Kg.K sgen 1  0.484  1  0.292  0.786KJ / K = 0.786 K Since Sgen >0, the device is possible. Such device actually exist and

called Vortex tube. Q.7 A heat pump operates between two Q.6 A mass of m kg of fluid at temperate identical bodies. In the beginning , me T1 is mixed with an equal mass both the bodies are at same temp T1 of game fluid at temperature푇2 . but operation of heat pump cools Prove that the resultant change of down one of the body to entropy of universe is temperatureT2. Show that for the TT 12 operation of heat pump the 2 minimum work input needed by the  2MEln TT12. heat pump for unit mass if given

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2 T1 Wmin   T2  2T 1 . Where C is T Solution: 2 Q = C dT specific heat of bodies. Solution: Let the final temperature attained by body to which heat is rejected is

W = 푄퐵 − 푄퐴 Heat transfer from body A,

A C(TQ 1 T)2 Heat absorbed by body B,

QTB C f T 1  300 2 Work input = Q   0.04T  dT 600 QB Q A  C T f  T 2  2T 1  300 0.04T3 Now entropy change of body A  3 T 600 dS   CIn 2 A T 0.04 33 1  300 600  Entropy change of body B 3 = -2520 kJ T dS  CIn f Now the change of entropy of the  B  T1 body For a reversible cycle entropy T2 dQTT2 CdT2 0.04T2 dT ds   change of universe = 0.  body    TTTTTT dS   0 1 1 1 universe dS   dS  0 0.04 AB 30022 600   5400J / k substituting the values, we get C 2  TT    Entropy change of sink ln ln2f  Cln   0 Q2 QW TT11    dS   2 TT T .T 00 Cln2f  0 For maximum work considering T2 1 dS  Universe = 0 T T .T 1 2 f dS dS 0  12   2 T1 QW Tf  or 5400   0 T2 T0 Substituting the value of Tf in 2520 W equation (i) or 5400   0 300 2 T1 so W = 900 kJ Wmin  C  T2  2T 1 T2

Q.8 A finite thermal system having heat capacity C=0.04T2J/K initially at 6oo K. Estimate the maximum work obtained from the thermal system if surrounding temperature is 300 K.

The second law of thermodynamics tells us The maximum work done obtained in the that it is not possible to convert all the heat cycle will be the work done in the absorbed by the system in to the work. reversible cycle. Only some fraction of heat can be So for the reversible cycle, converted on the work. This work will be W T  1  2 more valuable as compare to the heat. So th QT11 Work is high grade energy and heat is the T low grade energy. 2 W Q1 .  th  Q 1 . 1  T Example: High grade energy: 1 But the ambient temperature is T0. The sink 1) Mechanical work temperature will be T0. 2) electrical energy T0 3) water power Wmax  Q1 . 1   Q1  T0.  dS  T 4) wind power  1  5) kinetic energy of a jet Where dS is the entropy change of the 6) tidal power. system. The Carnot cycle and the available energy are shown in figure. The area 1-2-3- Example: Low grade energy: 4 represents the available energy. Suppose 1) Heat or thermal energy a finite body is used as a source. Let a large 2) heat derived from nuclear fission or number of differential Carnot engines be fusion. used with the given body as the source. 3) Heat derived from combustion of fossil fuels. 4) Solar energy. 4.1 AVAILABLE ENERGY The maximum amount of work that can be obtained by a system in a cycle is called as Available Energy (AE). Suppose a certain quantity of energy Q1 as heat can be received from a reservoir, at temperature T1.The sink temperature is T2. It rejects the heat to the surrounding at temperature of T2. The maximum possible work done by the system in the cycle is known as Available Energy. The maximum fig.: Available and unavailable Energy possible work can be achieved when the cycle is reversible cycle. Suppose a finite body is used as a source. Let a large number of differential Carnot engines be used with the given body as the source.

T0 dW dQ  dQ 1  T If the initial and final temperatures of the source are T1 and T2 respectively, the total Fig: heat engine

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission work done or the available energy is given TT00 by  AE   AE   Q1 . 1 2 TT TT22  2 1  T0 dQ w dQ  dQ 1   Q  To  TT    T ET  Q.T 1 2 TT11 1 0  TT1. 2 or w Q  T0 |  S| = T0. (dS) – T0. (dS!) So Decrease in available energy or increase in  T0  unavailable energy Available EnergyAE   Wmax  Q1 .1  T 1 T12 -T 1 =Q1 .T0    T.0  dS  – T0 . dS ----- (4.3) = Q1 –T0. (dS) ------(4.1) T12 .T From equation 3 it clears that when same 4.2 UNAVAILABLE ENERGY amount of heat is supplied at the lower temperature its value is low, because it The shaded area 4-3-B-A represents the produces the less amount of work as energy, which is discarded to the ambient compare to work done produced by the atmosphere and this quantity of energy, heat when it is supplied by higher cannot be converted into work and is temperature. called Unavailable energy. So according to first law of To thermodynamics heat at higher and lower UAE T0.  dS  Q1  . ----- (4.2) T1 temperature has the same significance. But according to second law of 4.3 LOSS IN AVAILABLE ENERGY thermodynamics heat at the higher temperature is more valuable as compare Suppose a certain quantity of energy Q is to heat at lower temperature. transferred from a body at constant So, temperature T1 to another body at  First law of thermodynamic is called as constant temperature T2 (T2

 T0  AE   Wmax Q1 .1 = Q1 –T0. (dS) 4.4 AVAILABILITY FUNCTION 1 T 1 Available energy, with the body at T2, The availability of a given system is defined Available Energy as the maximum useful work that can be T obtained in a process in which the system AE W Q. 1 0 Q–T.dS!  2 max 1 1 0   comes to equilibrium with the T2 surroundings or attains the dead state. It is given by maximum work done – atmospheric work.

A) AVAILABILITY FUNCTION FOR NON- FLOW PROCESS

Let P0 be the ambient pressure, V1 and V0 be the initial and final volumes of the system respectively. If in a process, the fig.: loss of available energy system comes into equilibrium with the Decrease in available Energy

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission surroundings, the work done in pushing The maximum power that can be obtained back the ambient atmosphere is P0 (V0-V1). in a steady flow process while the control volume exchanges energy as heat with the Availability WW–PVV.useful  max 0 0 1  Consider a system which interacts with the ambient at T0: The steady stet steady flow equation is ambient at T0. Then from the first law of thermodynamics given by 11 δQ ̶ δW = Du H mC22  mgZ   Q  H  mC  mgZ   w 1 221 1 2 2 2 cv Max work can be obtained when process is The subscript 1 and 2 refer to the entrance reversible process. and exit respectively. At the exit let the Let S and S be the entropies of the 1 2 system be in equilibrium with the systems at the entrance and exit of the environment at pressure P and device then (dS) = S –S . and (dS) o system 2 1 temperature To. Let symbols without surr = - (Q / To) subscripts refer to the entrance condition From the entropy principle of the system and changes in KE and PE are S21 –S  –  Q/To  0. negligible. Then useful work is given by For the reversible process, W = (H2-H1) + Q

S21 –S  –  Q / To   0 The greater the value of Q larger will be the useful work. Thus W will be maximum Therefore Q To S S 21  when Q is a maximum.

WUUTSSmax 1  0   0  1  0  Let S1 and S2 be the entropies of the systems at the entrance and exit of the UTS–UTS1  0 1   0  0 0  device then (dS)system = S2 – S1. and (dS)surr So Availability = Wuseful = (U1-T0 S1) – (U0-T0 = - (Q / To) S0)- P0(V0-V1) From the entropy principle = (U1+ P0V1-T0 S1) – (U0+P0V0-T0 S0) ( S2 –S1 ) – (Q /To) ≥ 0 = ɸ1- ɸ0 For the reversible process, Availability Function (=U+P0V-T0S---- ( S2 –S1 ) – (Q /To) = 0 (4.4) Therefore Q = To (S2- S1). Where ɸ = U+P0V-T0S is called the The useful work Wuseful = W = (H2-H1) + To availability function for the non flow (So-S) process. Thus, the availability on any given = (H1-ToS1) – (H2 – ToS1) state: ɸ 1- ɸ 0 So the maximum possible work done If a system undergoes a change of state Wmax = ψ1 - ψ2 from the initial state 1 (where the Where ψ is known as the availability availability is (ɸ 1- ɸ 0) to the final state 2 function for steady flow. (where the availability is (ɸ 2- ɸ 0), the ψ = H - ToS------(4.6) change in the availability or the change in maximum useful work associated with the 4.5 IRREVERSIBILITY process, is ɸ 1- ɸ2. So the work done by the system when The actual work done by a system is always system undergoes from state 1 to state 2 is less than the idealized reversible work, and given by the difference between the ideal work and

W1212  ff   UPVTS 10101     UPVTS 202   02  actual work is called the irreversibility of ------(4.5) the process. So irreversibility can be given as B) AVAILABILITY FUNCTION FOR FLOW I = Wmax – Wact PROCESS This is also some time referred to as degradation or dissipation. For a non flow process between the equilibrium states,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission when the system exchanges heat only with ii) dH = Tds + VdP the environment Gibbs function (G) and the Helmholtz I U1  U2 – To S1 S2 – U1 U2  Q    system   surr function (F) can be given as = To (S2-S1) – Q dG = dH – TdS = To (dS) system + To (dS)surr Gibbs function = To [(dS)system + (dS)surr ] Hence I> 0. dF = dU –TdS Similarly for the steady flow process Helmholtz function Or these functions can be written as I = Wmax – Wact = To (S2-S1) – Q iii) dF=-PdV - SdT − Helmholtz function = To (dS) system + To (dS)surr iv) dG = VdP - SdT - Gibbs function = To[ (dS)system + (dS)surr ] Thus same expression for irreversibility U, H, F and G are thermodynamic applies to both flow and non-flow process. properties and exact differential ,so The quantity To[ (dS)system + (dS)surr ] Appling the Exact differential rule equation, represents irreversibility. we get Maxwell equations: I = To(dS) = To[ (dS)system + (dS)surr ] TP    universe     ------(4.11) ------(4.7) VS sv   TV    THERMODYNAMIC RELATIONS     ------(4.12) PS    sP 4.6 EXACT DIFFERENTIAL EQUATIONS PS        ------(4.13) TV VT   Theorem No. 1: If a variable Z is a function VS    of the two variable x and y i.e.     ------(4.14) TP    If Z = f (x , y ) PT dZ = Mdx + Ndy These four equations are known as M and N are the function of x and y. Maxwell’s equations. The differential is exact differential if 4.8 T – DS EQUATIONS MN  ------(4.8) yx x y Let Entropy is function of temperature and volume S = f (T , V) Theorem No. 2: if a variable f is the Differentiating the equation function of the x, y, z and a relation exists SS    i.e. If f = f (x, y,z ) dS  dT   dV TV VT   x  y    z  multiplying the equation with temperature T     1 ------(4.9) yf   z ff   x  SS    Tds T  dT T   dV TV VT   Theorem No. 3: If a variable z is the We know that function of x, y i.e. S SP    If x = f (y, z ) ------(4.10) TC V and     T VT V  TV   4.7 MAXWELL’S EQUATIONS substituting the values in TdS equation, we get A pure substance in a single phase has only p Tds Cv dT T dv ------(4.15) two independent variables. T v According to combined first Law & Second This is known as first T.ds Equation. Law of thermodynamics i) dU = Tds – PdV

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Let the entropy is the function of V temperature T and Pressure P Tds CP dT T dP T S = f (T , P) P Differentiating the equation V d CP dT  T dP Vdp SS    T P ds  dT   dP TP  PT   v multiplying the equation with temperature dH Cp dT  V  T dP ------(4.19) T p SS     Tds T  dT T   dP For an ideal gas TP    PT PV = mRT We know that, V mRT S  SV     TV T  CP and      TPP T  PT     P T P Substituting the Value in equation 4.19, we substituting the values in TdS equation, we get get dH C dT  V  V  dP v P TdS Cp dT T dp ------(4.16) dH = Cv/dT---- (4.20) T p This is known as Second T -dS Equation 4.11 DIFFERENCE IN HEAT CAPACITIES

4.9 ENERGY EQUATION Equating the first and second T- dS Equation from equation 4.15 and 4.16 From combined first Law & second Law of VP    thermodynamics TdS=CPV dT T  dP C dT  T  dV TT    dU = TdS – P.Dv PV Substituting the first TdS equation in the PV    Or CPV C  dT  T  dV T   dP internal energy equation, TT VP   P Or dU  CV dT T dV P.dV T V P V T dV T dP or T T dT V P p CCCCPVPV dU Cv dT  T  p dv     (4.17) T v Or For an ideal gas TT    dT  dV   dP PV = mRT VP PV   P mRT P V or TP T T TVV T TT  T   V  and P  substituting the value in equation 4.17, we     CCVCCPPV  PVPV    get ------(4.21) dU CV dT  P  P  dV     (4.18) From the equation 4.21 dU =CV.dT PV    CCT PV     TT VP   4.10 ENTHALPY EQUATION PTV        But       1 From combined first and second Law of TVP VPT       2 thermodynamics, VP    dH = TdS + VdP CCTPV       ------(4.22) TV T  T

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission It indicates the following facts A gas is made to undergo continuous 2 V throttling process by a valve as shown in (a) Since is always positive and fig. Let P1, T1 be the arbitrarily chosen T  pressure and temperature before throttling P  and P2, T2 are the pressure and is negative so CCPV is always V T temperature after the throttling. These are positive. plotted on T-P diagram. (b) at absolute temperature (T=OK);

(C ) for an ideal gas

PV  mRT V MR V  TPTP P MRT  2 VV T V2  MRT Then the initial temperature and pressure CPV  C   T X22 X TV of the gas are set to new values, a family of CCR ------(4.23) isenthalpic is obtained for the gas as shown PV in fig. the locus of the points where μ is

zero is known as Inversion curve. The 4.11.1 VOLUME EXPANSIBILITY β   region inside the inversion curve when μ is negative is called as heating region. The 1V region inside the inversion curve when μ is        (4.24) positive is called as cooling region. VT p

4.11.2 ISOTHERMAL COMPRESSIBILITY

1U K  T       (4.25) VP substituting the values from equation 4.24 and 4.25 in equation 4.22, 2 1V TV  VT CC P PV 1V The enthalpy change of a gas is given by   V VP T dh Cp dT T V dP 2 T TVB p CC 4.26  PV K dh = 0 T  1  V  4.12 JOULE KELVIN COEFFICIENT     T    V    (4.28) P hp Cp  T  The slop of an isenthalpic Curve (throttling For an ideal gas, process) on T-P diagram is known as Joule V Kelvin Co-efficient (μ). T V 0 T p T  ------(4.27) So μ=0 P h So there are some important points:

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  For an ideal gas Joule Kelvin Co-  717.4KJ efficient is Zero. For m mass flow rate of gas  In the cooling region the value of Joule Heat given by gas=heat gain by Kelvin Co efficient is positive. water  In the heating region the value of Joule mg c pg T 2 T 1  m w  LH  Kelvin Co efficient is negative. m X1.08 1300 320 1X1662.5  At the inversion curve, value of Joule g  

Kelvin Co efficient is zero. mg  1.57kg

AE1  1.57  739.16  1161.1KJ Examples Loss of Availability = Agas - Aw Q.1 In a steam generator, water is 0 = (1161.1-717.4) evaporated at 26.0 c while the = 443.67 KJ KJ combustion gas (CP  1.08 .k) is kg Q.2 Air expands through a turbine from cooled from 13000 C to3200 c . The 500 Kpa, 5200C to 100 KPa , 0 surrounding are at30.00 c . 300 C . During expansion 10 KJ/kg Determine the loss of available of heat is lost to surrounding which 0 energy due to the above heat is at 98 KPa , 20 C. neglecting K.E. transfer per kg of water evaporated and P.E. changes, determine per kg (Latent heat of vaporization at of air (a) the decrease in availability (b) maximum work (c) the water at 2600 c 1662.5KJ / kg ). Irreversibility. For air, take Solution: c 1.005KJ / kg.k,h c T, where Availability decrease of gas p p AE H  T S  (H  T S ) Cp is constant. 1 1 0 1 2 0 2 H1  H 2  T 0 (S 1  S 2 ) Solution: T The change of entropy of the air is m c T  T  T m c ln 2 g p 1 2  0 g p  TP    T1 22 S2 S 1  mc p ln  mR ln   TP11    T2 mg c p T 1  T 2   T 0 ln  For unit mass of air T1 513 1 Given S2 S 1  1  1.005ln ( ) 1  0.287ln  793 5 T 1573k,T  320  273  12 0.4619 0.3267 593kT0  30  273  303k  0.1352KJ / kg k 593 Change in availability AE1g m X1.08 1573 593   303ln  1573 12101HTS  H 202 TS    HH 12  T(SS) 012   739.16m KJ g mCP T 1  T 2   T 0 (S 2  S 1 ) Availability increase of water 11.005520 – 300  293  0.1352   260.7 KJ / kg AE2 Q 1 T 0 ds Maximum work done

Tds10 T ds wmax  1  2  260.7KJ / kg But from S.F.E.E T10 T  ds Q hh w LH 12 T10 T  m [Neglecting K.E. and P.E.] T1 wactual  Q  (h1  h 2 ) 303 1  1662.5 1   10  1X1.005(520 300) 533

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q  (h12  h ) Let mass of air in the vessel is m kg PV 211.1 KJ/kg Mass of air (m)  11 So, RT1 Irreversibility I w w 1400 1 max actual  260.7 211.1 0.287 998  49.6KJ / kg 10.89kg Volume of gas after reaching the Q.3 0.2 kg of air at 3000 c is heated dead state reversible at constant pressure to mRT0 V0  2066 k. Find the available and P0 unavailable energies of the heat 10.89 287 298 0  added. Take TO  30 c and 100 3 cp  1.0017 KJ/kg.k. V0  9.31m Solution: Availability at initial state

Let the 0.2 kg of air is heated at cons AE1  1  2 leant pressure. So entropy change UUTSSP(VV)1  0  0 1  0   0 1  0 TP    22 mC T  T  T S2 S 1  mC P l n   mRln   V 1 2  0 TP 11     VP11    mCP ln  mCV ln    P0 V 1  V 0  2066 VP 0.2 1.0047ln  00    573   m0.718X 448 298   298  0.2577KJ / K 1 1400 Increase in available energy of air 1.005ln 0.718ln 931 100 AE H2  H 1   T 0  S 2  S 1   + 100(1-9.33) mC T  T   303  0.2577 P 2 1 9.31 150  103.52  831  0.2X1.0047 2066 573   78.084

1250.14 78.08 Q.5 A heat source at 6270 c transfer heat 1172.2KJ at rate of 3000 KJ/min. To a system Heat Input mC (T T ) P 2 1 maintained at 2870 c. A heat link is 0.2  1.0047(2066  573) available at 270 c. Assuming these 1250.24 temperatures to remain constant Unavailable Energy=Heat Supplied Find: Available energy i) Change in entropy of source 1250.24 1172  78.08KJ ii) Entropy production accompanying heat transfer Q.4 A pressure vessel has a volume of iii) The original available energy 1m 3 and contains air at 1.4mPa , iv) The Available energy after heat 1750 c . The air is coaled to 250 c transfer by heat transfer to surrounding at Solution: Temperature of source . Calculate the availability in T 6270 c  627  273  900K the initial and final states and the 1 irreversibility of this process. Take Temperature of a system 0 P0a 100KP . T2  287 c  287  273  560K Solution: Sink temperature

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 0 (ii) Entropy change of mixture T0  27 c  27  273  300K i) Change of entropy of the source Ds=Entropy change of 4 kg of Q 3000 water+Entropy change of b kg at ds  1   0.056KJ / Ksec water T 900X60 1 TTff    ii) Entropy production accompanying m1 c p ln  m2 c p ln   heat transfer heat transfer by TT12    source =Heat absorbed by System 349  349  ' 4  4.18ln  6  4.18ln   Tds12 T .ds 313  373  T .ds 1.82 1.66  0.15KJ / K ds'  1 T (iii)Unavailable energy with respect 2 0 3000 KJ to water at 40 c 0.089 UAE T ds system 560X60 Ksec 0   =29891.82)=542.36KJ iii) The original available energy Q.7 Using Maxwell’s relations, show that T0 300 Q1  1   3000 1  for a pure substance T 900 1 TdS C dT TV dV  2000KJ / min P β Where β is co-efficient of thermal iv) Available energy after heat expansion, K is co-efficient of transfer compressibility and C,CPV are specific heat at constant pressure T0 300 Q1  1   3000 1  and constant volume respectively. T 560 2  Solution: 1392.8KJ / min Let entropy of the pure substance S f (P,T) Q.6 4 kg of water at 400c mixed with 6 TS    kg of water at 1000c in steady flow TdS T  dβ T   dT _ (1) PT process.  TP   Calculate: But (i) Temperature a resulting mixture TT Slopof Isobaricprocess (ii) Change in entropy SCP P (iii) Unavailable energy with respect S to energy receiving water at So TC P T 100c . P Solution: And from Maxwell’s equation (i) The final temperature at the SV        mixture is Tf , then PT TP   Heat transfer by 4 kg at water = Substituting the values in equation(1) Heat transfer by 6 kg at water V TdS CP dT T dP m1 c T f T 1   m 2 c  T 2  T f   T P

4 4.18 Tff  40   6  9.18  100  T  Coefficient of thermal expansion T 40  1.5 100  T  1V ff B   VT P Tff 1.5T  150  10 0 TdS=CP dT-TVβdP Tf  T6 c

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission And entropy of pure substance again If U f T,V  S f (T,V) UU    SS    dU  dT   dV dS  dT   dV TV VT   TV    VT UUP        SS      CV ,    T    P TdS T  dT   dV  2  TTT v   T   v TV VT For isothermal process TT dU 0 Slopof isocoricprocess  T SC V V UP     TP   S VT TC V  TV   T V But from the relation And from Maxwell’s Relation SP        TT VV   PR Substituting the values in equation(2) 0 TVV P Tds CV dT  T dV    (B) P TR T T V TV And V P RT q 1 V  1  V  TPP    K   andβ   2     TVVV VPVT TP   Uq BPV    So  2  .   VV KVT    T TP q From cyclic property dU dV  T 2 PVT        V  .1      Integrating the equation VTP        22 TPV q dU dV PVP        2  .     11V VTT TP      V 11 BP U U  q  So  21  VV12 KT V dh U  U  P V  PV h  h  Substituting the values in equation(3)  T 2 1   2 2 1 1  2 1 B 11 TdS C dT T dV P V P V  q  V K 2 2 1 1   VV12

h2–h1   PV–PV2 2 1 1  q Q.8 Derive the expression for (dh)T for a T substance that obeys the equation of 11  state given by VV12 RT q P  VV2 Solution: We know that internal energy of pure substance P dU CV dT  T  P dV T V

5.1 PURE SUBSTANCE  Liquid molecular spacing is comparable to solids but their molecules can float A substance that has a fixed chemical about in groups. composition throughout is called pure  Gas molecules have weakest molecular substance. Examples-Water, helium carbon bond strength. Molecules in the gas dioxide, nitrogen Etc. phases are far apart, they have no It does not have to be a single chemical ordered structure. The molecules move element just as long as it is homogeneous randomly and collide with each other. throughout, like air. A mixture of phases of two or more substance is can still a pure 5.2 P- V DIAGRAM OF PURE SUBSTANCE substance if it is homogeneous, like ice and water or water and steam. Let the ice be heated slowly so that its temperature is always uniform.fig show the state change of a pure substance (other than water which contract due to conversion from ice to water).line passing through all the saturated solid states is called the saturated solid line. The line passing all the saturated liquid points is called as the saturated liquid line. The line passing all the saturated Vapour points is called as the saturated Vapour line. Where the two lines (saturated liquid line and saturated vapour line) meets with each other is called as critical point. The triple point on P-V diagram is a line where all the phases, i.e. solid liquid and gas phases exist in equilibrium.

(Not a pure substance because the composition of liquid air is different)

5.1.1 PHASE OF PURE SUBSTANCES

There are three principle phases – solid, liquid and gas, but a substance can have several other phases within the principle phase. Nevertheless, thermodynamics deals with the primary phases only. In general Fig.: P-V diagram of pure substance  Solids have strongest molecular bonds. Their molecules do not move relative to At the pressure below than triple point each other. substance can exist in liquid form. The

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission region below the triple point line is solid According to Gibbs phase rule, vapour mixture region. 2+C=F+P For Water Where Critical pressure (Pc) = 221.2 bar C = Number of chemically independent Critical Temperature (Tc) = 374.150C components F = Number of degrees of freedom of 5.3 TRIPLE POINT intensive properties P = Number of phases presented Triple point is state at which solid, liquid For a pure substance existing in a phase C = and gas phases exist in equilibrium. On a P- 1, P = 1 so, T diagram, this condition is a point. But on F = C + 2 – P = 1+2-1 = 2 P-V Diagram it is a line. The curve which It means there are two properties required shows the phase change of solid into liquid to b3 known to fix up the state of the or liquid into solid is called as fusion curve. system at equilibrium. The curve which shows the phase change of For a single component and two phase liquid into vapour or vapour into liquid is system (i.e. critical point), called as vaporization curve. C =1, P = 2 F = F = C + 2 – P = 1+2-2 = 1 It means only one property is required to fix up the state of the system at equilibrium. if all three exists in equilibrium (i.e. triple point) then C = 1, P = 3 F = C + 2 – P = 1+2-3 = 0 This state is unique for a substance. It means at triple point all the intensive properties are fixed.

5.5 PHASE CHANGE OF PURE SUBSTANCE

Fig.: Triple point on P- T diagram Thermodynamics deals only with liquid to gases to generate power. So it is important The curve which shows the phase change of to understand the conversion of liquid into solid into vapour directly or vapour into vapour or vice versa. solid is called as sublimation curve. The Consider water at room temperature slop of vaporization and sublimation curve (20°C) and normal atmospheric pressure as is positive. The slop of fusion curve for shown in fig. most of the substance is positive but for water is negative. For water, Triple point temperature (Tt) = 273.16 K Triple point pressure (Pt) = 0.6113 KPa

5.4 GIBBS PHASE RULE

Gibbs phase rule describes the number of degrees of freedom, or the number of Fig. formation of steam variables that must be fixed to specify the The water is in liquid phase and it is called composition of a phase. compressed liquid or sub cooled liquid.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Now heat is supplied to water, its given pressure process of phase change is temperature will increase. Due to the shown T-S diagram. increase in temperature, the specific volume v, specific enthalpy h and specific entropy s will increase. As a consequence, the piston will move slightly upward therefore maintaining constant pressure. Now if continue heat is added to the water, the temperature will increase further until 100°C. At this point, any additional addition of heat will not increase the temperature of water; it will utilise to vaporize some water. This specific point where water starts to vaporize is called saturated liquid point. And the condition of the Fig.:T-S diagram representing phase liquid is called as saturated liquid. change for water at constant pressure If we continue to add heat to water, more Let us consider the different pressures and and more vapour will be created, while the repeat process of phase change. On the temperature and the pressure remain different pressure we get the different constant (T=100°C and P=1 atm). These saturated liquid point and saturated conditions will remain the same until the vapour point. The locus of all saturated last drop of liquid is vaporized. At this liquid points is called as saturated liquid point, the entire cylinder is filled with curve. And the locus of all the saturated vapour at 100°C. This state is called vapour points at different pressure is called saturated vapour. as saturated vapour curve. The state between saturated liquid and Where the saturated liquid and vapour saturated vapour where two phases exist is curve meet with each other is called as called saturated liquid-vapour mixture or Critical Point. The same process is shown wet vapour. on T-s and P-s diagram. After the saturated vapour phase, any addition of heat will increase the temperature of the vapour; this state is called superheated vapour. This concept can be applied to pure substance other than water.

5.6 T-S DIAGRAM OF PURE SUBSTANCE

During the phase change of a pure substance as discussed in the previous topic first addition of heat is utilized to increase the temperature end entropy of the system. And then further addition of heat is to change the phase of liquid in which only entropy change temperature remains constant. But further addition of heat is to superheat the steam in which temperature and entropy both change. At a Fig. T-s diagram of pure substance

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 1-2 represents the sensible heat of water at 1 atmospheric pressure. At 1000C saturation point is attained.

Latent Heating

The addition of heat which is utilized to phase change is called as latent heating. In process 2-4, heat is added to water that is utilised to convert the phase of liquid in to vapour and it is called as latent heating. the latent heat decreases as pressure increases and at the critical point latent heat of vaporization is zero. Fig. P-v diagram of pure substance Critical Point 5.7. PROPERTIES OF PURE SUBSTANCE It is point at which liquid directly converts Saturation Temperature into vapour phase. At the critical point, latent heat is zero. For water at the critical The temperature at which water starts point, boiling depends on the pressure. In other Pcr = 220.8 bar words, water starts boiling at 100 ºC but Tcr = 374.140C only at 1 atm. At different pressures, water boils at different temperatures. Degree of Sub cooling

Saturation Pressure If the actual temperature of liquid is less than the saturated temperature of liquid at At a given pressure, the temperature at given pressure the liquid is called as sub which a pure substance changes phase is cooled liquid. The difference of saturated called the saturation temperature (Tsat). temperature and actual temperature of Likewise, at a given temperature, the liquid is called as degree of sub cooling. Let pressure at which a pure substance the saturated temperature is Tsat. and actual changes phase is called the saturation temperature is Tsub.,then pressure (Psat). Degree of sub cooling = Tsat -Tsub

Sensible Heating Degree of Sub cooling

If the actual temperature of vapour is greater than the saturated temperature of vapour at given pressure the vapour is called as superheated vapour. The difference of actual temperature of vapour and saturated temperature is called as degree of superheating. Let the saturated The addition of heat which is utilised to temperature is Tsat. and actual temperature increase the temperature is called as is Tsup., then sensible heating. In T- S diagram, Process Degree of superheating =Tsup - Tsat

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Wet Vapour compression refrigeration system. In most of the thermodynamic problem, these Wet vapour is mixture of liquid and vapour important phases can be existed. So it is phase. in wet vapour region liquid and important to determine the parameters in vapour phase exists in equilibrium. different region of pure substance. Dryness Fraction or quality of mixture (x) 1) Specific Volume, enthalpy and If vapour is made wet by liquid droplets in entropy of saturated liquid suspension, it is important to know the degree of wetness. Any mass of wet In the T-S and P-h diagram it is shown by consists of dry saturated liquid and dry point 1. Enthalpy of this point is the heat saturated vapour. supplied to increase the temperature from 0 The ratio of mass of dry saturated vapour 0 C to saturated temperature at given to total mass of water vapour is known as constant pressure. these all parameters can dryness fraction (x). It is also known as be found Corresponding to saturated temp. quality of steam. at given pressure with help of stem table. Consider specific enthalpy, entropy and Let the mass of dry saturated vapour is mv and the mass of dry saturated liquid in wet volume of saturated liquid at a given pressure and temperature are h s and v vapour is m , then dryness fraction f, f f. l specific enthalpy, entropy and volume of m x  v saturated vapour at a given pressure and mmlv temperature are hg, sg and vg. Then latent heat of vaporization hfg = hg - hf specific volume Vfg = Vg - vf , Specific entropy Sfg = Sg - Sf From steam table at given pressure Enthalpy of saturated liquid h1 = hf Entropy of saturated liquid s1 = sf Important points Volume of the saturated liquid  At saturated liquid line mass of vapour v1 = vf is zero so value of x is zero along saturated liquid line.  At saturated vapour line mass of liquid is zero, so dryness fraction is 1. So dryness fraction varies from 0 to 1.  All dryness fraction lines converge at original point.

5.8 SPECIFIC VOLUME, ENTHALPY AND Fig. T- S diagram ENTROPY OF DIFFERENT PHASES

The specific enthalpy entropy and specific volume are very important parameters to define the performance of any power producing system such as steam power plant working on Rankin cycle and power absorbing system such as vapour Fig. P- h diagram

T5 2) Specific Volume, enthalpy & entropy S5 S g c p l n  T of sub cooled liquid 4 Volume of superheated vapour In the T-S and P-h diagram it is shown by T5 point 2. v3= vg.  T Enthalpy of subcooled liquid 4

h2 h f  c p T 1  T 1  , PROPERTIES OF GAS MIXTURE

Entropy of subcooled liquid 5.9 EQUATION OF STATE

T1 S5 S f c p l n  The ideal gas equation of state is P.v = RT T 2 can be established for the real gas with two Volume of subcooled liquid can be found at important assumptions of the ideal gas. The particular temperature at the given ideal gas assumptions are: pressure from the steam table. a) There is no attraction between the molecules of the gas. 3) Specific Volume, enthalpy and b) The volume occupied by the molecules entropy of saturated vapour is negligible. In the T-S and P-h diagram it is shown by When the pressure is very low and point 4. temperature very large, the intermolecular From steam table at given pressure attraction and volume of molecules has no Enthalpy of saturated vapour more significance. As the pressure of gas

hh4g , increases, the intermolecular force forces Entropy of saturated vapour of attraction and repulsion increases and ss also volume of the molecules becomes 4g appreciable compare to total volume of the Volume of saturated vapour gas. v4 = vg 5.9.1 VANDER WAALS EQUATION 4) Specific Volume, enthalpy and entropy of wet vapour Vander Waals, by applying the law of mechanism of individual molecule In the T-S and P-h diagram it is shown by introduced two correction terms in ideal point 3. gas equation. This is valid for the real gases. Enthalpy of wet vapour  P (v  b)  RT h h x.h , 2 3 f fg v Entropy of wet vapour Where a and b are Vander Waals constant. LH s s  xs  s  x a is introduced to account the existence of 3 f fg f T mutual attraction between molecules. The Volume of wet vapour term a/v2 is called as force due to cohesion. v3= x.vg The term b is introduced to account the 5) Specific Volume, enthalpy and volume of molecules is known as co – entropy of superheated vapour volume.

In the T-S and P-h diagram it is shown by 5.9.2 PROPERTIES OF CRITICAL POINT point 5. Enthalpy of superheated vapour According to Vander Waals equation, h5 h g  c p T 5  T 4  a P+2  v-b  =RT or Entropy of superheated vapour v

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission RT a 5.9.3 COMPRESSIBILITY FACTOR P 2 ------(1) (v b) v The ideal gas equation can be given as Pv = RT If the gas is compressible (real gas), then pv Z  RT Z > 1 or Z < 1

5.9.4 COMPRESSIBILITY FACTOR AT Slop of P-v diagram at critical point is zero, CRITICAL POINT

so differentiating the equation with respect At the critical point the compressibility to v and putting equal to zero. factor is given by dp RT 2a PVcc 0  2  3 Zc  dv c vb  v RTc RT 2a  ------(II) Substituting the value of P, V and T at 2 v3 critical point, we get vbc   c a critical point is not only zero slop curve but X3b 2 3 at this point slop change so second Z 27b c 8a derivative of P with respect to v is also R X 8 zero. Differentiating the equation II and 27b R putting equal to Zero, 3 Zc  .375 d2 p 2RT 6a 8

240 3  dv v c vbc   c 5.10 PROPERTIES OF GAS MIXTURE

RT 3a  ------(III) 5.10.1 DALTON’S LAW OF PARTIAL 3 v4 vbc   c PRESSURES

Now dividing the equation II by equation Let consider a homogeneous mixture of III, we get inert ideal gas at pressure P, Volume V and 2 temperature T. Let us suppose there are n1 vcc b  v 3 numbers of molecules of gas A, n2 number Vc = 3b of molecules of gas B and n3 number of Substituting the value of vc in equation III, molecules of gas C. we get 2 2a 2b  8a RT  27b3 27b 8a Tc  27bR Substituting the value of Tc and vc in equation I, we get 8a RX a P 27bR 2b 9b2 a pc  2 27b fig. gas mixture

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Then the equation of state for mixture of In the same way internal energy, enthalpy the gas is and specific heats are given as PV = (n1+n2+n3) ṜT Ṝ = universal gas constant = 8.314 KJ/ kg Specific Internal Energy of the mixture mol K Partial pressure of gas A, B and C can be m u m u m u Ue  1 1 2 2 3 3 written as m1 m 2 m 3 nRR .T n .T nR .T P= 12 ,P = ,P = 3 1 VV2 3 V Specific enthalpy of the mixture According to Dalton law of partial pressure PPP P= m h m h m h 1 2 3 He  1 1 2 2 3 3 nRR .T n .T nR .T m1 m 2 m 3 P= 12++3 VVV Specific heat at constant pressure of the n R.T or P= K mixture V where, m Cp m Cp m Cp Cp  1 1 2 2 3 3 nk n  n1  n23  n  m1 m 2 m 3

5.10.2 MOLE FRACTION Specific heat at constant volume of the mixture The ratio of number of mole of a given gas to the total number of moles in the given m1 C v1 m 2 C v2 m 3 C v3 mixture of ideal gas is known as mole Cve  m m m fraction (x). 1 2 3

n1 n 2 n3 x1 = , x2 = and , x3 = Examples n n n    Q.1 Determine the specific enthalpy of Equivalent Gas constant (Re) steam at 2 MPa and with a temperature of 2500C. Ideal gas equation for the gas mixture in Solution: the term of masses can be written as From the steam table For gas A in the gas mixture At 2 MPa , T 212.40 C P V = m R T sat 1 1 1 It must be the condition of super For gas B in the gas mixture heating because the temperature of P V = m R T 2 2 2 steam is higher than saturated For gas C in the gas mixture steam. P V = m R T 3 3 3 So Degree of superheating Using Dalton law of partial pressure, Tsup -T sat =250-212.4=37.6K PV = P1V+ P2V+ P3V or PV = m1R1T + m2R2T + m3R3T Specific enthalpy of steam at 2MPa PV = (m1R1 + m2R2 + m3R3)T------(I) is given

Ideal gas equation for a gas mixture, hg  2797KJ / kg PV = (m1 + m2 + m3).Re.T------(II) So enthalpy Comparing the equation I and II, we get h hg  c p T sup  T sat  m1 R 1 m 2 R 2 m 3 R 3 Re  = 2797.2 + 2.09 × 37.5 m m m 1 2 3 = 2875.9 KJ/kg

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.2 Steam 0.95 dry at a pressure of 0.7 Enthalpy gained by water = MPa is supplied to a heater through Enthalpy lost by steam a pipe of 25mm internal diameter. m 376.8 79.8   81.9(2658.8  376.8) The velocity in the pipe is 12 m/sec. 81.9 2282 water enters the heater at 190C , the m  steam is blow into it and the mixture 297  of water and condensate leaves the m 629.28 kg / hr . heater at 900C. Calculate: a) mass of steam entering the heater Q.3 1.5 kg of steam originally at b) mass of water entering the heater pressure of 1 MPa and temperature Properties of steam are: 2250C is expended until the pressure becomes 0.28 MPa. The dryness fraction at the steam is 0.9. Determine the change of internal energy. Solution: At 1 MPa and 2250C, from steam Solution: table Specific volume of dry (0.95) at pr. 3 h =2886KJ/kg,v =0.2198m /kg 0.7 MPa is 11 1X106 V1g xv  0.95  0.273 So u1 =h 1 -p 1 v 1 =2886-3 X0.2198 = 0.259 m3/kg 10 a) Steam volume passing/sec = 2886 – 219.8 π = 2566.2 KJ/kg  dl2 At 0.28 MPa and dryness fraction 0.9 4 h2 h f 2 xh fg2 π 3 2 25 10   12 = 551.4 + 0.9 X 2170.1 4 = 551.4 + 1953.09 Steam volume passing / = 2504.49 KJ/kg π 2 hr   0.025  12 3600 Volume 4 v xv  0.9  0.646 mass of steam entering / hr 2 g2 3 π 2 = 0.581 m / kg 0.025  12  3600   Internal energy u h p v  4 2 2 2 2 0.259  2504.49 – (0.28103  0.5814 ) = 81.9 kg/hr = 2504.49 – 162.8 = 2341.69 KJ/kg b) Specific enthalpy of steam Hence change of internal energy entering heater is U2 U 1  m u 2  u 1  h2 h f  xh fg  697.1  0.95  2064.9 1.5 2341.69 – 2566.2 = 697.1 + 1961.7   = 2658.8 KJ / kg = - 489.77 KJ From the steam table at 900C Q.4 The dryness fraction of steam at a hf  376.8KJ / kg pressure of 2.2 MPa is measured And at 190C using throttling calorimeter. After h 79.8KJ / kg f throttling, the pressure in the So for the heater applying energy calorimeter is 0.13 MPa and the balance equation temperature is 1120C. Determine

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission the dryness fraction of steam at 2.2 Determine the dryness fraction of MPa. steam after throttling. Solution: - Solution: From the steam table For the throttling process

hf1 x 1fg1 h  h f2  x 2fg2 h

830.1+0.7×1957.7=928.8+X2+2250.8 1771.7 x  Property of Superheated steam 2 2250.8

Specific x2  0.787 enthalpy kJ/kg Pr Sat.temp  The steam becomes drier in this 0 0 MPa ( C) hg At 150 C case. 0.1 99.6 2675 2777 0.5 111.4 2693 2773 Q.6 a) Determine the volume occupied Enthalpy of this stem after throttling by 1 kg of steam at a pr. Of 0.85 can be found with help of linear MPa and having a dryness interpolation fraction of 0.97 . b) This volume is expanded adiabatically to a pressure of 0.17 MPa, law of expansion is PV1.3 = C. Determine i) Final dryness fraction of the steam ii) Change of internal energy of system during the expansion. 0 h 2 at the temp. of 112 C can be Solution: found by linear interpolation Properties of system from steam

TTsup sat table h2 h g  h 150  h g  3 TT At 0.85 MPa , vg = 0.2268 m /kg sup sat a) So =2685.2+(2774.6-2685.8) v =x v =0.97X0.2268=0.22m3 /kg (112 106.68) 1 1 g1

(150 106.68) b) (i) Final dryness fraction = 2685.2 + 10.9 = 2696.1 KJ/kg PVPV1.3 1.3 Now for the wet steam before steam 1 1 2 2 throttling 1 0.85 1.13 h =h +x h V2  0.22 1 f1 1 fg1 0.17 = 931 + x1 X 1870 KJ/kg V 0.22 4.15 For the throttling process 2 = 0.913 m3/kg hh 12 At 0.17 MPa , 931+(x1×1870) =2696.1 vg2 = 1.031 m3/kg 1765.1 x  The steam is wet, so dryness 1 1870 fraction is x 0.94 u 0.913 1 x 2   0.886 2 u 1.031 Q.5 Steams at 1.4 MPa and dryness g2 fraction 0.7 is throttled to 0.11 MPa. (ii) For the adiabatic expansion δQ = 0

U21  WU mixture gas constant of gas 1 PVPV R 8.314 1 1 2 2  R1    0.19 KJ / Kg.K UU21   M 44 n1 1 0.85 0.22  0.17  0.913  103 R 8.314  R1    0.30 KJ / Kg .K 1.13 1 M22 8 m R m R 44 0.19 28 0.30  246 KJ / kg Re 1 1 2 2  0.22 KJ / Kg.K m m 44 28 There is a loss of internal 12 energy. (d) The Equivalent specific heat at Q.7 A mixture of ideal gases consisting constant pressure R 0.19 of 3 kg of nitrogen and 5 kg of 1 Cv1    0.66 KJ/ K g.K carbon monoxide at a pressure of γ 1 1.286 1 0 300 KPa and temperature of 20 C . CP1 γCV1  1.286  0.6 6 Find (a) the mole fraction of = 0.85 KJ/Kg.K constituent (b) Equivalent R2 0.30 molecular weight of the mixture CV2   0.75 KJ / Kg.K γ 1 1.4 1 (c) Equivalent gas constant of the mixture, (d) the specific heat at CP2 γCV2  1.4  0.7 5 constant pressure and volume of the = 1.05 KJ/Kg.K

mixture. Take γ 1.286forCO2 and The equivalent specific heat of mixture at constant pressure γ 1.4forN2 . m1 C p1 m 2 C p2 Solution: Cpe  mm1  2 let the gas CO2 as the gas 1 and gas 44 0.85  28  1.05 N2 as the gas 2 is mixture.   0.928 KJ / Kg.K 44  28 The mole fraction for The equivalent specific heat of nm x11 and n mixture at constant volume 11n M m C m C  1 C  1 V1 2 V2 ve mm m1 5 1 2 M 44 0.66  28  0.75 x  1 44 0.52  1 m m 5 3 44 28 12  = 0.695 KJ/Kg.K MM1244 28

Similarly the mole fraction for Q.8 from an experimental determination m2 3 the specific heat ratio for acetylene M C H is found to 1.26. Find the two x  2 28 0.48 2 2 1 m m 5 3 specific heats. If heat is supplied to 12  increase the temperature from 300 C MM1244 28 0 (b) Equivalent molar weight of the to 60 C at constant pressure then mixture find the increase in the enthalpy and entropy of the system. Me = x1M1 + x2M2 = 0.52×44 +0.48 × 28 = 36.32 Kg/Kg Solution: mole Gas constant of acetylene R 8.3143 CH(R)   kJ / kgK 22 M 26

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission = 0.3198 kJ/kgK 2 W pdV  p V  V  27.266kJ As adiabatic index (γ) = 1.26, then 12  2 1  γ 1 We know that cRp  γ1 (b) Heat transferred 1.26 KJ  0.3198  1.55 K Q1 2 u 2  u 1  W 12 1.26 1 Kg mcv T 2  T 1   W 1 2  95.476kJ R And cv  γ1 (c) Entropy change of the gas 0.3196 Vp 1.23KJ / kgK S S  mc In22  mc In 1.26 1 2 1 p v Vp11 Change in enthalpy is given by V kJ mc In2 0.29543 .K h2 h 1  c p  T 2  T 1  p V1 kg KJ  .5530 4.1 65 Kg Change in enthalpy is given by T2   P2  SS2  c p1 ln  R ln   T1   P 1  T2 cp ln 0 T1 T2  Cp ln  T1  333  1.55 ln  303 = 0.146 KJ/Kg.K

Q.9 One kg of air in a closed system, initially at 50C and occupying 0.3 m3 volume, undergoes a constant pressure heating process to 1000C. There is no work other than Pdv work. Find (a) the work done during the process, (b) the heat transferred, and (c) the entropy change of the gas. Solutions: T1 = 278 K V1 = 0.3 m3 M = 1 kg P1 = 265.95 kPa T2 = 1000C = 373 K P2 = 265.95 kPa

mRT2 3 V2   0.40252m P2 (a) Work during the process

6.1 INTRODUCTION 6.1.3 REFRIGERANT In generally, refrigeration is defined as any process of heat removal. Refrigeration is In the refrigeration process, the substance defined as branch of Science that deals with employed as the heat absorber or coaling the process of reducing and maintaining agent is called the refrigerant. When the temperature of a space or material below absorbed heat causes an increase in the temperature of the surrounding. temperature of refrigerant, cooling process We see in nature that the heat is said to be sensible, whereas when the spontaneously flows of heat from a high absorbed heat causes a change in physical temperature body to a low temperature state of refrigerant, cooling effect is said to body. The reverse process to complete the be latent. Ice has been used successfully for thermodynamic cycle, in which heat Q will many years as a refrigerant. Not to many flow back from low temperature body to years ago ice was the only cooling agent high temperature body, is not possible. So available for use in domestic and small the logical conduction is that there must be commercial refrigerator. a process in which some work is done. 6.1.4 UNIT OF REFRIGERATION 6.1.1 NEED OF THERMAL INSULATION It is given by Tonnes of Refrigeration (TR). Since heat well always flow from a region 1TR is defined as refrigeration effect of high temperature to a region of lower produced by uniform melting of 1000 kg of 0 temperature. There is always a continuous ice from and at 0 C in 24 hours. flow of heat into the refrigerated region 1 TR = 1000×334 KJ refrigeration effect in 1000 334 from the warmer surroundings. To limit 24 hours  3.86 KJ / Sec. the heat flow into refrigerated region, it is 24 3600 usually necessary to isolate the region from  For practical problem it is taken as 3.5 its surrounding with a good insulation. KJ/Sec or 210 KJ/Minute.

6.1.2 REFRIGERATION LOAD 6.1.5 CO EFFICIENT OF PERFORMANCE The rate at which heat must be removed from refrigerated space or material in The co efficient of performance (COP) of order to produce and maintain the desired refrigerator is ratio of heat extracted in the temperature condition is called as refrigerator to the work done on the refrigeration load or cooling load. refrigerant working for the refrigeration The total refrigeration load is the sum of cycle. Heat extracting from refrigerated space Q heat gains from different sources:- COP   (1) Heat transmitted by conduction Work Done W through the insulated walls. (2) Heat that must be removed from warm 6.2 AIR STANDARD REFRIGERATION air that enters the space through CYCLE openings and closing doors. (3) Heat that must be removed from Air cycle refrigeration systems belong to refrigerated product to reduce the the general class of gas cycle refrigeration temperature of products to the storage systems, in which a gas is used as the temperature. working fluid. The gas does not undergo

any phase change during the cycle; low temperature TL during isothermal consequently, all the internal heat transfer expansion. processes are sensible heat transfer processes. Process 1-2: Isentropic compression of the  Gas cycle refrigeration systems find working fluid with the aid of external work. applications in air craft cabin cooling and The temperature of the fluid rises from TL  It also in the liquefaction of various gases. to TH.

6.2.1 AIR STANDARD CYCLE ANALYSIS Process 2-3: Isothermal compression of the working fluid during which heat is Air cycle refrigeration system analysis is rejected at constant high temperature TH. considerably simplified if one makes the following assumptions: Process 3-4: Isentropic expansion of the i. The working fluid is a fixed mass of air working fluid. The temperature of the that behaves as an ideal gas working fluid falls from TH to TL. ii. All the processes within the cycle are reversible, i.e., the cycle is internally reversible. iv. The specific heat of air remains constant throughout the cycle. First of all we study the ideal refrigeration cycle i.e. Reversed Carnot Cycle.

6.2.2 A REFRIGERATOR WORKING ON REVERSED CARNOT CYCLE:

Reversing the Carnot cycle does reverse the Fig. P-V and T-S Diagram of Reversed directions of heat and work interactions. A Carnot Cycle refrigerator or heat pump that operates on the reversed Carnot cycle is called a Carnot  The reversed Carnot cycle is the most refrigerator or a Carnot heat pump. efficient refrigeration cycle operating Reversed Carnot cycle is shown in Fig. It between two specified temperature consists of the following processes. levels. It sets the highest theoretical COP. The coefficient of performance for Carnot refrigerators and heat pumps are:

COP of Refrigerator

Heat absorbed COP   R Work supplied Heat absorbed

Heat rejected Heat absorbed

Q T (dS) Fig. a refrigerator working on reversed  L  L Carnot Cycle Q T HQ L H(dS) TL (dS) So, Process 4-1: Absorption of heat by the QTLL working fluid from refrigerator at constant COPR -----(7.1) QQTTHLHL

COP of Heat Pump There are four important processes in

Heat Rejected reversed Bray ton Cycle: COPHP  Process 1-2: Reversible adiabatic Work supplied compression in a compressor Heat Rejected  Process 2-3: Reversible isobaric heat Heat rejected Heat absorbed rejection in a heat exchanger Q T (dS) Process 3-4: Reversible adiabatic  H  H Q T expansion in a turbine HQ L H(dS) TL (dS) Process 4-1: Reversible isobaric heat QTHH absorption in a heat exchanger So COPHP  --(7.2) QQTTHLHL All the Processes are on P-V and T-S From the equation 7.1 and 7.2 diagram.

COPHP COP R 1 ------(7.3)

 Notice that turbine is used for expansion process between higher and lower temperature limit. Work done by the turbine helps supply some of the work required by the compressor from the external source.  Practically, the reversed Carnot cycle cannot be used for refrigeration Fig. Reversed Bray ton cycle on T-S diagram purpose as the isentropic process requires very high speed operation, Process 1-2: Gas at low pressure is whereas the isothermal process compressed isentropic ally from state 1 to requires very low speed operation. state 2. Applying steady flow energy equation and 6.3 REFRIGERATOR WORKING ON neglecting changes in kinetic and potential REVERSED BRAY TON CYCLE energy, W1-2 = m (h2-h1) = m. Cp. (T2-T1) This is an important cycle frequently But process 1-2 is reversible adiabatic employed in gas cycle refrigeration Process or isentropic process. So, systems. This may be thought of as a γ1 modification of reversed Carnot cycle, as TP  γ γ1 22 r γ    (7. 4) the two isothermal processes of Carnot  p  T11P cycle are replaced by two isobaric heat Where r = (P /P ) = pressure ratio transfer processes. This cycle is also called p 2 1 as Joule or Bell-Coleman cycle. Figure shows the schematic of a closed reverse Process 2-3: Hot and high pressure gas Bray ton cycle and also the cycle on T-s flows through a heat exchanger and rejects diagram. heat sensibly and isobaric to a heat sink. The enthalpy and temperature of the gas drop during the process due to heat exchange, no work transfer takes place and the entropy of the gas decreases. Q2-3 = m (h2-h3) = m. Cp. (T2-T3)

Process 3-4: High pressure gas from the heat exchanger flows through a turbine

Fig. Reversed Bray ton Refrigeration System undergoes isentropic expansion and

delivers net work output. The temperature 1 So, COP  of the gas drops during the process from T 3 1 r1p   to T . From steady flow energy equation:  4 From the above expression for COP, the W3-4 = m (h3-h4) = m. Cp. (T3-T4) But process 3-4 is reversible adiabatic following observations can be made: Process or isentropic process. So, γ1 γ γ1 TP33 γ   rp  ------(7.5) TP44 Where r = (P /P ) = pressure ratio p 3 4 T T So, 3  2 TT41

Process 4-1: Cold and low pressure gas from turbine flows through the low Fig. Comparison between Reversed Bray temperature heat exchanger and extracts ton Cycle and Carnot Cycle. heat sensibly and isobaric ally from a heat  source, providing a useful refrigeration For fixed heat rejection temperature (T ) and fixed refrigeration temperature effect. The enthalpy and temperature of the 3 (T ), the COP of reverse Bray ton cycle gas rise during the process due to heat 1 exchange, no work transfer takes place and is always lower than the COP of reverse the entropy of the gas increases. The Carnot cycle as shown in fig. refrigerating effect by the refrigerant can  COP of Bray ton cycle approaches COP be given as: of Carnot cycle as T approaches T 1 4 6.3.1 REFRIGERATING EFFECT (thin cycle), however, the specific refrigeration effect [c (T -T )] also Q = m (h -h ) = m. Cp. (T -T ) p 1 4 4-1 1 4 1 4 reduces simultaneously. COP of reversed Bray ton cycle can be  COP of reverse Bray ton cycle decreases given: as the pressure ratio r increases. COP  Refrigerating Effect /  NetWork Done p

Q  41 6.3.2 ACTUAL REVERSE BRAY TON CYCLE WW 1 2 3 4 The actual reverse Bray ton cycle differs m.c (T T )  p. 1 4 from the ideal cycle due to: m.cp. (T 2 T 1 ) m.c p. (T 3 T 4 )  Non-isentropic compression and (T T ) expansion processes.  14  Pressure drops in cold & hot heat exchangers. (T2 T 3 ) (T 1 T 4 ) Solving the equation

T4 T11  T1  T3  T4  T21 1   T  1   TT21    T 11 1   TT T γ1 21 2 1 γ r1p   T1 Fig. Actual Reversed Bray ton Cycle

Figure shows the ideal and actual cycles on operating cycle on T-S diagram. This is an T-s diagram. Due to this irreversibility, the open system. As shown in the T-S diagram, compressor work input increases and the outside low pressure and low turbine work output reduces. temperature air is compressed due to ram If η and η are the effect to ram pressure in the process 1-2. compressor ,isen turbine ,isen isentropic efficiencies of compressor and During this process its temperature turbine, respectively, these are defined as: increases from T1 to T2. This air is h h T T compressed in the main compressor in 2 1 2 1 process 2-3, and is cooled to temperature Copm. h'' h T T 2 1 2 1 T4 in the air cooler. Its pressure is reduced h3 h 4 T 3T 4 to cabin pressure in the turbine, as a result Turbien '' h3  h44T3 T its temperature drops from T4 to T5. The Thus the net work input increases due to cold air at stateT5 is supplied to the cabin. It increase in compressor work input and picks up heat as it flows through the cabin reduction in turbine work output. The providing useful cooling effect. The power refrigeration effect also reduces due to the output of the turbine is used to drive the irreversibility’s. As a result, the COP of fan, which maintains the required air flow actual reverse Bray ton cycles will be over the air cooler. considerably lower than the ideal cycles.  This simple system is good for ground Design of efficient compressors and cooling. turbines plays a major role in improving the COP of the system. In practice, reverse Bray ton cycles can be open or closed.

6.3.3 AIRCRAFT COOLING SYSTEMS

An aircraft, cooling systems are required to keep the cabin temperatures at a

comfortable level. Even though the outside fig: Simple air craft refrigeration system temperatures are very low at high altitudes, still cooling of cabin is required T ' γ1 due to: 2 1M2  Large internal heat generation due to T21 occupants, equipment etc. Where M is the Mach number, which is the  Heat generation due to skin friction ratio of velocity of the aircraft (V) to the caused by the fast moving aircraft sonic velocity C  At high altitudes, the outside pressure VV will be sub-atmospheric. When air at Mach number: M  C RT1 this low pressure is compressed and supplied to the cabin at pressures close 6.4VAPOUR COMPRESSION REFRIGERATION to atmospheric, the temperature CYCLE increases significantly. Vapour compression cycle is an improved  Solar radiation type of air refrigeration cycle in which a

suitable working substance, termed as 6.3.4SIMPLE AIRCRAFT REFRIGERATION refrigerant, is used. The refrigerant used, CYCLE does not leave the system, but is circulated

throughout the system alternately Figure shows the schematic of a simple condensing and evaporating. aircraft refrigeration system and the

In evaporating, the refrigerant absorbs its heat to the surrounding condensing latent heat from the solution which is used medium which is normally air or water. for circulating it around the cold chamber. In condensing, it gives out its latent heat to Receiver the circulating water of the cooler. The condensed liquid refrigerant from the Vapour compression refrigeration system condenser is stored in a vessel known as is now-a-days used for all purpose receiver from where it is supplied to the refrigeration. It is used for all industrial evaporator through the expansion valve or purposes from a small domestic refrigerant control valve. refrigerator to a big air conditioning plant. Expansion Valve

6.4.1 SIMPLE VAPOUR COMPRESSION It is also called throttle valve or refrigerant REFRIGERATION SYSTEM control valve. The function of the expansion valve is to allow the liquid refrigerant It consists of the following essential parts: under high pressure and temperature to pass at a controlled rate after reducing its pressure and temperature. Some of the liquid refrigerant evaporates as it passes through the expansion valve, but the greater portion is vaporized in the evaporator at the low pressure & temperature

Evaporator An evaporator consists of coils of pipe in which the liquid-vapour. refrigerant at low pressure and temperature is evaporated and changed into vapour refrigerant. In evaporating, the liquid vapour refrigerant absorbs its latent heat of vaporization from the medium which is to be cooled. Fig. Simple vapour compression refrigeration system 6.4.2 THEORETICAL VAPOUR COMPRESSION CYCLE Compressor A vapour compression cycle with dry The low pressure and temperature vapour saturated vapour after compression is refrigerant from evaporator is drawn into shown on T-S and P-h diagrams as shown the compressor through the suction valve in Figures. At point 1, let the temperature, A, where it is compressed to a high pressure and entropy of the vapour pressure and temperature. This high refrigerant is T1, p1 and S1 respectively. pressure & temperature vapour refrigerant The four processes of the cycle are as is discharged into the condenser through follows: the discharge valve B.

Condenser The condenser consists of coils of pipe in which the high pressure and temperature vapour refrigerant is cooled and condensed. The refrigerant, while passing through the condenser, gives up its latent

Some of the liquid refrigerant evaporates as it passes through the expansion valve, but the greater portion is vaporized in the evaporator. We know that during the throttling process, no heat is absorbed or rejected by the liquid refrigerant.

Fig: VCR cycle on T-s and P-h diagram Process 4-1-Vaporizing Process

Process 1-2-Compression Process The liquid-vapour mixture of the refrigerant at pressure P4=P1 and The vapour refrigerant at low pressure P1 temperature T4=T1 is evaporated and and temperature T1 is compressed changed into vapour refrigerant at constant isentropic ally to dry saturated vapour as pressure and temperature, as shown by the shown by the vertical line 1-2 on the T-s horizontal line 4-1 on T-s and p-h diagram and by the curve 1-2 on p-h diagrams. During evaporation, the liquid- diagram. The pressure and temperature vapour refrigerant absorbs its latent heat rise from P1 to P2 and T1 to T2 respectively. of vaporization. The heat which is absorbed The work done during isentropic by the refrigerant is called refrigerating compression per kg of refrigerant is given by effect and it is briefly written as RE. w = h2 – h1 The refrigerating effect or the heat Where h1 = Enthalpy of vapour refrigerant absorbed or extracted by the refrigerant at temperature T1, i.e. at suction of the during evaporation per kg of refrigerant is compressor, and h2=Enthalpy of the given by RE = h1 – h4 = h1 – hf3 vapour refrigerant at temperature T2. i.e. at Where hf3 = Sensible heat at temperature discharge of compressor. T3, i.e. enthalpy of liquid refrigerant Process 2-3- Condensing Process leaving the condenser.

The high pressure and temperature vapour CO Efficient of Performance refrigerant from the compressor is passed through the condenser where it is Coefficient of performance, C.O.P. = completely condensed at constant pressure (Refrigerating effect)/ (Work done) P2 and temperature T2 as shown by the hh hh O.P 14 1 f 3 horizontal line 2-3 on T-s and p-h h h h h diagrams. The vapour refrigerant is 2 1 2 1

changed into liquid refrigerant. The 6.5 EFFECT OF PARAMETERS ON COP OF refrigerant, while passing through the VAPOUR COMPRESSION REFRIGERATION condenser, gives its latent heat to the CYCLE surrounding.

1. Effect of Suction Pressure Process 3-4 - Expansion Process The suction or evaporator pressure

decreases due to the frictional resistance of The liquid refrigerant at pressure P = P 3 2 flow of the refrigerant. Let us consider a and temperature T3 = T2, is expanded by theoretical vapour compression cycle 1-2- throttling process (Isenthalpic Process) 3-4 when the suction pressure decreases through the expansion valve to a low from P to P ’ as shown on p-h diagram in pressure P4 = P and Temperature T4 = T1 s s 1 Figure. as shown by the curve 3-4 on T-s diagram It may be noted that the decrease in suction and by the vertical line 3-4 on P-h diagram. pressure:

a) Decreases the refrigerating effect from 6.5.1 EFFECT OF LIQUID SUB COOLING (h1 – h4) to (h11 – h4). b) Increases the work required for It is possible to reduce the temperature of compression from (h2–h1) to (h21– h11). the liquid refrigerant to within a few degrees of the temperature of the water entering the condenser. In some condenser designs it is achieved by installing a sub- cooler between the condenser and the expansion valve. The effect of sub-cooling of the liquid from t3 = tk to t31 is shown in Figure. It will be seen that sub-cooling reduces flashing of the liquid during

fig. Effect of suction pressure expansion and increases the refrigerating Since the C.O.P, of the system is the ratio of effect. But there is no effect on the work refrigerating effect to the work done, done by the compressor. And the COP of therefore with the decrease in suction the cycle is ratio of refrigeration effect to pressure, the net effect is to decrease the work done by the compressor. So the COP C.O.P. of the refrigerating system for the of the cycle increases in sub-cooling of the same refrigerant flow. Hence with the liquid. decrease in suction pressure  The refrigerating capacity of the system decreases.  The refrigeration cost increases.

2. Effect of Discharge Pressure

In actual practice, the discharge or condenser pressure increases due to frictional resistance of flow of the refrigerant. Let us consider a theoretical vapour compression cycle l-2-3-4 when the discharge pressure increases from PD to PD fig. Effect of suction pressure 1 as shown on p-h diagram in Figure resulting in increased compressor work 6.6VAPOURABSORPTION REFRIGERATION from (h2 – h1) to (h21 – h1) and decreasing SYSTEMS refrigeration effect from (h1 – h4) to (h1 – h41). So COP is ratio of refrigeration effect The Refrigeration effect by Vapour to work done. So COP of the cycle Compression Refrigeration System is an decreases. efficient method. But the input energy given in the VCR system is work i.e. high grade energy and therefore very expensive. So In Vapour Absorption Refrigeration Systems, the mechanical work is replaced by the heat. Hence these systems are also called as heat operated or thermal energy driven systems. Since these systems run on low-grade thermal energy, they are preferred when low-grade energy such as fig. Effect of suction pressure waste heat or solar energy is available.

Since conventional absorption systems use heat exchanger. The weak high natural refrigerants such as water or temperature ammonia solution from the ammonia they are environment friendly. generator is passed to the heat exchanger through the pressure reducing valve. The 6.6.1 WORKING OF SIMPLE VAPOUR pressure of the liquid is reduced to the ABSORPTION REFRIGERATION CYCLE absorber pressure by the throttle valve. Ammonia vapour is produced in the The systematic diagram of VAR system is generator at high pressure by a heating shown. In this, compressor is replaced by generator. The water vapour carried with absorber system (i.e. Absorber, Pump, Heat ammonia is removed in the rectifier and Exchanger Heat generator and rectifier). only the dehydrated ammonia gas enters Working of the system is almost same as into the condenser. High pressure NH3 the working of VCR system excepting the vapour is condensed in the condenser. The functionality of compressor. The working cooled NH3 solution is passed through a can be explained with help of sketch throttle valve and the pressure and diagram. temperature of the refrigerant are reduced The low temperature refrigerant enters the below the temperature to be maintained in evaporator and absorbs the required heat the evaporator. The refrigerant absorbs the from the evaporator and leaves the heat from the medium and converts in to evaporator as saturated vapour. Slightly the vapour phase and again it goes to low pressure NH3 vapour is absorbed by generator. the weak solution of NH3 which is sprayed in the absorber as shown in fig. 6.6.2 COP OF IDEAL VAR CYCLE

The working of VAR system is shown in fig. Let us consider the Heat QE is absorbed by the refrigerant in absorber at temperature of TE. In the heating generator, QG amount of heat is added at temperature TG to the refrigerant. The QC amount heat is rejected in condenser at temperature of TC. Pump work is WP. The COP of system is given as Refrigeration Effect COP   VAR Given input QQ EE Fig. Vapour Absorption Refrigeration QWQGPG System Considering a reversible refrigeration cycle, Weak NH3 solution (i.e. know as aqua– QC = QG + QE ------(1) ammonia) enters in to absorber and for a reversible cycle Clausius Inequality becomes strong solution after absorbing Law, δQ QQQ NH3 vapour. Then it is pumped to the ∮ 0, CG E  generator through the heat exchanger. The TTTTCEG pump increases the pressure of the strong Substituting the value from equation 1 solution. QGG QEQE Q The strong NH3 solution absorbs heat form  TTTC EG high temperature weak NH3 solution in the

QQEEQQGG  If the operating temperatures are above Or,    o TTTTCEGC 0 C, then pure water can also be used as

QECEGGC (T T ) Q (T T ) secondary refrigerant, The commonly Or,  used secondary refrigerants are: TTCG  The solutions of water and ethylene Q (T T ) T Or, E GC C glycol, Q T (T T ) G G CE  The solutions of propylene glycol or COP COP  VAR Carnot  Carnot calcium chloride.

6.7 REFRIGERANT 6.7.2 PROPERTIES OF REFRIGERANT

Refrigerant is the fluid used for heat There should be the following properties of transfer in a refrigerating system that the refrigerants: absorbs heat during evaporation from the a) Latent heat of vaporization: region of low temperature, and releases it should be as large as possible so that the heat during condensation at a region of required mass flow rate per unit cooling higher temperature. capacity will be small.

Due to several environmental issues such b) Isentropic index of compression: as ozone layer depletion and global It should be as small as possible so that the warming and their relation to the various temperature rise during compression will refrigerants used, the selection of suitable be small. refrigerant has become one of the most important issues in recent times. c) Vapour specific heat: For Example, Air used in an air cycle It should be large so that the degree of refrigeration system can also be considered superheating will be small. as a refrigerant. d) Thermal conductivity: 6.7.1 CLASSIFICATION OF REFRIGERANT Thermal conductivity in both liquid as well as vapour phase should be high for higher There are two types of refrigerants heat transfer coefficients. 1. Primary Refrigerant 2. Secondary Refrigerant e) Viscosity:

1. Primary Refrigerant Viscosity should be small in both liquid and vapour phases for smaller frictional Primary refrigerants are those fluids, which pressure drops. are used directly as working fluids, for example in vapour compression and f) Ozone Depletion Potential (ODP): vapour absorption refrigeration systems. According to the Montreal protocol, the ODP of refrigerants should be zero, i.e., they 2. Secondary Refrigerant should be non-ozone depleting substances. Secondary refrigerants are those liquids, Refrigerants having non-zero ODP have which are used for transporting thermal either already been phased-out (e.g. R11, R energy from one location to other. 12).

Secondary refrigerants are also known g) Global Warming Potential (GWP): under the name brines or antifreezes. Refrigerants should have as low GWP value Generally, the freezing point of brine will as possible to minimize the problem of be lower than the freezing point of its global warming. constituents. So these are used as Secondary refrigerants. h) Toxicity

The refrigerants used in a refrigeration R -500: Mixture of R 12 and R 152a system should be non-toxic. R -502: Mixture of R 22 and R 115 R-503: Mixture of R 13 and R 23 6.8 DESIGNATION OF REFRIGERANTS (4) Hydrocarbons: Propane (C H ) : R 290 There are following types of refrigerants. 3 8 The designation of these refrigerants can n-butane (C H ) : R 600 4 10 be done by following manner: Iso-butane (C H ) : R 600 a 4 10

1) Halocarbon compounds: Examples These refrigerants are derivatives of Q.1 In reversed Bray ton cycle, the alkenes (C H ) such as methane (CH ), n 2n+2 4 temperature at the end of heat ethane (C H ). These refrigerants are absorption and heat rejection are 2 6 0 0 designated by R-ABC 0 C and 30 C. The pressure ratio is 4 Where: and the pressure in the cooler is 4 A+1 indicates the number of Carbon (C) bar, determine the temperature at atoms all the states and volume flow rates B-1 indicates number of Hydrogen (H) at inlet of compressor and at exit of atoms, and Z indicates number of Fluorine turbine for 1 TR cooling capacity. (F) atoms Solution: T1 = 273 K Example: R- 22 T3 = 303 K A = 0 ⇒ No. of Carbon atoms = 0+1 = 1 Temperature at all points T γ1 B = 2 ⇒ No. of Hydrogen atoms = 2-1 = 1 2  r  γ C = 2 ⇒ No. of Fluorine atoms = 2 T1 The balance = 4 – no. of (H+F) atoms 0.4 T 273 41.4 405.67K = 4-1-2 = 1, So No. of Chlorine atoms = 1 2   γ1 T3  r  γ The chemical formula of R 22 = CHClF T 2 4 The commonly used refrigerant 303 T 203.9K i.e. R-134a = C H F 4 0.4 2 2 4 4 1.4

2) Inorganic refrigerants: These are designated by number 7 followed by the molecular weight of the refrigerant. Examples: (a) Ammonia: Molecular weight is 17, the designation is R -717

(b) Carbon dioxide: Molecular weight is 44, Volume flow rate of air for 1 TR or the designation is R -744 3.5KJ/sec (c) Water: Molecular weight is 18, Refrigeration effect m c T T  the designation is R -718 a p 1 4 3.53.5 3.5  mp X1.005(273 203.9) 3) Azeotropic Refrigerant: mp  0.05kg / sec Azeotropic mixtures are designated by 500 At the inlet of compressor density, series, these are the mixture of two or more 5 P1 30 halocarbon compound. P1   1.27kg / sec. Some Azeotropic mixtures: RT1 287X273

Volume flow rate 30 1  1.005 T23  T  V = P1 X ma = 1.27 X 0.05 60 = 0.064 m3/sec. = 0.5 × 1.005 (303 – 207.92) = 47.78 KW Q.2 The compressed air from main c) Refrigeration effect compressor of an air craft cooling RE m c T T1 system is bled off at 4.5 bar and a p cabin 3  2000C. It is then passed through the = 0.5 × 1.005 (298 – 207.92) heat exchange in which the ram air =45.27 KJ/s =45.27/3.5 = 12.9 TR is forced through for cooling purpose Q.3 A dense air refrigeration machine by fan driver by cooling turbine. The operating on Bell-Coleman cycle condition of the inlet of cooling operates between 3.4 bar and 17 bar. turbine is 4 bar an 300C. The air The temperature of air after the expands in cooling turbine is 4 bar cooler is 150C and after the an 300C. The air expands in cooling refrigerator is 60C. The refrigeration turbine up to 0.7 bar. The isentropic capacity is 6 Tonnes, calculate: efficiency of cooling turbine is 80% 1) temperature after the and flow rate through cooling turbine compression and expansion is 30 kg/min. Find 2) air circulation per minute a) Actual exit temperature from 3) work of compressor and cooling turbine expander b) The power delivered to ram air 4) Theoretical COP. blow fan Solution: 0 c) Tons of refrigeration, if T3  15 C 288K temperature of cabin cockpit T 60 C 279K area is 250C. 1 RE = 6 tonnes Solution:- (I) temp. After comp. & expansion 1) Exit temp. Of turbine γ1 γ1 γ 0.4 γ 0.4 TP 17 1.4 TP 0.7 1.4 22  22     T3  300 184.15K T11 P 3.4 TP33 4  T2=279×1.5=445.69 k V1 0.4 V 1.4 TP44 3.4    .62 T33 P 17 T4 = 288× 0.62 =180.2 K

T3 = 184.15 K Efficiency of the turbine TT 1 0.8  23 TT23 1 (II) Air circulation per minute T3  T 2  T 3  0.8  T2 1 Refrigeration effect= ma c p T 1 T 4  T3  303  0.8(303  184.15) 6 ×210 = ma ×1.005(279-180.2) T1  207.92k 3 Ma=12.7 kg/min. b) Power delivered to fan = work (III) Work of compressor:- output from turbine

n w m R T T  comp n1 a 2 1 1.4 12.7   287(445.29  279) 1.4 1 60 = 35.44 KW

Work of expander hh at -50C n 1g w m R T T  h 185.4kJ / kg turbine n1 a 3 4 1 1.4 12.7 Enthalpy of point 4   287(288  180.2) h h  h  69.5kJ 1.4 1 60 4 3 f 3 = 22.89 KW Refrigerant effect

(IV) Cop of cycle m h14  h   m  185.4  69.5  Heatabsorbed 6X3.5  Cop   m 115.9kJ / sec workdone 35.44 22.89 10 3.5  115.9  m COP = 1.67 Mass flow rate 35 m  0.3kg / sec. Q.4 A simple saturation cycle using R-12 115.9 is desired taking a load of 10 Tons. In process(1-2)i.e.isentropic process The refrigerating and ambient ss12 temperature are 00C and 300C. A T minimum temperature difference of sap sg1 s g2 c p l n  0 5 C is required in evaporator and Tsat condenser for heat transfer. Find Tsap a) Mass flow rate through system 0.6991 0.6839 cpn l  b)Power required in kw. The 308

properties of R-12 are following:- Tsup  312.9 k Sat. Sat. Specific volume m3/kg h = h + c T − T Temp. Pr. 2 g2 p sap sat 0C bar Sat (10-3) Sat. vapour = 201.5 + 0.95 (312.9-308) -50C 2.61 0.71 0.0650 = 206.25 KJ/Kg 00C 3.08 0.72 0.0554 (II) Power required by compressor  300C 7.45 0.77 0.0235 w m h21  h   350C 8.47 0.79 0.0206 0.3 (206.25  185.4)  6.3KW

Enthalpy kJ/kg Entropy kJ/kg. k Q.5 An ideal vapour compression Sat. Sat. Sat. Sat. refrigerator using R-12 operates Liquid vapour Liquid vapour between temperature limits of -100C 31.4 185.4 0.1251 0.6991 and 400C. The refrigerator leaves 36.1 187.5 0.1420 0.6966 the condenser dry saturated. The 64.4 199.6 0.2399 0.6854 rate of flow of refrigerant through 69.5 201.5 0.2559 0.6839 the unit is 150 Kg/hr. Specific heat of vapour R-12 is 0.95 Calculate the refrigerating effect per KJ/KgK kg of refrigerant , COP If Solution:- 1) refrigerant leaves the condenser Temp. Of refrigerant in evaporator as dry saturated is -50C and in the condenser is 350C. 2) refrigerant is sub cooled to 200C At low pressure before throttling.

Assume that enthalpy of refrigerant 1 h34 50.59 h before throttling is approximately Refrigerating effect = h1 – h4 equal to enthalpy of refrigerant at = 183.1– 50.59 0 under cooled temperature of 20 C = 132.51 KJ/kg and it as hf = 50.59 kJ/kg. hh 132.51 Cop  14   5.06 h21 h 209.6 183.19

Q.6 A two cylinder R-12 compressor has bore and stroke equal to 5.65 cm and 5 cm. Its rpm is 1450 rpm.If liquid refrigerant is throttled by throttle valve at 400C. Determine the mass of refrigerant circulated per minute and theoretical capacity when suction temperature is –100C. Assume γ = 1.13 , properties of R-12 are: Temp Pr.(bar) Volume Enthalpy vapour liquid kJ/kg m3/kg 0 Solution: -10 C 2.1928 0.07702 190.72 (1) if the refrigerant leaves dry 400C 9.5944 0.01863 239.29 saturated h1 = 183.19 kJ/kg s 7019kJ / kg.k Enthalpy Entropy Entropy vapour 1 vapour liquid kJ/kgk ss21 kJ/kg kJ/kgk 347.96 0.9656 1.5632 Tsap s1 s g c p l n  368.67 1.1324 1.5456 Tsat Solution: At 400C Sg = 0.6825 Theoretical piston displaced volume Cp = 0.609 kJ/kgk π 2 Tsap VP  d LNk s1n 0.6825 0.609l  4 313  π 2  5.65X1022   5 1450  2  10 Tsap 4 0.7019 0.6825 0.609ln  313 = 0.335 m3/min. 3 0 T 323.1K Vg = 0.07702 m /kg at -10 C sap So refrigerant mass circulated/min 0 h2 h g  c p T sap  T sat  at -10 C suction temperature = 203.20+ 0.609 (323.1-313) = 0.335/0.07702 = 4.6 kg/min h1 = 347.96 kJ/kg h2  209.37kJ / kg.k h4 = 239.29 kJ/kg h h  h  74.59kJ / kg 3 f 3 4 Refrigerating effect (i) Refrigerating effect = h1 – h4 = RE m h h  183.1 – 74.59 14 = 108.6 KJ/kg = 4.6 X (347.96 – 239.29) hh 108.6 = 500.8 kJ/min Cop  14   4.15 h h 209.6 183.1 21 Q.7 A R-12 refrigeration machine has (iii) off refrigerant is sub cooled saturated suction temperature of

50C and saturated discharge Q.8 Assume a simple saturated cycle temperature of 400C. Determine with suction saturation temp. - i) COP 100C and condensing saturation ii) Theoretical power. Per ton of temperature of 300C. If clearance refrigeration when the compression volume 15% of stroke volume. How is dry properties are: will the volumetric efficiency vary Temp. Pr. bar Volume hf for different refrigerant namely m3/kg (1) R-12 (2) R-22 (3)NH3? 0 5 C 3.6375 0.048 204.68 Solution: 400C 9.5909 0.0184 239.03 Volumetric efficiency can be given Enthalpy Entropy Entropy v vapour liquid vapour n 1  C  C suction kJ/kg kJ/kgk kJ/kgk vol. vdischarge 347.96 0.9656 1.5632 368.67 1.1324 1.5456 1) For R-12, from p-h chart m3 Solution: 3 vsuction  0.07813 , vdischarge 0.025m / kg Process 1-2 is isentropic process. kg 0.07813 SS12  So nvol.  1  0.05  0.05 0.025 = 0.894 = 89.4% For R-22, from p-h chart

Tsap 1.5569 Sg2 c p l n  Tsat T 1.5569 1.5569 0.762l sap n  v 0.07m33 / kgv 0.0225m / kg 313 suction discharge 0.07 Tsup  317.5k  So nvol.  1  0.05  0.05 0.0225 h2 h g2  c p T sap  T sat  = 1.05 – 0.0156 = 0.894 = 89.4% = 368.87 + 0.762 (317.5 – 313) = 372.3 kJ/kg (3) For NH3 , from p-h chart, h3f h at 313K =239.03 kJ/kg = h4 3 vsuction  0.4184m / kg,vdischarge Refrigerating effect m  h14 h   0.135m3 / kg 210 m  h14 h  0.4184 60 So nvol.  1  0.05  0.05 m  0.03kg / sec 0.135 Theoretical power per ton of = 0.895 = 89.5%

refrigeration Q.9 Catalogue data from R-12 P m h h 21  compressor shows that compressor 0.03 372.3  354.885  0.53KW / TR delivers 15 TR at saturation suction temperature of -50C and saturated hh14 354.885 239.03 COP    6.65 0 h h 372.3 354.885 condensing temperature of 40 C 21 and actual suction temperature is

150C. Liquid leaving the condenser is saturated. Calculate the mass flow rate of this compressor. Solution:

from the p-h diagram 3 hg1  353KJ / kg,vg 0.66m , 3 h11 363KJ / kg,v 0.072m / kgP The refrigeration effect

m  h14 h  15 210  m(363  239) 15 210 m 25.40 124 m  25.40kg / min. And volume flow rate V = 25.4 X 0.072 V = 1.828 m3/min.

Q.10 A geothermal wall at 1300C supplies heat at a rate of a 100500 kJ/hr to an absorption refrigeration system. The environment is at 300C and the refrigerated space is maintained at - 220C. Determine the maximum possible heat removed from refrigerated space. Solution:

TC  303kTE  251kTG  403k

QG  100500kJ / hr For a reversible absorption cycle QTTT COP EEX GC QTTTGCEG 251 403 303  303 251 403 100500 Q Q X1.19 X1.19 EG 3600

QE  32.22KW

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 7 PSYCHOMETRY Psychrometry is the branch of science Fig. Water vapour in air mixture which deals with study of the properties of the moist air i.e. mixtures of air and water 7.1.3 HUMIDITY RATIO vapour. Atmospheric air is a mixture of many gases plus water vapour and a The humidity ratio or specific humidity (W) number of pollutants. is the mass of water per kg of dry air in a The amount of water vapour and pollutants given mixture of dry air and water vapour. vary from place to place. A mixture of If mv is the mass of water vapour and ma is various gases that constitute air and water the mass of dry air. Then humidity ratio vapour is known as moist air. m W= v Constituent Molecular weight Mol fraction m Oxygen 32 0.2095 a Nitrogen 28.016 0.7809 We know that

Carbon dioxide 44.01 0.0003 PVVV V m R Tforthewatervapour

7.1 PROPERTIES OF MOIST AIR And Pa V m a R aV T for the dry air Substituting the values 7.1.1 MOIST AIR PVV The moist air can be thought of as a R T P Ra W VV mixture of dry air and moisture. Based on PV a PRaV the above composition the molecular RaT weight of dry air is found to be 28.966 and PVV 8.314 / 28 P the gas constant R is 0.287035 KJ/Kg.K. 0.622 P 8.314 /18 P P If the partial pressure of dry air and water a b V P vapour is Pa and PV, then according to W 0.622 V    (7.2) Dalton’s law of partial pressure PPbV

PPPb a v ------(7.1) If air is saturated at temperature T, then Where P is barometer reading of Specific humidity of saturated air b P atmospheric pressure. In general, Ws 0.622 s    (7.3) atmospheric pressure is taken as 760 mm PPbs of Hg. 7.1.4 DEGREE OF SATURATION 7.1.2 SATURATED AIR The degree of saturation is the ratio of the At a given temperature and pressure, the humidity ratio (W) of the mixture to the dry air can only hold a certain maximum humidity ratio of a saturated mixture (WS) amount of moisture. When the moisture at the same temperature and pressure. it is content in the moist air is maximum, the air denoted by μ. is known as saturated air. Degree of Saturation

0.622PV W PPP(PP) μ  b V  V b s 0.622P Ws s P.PPs b V 

PPbs P(PP) μ  V b s    (7.4) P.PPs b V 

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 7.1.5 RELATIVE HUMIDITY moisture presents in the air begins to condense is known as dew-point Relative humidity is defined as the ratio of temperature (DPT) of air. the mass of water vapour present in a given volume of air at temperature T to the mass 7.1.8 WET BULB TEMPERATURE (WBT) of water vapour when the same volume of air is saturated at the same temperature T It is temperature of air measured by and pressure. Relative humidity is normally thermometer when wet cloth is covered expressed as a percentage. When Φ is 100 over the thermometer. percent, the air is saturated. 7.1.9 DENSITY OF DRY AIR Relative humidity mass of water vapour in a given volume The density of air or air density is the mass of air at Temperature T of air per unit volume Air density decreases  mass of water vapour when same volume of air is with increasing altitude. It also changes saturated at Temperature T with variation in temperature or humidity. At sea level and at 15 °C, air has a density of m approximately 1.225 kg/m3.density of dry  V m air can be calculated using the ideal gas S law, expressed as a function Assuming the water vapour is an ideal gas, of temperature and pressure P V m R T VVV P ρ  a  7.6  And for the saturated air condition a R T P V m R T aD s s V Where ρ  Air density So from the equation () and Equation (), we P = Absolute pressure get a TD = Dry Bulb temperature mPVV  Ra = Specific gas constant for dry air

mPSS mP 7.1.10 ENTHALPY OF THE AIR So  VV  ------(7.5) mPSS The enthalpy of moist air is the sum of the enthalpy of the dry air and the enthalpy of for the dry air mV and PV are zero, so Φ = 0 the water vapour. Enthalpy values are for the dry air m andP are equal to always based on some reference value. The VV enthalpy of the mixture is given by m andP , so ss h  h Wh  C T  W h  1.88 t  t Φ = 100% a V p a  g W dp  So relative humidity varies from 0% to ------(7.7) 100%. 7.2 PSYCHROMETRIC CHART 7.1.6 DRY BULB TEMPERATURE A Psychrometric chart graphically DBT is the temperature of the air as represents the thermodynamic properties measured by a standard thermometer of moist air. Standard Psychrometric charts when it is unaffected by thermometer. are bounded by the dry-bulb temperature line (abscissa) and the vapour pressure or 7.1.7 DEW-POINT TEMPERATURE (DPT) humidity ratio (ordinate). The Left Hand Side of the Psychrometric chart is bounded If moist air is cooled at constant pressure, by the saturation line. Figure shows the then the temperature at which the schematic of a Psychrometric chart.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  Psychrometric charts are readily Dry bulb temperature is temperature of an available for standard barometric air sample, as determined by an ordinary pressure of 101.325 KPa at sea level thermometer. It is typically plotted as o and for normal temperatures (0-50 C). the abscissa (horizontal axis) of the graph.  The Relative humidity at the saturation The SI unit for temperature is degrees curve is 100 %. Celsius. These lines are drawn straight,  At the saturation curve, Dry bulb parallel to each other. Each line represents Temperature, Wet bulb Temperature a constant temperature. and Dew point temperature are same.

7.3.2 WET-BULB TEMPERATURE (WBT) LINE Wet-bulb temperature is temperature of an air sample after the air has passed over a large surface of liquid water in an insulated channel. When the air sample is saturated with water, the WBT will read the same as the DBT. These lines are oblique lines that differ slightly from the enthalpy lines. They are identically straight but are not exactly parallel to each other. These intersect the saturation curve at DBT fig. Psychrometric Chart point. The parameters of Psychrometric chart are relative humidity, dry bulb temperature, wet bulb temperature, dew point temperature, specific humidity, specific volume, specific enthalpy and vapour pressure. The Psychrometric chart allows all the

parameters of some moist air to be fig. Wet Bulb Temperature line determined from any three independent parameters, one of which must be the 7.3.3 DEW POINT TEMPERATURE (DPT) pressure. LINE

7.3 SIGNIFICANCE OF DIFFERENT LINES Dew point temperature (DPT) is the ON PSYCHROMETRY CHART temperature at which a moist air at the same pressure would reach water vapour 7.3.1 DRY-BULB TEMPERATURE LINES saturation. At this point further removal of

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission heat would result in water vapour typically plotted as the ordinate (vertical condensing into liquid water fog. From the axis) of the graph. For a given DBT there state point follow the horizontal line of will be a particular humidity ratio for constant humidity ratio to the intercept of which the air sample is at 100% relative 100% RH, also known as the saturation humidity. These are the horizontal lines on curve. The dew point temperature is equal the chart. to the fully saturated dry bulb or wet bulb temperatures.

fig. Dew Point temperature line Fig. Humidity ratio lines

7.3.4 Relative humidity (RH) Line 7.3.6 SPECIFIC ENTHALPY LINE Relative humidity (RH) is the ratio of the mole fraction of water vapor to the mole These are oblique lines drawn diagonally fraction of saturated moist air at the same downward from left to right across the temperature and pressure. RH is chart that are parallel to each other. In the dimensionless, and is usually expressed as approximation of ideal gases, lines of a percentage. Lines of constant RH reflect constant enthalpy are parallel to lines of the physics of air and water: they are constant WBT. determined via experimental measurement. Relative humidity: These hyperbolic lines are shown in intervals of 10%. The saturation curve is at 100% RH, while dry air is at 0% RH.

Fig. Specific Enthalpy Lines

fig. relative humidity lines 7.3.7 SPECIFIC VOLUME LINE

7.3.5 HUMIDITY RATIO LINE Specific volume is the volume of the

mixture containing one unit of mass of dry Humidity ratio is the proportion of mass of air. The SI units are cubic meters per water vapour per unit mass of dry air at the kilogram of dry air. These lines are a family given conditions. It is also known as the of equally spaced straight inclined lines moisture content or mixing ratio. It is that are nearly parallel.

7.4.2 SENSIBLE HEATING PROCESS

During this process, the moisture content of air remains constant and its temperature

fig. Specific Volume lines increases as it flows over a heating coil. The heat transfer rate during this process is 7.4 DIFFERENT PROCESS ON given by PSYCHROMETRIC CHART

7.4.1 SENSIBLE COOLING PROCESS

During this process, the moisture content of air remains constant but its temperature decreases as it flows over a cooling coil. For moisture content to remain constant the surface of the cooling coil should be dry and its surface temperature should be greater than the dew point temperature of air. If the cooling coil is 100% effective, then the exit temperature of air will be equal to the coil temperature. However, in practice, the exit air temperature will be higher than the cooling coil temperature. The heat transfer during the cooling process is given as

QC m a h O  h A   m a C pm  T O  T A  ----(7.8)

fig. Sensible Heating Process

Qh m a h b  h o   m a C pm  T b  T O      (7.9)

Where Cpm is the humid specific heat. ma is the mass flow rate of dry air (kg/s).

7.4.2.1 BY PASS FACTOR

In Fig., the temperature Td3 is the effective surface temperature of the heating coil, and is known as apparatus dew-point (ADP) temperature. Td1 and Td2 are the temperature of air at inlet and outlet of the coil. In an ideal situation, when all the air comes in perfect contact with the heating coil surface, then the exit temperature of air will be same as ADP of the coil.

When moist air is cooled below its dew- point by bringing it in contact with a cold surface as shown in Fig., some of the water But in actual case the exit temperature of vapour in the air condenses and leaves the air will always be lesser than the apparatus air stream as liquid, as a result both the dew-point temperature. So By -pass factor temperature and humidity ratio of air (BPF) is defined as the ratio of heat loss to decreases as shown. The heat and mass maximum possible heat transfer. transfer rates can be expressed in terms of

Heat Loss TTd3 d2 the initial and final conditions by applying BPF the conservation of mass and conservation Maximum Possible heat transfer Td3 T d1 ------(7.10) of energy equations. In the same way, Bypass factor for cooling coil can be given as Heat Loss TT BPF d2 d3 Maximum Possible heat transfer Td1 T d3 ------(7.11)

7.4.2.2 EFFICIENCY OF COIL Fig. Cooling

The efficiency of heating Coil is defined as the ratio of actual heat transfer to the cold air to maximum possible heat transfer to the cold air by heating coil. Efficiency Actual Heat transfer  Maximum Possible heat transfer TT Fig. dehumidification  d2 d1 ------(7.12) By applying mass balance for the water TT d3 d1 ma.Wo = ma.WC + mw In the same way efficiency for cooling coil By applying energy balance can be given as ma.ho = Qt +ma.hC + mw.hw The efficiency of cooling Coil is defined as From the above two equations, the load on the ratio of actual heat transfer to the cold the cooling coil, Qt is given by: air to maximum possible heat transfer to the cold air by heating coil. Qt=ma(ho.-hC)+ma.(Wo-WC).hw ------(7.15) Efficiency Where Wo and Wc are the humidity ration before and after the cooling process.mw is mass of water vapour separating from air. It can be observed that the cooling and de- TT d1 d2    (7.13) humidification process involves both latent TTd1 d3 and sensible heat transfer processes, hence, By comparing the equation (7.11) and the total, latent and sensible heat transfer equation (7.13) of Bypass factor and rates (Qt, Ql and Qs) can be written as efficiency of the heating and cooling coil, Qt = QL + QS we get QL = ma.(Wo- WC).hw  1  BPF ------(7.14) QS = ma (ho.- hC) = ma.Cpm. (To.- TC)

Qh = ma (hD-ho) - mw hw ------(7.16) By separating the total heat transfer rate Where Qh is the heat supplied through the from the cooling coil into sensible and heating coil. mw and hw is the mass and the latent heat transfer rates, a useful enthalpy of steam. Since this process also parameter called Sensible Heat Factor involves simultaneous heat and mass (SHF) is defined. transfer, we can define a sensible heat So SHF is defined as the ratio of sensible to factor for the process in a way similar to total heat transfer rate. that of a cooling and dehumidification Sensible Heat Transfer SHF  process. Total Heat Transfer Sensible Heat(SH) 7.4.5 COOLING & HUMIDIFICATION = Sensible Heat SH   Latent heat(LH) PROCESS

During this process, the air temperature 7.4.4 HEATING AND HUMIDIFICATION drops and its humidity increases. This PROCESS process is shown in Fig. This can be

achieved by spraying cool water in the air During winter it is essential to heat and stream. The temperature of water should humidifies the room air for comfort. This is be lower than the dry-bulb temperature of normally done by first sensible heating the air but higher than its dew point air and then adding water vapour to the air temperature to avoid condensation (T < stream through steam nozzles as shown in DPT Tw < TO). the fig. By this process, dry bulb It can be seen that during this process there temperature and humidity ratio increase. is sensible heat transfer from air to water Appling Mass balance of water vapour for and latent heat transfer from water to air. the control volume Hence, the total heat transfer depends upon m = m (W -W ) w a D o the water temperature. Where ma is the mass flow rate of dry air.

Fig. Cooling Process

Fig. Heating Process

Fig. Humidification process

 If the temperature of the water sprayed is equal to the wet‐bulb temperature of Fig. Humidification Process air, then the net transfer rate will be

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission zero as the sensible heat transfer from upon the state of the individual streams, air to water will be equal to latent heat the mixing process can take place with or transfer from water to air. The process without condensation of moisture. Fig is called as adiabatic dehumidification shows an adiabatic mixing of two moist air process. streams. As shown in the figure, when two  If the water temperature is greater than air streams at state points 1 and 2 mix, the WBT, then there will be a net heat resulting mixture condition 3 can be transfer from water to air. obtained from mass and energy balance.  If the water temperature is less than WBT, then the net heat transfer will be from air to water.

7.4.6 CHEMICAL DE‐HUMIDIFICATION PROCESS

This process can be achieved by using a hygroscopic material, which absorbs or adsorbs the water vapour from the moisture.

fig. Mixing of two air streams From the mass balance of dry air and water vapour:

ma1 m a2  m a3 From energy balance:

ma1 h 1 m a2 h 2 m a3 h 3  m a1  m a2  h 3

ma1 h 1 m a2 h 2 h3     (7.17) mma1 a2  From the above equation, it can be observed that the final enthalpy of mixture

Fig. Chemical De‐Humidification Process is weighted average of inlet enthalpies. If this process is thermally isolated, then Example: the enthalpy of air remains constant and the process is called as adiabatic Q.1 Atmospheric air at 760mm Hg dehumidification process. As a result the pressure and dry bulb and wet bulb temperature of air increases as its moisture temperature150 C and 120 C enter content decreases as shown in Fig. In the heating coil whose temperature general, the absorption of water by the is 410 C. The By pas factor of heating hygroscopic material is an exothermic coil is 0.5. Determine the dry bulb reaction, as a result heat is released during temperature, wet bulb temperature, this process, which is transferred to air and relative humidity of air leaving the the enthalpy of air increases. heating coil and also the sensible heat supplied per kg of dry air. 7.4.7 MIXING OF AIR STREAMS Solve the problem with help of psychrometric chart. Mixing of air streams at different states is Solution: commonly encountered in many processes, By pass factor of heating coil is given including in air conditioning. Depending as

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission TT pressure of vapour, relative humidity BPF  d3 d2 and dew point temperature. Assume TTd3 d1 41 T standard barometric pressure is 760 0.5  d2 mm of Hg. 41 15 Solution: 0 Td2 41  0.5 41  15   2 8 C Specific humidity or humidity ratio is given by P W 0.622 V PPbV P 0.016 0.622 V 760 PV

0.016 760 Pv  0.622 1.02575 = 19.05 mm of Hg. Relative humidity P RH  V1 PS1 Thus the actual condition at which Where Ps1 is saturated pressure air comes out of the heating coil is corresponding to the saturation 280C.With the help of the temperature equal to dry bulb psychrometric chart relative temperature of 27.50 C. From the humidity at this point is 28.5% Wet table it can be read as 27.535 mm of bulb temperature at this point is Hg. 160C. So, From psychrometric chart, PV1 19.05 Enthalpy at point 1 h1 = 31.7 KJ/Kg RH    0.694 PS1 27.535 Enthalpy at point 2 h2 = 44.9 KJ/Kg So the sensible heat supplied to the = 69.4 % Dew point temperature is the air Q =h2- h1 = 44.9-31.7 saturation temperature = 13.2 KJ/Kg of the dry air. corresponding to the vapour pressure of the 19.05 mm of Hg as 0 Q.2 The humidity ratio of an read from table. So, Tdp1 = 21.34 C atmospheric air at 27.50C is 0.016 o kg/ kg of dry air. determine partial Q.3 Air enters a chamber at 5 C dry bulb temperature (DBT) And 2.5o C wet

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission bulb temperature at the rate of ms ww 100m3 / min and the pressure of 1 32 mq bar. While passing through the 35  0.0035 0.0081 chamber the air absorbs 50 KW heat 3600 2.1 and pick’s up 35 kg/hr of saturated Enthalpy of living air : steam at 110oC. Show the process on ms a psychrometric chart and find the h3  h 2  hv  dry and wet bulb temp of the leaving mq air at 110oC, enthalpy of saturated hv =2691.3 kj/kg at 110 0C 35 steam is 2691.3 kj/kg. h 37.76  2691.3 Solution: 3 3600 2.104 For process 1-2 this is sensible =50.19ks/kg heating process At state (3): Heat absorbed = 50 kw t3 = 310C, tw2 =18.3 0C

m(h21  h ) i.e at state 1-2: Q.4 In an air conditioned space, 50 kg dry air/sec of fresh air at 45 0C DBT and 30% RH is introduced. The room air at 25 0C DBT and 50% RH is recalculated at 450 kg dry air/sec. The mixed air flows over a cooling coil which has apparatus dew point of 12 0C and by pass factor of 0.15. Determine the conditions at outlet of cooling coil, RSH, RLH, the cooling load of coil and the condensate rate .The saturation pressure of water of required temperature are Temperature (0C) Pws (bar) 12 0.14016 25 0.03166 W1=0.0035(from psychometric chart) 45 0.09584 Specific volume = 0.792m3/kg Enthalpy at temperature, h=1.005 T h1=14 kj/kg +W (2500+1.885) kg/kg dry air. Volume flow rate =100 m3/min =1.67 m3/s Solution: 1.67 Temperature of air after mixing Mass flow rate  0.792 Tx = 0.9 t2 +0.1t1 = 2.104 kg/s = 0.9×25 +0.4× 45 Now heat transferred:- Specific humidity: a) At outlet condition: Q m(h h ) 2.104 m( h 14) 2 1 2 Pv1 50 2.104 m(h  14) W1  0.622 2 Po Pv1  h = 37.76 kj/kg 2 Pv = 0.3×0.09584 Now at state 2, from psychometric 1 chart = 0.02875 bar 0.02875 W22  0.0035, h 37.76 kj/ kg W  0.622 1 1.0132 0.02875 After humidification by steam   injection = 0.01816 kg vapour/kg dry air

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission b) In the Room =37.06 KJ/kg dry air. Pv w 0.622  2 2  Condition at outlet Pv Pv2 Ts=14.25 0C Pv2 = 0.5 ×0.03166 = 0.01583 bar hs =37.06 KJ/kg dry air 0.622 0.01583 Ws=0.009 kg vapour/kg dry air W  Room sensible heat 2 1.0132 0.013 58   RSH = Ma Lp (T2-TS) W2 = 0.00987 kg vapour /kg dry air =500×1.0216 (25-14.25 Specific humidity at A = 5491.1 KW 0.622 0.014016 Room latent heat WA  0.0132 0.014016 RLH = Ma (w2-ws)+2500 = 0.008725 kg vapor /kg dry air =500 (0.00987-0.009) ×2500 Specific humidity at inlet to cooling = 1087.5 kw coil Load on cooling coil = ma (hx-hs) Wx = 0.9 W2+0.1 W1 =500(54.43-37.06) 8685 = 0.9×0.009870 + 0.1×0.01816 =8685 kw = 3.5 =0.0107 kg vapour /kg dry air = 2481.43 tones

Condensate rate = (wx-ws) Ma Enthalpy at X = (0.0107-0.009) ×500 hx=1.00×27+0.00107(25001.88+27) =0.85 kg vapour /sec = 54.43 kj/kg dry air =3060 kg vapour/hr

Enthalpy at 2 Q.5 Calculate all the psychometric H2=1.005×25 properties of air at 1 bar, 300C DBT +0.00987(25001.88×25) and 250C WBT. The following = 50.3 KJ/kg dry air properties of water may be B.P.F. of cooling coil assumed :- ts ta BPF  (tx ta) ts ta ts 12 0.15  (tx ta) 27 12 0 Ts  14.25 C ws-wa ws-0.008725 0.15= = (wx-wa) 0.0107-0.008725 Solution: Ws = 0.009 kg vapour/kg dry air The following expression may be used, if necessary P Pw  DBT  WBT  1.8 P  Pv v 2824 1.325(1.8 DBT 32) DBT=250C, WBT=150C P Pw  DBT  WBT  1.8 P   Pv  2854 1.325(1.8tDBT 32) 1 0.03166  25 15   1.8  0.017307 Enthalpy at supply point 2854 1.325(1.8  25  32) hs=1.005×14.25+0.009 17.43 v  0.017307 (2500+1.88×14.25) 2751.975

Pyschrometric properties:-

1. specific humidity pv W  0.622 p pv 0.01097 0.622 1 0.01097 W= 0.006899 kg vapor/kg of dry air

2. Relative humidity (ɸ) Pv  100 PP V 0.01097  100 0.03160 ɸ = 34.65%

3. Degree of saturation (µ) PP µ  ( S ) P  PV (1 0.03166)  0.3465 (1 0.00109) 0.96834 0.3465 0.9890 µ = 0.339 or 33.9%

4) Enthalpy of moist air (h) h = Cpa + W [2500 + 1.88] where Cpa =1.005 kj/kg.k h 1.005  25   0.006899 2500 1.88  25 h = 25.125+0.006899(2547) h 42.696KJ / Kg of dry air .

Power plant engineering is the branch of area covered by 2 – 3 process on T-S thermal science which deals with study of diagram is equal to the area covered by 4 – chemical energy of fuel is converted into 1 process. thermal energy and then utilises into the mechanical work done in the efficient 8.2 ERICSON CYCLE manner. When the thermal energy is Ericson cycle has two reversible isothermal converted into the mechanical energy it is heat additions & heat rejection process and impossible to convert whole amount of two reversible isobaric processes. It differs heat into work. So efficiency play an from cannot cycle in that two isentropic important role in power plant to check the processes of Carnot cycle are replaced by performance of the power plant. The main two constant pressure process. P-V and T-s important power cycles are gas power diagram of Ericson cycle is shown. cycle and steam power cycle. Gas power cycle works on Brayton cycle and steam power cycle works on Rankine cycle.

8.1 STIRLING CYCLE

Stirling cycle has two reversible isothermal heat additions & heat rejection process and two reversible isocoric processes. It differs from cannot cycle in that two isentropic processes of Carnot cycle are replaced by two constant volume process.

The efficiency of Stirling cycle is given as T η 1- 2 T1 Regenerative Ericson cycle has same efficiency as that at cannot cycle. And both Stirling and Ericson cycle has same efficiency it temperature limits are same. Gas turbine cycle with inter cooling, reheating & regeneration closely resemble The efficiency of Stirling cycle is given as the Ericson cycle. Both cycles are difficult in T T practice because heat transfer at constant η 1 3  1  min temperature requires either infinite long TT 1 max surface area or infinite long time. If cycle is regenerative cycle then it has same efficiency as that of cannot cycle i.e. 8.3 GAS POWER PLANT

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission In gas power cycle, the gas is a working The P-V and T-S diagram of Brayton cycle fluid. It does not go under the phase change has been shown. during the cycle. Gas power cycle works on Brayton cycle. There are two type of cycle: Open Cycle In this cycle, the working fluid renew after completion of the cycle. Closed cycle In closed cycle, chemical composition of working fluid remains same as heat is supplied through the heat transfer and not by direct combustion. Advantages of closed cycle over open cycle 1) Use of high Pr. Throughout cycle The efficiency of Brayton cycle is given by reduces size of plant Q CTT  η 1 R  1  P 4 1 2) Close cycle has higher thermal QCTT  efficiency. S P 3 2 3) Elimination of possible of turbine blades T4 T11  erosion T 1 1 ------(8.1) 4) Closed cycle can use any working fluid T3 or gas of higher value of 훾 to increases T12  T the power output and efficiency. 2 process 1-2 and process 3-4 are isentropic 8.3.1 BRAYTON CYCLE process, so γ1 Brayton cycle consists of four processes: γ γ1 TP T3 Process 1–2: Isentropic Compression 22 r  γ  TP T Process: In this process the heat transfer is 1 1 4 zero. Where r = compression ratio T T dQ = 0 3  4 ------(8.2) Process 2–3 : Constant pressure Heat TT21 addition Process: Heat addition in this substituting the values in equation 8.1, we process is given by get QS = m. CP T3 − T2 T η11 Process 3–4 : Isentropic Expansion Process: T In this process the heat transfer is zero. 2 1 dQ = 0 η1 γ1 Process 4–1 : Constant pressure Heat r  rejection Process: The heat is reject in γ constant pressure process in heat  Thus efficiency of Bray ton cycle exchanger. Heat rejection in this process is depends upon compression ration and given by nature of gas. QR = m. CP T4 − T1

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 8.3.2 ENERGY BALANCE EQUATION ON TT31 COMBUSTION CHAMBER wnet C P T 3   T2  T 1 T 2 for optimum condition differentiating the equation with respect to T2

dwnet TT3 1 0  Cp  0 2  1  0 dt T2 Heat released due to combustion of fuel is converted into enthalpy increases of gas in TTT2 1 3 combustion chamber. TT T 13 T .T Heat input = Enthalpy change of gas 4 1 3 TT13 m XCVXη m c  T T  f c.c g pg 3 2 T T .T m XCVXη m  m   c T  T  4 1 3 f c.c a f pg 3 2 Compression ratio for optimum work done γγ γ 1 2 γ 1  8.3.3 ANALYSIS OF GAS TURBINE CYCLE PT22  T3  r       PTT1 1  1  The upper limit of temperature is fixed due γ to metallurgical condition of turbine 2γ1  T3 blades. Though by increasing the Pressure r   T1  1 w C T  TT  TT  T Ratio, net efficiency 1 can be net P 3 1 3 1 3 1 γ1 γ 2 R p  w C T T  net max P 3 1 increased but net work will decrease. So there is a optimum limit of the compression 8.3.4 REGENERATION IN GAS TURBINE ratio at which optimum efficiency and work output can be obtained. Efficiency of gas can be increased by utilizing the energy of exhaust gases from the turbine exit in heating air leaving the comp. In heat exchanger, and this process is called as regeneration.

Net work done in gas power plant is given by

wnet w T w C

CTTTTP 3  4    2  1  ----- (8.3)

form the process 1-2 and process 3-4 T2 TTT3 1 3 8.3.5 EFFECTIVENESS OF HEART  T4  TTT1 4 2 TRANSFER substituting the value in the equation 8.3 It is ratio of actual temperature rise of air to maximum possible rise of the

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission temperature. It is also known as degree of 8.3.7 BRAY TON CYCLE WITH INTER regeneration. COOLING m C T T   a pa a 2 m m  C  T T  The efficiency of Brayton cycle can be a f pa 4 2 improved by use of stages compression If the mass of air is higher as compare to with inter cooling. mass of the fuel, so the effectiveness of heat exchanger is 8.3.7.1 EFFECT OF INTER COOLING Actualtemp.Rise TTa2  m  max possibleRise T42 T 1) Decrease in compressor work 2) No change in turbine work 8.3.6 EFFECT OF REGENERATION 3) Network increases

1) There is no change in turbine work. 2) There is no change in compressor work. 3) Net work done remains same 4) Heat Supplied is decreased so the

efficiency of the plant increase Efficiency of ideal regeneration cycle Q TT  η 1 R  1  b1 QTTS 3 a  for the perfect regeneration,

TTTTab42, Optimum Pressure Ratio for minimum substituting the values in equation, we get compressor work T  T  b 2 w w w T11 1  T  1  C C1 C2 TT11     11    CTTCTTP 2  1   P  1  3  Ta  T4  T33 1 T  1  For perfect inter cooling TT13 TT33    CTTCTT       γ1 P 2 1 P 4 1 γ T1 r1    η1  TT24 γ1 CP1 T   2 T3 TT 1r   γ 11 T γ1 γ-1 γ-1 η 11 r γ TPTP γγ   reg   2= i , 4 = 2 T3     TPTP1 1  1  i  So as compression ration increases the ηη efficiency of simple Brayton cycle increases PP    w C T i2  2 but the efficiency of regenerative Brayton C P 1     PP1i    cycle will decreases.  Differentiating the equation with respect to Pi, we get

pi p 1 p 2

substituting the value of pi in compressor work for perfect inter cooling γ1 P γ w C T 2 1 C1 P 1  P1 

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 8.4 STEAM POWER PLANT T4 wC2 C P T 4  T 3   C P T 3   1 T 3 In steam power plant, the working γ1 substance recalculates in two phases liquid P γ w C T 2 1 and vapour. The components of simple C2 P 1  P1 vapour plant are shown in figure. 

8.3.8 BRAY TON CYCLE WITH REHEATING

The efficiency of Brayton cycle can be increased by reheating the gas and passing it through the number of the turbines.

Simple vapour plant works on Rankine cycle. Heat is transferred to water in the boiler from external source. On P-h and T-S diagram, Rankine cycle can be shown.

Effect of reheating 1) Turbine work increases because const Pressure Line diverge 2) No change in compressor work 3) Network done increases

Conditions for perfect reheating

1) TT35

2) PPPi 1 2

3) wwT1 T2 4) w 2w T T1 Rankine cycle consists of four processes: 8.3.9 ISENTROPIC EFFICIENCY OF Process1-2:IsentropicCompressionProcess: COMPRESSOR AND TURBINE In this process woter is compressed isentropically in feed pump. Pump work Due to irreversibility turbine produce less can be given: work as compare to ideal work and Wp = h2 – h1 compressor requires more work as Process 2 – 3 : Constant pressure Heat compare to ideal work required for addition Process in the boiler: The heat is compressor. added to water in the boiler. It is given as efficiency of the compressor is given by  hQ 3S – h2 Isentropic work h h T T 2 1 2 1 Process 3 – 4 : Isentropic Expansion ηC  11  actual work h2 h 1 T 2 T 1 Process in the turbine: In this process the efficiency of the turbine is given by heat transfer is zero and work done by the

actual work h3 h 4 T 3 T 4 turbine is given as ηT  11  Isentropic work h3 h 4 T 3 T 4 WT h 3 – h 4

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Process 4 – 1 : Constant pressure Heat the steam from turbine and water just rejection Process in the condenser: The before entering the boiler. Regenerative heat is rejected at constant pressure cycle can be shown on T-S diagram. process in condenser. Heat rejection in this process is given by

 hQ 4R – h1 The efficiency of Rankine cycle is given by Q h4 – h1 η 1 R  1  QS h3 – h2

8.4.1 REHEAT RANKINE CYCLE

In reheating cycle, the expansion of steam from the initial state at turbine is carried out in two or more no. of re heater. The network out increases and hence steam rate decreases. P m w net Turbine work in regeneration is given by

wT h 1  h 2    1  x1  h 2  h 3  Energy balance equation on OFWH

xh2  1  x  h5  h 6

8.4.3EFFECT OF PARAMETERS ON 1) Work done by the turbine PERFORMANCE OF STEAM POWER PLANT w h  h  h  h T 3 4   5 6  1) Effect of Boiler pressure If the boiler 2) Work done by the turbine pressure is increased, the effect on the wp h 2  h 1  vdp performance can be checked with help of T- 3) Heat supplied in the boiler S diagram

QS h 3  h 2    h 5  h 4  ww η  Tp Qs  The main advantage of reheat factor is to increase dryness fraction at the turbine exit. 1) w increases 8.4.2 REGENERATION IN RANKINE CYCLE p

2) wT  same In order to increase the mean temperature 3) Q Decreases of the heat addition so that heat can be S transferred at higher temperature. So the 4) η increases efficiency of the cycle can be increased. In 5) Dryness fraction at outlet of turbine this cycle, it is possible by transferring the decreases

heat from vapour as it flows through the 2) Effect of Superheating turbine to the liquid flowing around the If steam is superheated in the boiler, the turbine. With the help of open feed water effect on the performance can be checked heater (OFWH) heat is transferred between with help of T-S diagram

lpgγl g ρ l g C differentiating the equation, we get dp dρ γ0 p ρ dp γp  1) wp same dρρS

2) w increases 2 γρRT T C  ρ 3) wnet increases So the velocity of sound in a medium is 4) Q increases S given by 5) η iincreses C  γRT 6) Dryness at the outlet of turbine Where R is the specific Gas constant. increases 8.5.5 MACH NUMBER 8.5 NOZZLE AND DIFFUSER Mach number is defined as the ratio of 8.5.1 NOZZLE actual velocity to the velocity of the sound or sonic velocity. V It is a device of varying cross section used M  to convert the Pressure Energy of working C fluid into kinematic energy. It is mainly #M>1, then V>C flow is supersonic flow. used to produce the jet of steam or gas of # M = 1 , then V = C flow is Sonic flow. high velocity to produce throat for Jet # M < 1 then V < C flow is Subsonic flow. propulsion to derive steam turbine or gas # M>5 then V>5C flow is hypersonic flow. turbine. 8.5.6 STAGNATION POINT AND STAGNATION PROPERTIES 8.5.2 DIFFUSER Stagnation point is state in the fluid flow It is a device of varying cross section where velocity of flow is become Zero and causing rise in Pressure Energy atthe cost kinetic energy converts into pressure of kinematic energy. These are used in energy. centrifugal and axial compressor. The Pressure, Density, temperature etc. of  The smallest section of nozzle is known stagnation point are known as stagnation as throat. Properties.

8.5.3 SPEED OF SOUND

It may be defined as velocity of Pressure Wave in fluid medium. dp C   dp s 8.5.4 VELOCITY OF SOUND IN IDEAL GAS for the ideal gas at stagnation point, 2 For an ideal gas, flow is isentropic V hh0  Pvγ  C 2 P V2  C CTCT ργ POP 2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission V2 dρ dA dV TT 0    O ρ A V 2CP γR dA dρ dP We know that C     2 P γ1 A ρ PV V2 γ1  dA dV dρV TT 1X 0 2γR  AVρ dV Substituting the value of dV from equation T V2 0 1 γ1   (8.7), we get T2γRT dA dP V2 2 T γ1  V 221 0 1. A ρV C T 2 C2 dA dV T γ1  M12 0 1M2 AV T2 γ 1) When M < 1, flow is Subsonic flow γ1 PT00   dA PT a) Converging passage Ve A

8.5.7 FLOW THROUGH NOZZLE & DIFFUSER

Assumption for flow 1) flow is the isentropic flow. 2) Stagnation enthalpy is constant hc 0 V2 hh 8.4  AVPM   ρ   0 2 dA differentiating the equation b) Diverging passage Ve dh VdV ------(8.5) A From combined first and second law of thermodynamic TdS dh VdP differentiating the equation, we get dP dh VdP ------(8.6) AVPM   ρ  ρ   dP or Vd 2) When M > 1 flow is supersonic flow ρ dρ a) ρV dV dP dV  ------(8.7) ρV

Continuity equation for the mass flow m  ρAV takinglogin both the sides AVPM   ρ    logm logρ  logA  logV dA differentiating the equation b) Diverging Passage Ve A

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission A compressor is a device in which work is done on the gas to raise the pressure of the gas. The work done by the compressor is negative in the compression process because the work is done on the system to raise the pressure of the gas.

AVPM   ρ   8.6.1 WORK DONE IN COMPRESSOR

Conclusion Process of compression in compressor is a) For subsonic flow and compressible shown in the fig. fluid convergent passage acts as a nozzle and divergent passage acts as a 1) Work done in adiabatic process diffuser. γ1 γ P γ b) For supersonic flow and compressible W  P V2  1 11 fluid convergent passage acts as γ 1 P1   diffuser & divergent passage acts as a nozzle. c) For an incompressible fluid converging passage always aits as nozzle and divergent as a diffuser. ρ dA dV    0 ρ A V dA dV 2) Work done in polytropic process  n1 AV n P n 2 W P11 V 1 8.5.8 MASS FLOW RATE PER UNIT AREA n 1 P 1 

mass flow rate per unit area is given by 8.6.2 ACTUAL DIAGRAM OF COMPRESSOR m 2 n 1 ρ 2C T rnn r 1 P 1     The actual work done will be more due to A2  For maximum flow rate resistance offered by suction value and discharge value. The shaded portion dm  0 represents the extra work to counter value dR A2 resistance. The actual diagram with the n clearance volume is shown in fig. n1 P2 2 r  P1  n 1 m will be max when Area is minimum and A it occurs at throat. So the critical condition

occurs at throat. So Critical pressure is Fig.: Actual P-V diagram of compressor given by n n1 8.6.3 CONDITION FOR MINIMUM WORK Pc 2   REQUIRED IN MULTISTAGE P n 1 1 COMPRESSOR WITH PERFECT INTER COOLING 8.6 COMPRESSOR For perfect inter cooling

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission TT13 The velocity triangle is shown in the fig. From the outlet of nozzle, steam flow at PPP2 1 3 absolute velocity V1. But due to blade For multi stage compressor velocity u, steam strikes at the blade with PPPP 2 3 4  n 1  C relative velocity Vr1. PPPP1 2 3 n Following are the parameters of the turbine. α=Angle at which steam enters or guide PN1 N  C blade angle P 1 θ= Vane angle at inlet ϕ= Vane angle at outlet β= Angle at exit of moving blade

8.7 STEAM TURBINE

Steam turbine is a prime mover which is used to convert the potential energy into fig.: velocity triangle the kinetic energy and then utilize it into V12 &V = Absolute velocity at entrance and the mechanical work. Main use of steam exit turbine is to produce the mechanical work V &V =Velocity of whirl at entry and at in steam power plant. w1 w2 exit or tan agent of absolute velocity 8.7.1 TYPES OF TURBINE Vr1 &V r2 Relative velocity at entry and exital blade

There are two types of turbines: Vf1 &V f 2  axial component of absolute 1) Impulse Turbine: In impulse turbine velocity pressure drops in nozzles but remains u = blade velocity same in moving blade. 2) Reaction Turbine: In reaction turbine 8.7.4 PERFORMANCE PARAMETERS OF pressure drops in fixed blade and it also TURBINE changes during the expansion of the steam in moving blades. 1) Velocity Ratio It is the ratio of blade velocity to the flow 8.7.2 COMPOUNDING IN TURBINE velocity at the inlet of the blade. It is given by ρ. In this type of compounding two shafts are u ρ  used to drive separate generator at V different RPM. There are two type of 1

compound turbine: 2)Blade or Diagram efficiency

It is ratio of net work done by turbine to 1) Pressure compounding- Rateua turbine kinetic energy at the inlet of the blade. 2) Velocity compounding - Curtis turbine 2u V V w1 w2  ηD  2 8.7.3 VELOCITY TRIANGLE OF TURBINE V1

If m a T 2θ  , Vr1  V R 2 then V w 1  V 1 cos 

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Vw2 V r 2 cos   u  Vr2 cos   u u Vw1 V w2  ηstage  V11cos   u   u  Vcosα2  u hh12 

2uVVVww1 2  2u 21 coscosαu ηD 22 Examples V V 11Q.1 Determine the minimum no. of stage 4ucos u2 η44ρcosα 4ρ2 required in an air comp. which takes D 2 0 VV11 air at 1 bar 27 C and delivers 180 η 4ρ cosα ρ  --- (8.8) bar. The max discharge temperature D is 1500C and considers the index of For maximum diagram efficiency, polytrophic as 1.25 and perfect inter differentiating the equation with respect to ρ. cooling b/w stages. dη D 0  4cosα  8ρ Solution: dp P N1  CN cosα P ρ  1 2 1.25 PT1.25 Substituting the value of ρ in equation 8.8, n 1 2 5.57N get  PT11 cos as 2 180 N ηDmax  4 as   cos   5.57 22 1 2 ηDmax  cos N= 3 For n number of stages (Curtis stage) cosα Q.2 A gas turbine plant operates on ρ  opt 2η Brayton cycle, between Tmin =300K and Tmax=1073K. Find the maximum η cos2 α Dmax work done per kg of air and cycle efficiency. Compare the efficiency 3)Blade or diagram efficiency of with the efficiency of Carnot cycle reaction turbine with same temperature limit. 22ρcosα ρ2  Solution: η  2 D 12ρcosα ρ2 Wpnetmax   TCMax T Min  For maximum diagram efficiency, 2 differentiating the equation with respect to  1.005  1073 30 0 , we get = 239.28 KJ/Kg dη Efficiency of the Brayton cycle D  0 1 dρ η1 ρ cosα γ1 r γ Substituting the value of in equation of T 300 diagram efficiency of reaction turbine, we 1 Min  1   47% get TMax 1073 2 2 2cos α Efficiency of the Carnot cycle ηDmax  2 1 cos α Tmin 300 ηC  1   1   72.1% Tmax 1073 4) Stage efficiency η 0.47 0.65 It is the ratio of the net work done to η 0.721 change of enthalpy at the inlet of the C turbine.

1.013 bar, 10°C and has a pressure exchanger = Cp(T5 –T2) ratio of 5.5. the maximum Actual Heat transfer = 0.7Cp(T5 –T2) temperature in the cycle is limited = 132 KJ/kg of air to 750°C. Compression is conducted in rotary compressor having an Cp(T3-T2) = (1+ m)×132 isentropic efficiency of 82%, and CpT3 = 634.43 +132.33 m expansion takes place in a turbine m = 8.68 × 10–3 kJ/kg of air with an isentropic efficiency of 85%. Heat addition (Q1)=Cp(T4-T3)=CpT4-

A heat exchanger with an efficiency CpT3 of 70% is fitted between the = 393.7 –132.33m compressor outlet and combustion = 392.6 kJ/kg of air chamber. For an air flow of 40 kg/s, find: WT = (1 + m) (h4 –h5) a) The overall cycle efficiency, = (1 + m) Cp (T4 – T5’) b) The turbine output, and =1.00868 × 1.005 × (1023 – 687.7) Solution: kJ/kg of air P1=101.3 KPa, T1=283 K, r=5.5, = 340 kJ/kg T4 = 750℃ = 1023 K Wc=(h2–h1)=Cp(T2‘–T1)=1.005× γ1 (499.6 – 283) TPγ γ1 1.4 1 22 r  γ  5.5  1.4 = 217.7 kJ/kg of air TP 11 Wnet = WT - Wc = 122.32 KJ/kg T2 = 460.6 K Networkdone 122.32 a)     31.16% Isentropic efficiency of compressor Heatsupplied 392.6

Isentropic work h2 h 1 T 2 T 1 b) Turbine output = (WT) = 122.32 ηC  11  actual work h2 h 1 T 2 T 1 kJ/kg of air= 4893 kW

TT21 460.6 283 0.82 11 T2 T 1 T 2 283 Q.4 A simple steam power cycle uses solar energy for the heat input. T2’ = 499.6K Water in the cycle enters the pump T γ 1 1.4 1 4 r  γ   5.5  1.4 as a saturated liquid at 40°C, and is T5 pumped to 2 bar. It then evaporates T5 = 628.6 K in the boiler at this pressure, and Efficiency of turbine enters the turbine as saturated actualwork TT vapour. At the turbine exhaust the η 45 conditions are 40°C and 10% T Isentropicwork T T1 45 moisture. The flow rate is 150 kg/h. 1023 638.6 determine: 0.85  1 a) the turbine isentropic efficiency, 1023T5 b) The net work output c) The cycle efficiency Solution From Steam Table

T1 = 120.23°C = 393.23 K, h3 =

2706.7 kJ/kg, s3 = 7.1271 kJ/kg – K At 40°C saturated pressure 7.384 T5’ = 687.7 K kPa

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission hf=167.57KJ/Kg, hfg=2406.7 KJ/Kg KJ sf 0.4764 .K sfg 7.9187KJ / Kg.K sf = 0.5725, sg = 8.2570 Kg ’ 3 h4 = hf + 0.9 × 2406.7 vf = 0.001005 m / Kg = 2333.6 kJ/kg At 8 bar

hf = 721.11KJ/Kg hfg = 2048KJ/Kg, KJ s 2.0462 .Ks 4.6166 f Kg fg KJ/Kg.K v = 0.001115 m3/ Kg f Regeneration cycle can be shown on For h4s if there is dryness fraction x T-s diagram 7.1271=0.5725+x×(8.2570– 0.5725) x = 0.853

h4 = 167.57 + 0.853 × 2406.7 = 2220.4 kJ/kg Isentropic efficiency, h3 – h4 2706.7 – 2333.6    isentropic h3 – h4' 2706.7 – 222 0 .7

= 76.72% Process 1-2 is isentropic process Net Turbine work output WT=h3 – h4 s1 = s2 , h1 = 3675.2 = 373.1 kJ/ kg 6.6502 = 2.0462 + 푥2 4.6166 Pump work, WP = v (P1 – P2) 푥2 = 0.997 –3 = 1.008 × 10 (200 – 7.384) kJ/kg Enthalpy of point 2 = 0.1942 kJ/kg h2 h f 2  x 2 h fg2  721.11  0.997  2048 Power = m × (W T - Wp) = 15.55 kW h 2763.54kJ / kg h3=2706.7kJ/kg, h2=167.76 kJ/kg 2 Process 1-3 is isentropic process Q1=(h3–h2)=(2706.7 – 167.76) kJ/kg = 2539 kJ/kg 푠1 = 푠3 WW 373.1 0.1942 6.6502 = 0.4764 + 푥3 푋 7.9187  TP14.69 % cycle 푥3 = 7796 Q1 2539 Enthalpy of Point 3

h h x h Q.5 A regeneration Rankin cycle as 3f 3 3 fg 3 steam entering turbine at 200 bar 137.82 0.7796X 2423.7 and 250℃ superheated state and  2027.45kJ / kg leaving at 0.05 bar. Considering feed Work done by pump 1 water to be open type. Determine 2 the efficiency of plant when feed wp1 =h 5 -h 4 =vdp=0.001005 8-0.05  ×10 water is operating at 8 bar. =0.798 Solution: h =0.798+137.82=138.618kJ/kg At 200 bar and 250℃ 5 Work done by pump 1 h=3675.2KJ/Kg, s=6.6502KJ/Kg.K, 2 3 w =h -h =0.001115 200-8 X10 푣푓 = 0.001005 m /Kg p2 7 6   At 0.005 bar hf = 137.82 KJ/Kg, = 22.08 푘퐽/푘푔

hfg = 2423.7KJ/Kg, h76 =h +22.08=723.19kJ/kg

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Energy Balance equation at open 0.2 M4 + M2 − 0.645 = 0 feed water heater Solving the equation, we get x h  1  x  h  h M = 0.76 2 5 6 T (γ 1)M22 (1.4 1)  0.76 0 11    xx2763.54  1    138.618  721.11 T 2 2 x = 0.2919 T = 277.78K

wnet h 1  h 2    1  x  h2  h 3  C= γRT 3 wwp1 p2  = 1.4 0.287  277.78  10 3675.2  2763.54   1  0.2919  = 334.08 m/sec V = c × M = 334.08×0.76 = 253.9 m/s 2763.54 2027.45  (798  22.08) γ1 1.4 1 γ 1.4 PT00 310 1462.4KJ / kg      P T  277.78  Qs =h 1 -h 7 =3675.2-723.19 P0 = 40 × 1.468 = 58.72 KPa = 2932.78 kJ/kg Efficiency of the plant Q.7 A simple open cycle gas turbine has w η= net a compressor turbine and a free Qs power turbine. It develops electrical 1462.4 power output of 250 MW .the cycle 49.86% 2932.78 takes in air at 1 bar and 288 K. The total compressor pressure ratio is Q.6 A stream of air flow in a duct of 14. The turbine inlet of compressor 100mm diameter at the rate of 1 is 1500 K. The isentropic efficiency kg/sec. The stagnation temperature of compressor and turbine is 0.86 is 370C. At one section of the duct and 0.89 respectively. The the static pressure is 40 KPa. mechanical efficiency of each shaft is Calculate the Mach number, velocity 0.98. Combustion effect is 0.98 and stagnation pressure at that while combustion pressure loss is section. 3% of compressor delivery Solution: pressure. Alternator efficiency is T0 = 37+ 273 =310 K, P= 40 KPa, and 0.98. Take calorific value of fuel is γ = 1.4 42000 KJ/kg. Cpa = 1.005 KJ/Kg K The mass flow rate per unit area is and Cpg = 1.15 KJ/Kg. K. Calculate mP i) Air –fuel ratio ρ  V   M γRT ii) Specific work output A RT Solution: γ PMT γ PM (γ-1)M2 = ×0 = × 1+ Electrical power =250mw RTRTT002 Power output from power turbine 250 2  1 1.4 40M (1.4-1)M alternator efficiency  mechanical efficiency 2 = × 1+ π0.1 0.287 310 2 250  = 260.3082 MW 4 0.98 0.98 1.4 127.39  40  1032MM 1  0.2 = 260308.2 KW 0.287 103 310 0.4 288 14  612.15 K 0.803 = M 1 0.2M2 1.4 Squaring both the sides ’ 1TT 21 1) T2 T 1  14  T2 T 2  0.645 = M22 (1 0.2M )  ηc

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission (1+0.0268) (1500 - T5) = 336.12 T5 = 1500 – 327.35 T5 = 1172.65 K Ma (1+0.0268) (T5 − 869) = 226355 Ma = 725.987 Ma = 726 kg/sec Special output =(1+F) Cpg(T5 − T6) = 1.0268 1.15 (1172.65 – 869) = 358.56 KW/Kg

612.15 288 288   664.92K Q.8 In a multi-stage parson’s reaction 0.86 turbine at one of the stages the rotor Compressor is run by compressor diameter is 125 cm and speed ratio turbine 0.72. The speed of the rotor is MaCpa(T2 - T1) = ma (1 + f) (T4 –T5 ) 3000rpm. Determine cpg × 0.98 1) The blade inlet angel if the blade 1.005 × (664.92-288) = (1+t) (tut5) × outlet angle is 22° 1.15 × 0.98 2) Diagram efficiency 250 329.4 Solution: 1 f  T – T   450.98xo. 98 0.98 Parson’s reaction turbine (1+F)(1500-T5) = 336.12 ------(1) D1=1.25m, N=300rpm Total efficiency of turbine = 0.89 ρ = u/v1=0.72m, β1=?, β2=22 T1.33 T1.33 For parson’s reaction turbine 4  6 α =β and α = β PP1.33 1 1.33 1 1 2 2 1 46 π D  N π  1.25  3000 0.33 u   1 1.33 60 60 1 P6 TT64 791k = 196.34 m/s P4 V1 = u/V1 = u/0.72 = 272.69 m/s T T 1500 T 4 6 6 Vr2 = V1 = 272.69 m/s 0.89  1  = (T66 T ) 1500 791 Inlet velocity triangle and Outlet T6 = 1500 – 0.89 (1500 – 791) velocity triangle T6 = 869K Power output from power turbine, Ma(1+F) Cpg (T5 − T6) = 260308.2

MTa5 1F  1.15 869   260308.2

Ma(1+F) (T5 − 869) = 226355 Heat balance equation in turbine,

Ma 1 F  CpgTM4 a CpaT 2

Ma CV combustion effie 1 Fncy  1.15  1 500 –1.005  664.92  41160 f 1725 – 668.25 = (41160 – 1725) F F = 0.0268 1 i) Air fuel ratio = = 37.32 Inlet velocity triangle outlet velocity F triangle ii) Specific work output Vw1 = 272.69 cos20 = 252.83 m/s Substituting the value of F in Vf1= 272.69 sin20 = 102.15 m/s equation 1

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Vf1 102.15 tan   1 vw1 v1 252.83 196.34 So, β1=61.05 and α2=β1-61.05 diagram efficiency power  d 1 m[V122 V22 Vf1 ] 2 power = mu (vw1+vw2) Inlet velocity triangle Outlet velocity u = 196.34 m/s triangle from velocity triangle vw1 = 252.83 m/s Vw1= v1 cosα= 550 cos17 =525.96 From outlet velocity triangle m/s vr2 cos β2 = u+Vw2 Vf1= v1 sinα = ssosinl7=160.80 m/s 272.69 cos 22=196.34+Vw2 vf1 tna 1 Vw2=56.49 m/s vw1 v1 Power = 1 × 196.34 × (25 + 56.49) 160.80 tan = 60.73KW 1 525.96 180 3 60 10 β1 = 24.92 d 1 2 2 2 vf 2 m[272.69 272.69 102.15 ] Tna  2 2 v2  0.90or 90 % vf1 160.80 tan  2 u 180 Q.9 A de level turbine has a mean blade β2 = 41.27° speed of 180 mps. The nozzles are Power output = m (Vw1+Vw2) u inclined at 170 to the tangent. The vw1 =. 525.96 m/s steam flow velocity through the vw2=0 (due to axial outlet condition) nozzle is 550 m/sec. For a mass flow u = 180 m/s of 3300 kg/hour and for axial exit 3300 condition: Find P  525.96 0   180 3600 1) The inlet and outlet angles of the P = 86.78 KW blade system. iii) Diagram efficiency 2) The power output. 2(vw1 vw2)u 3) Diagram efficiency.   2 100 Solution: V1 u= 180 m/s 2 (525.96  0)  180   10 0 Nozzle angle αͦ = 170, Mass flow rate (550)2 = 3300 kg/hr = 0.9167 kg/sec = 62.59 % V1 =550 m/s Inlet velocity triangle and outlet Q.10 A single acting two-stage air velocity triangle compressor deals with 4 m3/min of air at 1.013 bar and 15℃ with a speed of 250 RPM. The delivery pressure is 80 bar. Assuming complete inter -cooling, find the power required by compressor and bore and stroke of the compressor. Assume apiston speed of 3 m/s, mechanical efficiency 75% and volumetric efficiency of 80% per

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission stage. Assume the polytropic index of compression in both stage to be n = 1.25 and neglect clearance. Solution: Intermediate pressure for perfect intercooling

P2 = P1P3 = 1.013 × 90 = 9 bar Minimum power required by compressor n1 2n P n 1 W = P V2  1 11 n 1 P1 ηmech  1.25 1 2 1.25 1.013 100 4 9 1.25    1 1.25 1 0.75 60 1.013  1013 4  0.548  49.34KW 45 L is the length of stroke of the piston N 2L 3 m / s 60 90 100 L   36 cm 250 Effective swept volume Vs = 4/ 250 = 0.016 m3 π ×D2 ×L×η =0.016 4 LP Vol 0.016 4 D  LP 3.14 0.36 0.8 = 0.266 m = 26.6 cm Ideal gas equation PV PV 11 33 TT13 for perfect intercooling, T1 = T3 So. V P 1  3 VP31 π DL2 LP 9 4  π DL2 1.013 4 HP 1.013 D 0.266 HP 9 = 0.89 m = 89 cm

9.1 HEAT ENGINE due to high temperature in the combustion chamber caused by high compression. Heat engine is a device which transforms the chemical energy into heat energy and 9.2 APPLICATIONS OF IC ENGINES the utilized it into mechanical work in an efficient manner. Thus the thermal energy Mainly used as ‘prime movers’, e.g. for be is transforms into mechanical energy into the propulsion of a vehicle car, bus, truck, heat engine. There are two types of heat locomotive, marine vessel, or airplane. engine: Other applications include stationary saws, lawn mowers, bull-dozers, cranes, electric generators, etc.

9.3 CLASSIFICATIONS OF IC ENGINES

IC engines can be classified according to: 1. Number of cylinders – 1, 2, 3, 4, 5, 6 to 16 cylinder engines. The internal combustion engine is a heat 2. Arrangement of cylinders – Inline, V- engine that converts heat energy (Chemical type, Flat type, etc. energy of a fuel) into mechanical energy 3. Type of cooling – Air-cooled, Water- inside the cylinder of the engine (usually cooled, etc. made available on a rotating output shaft). 4. Number of strokes per cycle – 2-stroke, The External combustion engine is a heat 4-stroke engines. engine that converts heat energy (Chemical 5. Method of ignition – Spark Ignition (SI), energy of a fuel) into mechanical energy Compression Ignition (CI). outside the cylinder of the engine. The 6. Primary mechanical motion – internal combustion engine can be Reciprocating, rotary. categorized into two types: 9.4 BASIC COMPONENTS OF I.C. ENGINE 1) Spark Ignition (SI) Engine

An SI engine starts the combustion process in each cycle by use of a spark plug. The spark plug gives a high voltage electrical discharge between two electrodes, which ignites the air fuel mixture in the combustion chamber surrounding the plug. In early engine development, before the inventor of electric spark plug, many forms of torch holes were used to initiate combustion from an external flame.

2) Compression Ignition (CI) Engine The combustion process in a CI engine starts when the air-fuel mixture self-ignites Fig: Internal Combustion Engine

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Engine components: crankshaft of the engine. The purpose of The following is the list of major the flywheel is to store energy and furnish components found in most reciprocating a large angular momentum that that keeps internal combustion engines the engine rotating between power strokes and smoothes out engine operation. Block: Body of engine containing the cylinders made of cast iron or aluminium. Fuel pump: Electrically or mechanically In many older engines the valves and the driven pump to supply fuel from the fuel valve ports were contained in the block. On tank to the engine. air cooled engines the exterior surface of the block has cooling fins. 9.5 TERM SUSEDIN INTERNAL COMBUSTION ENGINE Camshaft: Rotating shaft used to push open valves at the proper time in the i) Top-Dead-Centre (TDC): Position of engine cycle either directly or through the piston when it stops at the furthest mechanical or hydraulic linkage. Most point away from the crankshaft.

modern automobile engines have one or ii) Bottom-Dead-Centre (BDC): Position more camshafts mounted in the engine of the piston when it stops at the point head. closest to the crankshaft. Carburettor: Venturi flow device that iii) Bore: Diameter of the cylinder or meters the proper amount of fuel into the diameter of the piston face, which is the air flow by means of pressure differential. same minus a very small clearance.

Catalytic converter: Chamber mounted in iv) Stroke: Movement distance of the exhaust flow containing catalytical material piston from one extreme position to the that promotes reduction of emission by other: TDC to BDC or BDC to TDC. It is chemical reaction. denoted by L.

Choke: Butterfly valve at carburettor v) Clearance volume: Minimum Volume intake, used to create rich fuel-air mixture in the combustion chamber with piston in intake system for cold weather starting. at TDC. It is given by Vc.

Combustion chamber: The end of the cylinder between the head and the piston vi) Displacement volume: Volume face where the combustion occurs. displaced by the piston as it travels through one stroke. It is also known as Piston: The cylindrically shaped mass that swept volume (Vs). reciprocates back and forth in the cylinder transmitting the pressure forces in the vii) Air Fuel Ratio: It is the ratio of mass combustion chamber rotating the crank shaft. air to mass of fuel input into engine. The top of the piston is called crown and the sides are called skirt. Pistons are made viii)Stroke to Bore Ratio: It is the ratio of of cast iron Aluminium or steel. the stroke length to the diameter of the Piston rings: Metal rings that fit into cylinder. circumferential groups around the piston L Stroke to bore ratio= and form a sliding surface against the d cylinder walls. The purpose of the rings is L  If 1It is known as Square Engine to form a seal between the piston and d cylinder walls. L  1It is known as Under Square Flywheel: Rotating mass with large d moment of inertia connected to the engine

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission L efficiency.  1It is known as Over Square actually thermal eff. d η  engine r Air standard efficiency ix) Compression ratio: It is the ratio of the total volume to clearance volume. 6) Mean effective pressure: M.E.P. is VT VVVCS S average pr. inside the cylinder of an r1P     VVVCC C internal combustion engine on power output P X A X L X NXK 9.6 DIFFERENCE BETWEEN FOUR STROKE I.P.  im watt AND TWO STROKE ENGINE 60

N Where N for4 stroke engine and 2 N  N for 2 – stroke engine the mean effective pressure can be given as area of indicator diagram P  im Length of indicator diagram

9.8AIR STANDARD CYCLE AND EFFICIENCY

9.7 PERFORMANCE PARAMETERS The analysis of actual cycle is very complicated. So the analysis of cycle is done 1) Indicated thermal efficiency: it is the on the basis of air standard cycle. The air ratio of the energy in the indicated standard cycle is based on such power to input fuel energy in the assumptions: appropriate unit. I. The working medium is perfect gas. I.P.(KJ / Sec) II. There is no change of the mass of the ηith  energy of fuel / sec working fluid. III. All the processes in the cycle are 2) Brake thermal efficiency: it is the reversible processes. ratio of the energy in the brake power to input fuel energy in the appropriate 9.8.1 OTTO CYCLE unit. B.P.(KJ / Sec) Nicolaus Otto proposed a constant volume η  heat addition and rejection process in place ith energy of fuel / sec of isothermal process of Carnot cycle. Otto

cycle is used now a day in spark ignition 3) Mechanical efficiency: it is the ratio of engine. the energy in the brake power to 4-stroke petrol engine operates on air indicated power in the appropriate unit. standard Otto cycle. It completes the Otto B.P..(KJ / Sec) ηm  cycle in 4 strokes (4 TDC to BDC movements I.P..(KJ / Sec) of the piston). In 4 stroke crank shaft rotates 7200. 4) Friction Power: friction power is the The four stroke of four stroke engine are: difference between indicated thermal 1) Suction Stroke, power and mechanical power or brake 2) Compression Stroke, power. 3) Power Stroke, F.P. =I.P.-B.P. 4) Exhaust Stroke. Otto cycle shown below consists of four 5) Relative efficiency: it is the ratio of the processes: actual thermal efficiency to air standard

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Process 1–2: Isentropic Compression process 1-2 is isentropic process so for the Process: In this process the heat transfer is isentropic process, zero. dQ = 0 γ1 TV γ1 Process 2 – 3 : Constant Volume Heat 21 r  TV addition Process: The fuel and air mixture 12 is burnt at constant volume process. Heat Process 3-4 is the isentropic process so for addition in this process is given by the isentropic process, γ1 QS m.C v T 3 T 2  T V γ1 3 4 r  Process 3–4 : Isentropic Expansion Process: TV43 In this process the heat transfer is zero. VV dQ = 0 14r VV Process 4–1 : Constant Volume Heat 23 T T rejection Process: The heat is reject in 2  3 constant volume process. Heat rejection in TT14 this process is given by Substituting the value in equation 9.1

QR m.C v T 4 T 1  T 4 1 P-V and T-S diagram for Otto engine is  TT11T1 shown in the fig. η 1  1  TT22T 3 1 T2 1  1  ------(9.2) r1 The thermal efficiency is the function of the compression ratio and the ratio of specific heat γ. if the is considered as constant the efficiency depends upon the compression ratio. The effect on efficiency of the Otto cycle can be understood with the help of the efficiency vs. compression ratio.

fig: Otto cycle: P-V and T- S diagram Efficiency of the cycle is given by Q η1 R QS m.C T T  1 V 4 1 m.C T T  fig: efficiency vs. compression curve V 3 2 TT  1 41 9.8.2 DIESEL CYCLE TT  32 In actual spark ignition engine, upper limit T of compression ratio is limited due to self- T 4 1 1  ignition temperature of the petrol. This can T1 1 ----- (9.1) be overcome by compressing the air T3 T2 1 separately. In diesel cycle fuel is burnt at T2 constant pressure. Other three processes are the same processes as in Otto cycle.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Diesel cycle shown below consists of four V Expansion ratio r  4 processes: e V3 Process 1–2: Isentropic Compression VV Process: In this process the heat transfer is 42 VV zero. 32 VV r dQ = 0 12   Process 2 – 3 : Constant pressure Heat VV2 3ρ C addition Process: The fuel is burnt at r p r constant pressure process. Heat addition in Ce Substituting the values of the temperature this process is given by in term of the compression ratio and cut off Q m.C T T S P 3 2  ratio, we get efficiency of diesel engine Process 3–4 : Isentropic Expansion Process: γ In this process the heat transfer is zero. 1 ρ1c  η1D   ------(9.4) dQ=0 γ1 r γ ρc  1  Process 4–1 : Constant Volume Heat rejection Process: The heat is rejected in The bracket term is always greater than constant volume process. Heat rejection in one, so efficiency of diesel cycle is always this process is given by less than efficiency Of Otto cycle for given Q m.C T T  compression ratio. The normal range of R v 4 1 compression ratio in petrol engine is from P-V and T-S diagram for diesel engine is 6–10 and in diesel 15–20. So efficiency of shown in the fig. diesel engine is more than efficiency of gasoline engine.

9.8.3 DUAL CYCLE

Dual cycle is also called as mixed cycle or limited pressure cycle. It is the compromise between the Otto cycle and Diesel cycle. Because for chemical reaction some time is required to so combustion does not occurs at constant volume and due to rapid uncontrolled combustion in diesel engines, combustion does not occur at constant pressure. Dual cycle shown below consists of five Efficiency of the cycle is given by processes: Q Process1-2: Isentropic Compression Process: η1 R Q In this process the heat transfer is zero. S dQ = 0 m.CV T 4 T 1  Process 2–3 : Constant volume Heat η1D    m.CP T 3 T 2  addition Process: The fuel is burnt at constant volume process. Heat addition in 1 T T  141      (9.3) this process is given by γ T32 T  QS1 m.C V T 3 T 2  V Compression Ratio r  1 Process 3–4: Constant pressure Heat V2 addition Process: The fuel is burnt at V constant pressure process. Heat addition in Cut off ratio ρ  3 C this process is given by V2

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission QS2 m.C P T 4 T 3  Substituting the values of temperatures in Process 4–5 : Isentropic Expansion Process: the term of compression ratio, pressure In this process the heat transfer is zero. ratio and cut off ratio, we get dQ = 0 Efficiency of Dual cycle γ Process 5–1: Constant Volume Heat 1 rpcρ1 η1  --- (9.5) rejection Process: The heat is rejected in DUAL γ1 r (rp  1) γrp ρ c  1  constant volume process. Heat rejection in

this process is given by 9.9 COMPARISON OF OTTO, DIESEL, DUAL QR m.C v T 5 T 1  CYCLE P-V and T-S diagram of Dual cycle is shown in the fig. 1) Same comp. ratio and heat addition Otto cycle 1-2-3-4, Diesel cycle 1-2-3’-4’ and Dual cycle 1-2-2’-3”-4” are shown on P-V and T-S diagram. From the T-S diagram it is clear that heat input i.e. area of T-S curve along S- axis is same.

V Efficiency of the Dual cycle is given by Q η1 R QS m.c T T  But the heat rejected in Otto cycle is 1 v 5 1 minimum and in Diesel cycle is m.cp T 4 T 3   m.c v  T 3  T 2  maximum So TT  1 51 γ T4 T 3    T 3  T 2  V So the efficiency of Otto cycle is Compression Ratio r  1 V maximum and the efficiency of diesel 2 cycle will be minimum in this case. P Pressure ratio rp  3      P otto Dual Deisel 2 V4 2) For Same compression Ratio and Cut off ratio ρC  V3 heat Rejection VVV Otto cycle 1-2-3-4, Diesel cycle 1-2-3’-4 Expansion ratio r  5 5 3 e VVV and Dual cycle 1-2-3”-4”-4 are shown 4 4 3 on P-V and T-S diagram. From the T-S V V r 1 3  diagram it is clear that heat rejection i.e. VV2 4ρ C area of T-S curve in process 4-1 along S-

r  c  re axis is same.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission V 1.95V ρ 3C  1.95 VV2C γ 1 ρ1C  η1 γ1  r  γ ρC  1  1 1.951.4  1 1  4  64.9% 20  1.4 1.95 1 

Q.2 4-Cylinder, 2-stroke IC engine has the following particulars: engine speed = 3000 rpm, bore = 120 mm, crank radius = 60 mm, mechanical efficiency = 90% and the engine But the heat supplied in Otto cycle is develops 75 BHP. Calculate the maximum and in Diesel cycle is swept volume and mean effective minimum So pressure (MEP). Q Qsdual Q S Deisel S otto Solution: Q and η1 R Mechanical efficiency, Q BrakePower bhp S  So the efficiency of Otto cycle is Indicated power Ihp maximum and the efficiency of diesel 75 0.9  , So, P = 83.33 hp cycle will be minimum in this case. P otto   Dual   Deisel We know that Engine Power, P = T ω Examples Here, P = 83.33 HP = 83.33 x 746 W

Q.1 A Diesel engine has a compression = 62166.67 W 2πN 2 π 3000 ratio of 20 and cut off takes place at    5% of stroke find air standard 60 60 efficiency γ 1.4 = 314.16 rad/s T =P/ ω = 62166.67/314.16 Solution: = 197.88 N.m Mean Effective Pressure (MEPorPmep) 2πN T MEP  C Vd Vs = K.(π/4).d2. L K = No. of the cylinder Stroke L = 2 x crank radius V1 r  20,V1  20V 2  20V 2 = 2 x 0.06 m V2 = 0.12 m V 19V 19V  S 2 C Vd  4   0.122  0.12  5.43  103 m 3 V3  0.05XVSC V 4 0.05X19V V Nc = No. of cycle per power stroke = CC 1 for a 2-stroke engine V3C 1.95V Therefore,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 2π 1 197.88 a 10 minutes test, the dynamometer MEP  0.050 43 reads 45kg and engine consumed 5 =228971.77 N/m2 kg of petrol of CV 45 MJ/kg. The = 228.97 KPa carburettor receives air at 29 0 C and 1 bar at the rate of 10kg/min. Calculate:- Q.3 Two engines are to operate on Otto 1) Brake power and Diesel cycles with the 2) Brake mean effective pressure following data: Maximum 3) Brake specific fuel consumption temperature 1400 K, exhaust 4) Brake specific air consumption temperature 700 K. State of air at 5) Brake thermal efficiency the beginning of compression 0.1 6) Air full ratio MPa, 300 K. Estimate the Solution: compression ratios, the maximum pressures, efficiencies, and rate of 6 cylinders, 4 strokes work outputs (for 1 kg/min of air) Bore =10cm =0.1m of the respective cycles. Stroke=0.12m Solution: n=6, N=4800rpm Arm of dynamometer=55cm= 0.55m T = 1400 K 3 Dynamometer reads 45 Kg in 10min T4 = 700 K P1= 100 KPa 1. Brake power T1 = 300 K B.P. = T×ω = 2πNT/60 RT 0.287 300 T= F r = 45× 9.81 × 0.55 v 1 1 = 242.79 Nm. P1 100 = 0.861 m3/kg B.P. = 2×3.14×4800×242.79/60 γ1 = 122.039KW γ1 TPγ V 2. Brake mean effective pressure 33  4 TPV pm l a N 4 4 3 Brake power = n γ1 60 2 1400 V  1  2 122.039 10  10  10  60  2 700 V Pm   2  π γ1 0.12  0.1  0.1  4800  6 TV 4 21  2   5.39bar T12 V  3. Brake specific fuel consumption. T 2  300  600K 2 mf V 1 Bsfc  rc1  2 γ1  5.657 B.P. V2 5 mf  60 Work done W= Q1–Q2=Cv (T3–T2)– 10 Cv (T4 – T1) = 30kg/hr =0.718[(1400–600)–(700– 30 300)]KJ/Kg  = 287.2 KJ/Kg. 122.039 = 0.245kg/KWh Q.4 A six cylinder ,four stroke spark ignition engine of 10cm 12cm 4. Brake specific air consumption ma (bore stroke) with a compression bsac  ratio of 6 is tested at 4800 rpm-on a bp dynamometer of arm 55 cm. during

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission kg 1) volumetric efficiency m =10×60 a h altual volume innaled / cycle v  600 sweet volume. =  4.92kg / KWh  122.039 1.00729 105 Pair 1.1738kg / m3 5. Brake thermal efficiency 287 299 brake power Now in orifice air enter per ηbth  m C.V. second, 122.039   3 2.54% Cd 2 Psair 5 45  10  10  10 π 0.6   0.0382  (2  1422  1.173) 10 60 4 6. Air fuel ratio = 0.03932 kg/s A   10  Sweet volume     20% F   0.5  2600  200066 10  2 60 Q.5 A sharp edged circular orifice of = 0.0433 m3/rev diameter 3.8 cm and co-efficient of Actual mass enter/sec discharge as 0.6 is used to measure = 0.03932 kg/s air consumption of a four stroke Actual volume enter/sec petrol engine. The pressure drop 0.03932  through the orifice is 145 cm of 1.1738 water and barometer reads 75.5cm = 0.3349 m3/s of Hg. The compression ratio of the 0.3349 engine is 6 and the piston ηv  77.36% 0.0433 displacement volume is 2000cm3. 2) air fuel ratio The temp of air is taken to be 260C m = 0.03932 kg/s at 2600 rpm, the engine brake a 0.14 power recorded is 29.5 KW. The fuel Sfc= 0.14 kg/min  kg / s consumption is 0.14 kg/min and the 60 calorific value of fuel used is 43960 =2.33×10-3 kg/s KJ/kg. Calculate following: A 0.03932 16.85 (1) Volumetric efficiency F 2.33 163 (2) Air – fuel radio 3) Brake mean effective Pressure (3) Brake mean effective pressure PmlaN (4) Brake thermal efficiency B.P.   60 2 Solution: 3 Diameter of orifice d =3.8 cm 60 2  29.5  10 6 6.8bar Coefficient of discharge=0.6 2600 2000 10 Pressure drop =145 mm of water 4) brake thermal efficiently BP = 1,422.45 Pa η  bth mf CV Barometer reading =75.5 cm of hg 29.5 103 = 1,00,729.08 Pa  2.33 1033  43960  10 r = 6, Vs =2000 cm3  28.86% Temperature of air =299k

B.P.=29.5 kw at 2600 rpm Q.6 The following data are know for a F.C.=0.14 kg/min four cylinder stroke petrol engine: CV=43960 kj/kg cylinder dimension = 11 cm bore, 13

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission cm stroke engine speed = 2250 rpm, C7 h 16 11O 2  7CO2  8H 2 O brake power=50kw, friction power If 10% more air is supplied =15 KW 79 fuel consumption rate =10.5 kg/h, C7 H 16  12.1O2  [12.1x( )]N2  calorific value of fuel= 50,000 KJ/kg, 21 air inhalation rate = 300 kg/h, 7CO2 8H 2 O  1.1O2  45.52N2 ambient condition =150C, 1.03 bar. Mass of O2 for 100 kg of fuel = 12.1 x Estimate: 32 = 387.3 kg of O2 1. Brake mean effective pressure Mass of air for 100 kg of fuel = 2. Brake thermal efficiency 1683.4 kg of air 3. Mechanical efficiency Mass of air required for 1 kg of fuel =16.82 kg Solution: Dry exhaust gas Analysis on volume 1. Brake mean effective pressure basis PLANK B.P  Moles % 60 2 Co2 7 11.36 % B.P. 60 2 O2 1.1 1.78 % P  LANK H2O 8 12.9 % N2 45.52 73.87 % 50 60 2  2 2250 4 π / 4  0.11  0.13 Q.8 Find the percentage increase in the = 5.396 bar efficiency of a Diesel engine having

2. Brake thermal efficiency a compression ratio of 16 and cut- brakepower off ratio (r) is 10% of the swept ηBTh  volume if C by 2% take C = o.717 mf CV V v KJ/Kg.K and γ=1,4 50   3 4.29% Solution: 10.5 / 3600  X50000 The efficiency of Diesel engine

1[ργ  1] 3. Mechanical Efficiency η1 BP = 50 KW r[γ1 γ(ρ 1)] FP = 15 KW Where IP = 50 + 15 = 65KW v1 BP 50 r 16 Mechanical efficiency  v2 IP 65 v3 cp ρ = cut of ratio ,   = 76.92% v2 cv

Q.7 A liquid fuel C7 h16 is burned with 10% more air than the stoichiometric air assuming complete combustion calculate: 1. The mass of air supplied per kg of fuel and 2. The volumetric analysis of the dry products of combustion. Cv = 0.717 KJ/kg k, γ = 1.4 Assume air contains 21 % O2 by Cp = γ× Cv = 1.0038 kj/kg k volume. R = Cp-Cv = 1.0038 – 0.717

Solution: = 0.2868 kj/kg k V3 – V2 = 0.1(V1 – V2) Equation for stoichiometric dividing the equation by V2, we get composition:

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission V3  V1  11  0.10   V2  V2  (ρ-1)=0.10(r - 1) (ρ- 1)=0.10(16 - 1) ρ=2.5 1[ργ  1] η1 r[γ1 γ(ρ 1)] 1 [2.51.4  1] 1 161.4 1 [1.4 (2.5 1)] = 0.5905 = 59.05 % When cv is decreases by 2% CV’ = 0.70266 R = 0.2868 = cp’-cv’ Cp’ = 0.98946 Cp' ’  1.4081 Cv' 1[ργ  1] η' 1 r[γ' 1 γ'(ρ 1)] 1 [2.51.408  1] 1 161.408 1 [1.408 (2.5 1)] = 0.5975 = 59.75 % Increase in efficiency of diesel cycle η' η 0.5975 0.5905  100  η 0.5905 = 1.185 %

GATE QUESTIONS

Topics

1. THERMODYNAMIC SYSTEM AND PROCESSES

2. FIRST LAW, HEAT, WORK AND ENERGY

3. SECOND LAW, CRANOT CYCLE AND ENROPY

4. AVALILABILITY AND IRREVERSIBILITY

5. PURE SUBSTANCES 6. POWER SYSTEM (RANKINE, BRAYTON, ETC.)

7. IC ENGINE

8. REFRIGERATION AND AIR CONDITIONING

Q.1 The following four figures have been Q.3 Match items from groups I, II, III, IV drawn to represent a fictitious and V. thermodynamic cycle, on the p -v and T -s planes.

a) F-G-J-K-M b) E-G-I-K-M E-G-I-K-N F-H-I-K-N c) F-H-J-L-N d) E-G-J-K-N E-H-I-L-M F-H-J-K-M [GATE–2006]

Common Data for Q.4 and Q.5 According to the first law of A thermodynamic cycle with an ideal gas as thermodynamics, equal areas are working fluid is shown below. enclosed by a) Figures 1 and 2 b) Figures 1 and 3 c) Figures 1 and 4 d) Figures 2 and 3 [GATE–2005]

Q.2 A reversible thermodynamic cycle containing only three processes and producing work is to be constructed. The constraints are Q.4 The above cycle is represented on T i) There must be one isothermal -s plane by process, a) b) ii) There must be one isentropic process, iii) The maximum and minimum cycle pressures and the clearance volume are fixed, and iv) Polytropic processes are not c) d) allowed. Then the number of possible cycles are a) 1 b) 2 c) 3 d) 4 [GATE–2007] [GATE–2005]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.5 If the specific heats of the working Q.10 Which of the following statements fluid are constant and the value of are TRUE with respect to heat and specific heat ratio is 1.4, the thermal work? efficiency (%) of the cycle is i) They are boundary phenomena a) 21 b) 40.9 ii) They are exact differentials c) 42.6 d) 59.7 iii) They are path functions [GATE–2007] a) both (i) and (ii) b) both (i) and (iii) Q.6 If a closed system is undergoing an c) both (ii) and (iii) irreversible process, the entropy of d) only (iii) the system [GATE -2016(1)] a) must increase Q.11 A mass m of a perfect gas at pressure p1 b) always remains constant and volume V1 undergoes an isothermal c) Must decrease process. The final pressure is p2 and d) can increase, decrease of remain volume is V2. The work done on the constant system is considered positive. If R is the [ME-GATE –2009] gas constant and T is the temperature, then the work done in the process is Q.7 Heat and work are V p a) p V ln 2 b) −p V ln 1 a) intensive properties 11 11 V1 p2 b) extensive properties c) point functions V p d) path functions c) RT ln 2 d) −mRT ln 2 [ME-GATE –2012] V1 p1

Q.8 A certain amount of an ideal gas is [ME-GATE -2017(1)] The volume and temperature of air initially at a pressure p1 and Q.12. (assumed to be an ideal gas) in a temperature T1. First, it undergoes a constant pressure process 1-2 such closed vessel is 2.87 m3 and 300K, respectively. The gauge pressure that T2 = 3T1/4. Then, it undergoes a constant volume process 2-3 such indicated by a manometer fitted to the wall of the vessel is 0.5bar. If the that T3 =T1/2. The ratio of the final volume to the initial volume of the gas constant of air is R = 287 J/kg. K ideal gas is and the atmospheric pressure is 1 a) 0.25 b) 0.75 bar, the mass of air (in kg) in the c ) 1.0 d) 1.5 vessel is [ME-GATE -2014(3)] a) 1.67 b) 3.33 c) 5. d) 6.66 Q.9 Two identical metal blocks L and M [ME-GATE -2018(2)] (specific heat = 0.4kJ/ kg. K), each Q.13 A n engine operates on the reversible having a mass of 5kg, are initially at cycle as shown in the figure. The 313K. A reversible refrigerator work output from the engine (in extracts heat from block L and kJ/cycle) is ______(correct to two rejects heat to block M until the decimal places). temperature of block L reaches [ME-GATE -2018(2)] 293K. The final temperature (in K) of block M is _____. [ME-GATE -2014(4)]

Q.14 Air is held inside a non-insulated cylinder using a piston (mass M=25 kg and area A=100 cm2) and stoppers (of negligible area), as shown in the figure. The initial

pressure Pi and temperature Ti of air inside the cylinder are 200 kPa and 0 400 C , respectively. The ambient

pressure P and temperature T are ∞ ∞ 0 100 kPa and 27 C , respectively. The temperature of the air inside the cylinder ( 0 C ) at which the piston will begin to move is ______(correct to two decimal places).

[ME-GATE-2018 (2)]

ANSWER KEY:

1 2 3 4 5 6 7 8 9 10 11 12 13

(a) (d) (d) (c) (a) (d) (d) (b) (b) (b) (c) 62.5 14 146

Q.1 (a) Given P-v diagram is clockwise. So From the first law of T-s diagram must be clockwise. thermodynamics for a cyclic process U = 0 Q.5 (a) And ∮∮δQ= δ W ∆ The symbol Q , which is called the cyclic integral of the heat transfer represents the∮ δ heat transfer during the cycle and W the cyclic integral of the work represents the work during the cycle.∮ δ We easily see that figure 1 and 2 satisfy the first law of thermo- This cycle shows the Lenoir cycle. dynamics, both the figure are in For Lenoir cycle efficiency is given by same direction (clockwise) and 1 satisfies the relation. r1γ − Q = W η= 1γ − p L − r1p  Q.2 (d)∮ δ ∮ δ Two cycles having constant volume p2 400 Where, r4p = = = process and two cycles having p1 100 constant pressure process can be

formed. Cp Therefore a total of four cycles can And γ= = 1.4(Given) C be formed. v So, 1  Q.3 (d) (4)1.4 − 1 η=−=− 1 1.4 1 0.789 L 41− Q.4 (c)  In the given p - v diagram, three = 0.211 processes are occurred. η= 21.1%; 21% i) Constant pressure (Process 1 – 2) L ii) Constant Volume (Process 2 – 3) iii) Adiabatic (Process 3 – 1) Q.6 (d) We know that, Constant pressure If a closed system is undergoing an and constant volume lines are irreversible process, the entropy of inclined curves in the T-s curve, and the system can increase, decrease or adiabatic process is drawn by a remain constant. vertical line on a T-s curve. Q.7 (d) Q. 8 (b) For constant pressure process (1to VV 2): 12= TT12

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  VP21 T21 3V = VV21= = VP12 T41 ∴ For⇒ constant volume process (2 to P1 3) w= P11 V ln P2 V ∴ Final volume3 V2 3 If the work done on the system is + ive, ∴ = = = = 0.75 Initial volume V11 V 4 then

P Q.9 (333 to 335) 1 w= − P11 V ln P2 Q.10 (b) Q.11 Sol: (b) Q.12 (c) PP= + P w= ∫ Pdv abs gauge atm

∵For ideal gas PV = mRT = const (c ) =+=50 100 150kPa

C By ideal gas eqn PV = C P = V Pabs V= mRT ⇒ V2 C 150× 2.87 W= dv = ∫ V m V1 .287× 300

= c[ln v]V2 m= 5kg V1 Q.13 (62.5) V w= c ln 2 For a reversible process, V1 PdV work output = ∫

Hence, for the given reversible cycle, Work Output = Area enclosed by the triangle

1 =0.5 250 = 62.5 kJ/cycle 2

Q.14 (146.03) V2 w= P11 V ln V1 ⇒ When work is done on the system (compression), VV< and PP> 21 21 For isothermal process

PV11= PV 2 2 = c

mg PP= + inside amb A 25 9.81 =100 +  10−3 100 10−4 2 P2 = 124.525 kN/m As mass and volume remainsconstant PP 12= TT12 200 124.525 = 673 T2 o T2 = 419.02 K = 146.03 C

Q.1 A small steam whistle (perfectly compression being 5000 kJ. During insulated and doing no shaft work) the process, heat interaction of 2000 causes a drop of 0.8 kJ/kg in the kJ causes the surroundings to be enthalpy of steam from entry to exit. heated. The changes in internal If the kinetic energy of the steam at energy of the gas during the process is entry is negligible, the velocity of the a) -7000 kJ b) -3000 kJ steam at exit is c) +3000 kJ d) +7000 kJ a) 4 m/s b) 40 m/s [GATE–2004] c) 80 m/s d) 120 m/s Common Data For Q.6 and Q.7 [GATE–2001] A football was inflated to a gauge pressure of 1 bar when the ambient temperature Q.2 A 2 kW, 40 liters water heater is was 15°C. When the game started next day, switched on for 20 minutes. The the air temperature at the stadium was 5°C. heat capacity C for water is 4.2 p Assume that the volume of the football kJ/kgK. Assuming all the electrical remains constant at 2500 cm3. energy has gone into heating the water, increase of the water Q.6 The amount of heat lost by the air in temperature in degree centigrade is the football and the gauge pressure a) 2.7 b) 4.0 of air in the football at the stadium c) 14.3 d) 25.25 respectively equal [GATE–2003] a) 30.6 J, 1.94 bar b) 21.8 J, 0.93 bar c) 61.1 J, 1.94 bar d) 43.7 J, 0.93 bar Common Data for Q.3 and 4 [GATE–2006] Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the Q.7 Gauge pressure of air to which the initial condition of 2 bar, 298 K and 1 m3. In ball must have been originally a particular process, the gas slowly inflated so that it would be equal 1 expands under isothermal condition, until bar gauge at the stadium is the volume becomes 2m3. Heat exchange a) 2.23 bar b) 1.94 bar occurs with the atmosphere at 298 K c) 1.07 bar d) 1.00 bars during this process. [GATE–2006] Q.8 Which of the following relationships Q.3 The work interaction for the is valid only for reversible processes Nitrogen gas is undergone by a closed system of a) 200 kJ b) 138.6 kJ simple compressible substance? c) 2 kJ d) -200 kJ (neglect changes in kinetic and [GATE–2003] potential energy) Q.4) The entropy changes for the a) Q = dU + W b) Tds = dU+pdv Universe during the process in kJ/K is c) Tds = dU + W d) Q = dU + pdv a) 0.4652 b) 0.0067 [GATE–2007] c) 0 d) -0.6711 [GATE–2003] Q.9 A gas expands in a frictionless piston-cylinder arrangement. The Q.5 A gas contained in a cylinder is expansion process is very slow, and compressed, the work required for is resisted by an ambient pressure

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission of 100 kPa. During the expansion a) is greater than 3500C process, the pressure of the system b) is less than 3500C (gas) remains constant at 300 kPa. c) is equal to 3500C The change in volume of the gas is d) may be greater than, less than, or 0.01m3. The maximum amount of equal to, 3500C depending on the work that could be utilized from the volume of the tank above process is [GATE–2008] a) 0 kJ b) 1 kJ c) 2 kJ d) 3 kJ. Q.12 In a steady state flow process taking [GATE–2008] place in a device with a single inlet and a single outlet, the work done Q.10 A balloon containing an ideal gas is per unit mass flow rate is given by initially kept in an evacuated and outlet W= vdp where v is the insulated room. The balloon inlet ruptures and the gas fills up the specific volume and p is the entire room. Which one of the pressure.− ∫ The expression for W following statements is TRUE at the given above end of above process? a) is valid only if the process is both a) The internal energy of the gas reversible and adiabatic decreases from its initial value, b) is valid only if the process is both but the enthalpy remains constant reversible and isothermal b) The internal energy of the gas c) is valid for any reversible increases from its initial value, process but the enthalpy remains constant d) is incorrect; it must be c) Both internal energy & enthalpy outlet

of the gas remain constant W= ∫ pdv d) Both internal energy and inlet [GATE–2008] enthalpy of the gas increase

[GATE–2008] Q.13 A frictionless piston-cylinder device Q.11 A rigid, insulated tank is initially contains a gas initially at 0.8MPa evacuated. The tank is connected and 0.015 m3. It expands quasi- with a supply line through which air statically at constant temperature to (assumed to be ideal gas with a final volume of 0.030 m3. The constant specific heats) passes at work output (in kJ) during this 1MPa, 3500C. A valve connected process will be with the supply line is opened and a) 8.32 b) 12.00 the tank is charged with air until the c) 554.67 d) 8320.00 final pressure inside the tank [GATE–2009] reaches 1MPa. The final temperature Q.14 A compressor undergoes a inside the tank. reversible, steady flow process. The gas at inlet and outlet of the compressor is designated as state 1 and state 2 respectively. Potential and kinetic energy changes are to be ignored. The following notations are used: v = Specific volume and p = pressure of the gas. The specific work

required to be supplied to the

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission compressor for this gas compression process is 2 2 a) ∫pdv b) ∫vdp 1 1 c) d) − − v1() p 21 -p pv21() v 2 [GATE–2009] Q.17 The density of air in kg/m3 at the

nozzle exit is Common Data for Q.15 and Q.16 a) 0.560 b) 0.600 The inlet and the outlet conditions of steam c) 0.727 d) 0.800 for an adiabatic steam turbine are as [GATE–2011] indicated in the figure. The notations are as usually followed. Q.18 The mass flow rate of air through the nozzle in kg/s is a) 1.30 b) 1.77 c) 1.85 d) 2.06 [GATE–2011]

Q.19 The contents of a well-insulated tank are heated by a resistor of 23 in which 10 A current is flowing. Consider the tank along with itsΩ Q.15 If mass rate of steam through the contents as a thermodynamic turbine is 20 kg/s, the power output system. The work done by the of the turbine (in MW) is system and the heat transfer to the a) 12.157 b) 12.941 system are positive. The rates of c) 168.001 d) 168.785 heat (Q), work (W) and change in [GATE–2009] internal energy ( U) during the

Q.16 Assume the above turbine to be part process in kW are of a simple Rankine cycle. The a) Q = 0, W = -2.3, ∆U = +2.3 density of water at the inlet to the b) Q = +2.3, W = 0, U = +2.3 pump is 1000 kg/m3. Ignoring c) Q = -2.3, W = 0, ∆U = -2.3 kinetic and potential energy effects, d) Q = 0, W = +2.3, ∆U = -2.3 the specific work (in kJ/kg) supplied ∆ [GATE–2011] ∆ to the pump is Common Data for Q.20 and Q.21 a) 0.293 b) 0.351 Air enters an adiabatic nozzle at 300 kPa, c) 2.930 d) 3.510 500 K with the velocity of 10 m/s. It leaves [GATE–2009] the nozzle at 100 kPa with a velocity of 180

2 Common Data For Q.17 and Q.18 m/s. The inlet area is 80 cm . The specific The temperature and pressure of air in a heat of air Cp is 1008 J/kgK.

large reservoir are 400 K and 3 bar Q.20 The exit temperature of the air is respectively. A converging-diverging nozzle a) 516 K b) 532 K 2 of exit area 0.005 m is fitted to the wall of c) 484 K d) 468 K the reservoir as shown in the figure. The [GATE–2012] static pressure of air at the exit section for isentropic flow through the nozzle is 50 Q.21 The exit area of the nozzle in cm2 is kPa. The characteristic gas constant and the a) 90.1 b) 56.3 ratio of specific heats of air are c) 4.4 d) 12.9 0.287kJ/kgK and 1.4 respectively [GATE–2012]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.22 For an ideal gas with constant values of specific heats, for Q.26 A mixture of ideal gases has the calculation of the specific enthalpy, following composition by mass a) it is sufficient to know only the temperature

b) both temperature and pressure If the Universal gas constant is 8314 are required to be known J/mol-K, the characteristic gas c) both temperature and volume constant of the mixture (in J/kg.K) is are required to be known ______. d) both temperature and mass are [GATE-2015 Set-3] required to be known [GATE-2015 Set-1] Q.27 An ideal gas undergoes a reversible Q.23 A well insulated rigid container of process in which the pressure varies volume 1m3 contains 1.0 kg of an linearly with volume. The conditions ideal gas [Cp = 1000J/kgK) and Cv = at the start (subscript 1) and at the 800J/kgK] at a pressure of 105 Pa. A end (subscript 2) of the process stirrer is rotated at constant rpm in with usual notation are: 1 = 100 3 the container for 1000 rotations and kPa, 1 = 0.2 m and 2 = 200 kPa, 2 the applied torque is 100 N-m. The = 0.1 m3 and the gas constant,𝑝𝑝 R = final temperature of the gas (in K) is 0.275𝑉𝑉 kJ/kg-K. The magnitude𝑝𝑝 of the𝑉𝑉 ______. work required for the process (in kJ) [GATE-2015 Set-1] is ______

Q.24 Work is done on adiabatic system Q.28 The internal energy of an ideal gas is due to which its velocity changes a function of from 10 m/s to 20 m/s, elevation a) temperature and pressure increases by 20m and temperature b) volume and pressure increases by 1 K. The mass of the c) entropy and pressure system is 10 kg, Cv=100 J/kgK and d) temperature only gravitational acceleration is 10 [GATE-2016 Set-2] m/s2. If there is no change in any Q.29 A piston-cylinder device initially other component of the energy of contains 0.4 m3 of air (to be treated the system, the magnitude of total as an ideal gas) at 100 kPa and 80 . work done (in kJ) on the system is The air is now isothermally _____. compressed to 0.1 m3. The work℃ [GATE-2015 Set-2] done during this process is ______kJ. (Take the sign convention such that Q.25 Steam enters a turbine at 30 bar, work done on the system is 300 (u = 2750 kJ/kg, h = 2993 negative) kJ/kg) and exits the turbine as [GATE-2016 Set-2] saturated℃ liquid at 15 kPa (u = 225 kJ/kg, h = 226 kJ/kg). Heat loss to Q.30 Steam at an initial enthalpy of 100 the surrounding is 50 kJ/kg of steam kJ/kg and inlet velocity of 100 m/s, flowing through the turbine. enters an insulated horizontal Neglecting changes in kinetic energy nozzle. It leaves the nozzle at 200 and potential energy, the work m/s. The exit enthalpy (in kJ/kg) is output of the turbine (in kJ/kg of ______steam) is______[GATE-2016 Set-3] [GATE-2015 Set-3]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.31 The molar specific heat at constant volume of an ideal gas is equal to 2.5 times the universal gas constant (8.314 J/mol.K). When the temperature increases by 100K, the change in molar specific enthalpy is ______J/mol [GATE-2017 Set-1]

Q.32 A frictionless circular piston of area 10−2 m2 and mass 100 kg sinks into a cylindrical container of the same area filled with water of density 1000 kg / m3 as shown in the figure. The container has a hole of area 10−3 m2 at the bottom that is open to the atmosphere. Assuming there is no leakage from the edges of the piston and considering water to be incompressible, the magnitude of the piston velocity (in m/s) at the instant shown is _____ (correct to three decimal places).

[GATE-2018 Set-2]

ANSWER KEY: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 (b) (c) (b) (a) (c) (d) (c) (d) (c) (c) (a) (c) (a) (b) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (a) (c) (c) (d) (a) (c) (d) (a) 4.5 2717 (d) 29 30 31 32 - - 29.09 1.456

EXPLANATIONS

Q.1 (b) Q.5 (c) CC22 W = -5000 kJ (Negative sign shows hh+=+12 1222 that work is done on the system) C0= Q = - 2000 kJ (Negative sign shows 1 that heat is rejected by the system) C2 hh−=2 From the first law of 122 thermodynamics, C2 Q = W + U 8 ×100. 3 = 2 2 So ∆U = ∆Q ∆W = 2000 C2 = 40m / s ( 5000) = kJ ∆ ∆ − ∆ − − Q.2 (c) − 3000 Heat Supplied Q.6 (d) H = Power × Time Given ( ) =2 × 1000 × 20 × 60 P1 = 1 bar Gauge =2400 × 1000 J (P1) But heat required in increasing the T1 =15 absolute = 1+1.013= 2.013 bar. water temperature T2=5 Vol. = Constant℃=288K = 2500 cm H= mCp T at constant℃=278K vol. (15 3) 2400×1000=1000×0.04 Q= mC dT ∆ v - ℃→5 ℃

We need to calculate the mass first, ×4.2×1000×∆T Q=Using m×718×(278 PV=mRT 288) ∆T=14.3℃ Q.3 (b) 56− For isothermal process, 2.013×× 10 2500 × 10 =×× m 287 288 3 V × 10 kg Work = 2 3 P11 V ln × 10− - V1 Qm = 6.088 -ve− sign indicates that 2 =×()()2 100 ××1 ln heatQ = lost 6.088 by the air×718 in the × ball) (278 288) 1 43.7 J ( T1 = P2 = ? T =288 Q.4 =(a) 138. 63 kJ 2 278

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Vol. = C (Heat) between system (room) and TP surrounding (atmosphere). 22= It means internal Energy dU = 0 and TP11 U = constant. 278 P2 = 2.013 × Now flow work pv must also remain 288 constant thus we may conclude that ⇒ during free expansion process pv i.e. = + Pabsol PP atm gauge product of pressure and specific =1.93 bar. (absolute) P= 0.93bar volume change in such a way that ⇒ gauge their product remains constant. ⇒ Q.7 (c) So, it is a constant internal energy We know that and constant enthalpy process. Since volume is constant PT Q.11 (a) 11= Given: p = 1 MPa, T = oC= PT 1 1 22 ∴ P = 1 bar (gauge) 2 For air = 1.4 350

We623 Kknow that final temperature Substituting∴ the values, we get (T ) inside𝛾𝛾 the tank is given by, P= 2.013 bars (absolute) 2 1 T = T = 1.4 × = .2 = = 1.07 bar (gauge) 2 1 599.2oC = 2.085 bars (absolute) T is greaterγ than 623oC 872 K Q.8 (d) 2

q = dU + pdv This equation Q.12 (c) 350 holds good for a closed system when

onlyδ pdv work is present. This is Q.13 (a) true only for a reversible (quasistatic) process.

Q.9 (c) Given: pa = 100 kPa, ps

Net pressure3 across the piston= 300 kPa, ∆v = 0.01 m p = ps pa = 100 = 200 kPa − 300 − P1 = 0. MPa V1 = 0.015 m V2 = 0.8 m3 Isothermal process3 Work output030 Maximum work that can be v2 utilized from the system is given by: W= P11 V ln  v W = p v = 200 × 0.01 = 2 kJ 1 3 0.030 =()0.8 × 10() 0.015 ln  Q.10 (c) ∆ 0.015 We know that enthalpy, =0. × 10 × 0.015 × ln(2) 3 Given that room is insulated. So 8 there is no interaction of Energy Q.14 (b)= 8.32 kJ

Q.16 (c)=12156.78 kW=12.157 MW

energy effects SFEENeglecting : Kinetic and potential h1 + q = h2 + wT (For Turbine) wT State 1 : Inlet wT = 600kJ/kg State 2: Outlet Specific⇒ 3200=2600+ work supplied to the pump energy change is⇒ are to be ignored. (Since the cycle works between the v=Potential Specific and volume Kinetic of the gas P= Pressure of the gas kPa) Specific work required to be pressure limits of 3 MPa and 70 supplied to the compressor for this Wp = dP = ( P1 P2) gas compression process is PP− 2 = 12� υ υ − w = 1 vdp ρ (Compressor is an open flow system) ()3000− 70 kN / m2 ∫ = 1000kg / m3 = 2.930kJ/kg Q.17 (c) ) at nozzle exit is 3 DensityPP of air in (kg/m 12= Q.15 (a) γγ ρρ12

1 γ WhereP ρ=density2 ρ21=  .ρ P1 1/1.4 5 Apply steady flow energy equation 50 3× 10 ρ2 = × [SFEE] 300 0.287×× 1000 400 VV22 ρ= 0.727kg / m3 h+++=+++12 gz q h gz w 2 122 12 2 q = 0; because it is an adiabatic Q.18 (d) steam turbine Given: 2 (100)2 2 , A2 = 0.005m , ×3 + + ×+ 3200 10 9.81 10 0 V =? 3 2 2 Forρ = isentropic0.787 kg/m expansion, (100)2 =⇒ 3 2600×10 + + 9.81 ×+ 6 W V2 = 2CTT− 2 p ()12 =×2 1.005 ×× 103 () 400 − 239.73 = ⇒ W=607839.24 J/kg Mass flow rate of steam through Mass flow rate at exit turbine= 607.84kJ/kg is 567.582 A2 m/sV2 =0.727×0.005× The power output of the turbine is 20kg/s. m = ρ 567.78 = 2.06 kg/s =20×607.84

Q.22 (a)

Q.23 (1283.4 to 1287.4)

Q.24 (4.5) Using SFEE 22 2 VV12 W= I R work is done on the Wm= − +−()() z12 zg +− h 1 h 2 system] 22 2 = −(10) ×[∵ 1022 20 W= 10 − +()() − 20 ×10 + 100 ×− 1 22 From− 1st law:23 - = − 2300W = −2.3kW  Q = 0 [∵ system is insulated] ∆Q = kW∆U + ∆W W = − 4.5 kJ ∆U + ∆W = 0 ∴ Work done on the system is 4.5 kJ. Q.20 (c)⇒ ∆U = −∆W = +2.3 Q.25 (2717) From energy balance for steady flow By Steady flow energy equation: system W = (h1 h2) q E = E 226) in out − 22   (Neglecting kinetic and potential VV12 W= (2993 − 50 = 2717 kJ/kg mh12+=  mh +  ...(i) energy changes) 22   As Equation (1) become Q.26 (274 to 276) VV22 CT+=12 CT + p122 p2 Q.27 (14.75 to 15.25) Given: Pressure varies linearly with VV22− 1022− 180 T=12 += T + 500 volume 21×× 2 Cp 2 1008 p = aV+b ------(A)

=−+16.02 500 at V1 = 0.2m , p1 = 100kPa ∴100 = 0.2a +3 b ----- (1)

at V2 = 0.1m , p2 = 200kPa Q.21 (d)= 483.98 ≃ 484 K 200∴ = 0.1 a + 3b ------(2) From Mass conservation From equation (1) & (2), we get min = mout VA VA 11= 2 2...(i) – 1000V V2 vv12 a = −1000 and b = 300 where v = specific volume of air = ∴Work equation = ∫pdV (A) ⇒ p = 300 RT V1 V2 P =∫ []300 − 1000V dV Therefore Eq. (1) becomes V1

PVA PVA V2 11 1= 2 2 2 V2 RT RT = 300V − 1000  12 2 PVAT××× V1 11 12 A2 = – 0.2] – 500 [0.12 – 0.22] PVT×× 2 21 Work = 15kJ (-ve sign implies 300××× 10 80 484 = work= 300[0.1 is done on the system) 100×× 180 500 −

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission CCpv−= R asked. ∵ in the question only magnitude is C= C += R 2.5R += R 3.5R pv Q.28 (d)∴ Work = 15kJ For ideal gas, Molar specific enthalpy change ( (∆ H) Q.29 (-55.6 to -55.4)

∆=H nCp dT

∆H ∆=H = C dT = 3.5RdT n p

=××3.5 8.314 100

Given:- = 2909.9J / mole P1 = 100 kPa Q.32 (1.456) V1 = 0.4 m 3 Given. Piston mass = 100 kg V2 = 0.1 m Work done3 in an isothermal process Pressure at point (1) is given by V 100 10 = 2 = = 52 P11 V ln  P1 −2 10 N/m V1 10  0.1 =100 ×0.4 ln 0.4 W = 55.45kJ

Q.30 (84 to 86)

Applying Bernoulli’s equation for By SFEE: 22 points (1) and (2): VV12 hh12+=+ [Neglecting 22 22 P1 V 1 PV 22 potential head, q = 0 & W = 0] + +=++ZZ12...... () 1 c.v γγ 22 ww22gg 1 100− 200 = +  h2 100 3 2 10 where P2 = 0 & Z2 = 0 (datum line) h2 = /kg From continuity equation: Q.31 (29.09) 85kJ A1 V1 = A2 V2 (discharge from hole=

Given that Cv = 2.5R volume swept by the piston per unit time)

From eq. (1) and (2)

2 5 2 10 V ()10V1 +1 +=0.5 103  10 2 10 2  10 99V2 or 10.5 = 1 20

⇒=V1 1.456 m/s

Q.1 A cyclic heat engine does 50 kJ of a) 12.50 b) 14.29 work per cycle. If the efficiency of c) 33.33 d) 57.14 the heat engine is 75%, the heat [GATE–2007] rejected per cycle is A cyclic device operates between 2 1 Q.5 a) 16 kJ b) 33 kJ three thermal reservoirs, as shown 3 3 in the figure. Heat is transferred 1 2 c) 37 kJ d) 66 kJ to/from the cycle device. It is 2 3 assumed that heat transfer between [GATE–2001] each thermal reservoir and the cyclic device takes place across Q.2 A Carnot cycle is having an efficiency of 0.75. If the temperature negligible temperature difference. of the high temperature reservoir is Interactions between the cyclic device and the respective thermal 7270C, what is the temperature of low temperature reservoir? reservoirs that are shown in the figure are all in the form of heat a) 230C b) -230C transfer. c) 00C d) 2500C [GATE–2002] Q.3 A solar collector receiving solar radiation at the rate of 0.6 kW/m2 transforms it to the internal energy of a fluid at an overall efficiency of 50%. The fluid heated to 350 K is The cyclic device can be used to run a heat engine which a) a reversible heat engine rejects heat at 315 K. If the heat b) a reversible heat pump or a engine is to deliver 2.5 kW power, reversible refrigerator the minimum area of the solar c) an irreversible heat engine collector required would be d) an irreversible heat pump or an irreversible refrigerator a) 83.33 m2 b) 16.66 m2 [GATE–2008] c) 39.68 m2 d) 79.36 m2 [GATE–2004] Q.6 An irreversible heat engine extracts heat from a high temperature Q.4 A heat transformer is device that source at a rate of 100 kW and transfers a part of the heat, supplied rejects heat to a sink at a rate of 50 to it at an intermediate temperature, kW. The entire work output of the to a high temperature reservoir heat engine is used to drive a while rejecting the remaining part reversible heat pump operating to a low temperature heat sink. In between a set of independent such a heat transformer, 100 kJ of isothermal heat reservoirs at 17°C heat is supplied at 350 K. The and 75°C. The rate (in kW) at which maximum amount of heat in kJ that the heat pump delivers heat to its can be transferred to 400 K, when high temperature sink is the rest is rejected to a heat sink at a) 50 b) 250 300 K is

[GATE–2009] u=CV T Q.7 One kilogram of water at room temperature is brought into contact Q.9 If the air has to flow from station P with a high temperature thermal to station Q, the maximum possible reservoir. The entropy change of the value of pressure in kPa at station Q universe is is close to a) equal to entropy change of the a) 50 b) 87 reservoir c) 128 d) 150 [GATE–2011] b) equal to entropy change of water c) equal to zero Q.10 If the pressure at station Q is 50 kPa, d) always positive the change in entropy ()SSQP− in [GATE–2010] kJ/kgK is Q.8 Consider the following two a) -0.155 b) 0 processes ; c) 0.160 d) 0.355 a) A heat source at 1200 K loses [GATE–2011]

2500 kJ of heat to a sink at 800 K Q.11 An ideal gas of mass m and b) A heat source at 800 K loses temperature T1 undergoes a 2000 kJ of heat to a sink at 500 K reversible isothermal process from Which of the following statements is an initial pressure P1 to final true ? pressure P2. The heat loss during the a) Process I is more irreversible process is Q. The entropy change s than Process II of the gas is b) Process II is more irreversible P P ∆ than Process I a) mRln2 b) mRln1 c) Irreversibility associated in both P1 P2 the processes are equal P2 Q d) Both the processes are reversible c) mRln− d) zero PT [GATE–2010] 11 [GATE–2013]

Common Data for Q.9 and Q.10 Q.12 Which one of the following pairs of In an experimental set up, air flows equations describes an irreversible between two stations P and Q adiabatically. heat engine? The direction of flow depends on the δQ pressure and temperature conditions a) ∮∮δQ>0and <0 maintained at P and Q. The conditions at T δQ station P are 150 kPa and 350 K. The b) ∮∮δQ<< 0and 0 temperature at station Q is 300 K. T The following are the properties and δQ c) ∮∮δQ>> 0and 0 relations pertaining to air: T Specific heat at constant pressure, CP δQ d) ∮∮δQ<> 0and 0 = 1.005 kJ/kgK; T Specific heat at constant volume, CV [GATE-2014 (3)] = 0.718 kJ/kgK; Characteristic gas constant, Q.13 A source at a temperature of 500 K R=0.287 kJ/kgK provides 1000 kJ of heat. The of Enthalpy, temperature environment is 27 . The maximum useful work (in kJ) h = CT P ℃

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission that can be obtained from the heat Q.18 One kg of air (R=287 J/kgK) source is ______. undergoes an irreversible process [GATE-2014 (3)] between equilibrium state1 (20 , Q.14 A reversible heat engine receives 2 0.9 m3) and equilibrium state2 kJ of heat from a reservoir at 1000 K (20 , 0.6 m3). The change ℃in and a certain amount of heat from a entropy s2 – s1 (in J/kg.K) is ______. reservoir at 800 K. It rejects 1 kJ of ℃ [GATE-2015 (2)] heat to a reservoir at 400 K. The net work output (in kJ) of the cycle is Q.19 The heat removal rate from a a) 0.8 b) 1.0 refrigerated space and the power c) 1.4 d) 2.0 input to the compressor are 7.2 kW [GATE-2014 (1)] and 1.8 kW, respectively. The coefficient of performance (COP) of Q.15 An amount of 100 kW of heat is the refrigerator is ______transferred through a wall in steady [GATE-2016 (2)] state. One side of the wall is maintained at 127 and the other Q.20 A reversible cycle receives 40 kJ of side at 27 . The entropy generated heat from one heat source at a (in W/K) due to the℃ heat transfer temperature of 127 and 37 kJ through the℃ wall is ____. from another heat source at 97 . [GATE-2014 (3)] The heat rejected (in ℃kJ) to the heat sink at 47 is ______℃ Q.16 A closed system contains 10 kg of [GATE-2016 (2)] saturated liquid ammonia at 10 . Q.21 A heat pump℃ absorbs 10 kW of heat Heat addition required to convert from outside environment at 250 K the entire liquid into saturated℃ while absorbing 15 kW of work. It vapour at a constant pressure is delivers the heat to a room that 16.2 MJ. If the entropy of the must be kept warm at 300K. The saturated liquid is 0.88 kJ/kg.K, the Coefficient of Performance (COP) of entropy (in kJ/kg. K) of saturated the heat pump is ______. vapor is ______[GATE-2017 (1)] [GATE-2014 Set-4] Q.22 One kg of an ideal gas (gas constant, Q.17 A Carnot engine (CE-1) works R = 400 J/kg.K; specific heat at between two temperature reservoirs constant volume, A and B, where TA = 900 K and TB = = cv 1000J/kg.K at 1 bar, and 300 K 500 K. A second Carnot engine (CE- is contained in a sealed rigid 2) works between temperature cylinder. During an adiabatic reservoirs B and C, where T = 300 C process, 100kJ of work is done on K. In each cycle of CE-1 and CE-2, all the system by a stirrer. The increase the heat rejected by CE-1 to in entropy of the system is ______reservoir B is used by CE-2. For one J/K. cycle of operation, if the net Q absorbed by CE-1 from reservoir A [GATE-2017 (1)] is 150 MJ, the net heat rejected to A n ideal gas undergoes a process reservoir C by CE-2(in MJ) is Q.23 from state 1 ______. T= 300 K,p = 100 kP )a( to state 2 [GATE-2015 (1)] 11

T22= 600 K,p = 500 kP )a( . The

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission specific heats of the ideal gas are : cp [GATE-2018 Set-2] = 1 kJ/kg-K and cv = 0.7 kJ/kg-K. The change in specific entropy of the ideal gas from state 1 to state 2 (in kJ/kg-K) is ______(correct to two decimal places).

[GATE-2018 Set-1]

Q.24 Steam flows through a nozzle at a mass flow rate of m = 0.1kg / s . with a heat loss of 5 kW. The enthalpies

at inlet and exit are 2500kJ/kg and 2350 kJ/kg, respectively. Assuming

negligible velocity at inlet (C01 ≈ ),

the velocity (C2 ) of steam (in m/s) at the nozzle exit is ______(correct to two decimal places).

[GATE-2018 Set-1]

Q.25 For an ideal gas with constant properties undergoing a quasi-static process, which one of the following represents the change of entropy ( ∆s ) from state 1 to 2?

TP  ∆−22 a) s = CP ln R ln  TP11 

TV  b)∆− s = C ln22 C ln VP  TV11 

  TP22 c)∆− s = CPV ln C ln  TP11 

TV21  d)∆+ s = CV ln R ln  TV12 

ANSWER KEY:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (b) (a) (d) (a) (c) (d) (b) (b) (c) (b) (a) 400 (c)

15 16 17 18 19 20 21 22 23 24 25 83.33 6.6 50 -16.36 4 64 1.67 287 0.21 447 (a)

Q.1 (a) Q.4 (d) W Given: T1 = 400K, T2 = 300 K, = η T = 350 K, Q = 100kJ QS Q Heat transferred to the source 50 1 0.75 = by the transformer Q S Q → Heat transferred to the sink = 2 QS 66.67 by the transformer → QR =∴ 66.67 − 50 =16.67kJ 2 =16 kJ 3

Q.2 (b) T η1= − L TH Applying energy balance on the TL 0.75 1−= system 1000 QQ=12 + Q TL = 0.25 =−=− ...(i) 1000 Q21 Q Q 100 Q 1 Apply Claudius inequality on the TL = 250K system. T= − 23℃ L QQQ 100 Q Q =+=12 = 1 + 2 T T T 350 400 300 Q.3 (a) 12 Internal energy of fluid after Substitute the value of Q2 from absorbing solar radiation equation (i) =0.5×600=300W/m2 100Q 100− Q Q 100 Q =+1 11 =+− 1 315 350 400 300 400 300 300 η1engine = − 350 100 100 1 1 =+−Q W 1  = 350 300 400 300 Q Solving the above equation, we get 2500 Q =57.14 kJ 0.1 = 1 Q1 Therefore the maximum amount of ∴=Q 25000W heat that can be transferred at 400 K 1 is 57.14 kJ Let A be the minimum area of collector Q.5 (a) ∴=×Q A 300 1 A heat engine cycle is a 25000 ∴=A thermodynamic cycle in which there 300 is a net heat transfer from higher A= 83.33m2 temperature to a lower temperature device. So it is a Heat Engine.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Applying Clausius inequality on the ()()∆SS +∆ > 0 system for checking the reversibility system surrounding

of the cyclic device Q.8 (b) We know from the clauses dQ ∮ = 0 Inequality, T dQ QQQ If ∮ = 0, the cycle is reversible 12+−=3 0 T TTT123 dQ 333 < 0 the cycle is irreversible 100××× 10 50 10 60 10 ∮ +−=0 T 1000 500 300 and possible 100+100–200 = 0 For case (I) Here, the cyclic integral of is zero. dQ 2500 2500 ∮ = − This implies, it is a reversible heat T 1200 800 engine. 25 25 =−=−1.041kJ / kg Q.6 (c) 12 8 For case (II) dQ 2000 2000 ∮ = − T 800 500 20 20 =−=−1.5kJ / kg 85 dQ Since, ∮ for the process II is T more in magnitude than process I. Therefore, process (II) is more irreversible than process (I)

Q.9 (b) Maximum possible value of pressure We know that coefficient of has been asked in the question, so performance of a Heat pump for the we take it as a case of reversible given system is, flow. Q (COP) = 3 H.P. W TP22  Since it is a reversible heat pump Tdsds= =C dhp ln − vdp− R ln  T P T 11  (COP) = H The maximum possible value of H.P. TT− HL pressure at station Q can be found 348 Q = 3 as follows: 348− 290 50 TPQQ  348× 50 Cp ln −=R ln  0 ∴=Q = 300K TP PP 3 58 300 PQ 1.005 ln = 0.287ln  Q.7 (d) 350 150 We know that, PQ = 87.43kPa Entropy of universe is always

increases. Q.10 (c) ∆>S0universe

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission By using Tds=dh–vdP, after QQQ Or 12+−=3 simplifying we get: 0 TTT123 TPQQ  21Q2 (SQp−= S ) C P ln − R ln  Or +−=0 TPPP  1000 800 400

300  50 Or Q2 = 0.4kJ =1.005ln − 0.287ln  350  150 From 1st law of thermodynamics for = 0.160 a cycle: Q12+=+ Q WQ 3 Q.11 (b) Or 2+0.4 = W+1 Or W = 1.4 kJ Q.12 (a) Clausius inequality for irreversible Q. 15 (83.33) dQ heat engine, ∮ < 0. Heat content T of irreversible heat engine dQ > 0

Q.13 (400) Maximum work can be derived by

using a reversible engine between 2 dQ the two temperature limits. −= + SS2 1∫ S gen 1 T WQmax= sη max 100 00 TL 0 +∴−= Sgen =Q1s  − 400 300 T H 11 1 Or S=−= kW /K Qs gen Wmax= QT s − L 3 4 12 TH 1000 W ∴= == Given, = Sgen 83.33W / K Qs 1000kJ 12 K

TL = 273 += 27 300K Q. 16 (6.6) TH = 500K h 16.2× 103 1000 = fg = = 57.24 kJ / KS ∴W = 1000 −× 300 fg max 500 T 283 = 400 kJ 57.24 s=fg = 5.724 kJ / kgK 10 entropy Q.14 (c) Since specific entropy = ) mass

sssg= fg + f =5.724 + 0.88 = 6.604 kJ/ kg K

Q.17 (50)

From Clausius inequality dQ ∮ = 0 , for a reversible cycle T

Q2 Also, η11 = − Q1

2 =∴ 83 MQ.33 J

TL2 300 Also, η2 =−=−= 1 1 0.4 T2 500

Q3 η12 = − Q2 =∴ 5Q 0 MJ 2 Q = 15 + 10 = 25 kw

Q.18 (-116.36) desired effect Cop = V2 0.6 work input s21−= s R In = 287ln V1 0.9 25 5 = = 15 3 Q.19 (4=−116.368) J/kg/K Cop = 1.67 QL COPref = Wi/p Q.22. By first law 7.2 COP= = 4 ref 1.8 δQ = du +δ w

Q.20 (64) 0= Cvf [T −− T i ] 100

1×−[] Tf 300 = 100

Tf = 400k

Entropy change of an ideal gas is given by

TVff S21−= S C v ln + R ln TVii

= Q VVfi By Clausius inequality T δQ ∴−=S S C ln f ∮ = 0 21 v T Ti [For reversible cycle] 40 37 Q 400 ⇒+−=0 =1000ln  400 370 320 300 Q = 64kJ −= S21 S 287.6J / k

Q.23 (0.21)

State−= 1 : T11 300K, P = 100kPa

State−= 2 : T22 600K, P = 500kPa

cp= 1kJ/kg-K, c pv −= c R, cv = 0.7kJ/kg-K,

⇒cpv −=− c 1 0.7 = R R = 0.3 kJ/kg-K Changein specific entropy

TP22 s21−= s c p ln − R ln TP11 600 500 =−=1 ln 0.3ln 0.21 kj/kg-K 300 100

Q.24 (447.213) . . m= 0.1 kg/s, Q = 5kW() heat loss Applying SFEE . . 1 2 mch1+ 11 + gz += Q 2

..1 2 mch2++ 22 gz + wcv 2

. c1 = 0 and wcv = 0

zz12= ()assume .. ..1 mh+= Q mh + m c2 1 222 . ..1 2 ⇒m c2 = mh() 12 −+ h Q 2 1 ⇒0.1 c23 10− = 0.1() 2500 −− 2350 5 2 2

c2 = 447.213 m/s

Q.25 (a)

Q.1 Considering the relationship to equilibrium with a reference Tds = dU + Pdv between the entropy environment, is called (s), internal energy (u), pressure a)Entropy b) Enthalpy (P), temperature (T) and volume (v), c) Exergy d) Rothalpy which of the following statements is [GATE-2014 (ME) Set-1] correct ? a) It is applicable only for a Q.5 One side of a wall is maintained at reversible process 400 K and the other at 300 K. The b) For an irreversible process, rate of heat transfer through the Tds > du + Pdv wall is 1000 W and the surrounding c) It is valid only for an ideal gas temperature is 25 . Assuming no d) It is equivalent to Ist law, for a generation of heat within the wall, reversible process the irreversibility (in℃ W) due to heat [GATE–2003(ME)] transfer through the wall is ______[GATE-2015 (ME) Set-3] Q.2 A steel billet of 2000 kg mass is to be cooled from 1250 K to 450 K. The Q.6 One kg of an ideal gas (gas constant heat released during this process is R = 287 J/kg.K) undergoes an to be used as a source of energy. The irreversible process from state-1 (1 ambient temperature is 303 K and bar, 300 K) to state -2 (2 bar, 300 K). specific heat of steel is 0.5 kJ/ kgK. The change in specific entropy (s2 – The available energy of this billet is s1) of the gas (in J/kg. K) in the a) 490.44 MJ b) 30.95 MJ process is ______c) 10.35 MJ d) 0.10 MJ [GATE–2004(ME)] [GATE-2017 Set-2] Q.7` A calorically perfect gas (specific Q.3 The pressure, temperature and heat at constant pressure 1000 velocity of air flowing in a pipe are 5 J/kg.K) enters and leaves a gas bars, 500 K and 50 m/s, turbine with the same velocity. The respectively. The specific heats of temperatures of the gas at turbine air at constant pressure and at entry and exit are 1100 K and 400 K. constant volume are 1.005 kJ/ kg K respectively. The power produced is and 0.718 kJ/ kg K, respectively. 4.6 MW and heat escapes at the rate Neglect potential energy. If the of 300 kJ/s through the turbine pressure and temperature of the casing. The mass flow rate of the gas surrounding are 1 bar and 300 K, (in kg/s) through the turbine is. respectively, the available energy in kJ/kg of the air stream is a) 6.14 b) 7.00 a) 170 b) 187 c) 191 d) 213 c) 7.50 d) 8.00 [GATE–2013(ME)] [GATE-2017 Set-2] Q.4 The maximum theoretical work obtainable, when a system interacts

ANSWER KEY:

1 2 3 4 5 6 7 (d) (a) (b) (c) 248.23 -198.9 (b)

Q.1 (d) interacts to equilibrium with dead TdS=du+Pdv state or reference environment. This equation holds good for any process reversible or irreversible, Q.5 (248.23) undergone by a closed system, since By Gauy-Studola theorem:

it is a relation among properties I= T0 ( ∆ s) uni which are independent of both. 1000 1000 (∆= s) − uni 300 400 Q.2 (a) = 0.8333 W/K = ∆ Q mCρ T - = (25 + 273) 0.833 Q=2000×(0.5kJ/kg K)×(1250-450) =248.23 W = 800000 kJ Q.6 (-∴ 198.9)Irreversibility: I T 1250 ∆=S mCln1 = 2000 × 0.5 × ln T 450 Change in specific entropy of ideal 2 gas =1021.165 A.E(= Q − T ∆s) TP 0 S−= S C lnff − R ln =490439.67 k fi p TPii =490.44 MJ 2 Sfi−=− S 287ln  Q.3 (b) 1 Available energy 2 2 S−=− S 198.93J / kg.k C1 C0 fi = ()()h1− h 0 − Ts 01 −+ s 0  − 22 ….. (i) Q.7 (b)

TP11 s10−= s C p ln − Rln TP00 500   5 =1.005 ln  − 0.287ln  300   1 =0.05147

Available energy =∴1.005 eq.(i) (500 – 300) 300(0.05147) (50)2 +×10−3 2 By SFEE =186.8 kJ/ kg CC22 = 187 kJ/ kg h+12 + zg += q h + + zg + w 122 12 2 Q.4 (c) = (given) Exergy is also known as available CC12 energy and it is maximum theoretical work obtainable, when a system zz12= (assume)

WQ +  m = Cp2 [T− T 1 ]

4600+ 300 = 1×− [1100 400]

4900 m = = 7kg / sec 700

Q.1 Which combination of the following a) -Rln2 b) 0 statements is correct? c) Rln2 d) Rln4 P: A gas cools upon expansion only [GATE–2008] when its Joule-Thomson coefficient is positive in the Common Data For Q.4, 5 and 6 temperature range of expansion. In the figure shown, the system is a pure Q: For a system undergoing a substance kept in a piston-cylinder process, its entropy remains arrangement. The system is initially a two- constant only when the process phase mixture containing 1 kg of liquid and is reversible. 0.03 kg of vapour at a pressure of 100 kPa. R: The work done by closed system Initially, the piston rests on a set of stops, in an adiabatic is a point function. as shown in the figure. A pressure of 200 S: A liquid expands upon freezing kPa is required to exactly balance the when the slope of its fusion weight of the piston and the outside curve on pressure- Temperature atmospheric pressure. Heat transfer takes diagram is negative. place into the system until its volume a) R and S b) P and Q increases by 50%. Heat transfer to the c) Q, R and S d) P, Q and R system occurs in such a manner that the [GATE–2007] piston, when allowed to move, does so in a very slow (quasi-static/quasi-equilibrium) Q.2 Water has a critical specific volume process. The thermal reservoir from which of 0.003155m3/kg. A closed and heat is transferred to the system has a rigid steel tank of volume 0.025m3 temperature of 400°C. Average contains a mixture of water and temperature of the system boundary can be steam at 0.1MPa. The mass of the taken as 175°C. The heat transfer to the mixture is 10 kg. The tank is now system is 1 kJ, during which its entropy slowly heated. The liquid level increases by 10 J/K. inside the tank a) will rise b) will fall c) will remain constant d) may rise or fall depending on the amount of heat transferred [GATE–2007] Q.3 2 moles of oxygen are mixed adiabatically with another 2 moles

of oxygen in mixing chamber, so that Specific volume of liquid (vf) and vapour the final total pressure and (vg) phases, as well as values of saturation temperature of the mixture become temperatures, are given in the table below. same as those of the individual constituents at their initial states. The universal gas constant is given as R. The change in entropy due to mixing, per mole of oxygen, is given by

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.4 At the end of the process, which one J m3 of the following situations will be true? a) b) kg.K kg a) superheated vapour will be left m5 Pa in the system c) d) b) no vapour will be left in the kgs2 kg system [GATE-2015 (2)] c) a liquid + vapour mixture will be left in the system Q.10 A rigid container of volume 0.5 m3 d) the mixture will exist at a dry contains 1.0 kg of water at 120 saturated vapour state 33 υfg=0.00106 m / kg, ν = 0.8908 m( /kg) [GATE–2008] ℃ The state of water is Q.5 The work done by the system during a) Compressed liquid the process is b) Saturated liquid a) 0.1 kJ b) 0.2 kJ c) A mixture of saturated liquid and c) 0.3 kJ d) 0.4 kJ saturated vapour [GATE–2008] d) Superheated vapor [GATE-2015 (3)] The net entropy generation Q.6 (considering the system and the dp thermal reservoir together) during Q.11 For water at 25 , s = 0.189 dT the process is closest to s a) 7.5 J/K b) 7.7 J/K kPa/K (ps is the saturation℃ pressure c) 8.5 J/K d) 10 J/K in kPa and s is the saturation [GATE–2008] temperature in K) and the specific volume of dry𝑇𝑇 saturated vapour is 3 Q.7 A pure substance at 8 MPa and 43.38 m /kg. Assume that the 400 is having a specific internal specific volume of liquid is energy of 2864 kJ/kg and a specific negligible in comparison with that of volume℃ of 0.03432 m3/kg. Its vapour. Using the Clausius- specific enthalpy (in kJ/ kg) is ____. Clapeyron equation, an estimate of [GATE-2014 (2)] the enthalpy of evaporation of water at 25 (in kJ/kg) is ______Q.8 1.5 kg of water is in saturated liquid [GATE-2016 (1)] 3 ℃ state at 2bar (Vf=0.001061 m /kg, uf=504.0 kJ/kg, hf = 505kJ/kg). Heat Q.12 The INCORRECT statement about is added in a constant pressure the characteristics of critical point of process till the temperature of a pure substance is that water reaches 400 (v=1.5493m3/ a) there is no constant temperature kg, u=2967.0kJ/kg, h=3277.0 kJ/kg). vaporization process The heat added (in kJ)℃ in the process b) it has point of inflection with is ______. zero slope [GATE-2014 (1)] c) the ice directly converts from solid phase to vapor phase Q.9 The Vander Waals equation of state d) saturated liquid and saturated a vapor states are identical is p +()υ −= b RT , where p is [GATE-2016 (3)] υ2 Q.13 Which one of the following statements is correct for a temperature and R is characteristic superheated vapour? gaspressure, constant. υ is Thespecific SI unit volume, of a is T is

(B) Its temperature is less than the saturation temperature at a given pressure.

(D) Its enthalpy is less than the enthalpy of the saturated vapour at a given pressure.

[GATE-2018 Set-1]

Q.14 A vehicle powered by a spark ignition engine follows air standard Otto cycle ( γ =1.4). The engine generates 70 kW while consuming 10.3 kg/hr of fuel. The calorific value of fuel is 44,000 kJ/kg. The compression ratio is ______(correct to two decimal places).

[GATE-2018 Set-1]

Q.15 A tank of volume 0.05 m3 contains a mixture of saturated water and saturated steam at. The mass of the liquid present is 8 kg. The entropy (in kJ/kg K) of the mixture is ______(correct to two decimal places).

Property data for saturated steam and water are:

0 At 200 C,psat = 1.5538 MPa 33 vfg= 0.001157 m / kg,v = 0.12736 m / kg

sfg= 4.1014kJ / kg K, s f = 2.3309kJ / kg K

[GATE-2018 Set-1]

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(a) 7.61 2.5 (a) (a) (b) (a) (d) (c) 3138 4158 (c) (c) 2443 (c)

EXPLANATIONS

Q.1 (a) Given: T1= Tp 2, 1 = p 2 Universal Gas constant = R Q.2 (a) Here given oxygen are mixed Given adiabatically = 3 υcri 0.003155m / kg, So, dQ= 0 V= 0.025m3 ,P = 0.1MPa and dQ We know ds= = 0 (since dQ=0) M=10 kg. T We know, Rigid means volume is constant. Q.4 (a) Specific volume, When the vapour is at a V 0.025 temperature greater than the υ= = = 0.0025m3 / kg s m 10 saturation temperature, it is said to We see that the critical specific exist as superheated vapour. The volume is more than the specific pressure and temperature of super volume and during the heating heated vapour are independent process, both the temperature and properties, since the temperature the pressure remain constant, but may increase while the pressure the specific volume increases to the remains constant. critical volume (i.e. critical point). Here vapour is at 400 and The critical point is defined as the saturation temperature is 200 point at which the saturated liquid So, at 200 kPa pressure,℃ and saturated vapour states are superheated vapour will be ℃left in identical. the system.

Q.5 (d)

Given p12= 100KPa,p = 200kPa

m =1kg and mvl = 0. 03 kg 0.03 x== 0.029 1 1.03

v1f1fg= v + xv = 0.001 + 0.0029(0.1 – 0.001) So, point (B) will touch the = 0.0038 saturated liquid line and the liquid V= mv = 1.03v = 0.0039 line will rise the Point O. 11 1 Now, given that heat transfer takes Q.3 (b) place into the system until its volume increases by % 50

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Therefore, V21= 1.5 = 0 85V.005 Psat @ T 1→ saturation pressure at T 1 temperature Now Work done = − P(22V V) 1 P1 → pressure of superheated vapour at state 1. = 200(0.00585 - 0.0039) P< P @T =0.4 kJ 1 sat 1

Q.14 (7.61) Q.6 (c) −1000 γ = 1.4 Sgen += 10 673 B.P. = 70kN = 8.514 J/K . mf = 10.3kg/hr Q.7 (3138.56) h = u + pv CV = 44000 kJ/kg h = 2864 + 8 ×103×0.03432 BP. . 70 η =. =  100 % h = 3138.56 kJ/kg  10.3 mf CV  44000 3600 Q.8 (4158) η = 0.556 Q = m (h ) 2 1 1 ηotto =−=1γ −1 0.556 = 4158 kJ−h ()r =1.5(3277−505) 1 ⇒− = Q.9 (c) 11.4− 1 0.556 ()r

1 Q.10 (c) = 0.4439 0.5 0.4 v= m33 / kg = 0.5m / kg ()r 1 r = 7.61 Since v<< vv the state of water fg is mixture of saturated water and saturated vapour. Q.15 (2.49)

Q.11 (2443.24) By Causius-Clapeyron equation dP ss− h s =g f = fg dTs v g T sat× v g

∴=hfg 0.189 × 298 × 43.38

hfg = 2443.24kJ / kg

Q.12 (c) Q.13 (a)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Volume of liquid = mff V =8 0.001157 = 9.256 10−33m = 0.009256 m3 So volume of steam = 0.05− 0.009256 = 0.040744m3 Volume of steam Mass of steam = vg 0.040744 = 0.12736

= 0.319912 kg = ms

Totalmassof mixture = mfs+ m =8 + 0.319912 = 8.319912kg m dryness fraction = x= s mmfs+ 0.319912 = = 0.03845 8.319912 So, the entropy of the mixture is given by

s = sf+ x s fg =2.3309 + 0.03845 4.1014 = 2.488kg/kgK

Q.1 The Rateau turbine belongs to the Q.5 In a gas turbine, hot combustion category of` products with the specific heats Cp = a) pressure compounded turbine 0.98 kJ/kgK, and Cv = 0.7538 kJ/kgK b) reaction turbine enters the turbine at 20 bar, 1500 K c) velocity compounded turbine and exits at 1 bar. The isentropic d) radial flow turbine efficiency of the turbine is 0.94. The [GATE–2001] work developed by the turbine per kg of gas flow is Q.2 The efficiency of superheat Rankine cycle is higher than that of simple a) 689.64 kJ/kg b) 794.66 kJ/kg Rankine cycle because c) 1009.72 kJ/kg d) 1312.00 kJ/kg a) the enthalpy of main stream is [GATE–2003] higher for superheat cycle Q.6 The compression ratio of a gas b) the mean temperature of heat power plant cycle corresponding to addition is higher for superheat maximum work output for the given cycle temperature limits of Tmin and Tmax c) the temperature of steam in the will be condenser is high γ γ d) the quality of steam in the T 2()γ1− T 2()γ1− a) max b) min condenser is low.   Tmin Tmax [GATE–2002] γ1− γ1− γ γ Q.3 In Rankine cycle, regeneration Tmax Tmin c)  d)  results in higher efficiency because Tmin Tmax a) pressure inside the boiler [GATE–2004] increases b) heat is added before steam Common Data for Q.7 and Q.8: enters the low pressure turbine Consider a steam power plant using a c) average temperature of heat reheat cycle as shown. Steam leaves the addition in the boiler increases boiler and enters the turbine at 4 MPa , 0 d) total work delivered by the 350 C (h3 = 3095 kJ/kg). After expansion in turbine increases the turbine to 400 kPa (h4 = 2609 kJ/kg), [GATE–2003] the steam is reheated to 350 (h = 3170 kJ/kg) and then expanded in a low Q.4 Considering the variation of static 5 pressure turbine to 10 kPa (h ℃ = 2165 pressure and absolute velocity in an 6 kJ/kg). The specific volume of liquid impulse steam turbine, across one handled by the pump can be assumed to be row of moving blades a) both pressure and velocity decreases b) pressure decreases but velocity increases c) pressure remains constant, while velocity increases d) pressure remains constant, while velocity decreases [GATE–2003]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.7 The thermal efficiency of the plant Q.10 In the velocity diagram shown neglecting pump work is below, u = blade velocity, C = a) 15.8% b) 41.1% absolute fluid velocity and W = c) 48.5% d) 58.6% relative velocity of fluid and the [GATE–2004] subscripts 1 and 2 refer to inlet and outlet. This diagram is for Q.8 The enthalpy at the pump discharge (h2) is a) 0.33 kJ/kg b) 3.33 kJ/kg c) 4.0 kJ/k d) 33.3 kJ/kg [GATE–2004]

a) an impulse turbine Q.9 A p -v diagram has been obtained b) a reaction turbine from a test on a reciprocating c) a centrifugal compressor compressor. Which of the following d) an axial flow compressor represents that diagram? [GATE–2005]

Common data for Q.11 and Q.12 In two air standard cycles-one operating in the Otto and the other on the Brayton cycle air is isentropically compressed from 300 a) to 450K. Heat is added to raise the temperature to 600K in the Otto cycle and to 550K in the Brayton cycle.

Q.11 In and are the efficiencies of the Otto and Bray ton cycles, then a) η o = 0.25,ηB = 0.18 b) = = 0.33 b) o B c) η = 0.5, η = 0.45 o B d) itη is notη possible to calculate the efficienciesηo ηB unless the temperature after the expansion is given [GATE–2005]

c) Q.12 If Wo and WB are work outputs per unit mass, then a) Wo > WB b) Wo < WB c) Wo = WB d) it is not possible to calculate the work outputs unless the temperature after the expansion d) is given [GATE–2005] [GATE–2005]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Common Data for Q.13 and Q.14: Reason (R): The regenerative feed The following table of properties was water heating raises the average printed out for saturated liquid and temperature of heat addition in the saturated vapour of ammonia. The titles for Rankin cycle. only the first two columns are available. All a) Both (A) and (R) are true and (R) that we know that the other columns is the correct reason for (A) (column 3 to 8) contain data on specific b) Both (A) and (R) are true but (R) properties, namely, internal energy (kJ/kg), is NOT the correct reason for (A) enthalpy (kJ/kg) and entropy (kJ/kgK) c) Both (A) and (R) are false d) (A) is false but (R) is true [GATE–2006]

Q.17 Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion (A): Condenser is an Q.13 The specific enthalpy data are in essential equipment in a steam columns power plant. a) 3 and 7 b) 3 and 8 Reason (R): For the same mass flow c) 5 and 7 d) 5 and 8 rate and the same pressure rise, a [GATE–2005] water pump requires substantially less power than a steam compressor. Q.14 When saturated liquid at 400C is a) Both (A) and (R) are true and (R) throttled to – 20°C, the quality at is the correct reason for (A) exit will be b) Both (A) and (R) are true and (R) a) 0.189 b) 0.212 is NOT the correct reason for (A) c) 0.231 d) 0.788 c) Both (A) and (R) are false [GATE–2005] d) (A) is false but (R) is true [GATE–2006] Given below is an extract from Q.15 Q.18 Group I shows different heat steam tables. addition process in power cycles. Likewise, Group II shows different heat removal processes. Group III lists power cycles. Match items from Groups I, II and III. Group I Group II Group III P. Pressure S. Pressure 1.Rankine Cycle Specific enthalpy of water in kJ/kg Constant Constant at 150 bar and 45°C is Q. Volume T. Volume 2. Otto cycle Constant Constant a) 203.60 b) 200.53 R. U. 3. Carnot cycle c) 196.38 d) 188.45 Temperature Temperature Constant Constant [GATE–2006] 4. Diesel cycle Q.16 Determine the correctness or 5.Brayton cycle otherwise Assertion (A) and the a) P-S-5, R-U-3, P-S-1 , Q-T-2 Reason (R) Assertion (A): In a b) P-S-1, R-U-3, P-S-4, P-T-2 power plant working on a Rankin c) R-T-3, P-S-1, P-T-4, Q-S-5 cycle, the regenerative feed water d) P-T-4, R-S-3, P-S-1, P-S-5 heating improves the efficiency of [GATE–2006] the steam turbine.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.19 Which combination of the following subscripts f and g denote saturated liquid statements is correct? The state and saturated vapour state. incorporation of reheater in a steam power plant: P: always increases the thermal efficiency of the plant. Q: always increases the dryness

fraction of steam at condenser inlet Q.21 The network output (kJ/kg) of the R: always increases the mean cycle is temperature of heat addition. a) 498 b) 775 S: always increases the specific c) 860 d) 957 work output. [GATE–2010] a) P and S b) Q and S c) P, R and S d) P, Q, R and S Q.22 Heat supplied (kJ/kg) to the cycle is [GATE–2007] a) 2372 b) 2576 c) 2863 d) 3092 Q.20 A thermal power plant operates on a [GATE–2010] regenerative cycle with a single open feed water heater, as shown in Q.23 The values of enthalpy of steam at the figure. For the state points the inlet and outlet of a steam shown, the specific enthalpies are: turbine in a Rankine cycle are 2800 h1 = 2800 kJ/kg and h2 = 200 kJ/kg. kJ/kg and 1800 kJ/kg respectively. The bleed to the feed water heater is Neglecting pump work, the specific 20% of the boiler steam generation steam consumption in kg/kWh is rate. The specific enthalpy at state 3 is a) 3.60 b) 0.36 c) 0.06 d) 0.01 [GATE–2011]

Q.24 An ideal Brayton cycle, operating between the pressure limits of 1 bar and 6 bar, has minimum and maximum temperature of 300 K and 1500 K. The ratio of specific heats of the working fluid is 1.4. The approximate final temperatures in a) 720 kJ/kg b) 2280 kJ/kg Kelvin at the end of compression & c) 1500 kJ/kg d) 3000 kJ/kg expansion processes are respectively [GATE–2008] a) 500 and 900 b) 900 and 500 c) 500 and 500 d) 900 and 900 Common Data For Q.21 and Q.22 [GATE–2011] In a steam power plant operating on the Rankine cycle, steam enters the turbine at Q.25 Specific enthalpy and velocity of 4MPa, 350° C and exists at a pressure of 15 steam at inlet and exit of a steam kPa. Then it enters the condenser and exits turbine, running under steady state, as saturated water. Next, a pump feeds are as given below: back the water to the boiler. The adiabatic efficiency of the turbine is 90%. The thermodynamic states of water and steam are given in table h is specific enthalpy, s is specific entropy and the specific volume;

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The rate of heat loss from the 1 1 a)1− b) 1− turbine per kg of steam flow rate is rγ1− rγ 5 kW. Neglecting changes in p p 1 1 potential energy of steam, the − − c) 1 1/γ d) 1 (γ− 1)/γ power developed in kW by the rp rp steam turbine per kg of steam flow [GATE-2014 (2)] rate is a) 901.2 b) 911.2 Q.30 For a gas turbine power plant, c) 17072.5 d) 17082.5 identify the correct pair of statements. [GATE–2013] P. Smaller in size compared to Common Data For Q.26 and 27 steam power plant for same In a simple Brayton cycle, the pressure power output ratio is 8 and temperatures at the entrance Q. Starts quickly compared to of compressor and turbine are 300 K and steam power plant 1400 K, respectively. Both compressor and R. Works on the principle of gas turbine have isentropic efficiencies Rankine cycle equal to 0.8. For the gas, assume a constant S. Good compatibility with solid fuel value of Cp (specific heat at constant a) P and Q b) R and S pressure) equal to 1 kJ/kg-K and ratio of c) Q and R d) P and S specific heats as 1.4. Neglect changes in [GATE-2014 (3)] kinetic and potential energies. Q.31 An ideal reheat Rankine cycle Q.26 The power required by the operates between the pressure compressor in kJ/kg of gas is limits of 10 kPa and 8 MPa, with a) 194.7 b) 243.4 reheat being done at 4 MPa. The c) 304.3 d) 378.5 temperature of steam at the inlets of [GATE–2013] both turbines is 500 and the enthalpy of steam is 3185 kJ/kg at Q.27 The thermal efficiency of the cycle in the exit of the high pressure℃ turbine percentage (%) is and 2247 kJ/kg at the exit of low a) 24.8 b) 38.6 pressure turbine. The enthalpy of c) 44.8 d) 53.1 water at the exit from the pump is [GATE–2013] 191 kJ/ kg. Use the following table Q.28 In a power plant, water (density for relevant data =1000 kg/ m3) is pumped from 80 kPa to 3 MPa. The pump has an isentropic efficiency of 0.85. Assuming that the temperature of the water remains the same, the specific work (in kJ/ kg) supplied to Disregarding the pump work, the the pump is cycle efficiency (in percentage) is a) 0.34 b) 2.48 ____ c) 2.92 d) 3.43 [GATE-2014 (1)] [GATE-2014 (1)] Q.32 In an ideal Brayton cycle, Q.29 The thermal efficiency of an air- atmospheric air (ratio of specific standard Brayton cycle in terms of heats,C /C = 1.4, specific heat at pressure ratio r C /C ) is constant pressure = 1.005 kJ/kgK) given by p V p and γ (= p v at 1 bar and 300 K is compressed to

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 8 bar. The maximum temperature in the remaining steam expands the cycle is limited to 1280 K. If the isentropically to 9 kPa heat is supplied at the rate of 80 Inlet to turbine: P = 14MPa, T = MW, the mass flow rate (in kg/s) of 560 , h= 3486 kJ/kg, s = 6.6 kJ/kgK air required in the cycle is Intermediate stage: h = 2776 kJ/kg [GATE-2014 (2)] Exit℃ of turbine: P = 9kPa,

hf =174 kJ/kg, hg =2574 kJ/kg, Q.33 Steam at a velocity of 10m/s enters s = 0.6kJ/kgK, s = 8.1kJ/kgK the impulse turbine stage with If the flow rate of steam entering the symmetrical blading having blade f g turbine is 100 kg/s, then the work angle 30°. The enthalpy drop in the output (in MW) is ____. stage is 100 kJ. The nozzle angle is 20°.The maximum blade efficiency [GATE-2015 (1)] (in percent) is ______. Q.37 In a Rankine cycle, the enthalpies at [GATE-2014 (2)] turbine entry and outlet are 3159

kJ/kg. and 2187 kJ/kg respectively. Q.34 Steam with specific enthalpy (h) If the specific pump work is 2 kJ/kg, 3214 kJ/kg enters an adiabatic the specific steam consumption (in turbine operating at steady state kg/kW-h) of the cycle based on the with a flow rate 10 kg/s. As it net output is _____. expands, at a point where h is 2920 kJ/kg, 1.5 kg/s is extracted for [GATE-2015 (2)] heating purposes. The remaining 8.5 The INCORRECT statement about kg/s further expands to the turbine Q.39 regeneration in vapor power cycle is exit, where h = 2374 kJ/kg. that Neglecting changes in kinetic and a) it increases the irreversibility by potential energies, the net power adding the liquid with higher output (in kW) of the turbine is energy content to the steam ______. generator [GATE-2014 Set-4] b) heat is exchanged between the

expanding fluid in the turbine Q.35 Which of the following statements and the compressed fluid before regarding a Rankine cycle with heat addition reheating are TRUE? c) the principle is similar to the i) increase in average temperature principle of Stirling gas cycle of heat addition d) it is practically implemented by ii) reduction in thermal efficiency providing feed water heaters iii) drier steam at the turbine exit

a) only (i) and (ii) are correct [GATE-2016(1)]

b) only (ii) and (iii) are correct In a steam power plant operating on c) only (i) and (iii) are correct Q.40 an ideal Rankine cycle, superheated d) (i),(ii) and (iii) are correct steam enters the turbine at 3 MPa [GATE-2015 (2)] and 350 . The condenser pressure

is 75 kPa. The thermal efficiency of Q.36 Steam enters a well insulated the cycle℃ is ______percent. turbine and expands isentropically Given data: throughout. At an intermediate For saturated liquid, pressure, 20 percent of the mass is extracted for process heating and

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission At P=75kPa, h =384.39kJ/kg, =0.001037m3/kg, s =1.213kJ/kg-K f Q.44 In the Rankine cycle for a steam Atνf 75 kPa, h = 2278.6 kJ/kg, s = power plant the turbine entry and f 6.2434 kJ/kg-K fg fg exit enthalpies are 2803 kJ/kg and At P=3MPa & T=350 (superheated 1800 kJ/kg, respectively. The steam), h = 3115.3 kJ/kg, s = 6.7428 enthalpies of water at pump entry kJ/kg-K ℃ and exit are 121 kJ/kg and 124 [GATE-2016(1)] kJ/kg, respectively. The specific Q.41 Consider a simple gas turbine steam consumption (in kg/k W.h) of (Brayton) cycle and a gas turbine the cycle is ______cycle with perfect regeneration. In both the cycles, the pressure ratio is [GATE-2017 Set-2] 6 and the ratio of the specific heats of the working medium is 1.4. The ratio of minimum to maximum temperatures is 0.3 (with temperatures expressed in K) in the regenerative cycle. The ratio of the thermal efficiency of the simple cycle to that of the regenerative cycle is ______[GATE-2016(2)]

Q.42 In a 3-stage air compressor, the inlet pressure is 1, discharge pressure is 4 and the intermediate pressures are 2 and 𝑝𝑝 3 (p2 < 3). The total pressure𝑝𝑝 ratio of the compressor is 10 and𝑝𝑝 the 𝑝𝑝pressure 𝑝𝑝ratios of the stages are equal. If 1 = 100 kPa, the value of the pressure 3 (in kPa) is ______𝑝𝑝 [GATE𝑝𝑝 -2016(3)] Q.43 The Pressure ratio across a gas turbine (for air, specific heat at constant pressure,

c= 1040J/kg.K and ratio of specific p heats, γ = 1.4 ) is 10. If the inlet temperature to the turbine is 1200K and the isentropic efficiency is 0.9, the gas temperature at turbine exit

is ______K.

[GATE-2017 Set-1]

ANSWER KEY:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) (b) (c) (d) (a) (a) (b) (d) (d) (b) (b) (a) (d) (b) 15 16 17 18 19 20 21 22 23 24 25 26 27 28 (d) (a) (b) (a) (b) (a) (c) (c) (a) (a) (a) (c) (a) (d) 29 30 31 32 33 34 35 36 37 38 39 40 41 42 (d) (a) 40.73 108.07 88.3 7581 (c) 125.56 (a) 3.7 26.01 0.8 464.14 43 44 679 3.6

Q.1 (a) Q.4 (d) The Rateau turbine is a pressure In impulse turbine, the steam compounded turbine. pressure remains constant while it flows through the moving blades, Q.2 (b) where only the kinetic energy converts into mechanical energy.

The average temperature at which heat is transferred to steam can be increased without increasing the boiler pressure by superheating the steam to high temperatures. The effect of superheating on the performance of vapour power cycle is shown on a T - s diagram. Thus both the net work and heat input increase as a result of superheating Q.5 (a) the steam to a higher temperature. T = 1500 K The overall effect is an increase in C = 0.98 kJ/kg K 3 thermal efficiency. C∴ Given,= 0.7538 kJ/ kg K. p Since, the average temperature at C which heat is added increases. V p =1.3 CV Q.3 (c) ∴ γ =

γ1− The object of the regenerative feed r TP44 heating cycle is to supply the Now, =  TP33 working fluid to the boiler at same 0.3 state between 2 and 2’ (rather than 1.3 T4 1 at state 2) there by increasing the =  1500 20 average temperature of heat T = 751.37K addition to the cycle. ⇒ Now turbine efficiency; 4

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission TT− ' 1 η = 34 T 2x = 3 TT34− rp  T1 1500− T' 4 ∴ γ 21()γ− 1500− 751.34 Tmax rp =  ⇒0.94 T = 796.25 K Tmin Turbine′ work; 4 ⇒ W∴ = c (T T ) Q.7 (b) – ′ Power obtained from plant t p 3 4 − = (h h ) + (h h ) = 0.98 (1500 796.25) Heat supplied to the plant. 3 − 4 5 − 6 Q.6 =(a) 689.67 kJ/kg K = (h h ) (h h ) Thermal efficiency 3 1 5 4 Power− obtained− −

Heat supplied = ()3095−+− 2609 (3170 2165)

()3170−+− 2609 (3095 29.3) =

Q.8 =0.4111=41.11%(d) Work output during the cycle Enthalpy at exit of pump must be =mCp3 (T −− T 2 ) mC p4 (T − T 1 ) greater than enthalpy at inlet of

=mCp3 (T −− T 4 ) mC p2 (T − T 1 ) pump i.e. h2 must be greater, then h  1 TT42 options only one option is greater =mCp3 T 1 −− T 1 − 1 TT31 than=29.3kJ/kg. h Among the given four Since . 1 γ−1 correct. =29.3kJ/kg, which is T3 T2 γ = = = ()rp 33.3kJ/kg. Hence option (D) is TT 41 1 (Boiler pressure Condenser γ−1 Or Pumpρ work = vdP Let= x, mCp = K γ =pressure) 1 (4000 3 1 T  10 ∴ KTWork/cycle 1−−  T4 − 1  31xx   = h2 10)=3.99 kJ/kg rrpp    = For maximum power, differentiate Q.9 (d) =29.3 + 3.99 = 33.3 kJ/kg with respect to  dW x − =K T ×− T xr()x1 = 0 dr 3()x1− 1p p rp (For⇒ max.) xT ()x1− 3 = ()x1+ Tx1p() r rp ⇒ 2x T3 rp = T1 ⇒

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission γ1− From above figure, we can easily see TV that option (d) is same. 21=  TV Q.10 (b) 12 Velocity of flow is constant T3 T2 throughout the stage and the So, =  TT41 diagram is symmetrical hence it is a diagram of reaction turbine. T3 600 T41= ×= T × 300 = 400K T2 450 Q.11 (b) We know that efficiency. W0 V (600 – CV (400– T And V – V V ...(ii) = = − 1 ηηOtto Brayton 1 From= C P v 450)diagram − of Brayton300) T2 cycle,= 150 workC 100 done, C =is, 50 C 300 ηηOtto= Brayton = 1 − WB – Q−2 Cp(T –T2) – Cp (T4 – T ) 450 6 1 3 1 =−= = TQ = 300 1 0.33 T=1 ×= T × 550 = 366.67K 9 And43 T 450 So, η= η = 33% 2 Otto Brayton W= C()() 550 −− 450 366.67 − 300 sp Q.12 (a) = 33.33 c ... (iii)

p WOv 50C 50 Dividing= equation = (ii) by (iii), we get WBp 33.33C 33.33γ C p = γ and γ=1.4 Cv 50 50 = >1 33.33× 1.4 46.662

=From this, we see that, W > W

O B Q.13 (d) From saturated ammonia table

enthalpy date column. column 5 and 8 are the specific

From the previous part of the Q.14 (b) question The enthalpy of the fluid before throttling is equal to the enthalpy of T ( ) = 600 K, T ( ) = fluid after throttling because in 550 K, 3 Otto 3 Brayton throttling process enthalpy remains From the P v diagram of Otto constant. cycle, we have h = h W Q –Q −(T – T (T –T ) 0 2 V 2 V 4 – 1 3 ...(i) 1 1 2 = = C– 4, ) − C h=+− hf xh() gf h γ1− γ1− 371.43 = 89.05 + x (1418 89.05) T VV For3 = process41 3 = = =  V4 V,V 13 V 2 282.38 TV45  V 2 =x 89.05= + x (1328.95) = 0.212 – 2, 1328.95

For process 1

Q.15 (d) When the temperature of a liquid is and the same pressure rise, water less than the saturation temperature pump require very less power at the given pressure, the liquid is because the specific volume of liquid called compressed liquid (state 2 in is very less as compare to specific figure).The pressure and temperature volume of vapour. of compressed liquid may vary independently and a table of Q.18 (a) properties like the superheated vapour table could be arranged, to Q.19 (b) give the properties at any p and T. Whether the cycle efficiency The properties of liquids vary little increases or not depends upon the with pressure. Hence, the properties mean temperature of heat addition. are taken from the saturation table In practice the use of reheat only at the temperature of the compressed gives a small increase in cycle efficiency, but it increases the net 0 C, Specific enthalpy of water work output by making possible the liquid. So, from the given table at T = use of higher pressures, keeping the 45 quality of steam at the outlet of the =188.45 kJ/kg. turbine within permissible limits.

Q.20 (a) kg h2 kJ/ kg From the given diagram of thermal Given: 2800kJ/ =200 the Boiler to the open feed water heaterpower and plant, point point 2 is 1 directed is directed by the by pump to the open feed water heater. Q.16 (a) The bleed to the feed water heater is

Q.17 (b) (a) Condenser is essential 20% of the boiler steam generation 1 equipment in a steam power plant i.e. 20% of h because when steam expands in the turbine and leaves the turbine in the form of super saturated steam. It is not economical to feed this steam directly to the boiler. So, condenser is used to condense the steam into water and it is an essential part So,h2 h of h2 (equipment) in steam power plant. × × ) is correct. 1 = 20% of +80% (b) The compressor and pumps Q.21 (c)= 0.2 2800 + 0.8 200 = 720 kJ/kg requireAssertion power (a input. The compressor h= 3092. k g5kJ/ is capable of compressing the gas to 1

very high pressures. Pump work h2=+− h f xh() gf h very much like compressor except Now, s==+− s s xs s that they handle liquid instead of 1 2 f2() g2 f2 gases. Now for same mass flow rate

𝛾𝛾 = 1.4

x = 0.8033 Dryness fraction at exit of turbine

γ1− TPγ 22=  TP11 0.4 T 6 1.4 Hence, 2 =  h2 - 300 1 h2 T = 500.5K 0.4 =225.94+0.8033(2599.1 h - h2 225.94) T2 6 1.4 =2132.3kj/kg 3 – 1 =  Turbine work = T14  =3092.5 2132.3 T = 900K WWact= th ×η isen =960.2 kJ/kg 4 ⇒ Q.25 (a)

= 0.9×960.2 vdP 22 = 864.2kJ/kg – mV12 mV Applyingmh12++=++ SFEE Q mh W Pump work = 22 = 0.001014 (4000Turbine 15) work – Pump 18022 5 = 4.04079 kJ/kg 3250+−=++ 5 2360 W work 2000 2000 Net work=– 4.04 ⇒

= 864.2 (Since m = 1 kg/s) = 860.16 kJ/kg Q.22 (c) Q.26 ⇒(c) W = 901.2 kW Heat h -h4

h4 Pump work1 supplied = 3 =h + =225.94 + 4.04078h -h4 = 229.98 kJ/kg 1 Heat supplied = = 3092.5 229.98 Q.23 (a)= 2862.519 kJ/kg W = C (T2 ) Specific steam consumption γ1− TC p 1 (kg/kWh) 2 = (r ) γ ’−T T p 3600 3600 1 = = 0.4 Network Done 2900− 1900 1.4 () T2 = 300 × (8) T = 543.43K Efficiency⇒ of compressor 2 Q.24 (a)=3.6kg/kWh TT− = 21 P Tminimum ηc ' TT21− P12 Tmaximum = 1bar = 300 K = 6 bar = 1500 K

' TT21− Q.31 (40.73) T0.8== 604.28k W′ = 1(604.28 300) W2 = 304.28 kJ/kg c Power required− by the compressor c

Q.27 (a) γ1− T3 γ = (γ)p T4 1400 T4 =0.4 = 772.86K

(8)1.4 ⇒ Efficiency of turbine TT− ' PGiven, η = 34 2 T P TT34− P =10kPa − ' 1 1400 T4 =8kPa 0.8 = h32s TT34− h4s=4kPa ' h =3185 kJ/kg 1400− T4 h = 2247 kJ/kg 1400− 772.86 1 h6s=3399 kJ/kg 0.8T = 3 Work of turbine, WT=3446 kJ/kg2s 4s) 4 Q =191 kJ/kg ) W′ = 898.288 C (T KT ) S 1 6s 3 2s W= (h − h ) + (h − h – T 1 3 T p 3 4 = (h − h ) + (h − h − ′ Q∴ = (3399 − 3185) + (3446 − 2247) Thermal= 1 (1400 efficiency 898.288) of the cycle S =1413 kJ/kg = 501.712WW− kW/ kg η = TC = W(3399−191)1413 + (3446−3185) T Q η=T = =3469Q kJ/kg 3469 501.712− 304.28 s = = 24.8% 1(1400− 604.28) Or = 0.4073 Q. 28 (d) Q.32 (108.071= 40.73%)

workisen = vdp 1 ()PP− ρwater 21 = ()3000− 80 = 1000 2920 = = 2.92kJ / kg C 1000 p =γ=1.4 w 2.92 Cv workact = = = 3.43 η 0.85 Given,Cp P = 1.005 kJ/ kgK Q.29 (d) T1 = 1bar P21 = 300K = 8bar

3 -2, Q.36 (125.56) = T =γ− 12801 K γ 0.4 ForTP22 process 1 = = 81.4 TP11

T2 = 1.8114 s = s We∴ know= 300×1.8114 that Qs= mC T − T p3() 2 x = 543.43K 1 3 x = 0.8 3 80× 106 h⇒ 6.6= 174 = 0.6+ + 0.8 [[7.5]2574 174] = 3 =m 80MW (given)3 1.005×− 10() 1280 543.43 3 − [h h ] + 80[h h ] = 2094 kJ/kg- - 1 2 2 3 Q.33 =(88.3 108.071) kg/s ∴ W = 100 − − We know that for maximum blade =100[3486 2776]+80[2776 2094] efficiency, Q.37 *= 125560 kW = 125.56 MW cos α ρ= 2 Q.38 (3.7) u 3600 where ρ= Wnet V1

Now,Specific Whh steamT= 21 consumption − =

Andη=α maximumcos2 balde efficiency is ()b max WT = 972kJ / kg given by: 2 = 3159 2187 =o = W= 2kJ / kg ()cos 20 0.883 ∴ p Thus specific steam consumption And 3600 = Q.34 =(7581 88.3%) 972− 2 kg/ kWh= 3.7815 kg/ kWh Q.39 (a)

Q.40 (26.01)

m1 = 10kg/s m32 h = 1.5kg/s = 8.5kg/s h12 h = 3214kJ/kg -

WT= 2920kJ/kg(h 2 2 (h2 ) ss12= 3 For process 1→ 2: WT= 2374 kJ/kg (2920 1 1 3 x2 = m − h ) + m − h x2 = 0.8857 WT =10(3214−2920)+8.5 6.7428 = 1.213 + (6.2434) 2374) h2 = Q.35 (c) = 7581kW h= 2408.54kJ / kg ∴ 2 384.39 + 0.8857[2278.6]

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Where N-Number of stages WT= hh 12 − r= 3 10 ()p s = WT 712.76kJ / kg r= 2.1544 = 3115.3 2402.54 ∴() p s WP= vP f2[] − P 1 p2 = 2.1544 p1 WP = 3.033kJ / kg = 0.001037 [3000 75] p2 = 2.1544 × 100 =−= Wnet W T W P 709.727kJ / kg

Q3= hh 14 − ps = 215.44 kPa= 2.1544 Whhp= 42 − p2 h= Wh + Alsop= 2.1544 × 215.44 Also, 4 p2 3

∴ h= 387.423kJ / kg p= 464.14kPa 4 ∴ 3 = 3.033+384.3 Qs = 3115.3 − 387.423 Q.43 (679.38)

∴ W η= net = 2727.87th kJ/ kg Qs 709.727 η= =0.2601 th 2727.87 η=26.01% ∴ th

Q.41 (0.8) TT− Efficiency - η=34 1 isen T− T' η=− 34 th 1 γ−without1/ γ regeneration: ()rp P2 P3 1 Q = =10 PP η=−th 10.4/1.4 = 0.4006 14 ()6 r1− - r T'44 P γ−1 T =  η=−γ min TP Efficiency()threg 1r with() p regeneration: 33 Tmax =−×0.4/1.4 1.4− 1 1() 6 0.3 1 1.4 T4 '= 1200 10 η 0.4 Ratio =th = = 0.4994 η≈ 0.5 T '= 621.53k ()th reg 0.5 4

0.9[1200−=− 621.53] T T 34 Q.42 (464.14)= 0.8 T= 679.38k We know that in multi-stage 4 compression N Q.44 (3.6) rr= ()()pp0s 1/ N r= () rp ()p s 0

Givenh1 = 2803kJ / kg

h2 = 1800kJ / kg

h3 = 121kJ / kg

h4 = 124kJ / kg

WT=−= h 12 h 2803 − 1800

WT = 1003kJ / kg

Whhp= 43 −

=124 − 121

Wp = 3kJ / kg

Wnet= WW T − p

=1003 − 3

Wnet = 1000kJ / kg

3600 kg Q SSC = Wnet kwh

3600 SSC= = 3.6kg / kwh 1000

Q.1 A single-acting two-stage Q.5 A diesel engine is usually more compressor with complete inter efficient than a spark ignition engine cooling delivers air at 16 bar. because Assuming an intake state of 1 bar at a) diesel being a heavier 15°C, the pressure ratio per stage is hydrocarbon, releases more heat a) 16 b) 8 per kg than gasoline c) 4 d) 2 b) the air standard efficiency of [GATE–2001] diesel cycle is higher than the Otto cycle, at a fixed Q.2 In a spark ignition engine working compression ratio on the ideal Otto cycle, the c) the compression ratio of a diesel compression ratio is 5.5. The work engine is higher than that of an output per cycle (i.e., area of the P-v SI engine diagram) is equal to d) self ignition temperature of 23.625×105×Vc(in Joules), where Vc diesel is higher than that of is the clearance volume in m3. The gasoline indicated mean effective pressure is [GATE–2003] a) 4.295 bar b) 5.250 bar c) 86.870 bar d) 106.300 bar Q.6 An automobile engine operates at a [GATE–2001] fuel air ratio of 0.05, volumetric efficiency of 90% and indicated Q.3 An ideal air standard Otto cycle has thermal efficiency of 30%. Given a compression ratio of 8.5. If the that the calorific value of the fuel is ratio of the specific heats of air ( ) is 45 MJ/kg and the density of air at 1.4, what is the thermal efficiency in intake is 1 kg/m3, the indicated percentage of the Otto cycle ? 𝛾𝛾 mean effective pressure for the a) 57.5 b) 45.7 engine is c) 52.5 d) 95 a) 6.075 bar b) 6.75 bar [GATE–2002] c) 67.5 bar d) 243 bar [GATE–2003] Q.4 For a spark ignition engine, the equivalence ratio ( ) of mixture Q.7 For an engine operating on air entering the combustion chamber standard Otto cycle, the clearance has values ∅ volume is 10% of the swept volume. a) < 1 for idling and > 1 for The specific heat ratio of air is 1.4. peak power conditions The air standard cycle efficiency is b) ∅> 1 for both idling ∅and peak a) 38.3% b) 39.8% power conditions c) 60.2% d) 61.7% c) ∅ > 1 for idling and < 1 for [GATE–2003] peak power conditions d) ∅ < 1 for both idling ∅and peak Q.8 At the time of starting, idling and power conditions low speed operation, the carburetor ∅ [GATE–2003] supplies a mixture which can be termed as

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission a) Lean Q.12 Which one of the following is NOT a b) slightly leaner than stoichiometric necessary assumption for the air- c) stoichiometric standard Otto cycle? d) rich a) All processes are both internally [GATE–2004] as well as externally reversible. Q.9 During a Morse test on a 4 cylinder b) Intake and exhaust processes are engine, the following measurements constant volume heat rejection of brake power were taken at processes. constant speed. c) The combustion process is a All cylinders firing 3037 kW constant volume heat addition Number 1 cylinder not firing 2102 process. kW d) The working fluid is an ideal gas Number 2 cylinder not firing 2102 with constant specific heats kW [GATE–2008] Number 3 cylinder not firing 2100 kW Q.13 In an air-standard Otto-cycle, the Number 4 cylinder not firing 2098 compression ratio is 10. The kW condition at the beginning of the The mechanical efficiency of the compression process is 100 kPa and engine is 27°C. Heat added at constant a) 91.53% b) 85.07% volume is 1500 kJ/kg, while 700 c) 81.07% d) 61.22% kJ/kg of heat is rejected during the [GATE–2004] other constant volume process in the cycle. Specific gas constant for Q.10 An engine working on air standard air = 0.287 kJ/kg K. The mean Otto cycle has a cylinder diameter of effective pressure (in kPa) of the 10 cm and stroke length of 15 cm. cycle is The ratio of specific heats for air is a) 103 b) 310 1.4. If the clearance volume is 196.3 c) 515 d) 1032 cc and the heat supplied per kg of [GATE–2009] air per cycle is 1800 kJ/kg, the work output per cycle per kg of air is Q.14 A turbo-charged four-stroke direct a) 879.1 kJ b) 890.2 kJ injection diesel engine has a c) 895.3 kJ d) 973.5 kJ displacement volume of 0.0259 m3 [GATE–2004] (25.9 litres). The engine has an output of 950 kW at 2200 rpm. The Q.11 The stroke and bore of a four stroke mean effective pressure (in MPa) is spark ignition engine are 250 mm closest to and 200 mm respectively. The a) 2 b) 1 clearance volume is 0.001m3. If the c) 0.2 d) 0.1 specific heat ratio = 1.4, the air- [GATE–2010] standard cycle efficiency of the engine is � Q.15 The crank radius of a single-cylinder a) 46.40% b) 56.10% I.C. engine is 60 mm and the c) 58.20% d) 62.80% diameter of the cylinder is 80 mm. [GATE–2007] The swept volume of the cylinder in cm3 is a) 48 b) 96 c) 302 d) 603

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission [GATE–2011] If and T denote the specific heat ratio and temperature, respectively, Q.16 In an air-standard Otto cycle, air is the𝛾𝛾 efficiency of the cycle is supplied at 0.1 MPa and 308 K. The TT− TT− a)1− 41 b) 1− 41 TT32− γ[T32− T ] specific gas constant (R) of air are γ[T− T ] ()TT− 1.4ratio and of the288.8 specific J/kgK heats respectively. (γ) and the If c) 1− 41 d) 1− 41 the compression ratio is 8 and the TT32− (γ−− 1)() T32 T maximum temperature in the cycle [GATE-2015 (3)] is 2660 K, the heat (in kJ/kg) supplied to the engine is ______Q.21 For the same values of peak [GATE-2014 (1)] pressure, peak temperature and Q.17 A diesel engine has a compression heat rejection, the correct order of ratio of 17 and cut-off takes place at efficiency for Otto, Dual and Diesel 10% of the stroke. Assuming ratio of cycles is – a) > > standard efficiency (in percent) is b) > > Otto Dual Diesel ______.specific heats (γ) as 1.4, the air c) η > η >η Diesel Dual Otto [GATE-2014 (3)] d) η > η > η Dual Diesel Otto η η [GATEη -2015 (2)] Diesel Otto Dual Q.18 In a compression ignition engine, Q22 Air containsη 79%η N2 andη 21% O2 on a the inlet air pressure is 1 bar and molar basis. Methane (CH4) is burned the pressure at the end of isentropic with 50% excess air than required compression is 32.42 bars. The stoichiometrically. Assuming expansion ratio is 8. Assuming ratio complete combustion of methane, the the air molar percentage of N2 in the standard efficiency (in percent) is products is ______.of specific heats (γ) as 1.4, [GATE-2014 Set-4] [GATE-2017 Set-1] Q.23 An engine working on air standard Q.19 Air enters a diesel engine with a Otto cycle is supplied with air at 0.1 density of 1.0 kg/m3. The compression ratio is 21. At steady MPa and 350 C . The compression ratio state, the air intake is 30×10 kg/s is 8. The heat supplied is 500 kJ/kg. and the net work output is 15−3 kW. Property data for air: cp = 1.005 kJ/kg The mean effective pressure (kPa) is ______. K, cv = 0.718kJ/kg K, R = 0.287 kJ/kg [GATE-2015 (1)] K. The maximum temperature (in K) of the cycle is ______(correct to one Q.20 An air-standard Diesel cycle consists decimal place). of the following processes: 1-2 : Air is compressed [GATE-2018 Set-1] isentropically. 2-3 :Heat is added at constant pressure. 3-4 :Air expands isentropically to the original volume.

4-1:Heat is rejected at constant volume.

ANSWER KEY:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 (c) (b) (a) (b) (c) (a) (d) (d) (c) (d) (c) (b) (d) (a) 15 16 17 18 19 20 21 22 23 (d) 525 (b) (b) 73.8 1403

Q.1 (c) For same compression ratio and the Given that the intercooling is perfect same heat supplied, Otto cycle is P = P P most efficient and diesel cycle is least efficient. P =2 1 ×1163 P∴ = 4 bar� In practice, however, the 2 compression ratio of the Diesel √ P P Pressure2 ratio per stage =2 =3 = 4 engine ranges between 14 and 25 PP12 whereas that of the Otto engine between 6 and 12. Because of it the Q.2 (b) efficiency of Diesel cycle is higher WW mep = = efficiency than that of Otto engine. Vs VV 1C− 23.625×× 105 V Q.6 (a) = C actual volume 5.5VCC− V Volumetric efficiency = 5 swept volume 23.625× 10 VC = Va 4.5V = = 0.9 C V 5 s =5.25× 10 Pa V = 0.9V =5.250 bar Mass of air, a s ∴m =ρ V = 0.9V Q.3 (a) a air s s γ1− m = 0.05 × 0.9V = 0.045V 1 P× LAN η1= −  f mep s s r ηthermal = m× C.V 0.4 f 1 PV× =1 −  mep s 8.5 0.045v×× 45 106 = 0.5751 or 57.51% s ⇒P 0.3mep == 6.075bar

Q.4 (b) ∴ Equivalence ratio is defined as the Q.7 (d) actual fuel air ratio to stoichiometric γ1− vc fuel air ratio η1Otto = −  v1 F γ1−   A Vc φ = actual =1 −  F VVcs+  A Stoi Where V = clearance vol For both idling and peak power = 10% of V =0.1V c γ1− conditions the actual fuel-air ratio 0.1V =s − s s required is more than the η1Otto  0.1Vss+ V stoichiometric fuel-air ratio. 1.4− 1 ∴ 0.1 = 1−  Q.5 (c) 1.1 =0.6167=61.67%

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 3 Cp vc = 0.001m , γ = = 1.4 Q.8 (d) C Q.9 (c) v When cylinder 1 is not firing then Swept Volume power is 2102 kW and when all π 2 Vs =×= A L (D) × L cylinders are firing then power is 4 π =(0.2)23 ×= 0.25 0.00785m Hence power supplied by cylinder 4 3037 kW. - Compression ratio Similarly power supplied by vr vvcs+ No.1= 3037 2102 = 935 kW r = = cylinder - vv Similarly power supplied by cc 0.001+ 0.00785 cylinder No. 2 = 3037-2102 = 935 kW = Similarly power supplied by 0.001 cylinder No. 3 = 3037- 2100 = 937 kW = 8.85 Air standard efficiency IPtotal =+++= 935 935 937 939 3746 No.4 = 3037 2098 = 939 kW 11 =−=− 3037 η1γ−− 1 1 1.4 1 Hence ηmech = (r) (8.85) 935+++ 935 937 939 1 BP =−=−= η = 1 1 0.418 0.582 mech IP 2.39 or 58.2 % 3037 = = 0.8107 3746 Q.12 (b)

=81.07%

mech Q.10 (d)∴ η r1− Vc η1Otto = −  VVcs+ V = 196.3cc Assumptions of air standard otto ππ cycle C 22 Vs =×=×× D L 10 15 a) All processes are internally 44 reversible. =1178.097cc b) Air behaves as ideal gas =1.4 c) Specific heats remains constant 1.4− 1 196.3 (Cp & Cv) 𝛾𝛾 η1Otto = −  196.3+ 1178.097 d) Intake process is constant ∴=0.5408 volume heat addition process = 54.08% and exhaust pro-cess is constant volume heat rejection process. =∴ ηOtto× (Heat supplied) Intake process is a constant =∴ 0.5408Work output ×1800 volume heat addition process, Otto from the given options; option η =973.58kJ (2) is incorrect. Q.11 (c) Given L=250 mm = 0.25 m, D = 220 Q.13 (d) m = 0.2 m.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission P LAn Power = m kw 60,000 As four stroke N n = =1100rpm 2 950× 60000 P = P = 100 kPa m 0.0259× 1100 Compression ratio, = 2 ×10 N/m = 2 MPa 1 r = 10 6 2 T = Q.15 (d) Heat added, Given : r = 60 mm, D = 80 mm= 8 cm 1 Q 27 + 273 = 300K Stroke length, Heat rejected, L = 2r = 2 × 60 = 120 mm =12 cm S Q = 1500kJ/kg Swept Volume, V = A × L Mean effective pressure R ππ22 Work=700kJ/kg done per cycle = D×= L (8)s × 12 = 44 Swept volume = 603.18 603 cm Compression ratio, 3 r = V /V = 10 Q.16 (1400 to 1420)≃ V = 10V Swept1 volume2 =1 V V2 V = V1 −=12 0.9V 11−10 For initial air P V = RT Given, RT P = 0.1 MPa 1 1 11 1 V1 = P1 T1 ⇒0.287kJ / kgK× 300K = 1.4 = = 308 K 100kPa r𝛾𝛾 = 8 = 0.861m /kg TR = 288.8 J/kg K Swept volume3 = 0.9 V R 288.8 = 0.9(0.861) C3 = = 1 v= 2660 K = 0.7749m /kg γ− 1 0.4 Work done3 in cycle, γ1− W = Q Q TV21 0.4 = 722 J/kgK == ()8 net out supply rej TV12 Mean effective pressure− = ×=0.4 = 1500W − 700 = 800 kJ/kg T2 308 8 707.6K = net Swept volume Qs = C = 722(2660 707.6) 800 v ∆T Pmep. = 0.7749 − P . = 1032.39 kPa Q.17 (58= 1409.6 to 62) kJ/kg mep Q.14 (a)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission γ VP 12= = 32.42 VP21 ∴ V Or 1 =(32.42)1/1.4 = 11.999 ≈ 12 V2 V V 12 =3 =×=1 rc 1.5 V21 8V V1 =17 γ 1 r1− V2 = − c η1d γ1−  C r γ(rc − 1) p =γ = 1.4 C 1 1.51.4 − 1 v = 1− 0.4 × or V32−= V 0.1(V 12 − V ) 12 1.4 0.5 = 0.596 or 59.6 % V3 V1 or −=1 0.1 − 1 VV 22 Q.19 (525) V or 3 =0.1 × 16 += 1 2.6 V2 Q.20 (b) γ Heat applied,Q = C (T T ) 1 r1− η1= − c Heat rejected, Q = C (T T ) Diesel γ1−  s p 3 2 r γ(rc − 1) − r v 4 1 1 2.61.4 − 1 − = − 1 0.4  17 1.4(2.6− 1) 1 3.81− 1 = 1−  170.4  1.4× 1.6 = 0.596 or 59.6%

Q1 (T− T ) Q.18 (59 to 61) η =1 − r=1 − 41 Qsγ (T 32− T )

Q.21 (b) Q.22 (73.83)

100 Ltr air contains 79 ltr of N2 and 21 ltr of O2

The combustion equation with 50% excess air is Given CH+× 1.5 2[3.762N + O ] P1= 1 bar, P2 4 22 → + +× + Cp 2H2 O CO 2 3 3.3762N 22 O =1.4=32.42bar C V It is assumed that nitrogen is insert γVV = 41= = 8 and does not participate in the VV33 reaction. Whatever nitrogen is there in reactants, same will come out in PVγγ= PV the products. for11 process 2 2 1 ⟶2, On the basis of Molar or volume

N2 = 73.83%

Q.23 (1403.9)

o P11= 0.1 MPa, T = 35 C = 308K V 1 =n8 = V2

Qs = 500 kJ/kg

cp = 1.005kJ/kgK

cv = 0.718 kJ/kgK R = 0.287 kJ/kgK c γ =p = 1.399 1.40 cv

TT3= max = ? For process1− 2

For process 1− 2 γγ PV11= PV 22 γ V1 1.4 P21= P = 0.1 () 8 V2

P2 = 1.8379 MPa

PV11 PV 22 PV 22 and= ⇒= T21 T T1 T 2 PV 11 1.8379 1 T=  308 2 0.1 8

T2 = 707.6K For process 2→ 3

Qs= C v3() T −= T 2 500 kJ/kg

Q.1 An industrial heat pump operates Q.4 The COP of the refrigerator is a) 2.0 b) 2.33 between the temperatures of 270 C and c) 5.0 d) 6.0 −130 C . The rates of heat addition and heat [GATE-03] rejection are 750 W and 1000 W, respectively. Q.5 In the window air conditioner, the The COP for the heat pump is expansion device used is a) 7.5 b) 6.5 a) capillary tube c) 4.0 d) 3.0 b) thermostatic expansion valve [GATE-03] c) automatic expansion valve d) float valve Q.2 For air with a relative humidity of [GATE-04] 80% a) the dry bulb temperature is less than Q.6 During the chemical the wet bulb temperature dehumidification process of air b) the dew point temperature is less a) dry bulb temperature and specific than wet bulb temperature humidity decreases c) the dew point and wet bulb b) dry bulb temperature increases and temperature are equal specific humidity decreases d) the dry bulb and dew point c) dry bulb temperature decreases and temperature are equal specific humidity increases d) dry bulb temperature and specific Common Data For Q.3 and Q.4 humidity increases A refrigerator based on ideal vapour [GATE-04] compression cycle operates between the 0 0 temperature limits of −20 C and 40 C . Q.7 Environment friendly refrigerant The refrigerant enters the condenser R134 is used in the new generation as saturated vapour and leaves as domestic saturated liquid. The enthalpy and refrigerators. Its chemical formula is entropy a) CHClF2 b) C2Cl3F3 values for saturated liquid and vapour c) C2Cl2F4 d) C2H2F4 at these temperatures are given in the [GATE-04] table below. Q.8 A heat engine having an efficiency of 70% is used to drive a refrigerator having Q.3 If refrigerant circulation rate is a coefficient of performance of 5. The 0.025 kg/s, the refrigeration effect is energy absorbed from low temperature equal to reservoir by the refrigerator for each kJ a) 2.1 kW b) 2.5 kW of energy absorbed from high c) 3.0 kW d) 4.0 kW temperature [GATE-03] source by the engine is a) 0.14 kJ b) 0.71 kJ

c) 3.5 kJ d) 7.1 kJ [GATE-04]

Q.9 Dew point temperature of air at one atmospheric pressure (1.013 bar) is 180 C . The air dry bulb temperature is 0 . 30 C The saturation pressure of water at a) × −33 b) × −33 0 6.35 10 m / s 63.5 10 m / s 18 C c) × −33 d) × −33 0 635 10 m / s 4.88 10 m / s and 30 C are 0.02062 bar and 0.04241 [GATE-04] bar respectively. The specific heat of air and water vapour respectively are 1.005 and 1.88 kJ/kg K and the latent heat of Q.11 For a typical sample of ambient air 0 vaporization of water at 0C is 2500 (at 350 C , 75% relative humidity and kJ/kg. The specific humidity (kg/kg of standard dry atmosphere pressure), the amount of air) and enthalpy (kJ/kg or dry air) of moisture in kg per kg of dry air will be this moist air respectively, are approximately a) 0.01051, 52.64 b) 0.01291, a) 0.002 b) 0.027 63.15 c) 0.25 d) 0.75 c) 0.01481, 78.60 d) 0.01532, [GATE-05] 81.40 [GATE-04] Q.12 Water at 420 C is sprayed into a stream of air at atmospheric pressure, Q.10 A R-12 refrigerant reciprocating dry bulb compressor operates between the temperature of 0 and a wet bulb condensing 40 C 0 0 temperature of . The air leaving temperature of 30 C and evaporator 20 C the temperature of − 0 . The clearance 20 C spray humidifier is not saturated. Which volume ratio of the compressor is 0.03. of the following statements is true ? Specific heat ratio of the vapour is 1.15 a) Air gets cooled and humidified and the specific volume at the suction is b) Air gets heated and humidified 0.1089 m3/kg. Other properties at c) Air gets heated and dehumidified various states are given in the figure. To d) Air gets cooled and dehumidified realize 2 tons of refrigeration, the actual [GATE-05] volume displacement rate considering

the effect of clearance is Q.13 The vapour compression refrigeration cycle is represented as shown in the figure below, with state 1 being the exit of the evaporator. The coordinate system used in this figure is

a) p-h b) T -s c) p-s d) T –h [GATE-05]

Q.14 Various psychometric processes a) 167 b) 100 are shown in the figure below c) 80 d) 20 [GATE-05]

Q.16 Dew point temperature is the temperature at which condensation begins when the air is cooled at constant a) volume b) entropy c) pressure d) enthalpy [GATE-06] Q.17 The statements concern psychometric chart. 1. Constant relative humidity lines are uphill straight lines to the right 2. Constant wet bulb temperature lines are downhill straight lines to the right The matching pairs are 3. Constant specific volume lines are a) P-(i), Q-(ii), R-(iii), S-(iv), T-(v) downhill straight lines to the right b) P-(ii), Q-(i), R-(iii), S-(v), T-(iv) 4. Constant enthalpy lines are coincident c) P-(ii), Q-(i), R-(iii), S-(iv), T-(v) with constant wet bulb temperature d) P-(iii), Q-(iv), R-(v), S-(i), T-(ii) Lines Which of the statements are [GATE-05] correct ? a) 2 and 3 b) 1 and 2 Q.15 A vapour absorption refrigeration c) 1 and 3 d) 2 and 4 system is a heat pump with three [GATE-06] thermal reservoirs as shown in the figure. A Q.18 A building has to be maintained at refrigeration effect of 100 W is required 0 0 21 C (dry bulb) and 14.5 C (wet bulb). at The dew point temperature under these 250 K when the heat source available is conditions is 0 . The outside at 400 K. Heat rejection occurs at 300 K. 10.17 C

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission temperature is −230 C (dry bulb) and bulb temperature of 200 C is humidified the internal and external surface heat in an air washer operating with transfer coefficients are 8 W/m2 K and continuous water recirculation. The wet 23 W/m2 K respectively. If the building bulb depression (i.e. the difference wall has a thermal conductivity of 1.2 between the dry and wet bulb W/m K, the minimum thickness (in m) temperature) at the exit is 25% of that of at the inlet. The dry bulb the wall required to prevent temperature at the exit of the air washer condensation is is closest to a) 0.471 b) 0.407 a) 100 C b) 200 C c) 0.321 d) 0.125 c) 250 C d) 300 C [GATE-07] [GATE-08]

Q .19 Atmospheric air at a flow rate of 3 Q.22 In an ideal vapour compression kg/s (on dry basis) enters a cooling and refrigeration cycle, the specific enthalpy dehumidifying coil with an enthalpy of of refrigerant (in kJ/kg) at the following 85 kJ/ kg of dry air and a humidity ratio states is given as: of 19 grams/kg of dry air. The air leaves Inlet of condenser :283 the coil with an enthalpy of 43 kJ/kg of Exit of condenser :116 dry air and a humidity ratio of 8 Exit of evaporator :232 grams/kg of dry air. If the condensate The COP of this cycle is water a) 2.27 b) 2.75 leaves the coil with an enthalpy of 67 c) 3.27 d)3.75 kJ/kg, the required cooling capacity of the [GATE-09] coil in kW is a) 75.0 b) 123.8 Q.23 A moist air sample has dry bulb 0 c) 128.2 d) 159.0 temperature of 30 Cand specific [GATE-07] humidity of 11.5 g water vapour per kg dry air. Assume molecular weight of air as Q.20 Moist air at a pressure of 100 kPa 28.93. If the is compressed to 500 kPa and then saturation vapour pressure of water at 0 cooled to 350 C in an aftercooler. The air 30 C is 4.24 kPa and the total pressure at the entry to the aftercooler is is unsaturated and becomes just saturated 90 kPa, then the relative humidity (in at the exit of the aftercooler. The %) of air sample is saturation pressure of a) 50.5 b) 38.5 water at 350 C is 5.628 kPa. The partial c) 56.5 d) 68.5 pressure of water vapour (in kPa) in the [GATE-10] moist air entering the compressor is Q.24 If a mass of moist air in an airtight closest to vessel is heated to a higher temperature, a) 0.57 b) 1.13 then c) 2.26 d) 4.52 a) specific humidity of the air increases [GATE-08] b) specific humidity of the air decreases c) relative humidity of the air increases Q.21 Air (at atmospheric pressure) at a d) relative humidity of the air decreases 0 dry bulb temperature of 40 C and wet [GATE-11]

Q.25 The rate at which heat is extracted, in kJ/s from the refrigerated space is relevant properties are given below. The a) 28.3 b) 42.9 enthalpy (in kJ/kg) of the refrigerant at c) 34.4 d) 14.6 isentropic compressor discharge is ______[GATE-12] Q.26 The power required for the compressor in kW is a) 5.94 b) 1.83 c) 7.9 d) 39.5 [GATE-12] Q.27 The pressure, dry bulb temperature and relative humidity of air in a room are 0 1 bar, 30 C and 70%, respectively. If the [GATE- 0 saturated pressure at 30 C is 4.25 kPa, 2014 (2)] the specify humidity of the room air in kg water vapour/kg dry air is Q. 31 A reversed Carnot cycle a) 0.0083 b) 0.0101 refrigerator maintains a temperature of c) 0.0191 d) 0.0232 − 5°C. The ambient air temperature is 35°C. The heat gained by the [GATE-13] refrigerator at a continuous rate is 2.5 Q.28 which one of the following is a kJ/s. The power (in watt) required to CFC refrigerant ? pump this heat out continuously is ____ . (a) R744 (b) R290 [GATE-2014 (4)] (C) R502 (d) R718 Q.32 Air in a room is at 35°C and 60% relative humidity (RH). The pressure in [GATE-2014 (1)] the room is 0.1 MPa. The saturation Q.29 A sample of moist air at a total pressure of water at 35°C is 5.63 kPa. pressure of 85 kPa has a dry bulb The humidity ratio of the air (in 0 temperature of 30 C (saturation vapour gram/kg of dry air)is______. pressure of water = 4.24 kPa). If the air sample has a relative humidity of 65%, [GATE-2015(3)] the absolute humidity (in gram) to Q.33 Refrigerant vapor enters into the water vapour per kg of dry air is______. Compressor of a standard vapor o [GATE-2014 (3)] compression cycle at - 10 C (h = 402 kJ/kg) and leaves the compression at Q.30 A heat pump with refrigerant R22 50°C (h = 432 kJ/kg). It leaves the is used for space heating between condenser at 30°C (h = 237 kJ/kg). The temperature limits of − 20°C and 25°C. COP of the Cycle is ______The heat required is 200 MJ/h. Assume specific heat of vapour at the time of [GATE-2015(3)] discharge as 0.98 kJ/kg-K. Other

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission Q.34 The thermodynamic cycle shown kg of dry air (gw/kgda) is ______. figure (T- diagram) indicates [GATE-2016 (3)] Q.38 A refrigerator uses R-134 a as its refrigerant and operates on a ideal vapour-compression refrigeration cycle between 0.14 MPa and 0.8 MPa. If the mass flow rate of the refrigerant is 0.05 kg/s the rate of heat rejection to the environment is ____kW. Given data : At P = 0.14 MPa, h 236.04 kJ/kg, s = 0.9322 kJ/kgK At P = 0.8 MPa, h = 272.05 kJ/kg (a) reversed Carnot cycle (superheated vapour) (b) reversed Brayton cycle At P = 0.8 MPa, h = 93.42 kJ/kg(saturated liquid) (c) vapor compression cycle [GATE-2016 (2)] (d) vapor absorption cycle

[GATE-2015(3)] Q.39 In the vapour compression cycle Q.35 The COP of a Carnot heat pump shown in the figure, the evaporating and operating between 6°C and 37°C condensing temperatures are 260 K and is ______310K, respectively. The compressor [GATE-2015(2)] takes in liquid-vapour mixture (state 1) and isentropically compresses it to a dry saturated vapour condition (state 2). Q.36 A stream of moist air (mass flow The specific heat of the liquid rate 10.1 kg/s) with humidity ratio of refrigerant is 4.8 kJ/kgK and may be 0.01 kg/kg dry air mixes with a second treated as constant. The enthalpy of stream of superheated water vapour evaporation for the refrigerant at 310 K flowing at 0.1 kg/s. Assuming proper is 1054 kJ/kg. and Uniform mixing with no condensation, the humidity ratio of the final stream in (kg/kg dry air) is ______. [GATE-2015(1)] Q.37 In a mixture of dry air and water vapor at a total pressure of 750 mm of Hg, the partial pressure of water vapor is 20 mm of Hg. The humidity ratio of the air in grams of water vapor per

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission The difference between the enthalpies b) relative humidity increases at state point 1 and 0 (in kJ/kg) c) specific humidity increases is______. d) specific humidity decreases [GATE-2016 (3)]

Q.40 The partial pressure of water [GATE-2017 (2)] vapour in a moist air sample of relative humidity Q.43 Ambient air is at a pressure of 100 kPa, dry bulb temperature of 30 0 C 70% is 1.6 kPa, the total pressure being and 60% relative humidity. The 101.325 kPa. Moist air may be treated as saturation pressure of water at 30 0 C is an ideal gas mixture of water vapour 4.24 kPa. The specific humidity of air (in and dry air. The relationbetween g/kg of dry air) is ______(correct to two saturation temperature (Ts in K) and decimal places). saturation pressure (Ps in kPa) for water is given [GATE-2018 (1)] Q.44 A standard vapor compression p 14.317− 5304 by ln s = refrigeration cycle operating with a 0 pT0 s condensing temperature of 35 C and an evaporating temperature of -10 0 C where p0 = 101.325 kPa. The dry bulb develops 15 kW of cooling. temperature of the moist air sample (in The p-h diagram shows the enthalpies at oC) is______various states. If the isentropic efficiency of the compressor is 0.75, the [GATE-2016 (2)] magnitude of compressor power (in Q.41 Moist air is treated as an ideal kW) is ______(correct to two decimal places). gas mixture of water vapor and dry air [GATE-2018 (2)] (molecular weight of air = 28.84 and molecular weight of water = 18). At a location, the total pressure is 100 kPa, the temperature is 30°C and the relative humidity is 55%. Given that the

saturation pressure of water at 30°C is

4246 Pa, the mass of water vapor per kg of dry air is ______grams.

[GATE-2017 (1)]

Q.42 If a mass of moist air contained in

a closed metallic vessel is heated,

then its a) relative humidity decreases

ANSWER KEY:

1 2 3 4 5 6 7 8 9 10 11 12 13 (c) (b) (a) (b) (a) (b) (d) (c) (b) (a) (b) (b) (a) 14 15 16 17 18 19 20 21 22 23 24 25 26 (b) (c) (c) (a) (b) (c) (b) (c) (a) (b) (d) (a) (c) 27 28 29 30 31 32 33 34 35 36 37 38 39 (c) (b) 20.84 (b) 433.3 373.13 21.74 5.5 (b) 10 0.02 17.04 8.93

27 28 29 30 31 32

1103.51 19.89 14.9 (a) 16.236 10

Q.1(c)

Q1 1000 So, (COP)H.P. = = = 4 Q12−− Q 1000 750

Q.2 (b) We know that for saturated air, the relative humidity is 100% and the dry bulb temperature, wet bulb temperature and dew point temperature is same. But when air is not saturated, dew point temperature is always less than the wet bulb temperature. DPT < WBT

Q.3(a)

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission 0 Given : T14 = T =- 20 C = (- 20 + 273)K = 253 K, • m = 0.025 kg/ sec 0 T23 =T = 40 C = (40 + 273)K = 313 K From the given table, 0 At, T22 = 40 C, h = 200 kJ/kg

And h34 = h = 80 kJ/kg From the given T s - curv e

s12 = s

s2 = s f + xs fg

{s22 is taken 0.67 because s at the temperature 400 C & at 2 high temperature In the process of chemical dehumidification of air , the air is passed and pressure vapour refrigerant exist.} over chemicals which have an affinity for 0.67 = 0.07 + x(0.7366 - 0.07) moisture and the moisture of air gets x=0.90 condensed out and gives up its latent And Enthalpy at point 1 is, heat. Due to the condensation, the specific humidity decreases and the heat h1 = h f + xh fg of condensation supplies sensible heat for = 20 + 0.90(180 - 20) = 164 kJ/kg heating the air and thus increasing its dry Now refrigeration effect is bulb temperature. produce in the evaporator. So chemical dehumidification increase dry Heat extracted from the bulb temperature & decreases specific humidity evaporator or refrigerating effect,

•

RE = m (h1- h4 ) Q.7 (d) If a refrigerant is written in the = 0.025(164 - 80) = 2.1 kW form of Rabc. The first digit on the right (c) is the

number of fluorine (F) atoms, the second Q.4 (b) digit from the right (b) is one more than Refrigerating effect (COP)refrigerator = the number of hydrogen (H) atoms Work done required & third digit from the right (a) is hh− 164− 80 =14 = = 2.33 one less than the Number of carbon h21−− h 200 16 (C) atoms in the refrigerant. So, For R134 [GATE-03] First digit from the Right = 4 = Number of Fluorine atoms Q.5 (a)Air conditioner mounted in a Second digit from the right = 3 - 1 = 2 = window or through the wall are self- Number of hydrogen atoms contained units of small capacity of 1 TR Third digit from the right = 1 + 1 = 2 = to 3 TR. The capillary tube is used as an Number of carbon atoms expansion device in small capacity Hence, Chemical formula is C2H2F4 refrigeration units

Q.6 (b)

h= 1.022 Tdb ++ω () h fgdp 2.3 tdp KJ h= 63.15 kg

Q.10 (a) Q.11 (b) From steam table, saturated air pressure corresponding to dry bulb temperature ()()COP= 5, η = 70% = 0.7 0 ref H.E of 35 C is ps = 0.05628 bar. Q Relative humidity COP=3 = 5...... (i ) ()ref P W Relative humidi t y φ= v W Pvs = = ()ηH.E 0.7...... (ii ) Q1 Pv 0.75= ⇒= Pv 0.04221 bar Multiplying ()()i and ii Pvs

Q W pv 3 =5 × 0.7 Specific humidity ω = 0.622 ppbv− WQ1 0.04221 Q3 ω = 0.622  = 3.5 1.01− 0.04221 Q1 = Hence, Energy absorbed (Q3) 0.0271 kg/kg of dry air from low temperature reservoir by the refrigerator for each kJ of Q.12 (b)Given : energy absorbed (Q1) from high Hence air gets heated, Also water is added to temperature source by the it, so it gets humidified. engine= 3.5 kJ 00 0 tsp= 42 C, t db = 40 C, t wb = 20 C

Here we see that t> t sp db Q.9(b) 0 Given : tdp = 18 C = (273 + 18)K Q.13 (a)Given curve is the theoretical p-h = 291K, curve for vapour compression p = patm = 1.013 bar refrigeration cycle. 0 tdb = 30 C = (273 + 30)K = 303 K pv = 0.02062 bar (for water Q.14(b) vapour at dew point). cair = 1.005 kJ/kg K, Q.15(c) cwater = 1.88 kJ/kg K Latent heat of vaporization of Q.16(c) water at 00C. hfgdp = 2500 kJ/kg p Specific humidity,W = 0.622 × v pp− v kg W = 0.01291 kg of dry air

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission limiting condition to prevent condensation 0 Ts1=10.17 C Here Ts1 & Ts2 are internal & external wall surface temperature of building. Hence, heat flux per unit area inside the building, Q It is the temperature of air recorded by a qi= = hT1() DBT 11 − T s

thermometer, when the moisture (water A …….(i) = 2 vapour) present in it begins to condense. qi 86.64 W/m If a sample of unsaturated air, containing superheated water vapour, is cooled at Heat flux per unit area outside constant pressure, the partial pressure the building is = ( - ) (pv) of each constituent remains constant q0 h2 Ts2 TDBT2 until the water vapour reaches the = 23(Ts2+ 23) …....(ii) saturated state as shown by point B. At Heat flow will be same at inside & outside the building. So from this equation (i) & (ii) point B the first drop of dew will be qi = q0 formed and hence the temperature at 0 Ts2 = -19.23 C point For minimum thickness of the B is called dew point temperature wall, use the Fourier’s law of conduction for the building. Heat Q.17(a) flux through wall,

kT− T Q.18 (b) ()ss12 q = x x = 0.407 m

Q.19 (c) Given : ma = 3 kg/sec, Using subscript 1 and 2 for the inlet and outlet of the coil

Let h1 & h2 be the internal and respectively. external surface heat transfer h1 = 85 kJ/kg of dry air, coefficients respectively and W1 = 19 grams/kg of dry air building wall has thermal =19 ×10-3 kg/kg of dry air conductivity k . h2 = 43 kJ/kg of dry air, Given : h1= 8 W/m K W2 = 8 grams/kg of dry air 2 -3 h2 =23W/m K, = 8 ×10 kg/kg of dry air k=1.2 W/m K, h3 = 67 kJ/kg 0 TDPT=10.17 C Mass flow rate of water vapour at Now to prevent condensation, the inlet of the coil is, -8 temperature of inner wall should mv1 = W1 ×m1 = 57 ×10 kg/sec be more than or equal to the dew And mass flow rate of water point temperature. It is the vapour at the outlet of coil is, mv2 = W2 ×ma

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission -8 mv2 =24 ×10 kg/sec tWBT(inlet) = tWBT(outlet) So, mass of water vapour So, tDBT(exit)- 20 = 0.25*20 0 condensed in the coil is, tDBT(exit) = 20 + 5 = 25 C Therefore, required cooling capacity of the coil = change in enthalpy of dry air + Q.22 (a) change in enthalpy of condensed p-h curve for vapour water compression refrigeration cycle = (85 - 43) ×3 + 67 ×33 ×10-3 is as follows = 128.211 kW

Q.20 (b) Given : p1 = 100 kPa, p2 = 500 kPa, pv1 = ? pv2 = 5.628 kPa (Saturated pressure at 350C) We know that, p = × v Specific humidity,W 0.622 The given specific enthalpies are pp− v Inlet of condenser h2 = 283 kJ/kg For CASE-II kJ 0.628 Exit of condenser h3 = 116 W = 0.622 kg 500− 5.628 =h4 W=7.08 × 10−8 kg / kg of dry air (from p-h curve) For saturated air specific Exit of evaporator h1 = 232 kJ/kg humidity remains same. So, for Now, Refrigeration Effect case (I) : COP= p Work Done W =0.622 × v1 − hh− pp11v =14 = 2.27 On substituting the values, we hh21− get p=1.13 kPa v1 Q.23 (b)

0 Given : tDBT = 30 C, W = 11.5 g water vapour/kg dry Q.21 (C) air 0 Given: At inlet tDBT =40 C ps = 4.24 kPa, p = 90 kPa 0 tWBT=20 C We know that, Wet bulb depression = tDBT -tWBT = 40 - 20 = 20 0C And given wet bulb depression at the exit = 25% of wet bulb depression at inlet This process becomes adiabatic saturation and for this process,

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission p mass flow rate of the refrigerant is 0.2 = × v Specific humidity,W 0.622 kg/s. Properties for R134a are as follows : pp− v

− p 11.5×= 108 0.622v 90 − pv

pv =1.634 kPa p Relative humidity,()φ = v ps 1.634 = 4.24 Q.25 (a) φ = 0.385 = 38.5% Q.4 (d) T-S From the given curve, we easily see that relative humidity of air decreases, when temperature of moist air in an airtight vessel increases. So, option (d) is correct. Specific humidity remain constant with temperature increase, so option a & b are incorrect

T-S Diagram for given Refrigeration cycle is given above Since Heat is Extracted in Evaporation process.

So rate of heat extracted = mh()14− h

From above diagram (h3 = h4) for throttling process, so Heat extracted = m (h1- h3) From given table h1 = hg at 120 kPa, hg = 237 kJ/kg h3 = hf at 120 kPa, hf = 95.5 kJ/kg Hence Heat extracted = m (hg - hf) = 0.2(237 - 95.5) = 28.3 kJ/s

A refrigerator operates between 120 kPa Q.26 (c) and 800 kPa in an ideal vapour Since power is required for compressor compression cycle with R-134a as the in refrigeration is in compression cycle refrigerant. The refrigerant enters the (1-2) compressor as saturated vapour and Hence, Power required = h2-h1 leaves the condenser as saturated liquid. = h2-hf The Since for isentropic compression process.

© Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission s1 = s2 from figure. = 0.95 Total pressure : p = 85kPa For entropy s = 0.95 the enthalpy Dry bulb temperature, h = 276.45 kJ/kg = o h = h2 = 276.45 (From table) Tdb 30 C Hence Power = 0.2(276.45 - 237) ps = 4.24kPa = 7.89 = 7.9 kW φ =65 % = 0.65 Q.27 (c) Relative humidity, Specific Humidity is given by p φ = v p ps ω =0.622 × v …...(i) pp− p av 0.65 = v 4.24 Where, or p =0.65 4.24 = 2.756 kPa Relative  Saturated  v pv =  ×  humidity  steam pressure  Absolute humidity, =×=×φ p 0.7 0.0425 0.622 p s ω = v = 0.02975 bar pp− v = 0.02084 kg of w.v/kg of dry air So that from equation (i), we have = 20.84 gramof w.v./kg of dry air 0.02975 ω =×= 0.622 Pa 1 bar ω <ω Dehumidifier 1− 0.02975 21 Q = 0.0191 kg/kg of dry air Q.30 Q.28

R744-CO2 R290-C H() Propane 33

R502-CHCIF3+ CCIF 23 CF R718-Water

Q.29

T2 ss22−=' c p log e T2'

ss12= ()for isentropic process

s2 = 1.7841 kJ/kgK s= 1.7183 kJ/kgK 2

cp = 0.98 kJ/kgK

T2ss 22− ' ∴=loge Tc2p'

T2 1.7841− 1.7183 loge = = 0.0671 T2' 0.98

= 318.695K Tdb = 35°C

h2−= h 2'' cT p2() − T 2 φ = 60 % = O.60 h=+− 413.02 0.98 318.695 298 2 () p = 0.1 MPa = 100 kPa

h2 = 433.3 kJ/kg ps = 5.63 kPa at 35°C

Q.31 Relative humidity, Given data : p φ = v T2 =−° 5 C =() −+ 5 273 K = 268 K ps

T= 35 °= C 35 + 273 K = 308 K pv 1 () 0.60 5.63

pv =0.60 5.63 = 3.378 kPa Humidiy ratio, 0.622 p 0.622 3.378 ω== v pp−−v 100 3.378

= 0.02174 kg of w.v./kg of dry air

= 21.74 gram of w. v./kg of dry air

Q.33 Given data :

h12=402kJ/kg, h = 432kJ/kg

h34= h = 237 kJ/kg

Q= 2.5kJ/s = 2.5 kW = 2500 W 2 T = 2 ()COP R TT12− 268 ()COP= = 6.7 R 308− 268 Q also() COP = 2 W hh− 402− 237 165 2500 COP = 14= = ∴=6.7 hh−−432 402 30 W 21 = 2500 5.5 or W == 373.13 watt 6.7 Q.34 (b)

Q.32 Reversed Brayton cycle is shown in the figure

10.1= m + 0.01 m aa 10.1 ma = 10.1

10.1 Mass of dry air m = = 10 kg/s a 1.01 Mass of water vapour,

mv1 = 10.1 −= 10 0.1 kg/s m= 0.1 kg/s Q.35 v2

()mv=+= m v1 mv2 0.2 kg/s Given data : total o T= 6 C =() 6 + 273 K= 279K mv 0.2 2 Humidity ratio ω=final = o ma 10 T1 = 37 C =() 37 + 273 K= 310K = 0.02 kg/kg of dry air

Q.37

Given data :

Total pressure,

p = 750mm of Hg

Partial pressure, V

Pv = 20mm of Hg

We know that humidity ratio,

0.622 p ω= v kg. w.v./kg d.a. pp− v 0.622 20 = 750− 20 = 0.01704 kg w.v./kg of d.a. T 310 = 17.04 g of w.v./kg of d.a. ()COP =1 = HP −− T12 T 310 279 = 10 Q.38

Given data: Q.36 m = 0.05kg/s Mass flow rate of moist air = 10.1 kg/s Pe = 0.14 MPa m Humidity ratio : ω=v =0.01 kg/s h1 = 236.04 kJ/kg ma

T3 h3 = h4 = hf = 93.42 kJ/kg s= s + c ln 3T T 310 = s + 4.8 ln T T Similarly, 260 s= s + c ln 0T T 260 = s + 4.8 ln T T

hfg 1054 ∴ss23 −= = h2 = 272.02 kJ/kg T 310 1054 Rate of heat rejection to the environment, ss= + 23310

Q2-3=mh 2 −= h 3 0.05 [] 275.02 − 93.42 Substitute the value of s  3 = 8.93 kW in above equation, we get ss= Q.39 21 310 1054 Given data : s1T=++ s 4.8 ln T 310

Te = 260K ∴ −= − h1h 0 Tse  10 s

TC = 310K 310 1054 260 =260 ss + 4.8 ln + −− 4.8ln TTT 310 T cpl = 4.8 kJ/kgK 310 T 1054 hfg = 1054 kJ/kg at TC = 310 K =260 4.8ln  + T 260 310 = 1103.51 kJ/kg

Q.40

Given data :

Relative humidity,

φ = 70 % = 0.70

Partial pressure, Taking reference temperature T, at which entropy pv = 1.6 kPa

is ST. Total pressure,

p0 = 101.325 kPa

We know that,

1.6 mv 0.70 = Q ω= p ma s 1.6 = ω or p = = 2.2857 kPa mmva s 0.70 =0.0149 ×× 103 1 Temperature corresponding to saturation

pressure mv = 14.9grm

is dry bulb temperature, Q.42 (a) For closed metallic vessel sp. Humidity ps 5304 ln= 14.317 − (ω) remain const po Ts 2.2857 5304 ln= 14.317 − 101.325 Ts 5304 −=−3.79166 14.317 Ts 5304 =14.317 + 3.79166 Ts 5304 = 18.10866 T s or T= 292.89K s From the figure =() 292.89− 273o C φ <φ = 19.89o C 21 Q.43

P Relative humidity φ = v Q.41 Ps

But,φ = 60 % = 0.6 ,Ps = 4.24 kPa Pv Relative humidity φ= ⇒=pasφ  p = 0.6 4.24 = 2.544 kPa Pvs p Specific humidity ω = 0.622 v − P ppbv 0.55 = v 4.246 where pb = ambient air pressure = 100 kPa = 2.544 Pv 2.335 ⇒ω=0.622   100− 2.544 M  HO2 P ⇒ω=0.016236kg/kg of dry air ω=v Mair PP t− v ⇒ω=16.236 g/kg of dryair

18 2.335 ω=  28.84 100− 2.335

. RC = m ()()hh14−

. RC = m() 400− 250

. m   kg/sec

Wisentropic=−=() hh 2 1 ()475 − 400 = 75kJ/kg

Wisentropic ηC = Wactual

. WWactual=m actual = 0.1 100

. = 10 kW Pm