TURBO MACHINES (Unit 1 & Unit 2)

DR. G.R. SRINIVASA

TURBO MACHINES VTU Syllabus

Subject Code : 06ME55 IA Marks : 25 No. of Lecture Hrs./Week : 04 Exam Hours : 03 Total No.of Lecture Hrs. : 52 Exam Marks : 100

PART – A

UNIT – 1 INTRODUCTION: Definition of a Turbomachine; parts of a Turbomachine; Comparison with positive displacement machine; Classification: Application of First and Second Laws to Turbomachines, Efficiencies. Dimensionless parameters and their physical significance; Effect of Reynolds number; Specific speed; Illustrative examples on dimensional analysis and model studies. 6 Hours

UNIT – 2 ENERGY TRANSFER IN TURBO MACHINE: Euler Turbine equation; Alternate form of Euler turbine equation – components of energy transfer; Degree of reaction; General analysis of a Turbo machine – effect of blade discharge angle on energy transfer and degree of reaction; General analysis of centrifugal pumps and compressors – Effect of blade discharge angle on performance; Theoretical head – capacity relationship 6 Hours

UNIT – 3 GENERAL ANALYSIS OF TURBO MACHINES: Axial flow compressors and pumps – general expression for degree of reaction; velocity triangles for different values of degree of reaction; General analysis of axial and radial flow turbines – Utilization factor; Vane efficiency; Relation between utilization factor and degree of reaction; condition for maximum utilization factor – optimum blade speed ratio for different types of turbines 7 Hours

UNIT – 4 THERMODYNAMICS OF FLUID FLOW AND THERMODYNAMIC ANALYSIS OF COMPRESSION AND EXPANSION PROCESSES: Sonic velocity and Mach number; Classification of fluid flow based on Mach number; Stagnation and static properties and their relations; Compression process – Overall isentropic efficiency of compression; Stage efficiency; Comparison and relation between overall efficiency and stage efficiency; Polytrophic efficiency; Preheat factor, Expansion Process – Overall isentropic efficiency for a turbine; Stage efficiency for a turbine; Comparison and relation between stage efficiency and overall efficiency for expansion process, polytrophic efficiency of expansion; Reheat factor for expansion process. 7 Hours

PART – B UNIT – 5 CENTRIFUGAL COMPRESSORS: Classification; Expression for overall pressure ratio developed; Blade angles at impeller eye root and eye tip; Slip factor and power input factor; width of the impeller channel; Compressibility effect – need for pre-whirl vanes; Diffuser design: Flow in the vaneless space, determination of diffuser inlet vane angle, width and length of the diffuser passages; Surging of centrifugal compressors; AXIAL FLOW COMPRESSORS: Classification; Expression for Pressure ratio developed per stage – work done factor, radial equilibrium conditions. 6 Hours

UNIT – 6 CENTRIFUGAL PUMPS: Definition of terms used in the design of centrifugal pumps like manometric head, suction head, delivery head, pressure rise, manometric efficiency, hydraulic efficiency, volumetric efficiency, overall efficiency, multistage centrifugal pumps, minimum starting speed, slip, priming, cavitation, NPSH. 6 Hours

UNIT – 7 STEAM TURBINES: Classification, Single stage impulse turbine; Condition for maximum blade efficiency, stage efficiency, Compounding – Need for compounding, method of compounding. Impulse Staging – Condition of maximum utilization factor for multi stage turbine with equiangular blades; effect of blades and nozzle losses. Reaction turbine; Parson’s reaction turbine, condition for maximum blade efficiency, reaction staging. 7 Hours

UNIT – 8 HYDRAULIC TURBINES: Classification; Pelton Turbine-velocity triangles, Design parameters, turbine efficiency, volumetric efficiency; Francis turbine – velocity triangles, runner shapes for different blade speeds, Design of Francis turbine; Function of a Draft tube, types of draft tubes; Kaplan and Propeller turbines – Velocity triangles and design parameters. 7 Hours

TEXT BOOKS: 1. An Introduction to energy conversion, Volume III – Turbo machinery, V. Kadambi and Manohar Prasad, New Age International Publishers (P) Ltd. 2. “Turbines, Compressors & Fans”, S.M. Yahya, Tata-McGraw Hill Co., 2 nd Edition (2002).

REFERENCE BOOKS: 1. “Principles of Turbo Machinery”, D.G. Shepherd, The Macmillan Company (1964). 2. Fundamentals of Turbomachinery: William W. Perg, John Wiley & Sons, Inc. 2008. 3. A Text book of Turbo Machines - M.S.Govindgouda & A.M. Nagaraj-M.M.Publications-IV Edition-2008 4. “Fluid Machinery” B.K. Venkanna, PHI.

In these lectures, we will learn different type of turbo machines, their action, as power generating turbo machines or power absorbing turbo machines.

You will be shown schematic diagrams of various turbo machines with flow directions.

DEFINITION: A turbo machine is a device in which energy transfer occurs between a flowing fluid and rotating element due to dynamic action. This results in change of pressure and momentum of the fluid.

TYPE: If the fluid transfers energy for the rotation of the impeller, fixed on the shaft, it is known as power generating turbo machine. If the machine transfers energy in the form of angular momentum fed to the fluid from the rotating impeller, fixed on the shaft, it is known as power absorbing turbo machine.

Examples of a turbo machine: The figures 1 & 2 show a typical turbo charger used in diesel engines to improve its thermal efficiency by increasing the pressure of air pumped into engine combustion chamber.

Fig. 1

Fig. 2

PARTS OF A TURBO MACHINE

Fig. 3(a) – Schematic diagram showing parts of a steam turbine

Fig. 3(b) - Schematic diagram of an impulse water turbine (Tangential flow)

PARTS OF A TURBO MACHINE

The principle components of a turbo machine are:

1. Rotating element (vane, impeller or blades) – operating in a stream of fluid. 2. Stationary elements – which usually guide the fluid in proper direction for efficient energy conversion process. 3. Shaft – which either gives input power or takes output power from fluid under dyn amic conditions and runs at required speed. 4. Housing – to keep various rotating, stationery and other passages safely under dynamic conditions of the flowing fluid. E.g. Steam turbine parts and Pelton turbine parts.

Fig. 3(c) Axial flow turbo machine

Fig. 3(d) Radial flow turbo machine

CLASSIFICATION OF TURBO MACHINES

1. Based on energy transfer a) Energy is given by fluid to the rotor - Power generating turbo machine E.g. Turbines b) Energy given by the rotor to the fluid – Power absorbing turbo machine c) E.g. Pumps, blowers and compressors

2. Based on fluid flowing in turbo machine a) Water b) Air c) Steam d) Hot gases e) Liquids like petrol etc.

3. Based on direction of flow through the impeller or vanes or blades, with reference to the axis of shaft rotation a) Axial flow – Axial pump, compressor or turbine b) Mixed flow – Mixed flow pump, Francis turbine c) Radial flow – Centrifugal pump or compressor d) Tangential flow – Pelton water turbine

4. Based on condition of fluid in turbo machine a) Impulse type (constant pressure) E.g Pelton water turbine b) Reaction type (variable pressure) E.g. Francis reaction turbine

5. Based on position of rotating shaft a) Horizontal shaft – Steam turbines b) Vertical shaft – Kaplan water turbines c) Inclined shaft – Modern bulb micro -hydel turbines

Fig. 4 (a) – Single stage axial flow pump or compressor

Fig. 4(b) – Kaplan turbine (axial flow) Fig. 4(c) – Mixed flow pump

Fig. 4 (d) – Modern Francis turbine (mixed flow type)

Fig. 4(e) – Centrifugal compressor or pump

Fig. 4 (f) – Bulb turbine (inclined shaft)

APPLICATION OF FIRST AND SECOND LAWS TO TURBO MACHINES

Fig. 5 – Steady flow energy process in turbo machine

STEADY FLOW ENERGY EQUATION – I Law of Thermodynamics Taking unit mass flow rate entering the turbo machine (1 Kg) at section 1 -1 and leaving at section 2 -2 through control volume as shown in figure.

2 2 u1+p 11+V 1 /2+gZ 1+q = w+u 2+p 22+V 2 /2+gZ 2 ...... (1) Where, u = Internal Energy (J/Kg) p = Pressure Intensity (N/m 2) v = Specific Volume (m 3/Kg) V = Velocity of the fluid (m/sec) Z = Potential head from datum (m) g = Acceleration due to gravity (m/sec 2) q = Heat transfer throu gh control volume (J/Kg) w = Work done (Nm/Kg)

In a Turbo machine, during the flow process, it is assumed to be adiabatic, i.e. no heat enters or leaves the system, hence heat transfer can be neglected i.e., q = 0.

Taking pv = RT and u = C vT, the equation (1) becomes

2 2 (C vT1+RT 1)+V 1 /2+gZ 1 = w+(C vT2+RT 2)+V 2 /2+gZ 2 ...... (2)

2 2 T1 (C v+R)+V 1 /2+gZ 1 = w+T 2 (C v+R)+V 2 /2+ gZ 2 ...... (3)

If the flow through the turbo machine is horizontal as shown in figure and aligned, Z 1 = Z2

Hence, rearranging equation (3) it becomes

2 2 w = (C v+R) (T 1 – T2)+(V 1 -V2 /2) ...... (4)

Taking C p – Cv = R, the above equation becomes

2 2 w = C p (T 1 – T2)+(V 1 -V2 /2) ...... (5) Taking enthalpy h = C pT, 2 2 w = (h 1 – h2)+(V 1 -V2 /2) per unit mass rate

2 2 w = (h 1+ V1 /2) – (h 2+V 2 /2) ...... (6) where, h = Static enthalpy V2/2 = Kinetic energy

Taking stagnation enthalpy = Static enthalpy + Kinetic energy, i.e. 2 h0 = h +V /2

The equation (6) becomes

w = ( h01 – h02 ) = - ∆h0 [Stagnation enthalpy change] Thus, in a turbo machine, we assume that there will be kinetic energy (high velocity) during flow and normally stagnation enthalpy change is considered under dynamic conditions.

In power generating turbo machines, 'w' will be positive and h0 will be negative i.e., stagnation enthalpy will be decreasing from inlet to outlet of a turbo machine rotor.

In a power absorbing turbo machine, 'w' will be negative and ∆h0 will be positive i.e., stagnation enthalpy will be increasing from inlet to outlet of a turbo machine rotor. It should be understood that work done or enthalpy change will occur only during transfer of energy through impellers or rotors and not through stators or fixed passages. Only pressure change, kinetic energy change or potential energy change will occur through stationery or stator passages depending on shape during the dynamic action of flow in the turbo machine.

COMPARISON BETWEEN POSITIVE DISPLACEMENT MACHINES AND TURBO MACHINES

Action: A positive displacement machine creates thermodynamic and mechanical action between near static fluid and relatively slow moving surface and involves in volume change and displacement of fluid as in IC engines.

A turbo machine creates thermodynamic and dynamic action between flowing fluid and rotating element involving energy transfer with pressure and momentum changes as shown in gas turbines.

Operation: The positive displacement machine commonly involves reciprocating motion and unsteady flow of fluids like in reciprocating IC engines or slow rotating fluids like in gear pumps. A turbo machine involves steady flow of fluid with pure rotary motion of mechanical elements. Only unsteadiness will be there during starting, stopping and changes in loads on the machine.

Mechanical features: A positive displacement machine commonly work at low speeds and involves complex mechanical design. It may have valves and normally will have heavy foundation. A turbo machine works at high speeds, simpler in design, light in weight, have less vibration problems and require light foundation.

Efficiency of energy conversion: A positive displacement machine gives higher efficiency due to energy transfer near static conditions either in compression or expansion processes. A turbo machine gives less efficiency in energy transfer. The energy transfer due to dynamic action will be less during compression process of fluid like pumps and compressors and will be slightly more during expansion processes like in turbines but still lower than reciprocating machines.

Volumetric efficiency: The volumetric efficiency of a positive displacement machine is low due to closing and opening of the valves during continuous operation. In turbo machines, since there are no valves under steady flow conditions, the volumetric efficiency will be close to 100 per cent. A turbo machine has high fluid handling capacity.

Weight to mass flow rate: A reciprocating air craft IC engine power engine developing 300 KW handles 2 kgs/sec of air weighs around 9500 N. Whereas, a rotary gas turbine of an air craft for same 300 KW power can handle 22 kgs/sec of air and weighs only 8000N handling more mass of air/sec. In stationary power plants, the specific weight of reciprocating power plants will be 10-15 times higher than the turbo power plants.

Fluid phase: Turbo machines have the phase changes occurring in fluid like cavitation in hydraulic pumps and turbines and surge and stall in compressors, blowers and fans if the machines are operated at off design condition leading to associated vibrations and stoppage of flow and damage to blades. Positive displacement machines have no such problems

EFFICIENCIES

Efficiency = Output (in same units) as percentage Input Power generating turbo machine

Efficiency adiabatic = Mechanical energy supplied to the rotor isentropic Hydrodynamic energy available from fluid hydraulic

Efficiency mechanical = Work output of the shaft Mechanical energy supplied to the rotor

Overall Efficiency overall = mechanical x adiabatic

adiabatic = t – t = ho1 – ho2 isentropic ho1 – ho2s hydraulic Total – Total (Stagnation – Stagnation) adiabatic = s – s = h1 – h2 isentropic h1 – h2s hydraulic

Static – Static If V 1 = V 2 t – t = s – s If V 1 V 2 t– t s – s

Fig. 6 (a) h–s diagram of power generating turbo machine (Expansion process)

adiabatic = ho2s – ho1 = t – t isentropic ho2 – ho1

hydraulic Total – Total (Stagnation – Stagnation)

adiabatic = h2s – h1 = s – s isentropic h2 – h1 hydraulic Static – Static

If V 1 = V 2 tt = ss If V 1 V 2 tt ss

ds is always positive as process is irreversible If ds 0 loss 0 If ds = 0 (Ideal) Process is reversible and is not possible as is violates II law of thermodynamics, i.e.

t – t cannot be equal to 100% and is always less than100% depending on the process. Fig. 6 (b) h–s diagram of power absorbing turbo machine (Compression process)

Problem No. 1 Air flows through an air turbine rotor where the stagnation pressure drops at a rate 5:1. Total to total efficiency is 0.8 and the air flow rate is 5 kgs/sec. If the total power output is 400 KW, = 1.4, and C p = 1004 J/kgK, find: i) Inlet total temperature ii) Actual exit total temperature if exit velocity is 100 m/sec and total to static efficiency of the rotor.

Solution Since this is an expansion process drop in heat or drop in temperature will give work to the rotor i.e.

– ∆h0 = w (J/kg = Nm/kg) Mass rate ṁ = 5 kg/sec ∴ w = P/ ṁ = 400/5 = 80 KJ/kg or KNm/kg w = – ∆h0 = – Cp (T 01 – T02 )

For isentropic expansion process, (γ - 1)/ γ 0.4/1.4

T01 = p 01 = 1 = 0.631 .... (1) T02s p02 5

∴ T01 – T02 = 80/1.004 = 79.681 K .... (2)

tt = (T 01 – T02 ) (T 01 – T02s )

∴ T01 – T02 = tt (T 01 – T02s ) = tt T01 (1 – 0.631)

∴ = 0.8 (1 – 0.631) T 01 = 0.295 T 01 .... (3)

From eqn. 1, 2 and 3

0.295 01 = 79.681 K

∴T01 = 79.691 / 0.295 = 270.105 K

∴ T02 = 270.105 – 79.681 = 190.424 K

2 Hence, T 2 = T 02 – V2 2C p = 190.424 – 100 2 / 2x 1004 = 190.424 – 4.980 = 185.444K

t –s = T 01 –T02 = 79.681

T01 – T2 (270.105 – 185.444)

= 0.941

DIMENSIONLESS GROUPS AND THEIR SIGNIFICANCE

The Performance of a turbomachine like pumps, water turbines, fans or blowers for incompressible flow can be expressed as a function of: (i) density of the fluid ρ (ii) Speed of the rotor N (iii) Characteristic diameter D (iv) Discharge Q

(v) Gravity head (gH) (vi) Power developed P and (vii) Viscosity µ. Obtain dimensionless groups and explain their significance.

Solution Using Buckingham π theorem Turbomachine = f [ ρ, N, D, Q, gH, P, µ] Performance Taking N, D as repeating variables and grouping with other variables as non dimensional groups a1 b1 c1 π1 = [ ρ N D Q] a2 b2 c2 π 2 = [ ρ N D gH] a3 b3 c3 π 3 = [ ρ N D P] a4 b4 c4 π 4 = [ ρ N D µ]

Parameter SI Unit MLT Units ρ kg/m 3 ML – 3 N rpm T – 1 D m L Q m3/s L3T – 1 gH m2/s 2 L2T – 2 P Nm/sec ML 2T – 3 (N=kgm/s 2) µ Ns/m 2 or ML – 1T – 1 kg/ms

Equating powers of mass, length and time in the LHS and RHS of the π terms, we obtain 3 π1 = Q/ND = Φ, Flow coefficient 2 2 π2 = gH/N D = Ψ Head coefficient 3 5 π3 = P / ρN D = ṁ Power coefficient 2 π4 = µ /ρND = R e Reynolds number

For model studies for similar turbomachines, we can use

Q1 = Q2 3 3 N1D1 N2D2

gH 1 = gH 2 2 2 2 2 N1 D1 N2 D2

P1 = P2 3 5 3 5 ρ1 N1 D1 ρ2 N2 D2

µ 1 = µ 2 2 2 ρ1 N1D1 ρ2 N2D2

SIGNIFICANCE OF NON DIMENSIONAL GROUPS

Loss in efficiency and head in a moderate CF pump (With reduced Reynolds number)

DIFFERENT SPECIFIC SPEED TURBO MACHINE ROTORS

1. Pelton Wheel 4. Steam turbine 2. Pelton Wheel 5. Centrifugal pump 3. Francis turbine 6. Radial compressor

7. Steam turbine

8. Steam turbine

9. Centrifugal pump

10. Gas turbine

11. Radial Compressor

12. Axial Compressor

13. Steam turbine

14. Francis turbine

15. Francis turbine

16. Mixed flow pump

17. Kaplan turbine

18. Axial compressor

19. Axial compressor

20. Propeller pump

21. Axial blower

22. Propeller pump

23. Kaplan turbine

SIGNIFICANCE OF NON DIMENSIONAL πππ GROUPS

1. Discharge coefficient or capacity coefficient

3 π1 = Q/ND = C 1 is capacity coefficient or flow coefficient for similar Turbo machines. 2 2 2 π1 = Q α D V α V as Q = AV = π D V ND 3 ND 3 ND 4

∴ π1 = V = 1 U Φ ∴Where, Φ is called speed ratio = U = tangential velocity of runner V theoretical jet velocity of runner

For any giver turbo machine, speed ratio is fixed.

For a given pump or fan of certain diameter running at various speeds the discharge is proportional is speed.

2. Head coefficient

2 2 π2 = gH/N D is called head coefficient.

π2 = gH = gH as U α ND N2D2 U2

∴ π2 = H = Ψ = head coefficient dimensionless U2 g The above ratio shows that head coefficient will be similar for same type of pumps or turbines for given impeller size and head varies as square of the tangential speed of the rotor.

3. Power coefficient 3 5 ∴ π3 = P / N D is called power coefficient or Specific power. For a given pump or a water turbine, the power is directly proportional to the cube of the speed of runner.

SPECIFIC SPEED – Pumps, fans, blowers and compressors Specific Speed ( Ω) is a dimensionless term of great importance in pumps, fans, blowers and compressors

Where

1/2 1/2 2 2 3/4 1/2 π5 = ( π1) = Q x (N D ) = NQ = ΩΩΩ 3/4 3 1/2 3/4 3/4 (π2) (ND ) (gH) (gH)

Dimensionless ∴ Ω = N √√√ Q where N = radians / sec (gH) ¾ Q = m 3/sec H = meters g = m/ sec 2

SPECIFIC SPEED OF A TURBINE

The specific speed of a turbine ( Ω) is obtained by the combination of head coefficient and power coefficient as follows. 1/2 3 5 π6 = ( π3) where π3 = P / N D 5/4 2 2 (π2) π2 = gH/N D

= P 1/2 x (N 2D2)5/4 = NP ½ (N 3D5 )1/2 (gH) 5/4 ½ (gH) 5/4

Ω = N √ P/ (Dimensionless) where, N = radians / sec (gH) 5/4 P = watts = kg / m 3 g = m/ sec 2 H = meters

SN Turbo machine ΩΩΩ Equation (Power absorbing) (Non dimensional)

1 Centrifugal pump (Slow – fast speed) 0.24 – 1.8 ΩΩΩ = N √√√ Q 3/4 (gH) 2 Mixed flow pump 1.8 – 4.0 where N = radians / sec 3 Axial flow pump, Propeller pump 13.2 – 5.7 3 Q = m /sec H = meters 4 Radial flow compressor, blower, etc. 0.4 – 1.4 2 g = m/ sec 5 Axial flow blower, compressor, etc. 1.4 – 20

SN Turbo machine ΩΩΩ Equation (Power generating) (Non dimensional)

6 Pelton turbine 0.02 – 0.19 ΩΩΩ = N √√√ P/ Single Jet 0.1 – 0.3 5/4 Double Jet 0.14 – 3.9 (gH) Multi Jet Where, N = radians / sec 7 Francis turbine 0.39 – 0.65 P = watts 3 Radial flow – slow speed 0.65 – 1.2 = kg / m Mixed flow – medium 1.2 – 2.3 2 Mixed flow – fast g = m/ sec H = meters 8 Propeller turbine – axial 1.6 – 3.6

9 Kaplan turbine – axial 2.7 – 5.4

10 Axial flow steam and gas turbines 0.35 – 1.9

Even though non dimensional specific speed Ω = N √ Q /(gH) 3/4 for pumps and compressors and

Ω = N √ P/ (gH) 5/4 for turbines gives relationship for blade shapes in various turbo machines, in practice it is not used.

Specific speed based on model studies only for hydraulic pumps and turbines are adapted in design practice for incompressible turbo machines.

Effect of Reynolds Number

Just like flow in pipes with friction, with decreasing Reynolds number, the loss factor increases at first slowly, then more and more rapidly in Turbo machines. The majority of ordinary turbo machines, (handling water, air, steam or gas) are found to operate in fully rough region. The critical Reynolds number at which the flow becomes fully rough, varies with the size of the machine (it depends on relative roughness) and its exact location for a given machine is difficult to predict.

The understanding of boundary layer and its separation is of importance in loss effects. The graph shows the loss factor in head and efficiency of a moderate size centrifugal pump.

Specific Speed of Pump (Ns) based on Model Studies

The specific speed Ns of a pump is defined as the speed of a geometrically similar pump of such size and dimensions that it will develop unit head (1m) and gives unit discharge (1 m3/sec).

From head coefficient, gH / N 2D2 is common for similar pumps i.e.

H1 = H 2 for similar pumps for model studies 2 2 N1D1 N2D2

N2D2 α H Therefore, D 2 α H/N 2 OR D α √ H / N ... (1)

From flow coefficient, Q / ND 3 is constant for similar pumps.

Q1 = Q2 for model studies. 3 3 N1D1 N2D2

Therefore, Q α ND 3 Q α N ( √ H/N) (H/N 2) as D α √ H/N

Therefore, Q α H3 /2 ... (2) N2 OR Q = C H3 /2 N2 where, C is a constant

2 C = N Q taking roots, C 1 = N √Q H3/2 H3/4

Hence, N 1√Q1 = N 2 √Q2 for similar pumps for model studies 3/4 3/4 H1 H2

Form definition of specific speed for a model pump, when H = 1m and Q = 1m 3/sec, we have

N√Q = Ns √1 H 3/4 13/4

∴∴∴ Ns = N √√√Q where, Ns is in rpm. H 3/4 Ns is same for similar pumps of different sizes

Ns = N √√√Q rpm SN Pumps 3/4 H

Centrifugal pump 1 12 – 95 Slow – Fast speed

2 Mixed flow pump 95 – 210

Axial flow pump 3 172 -320 Propeller pump

Specific Speed of a water turbine based on Model Studies

Definition The specific speed Ns of a water turbine is defined as the speed of a geometrically similar turbine of such size and shape that it will produce 1kW power under a head of 1m. From power coefficient,

3 5 π3 = P / N D = constant for similar water turbines. ∴ P α N3 D5

Since D α √ H/N P α N3 H 5/2 Taking roots, N2 √ P α H 5/4 N

N√P = constant for similar turbines. H 5/4

Form the definition of specific speed of a water turbine when P = 1kW and H = 1m N √P = Ns √1 H 5/4 15/4 ∴ Ns = N √√√P where Ns is in rpm of the model turbine. H 5/4 Ns is same for similar turbines of different sizes.

SN Water Turbines Ns = N √√√P rpm Equation (Power generating) 5/4 H

1 Pelton turbine Ns = N √√√P 3 – 30 Single Jet 5/4 17 – 50 Double Jet H 24 – 70 Multi Jet rpm Where, 2 Francis turbine N in rpm 60 – 102 Radial flow – slow speed P in kW 102 – 188 Mixed flow – medium H in meters 188 – 368 Mixed flow – fast

3 Propeller turbine 256 – 578 axial fixed blades

4 Kaplan turbine 428 – 856 axial adjustable blades

Problems in model studies

1. A model of a water turbine 0.5 m diameter develops 10 kW running at 800 rpm under a head of 20 m. The prototype has to work under a head of 180 m at 200 rpm. If the efficiencies are same, find : i) Diameter of the prototype ii) specific speed iii) power developed and name the type of tu rbine runner.

Solution Data Suffix 1 Model Suffix 2 Prototype

D1 = 0.5 m; N1 = 800 rpm; H 1 = 20 m; P 1 = 10 kW; D 2 =? ; N 2 = 200 rpm; H 2 = 180 m; P 2 =? 1 = 2 Type of turbine runner =?

Ns 1 = N 1√P1 = 800 √ 10 = 59.81 rpm 5/4 5/4 H1 20

From model studies, √H1 = √H2 N1D1 N2D2

∴ D2 = √H2 x N 1 x D 1

√H1 N2

= √180 x 800 x 0.5 = 6 m √20 200

As 1 = 2 , Ns 1 = Ns 2

∴ Ns 2 = N 2√P2 5/4 H2

5/4 5/4 ∴ √P2 = Ns 2 (H 2) = 59.81(180)

N2 200

∴ P2 = 38874.86 kW,

As Ns = 59.81 rpm, it must be a slow speed Francis turbine runner.

Problem 2 An axial flow pump with an impeller rotor diameter of 30 cm handles water at a rate of 2.7 m 3/minute running at 1500 rpm. The energy input is 125 J/kg and Total to Total efficiency is 0.75. If a geometrically similar pump has a diameter of 20 cm running at 3000 rpm, find its i) Flow rate ii) change in total pressure iii) input power.

Solution Data

Pump 1: D1 = 0.3 m Pump 2: D 2 = 0.2 m 3 Q1 = 2.7 m /minute Q2 = ? N1 = 1500 rpm N2 = 3000 rpm

E1 = 125 J/kg ∆p0 = ?

t – t = 0.75 P2 = ?

From model studies,

Q1 = Q 2 3 3 N1D1 N2D2

3 ∴ Q2 = N2 x D2 x Q1 3 N1 D1 = 3000 x (0.2) 3 x 2.7 1500 (0.3) 3 = 1.6 m 3/minute

Similarly,

gH 1 = gH 2 2 2 2 2 N1 D1 N2 D2

Nm/kg = J/kg = kg m m = m 2/sec 2 sec 2 kg

∴ Energy input E = gH = J/kg = (m 2/sec 2)

∴ E1 = E 2 2 2 2 2 N1 D1 N2 D2

2 2 ∴ E2 = N2 x D2 x E1 2 2 N1 D1

= (3000) 2 x (0.2) 2 x 125 (1500) 2 (0.3) 2

∴ E2 = 222.22 J/kg (actual)

t – t = E 2s = ∆h0s

E2 ∆h0

E2s = tt E2 = 0.75 x 222.22 = 166.65 J/kg (Ideal)

∆h0s = ∆p0 ρ

Hence, ∆p0 = ∆h0s x ρ = 166.67 x 1000 =1.667 x 10 5 N/m 2 = 1.667 bar

Input power P2 = ṁ x E 2 [kgs/sec x J/kg = Watts]

= ρQ x E 2 = 1000 x 1.6 x 222.22 60 x 1000 = 5.926 kW

Problem 3 A small scale model of a hydraulic turbine runs at 360 rpm under a head of 22 m and produces 10 kW output. Determine its: i) Unit discharge ii) Unit speed iii) Unit power, assuming turbine Total to Total efficiency = 0.8

If a prototype turbine is 12 times the model size and its efficiency is given by the Moodys’s formula p = 1–(1 – m) 0.2 [D m/D p] , iv) what is output power of the prototype v) type of turbine runner if head available for prototype is 250 m.

Solution Data Model Prototype

Nm = 360 rpm Dp/D m = 12 Hm = 22 m Hp = 250 m Pm = 10 kW Pp = ? m = 0.8 Type of turbine = ? Qu m = ? Nu m = ? Pu m = ?

Discharge through the model turbine

tt = P m /ρρρQm (gH m)

∴ Qm = 10 x 1000 0.8 x 1000 x 9.81 x 22

= 0.058 m 3/sec

i) Unit discharge Qu m = Q m /√√√Hm = 0.058 / √22 = 0.012 m 3/sec ii) Unit speed Nu m = N m / √√√Hm = 360 / √22 = 76.752 rpm 3/2 iii) Unit power Pu m = P m /(H m) = 10 /(22) 3/2 = 0.097 kW

Efficiency of the prototype 0.2 p = 1–(1 – m) [D m/D p] = 1 –(1 – 0.8) [1/12] 0.2 = 0.878 From model studies,

Hm = Hp 2 2 2 2 Nm Dm Np Dp

2 2 2 2 2 ∴ Np = Hp Dm Nm = 250 1 x 360 = 10227.273 2 Hm Dp 22 12

∴ Np = 101 .13 rpm

Similarly,

Qm = Qp 3 3 NmDm NpDp 3 3 ∴ Qp = Dp Np x Q m = 12 101 .13 x 0.058

Dm Nm 1 360 3 ∴ Qp = 28.155 m /sec

p = P p

ρ g Q p Hp

iv) Output power of the prototype

∴ P p = 1000 x 9.81 x 28.155 x 250 x 0.878 1000

= 60626 .021 kW

(Ns) p = N p√ P p = 101.13 x √ 60626.021 5/4 5/4 (H p) (250)

Hence (Ns) p = 25.049 rpm 25 rpm

Therefore, pelton turbine rotor with two jets is ideal.

THERMAL TURBO MACHINES - DIMENSIONLESS GROUPS As applicable to Compressible flow turbo machines like compressors, steam turbines, gas turbines

The performance parameters ∆h0s, , P for a thermal turbo machine can be expressed as

∆∆∆h0s , , P = f [ µµµ, N, D, ṁ , ρρρ01 , a 01 , γγγ]

As stagnation density and stagnation sound velocity change through the turbo machine at high velocity, ρ01 and a 01 at inlet conditions are selected.

∴ Turbo machine performance = f [ ∆∆∆h0s , , P , µµµ, N, D, ṁ , ρρρ01 , a 01 , γγγ]

Using Bukingham’s π theorem, to obtain dimensionless groups, we get by taking ρ01 , N and D as repeating variables and grouping with other variables and following the procedure as usual,

Parameter SI units MLT units 2 – 2 ∆h0s J/kg OR Nm/kg L T -- Dimensionless group 2 – 3 P J/sec OR Nm/sec ML T 2 – 1 – 1 µ Ns/m ML T N RPM T–1 D Metre L – 1 ṁ kgs/sec MT 3 – 3 ρ01 kg/m ML – 1 a01 m/sec LT

γ = C p/C v -- Dimensionless

We get dimensional groups as follows

2 ∆∆∆h0s , , P = f ṁ , ρρρ01 ND , ND , γγγ ... (1) 2 2 3 5 3 N D ρρρ01 N D ρρρ01 ND µµµ a01

The term ṁ = ṁ ... (2) 3 2 ρ01 ND ρ01 a01 D

As ND α a01 , stagnation sound velocity and ND/a 01 is taken as blade Mach number.

Also a 01 =√ γ RT 01

Using the laws of perfect gases and from I principles, Isentropic enthalpy rise

∆h0s = Cp (T 02s – T01 ) ... (3)

Using relationship P/ ρ γγγ = C P/ ρ = RT

∆h = C pT

For stagnation conditions, (γ - 1)/ γ

T02 = p 02 ... (4) T01 p01

Using eqn. 4 in eqn. 3, (γ - 1)/ γ ∆h0s = Cp T01 p02 –1

p01

Since C p = γ R and (γ - 1)

2 a01 = γ R T 01

We can write

∆h0s α f p 02 ...(5) 2 a01 p01

From flow coefficient,

ṁ = ṁ RT 01 = ṁ √RT 01 ... (6) 2 2 2 ρ01 a01 D p01 (√ γRT 01 )D D p01 √ γ

From power coefficient,

P = ṁ Cp ∆T0 = C p∆T0 ∆T0 ... (7) 3 5 2 2 2 ρ01 N D [ρ01 D (ND)] (ND) (ND) T01

By collecting the above non dimensional groups and putting in eqn. 1, p02 , , ∆T0 = f ṁ√RT 01 , ND , Re , γ ... (8) 2 p01 T01 D p01 √RT 01 For a given turbo machine of specific size D and constant R and γ for gas neglecting R e (viscous effect ) eqn. 8 can be written in simplified form as

p02 , , ∆∆∆T0 = f ṁ √√√T01 N ... (9) p01 T01 p01 , √√√T01

where, ṁ √T01 / p 01 and N/ √T01 are no longer dimensionless.

Operating characteristics of Compressor

Operating characteristics of a Steam Turbine

The graph shows that the compressor performance strongly depends on N/ √T01, whereas for a turbine, the dependence is weak. The operating characteristics of a compressor shows that if mass rate is reduced below certain value for given speed, the machine will surge leading to machine vibrations and stall. For a steam turbine, increasing the mass flow rate for various speeds will lead to choking of the flow for given pressure drop.

* * * * *

UNIT – 2

ENERGY TRANSFER IN TURBO MACHINES

Referring to figure, 1 is inlet; 2 is outlet of rotor

V = Absolute velocity of fluid (m/s) R = Radius of the wheel (m) ω = Angular velocity of rotor (rad/s) N = Speed of rotor (rpm) U = Linear velocity of vane (m/s) ṁ = Mass flow rate of fluid (kg/s)

Tangential momentum of fluid at inlet

ṁ Vu1 (N) Momentum of momentum OR Angular momentum of fluid at inlet

ṁ . V u1 . r 1 (Nm) Angular momentum of fluid at outlet

ṁ . V u2 . r 2 (Nm) Torque on the wheel = Change in angular momentum

∴ T = ṁ (V u1 r1 – Vu2 r2 ) (Nm)

∴ Work done/sec = Torque x angular velocity = T x ω

Taking ω r1 = 2 π r1/N = U 1

ω r2 = 2 π r2 /N = U 2 Work done/sec = ṁ [V u1 U1 – Vu2 U2] Nm/s or Watts Work done/Unit mass when m = 1kg WD/kg = Energy transfer 2 2 = [V u1 U1 – Vu2 U2] ...[Nm/kg = m /sec ] The above equation is known as EULER’S TURBINE EQUATION.

If V u1 U1 >> V u2 U2

Then, [V u1 U1 – Vu2 U2] is +ve It is applicable to Power Generating Turbo Machines or Turbines.

If V u1 U1 << V u2 U2

Then, [V u1 U1 – Vu2 U2] is –ve It is applicable to Power Absorbing Turbo Machines like pump, fans, blowers and compressors.

In a turbine if V u1 U1 >> V u2 U2 and

Vu2 is in opposite direction to rotation of wheel, then work done will be greater.

Work done/kg

WD/kg = [V u1 U1 – (–Vu2 ) U 2] = V u1 U1 + V u2 U2 (Nm/kg) Generally, for a turbine, work done/kg

WD/kg = [V u1 U1 ± Vu2 U2] (Nm/kg) where, V u1 U1 > V u2 U2

For pumps, fans, blowers and compressors

Work done/kg = [V u2 U2 – Vu1 U1] where V u2 U2 > V u1 U1 If ṁ = mass rate of flow in kgs/s Power developed in a turbine

P = ṁ [V u1 U1 ±±± Vu2 U2] Watts or Nm/s or J/s Power given to fluid in pumps, fans, blowers and compressors P = ṁ [V u2 U2 – Vu1 U1] Watts or Nm/s or J/s

ALTERNATE FORMS OF EULER’S TURBINE EQUATION

Radial inward flow Francis Water Turbine

Inlet and Outlet Velocity Triangles

Referring to velocity triangles 1 – inlet , 2 – outlet

V1 = Absolute velocity of the fluid at inlet (before entering the rotor vanes) Vr1 = Relative velocity of the fluid at rotor inlet Vu1 = Tangential component of absolute velocity

Whirl component of velocity at inlet

Vf1 = Flow component of absolute velocity at inlet Vru1 = Whirl component of relative velocity at inlet

U1 = Linear rotor vane velocity at inlet

α1 = Absolute jet angle at inlet

β1 = Vane (blade) angle at inlet

Referring to outlet velocity triangle 2 – outlet

V2 = Absolute velocity of the fluid at outlet after leaving the rotor vanes.

Vr2 = Relative velocity of the fluid rotor outlet (Just about to leave the rotor)

Vu2 = Whirl component of absolute velocity at outlet Vf2 = Flow component of absolute velocity at outlet

Vru2 = Whirl component of relative velocity at outlet U2 = Linear rotor velocity at outlet

α2 = Fluid or jet angle at outlet (To the direction of wheel rotation)

β2 = Vane (blade) angle at outlet (To the direction of wheel rotation)

From inlet velocity triangle 2 2 2 Vf1 = V 1 – Vu1 2 2 2 Vr1 = V f1 + V ru1 2 2 2 2 Vr1 = V 1 – Vu1 + (V u1 – U1) 2 2 2 2 = V 1 – Vu1 + V u1 – 2V u1 U1 + U 1

Rearranging 2 2 2 2V u1 U1 = V 1 + U 1 – Vr1

2 2 2 2 2 Vu1 U1 = V 1 + U 1 – Vr1 m /s OR Nm/kg... (1) 2

From outlet velocity triangle 2 2 2 Vr2 = V ru2 + V f2

2 2 2 = (U 2 – Vu2 ) + (V 2 – Vu2 )

Taking V ru2 = (U 2 – Vu2 ) in magnitude only and not in directions 2 2 2 2 2 Vr2 = U 2 – 2V u2 U2 + V u2 + V 2 – Vu2

2 2 2 2 2 ∴∴∴ Vu2 U2 = V 2 + U 2 – Vr2 m /s OR Nm/kg... (2) 2

CASE 1 :

Taking direction of rotation as positive Vu1 +ve and V u2 also +ve. Work done/kg or Energy transfer in Turbine

Work done/kg = (V u1 U1 – Vu2 U2)

2 2 2 2 2 2 Energy Transfer (E) = V 1 + U 1 –Vr1 – V2 + U 2 – Vr2 2 2

2 2 2 2 2 2 = V 1 – V2 + U1 – U2 + Vr2 – Vr1 2 2 2

COMPONENTS OF ENERGY TRANSFER 2 2 2 2 1) V 1 – V2 is change in absolute kinetic energy in m /s or Nm/kg 2 2 2 2) U 1 – U2 is change in centrifugal energy of fluid felt as 2 static pressure change in rotor blades in m2/s 2 or Nm/kg 2 2 3) V r2 – Vr1 is change in relative velocity energy felt as 2 static pressure change in rotor blades in m 2/s 2 or Nm/kg

CASE 2:

If V u2 is – ve

E = WD/kg = V u1 U1 + V u2 U2 (Work done will be more)

2 2 2 2 2 2 = V 1 – V2 + U1 – U2 + Vr2 – Vr1 2 2 2

CASE 3:

If V u2 = 0 No whirl at outlet commonly used in high capacity turbines

E = WD/kg = (V u1 U1) Nm/kg only

Degree of Reaction R

Degree of Reaction R is the ratio of Energy Transfer due to Static Enthalpy change to Total Energy Transfer due to Total Enthalpy change in a rotor.

R = Static head = Static enthalpy change = ∆∆∆h

Total head Total enthalpy change ∆∆∆h0

2 2 2 2 ∆h = (U 1 – U2 ) + (V r2 – Vr1 ) 2 2

2 2 2 2 2 2 ∆h0 = (V 1 – V2 ) + (U 1 – U2 ) + (V r2 – Vr1 ) 2 2 2

2 2 2 2 R = (U 1 – U2 ) + (V r2 – Vr1 ) ... (1) 2 2 2 2 2 2 (V 1 – V2 ) + (U 1 – U2 ) + (V r2 – Vr1 )

2 2 2 2 Taking (U 1 – U2 ) + (V r2 – Vr1 ) = ‘S’ as static component 2 2

2 2 and (V 1 – V2 ) = ‘KE’ Kinetic Energy Component (Absolute velocity change energies) 2

R = S = 1 KE + S 1 + KE/S

KE + 1 = 1 S R

KE = 1 – 1 = 1 – R S R R

S = R KE 1 – R when S = Static energy felt by rotor

KE = Kinetic energy change in rotor (in terms of V 1 and V 2, Absolute velocities)

Examples

1. For axial flow turbo machines, centrifugal forces can be neglected as U 1 = U 2

2 2 R = (V r2 – Vr1 ) 2 2 2 2 (V 1 – V2 ) + (V r2 – Vr1 ) (blade shapes become very important for energy transfer)

2. If U 1 = U 2 and there is no relative velocity energy change in the rotor, then Static Pressure (S) is 0. ∴∴∴ R = 0 which is known as Impulse Turbo machine.

If R 0, then the machine is a Reaction Turbo machine. The turbo machine must be running under controlled flow conditions inside the casing and flow passages for pressure changes.

STEADY FLOW ENERGY EQUATION

I Law of Thermodynamics

2 2 q + ṁ [h 1+V 1 /2 +gz 1] = WD + ṁ [h 2+V 2 /2+gz 2]

Taking 1 as inlet and 2 as outlet conditions q = Rate of heat transfer WD = Work done 2 V2 /2 = Kinetic Energy gz = Potential Energy

h0 = Stagnation Enthalpy

∆h0 = Total enthalpy change 2 Where h 0 = h + V /2 + gz

When q = 0 for isentropic flow

WD = – (h 01 – h02 ) = – ∆h0

In differential form, – dh 0 = + dw 2 2 2 2 2 2 ∴ – J = V 1 – V2 + U 1 – U2 + V r2 – Vr1 = dw (Nm/kg) kg 2 2 2 is known as Ideal Euler’s Work.

Effect of Blade Discharge Angle βββ2 on Energy Transfer E and Degree of Reaction R

Consider an outward radial flow turbo machine as shown in figure, where 1 is inlet, 2 is outlet Assumptions 1) Radial velocity of flow is constant, i.e.

Vf1 = V f2 = V f 2) No whirl component at inlet

Vu1 = 0

3) Diameter at outlet is twice as at inlet, i.e.

D2 = 2D 1 or U 2 = 2U 1 0 4) Blade angle at inlet β1 = 45 , V1 = V f1 = U 1

Assuming Turbine Equation

E = WD/kg = (V u1 U1 – Vu2 U2) Vu1 = 0 as there is no whirl at inlet

∴ Vu1 U1 = 0 2 2 E = [– Vu2 U2] ... Nm/kg or J/kg or m /sec

Considering outlet velocity triangle,

E = – U2 [U 2 – Vru2 ] as Cot β2 = Vru2

Vf2

= – U2 [U 2 – Vf2 Cot β2]

From assumptions

Vf1 = V f2 = V f U2 = 2U 1= 2V f

E = – 2V f [2V f – Vf Cot β2] 2 = – 2V f [2 – Cot β2] 2 E = 2V f [Cot β2 – 2] taking V f = 1 (unity) for all β2

E = 2 [Cot β2 – 2] Nm/kg or J/kg

Considering outlet velocity triangle 2 2 2 Vr2 = V f2 + V ru2 2 2 = V f2 + (V f2 Cot β2) 2 2 = V f2 [1 + Cot β2]

From inlet velocity triangle 2 2 2 2 Vr1 = V f1 + V f1 = 2 V f Degree of reaction R is given by

2 2 2 2 R = (U 1 – U2 ) + (V r2 – Vr1 ) 2 x E Transfer

Substituting for V f

2 2 2 2 2 R = [(V f – 4V f ) + V f (1+ Cot β2) – 2V f ]

2 x 2V f (Cot β2 – 2)

2 2 2 2 = – 5V f + V f + V f Cot β2 2 4V f (Cot β2 – 2)

2 2 = V f [Cot β2 – 4] Taking V f = Unity 2 4V f (Cot β2– 2)

2 = Cot β2– 4

4 (Cot β2 – 2)

= (Cot β2– 2) (Cot β2 +2)

4 (Cot β2 – 2)

R = Cot β2 + 2 4

INFERERNCE

WD/kg or E = 2 (Cot β2 – 2)

R = 2 + Cot βββ2 4

0 For βββ2 = 10 E = + 7.343 R > 1 1.918

The machine is a Reaction Turbine and V u2 is to direction of rotation

0 For βββ2 = 26.5 E = 0 R = 1

The machine is rotating and Transferring no energy as V u2 = 0

0 For βββ2 > 26.5 and E = - ve < 153.5 0 R = + ve

The machine is Power Absorbing like pump or compressor, V u2 in same Direction of rotation.

0 For βββ2 = 153.5 E = - ve

The machine works as R = 0 power absorbing impulse type, V u2 and rotation of wheel are in same direction.

0 For βββ2 > 153.5 E = - ve R = - ve

The machine is power absorbing reaction type, V u2 is very high. Static head is less at outlet than at inlet.

WD/kg or E = 2 (Cot βββ2 – 2)

R = 2 + Cot βββ2 4

Problem 1

In an inward flow radial turbine, water enters at an angel of 22 0 to the direction of rotation and leaves axially without whirl at outlet. The inlet and exit diameters are 0.6 m and 0.3 m respectively. The rotor speed is 300 rpm. The flow

velocity is 3 m/s and constant throughout. The width of the wheel at inlet is 15cms. Neglecting thickness of blades, calculate: (1) Rotor blade angles at inlet and outlet (2) Power developed

Solution

Data: 0 0 d1 = 0.6 m ; d 2 = 0.3 m ; b 1 = 15 cm; N = 300 rpm; α1 = 22 ; V f1 = V f2 = 3 m/s; α2 = 90 ; V u2 = 0; V 2 = V f2 ; K1 = 1 (blockage by blades neglected).

U1 = π d1 N = π x0.6x300 = 9.425 m/s 60 60

U2 = π d2 N = π x0.3x300 = 4.713 m/s 60 60

V1 = V f1 = 3 = 8.008 m/s 0 sin α1 sin 22

Vu1 = V f1 = 3 = 7.425 m/s 0 tan α1 tan 22

U1 > V u1 and hence the velocity triangles at inlet is as indicated

Vru1 = U 1 – Vu1 = 9.425 – 7.425 = 2 m/s

tan (180 – β1) = V f1 / V ru1 tan β1 ’ = V f1 = 3 (U 1 – Vu1 ) 2

0 tan β1 ’ = 56.13

Blade angle at inlet 0 β1 = 180 – 56.13 = 123.87

From outlet velocity triangle, tan β2 = V f2 / U 2

– 1 β2 = tan Vf2

U2 = tan – 1 [3/4.713] = 32.478 0

Mass flow rate ṁ = ρ Q

= K 1 ρ π d1 b1 Vf1 Taking K 1 = 1 (no blockage by blades) ṁ = 1000 x π x 0.6 x 0.15 x 3 = 848.23 kgs/sec of water

P = ṁ [V u1 U1 ± Vu2 U2] Nm/s or Watts

but V u2 = 0 for no whirl at outlet

P = ṁ [V u1 U1] =848.23 x 7.42 x 9.425 = 59319.693 watts = 59.32 Kw

Problem

In a certain turbo machine, the inlet whirl velocity is 15 m/s, inlet flow velocity is 10 m/s, blade speeds are 30 m/s and 8 m/s at inlet and outlet respectively. Discharge is radial with absolute velocity of 15 m/s. If water is the working fluid flowing at a rate of 1500 liters/sec, calculate: (1) Power in KW (2) Change in total pressure in bar (3) Degree of reaction (4) Utilization factor (VTU Dec, 2010)

Solution

Data: Vu1 = 15 m/s V2 = 15 m/s = V f2 Vf1 = 10 m/s Q = 1000 liters/sec U1 = 30 m/s P =? Kw U2 = 8 m/s R =?

Vf1 = V f2 ε =?

U1 > V u1

∴ Vru1 = U 1 – Vu1 = 30 – 15 = 15 m/s

– 1 β1’ = tan Vf1 / V ru1 = tan – 1 10/15 0 β1’ = 33.69

Blade angle at inlet β1 = 180 – β1’

β1 = 188 – 33.69 = 146.31 0

Vr1 = V f1 / β1’ = 10 / sin 33.69

Vr1 = 18.028 m/s

2 2 V1 = √(V u1 + V f1 ) = √(15 2 + 10 2) = 18.028 m/s

Taking discharge as radial, V f2 = V 2 tan β2 = (15/8) – 1 β2 = tan (15/8) = 61.928 0

– 5 – ∆p0 = ρ(U 1 Vu1 ) x 10 bar 2 2 2 2 2 2 5 – ∆p0 = ρ/2 [U 1 – U2 + V r2 – Vr1 + V 1 – V2 ] x [1/10 ]

= 1000/2 [30 2 – 82 + 17 2 – 18 2 +18 2 –15 2] x [1/10 5] = – 4.5 bar (1 bar = 10 5 N/m 2)

Work done = V u1 U1 Nm/kg

P = ṁ Vu1 U1 = ρQV u1 U1 = 1000 x (1500/1000) x 15 x 30 = 675000 watts = 675 Kw

2 2 2 2 R = (U 1 – U2 ) + (V r2 – Vr1 ) 2 x V u1 U1 = 30 2 – 82 + 17 2 – 18.028 2 x [15 x 30] = 0.889

Utilization factor ε ε = WD 2 WD + V 2 /2

= V u1 U1 2 Vu1 U1 + (V 2 / 2)

= 15 x 30 (15 x30) + 15 2 /2

= 460 562.5

ε = 0.80

Problem on Water Turbine

A hydraulic reaction turbine of radial inward flow type, works under a head of 160 m of water. At a point at entry, the rotor blade angles are 119 0 and the diameter of the runner 3.65 m. At the exit the diameter is 2.45 m. The absolute velocity of the flow is radially directed with a magnitude of 15.5 m/s and the radial component of velocity at inlet is 10.3 m/s. Determine: (1) Power developed by the machine for a flow rate of 110 m 3/s (2) Degree of reaction (3) Utilization factor

Solution Inward flow 0 1 – inlet; 2 – outlet; H = 160 m; β1 = 119 d1 = 3.65 m d2 = 2.45 m V2 = V f2 = 15.5 m/s Vf1 = 10.3 m/s P =? Q = 110 m2/s R = ? ε = ?

V1 = √2gH = √2x9.81x160 = 56.029 m/s

2 2 Vu1 = √(V 1 – Vf1 ) =√(56.029 2 – 10.3 2) = 55.074 m/s

– 1 α1 = tan (V f1 / V u1 ) = tan – 1 (10.3/55.074) = 10.593 0

0 β1 = 119 indicates U1 > V u1 0 0 0 β1’ = 180 – 119 = 61

tan β1’ = V f1 / V ru1

∴ Vru1 = V f1 / tan β1’ = 10.3 / tan 60 0 = 5.709 m/s

2 2 Vr1 = √(V f1 + V ru1 ) = √(10.3 2 + 5.709 2) = 11.775 m/s

U1 = [V u1 + V ru1 ] = [55.074 + 5.709] = 60.783 m/s

U1 = π d1N / 60

∴ N = (60 x U 1) / πd1

N = 60 x 60.783 π x 3.65

= 318.047 rpm

U2 = π d2N / 60

= π x 2.45 x 318.047 60

U2 = 40.8 m/s

0 For radial discharge V u2 = 0, d 2 = 90 V2 = V f2 = 10.5 m/s

tan β2 = V f2 / U 2 – 1 ∴ β2 = tan (10.5 / 40.8) = 14.432 0

2 2 Vr2 = √(V f2 + U 2 ) = √(10.5 2 + 40.8 2) = 42.129 m/s

Given 3 3 Q = 110 m /s and ρw = 1000 kg/m

Power developed

P = ρρρ Q V u1 U1 watts = [1000 x 110 x 55.074 x 60.783] watts = 368231.924 Kw or P = 368.232 MW

2 2 2 2 R = (U 1 – U2 ) + (V r2 – Vr1 ) 2 x V u1 U1

= (60.783 2 – 40.8 2 + 42.129 2 – 11.775 2) 2 x 55.074 x 60.783

= 0.548

Utilization Factor ε = WD 2 [WD + V 2 /2]

= V u1 U1 2 Vu1 U1 + (V 2 /2)

= 3347.563 3347.563 + 55.125

ε = 0.984

GENERAL ANALYSIS OF PUMPS & COMPRESSORS

In compressors and pumps work is done on the fluid

Vu2 U2 > V u1 U1 2 2 ∴ Work done = [V u2 U2 – Vu1 U1] Nm/kg or m /sec

2 2 2 2 2 2 = (V 2 – V1 ) + (U 2 – U1 ) + (V r1 – Vr2 ) 2 2 2

Assuming no whirl at inlet 0 α 1= 90 , V u1 = 0 Head developed (m)

H = V u2 U2 g

H = U 2 [U 2 – Vru2 ] / g

= U 2 [U 2 – Vf2 Cot β2] g

For flow rate in m 3/s

Q = A2 Vf2 ∴ Vf2 = Q / A 2

H = U 2 [U 2 – Q Cot β2] g 2

For a given pump or compressor D2 , N , A 2 and β2 are fixed, H and Q are variable.

2 H = U2 [U 2 – Q Cot β2]

g A2

2 H = U 2 – (U 2Cot β2) Q

g gA 2

H = K 1 – K2 Q 2 where K 1 = U 2 /g

K2 = U 2Cot β2 / gA 2

For backward curved vanes 0 β2 < 90

K2 is +ve [HQ] curve has –ve slope

For Radial vanes 0 β2 = 90

K2 = 0 2 H = K 1 = U 2 /g = Constant for all Q

For Forward curves vanes

K2 = – ve [HQ] curve has +ve slope

0 0 Majority of centrifugal pumps will have β2 25 to 45 (Backward curved vanes) 0 Radial flow compressors will have β2 = 90

CENTRIFUGAL PUMP & COMPRESSORS

Basic Analysis

ṁ = ρ1Q1 = ρ2 Q2 Q1 = A 1Vf1 Q2 = A 2Vf2

A1 = πD1B1

A2 = πD2B2 When Velocity of flow in constant i.e.

Vf1 = V f2 = Vf

For given mass rate of flow

ṁ = ρ πD1B1 Vf1 = ρπ D2B2Vf2 or D 1B1 = D 2B2 or D 2 = B 1 D1 B2

If V u1 = 0 i.e. No whirl at inlet Theoretical work done = WD/kg = U 2 Vu2 2 = U 2 Vu2 ... (1) U2

For constant velocity of flow

V1 = V f1 = V f2 = U 1tan β1

0 From exit velocity triangle for β2 < 90 Vu2 = U 2 – Vru2

= U 2 – Vf2 Cot β2

∴ Vu2 = 1 – (V f2 Cot β2)

U2 U2

∴ Putting in theoretical work done equation 2 WD/kg = U 2 [1 – (V f2 Cot β2)]

From sine rule in outlet velocity triangle

V2 = U 2

sin β2 sin ( α2 + β2)

Vu2 = V 2 Cos α2 = U 2 sin β2 . Cos α2

sin ( α2 + β2)

Vu2 = sin β2 . Cos α2

U2 (sin β2+ Cos α2) + (Cos α2 + sin β2)

Vu2 = tan β2

U2 tan α2 + tan β2

Power developed – Theoretical

P = ṁ WD/kg = ṁ U2 Vu2 Nm/s or J/s or Watts

= ṁ ∆h0

= ṁ Cp ∆T0 For isentropic process, Total Energy or Stagnation Energy Transfer

∆h0 = ( ∆p) 0 ρ or ∆p0 = ( ∆h0) ρ 2 = ρ U2 [1 – (V f2 Cot β2)]

Taking static pressure rise due to centrifugal forces and relative velocity of flow known as diffusion effect.

(p 2 – p1) = ( ∆p) static

2 2 2 2 = ρ [U 2 – U1 ] + ρ [V r1 – Vr2 ] 2 2

2 2 2 2 2 2 (p 02 – p01 ) = ( ∆p) 0 = ρ[ (U 2 – U1 ) + (V r1 – Vr2 ) + (V 2 – V1 )] 2 2 2

2 2 ∴ ∆p0 = (p 2 – p1) + ρ (V 2 – V1 ) N = kg . m . m 2 m2 m3 sec 2

Degree of Reaction R

R = ( ∆p) static

(∆p) stagnation

2 2 R = V u2 U2 – (V 2 – V1 ) 2

Vu2 U2

2 2 2 R = V u2 U2 – [(V f2 + V u2 ) – Vf1 ] 2

Vu2 U2

2 = 1 – Vu2 2V u2 U2

R = 1 – Vu2 2U 2

From velocity triangle at outlet for various βββ2

0 1. When βββ2 < 90 backward curves vane

Vu2 < U 2 ∴ R < 1 > 0.5

0 2. For Radial blades βββ2 = 90

Vu2 = U 2 ∴ R = 1 – ½ = 0.5

0 3. For forward curved vanes βββ2 > 90

Vu2 > U 2 ∴ R < 0.5

Problem on Centrifugal Pump

The internal and external diameters of a Centrifugal pump are 20 cm and 40 cm respectively. The pump is running at 1200 rpm. The vane angles at inlet is 20 0. Water enters the impeller radially and velocity of flow is constant. Find the work done by the impeller per kg of water for the following conditions 0 0 0 (a) 2 = 30 (b) 2 = 90 (c) 2 = 100

Solution

Data

Internal diameter d 1 = 0.2m Outer diameter d 2 = 0.4 m Speed N = 1200 rpm 0 Vane angle at inlet 1 = 20 0 Water enters radially Vu 1 = 0, 1 = 90 , V 1=V f1 Flow velocity constant V f1 = V f2 = V f

U1 = πd1N = π x 0.2 x 1200 = 12.5666 m/s 60 60

U2 = πd2N = π x 0.4 x 1200 = 25.133 m/s 60 60

From inlet velocity triangle and constant velocity of flow

V1 = V f1 = V f2 = V f = U 1 tan 1

0 Vf = 12.566 x tan 20 = 4.574 m/s

0 (a) 2 = 30

tan 2 = V f2 = V f2 Vru2 (U 2 – Vu2 )

∴ Vu2 = U 2 – Vf2

tan 2

= 25.133 – 4.574 tan 30

= 17.211 m/s

WD/kg = [V u2 U2] = 17.211 x 25.133 = 432.554 Nm/kg

0 (b) 2 = 90

U2 = V u2 = 25.133 m/s 2 2 WD/kg = U 2 = (25.133) 2 2

= 631.666 Nm/kg

From velocity triangle at outlet 0 0 (c) 2 > 90 = 100

tan 2’ = V f2 Vru2 tan 2’ = V f2 Vru2

or tan (180 – 2) = Vf2 (V u2 – U2) tan 80 0 = 4.574

[V u2 – 25.133]

∴Vu2 = 25.133 + 4.574 tan 80 0 = 25.921 m/s

WD/kg = V u2 U2 = 25.921 x 25.133 E = 651.48 Nm/kg (J/kg)

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