Thermodynamics Introduction and Basic Concepts
by
Asst. Prof. Channarong Asavatesanupap Mechanical Engineering Department Faculty of Engineering Thammasat University 2 What is Thermodynamics?
Thermodynamics is the study that concerns with the ways energy is stored within a body and how energy transformations, which involve heat and work, may take place. Conservation of energy principle , one of the most fundamental laws of nature, simply states that “energy cannot be created or destroyed” but energy can change from one form to another during an energy interaction, i.e. the total amount of energy remains constant. 3 Thermodynamic systems or simply system, is defined as a quantity of matter or a region in space chosen for study. Surroundings are physical space outside the system boundary. Surroundings
System
Boundary
Boundary is the surface that separates the system from its surroundings 4 Closed, Open, and Isolated Systems The systems can be classified into
(1) Closed system consists of a fixed amount of mass and no mass may cross the system boundary. The closed system boundary may move. 5
(2) Open system (control volume) has mass as well as energy crossing the boundary, called a control surface.
Examples: pumps, compressors, and water heaters. 6
(3) Isolated system is a general system of fixed mass where no heat or work may cross the boundaries.
mass No
energy No
An isolated system is normally a collection of a main system and its surroundings that are exchanging mass and energy among themselves and no other system. 7 Properties of a system Any characteristic of a system is called a property. Some familiar properties are volume V, mass m, density r, pressure P, temperature T and etc. Density is defined as mass per unit volume
Water@ 20 C , 1 atm r = 998 kg/m3
The reciprocal of density is the specific volume, which is defined as
Specific Gravity SG is defined as the ratio of the density of a substance to the density of some standard substance at a specified temperature (usually water at 4C). Temperature (T) is a measure of the average energy of motion, or kinetic energy, of particles in matter. (or a measure of hotness and coldness) Temperature scales Common scale:
Celsius scale °C SI unit
Fahrenheit scale °F English unit
The Celsius scale is related to the Fahrenheit scale by
Thermodynamic scale(Absolute scale):
Kelvin K SI unit
Rankine R English unit Temperature scales
The common scales are related to the absolute scale by
SI unit
English unit
Example: Water boils at 100C at one atmosphere pressure. At what temperature does water boil in F, K and R.
T(°F) = 100x1.8 + 32 = 212 °F
T(K) = 100 + 273.15 = 373.15 k
T(R) = 212 + 459.67 = 671.67 F 11
Pressure(P) is the force per unit area applied in a direction perpendicular to the surface of an object
N P = F 2 (Pa) A m
For English system, 12
Pressure scales Absolute scale:
Absolute pressure is the pressure that is measured relative to absolute zero pressure (absolute vacuum).
Gage scale:
Gage pressure is the pressure that is indicated on a pressure-measuring device (called a pressure gage). Generally, the device is calibrated to read zero in the atmosphere. 13
Vacuum pressure
Pressures below atmospheric pressures are called vacuum pressures. A device that is used to measure vacuum pressure is called a vacuum gage.
Pressure Symbol
Absolute Pa
Gage Pg
Pressure Eng. unit Absolute psia Gage psig 14 Example A pressure gage connected to a valve stem of a truck tire reads 240 kPa at a location where the atmospheric pressure is 100 kPa. What is the absolute pressure in the tire, in kPa and in psia?
Pabs P atm P gage 100kPa 240 kPa 340 kPa The pressure in psia is 14. 7 psia P340 kPa 49. 3 psia abs 1013. kPa
What is the gage pressure of the air in the tire, in psig?
Pgage P abs P atm 49.. 3psia 14 7 psia 34. 6 psig 15 Intensive and Extensive properties
• Intensive properties are those that are independent of the mass of a system.
• Extensive properties are those whose values depend on the size—or extent—of the system. 16 Internal energy (U) is defined as the sum of all the microscopic forms of energy of a system. It is related to the molecular structure and the degree of molecular activity and can be viewed as “the sum of the kinetic and potential energies of the molecules”.
Properties Symbol Unit Extensive U J [Joule] Intensive u=U/m J/kg
Gas: U = CV T = mcV T [kJ/kg]
where = heat capacity at constant volume [kJ/K] CV = specific heat capacity [kJ/kg-K] cV 17 Enthalpy (H) is a measure of the total energy of a thermodynamic system and is defined as “the summation of the internal energy and the flow work (PV); H = U + PV”.
Properties Symbol Unit Extensive H J [Joule] Intensive h=H/m J/kg
Gas: H = CP T = mcP T [kJ/kg]
where = heat capacity at constant pressure [kJ/K] CP = specific heat Capacity [kJ/kg-K] cP 18
Specific heats (c)
is defined as the energy required to raise the temperature of a unit mass of a substance by one degree. In thermodynamics, we are interested in two kinds of specific heats: specific heat at constant volume cv and specific heat at constant pressure cp.
1 kg Water 1 kg air 1 kg air
DT = 1°C DT = 1°C DT = 1°C
4.18 kJ 0.72 kJ 1.00 kJ 19
How to identify the state of a substance?
1.Equations of state
2.Property tables
3.Property diagrams 20
Ideal gas law
PV = nRuT
Pv = RT where P and T are absolute pressure and temperature, respectively. 3 R is a gas constant = Ru/M [kJ/kg.K or kPa.m /(kg.K)] Ru is a universal gas constant = 8.314 kJ/(kmol.K) n is the number of moles = m/M M is Molar mass 21
Example 1 A room of the size 4m x 5m x 6m contains air at P = 100 kPa and T = 25°C. Determine the mass of air inside the room. Assume 3 that Ra = 0.287 kPa.m /kg.K
PV 100kPa 120m3 m 3 140.3kg RaT 0.287kPa m / kg K (25 273)K 22
Example 2 The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire when the air temperature in the tire rises to 50°C 23
Liquid and vapor phases of a substance • Compressed liquid (subcooled liquid): A substance that it is not about to vaporize. • Saturated liquid: A liquid that is about to vaporize. • Saturated vapor: A vapor that is about to condense. • Saturated liquid–vapor mixture: The state at which the liquid and vapor phases coexist in equilibrium. • Superheated vapor: A vapor that is not about to condense (i.e., not a saturated vapor). 24
Examples of property tables
.
g = gaseous phase (saturated vapor) f = liquid phase (saturated liquid) 25
Examples of property tables
. 26
Examples of property tables 27
Example 3 Find the the density, specific internal energy, and specific enthalpy of water at T = 25 C and P = Patm. From table A-4:
1 1 3 rw 3 997.01kg / m vw 0.001003m / kg
uw u f @25C 104.83kJ / kg
hw hf @25C 2441.7kJ / kg 28 Example 4 Determine the amount of energy required to increase the temperature of 1-kg water in example 2 to 150C.
Phase-change process
From table A-4: 1. Liquid water at 25C hw,1 h f @25C 104.83kJ / kg Q12 m Dh 1(419.17 104.83) From table A-4: 314.34kJ 2. Liquid water at 100C hw,2 h f @100C 419.17kJ / kg
Q23 m Dh 1(2,675.6 419.17) From table A-4: 3. Vapor water at 100C 2,256.43kJ hw,3 hg @100C 2675.6kJ / kg 29
Example 4 Determine the amount of energy required to increase the temperature of 1-kg water in example 3 to 150C. Phase-change process
3. Vapor water at 100C From table A-4: hw,3 hg @100C 2675.6kJ / kg Q34 m Dh 1(2,776.6 2,675.6) 4. Super heated water at 150C 101.0kJ From table A-6: hw,4 hsuperheat@150C 2776.6kJ / kg 30
Example of property diagrams
.
Water 31 Psychrometric chart
Air 32
Moist air ( Air/Water vapor mixture) properties • Dry-bulb temperature: The air temperature indicated by a standard thermometer. • Wet-bulb temperature: The air temperature indicated by a thermometer with a wet wick attached to it bulb. • % Relative Humidity: The amount of water vapor held in the air as a percent of the maximum amount of water vapor the air can hold at a specific temperature. • Enthalpy: The total heat contained in the air. • Dew point: The air temperature at which condensation begins. • Humidity ratio: The mass of water vapor held in 1 kilogram of dry air. 33 Example 5 Find the the density, specific enthalpy, humidity ratio and dew-point temperature of air at T = 25 C and %RH = 50%.
ha 50kJ / kg
a 0.10kgw / kga
Tdp 14C
1 1 3 ra 1.16kg / m va 0.86 34
Example 6 Condensation within the 1 m2 wall
2 Gypsum 2 hi = 10 W/(m .K) Brick Fiberglass ho = 40 W/(m .K)
25C, 50% 35C, 48%
kB= 0.68 W/(m.K)] kF= 0.038 W/(m.K)] 0.1m 0.15m 0.01m kG= 0.48 W/(m.K)] 35
Example 6 (cont.)
Dx 1 Rwall Rair KA hA
Rth,tot Rtot Ri Rb R f Rg Ro
2 Rth,tot 0.1 0.147 3.947 0.021 0.03 4.24 (m K) /W
The heat flux is
DT (35 25) K q 2.35W / m2 2 Rth,tot 4.24 (m .K) /W 36
Fiberglass Example 6 (cont.)
1 35C, 70% 2
28C
DT q R 25C, 50% 3 th,i 4
DTa 2.35 0.03 0.07C T1 34.93C
DTg 2.35 0.02 0.05C T2 34.43C
Condensation forms DT f 2.353.947 9.27C T3 25.15C within the fiberglass layer. 37
Example 7 Preventing condensation on cold air ducts.
Ambient air 25C, 50%
Tsurface = 12 C
Cold air 38
Example 7 (cont.) Preventing condensation on cold air ducts.
Ambient air 25C, 50% From the example 1, the dp temperature = 16 C
Tsurface < dp temperature
Tsurface = 12 C
Cold air
Water drops form on the cold duct wall where the surface temperature is below the dewpoint temperature. 39
Example 7 (cont.) Preventing condensation on cold air ducts.
Ambient air 25C, 50% From the example 1, the dp temperature = 16 C
q conv Before
q h(T T ) h = 30 W/(m2K) conv ambient surface 0 Tsurface = 12 C 30(25 12) 390W / m2
Cold air 40
Example 7 (cont.) Preventing condensation on cold air ducts.
To prevent condensation, Insulation is needed on the duct.
Ambient air 25C, 50%
Ts = ? C Insulator b =?
Fiberglass is used, Cold air k= 0.038 W/(m.K)] 41
Example 7 (cont.) Preventing condensation on cold air ducts. Ambient air 25C, 50%
Ts = ? C Insulator h =? After
(Ts Tsurface) Cold air q k h(Tambient Ts) b (16 12) 0.038 30(25 16) b
b 0.00056 m
Heat gain q 270W / m2 ???? 42
Example 7 (cont.) Preventing condensation on cold air ducts. What will happen, if h = 2.5 cm ?
(Ts 12) 0.038 30(25 Ts ) 0.025
Ts 24.37C (no condensation)
2 Heat gain q 18.9W / m (-93%) 43 First Law of Thermodynamics is an expression of the conservation of energy principle.
where DE = E –E Ein –Eout = DEsys sys final init 44 1st Law for closed systems (1) Energy transferred across the boundary of the closed system can be divided into 2 forms: Heat and Work
(2) Energy stored in the closed system is represented by the total internal energy (U)
Therefore, the first law of thermodynamics is written as
(Qin + Win) – (Qout +Wout) = DU+DEp+DEk
DQ – DW = DU+DEp+DEk 45 1st Law for closed systems If the system does not move with a velocity and has no change in elevation, the conservation of energy equation reduces to where 1 represents initial state DQ – DW = U2-U1 [kJ] 2 represents final state
dU Q W [kW ] dt
No motion (DKE=0) + No elevation change (DPE=0) 46
Example 7 A closed tank has a volume of and is filled with 200 kg of water at the temperature of 30 C. The water is heated by a 30kW electric heater. How long does it take for water to reach 45C. dU Q W dt
(u2 u1) Water 20kW 0W mw 200 kg. Dt (188.43125.73)kJ / kg 20 kW heater Dt 200kg 30kW Dt 418 s 47
Note:
(u u ) 20kW 0W m 2 1 w Dt c (T T ) m v 2 1 w Dt 4.178(45 30)kJ / kg Dt 200kg 30kW 417.8 s 48 1st Law for Open systems Energy and material transfer into or out of the system boundary. For steady flow,
m m [kg / s] in out
V 2 V 2 Q W m (h gz) m (h gz) [kW ] out2 in 2 49
Example 8 Determine the amount of energy removed for space cooling if the infiltration mass flow is 0.35 kg/s
To = 35 C %RH = 70
Energy removed Ti = 25 C for space cooling, %RH = 50 (Qinf) 50
Example 6 (cont.) Mass balance:
m in m out m inf 0.35 kg / s
Energy balance: 0 0 0 V 2 Q W m (Dh D Dgz) inf inf 2
Qinf m inf (hin hout ) 51
HW#1 1. student living in a 4-m x 6-m x 6-m dormitory room turns on her 150-W fan before she leaves the room on a summer day, hoping that the room will be cooler when she come back in the evening. Assume all the doors and windows are tightly closed and disregarding any heat transfer through the walls and the windows, determine the temperature in the room when she comes back 10 h later. Use specific heat values at room temperature, and assume the room to be at 100 kPa and 15 C in the morning when she leaves. 2. The air in a room is 20C and 50% relative humidity. Will moisture condense on a window whose surface is 7C? If the room is 4.5m2 and 2.5 m high, how much water is contained in the room? 3. A chilled-water line carries chilled water at 7C through a room at 21C and 60%RH. How much fiber glass insulation is needed on the pipe to avoid condensation?