1. Thermodynamics Lecture Series
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Thermodynamics Lecture Series
Assoc. Prof. Dr. J.J.
Thermodynamics Properties Relations: Clapeyron, Maxwell & Joule-Thompson
Applied Sciences Education Research Group (ASERG)
Faculty of Applied Sciences
Universiti Teknologi MARA
Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

2. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
CHAPTER 4
Thermodynamics Property Relations
Goal: Determine Unknown Thermodynamic Properties Which Are Not Directly Measurable (u, h, s) in Terms Measurable Properties: How to develop property table
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3. Learning Outcome Objectives:
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Learning Outcome
Objectives:
Develop fundamental relations between commonly encountered thermodynamic properties and express the properties that cannot be measured directly in terms of easily measurable properties .
Review and use partial derivatives in the development of thermodynamic property relations .
Develop the Maxwell relations, which form the basis for many thermodynamic relations .
Develop the Clapeyron equation and determine the enthalpy of vaporization from P, v, and T measurements alone .
Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

4. Learning Outcome Objectives: Discuss the Joule-Thomson coefficient .
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Learning Outcome
Objectives:
Develop general relations for Cv, Cp, du, dh, and ds that are valid for all pure substances under all conditions .
Discuss the Joule-Thomson coefficient .
Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

5. Energy Transfer – Work Done
Mechanical work:
Piston moves up
Boundary work is
done by system
Electrical work is done on system
H2O:
Super
Vapor
No heat transfer
T increases
after some time
Wpw,in ,kJ
H2O:
Sat. liquid i
Voltage, V
We,in = Vit/100, kJ
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6. First Law – Energy Transfer
Properties will change indicating change of state
How to relate changes to the cause
System
E1, P1, T1, V1 To
Mass in
Mass out
E2, P2, T2, V2
Mass transfer as a cause (agent) of change
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7. First Law – General Energy Balance
Energy Entering a system -
Energy Leaving a system =
Change of system’s energy
Energy Balance
Ein – Eout = Esys, kJ or
ein – eout = esys, kJ/kg or
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8. First Law – Mass Balance
Mass Entering a system -
Mass Leaving a system =
Change of system’s mass
Mass Balance
min – mout = msys, kg or
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9. First Law – General Energy Balance
Energy Balance –General system
Qin + Win + Emass,in – Qout – Wout - Emass,out
= DU+ DKE + DPE, kJ
qin + win + qin – qout – wout – qout
= Du+ Dke + Dpe, kJ/kg
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10. First Law – Steady - flow
Energy Balance – Control Volume Steady-Flow
Steady-flow is a flow where all properties within boundary of the system remains constant with time
DEsys= 0, kJ; Desys= 0 , kJ/kg,
DVsys= 0, m3; Dmsys= 0
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11. First Law - Single Stream
Mass & Energy Balance–Steady-Flow: Single Stream
Mass balance
Energy balance
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12. First Law Energy balance CV
Mass & Energy Balance–Steady-Flow: Single Stream
Mass balance:
Energy balance:
qin – qout+ win – wout = qout – qin, kJ/kg
= h2 – h1 + ke2 – ke1 + pe2 – pe1, kJ/kg
= Dh + Dke + Dpe, kJ/kg
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

13. First Law - Closed System
Energy Balance Closed Stationary system
qin – qout+ win – wout = Du, kJ/kg
For pure substances, use property table and mathematical manipulations to determine u2 and u1. Then Du = u2 – u1.
And h2 and h1, . Then Dh = h2 – h1.
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

14. Property Table Saturated water – Pressure table
P, kPa 10 50
P, MPa
0.100
1.00
22.09
Sat. temp.
Tsat, C
45.81
81.33
99.63
179.91
311.06
374.14
Specific volume, m3/kg
f, m3/kg
g, m3/kg
14.67
3.240
1.6940
Specific internal energy, kJ/kg
uf, kJ/kg
ufg, kJ/kg
ug, kJ/kg
191.82
2246.1
2437.9
340.44
2143.4
2483.9
417.36
2088.7
2506.1
761.68
1822.0
2583.6
1151.4
2544.4
2029.6

15. First Law –Specific Heat
Energy Balance Closed Stationary system C  , Cp,
Specific heats to find DU and DH
Specific Heat Capacity
C  at constant volume, Cp, at constant pressure
Amount of heat necessary to increase temperature of a unit mass by 1K or 1degree Celcius
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16. First Law –Specific Heat
Energy Balance Closed Stationary system C  , Cp,
For Ideal Gases: Estimate internal energy change and enthalpy change
Assume smooth change of C with T, & approximate to be linear over small DT (approx. a few hundred degrees)
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

17. First Law –Specific Heat
Energy Balance Closed Stationary system C  , Cp,
For Ideal Gases : Estimate internal energy change and enthalpy change
Assume smooth change of C with T, & approximate to be linear over small DT (approx. a few hundred degrees)
C , avg is determined using interpolation technique & use Table A-2b
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005

18. Second Law – Will a Process Happen
Carnot Principles
For heat engines in contact with the same hot and cold reservoir
P1: 1 = 2 = 3 (Equality)
P2: real < rev (Inequality)
Processes satisfying Carnot Principles obeys the Second Law of Thermodynamics
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19. Second Law – Will a Process Happen
Clausius Inequality :
Sum of Q/T in a cyclic process must be zero for reversible processes and negative for real processes
reversible
real
impossible
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20. Entropy – Quantifying Disorder
Source
Entropy
qin = qH
Entropy Change in a process
Steam Power Plant
net,out
qout = qL
Sink
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21. Entropy – Quantifying Disorder
Entropy Balance –Closed system
Energy Balance:
Entropy Balance:
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22. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Derivatives
State postulate: Need 2 independent intensive property to define a system’s state
Animation
Other properties: expressed in terms of the 2 specified properties. z = z(x,y)
Let a function f depends only on x.
f=f(x). Slope of curve shows degree of
dependence of f on x. Then;
The derivative of the function f(x), with respect to x is the rate of change of f with x.
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23. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Derivatives
Animation
Example 11-1: Determine Cp(T) of ideal gases at 300K where Cp(T)=dh(T)/dT. Since h(T) is not available, then approximate by replacing differentials by differences near 300K.
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24. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Partial Derivatives
What if a function depends on 2 variables such as P=P(T,)? Generally, z=z(x,y): Before we can have total derivatives, we must find the partial derivatives such as:
X2+Y, partial x
X2+Y, partial y
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25. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Partial Derivatives
X2+Y, partial x
X2+Y, partial y
For independent variables
For dependent variables,
Since z = z(x,y) and if both variables are to change at the same time, then
Add & subtract z(x,y+y),
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26. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Partial Derivatives
X2+Y, partial x
X2+Y, partial y
Since z = z(x,y) and if both variables are to change at the same time, then
Add & subtract z(x,y+y),
In the limit that x→0, y→0, total derivative is
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27. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Partial Derivatives
X2+Y, partial x
X2+Y, partial y
In the limit that x→0, y→0, total derivative is or
where
Derivative of M & N are
Since the order of differentiation is not important, then,
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28. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Maxwell Relations
Maxwell relations are equations relating partial derivatives of P, , T & entropy s for a simple compressible system which are obtained from the 4 Gibb’s relations:
From the enthalpy h= u+ P:
Defining Hemholtz function as a = u - Ts & Gibb’s functions as g = h - Ts, then;
All four of the Gibb’s relations are of the form Mdx+Ndy, where M & N are:
and
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29. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Maxwell Relations
Maxwell relations are equations relating partial derivatives of P, , T & entropy s for a simple compressible system which are obtained from the 4 Gibb’s relations:
Relates entropy to measurable properties P, T & 
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30. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Clapeyron Equation
Can determine enthalpy change during phase change such as hfg, based only on P, , T
During phase change Psat=f(Tsat) & independent of the volume. Then
Slope of sat. curve & can be treated as a constant during integration
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31. Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005
Clapeyron Equation
Can determine enthalpy change during phase change such as hfg, based only on P, , T
During the phase change, P also remains constant. Then using dh=Tds+dP;
Rephrase
Clapeyron Equation
Then
Applicable to any 2-phase system
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32. Clapeyron-Clausius Equation
Can determine enthalpy change during phase change such as hfg, based only on P, , T
For liquid-vapor phase change, approximate: fg ≈ g since g » f.. For vapor, treat it as ideal gas, then use Pg=RT
Then
Then
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33. Clapeyron-Clausius Equation
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Clapeyron-Clausius Equation
Can determine enthalpy change during phase change such as hfg, based only on P, , T
For small temperature intervals, hfg can be treated as constant at some average value;
Then
Use to determine the change in saturation pressure when saturation T changes. Can also use hig, enthalpy of sublimation
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Copyrights DR JJ, ASERG, FSG, UiTM Shah Alam, 2005