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# Mathematical Thermo

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```									Mathematical Thermo
Thermo and Math
• Thermodynamic properties can be treated as
pure mathematical functions and this can help
one find ways to compute changes in them.
– One mathematical fact is that thermodynamic
properties depend on exactly two other properties.
They are said to be exact functions of two others.
What this means to the engineer is that within a single
phase, there is one and only one value of U (internal
energy), or H, or S, or V at a given T and P.
Expressing Property
Changes Mathematically
• “Exactness” means it is possible to
mathematically express a change in one of
the properties in terms of the changes in
two others. For example one important
change is mathematically written:
~           ~
~ ~        ~      ~  U          U  ~
U  U (T , V )   dU       dT   ~  dV
 T  ~     V 
    V         T
How would you mathematically express the change in the
enthalpy in terms of the change in temperature and pressure?

~           ~
~ ~                      ~  H          H 
H  H (T , P)           dH       dT  
 T 
 dP
 P 
    P         T
How would you mathematically express the change in the
entropy in terms of the change in temperature and pressure?
~           ~
~ ~
S  S (T , P )          ~  S          S 
dS       dT    dP
 T        P 
    P      T
You would “say” these equations: “The change in the enthalpy
of a substance is the sum of the change in the enthalpy with
respect to temperature at constant pressure times the change
in temperature and the change in the enthalpy with respect to
pressure at constant temperature times the change in
pressure”.
If two things determine a property, then the change in the
property is a result of changes in each of the two things.
The Secret
The secret to thermodynamic calculations is to understand
how the partial differentials (the coefficients) depend on
things one can measure. If the coefficients were known
then one could always mathematically determine a finite
change in a property by integrating each term on the RHS.

~           ~
~    H        H 
dH                
 T  dT   P  dP
    P         T
~             ~
~     ~
2
~       H 
T2          P2
 H 
H 2  H 1   dH    T  dT    P  dP
            
1      T 1    P     P1    T
Heat Capacity
• The “constant pressure” heat capacities of fluids
have been experimentally measured by noting
how much a joule of electrical work addition
raised the temperature.
~         This is measured
~     H 
Cp      
and tabulated
 T 
    P
The pressure used is low (at or below 1 atm). This
is called the ideal gas heat capacity.
Tabulated Cp’s

This means B and C are really small numbers!
Heat capacity
• The heat capacity of a substance
behaving as an ideal gas may change with
temperature but does not change with
pressure.
• Either the heat capacity is tabulated or it
can be computed from group contribution
models. (See Perry’s Manual)
C po  A  BT  CT 2
The Trick
• The trick is to manipulate the mathematical
equations using thermodynamic definitions
of the property changes.
– This requires one to express and use some
relationships called the Maxwell Equations
– This requires one to know how to prepare
partial differentials from total differentials.
Thermodynamic
Definitions
~       ~      ~
dU  Td S  PdV        Most basic definition

~ ~         ~           ~    ~     ~ ~
H  U  PV             dH  dU  PdV  VdP
~       ~ ~
dH  Td S  VdP
~ ~        ~             ~    ~     ~ ~
G  H  TS              dG  dH  TdS  S dT
~     ~      ~
dG   S dT  VdP
~ ~       ~             ~    ~      ~ ~
A  U  TS             dA  dU  Td S  S dT
~     ~          ~
dA   S dT  PdV
Exactness Property
These “thermodynamic” equations help one simplify the
mathematical equations:

Since U, H, G, and A are exact functions “the derivative
of the first coefficient with respect to the second variable
holding the first variable constant is equal to the
derivative of the second coefficient with respect to the
first variable holding the second variable constant”.
variables
~      ~     ~
dU  Td S  PdV

Coefficients
The Maxwell Equations
~      ~     ~       ~      ~ ~
dU  Td S  PdV     dH  Td S  VdP
~
 T     P        T      V 
 ~    ~              ~
 V  S
~  S V~     P  S  S  P
~
    
~     ~      ~      ~     ~        ~
dG   S dT  VdP   dA   S dT  PdV
~
~
 S     ~
 V           S     P 
                 ~        
 P   T           V 
 T       P             T  T V~
Differentiation Rule
~      ~ ~
dH  Td S  VdP

~          ~
 H       S    ~ P 
          
 P   T  P   V  P 
 T
    T     T

~         ~
 H      S  ~
      T  V
 P      P 
    T    T
Using Maxwell
Relationship with
~
Differential Rule
~
 H      S  ~
      T  V
 P      P 
    T    T

~           ~
 H        V    ~
              
 P   T  T   V
    T         P

Note: All these terms depend on
P,V, and T and can be found if
the EOS is known
Differentiation Rules
~      ~     ~
dU  Td S  PdV

~        ~
 U      S     V 
 ~   T  ~   P ~ 
 V     V 
 V  T      T      T

~
 U     S 
 ~   T ~   P
 V 
 V  T     T
Including Maxwells
~
 U     S 
 ~   T ~   P
 V 
 V  T     T

 U     P 
 ~   T     P
 V  T  T V~

Note: All these terms depend on
P,V, and T and can be found if
the EOS is known
Joule Thompson Coef
Show how the J-T coefficient depends on Cp and PVT
behavior
 T 
H       
 P  H
~
Joule Thompson Coef
Show how the J-T coefficient depends on Cp and PVT
behavior
 T 
H       
 P  H
~
~           ~                                         ~
~    H        H                        ~ ~           H 
dH                
 T  dT   P  dP                   dH  C p dT      
 P  dP
    P         T                                       T
~                   ~
 H    ~ T       H   P                              ~
      Cp
                               T     1  H 
 T  ~             P   T  ~
 T  H 
      ~      
 P 
    H           ~
T    H             P  H
~  Cp     T
~
 T     1   V     ~  Note: All these terms depend on
      ~ T      
 T   V  P,V, and T and can be found if
 P  H C p  
~
      P     the EOS is known

JT coefficient
Find the J-T coefficient for an ideal gas and for a “Clausius
Gas”

Clausius                   RT
Gas                    P ~
V b

Explain why a Clausius gas would always rise in temperature
as it flowed through a value.
J-T Coefficient
Show that if V is a constant (Like a liquid) that the temperature
will go up when the pressure drops through a valve.

In a valve, the pressure drops but the enthalpy
stays constant. The question is what the
temperature will do.

~
  V 
 T        1             ~
          ~ T     
 T   V 
 P  H
~     Cp  
      P    

Heat capacity
relationship
Show how the difference in Cp and Cv depends on PVT
relationships
~                         ~
~     H                ~     U 
Cp      
 T                CV  
~       
 T  ~
    P                       V
~               ~          ~            ~
 H             H       U        V       ~ P 
 V 
~ ~
dH  C p dT         dP                     P                   
 P             T  ~  T  ~       T  ~      T V
     T             V        V          V             ~

~                         ~
 H      ~  T         H   P 
                             
 T  ~  C p  T  ~   P   T  ~        ~              V  ~
     V           V         T   V      H       ~                    ~  P 
                         
 T  ~  C p   T  T   V  T  ~
     V          
          P     
      V
~      ~       ~      ~       ~ ~
dH  dU  dPV  dU  PdV  VdP
Heat capacity
relationship
~               V ~
 H      ~                   ~  P 
                        
 T  ~  C p   T  T   V  T  ~
    V           
         P    
     V
~        ~         ~
 H     U      V    ~ P 
            P     V 
    
 T  ~  T  ~   T  ~   T V
    V      V       V          ~
~
 H         ~ P 
  CV  V 
~
                   
 T  ~   ~
 T V
    V                ~
~
~     ~       V   P 
C p  CV  T 
~         
 T   T  ~
    P    V
Show how Cp and Cv are related for an ideal gas:

~
~     ~       V   P 
C p  CV  T 
~         
 T   T  ~
    P    V
Show how Cp and Cv are related for a constant volume
fluid:

~
~     ~       V   P 
C p  CV  T 
~         
 T   T  ~
    P    V
Ideal gases
~
  V 
~ ~                       ~

dH  C p dT   T      V  dP

  T  P
              


~ ~           P       ~
dU  CV dT  T 
~               P  dV
  T V~


Show that dH = CpdT for an ideal gas
Show that dU = CvdT for an ideal gas
Show that dS = CpdT/T – RdP/P for an ideal gas
Liquids
~
  V 
~ ~                       ~

dH  C p dT   T      V  dP

  T  P
              


~ ~           P       ~
dU  CV dT  T 
~               P  dV
  T V~


Show that dH = CpdT + VdP for a constant specific volume
fluid (approx by a liquid)
Show that dU = CvdT for a constant volume fluid
Show that dS = CpdT/T for a constant volume fluid
van der Waals gas
For a vdW gas, it is possible to calculate the difference in the
Internal energy between two temperatures and pressures
fairly easily.

Problem 6
For a gas obeying the van der Waals equation:

RT   a
P ~    ~2                       Find an expression for the
V b V                         difference U1-U2.

~
~               ~
T1          V1
  P       ~
U 1 (T1 ,V1 )  U 2 (T2 ,V2 )   CV dT    T 
 T ~   P dV

V 2     V    
~
T2
3 Step Process
• Step 1 – Calculate the difference between
the internal energy at starting condition
and the internal energy when the volume
is large (ideal gas behavior) at the same
temperature.
~
~ ~
V1
  P        ~    a a
U 1  U 1 *   T       P dV   ~  
V  
    T V
~
       1   
Evaluated at T1
3 Step Process
• Step # 2 – Calculate the difference in
internal energy between T1 and T2 at
constant V. (V = infinity)
T1
~                ~                  ~
U 1 * (T1 , )  U 2 * (T2 , )   CV * dT
T2
3 Step Process
• Calculate the difference between the
internal energy at T2 and large volume
(ideal gas) and at T2 and the final volume

~      ~

  P       ~    a a     
U 2 * U 2 
V
2T      P dV    ~
~   T  V~

 V
    2




Evaluated at T2
Problem 7. A heat exchanger is used to cool 2 kg/s of 50
bar HFC 134a from 140 C to 50 C using 25 C air. Use van
der Waals equation and Cp = 100 J/mole-K to calculate
the rate of heat release to the surroundings.

Steps:   1. Energy balance gives heat release = m(Hin-
Hout)
2. Hin – Hout = Uin – Uout + PinVin-PoutVout
3. Calculate Uin – Uout
4. Solve vdW for Vin and Vout and then
calculate PinVin -PoutVout
Summary
• Thermodynamic properties have both
mathematical and thermodynamic definitions
• The mathematical relationships can be
manipulated so that property changes can be
related to heat capacity data and PVT
relationships.
• The heat capacity and EOS are used to solve for
the way H,U, and S are related to changes in
T,P, and V
Example calculations using vdW equation

Problem 1.
Rework Problem 6 of worksheet 7 step by step to find an
expression for the difference in internal energy between
condition 1 U1(T1,V1) and condition 2 U2(T2,V2) for a van
der Waals gas that depends on (only) the ideal gas constant
volume heat capacity (Cv) and the van der Waals
parameters a and b.
Problem 2.
Use the van der Waals equation to compute the specific
volumes of methane at 50 bar and 330 K (condition 1) and
50 bar, 200 K (condition 2).
(Note: Find the values of a and b using either CRC or,
failing that, the van der Waals equations for a and b given
in the notes)
Problem 3.
Use the van der Waals equation and an ideal gas
constant pressure heat capacity (Cp) value (35
J/mole-K) to compute the internal energy difference
for methane between 50 bar, 330 K (condition 1) and
50 bar, 200 K (Condition 2).
(How is Cp related to Cv in the ideal gas state?)
Problem 4
Use the results from Problem 2 and 3 to compute the
enthalpy difference for methane between 50 bar, 330 K
(condition 1) and 50 bar, 200 K (Condition 2).
(What is the relationship between U and H?)

```
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