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Property
Relationships
         Chapter 6




                     1
: The T-ds relations
Apply the differential form of the first law for a closed
stationary system for an internally     reversible
process

   δQint rev − δWint rev = dU
δQint rev = TdS           δWint rev = PdV


           TdS − PdV = dU                                   2
Divide by the
mass, you get     Tds = du + Pdv
This equation is known as:
First Gibbs equation or First Tds relationship
Divide by T, ..          du Pdv
                    ds =   +
                         T   T
Although we get this form for internally reversible process,
we still can compute ∆s for an irreversible process. This is
because S is a point function.


                                                               3
Second T-ds (Gibbs) relationship
Recall that…   h = u + Pv
  Take the differential for both sides

   dh = du + Pdv + vdP
  Rearrange
  to find du         du = dh − Pdv − vdP
 Substitute in the
 First Tds
 relationship   Tds = du + Pdv
                Tds = dh − vdP
   Second Tds relationship, or Gibbs equation
                                                4
dh vdP
Divide by T, ..   ds =   −
                       T   T
Thus We have two equations for ds
      du Pdv                         dh vdP
 ds =   +                       ds =   −
      T   T                          T   T
  To find ∆s, we have to integrate these equations. Thus
  we need a relation between du and T (or dh and T).
  Now we can find entropy change (the LHS of the
  entropy balance) for liquids and solids


                                                           5
Entropy Change of Liquids- 2
and Solids
Solids and liquids do not change specific volume appreciably
with pressure. That means that dv=0, so the first equation is
the easiest to use.
                          0
     du Pdv                   Thus For           du
ds =   +                      solids and    ds =
     T   T                    liquids
                                                 T
, Recall also that For solids and liquids   du = CdT
               du CdT
        ∴ ds =   =
               T   T                                       6
Integrate to             T2     Only true
give…          ∆s = C ln 
                        T       for solids
                         1      and
                                  liquids!!
What if the process                    T2 
                        ∆s = 0, ⇒ C ln  = 0
                                      T 
?is isentropic
                                       1

                                 T1 = T2
The only way this expression
can equal 0 is if,


 Hence, for solids and liquids, isentropic
 processes are also isothermal.                 7
Example(6-7): Effect of Density of a Liquid on
Entropy
   Liquid methane is commonly used in various cryogenic
   applications. The critical temperature of methane is 191 K
   (or -82oC), and thus methane must be maintained below
   191 K to keep it in liquid phase. The properties of liquid
   methane at various temperatures and pressures are given
   next page.
   Determine the entropy change of liquid methane as it
   undergoes a process from 110 K and 1 MPa to 120 K and 5
   MPa
(a) using actual data for methane and
(b) approximating liquid methane as an incompressible
    substance. What is the error involved in the later case?


                                                               8
∆s = s 2 − s 2 = 5.145 − 4.875 = 0.270kj / kgK

          T2         T2           120 
∆s = C ln  = Cavg ln  = 3.4785 ln
         T          T                   = 0.303kj / kgK
          1          1            110                    9
Example (6-19): Entropy generated when a hot
block is dropped in a lake

A 50-kg block of iron casting at 500 K is dropped in a large
lake that is at 285 K. The block reaches thermal equilibrium
with lake water.
Assuming an average specific heat of 0.45 kJ/kg.K for the
iron, determine:
(a) The entropy change of the iron block,
(b) The entropy change of the water lake,
(c) the entropy generated during this process.




                                                            10
(a) The entropy change of the iron block,

                   T2                285 
  ∆Siron            = 50 × 0.45 × ln
           = mC ln                         = −12.65 kj / K
                   T1                500 
   we need also to find Q coming out of the system.

    Q − W = ∆U          Q − 0 = mC (T2 − T1 )
    Q = 50 × 0.45(285 − 500) = −4838 kJ
                                                      Tsurr= 285 K
  (b) The entropy change of the water lake,
                                                           T=500K
                   Q      4838
       ∆Slake =         =      = 16.97 kJ / K
                  Tlake    285




                                                                     11
(c) the entropy generated during this process.

Choose the iron block and the lake as the system
and treat it is an isolated system.



                                           Tsurr= 285 K
Thus Sg = ∆Stot = ∆Ssys + ∆Slake
                                                T=500K

Sg = ∆Stot = -12.6 + 16.97
               = 4.32                    System
                                        boundary

                                                          12
The Entropy Change of Ideal- 3
Gases, first relation
First relation
The entropy change of an ideal gas can be obtained by
substituting du = CvdT and P /T= R/υ into Tds relations:

Tds = du + pd υ         du Pd υ          dT    dυ
                   ds =   +     ds = C v    +R
                        T   T            T     υ


  integrating                 2
                                       dT       υ2
                             1
                              ∫
                  ⇒ s2 − s1 = Cv ( T )
                                       T
                                          +R ln
                                                υ1
                                                       13
Second relation
A second relation for the entropy change of an ideal gas
for a process can be obtained by substituting dh = CpdT
and υ /T= R/P into Tds relations:
                       dh vdp                  dT    dp
Tds = dh −vdp     ds =   −            ds = C p    −R
                       T   T                   T      p

   integrating
             2
                       dT        P2
            1
             ∫
   s2 − s1 = C p ( T )
                       T
                          − R ln
                                 P1

                                                      14
2
                          dT       υ2
     s2 − s1 = ∫ Cv ( T )    +R ln
               1
                          T        υ1
                2
                         dT        P2
              1
               ∫
     s2 − s1 = C p ( T )
                         T
                            − R ln
                                   P1



     The integration of the first term on the RHS can be done via
      two methods:

1.    Assume constant Cp and constant Cv (Approximate Analysis)


2.    Evaluate these integrals exactly and tabulate the data (Exact
      Analysis)
                                                                      15
Method 1: Constant specific
)heats (Approximate Analysis
First relation
         2
                     dT      υ2            T2      v2 
s2 − s1 = ∫ Cv ( T )    +R ln ⇒ ∆s = Cv ln  + R ln 
                                          T       v 
          1
                     T       υ1            1       1
Only true for ideal gases, assuming constant heat capacities

Second relation
         2
                      dT       P2             T2      P2 
s2 − s1 = ∫ C p ( T )    − R ln ⇒ ∆s = C p ln  − R ln 
                                             T       P
          1
                      T        P1             1       1
Only true for ideal gases, assuming constant heat capacities
                                                                16
Sometimes it is more convenient to calculate the
change in entropy per mole, instead of per unit mass



                     T2       v2 
∆s = s2 − s1 = Cv ln  + Ru ln 
                    T        v             kJ/kmol. K
                     1        1
                      T2       P2 
∆s = s2 − s1 = C p ln  − Ru ln 
                     T        P 
                                                kJ/kmol. K
                      1        1

      Ru is the universal gas constant

                                                         17
Method 2: Variable specific
heats (Exact Analysis)
                              2   C p dT        P2
We use the
second relation
                   ∆s =   ∫
                          1         T
                                         − R ln
                                                P1

 Wecould substitute in the equations for Cv
 and Cp, and perform the integrations
    Cp = a + bT + cT2 + dT3
    But this is time consuming.
 Someone   already did the integrations and
 tabulated them for us (table A-17)
    They assume absolute 0 as the starting point
                                                     18
The integral is expressed as:
    T2            dT         T2            dT    T1        dT
∫T1
         C p (T )
                  T
                     =   ∫
                         0
                                  C p (T )
                                           T
                                              − ∫ C p (T )
                                                 0         T
      T2
                   dT
∫                     = s2 − s1
                          0     0
           Cp( T )
    T1             T
                                  T
                                              dT
                  s =        ∫
                    0
Where                                 Cp( T )
                              0               T
is tabulated in Table A-17

             Therefore
                                  P2
             ∆s = s − s − R ln          0
                                        2
                                                   0
                                                   1
                unit : kJ / kg .K P1                            19
Is s = f (T) only? like u for an
.ideal gas Let us see
       Temperature dependence
                                                 Pressure

                                 P2             dependence

               ∆s = s − s − R ln
                         0
                         2
                                0
                                1
                                 P1

    From this equation, It can be seen that the
     entropy of an ideal gas is not a function only of
     the temperature ( as was the internal energy) but
     also of the pressure or the specific volume.
    The function s° represents only the temperature-
     dependent part of entropy
                                                      20
How about the other relation
                    2
                           dT        v2
         ∆s =   ∫
                1
                        Cv
                           T
                              + R ln
                                     v1

 We can develop another relation for the
 entropy changed based on the above relation
 but this will require the definition of another
 function and tabulating it which is not
 practical.            T
                                dT
                        ?=   ∫
                             0
                                 Cv ( T )
                                            T
                                                    21
6-4   Isentropic Processes
 The entropy of a fixed mass can be
  changed by
 1. Heat transfer,

 2. Irreversibilities
 It follows that the entropy of a
  system will not change if we have
 1. Adiabatic process,

 2. Internally reversible process.
 Therefore, we define the following:
                                        22
Isentropic Processes of Ideal
Gases
 Many   real processes can be modeled as
  isentropic

 Isentropic
           processes are the standard against
  which we should measure efficiency

 We   need to develop isentropic relationships
  for ideal gases, just like we developed for
  solids and liquids
                                                  23
)Constant specific heats (1st relation
Recall                 T2      v2 
            ∆s = Cv ln  + R ln 
                      T       v 
                       1       1
For the isentropic case, ∆S=0. Thus
                                                                  R
      T2        v2                  T2  R  v2     v1    Cv
Cv ln  = − R ln 
     T         v                 ln  = − ln  = ln 
                                       T 
      1         1                   1   Cv  v1 
                                                  
                                                         v 
                                                          2
Recall also from ch 2, the following relations..…

  R = C p − Cv ⇒ R / Cv = C p / Cv − 1 = k − 1
                              k −1        Only applies to
      T2   v1 
    ∴  =  
     T  v 
                                          ideal gases,
                                          with constant
      1  2                            specific heats           24
Constant specific heats (2nd
 )relation        T2   P2 
                  ∆s = C p ln  − R ln  = 0
                             T       P 
                              1       1
                                    R
     T2  R  P2       P2       Cp
  ln  =
     T  C ln P  = ln P 
                       
     1    p  1       1
                                             k −1
Recall..…   R / Cv = k − 1    or    R /C p =
                                              k
                             k −1
                                         Only applies to
   T2   P2                k
                                         ideal gases,
 ∴  =  
  T   P                              with constant
   1  1                              specific heats
                                                           25
…Since
                 k −1                                    k −1
 T2   v1                          T2   P2         k
 = 
T  v                      and     = 
                                     T   P 
 1  2                             1  1
                                           k −1
                          k −1
  HENCE           v1             P2     k
                  
                 v             = 
                                  P
                  2              1

                          k
Which can be      v1   P2                     Third
simplified to…     = 
                 v   P                        isentropic
                  2  1                        relationship
                                                                 26
Full form of Isentropic relations of Ideal Gases
                  k −1                   k −1        k
  T2   v1             T2   P2     k      v1   P2 
  = 
 T  v                 =                   = 
                                                v   P 
                         T   P 
  1  2                1  1               2  1
Compact form
                            1− k

Tv   k −1
            = constant TP    k
                                   = constant    Pv k = constant
  Valid for only for
  1- Ideal gas

  2- Isentropic process

  3- Constant specific heats
                                                               27
That works if the specific heat constants can
be approximated as constant, but what if
?that’s not a good assumption


We need to use the
exact treatment           0               P2 
                        ∆s = s − s − R ln 
                              0
                              2
                                   0
                                   1     P
                                          1
This equation is a
good way to evaluate                 P2 
property changes,       s = s + R ln 
                         0
                         2
                              0
                              1     P
but it can be tedious
if you know the                      1
volume ratio instead
of the pressure ratio                             28
 P2 
s = s + R ln 
 0
 2
      0
      1     P
                           s20 is only a function
             1           of temperature!!!

s −s
 0    0
          P2               Rename the exponential term
 2
     = ln 
      1
         P                 as Pr , (relative pressure)
  R       1                which is only a function of
                             temperature, and is tabulated
P2       s2 − s10 
           0
                             on the ideal gas tables
   = exp
         R       
P1                
         s2 
             0
                           s2 
                               0
     exp 
P2            R÷
                
                       exp 
                                R ÷ Pr 2
                                   =               P2 Pr 2
   =                                                  =
P1        s10             s10  Pr 1             P1 Pr 1
     exp              exp 
               R÷               R÷
                                                            29
You can use either of the following 2 equations
  P2 Pr 2                           P2 
     =                 s = s + R ln 
                        0
                        2
                              0
                              1    P
  P Pr1
   1                                1
 This is good if you know the pressure ratio but how about
 if you know only the volume ratio
In this case, we use the ideal gas law
P v1 P2 v2             v2 T2 P      T2 Pr1      T2 Pr1         vr 2
 1
    =      ⇒             =    1
                                  =           =              =
T1    T2               v1 T1 P2     T1 Pr 2     Pr 2 T1        vr1
                             v2 vr 2
where   vr = T / P r        ∴ =
                             v1 vr1
  Remember, these relationships only hold for ideal gases and
  isentropic processes                                     30
Example (6-10):
Isentropic Compression of Air in a Car Engine
Air is compressed in a car engine from 22oC and 95
kPa in a reversible and adiabatic manner. If the
compression ratio V1/V2 of this piston-cylinder device
is 8, determine the final temperature of the air.
<Answer: 662.7 K>
Sol:




                                                         31

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Ch06b entropy

  • 2. : The T-ds relations Apply the differential form of the first law for a closed stationary system for an internally reversible process δQint rev − δWint rev = dU δQint rev = TdS δWint rev = PdV TdS − PdV = dU 2
  • 3. Divide by the mass, you get Tds = du + Pdv This equation is known as: First Gibbs equation or First Tds relationship Divide by T, .. du Pdv ds = + T T Although we get this form for internally reversible process, we still can compute ∆s for an irreversible process. This is because S is a point function. 3
  • 4. Second T-ds (Gibbs) relationship Recall that… h = u + Pv Take the differential for both sides dh = du + Pdv + vdP Rearrange to find du du = dh − Pdv − vdP Substitute in the First Tds relationship Tds = du + Pdv Tds = dh − vdP Second Tds relationship, or Gibbs equation 4
  • 5. dh vdP Divide by T, .. ds = − T T Thus We have two equations for ds du Pdv dh vdP ds = + ds = − T T T T To find ∆s, we have to integrate these equations. Thus we need a relation between du and T (or dh and T). Now we can find entropy change (the LHS of the entropy balance) for liquids and solids 5
  • 6. Entropy Change of Liquids- 2 and Solids Solids and liquids do not change specific volume appreciably with pressure. That means that dv=0, so the first equation is the easiest to use. 0 du Pdv Thus For du ds = + solids and ds = T T liquids T , Recall also that For solids and liquids du = CdT du CdT ∴ ds = = T T 6
  • 7. Integrate to  T2  Only true give… ∆s = C ln  T  for solids  1 and liquids!! What if the process  T2  ∆s = 0, ⇒ C ln  = 0 T  ?is isentropic  1 T1 = T2 The only way this expression can equal 0 is if, Hence, for solids and liquids, isentropic processes are also isothermal. 7
  • 8. Example(6-7): Effect of Density of a Liquid on Entropy Liquid methane is commonly used in various cryogenic applications. The critical temperature of methane is 191 K (or -82oC), and thus methane must be maintained below 191 K to keep it in liquid phase. The properties of liquid methane at various temperatures and pressures are given next page. Determine the entropy change of liquid methane as it undergoes a process from 110 K and 1 MPa to 120 K and 5 MPa (a) using actual data for methane and (b) approximating liquid methane as an incompressible substance. What is the error involved in the later case? 8
  • 9. ∆s = s 2 − s 2 = 5.145 − 4.875 = 0.270kj / kgK  T2   T2   120  ∆s = C ln  = Cavg ln  = 3.4785 ln T  T   = 0.303kj / kgK  1  1  110  9
  • 10. Example (6-19): Entropy generated when a hot block is dropped in a lake A 50-kg block of iron casting at 500 K is dropped in a large lake that is at 285 K. The block reaches thermal equilibrium with lake water. Assuming an average specific heat of 0.45 kJ/kg.K for the iron, determine: (a) The entropy change of the iron block, (b) The entropy change of the water lake, (c) the entropy generated during this process. 10
  • 11. (a) The entropy change of the iron block,  T2   285  ∆Siron   = 50 × 0.45 × ln = mC ln   = −12.65 kj / K  T1   500  we need also to find Q coming out of the system. Q − W = ∆U Q − 0 = mC (T2 − T1 ) Q = 50 × 0.45(285 − 500) = −4838 kJ Tsurr= 285 K (b) The entropy change of the water lake, T=500K Q 4838 ∆Slake = = = 16.97 kJ / K Tlake 285 11
  • 12. (c) the entropy generated during this process. Choose the iron block and the lake as the system and treat it is an isolated system. Tsurr= 285 K Thus Sg = ∆Stot = ∆Ssys + ∆Slake T=500K Sg = ∆Stot = -12.6 + 16.97 = 4.32 System boundary 12
  • 13. The Entropy Change of Ideal- 3 Gases, first relation First relation The entropy change of an ideal gas can be obtained by substituting du = CvdT and P /T= R/υ into Tds relations: Tds = du + pd υ du Pd υ dT dυ ds = + ds = C v +R T T T υ integrating 2 dT υ2 1 ∫ ⇒ s2 − s1 = Cv ( T ) T +R ln υ1 13
  • 14. Second relation A second relation for the entropy change of an ideal gas for a process can be obtained by substituting dh = CpdT and υ /T= R/P into Tds relations: dh vdp dT dp Tds = dh −vdp ds = − ds = C p −R T T T p integrating 2 dT P2 1 ∫ s2 − s1 = C p ( T ) T − R ln P1 14
  • 15. 2 dT υ2 s2 − s1 = ∫ Cv ( T ) +R ln 1 T υ1 2 dT P2 1 ∫ s2 − s1 = C p ( T ) T − R ln P1  The integration of the first term on the RHS can be done via two methods: 1. Assume constant Cp and constant Cv (Approximate Analysis) 2. Evaluate these integrals exactly and tabulate the data (Exact Analysis) 15
  • 16. Method 1: Constant specific )heats (Approximate Analysis First relation 2 dT υ2  T2   v2  s2 − s1 = ∫ Cv ( T ) +R ln ⇒ ∆s = Cv ln  + R ln  T  v  1 T υ1  1  1 Only true for ideal gases, assuming constant heat capacities Second relation 2 dT P2  T2   P2  s2 − s1 = ∫ C p ( T ) − R ln ⇒ ∆s = C p ln  − R ln  T  P 1 T P1  1  1 Only true for ideal gases, assuming constant heat capacities 16
  • 17. Sometimes it is more convenient to calculate the change in entropy per mole, instead of per unit mass  T2   v2  ∆s = s2 − s1 = Cv ln  + Ru ln  T  v  kJ/kmol. K  1  1  T2   P2  ∆s = s2 − s1 = C p ln  − Ru ln  T  P  kJ/kmol. K  1  1 Ru is the universal gas constant 17
  • 18. Method 2: Variable specific heats (Exact Analysis) 2 C p dT P2 We use the second relation ∆s = ∫ 1 T − R ln P1  Wecould substitute in the equations for Cv and Cp, and perform the integrations  Cp = a + bT + cT2 + dT3  But this is time consuming.  Someone already did the integrations and tabulated them for us (table A-17)  They assume absolute 0 as the starting point 18
  • 19. The integral is expressed as: T2 dT T2 dT T1 dT ∫T1 C p (T ) T = ∫ 0 C p (T ) T − ∫ C p (T ) 0 T T2 dT ∫ = s2 − s1 0 0 Cp( T ) T1 T T dT s = ∫ 0 Where Cp( T ) 0 T is tabulated in Table A-17 Therefore P2 ∆s = s − s − R ln 0 2 0 1 unit : kJ / kg .K P1 19
  • 20. Is s = f (T) only? like u for an .ideal gas Let us see Temperature dependence Pressure P2 dependence ∆s = s − s − R ln 0 2 0 1 P1  From this equation, It can be seen that the entropy of an ideal gas is not a function only of the temperature ( as was the internal energy) but also of the pressure or the specific volume.  The function s° represents only the temperature- dependent part of entropy 20
  • 21. How about the other relation 2 dT v2 ∆s = ∫ 1 Cv T + R ln v1  We can develop another relation for the entropy changed based on the above relation  but this will require the definition of another function and tabulating it which is not practical. T dT ?= ∫ 0 Cv ( T ) T 21
  • 22. 6-4 Isentropic Processes  The entropy of a fixed mass can be changed by 1. Heat transfer, 2. Irreversibilities  It follows that the entropy of a system will not change if we have 1. Adiabatic process, 2. Internally reversible process.  Therefore, we define the following: 22
  • 23. Isentropic Processes of Ideal Gases  Many real processes can be modeled as isentropic  Isentropic processes are the standard against which we should measure efficiency  We need to develop isentropic relationships for ideal gases, just like we developed for solids and liquids 23
  • 24. )Constant specific heats (1st relation Recall  T2   v2  ∆s = Cv ln  + R ln  T  v   1  1 For the isentropic case, ∆S=0. Thus R  T2   v2   T2  R  v2   v1  Cv Cv ln  = − R ln  T  v  ln  = − ln  = ln  T   1  1  1 Cv  v1    v   2 Recall also from ch 2, the following relations..… R = C p − Cv ⇒ R / Cv = C p / Cv − 1 = k − 1 k −1 Only applies to  T2   v1  ∴  =   T  v  ideal gases, with constant  1  2 specific heats 24
  • 25. Constant specific heats (2nd )relation  T2   P2  ∆s = C p ln  − R ln  = 0 T  P   1  1 R  T2  R  P2   P2  Cp ln  =  T  C ln P  = ln P       1 p  1  1 k −1 Recall..… R / Cv = k − 1 or R /C p = k k −1 Only applies to  T2   P2  k ideal gases, ∴  =   T   P  with constant  1  1 specific heats 25
  • 26. …Since k −1 k −1  T2   v1   T2   P2  k  =  T  v  and  =  T   P   1  2  1  1 k −1 k −1 HENCE  v1   P2  k   v  =  P  2  1 k Which can be  v1   P2  Third simplified to…   =  v   P  isentropic  2  1 relationship 26
  • 27. Full form of Isentropic relations of Ideal Gases k −1 k −1 k  T2   v1   T2   P2  k  v1   P2   =  T  v   =    =  v   P  T   P   1  2  1  1  2  1 Compact form 1− k Tv k −1 = constant TP k = constant Pv k = constant  Valid for only for  1- Ideal gas  2- Isentropic process  3- Constant specific heats 27
  • 28. That works if the specific heat constants can be approximated as constant, but what if ?that’s not a good assumption We need to use the exact treatment 0  P2  ∆s = s − s − R ln  0 2 0 1 P  1 This equation is a good way to evaluate  P2  property changes, s = s + R ln  0 2 0 1 P but it can be tedious if you know the  1 volume ratio instead of the pressure ratio 28
  • 29.  P2  s = s + R ln  0 2 0 1 P s20 is only a function  1 of temperature!!! s −s 0 0  P2  Rename the exponential term 2 = ln  1 P  as Pr , (relative pressure) R  1 which is only a function of temperature, and is tabulated P2  s2 − s10  0 on the ideal gas tables = exp  R   P1   s2  0 s2  0 exp  P2  R÷  exp   R ÷ Pr 2 = P2 Pr 2 = = P1  s10   s10  Pr 1 P1 Pr 1 exp  exp  R÷  R÷   29
  • 30. You can use either of the following 2 equations P2 Pr 2  P2  = s = s + R ln  0 2 0 1 P P Pr1 1  1 This is good if you know the pressure ratio but how about if you know only the volume ratio In this case, we use the ideal gas law P v1 P2 v2 v2 T2 P T2 Pr1 T2 Pr1 vr 2 1 = ⇒ = 1 = = = T1 T2 v1 T1 P2 T1 Pr 2 Pr 2 T1 vr1 v2 vr 2 where vr = T / P r ∴ = v1 vr1 Remember, these relationships only hold for ideal gases and isentropic processes 30
  • 31. Example (6-10): Isentropic Compression of Air in a Car Engine Air is compressed in a car engine from 22oC and 95 kPa in a reversible and adiabatic manner. If the compression ratio V1/V2 of this piston-cylinder device is 8, determine the final temperature of the air. <Answer: 662.7 K> Sol: 31