Example 5.1
Derive an expression for enthalpy change of a gas during an isothermal process assuming using
the following EOS: P (V − b) =
RT

 ∂V  
dH= C p dT + V − T 
  dP
 ∂T  P 

P (V-b) = RT
∂V 
R
 =
 ∂T  P P
⇒ 


dH = C p dT +  V −
∆
=
H
RT 
C p dT + bdP
 dP =
P 
∫ C dT + b( P − P )
P
2
1
Example 5.2
Derive an expression for enthalpy change of a gas during an isothermal process assuming
using the following EOS: Z = 1 + AP r / T r
r
 ∂Z  dPr
HR
= −Tr 2 ∫ 

∂Tr  P Pr
RTc
0 
r
P
r
AP  dP
HR
2 
=
−Tr ∫  − 2r  r =
APr
RTc
Tr  Pr
0 
P
∆H =
∆H ig + H 2R − H1R
=
∆H ig + A( Pr 2 − Pr1 ) RTc
Example 5.3
Derive expressions for HR, SR from RK-EOS.
General expressions are:
H= RT ( Z − 1) +
R
  ∂P 

 − P  dV
T 
∂T V

V =∞  
V
∫
 ∂P 
R
 −  dV

∂T V V 
V =∞ 
..(1)
V
SR =
R ln Z +
∫
RT
a
RT
a′
= −
P = −
V − b V (V + b) V − b V (V + b) T
..(2)
..(3)
=
Where, a′
0.42748 R 2TC2.5
0.08664
;b
=
RTC
PC
PC
 R

a'
 ∂P 
−P T 
+
∴ T
=
3/2 
 ∂T V
V − b 2V (V + b)T 
∴
-
 RT

a'
3a '
V − b − V (V + b)T 1/2  = 2V (V + b)T 1/2


V
3a '
H
RT ( Z − 1) +
∴=
2 T
dV
V (V + b)
V =∞
∫
R
Putting Z = PV / RT
H R PV – RT −
=
3a V + b 
ln
2b  V 
Replacing P we get: H R =
bRT
a
3a V + b 
−
− ln 
(V − b ) (V + b ) 2b  V 
Similarly using (2) and (3) one may show:
S R =R ln
( PV ) + R ln  V − b  −
RT

 V
a
V +b 
ln 


 2bT  V 
Example 5.4
Carbon dioxide at upstream conditions T 1 = 350 K and P 1 = 80 bar is throttled to a
downstream pressure of 1.2 bar. Estimate the downstream temperature and ΔS of the gas.
1.157
−3
For CO 2 : C ig
× 105
p / R = 5.457 + 1.045 × 10 T −
2
T
T 1 = 350K, P 1 = 80 bar, T C = 304.2K, P C = 73.8 bar and ω = 0.224
For the process H 2 – H 1 = 0 (Isenthalpic, from energy balance)
Now H 2 – H 1 = H 2R − H1R + ∆H ig ...........( A)
Tr 1 = T 1 /T c = 1.151, Pr 1 = P 1 /P c = 1.084
If one assume that at 1.2 bar (at exit), the gas is ideal, then:
ig
H 2R  0; then ∆H=
T2
∫C
T1
ig
p
=
dT H1R ….. (B)
Use generalized correlations for residual properties, and read from relevant figures for
residual properties to find H1R at given Tr 1 , & Pr 1 and then solve Equation (B) by trial &
error to get T 2  280K

T
P 
Thus: ∆S
= S 2R − S1R + ∆S ig  − S1R + C p ln 2 − R ln 2   31.5 J / molK
T
P

1
1

Example 5.5
Estimate the final temperature and the work required when 1 mol of n-butane is compressed
isentropically in a steady-flow process from 1 bar and 50oC to 7.8 bar.
ω = 0.2, T c = 425.1 K, P c = 37.96 bar
T 1 = 323 K, P 1 = 1 bar, P 2 = 7.8 bar
For the process, ∆S =
0
T r2 = 0.76017, P r2 = 0.2052,
P r1 = 0.02639; hence we assume that at state 1 the residual properties are zero as the gas is at
ideal state.
For n-butane : C ig
1.9 + 36.9 × 10-3 T − 11.4 × 10-6 T 2
p / R =
Using generalized correlations for state 2 and reading from relevant figures for residual
properties one finally obtains the following residual property values at ‘2’:
(H )
From generalized correction at point ‘2’
R o
RTC
=
∆S
∫
T2
T1
Cp
(S )
R o
= −0.5679 ,
RTC
= −0.05210
P
dT
− R ln 2 + S 2R − S1R
T
P1
Assume T 2 and solve iteratively with ∆S = 0 to obtain the final value of T 2 which is = 381K.
Next use generalized correlation figures for residual enthalpies; whereby at point ‘2’:
HR
= −0.30330
RTc
W = ∆H =
∆H ig + RTc ( H 2R − H1R )
=
T2
∫C
ig
p
dT + RTc ( H 2R − H1R )
T1
Finally, W = 5678 J/mol
Example 5.6
Calculate the changes in enthalpy and entropy per mole when a mixture of 70 mole %
ethylene (1) and 30 mole% propylene (2) at 323K and 10 bar is taken to 60 bar and 600 K
using the generalized compressibility factor approach.
4.196 + 154.565 x10−3 T − 81.076 x10−6 T 2 + 16.813 x10−9 T 3 ;
C igp1 =
3.305 + 235.821x10−3 T − 117.58 x10−6 T 2 + 22.673 x10−9 T 3
C igp2 =
Use pure species data to compute those of mixture.
Thus,
(1 = ethylene, 2 = Propylene)
Similarly
At state 1,
Now,
From Tables at given T r,mix and P r,mix
At state 1
and
Thus,
Similarly on repeating the calculation for state 2, T = 600K, P = 60 bar, we obtain
Now,
Thus,
Where,
For component 1,
Similarly,
Similarly,
Hence,
Now,
Final answer on substitution of all parameters
Example 5.7
A certain gas is compressed adiabatically from 293 K and 135 KPa to 550 KPa. What is the
work needed? What is the final T 2 ? Assume ideal gas behavior. Compressor η = 0.8.
For the gas: C p ig = 1.65 + 8.9 x 10-3T – 2.2 x 10-6 T2
For rev. process ∆S = S 2 -S 1 = 0
Flow process thus lead to:
(1)
(2)
∆S =
T2
ig
∫ Cp
T1
P
2
dT
dP
∴∆
−R∫
T
P
P1

W
s
∴
T2
∫
(1.65 + 8.9 x 10-3 T – 2.2 x 10-6 T2)
293
⇒ T 2 reversible (by iteration) ~ 395 K
dT
= 8.314 ln
T
 550 


 135 
By first Law Q (=0) + W s = ∆H
∴ W s (isentropic, reversible) = H 2 – H 1 =
T2
∫C
ig
p
dT
T1
395
∴W
reversible
S
−3
−6 2
=
∫ [1.65 + 8.9 x 10 T − 2.2 x 10 T ) dT ~ 3960 J/mol.
293
∴ Thus actual work needed = WSrev η
= 3960/0.8 = 4950 J/mol
∴ W
T2
rev
S
= ∆H = ∫ C igp dT
T1
∴ 4950 =
T2irrev
∫
[1.65 + 8.9 x 10−3 T − 2.2 x10−6 T 2 ] dT
293
⇒ T 2 (actual) ~ 420oK ⇒ 147oC