Table 3.1
αTrω,
()
1 0.480 1.574ω+ 0.176ω2
−
()
1T
r
2
−
⎛
⎜
⎝
⎞
⎠
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
:=
B10.139 0.172
Tr4.2
−:= B10.02=
(c) For Redlich/Kwong EOS:
σ1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1
αTr() T
r0.5−
:= Table 3.1 qT
r
()
Ψα Tr
()
⋅
ΩTr
⋅
:= Eq. (3.54)
βTrPr
,
()
ΩPr
⋅
Tr
:= Eq. (3.53)
41
Guess: Z 0.9:=
Given Eq. (3.52)
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅
⋅−=
3.35 T 523.15 K⋅:= P 1800 kPa⋅:=
(a) B 152.5−cm3
mol
⋅:= C 5800−cm6
mol2
⋅:= VRT⋅
P
:= (guess)
Given PV⋅
RT⋅1B
V
+C
V2
+=V Find V():=
Calculate Z Guess: Z 0.9:=
Given Eq. (3.52)
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅
⋅−=
(e) For Peng/Robinson EOS:
σ12+:= ε 12−:= Ω 0.07779:= Ψ 0.45724:= Table 3.1
Table 3.1
αTr ω,
()
1 0.37464 1.54226ω+ 0.26992ω2
−
()
1T
r
1
2
−
⎛
⎜
⎝
⎞
⎠
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
2
:=
qT
r
()
Ψα Trω,
()
⋅
ΩTr
⋅
:= Eq. (3.54) βTrPr
,
()
ΩPr
⋅
Tr
:= Eq. (3.53)
Calculate Z
42
3.37 B 53.4−cm3
mol
⋅:= C 2620 cm6
mol2
⋅:= D 5000 cm9
mol3
⋅:= n mol:=
T 273.15 K⋅:=
Given PV⋅
RT⋅1B
V
+C
V2
+D
V3
+=fPV,( ) Find V():=
i010..:= Pi10 10−20 i⋅+
()
bar⋅:= Vi
RT⋅
:= (guess)
(b) Tc647.1 K⋅:= Pc220.55 bar⋅:= ω 0.345:=
Tr
T
Tc
:= Pr
P
Pc
:= B00.083 0.422
Tr1.6
−:=
Tr0.808=Pr0.082=B00.51−=
B10.139 0.172
Tr4.2
−:= B10.281−= Z1B
0ωB1
⋅+
()
Pr
Tr
⋅+:=
(c) Table F.2: molwt 18.015 gm
mol
⋅:= V 124.99 cm3
gm
⋅molwt⋅:=
43
Zi
1
0.953
0.906
=Z1i
1
0.953
0.906
=Z2i
1
0.951
0.895
=
-10
1·10
20
40
60
0.7
0.9
1
Z2i
44
Eq. (3.53)
Calculate Z for liquid by Eq. (3.56) Guess: Z 0.01:=
Given
ZβTrPr
,
()
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅1βTrPr
,
()
+Z−
qT
r
()
βTrPr
,
()
⋅
⎛
⎜
⎝
⎞
⎠
⋅+=
Given
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
ZZ βTrPr
,
()
+
()
⋅
⋅−=
3.38 (a) Propane: Tc369.8 K⋅:= Pc42.48 bar⋅:= ω 0.152:=
T 313.15 K⋅:= P 13.71 bar⋅:=
Tr
T
Tc
:= Tr0.847=Pr
P
Pc
:= Pr0.323=
For Redlich/Kwong EOS:
σ1:= ε 0:= Ω 0.08664:= Ψ 0.42748:= Table 3.1
αTr() T
r0.5−
:= Table 3.1 qT
r
()
Ψα Tr
()
⋅
ΩTr
⋅
:= Eq. (3.54)
βTrPr
,
()
ΩPr
⋅
Tr
:=
45
B10.207−=
B10.139 0.172
Tr4.2
−:=
B00.468−=
B00.083 0.422
Tr1.6
−:=
For saturated vapor, use Pitzer correlation:
Zc0.276:=Vc200.0 cm3
mol
⋅:=
Tr0.847=Tr
T
Tc
:=
Rackett equation for saturated liquid:
46
Parts (b) through (t) are worked exactly the same way. All results are
summarized as follows. Volume units are cu.cm./mole.
R/K, Liq. R/K, Vap. Rackett Pitzer
(a) 108.1 1499.2 94.2 1537.8
(b) 114.5 1174.7 98.1 1228.7
(c) 122.7 920.3 102.8 990.4
(i) 153.2 1330.3 133.9 1405.7
(j) 164.2 1057.9 140.3 1154.3
(k) 179.1 835.3 148.6 955.4
(l) 201.4 645.8 160.6 795.8
47
ZβTrPr
,
()
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅1βTrPr
,
()
+Z−
qT
r
()
βTrPr
,
()
⋅
⎛
⎜
⎝
⎞
⎠
⋅+=
Calculate Z for vapor by Eq. (3.52) Guess: Z 0.9:=
Given
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅
⋅−=
3.39 (a) Propane Tc369.8 K⋅:= Pc42.48 bar⋅:= ω 0.152:=
T 40 273.15+()K⋅:= T 313.15 K=P 13.71 bar⋅:=
Tr
T
Tc
:= Tr0.847=Pr
P
Pc
:= Pr0.323=
From Table 3.1 for SRK:
48
Parts (b) through (t) are worked exactly the same way. All results are
summarized as follows. Volume units are cu.cm./mole.
SRK, Liq. SRK, Vap. Rackett Pitzer
(a) 104.7 1480.7 94.2 1537.8
(e) 142.1 1487.1 125.4 1577.0
(f) 150.7 1189.9 130.7 1296.8
(g) 161.8 947.8 137.4 1074.0
(m) 61.2 1248.9 53.5 1276.9
(n) 63.5 1003.2 55.1 1038.5
(o) 66.3 810.7 57.0 853.4
(p) 69.5 657.4 59.1 707.8
49
ZβTrPr
,
()
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅1βTrPr
,
()
+Z−
qT
r
()
βTrPr
,
()
⋅
⎛
⎜
⎝
⎞
⎠
⋅+=
Calculate Z for vapor by Eq. (3.52) Guess: Z 0.6:=
Given
Z1βTrPr
,
()
+qT
r
()
βTrPr
,
()
⋅ZβTrPr
,
()
−
Zεβ TrPr
,
()
⋅+
()
Zσβ TrPr
,
()
⋅+
()
⋅
⋅−=
3.40 (a) Propane Tc369.8 K⋅:= Pc42.48 bar⋅:= ω 0.152:=
T 40 273.15+()K⋅:= T 313.15 K=P 13.71 bar⋅:=
Tr
T
Tc
:= Tr0.847=Pr
P
Pc
:= Pr0.323=
From Table 3.1 for PR:
50
Parts (b) through (t) are worked exactly the same way. All results are
summarized as follows. Volume units are cu.cm./mole.
PR, Liq. PR, Vap. Rackett Pitzer
(a) 92.2 1454.5 94.2 1537.8
(e) 125.2 1453.5 125.4 1577.0
(f) 132.9 1156.3 130.7 1296.8
(g) 143.0 915.0 137.4 1074.0
(h) 157.1 715.8 146.4 896.0
(i) 129.4 1271.9 133.9 1405.7
(n) 56.0 987.3 55.1 1038.5
(o) 58.4 794.8 57.0 853.4
(p) 61.4 641.6 59.1 707.8
51
Pr2.282=
3.42 Assume validity of Eq. (3.38).
P11bar:= T1300K:= V123000 cm3
mol
:=
Z1
P1V1
⋅
RT
1
⋅
:= Z10.922=BRT
1
⋅
P1
Z11−
()
⋅:= B 1.942−103
×cm3
mol
=
3.41 (a) For ethylene, molwt 28.054 gm
mol
:= Tc282.3 K⋅:= Pc50.40 bar⋅:=
ω0.087:= T 328.15 K⋅:= P 35 bar⋅:=
Tr
T
Tc
:= Pr
P
Pc
:= Tr1.162=Pr0.694=
From Tables E.1 & E.2: Z00.838:= Z10.033:=
ZZ
0ωZ1
⋅+:= Z 0.841=
(b) T 323.15 K⋅:= P 115 bar⋅:= Vtotal 0.25 m3
⋅:=
Tr
T
Tc
:= Tr1.145=Pr
P
Pc
:=
52
P 16 bar⋅:= Tc369.8 K⋅:= Pc42.48 bar⋅:=
ω0.152:= Vc200 cm3
mol
⋅:= Zc0.276:= molwt 44.097 gm
mol
:=
Tr
T
Tc
:= Tr0.865=Pr
P
Pc
:= Pr0.377=
Vliq VcZc
1T
r
−
()
0.2857
⎡
⎣
⎤
⎦
⋅:= Vliq 96.769 cm3
mol
=
3.43 T 753.15 K⋅:= Tc513.9 K⋅:= Tr
T
Tc
:= Tr1.466=
P 6000 kPa⋅:= Pc61.48 bar⋅:= Pr
P
Pc
:= Pr0.976=
ω0.645:= B00.083 0.422
Tr1.6
−:= B00.146−=
B10.139 0.172
Tr4.2
−:= B10.104=
3.44 T 320 K⋅:=
53
VRT⋅
PB0ωB1
⋅+
()
R⋅Tc
Pc
⋅+:= V 9.469 103
×cm3
mol
=
3.46 (a) T 333.15 K⋅:= Tc305.3 K⋅:= Tr
T
Tc
:= Tr1.091=
P 14000 kPa⋅:= Pc48.72 bar⋅:= Pr
P
Pc
:= Pr2.874=
ω0.100:= Vtotal 0.15 m3
⋅:= molwt 30.07 gm
mol
:=
From tables E.3 & E.4: Z00.463:= Z10.037−:=
3.45 T 298.15 K⋅:= Tc425.1 K⋅:= Tr
T
Tc
:= Tr0.701=
P 2.43 bar⋅:= Pc37.96 bar⋅:= Pr
P
Pc
:= Pr0.064=
ω0.200:= Vvap 16 m3
⋅:= molwt 58.123 gm
mol
⋅:=
B00.083 0.422
Tr1.6
−:= B00.661−=
B10.139 0.172
Tr4.2
−:= B10.624−=
54
3.47 Vtotal 0.15 m3
⋅:= T 298.15 K⋅:=
Tc282.3 K⋅:= Pc50.40 bar⋅:= ω 0.087:= molwt 28.054 gm
mol
:=
VVtotal
40 kg⋅
molwt
⎛
⎜
⎝
⎞
⎠
:= PV⋅PrPc
⋅V⋅=ZR⋅T⋅=
or PrαZ⋅=where αRT⋅
PcV⋅
:= α 4.675=
Whence Pr4.675 Z⋅=at Tr
T
Tc
:= Tr1.056=
ZZ
0ωZ1
⋅+:= Z 0.459=VZR⋅T⋅
P
:= V 90.87 cm3
mol
=
(b) VVtotal
40 kg⋅
:= P 20000 kPa⋅:= PV⋅ZR⋅T⋅=ZR⋅Tr
⋅Tc
⋅=
or Tr
α
Z
=where αPV⋅
RT
c
⋅
:= α 29.548 mol
kg
=
Whence Tr
0.889
Z
=at Pr
P
Pc
:= Pr4.105=
This equation giving Tr as a function of Z and Eq. (3.57) in conjunction with
Tables E.3 & E.4 are two relations in the same variables which must be
satisfied at the given reduced pressure. The intersection of these two
relations can be found by one means or another to occur at about:
55
Pr1
P1
Pc
:= Pr1 0.452=
Vtotal 0.35 m3
⋅:= ω 0.100:=
From Tables E.1 & E.2: Z0.8105:= Z10.0479−:=
Assume Eq. (3.38) applies at the final state.
B00.083 0.422
Tr21.6
−:= B00.113−=
B10.139 0.172
Tr24.2
−:= B10.116=
3.48 mwater 15 kg⋅:= Vtotal 0.4 m3
⋅:= VVtotal
mwater
:= V 26.667 cm3
gm
=
3.49 T1298.15 K⋅:= Tc305.3 K⋅:= Tr1
T1
Tc
:= Tr1 0.977=
P12200 kPa⋅:= Pc48.72 bar⋅:=
56
3.51 Basis: 1 mole of LIQUID nitrogen
Tn77.3 K⋅:= Tc126.2 K⋅:= Tr
Tn
Tc
:= Tr0.613=
P 1 atm⋅:= Pc34.0 bar⋅:= Pr
P
Pc
:= Pr0.03=
ω0.038:= molwt 28.014 gm
mol
⋅:= Vliq 34.7 cm3
⋅:=
B00.083 0.422
Tr1.6
−:= B00.842−=
B10.139 0.172
Tr4.2
−:= B11.209−=
Z1B
0ωB1
⋅+
()
Pr
Tr
⋅+:= Z 0.957=
3.50 T 303.15 K⋅:= Tc304.2 K⋅:= Tr
T
Tc
:= Tr0.997=
Vtotal 0.5 m3
⋅:= Pc73.83 bar⋅:= ω 0.224:= molwt 44.01 gm
mol
⋅:=
B00.083 0.422
Tr1.6
−:= B00.341−=
B10.139 0.172
Tr4.2
−:= B10.036−=
VVtotal
10 kg⋅
molwt
⎛
⎜
⎝
⎞
⎠
:= V 2.2 103
×cm3
mol
=
57
3.52 For isobutane: Tc408.1 K⋅:= Pc36.48 bar⋅:= V11.824 cm3
gm
⋅:=
T1300 K⋅:= P14 bar⋅:= T2415 K⋅:= P275 bar⋅:=
Tr1
T1
Tc
:= Pr1
P1
Pc
:= Tr2
T2
Tc
:= Pr2
P2
Pc
:=
Tr1 0.735=Pr1 0.11=Tr2 1.017=Pr2 2.056=
nvapor
PV
liq
⋅
ZR⋅Tn
⋅
:= nvapor 5.718 10 3−
×mol=
Final conditions:
ntotal 1 mol⋅nvapor
+:= V2V
liq
⋅
ntotal
:= V 69.005 cm3
mol
=
Use Redlich/Kwong at so high a P.
Ω0.08664:= Ψ 0.42748:= α Tr() Tr
.5−
:= α Tr
()
0.651=
aΨα Tr
()
⋅R2
⋅Tc2
⋅
Pc
:= Eq. (3.42) bΩR⋅Tc
⋅
Pc
:= Eq. (3.43)
a 0.901 m3bar cm3
⋅
mol2
=b 26.737 cm3
mol
=
58