SOLUTION THERMODYNAMICS
CHAPTER 6
6.1 Starting with Eq.(6.8), show that isobars in the vapor region of a Mollier (H S)
diagram must have positive slope and positive curvature.
SOLUTION:
{∂∂HS } =T
Proses Isobar dan memiliki slope bernilai
P
positif
Membedakan persamaan di atas:
{ } { }
2
∂ H
∂T
=
2
∂S P ∂S
P
Jika di substitusikan dengan contoh 6.17
{ }
∂2 H
T
=
2
∂ S P Cp
Proses isobar dan memiliki kelengkungan
positif
6.2 (a) Making use of the fact that Eq. (6.20) is an exact differensial expression, show
that :
(∂ Cp/∂ P)
T
= -T ( ∂
V/ ∂ T2)P
2
What is the result of application of this equation to an ideal gas?
(b) Heat capacities Cv and Cp are defined as temperature derivatives respectively
of U and H. Because these properties are related, one expects the heat
capacities also to be related. Show that the general expression connecting Cp to
Cv is :
Cp=Cv+ T (
∂P
)
∂T
∂V
¿
V( ∂T
P
SOLUTION :
(a) Aplikasikan persamaan 6.12 ke persamaan 6.20 :
∂V
∂ V −T (
)P
∂T
∂Cp
=
∂P T
∂T
P
[
( )
{
}
]
2
( ) ( ) ( ) ( )
∂Cp
∂V
=
∂ P T ∂T
−T
P
∂ V
∂V
−
2
∂T P ∂T
P
Sehingga ,
∂Cp
∂2 V
=−T
∂P T
∂T2
( )
( )
P
Untuk gas ideal :
∂V
∂T
( )
=
P
R
∂2 V
dan
=0
P
∂T 2 P
( )
(b) Persamaan 6.21 dan 6.33 adalah persamaan umum untuk dS dan untuk mengubah
kedua persamaan harus diberikan nilai dS yang sama. Oleh karena itu persamaan
menjadi:
Cp−Cv
∂P
∂V
=(
dV + (
dP
( ∂V
)
)
∂T
∂T
∂T )
V
P
Dalam kondisi P konstan, menjadi :
∂ P ∂V
Cp=Cv+ T
∂T V ∂T P
( )( )
Dengan contoh persamaan 3.2 dan 6.34 :
( ∂V∂T ) =βV dan( ∂∂ PT ) = Kβ
P
P
Substitusikan dengan persamaan pada P konstan :
Cp−Cv=βTV
( Kβ )
6.3 If U is considered a function of T and P, the natural heat capacity is neither Cv
nor Cp, but rather the derivate
(∂ U /∂ T ) p . Develop the following connections
between (∂ U /∂ T ) p , Cp, and Cv :
∂V
(∂ U /∂ T )P = Cp−P
∂T
( ) =Cp−βPV
P
[ ( ) ]( )
Cv+ T
=
∂P
∂V
β
v−P
p=Cv+ ( βT −kP ) V
∂T
∂T
k
To what do these equations reduce for an ideal gas? For an incompressible liquid?
SOLUTION :
Untuk definisi H,U=H=PV, turunannya:
∂H
∂V
=(
−P (
( ∂U
)
)
∂T
∂T
∂T )
P
P
P
or
( ∂U
∂T )
P
=
Cp−P
( ∂∂ VT )
P
Substitusi turunan akhir dari persamaan 3-2 definisi dari
(∂ V /∂ T )P =Cp−βPV
β
Pembagian persamaan 6.32 dengan dT dan terbatas untuk P konstan hasilnya adalah
∂P
∂V
p=Cv+ T (
v −P (
( ∂U
)
)
[ ∂T ] ∂T ) p
∂T
Pemecahan untuk turunan kedua dari persamaan 6.34 dan 3.2 substitusinya adalah :
β
p=Cv+ ( β T −kP ) V
( ∂U
)
∂T
k
6.4 The PVT behavior of a certain gas is described by the equation of state:
P(V-b) = RT
Where b is constant. If in addition Cv is constant, show that:
a U is a function of T only
b ɣ = constant
c For a mechanically reversible process, P(V-b)ɣ = const.
SOLUTION :
a
Berdasarkan persamaan 6.32:
dU =
[
C v dT + T (
∂P
) −P
∂T V
]
oleh persamaan keadaan: P =
RT
(V −b)
∂P
∂T
dimana (
R
P
¿ ¿V =
=
V −b T
dengan mensubstitusikan persamaan keadaan ke persamaan 6.32, maka diperoleh:
dU =
[
C v dT + T (
R
RT
) −
V −b V V −b
persamaan akhir menjadi:
]
dU =
fungsi T)
b
dH = dU + d(PV)
d(PV) = R d T + b d P
…… (1)
…… (2)
⟹
C v dT
dU =
[
C v dT + (
RT
RT
) −
V −b V V −b
]
, (terbukti bahwa U hanya merupakan
kombinasikan persamaan (1) dan (2) dengan persamaan akhir dari jawaban (a),
diperoleh:
C
(¿¿
v dT ) + (R d T + b d P)
dH =
¿
⟹
C
¿
¿
¿
integralkan persamaan (3), gunakan persamaan
+ R) dT + b d P
…… (3)
∂H
=C V + R
∂T
hasil akhir menjadi: Cp = Cv + R, karena Cv konstan, maka:
C
γ ≡ P = konstan
CV
c
Untuk persamaan mekanik proses adiabatic reversibel, dU = dW , sesuai dengan
persamaan keadaan: dU =
C v dT =−P dV =
, maka persamaan dU = dW berubah menjadi:
d ( V −b )
−RT
dV =−RT
V −b
V −b
Disederhanakan menjadi:
dari jawaban (b),
C v dT
γ≡
CP
CV
dT −R
=
d ln ( V −b )
T
CV
R C P−C V
=
=γ−1 ,
CV
CV
= konstan, diubah menjadi
maka:
d lnT = - ( γ −1 ) d ln(V-b)
⟹
hubungkan dengan persamaan: T
d lnT + d ln (V −b)
(V −b) γ−1
γ−1
=0
= konstan, substitukan dengan
persamaan keadaan:
P ( V −b ) (V −b)γ−1
=konstan , maka: P (V −b) γ = const (TERBUKTI)
R
6.5 A pure fluids is described by the canonical equation of state : G = Γ (T) + RT ln P,
where Γ (T) is a substance-specific function of temperature. Determine for such a
fluid expressions for V, S, H, U, Cp, and Cv. These results are consistent with those
for an important model of gas-phase behavior. What is the model ?
SOLUTION :
It follows immediately from Eq. ( 6.10 ) that :
∂G
¿
V=( ∂P
T
and
∂G
¿
S = - ( ∂T
P
Differentation of the give equation of state yields:
V=
RT
P
and
S=-
d Γ(T)
dT
- R ln P
Once V and S ( as well as G ) are known, we can apply the equations :
H = G + TS
and
U = H-PV = H – RT
These become :
H=
Γ (T )
-T
dΓ (T )
`
dT
and
Γ (T )
U=
-T
dΓ (T )
dT
– RT
By Eq.
∂H
¿
Cp = ( ∂ T
Because
Γ
P
and
∂U
¿
Cv = ( ∂ T
T
is a function of temperature only, these become:
2
d T
Cp = -T dT 2
2
and
d T
Cv = -T dT 2
- R = Cp – R
The equation for V gives the ideal-gas value. The equations for H and U show these
properties to be functions of T only, which conforms to ideal-gas behavior. The equation
for S shows its relation to P to be that of an ideal gas. The equations for CP and CV show
these properties to be functions of T only, which conforms to ideal-gas behavior, as does
the result, CP = CV + R. We conclude that the given equation of state is consistent with the
model of ideal-gas behavior.
6.6 A pure fluid, described by the canonical equation of state : G = F(T) + KP where
F(T) is a substance specific function of temperature and K is a substance specific
constant. Determine for such a fluid expressions for V, S, H, U, Cp, and Cv. These
results are consistent with those for an important model of liquid phase behavior.
What is the model ?
SOLUTION :
Dilihat dari persamaan 6.10 :
G = F (T) + K P
Diferensiasi dari persamaan yang diberikan :
 Dalam V
( ∂G
∂T )
;
V =K
 Dalam S
∂G
S=−
∂T
;
S=
V=
T
( )
P
−dF (T )
dT
Maka V dan S telah diketahui dalam persamaan :

H =G+TS
dan
U = H – PV
dan
U= H – PK
Maka persamaan menjadi,

H=F ( T ) + KP−T
dF (T )
dT
Sehingga,
U=F ( T ) + KP−T
U=F ( T )−T
dF (T )
−PK
dT
dF(T )
dT
Sedangkan dari persamaan (2.16) dan (2.20) :
Cp=
( ∂∂HT )
P
Cv=
;
d F
Cv=−T
2
dT
2
d F
Cp=−T
2
dT
( ∂∂UT )
;
V
2
F(T) dianggap konstan
Maka kesimpulannya adalah Cp = Cv
Persamaan untuk V menunjukkan hal itu akan konstan dari kedua T dan P. Ini adalah
definisi dari
cairan termampatkan H dipandang
sebagai fungsi
dari
kedua T
dan P
sedangkan U, S, Cp, dan Cv adalah fungsi T saja. Kami juga memiliki hasil yang Cp = Cv.
Semua ini konsisten dengan model, cairan mampat.
6.7 Estimate the changes in enthalpy and entropy when liquid ammonia at 270K is
compressed from its saturation pressure of 381 kPa. For saturated liquid
ammonia at 270K, Vt = 1.551 x 10-3 m3 kg-1, and
β=2.095 x 10−3 K−1 .
SOLUTION :
At constant temperature Eqs. (6.25) and (6.26) can be written :
dS = −β .V dP
and dH = 1−( β . T . V . Dp)
For an estimate, assume properties independent of pressure
T = 270 K
P1 = 381 kPa
P2 = 1200 kPa
V = 1. 551 x 10-3 m3 kg-1
β=2.095 x 10−3 K−1
∆S = - β.V (P2 – P1)
∆H = 1 - β.T.V (P2 – P1)
= -2.661 J / KG. K
= 551.7 J / kg
6.8 Liquid isobutane is throttled through a valve from an initial state of 360 K and
4000 kPa to a final pressure of 2000 kPa. Estimate the temperature change and
the entropy change of the isobutane. The specific heat of liquid isobutane at 360
K is 2.78 J g-1 ○c-1.
Estimates of V and
β
may be found from Eq. (3.63).
SOLUTION :
Isobutane:
Tc
= 408.1K
Zc = 0.282
Cp = 2.78 J / gm K
P1
= 4000 kPa
P2
= 2000 kPa
mol (wt)
= 58.123 gm / mol
Vc = 262.7 cm3 / mol
Eq. (3.63) for volume of a saturated liquid may be used for the volume of a compressed
liquid if the effect of pressure on liquid volume is neglected.
T = ( 359 ; 360 ; 361 ) K
Tr = T / Tc Tr = ( 0.88 ; 0.882 ; 0.885 )
(The elements are denoted by subscripts 1, 2, & 3)
V = Vc . Zc ( 1 – Tr )0.2857
V = (131.662 ; 132.113 ; 132.767 ) cm3 / mol
V = 262,7 cm3/mol x (0,282)0,5457 = 131,662 cm3/mol
V = 262,7 cm3/mol x (0,282)0,5430 = 132,113 cm3/mol
V = 262,7 cm3/mol x (0,282)0,5391 = 132,767 cm3/mol
Assume that changes in T and V are negligible during throtling.
Then Eq.(6.8) is integrated to yield:
H = TS VP
S
H=0
Then at 360 K,
= V1P2P1T1
=
=
S
but
( −1,31 x 662 x 10−4 ) m3 . mol−1 . (−2 x 106 ) Pa
359 k
263,32 x 4
359
Pa m3/mol K
= 0.733 J / mol K.
We use the additional values of T and V to estimate the volume expansivity:
V
= 1.105 cm3 / mol
T1 → T2 K
= (132,767 – 131,662) cm3/mol
= (361 – 359) = 2K
= 1,105 cm3/mol
(V1 ) (V / T ) → 4.098755 x 103 K-1
=(
1
131,662 cm 3 mol−1
3
). (
1,105 cm mol
2K
−1
)
=
1,105
131,662 .2 K
=
1,105
263,324 K
= 4, 196351 x 10-3 K-1
Assuming properties independent of pressure, Eq. (6.29) may be integrated to give
= Cp . ( ∂ T / T ) – ( ∂ V/ ∂ P)
S
= 2,78 j/gmk . 5,55x10-3 – 4,196351x10-3 K-1 . 1,32181x10-4 m3/mol . -2x106 Pa
= (0,05429 + 1,11) J/ mol K
= 1,125429 J/mol K
Whence T = (V / Cp ) . [ ( ∂ S  ∂ V1 ∂ P) / mol wt ]
=
1,125429
(359
K/2,78
J/gm
K)
.
j
−3 −1
−4 3
−1
6
K + 4,196351 x 10 K . 1,31662 x 10 m mol .−1 x 10 Pa
mol
58,123 gm .mol−1
= (129,14) . ( -
1,105
58,123
)K
6.9 One kilogram of water ( V1=1003 cm3kg-1 ) in a piston / cylinder device at 25 0C and
1 bar is compressed in a mechanically reversible, isothermal process to 1500 bar.
Determine Q, W, ΔU, ΔH, and ΔS given that β = 250 x 10 -6K-1 and k = 45x10-6 bar1
.
SOLUTION :
V2
= V1.EXP[-k(P2-P1)]
= 1003
c m3
kg
. EXP[(-45x10-6 bar-1)( 1500 bar-1 bar )]
= 1003
= 1003
c m3
kg
= 1003
cm
kg
=
. EXP[(-45x10-6 bar-1)(1499 bar)]
. EXP( -0,067455)
3
. 0,9348
c m3
kg
= 937,604
Vave
3
cm
kg
V 1+V 2
2
= 1003
cm
kg
3
+ 937,604
cm
kg
3
2
= 970,302
ΔH
c m3
kg
= Vave . (1 – β.T) . (P2 – P1)
= 970,302
c m3
kg
( 1 – (250 x 10-6K-1.298,15K)).(1500 bar – 1 bar)
= 970,302
c m3
kg
(1 – 0,0745375).(1499 bar)
¯
c m3 . ¿
kg
= 1454482,698
¿
¯
c m3 . ¿
kg
= 1346069,194
¿
= 134606,9194
J
kg
. (0.9254625)
kJ
kg
= 134,6069194
= 134,607
ΔU
ΔS
kJ
kg
= ΔH – (P2.V2 – P1.P1)
= 134,607
kJ
kg
= 134,607
kJ
kg
= 134,607
kJ
kg
– (140,6406
= 134,607
kJ
kg
– 140,5403
= -5,9333
kJ
kg
– [(1500 bar . 937,604
3 ¯
cm . ¿
kg
– (1406406
¿
– 1003
– 0,1003
c m3
kg )
kJ
kg )
kJ
kg
= -β . Vave (P2 – P1)
-6
-1
= (-250 x 10 K )(970,302
3
= -363,6207
cm .
= -0,03636207
= -0,0364
= T. ΔS
c m3
kg )(1500 bar – 1 bar)
c m3
kg . K )(1499 bar)
= (-0,2425755
Q
kJ
kg
c m3
kg ) – (1 bar . 1003
¯¿
kg . K
¿
kJ
kg . K
kJ
kg . K
c m3
kg )]
= 298,15K.( -0,0364
)
kJ
kg
= 4,91936
W
kJ
kg . K
= ΔU – Q
= (-5,9333
kJ
kg ) – (-10,85266
= 4,91936
kJ
kg
kJ
kg )
6.10 Liquid water at 250C and 1 bar fills a rigid vessel. If heat is added to the water
until its temperature reaches 500C. What pressure is developed? The avarege
β
value of
between 25 and 500C is 36.2 x 10-5 K-1. The value of
ĸ
at 1 bar
and 500C is 4.42 x 10-5 bar-1 and may be assumed is independent of P. The
specific volume of liquid water at 250C is 1.0030 cm3g-1.
SOLUTION :
Berdasarkan persamaan 3.5 (perubahan volume konstan)
β.∆T – ĸ .∆ P = 0
β (T – T ) – ĸ
2
1
β (T – T ) = ĸ
2
1
P2
=
=
β
ĸ
(P2 – P1) = 0
(P2 – P1)
(T2 – T1) + P1
−5 −1
4.42 X 10 ¿¯
36.2 x 10−5 K−1
¿
= 205.75 bar
(323.15 K – 298.15 K) + 1 bar
6.11 Determine expressions for GR,HR,and SR implied by the three-term virial
equation in volume,Eq (3.39)
SOLUTION :
From Eq. 3.39:
Z=
PV
RT
B C
+
= 1+ V V 2
Vig = gas ideal,dimana : P.V = n.R.T
Vig =
Z=
RT
P
PV
RT
V=
ZRT
P
Sehingga,
Vr = V- Vig
=V-
=
Vr =
RT
P
ZRT RT
−
P
P
RT
P
( Z - 1)
Volum Residu :
R
V =
RT
P
(Z-1)
Dimana : dG = VdP -
… (1)
∫ dT
T konstan sehingga :
dG = V.Dp
Integrasi dari nol menuju P >
P
GR =
∫ VR dP
0
… (4)
Dibagi RT
GR
RT
P
=
dP
∫ VR
RT
=
∫ VRT
… (5)
0
Maka,
GR
RT
P
P
=
∫
0
R
dP
0
RT
( Z−1)
P
RT
dP
P
=
∫ 1P ( Z −1 ) dP
0
Finally, the answer is
R
P
G
dP
=∫ ( Z−1 )
RT 0
P
Substitusi persamaan (3) ke persamaan (5)
GR
RT
P
∫ ( Z−1 ) dP
P
=
… (6)
0
Substitusi persamaan (2) ke persamaan (6)
GR
RT
P
dZ
+
∫ ( Z−1 ) dP
P Z
=
0
(Z-1)
Diferensiasi kondisi 2 menjadi :
GR
RT
¿
P
∫ ( Z−1 )
=
0
dP
+¿ Z-1) – ln Z … (7)
P
Persamaan (4) dapat ditulis
d
G
RT
=
1
RT
dG -
G
RT 2
Dt
dimana dG disubstitusi dengan persamaan (4) dan G dengan
G = H – TS menjadi (dalam keadaan T konstan)
GR
d RT ) =
¿
VR
RT dP -
HR
dT = (Z-1)
RT 2
HR
RT 2 dT
dP
P
GR
– d ( RT )
… (8)
Persamaan (8) dibagi dT dan dalam P konstan, maka :
HR
RT 2
GR
∂(
)
∂P
RT ]
( ∂T )P – [
P
∂T
Z−1
P
=
Diferensiasi persamaan (1) menyediakan hasil di kanan dan diferensiasi persamaan (7)
menyediakan hasil kedua. Substitusinya menghasilkan :
HR
RT
= -T
∂Z
∂T
¿
P
∫¿
dP
P
)P
+ Z-1 … (9)
0
Dan dari persamaan 3.39 didapatkan = V = 1/ρ
Z-1 = Bρ + Cρ2
Pensubstitusian persamaaan (7) dan persamaan (9) akan menghasilkan :
GR
RT
HR
RT
3
2
2 Cρ – ln Z
= 2Bρ +
=
B
T[( T
dB
C
)]ρ
+
(
dT
T
-
…1
-
1
2
dC
2
dT )ρ
Persamaan Gibbs umum :
G = H – TS
Untuk residual
GR = HR - TSR
Dan entropinya adalah
SR
R
=
HR
RT
-
GR
RT
…3
…2
6.12 Nilai parameter dari persamaan van der walls telah ditunjukkan pada baris
pertama tabel 3.1, halaman 99. Pada bagian bawah halaman 214 ditunjukkan
pada persamaan I = β/Z. Maka persamaan menjadi :
Gr
RT
= Z − 1 − ln(Z − β) – qβ/Z
Telah diberikan T, V, dan P pada fase vapor dan pada fase liquid denganσ = € = 0. Maka
persamaan van der wall menjadi
β=
Pr
8Tr
dan
q =
27
8Tr
maka dapat disubstitusikan pada persamaan tersebut menjadi
Hr
RT
=Z−1−
qβ
Z
dan
Sr
RT
= ln(Z − β)
6.13. Determine expressions for GR, HR, and SR implied by the Dieterici equation:
Here, parameters a and b are functions of composition only.
SOLUTION :
According to eq6.56
P = ZρRT
I.
eq. 6.63
GR
=( z−1−ln ( z−β ) . ql)
R.T
GR = ( z−1−ln ( z −β ) . ql) RT
Substitusi eq. 6.56 and eq. 6.63b to eq 6.13
R
ZρRT =
RT
−G
exp (
)
R
VRT
V −G
ZρRT =
−( z−1−ln ( z −β ) . ql) RT
VRT
RT
exp ¿
v−( z−1−ln ( z− β ) . ql) RT
ZρRT =
−( z−1−ln ( z− β ) . ql )
1
exp (
)
V
v−( z−1−ln ( z− β ) . ql)
Z=
−( z−1−ln ( z −β ) . ql )
1
exp (
)
V
(v −( z−1−ln ( z −β ) . ql ) ) ρRT
II. Eq. 6.65
R
d ln α ( Tr )
S
=z−1+
−1 ql
R
d ln Tr
(
S R =z−1+
)
( ddlnαln Tr(Tr ) −1) qlR
Substitusi eq. 6.56 and eq. 6.65 to eq 6.13
ZρRT =
−S R
VRT
)
RT
exp ¿
R
v−S
−z−1+
ZρRT =
( d dlnlnαTr( Tr ) −1) qlR
VRT
RT
exp ¿
d ln α ( Tr )
v−z−1+
−1 qlR
d ln Tr
(
)
−z−1+
ZρRT =
( d dlnlnαTr( Tr ) −1) ql
VT
T
exp ¿
d ln α ( Tr )
v−z−1+
−1 ql
d ln Tr
(
−z−1+
Z=
)
( ddlnlnαTr( Tr ) −1) ql
VT
T
exp ¿
d ln α ( Tr )
(v −z−1+
−1 ql) ρRT
d ln Tr
(
)
−z−1+
( d dlnlnαTr( Tr ) −1) ql
VT
Z=
1
(
))
d ln α ( Tr )
v−z−1+
−1 ql ρR
d ln Tr
(
exp ¿
III. Eq. 6.64
d ln α ( Tr )
HR
=Z−1+
−1 ql
RT
d ln Tr
(
H = Z −1+
R
(
)
d ln α ( Tr )
−1 qlRT
d ln Tr
)
Substitusi eq. 6.56 and eq. 6.64 to eq 6.13
R
ZρRT =
−H
VRT
)
RT
exp ¿
R
v−H
( ddlnlnα Tr(Tr ) −1) qlRT
−Z−1+
ZρRT =
VRT
)
RT
exp ¿
d ln α ( Tr )
v−Z−1+
−1 qlRT
d ln Tr
(
Z=
)
1
exp (
d ln α ( Tr )
v−Z−1+
−1 qlRTρ
d ln Tr
(
( d dlnlnα Tr( Tr ) −1) ql )
−Z−1+
)
VRTρ
14.Calculate Z,H and R by the Redlich/Kwong equation for one of the following and
compare result with values found from suitable generalized correlations:
a.Acetylene at 300K and 40 bar
b.Argon at 175 K and 75 bar
c.Benzene at 575 K and 30 bar
d.n-Butane at 500 K and 30 bar
e.Carbon dioxide at 325 K and 60 bar
f.Carbon monoxide at 175 K and 60 bar
g.Carbon tetrachloride at 575 K and 35 bar
h.Cyclohexane at 650 K and 50 bar
i.Ethylene at 300 K and 35 bar
j.Hydrogen sulfide at 400 K and 70 bar
k.Nitrogen at 150 K and 50 bar
l.n-Octane at 575 K and 15 bar
m.Propane at 375 K and 25 bar
n.Propylene at 475 K and 75 bar
SOLUTION :
Redlich/kwong equation :
Eq(3.53)
Guess Z 1
Given Z=1 Eq(3.52)
Z
q
I
114
HRi RTi Ziqi 1 1.5qili Eq (6.67) The derivative in these
SRi Rln Ziqi I 0.5qili Eq (6.68) Equations equal-0.5
Z i qI
0.695
Sri
0.605
-5.461
-2.302·103
0.772
-4.026
-2.068·103
0.685
-6.542
-3.319·103
0.729
-5.024
-4.503·103
0.75
-5.648
-2.3·103
0.709
-5.346
-1.362·103
0.706
-5.978
-4.316·103
0.771
-4.12
-5.381·103
0.744
-4.698
-1.764·103
0.663
-7.257
-2.659·103
0.766
-4.115
-1.488·103
0.775
-3.939
-3.39·103
0.75
-5.523
-2.122·103
-8.767
HRi
-3.623·103
6.15 Calculate ZR,HR, and SR by the Soave/Redlich/Kwong equation for the substance
and conditions given by one of the parts of Pb.6.14 and compare results with
values found from suitable generalized correlations.
SOLUTION :
Soave/Redlich/Kwong equation:
 = 0,08664
 = 0,42748
C = (0,480 + 1,574. - 0,176. 2)
α = [ 1+c.1- Tr0,5]2
Pr
Pr
¿
β = (. Tr ), maka β = ( Tr
Eq.(3.50)
q=
α(Tr )
Tr
Guess:
z=1
Given:
Z = 1+ β – q.
The derivative
-ci.
(
Tri
)
∝i
-2,595.103
-2,090.103
-3,751.103
-4,821.103
-2,585.103
-1,406.103
-4,816.103
-5,806.103
-1,857.103
-2,807.103
-1,527.103
-4,244.103
-2,323.103
-3,776.103
E.q.(3.51)
z−β
β z .( z + β)
E.q.(3.49)
in the following equations equals;
0,5
dimana,
i=1,.....,14
Ii = ln
βi , qi+ βi
¿
βi , qi
Z (¿)
Z¿
¿
¿
HRi= R.Ti [ Z ( βi , qi )−1−¿ [ci.
Sri = R [ ln (Z) ( βi ,qi )− βi [ci.
Tri
0,5
∝ i ) +1].qi,Ii]
Tri
0,5
∝ i ) .qi,Ii]
So,
0,691
0,606
0,774
0,722
0,741
0,768
0,715
0,741
0,774
0,749
0,673
0,769
0,776
0,787
Z ( βi, qi ) =
J/mol.K
-6,412
-8,947
-4,795
-7,408
-5,974
-6,02
-6,246
-6,849
-4,451
-5,098
-7,581
-5,618
-4,482
-6,103
HRi=J/mol
SRi =
6.17 Estimate the change in enthalpy and entropy when liquid ammonia at 270 K is
compressed from its saturaration pressure of 381 kPa to 1,200 kPa. For saturated
liquid ammonia at 270 K, Vt = 1.55 x 10 -3m 3 kg-1, and β = 2.095 x 10-3 K-1.
SOLUTION :
T
= 323.15 K
t=
T
K
273.15
t = 50
The pressure is the vapor pressure given by the Antoine equation:
P(t) = exp
d
dt
(13.8858− t+2788.51
220.79 )
P(t) 1.375
P(50) = 36.166
P = 36.166 kPa
dPdt =1.375
kPa
K
a. The entropy change of vaporization is equal to the latent heat divided by
the temperatur.
For the Clapeyron equation, Eq. (6.69), we need the volume change of vaporization.
For this we estimate the liquid volume Eq.(3.63) and the vapor volume by the
generalized virial correlation.
For benzene:
ω = 0.210
Tc = 562.2 K Pc = 48.98 bar
Vc = 295 cm3/mol
Tr = T / Tc
Tr = 0.575
Zc = 0.271
Pr = P/Pc
By Eqs. (3.65), (3.66), (3.61), & (3.63)
B0 = 0.083-0.422/Tr 1.6
= -0.941
B0 = 0.139-0.172/Tr 4.2
= -1.621
V vap = R .T / P (1 + (B0 +ω. B1) Pr/Tr)
= 7.306 x 104 cm3/mol
By Eq. (3.72), Vliq = Vc . Zc (1 – Tr 2/7)
= 93.151 cm3/mol
Pr = 0.007
Solve Eq. (6.72) for the latent heat and divide by T to get the entropy change
of vaporization:
∆S = dPdt. (Vvap – V liq)
= 100.34 J / mol. K
(a) Here for the entropy change of vaporization:
∆S = R. T / P x dPdt
= 102.14 J / mol.
6.18 Let P1sat and P2sat be values of the saturation vapor pressure of a pure liquid at
absolute temperature T1 and T2. Justify the following interpolation formula for
estimation of the vapor pressure Psat at intermediate temperature T :
ln Psat = ln P1sat +
T ₂ ( T −T ₁ )
T ( T ₂−T ₁ )
ln
P 2sat
P 1sat
SOLUTION :
Ln P2sat = A –
B
T2
B
T
Ln Psat = A –
Ln P1sat = A –
………………A
…………………..B
B
T 1 ………………..C
Eliminasi C dari A
B
B
−
Ln P2sat – ln P1sat = A – A – ( T 2 T 1 )
Ln
P 2sat
P 1sat
Ln
P2
sat
P1
1
1
−
= B( T 1 T 2 )
Ln
P 2sat
P 1sat
= B(
sat
=0+
Eleminasi C dari B
B
B
−
T 1 T2
T 2−T 1
T 1. T 2 )………………………..pers.1
B B
−
Ln P – ln P1 = A – A – ( T T 1 )
sat
sat
Ln
P sat
P 1sat
Ln
P
P 1sat
Ln
P sat
P 1sat
sat
B B
−
T1 T
=0+
1
1
−
= B( T 1 T )
T −T 1
T 1. T )…………………………pers.2
= B(
Bandingkan pers.1 dan pers.2
Ln
P sat
P 1sat
Ln
P
sat
P1
Ln
P sat
P 1sat
Ln
P
: Ln
P 2sat
P 1sat
: Ln
P2
sat
P1
sat
sat
sat
sat
=
- ln
sat
ln P = ln P1 +
=
T ₂ ( T −T ₁ )
T ( T ₂−T ₁ )
P1
sat
T −T 1
T 1. T ) : B(
= B(
=
T ₂ ( T −T ₁ )
T ( T ₂−T ₁ )
. Ln
P 2sat
P 1sat
T ₂ ( T −T ₁ )
T ( T ₂−T ₁ )
T ₂ ( T −T ₁ )
T ( T ₂−T ₁ )
T 2−T 1
T 1. T 2 )
ln
P 2sat
P 1sat
sat
. Ln
P2
sat
P1
…………………(terbukti)
6.19 Assumsing the validity of Eq. (6.70), derive Edmister’”s formula for estemation
of
the
acentric
factor:
Where Ѳ ≡ Tn/Tc, Tn is the normal boiling point, and Pc is an (atm)
SOLUTION :
Tuliskan persamaan (6.70) dalam log10 sehinggga menjadi : log P sat = A – B/T...(A)
Masukan titik kritis : log Pc = A – B/Tc............................(B)
masukan perbedaan T, log P sa = B(i/Tc – 1/T) = B ((Tr – 1)/T).....................(C)
jika P sat dalam (atm), lau aplikasikan (A) pada area titik didih normal:
log 1 = A – B/Tn
or
A = B/Tn
dengan θ ≡ Tn/Tc, Eq. (B) sekarang dapat ditulis menjadi ::
Dimana:
Persamaan menjadi:
Masukan Tr = 0.7, maka:
Dari persamaan (3.54)
ω = −1.0 − log(Psatr )Tr=0.7
Jadi,
6.21 The state of 1 (lb m) of steam is changed from saturated vapor at 20 (psia) to
superheated
vapor at 50 (psia) and 1,000 ( 0F). What are the enthalpy and
entropy changes of the steam? What would the enthalpy and entropy changes be
if steam were an ideal gas?
SOLUTION :
Tabel F.4 :
H1 = 1156.3
BTU
lbm
H2 = 1533.4
BTU
lbm
BTU
S1 = 1.7320 lbm . rankine
BTU
S2 = 1.9977 lbm . rankine
Sehingga,
∆ H=¿ H – H
2
1
= 377.1
BTU
lbm
∆ S=¿ S – S
2
1
BTU
=0.266 lbm . rankine
Berdasarkan persamaan 4.9 dan 5.18 (uap sebagai gas ideal)
T0 = (227.96 + 459.67)rankine
T1 = (1000+459.57)rankine
P1 = 20 psi
P2 = 50 psi
T0 = 382.017 K
T1 = 810.928 K
τ=
T1
T0
=
810.928 K
382.017 K
(Cp) H
=A+
R
= 2.123
B
C 2 2
D
T 0 ( τ +1 ) + T 0 ( τ +τ +1 ) + 2
2
3
τT0
(Cp) H
=3.470 +
R
1.450 . 10−3
0.121 . 105
(
)
382.017 2.123+1 +
2
2
2.123(382.017)
= 3.51391
Cp
( ¿ ¿H
= 3.51391 x R
= 3.51392 x 8.314 J/molK
= 29.215 J/molK
Cp
¿
¿
∆ H=¿
= 29.215 J/molK(428.911 K)
=12530.635 J
(Cigp )s
=A+
R
[ (
[
B T 0+ C T 20 +
= 3.470 +
D
τ T 20
2
() τ +12 )]( τ−1
ln τ )
1.450 . 10−3 .382.017+
(
0.121 .105
2.123+ 1
2
2
2
2.123 (382.017)
)(
)]( 2.123−1
0.753 )
= 4.338
∆S
R
ig
=
(C p )s T 1
P1
ln −ln
R
T0
P0
= 4.338(0.753) – 0.916
= 2.350514
∆ S=2.350514 X 8.314
= 19.5422 J/K
6.22. A two-phase system of liquid water and water vapor in equilibrium at 8,000 kPa
consist of equal volumes of liquid and vapor. If the total volume V’=0.15m³,
what is the total enthalpy H’ and what is total entropy s’?
SOLUTION :
Data, table F.2 pada 8.000 kPa
Vliquiid = 1.384.
J
gm . K
cm³
gm
J
Hliquid = 1317.1. gm
Sliquid = 3.2076.
Vvapor = 23,525.
cm³
gm
J
Hvapor = 2759.9. . gm
Svapor = 5.7471
J
gm . K
1. M liquid
0.15 . 106
cm ³
2
=Vliquid
=
0.15 . 106
cm ³
2
cm ³
1,384
gr
=
150000
2,768 gr
= 54191 gr = 54,191 kg
2. Mvapor
3. Htotal
=-
0.15 . 106
cm ³
2
Vvapor
=
0.15 . 106
c m3
2
c m3
23,525
gr
=
150000
47,05 gr
= 3188 gr = 3,188 kg
= Htotal = - m liquid. Hliquid + m vapor. Hvap
= 54,191 kg. 1317,1
4. Stotal
KJ
kg
+ 3,188 kg. 2759
KJ
kg
= 71374,96 KJ + 8798,56 KJ
= 80173,5 KJ
KJ
= 54,191 kg. 3,2076 kg/ K + 3,188 kg. 5,7471
KJ
kg/ K
= 173,82 KJ/K + 18,321 KJ/K
= 192,145 KJ/K
6.23 A vessel contains 1kg og H2O as liquid and vapor and equilibrium 100 kPa. If the
vapor occupies 70% of the volume of the vessel, determine H dan S for the 1kg of H2O
SOLUTION :
Data dari tabel F.2 at 1000 kPa:
Vliq = 1.127 cm3/gm
Hliq = 762.605 J/gm
Huap =14.29 cm3/gm
Hvap = 2776.2 J/gm
Mencari x = fraksi masa dari vapor
x .Vvap
70
=
(1 – x) .Vliq 30
x  .194.29 cm 3/gm
70
=
(1 – x) .1.1 27 cm 3/ gm 30
X = 0.013
 H = (1 – x) . Hliq + x. Hvap
= (1 – 0.013) 762.605 J/gm + 0.013 . 2776.2 J/gm
= 789,495 J/gm
 S = (1 – x) . Sliq + x.Svap
= (1 – 0.013) 2.1382 J/gm K + 0.013 . 6.5828J/gmK
= 2.198 J/gm K
6.24 Reaktor bertekanan mengandung liquid air dan uap air di keadaan setimbang
pada suhu 350 °F total massa dari liquid dan uap adalah 3 lbm. Jika volum dari
uap adalah 50 kali volum liquid, berapakah total entalpi dari reaktor?
SOLUTION :
Data dari tabel F.3 pada 350 °F diketahui:
Vliq = 0, 01799
Hliq = 321,76
ft
lbm
BTU
lbm
Vvap = 3, 342
Hvap = 1192,3
ft
lbm
BTU
lbm
mliq + mvap = 3 lbm mvap ×Vvap = 50. mliq. Vliq mliq +
50. mliq .Vliq
+ = 3 lbm
Vvap
3× lbm
Vliq
1+50
Vuap
mliq =
3 lbm
ft
lbm
1+50
ft
3,342
lmb
0,01799
= 3 lbm
= 2, 364 lb
mvap = 3 lbm mvap - 2, 364 lb mvap = 0, 636 lb
Htotal = mliq×Hliq + mvap ×Hvap
= (2, 364 lb x 321, 76
BTU
lbm
) + (0,636 lb x 1192,3
BTU
lbm )
Htotal = 1519,1 BTU
6.25 Wet steam at 2300C has a density of 0,025 g cm-3. Determine x, H, and S.
SOLUTION :
ρ=
m
V
ρ=
1
V
V=
1 cm3
0,025 gr
 Data pada Tabel F.1 pada suhu 230oC
3
V liquid =1,209 cm /gr
3
V vapor=71,45cm / gr
H liquid =990,3 J / gr
H vapor=2802,0 J /gr K
S liquid=2,61202 J /gr K
S vapor =6,2107 J /gr K
 Mencari x
V =( 1−x ) V liquid+ x uap
x=
V −V liquid
V uap−V liquid
1 cm 3
cm 3
−1,209
0,025 gr
gr
x=
3
cm
cm 3
71,45
−1,209
gr
gr
1−0,030225 cm3
0,025 gr
x=
cm3
70,241
gr
x=
38,791
70,241
x=0,552
 Mencari H
H=( 1−x ) H liquid + x H uap
H=( 1−0,552 ) 990,3
H=443,6554
J
J
+0,552 . 2802,0
gr
gr
J
J
+ 1546,704
gr
gr
H=1990,3584
J
gr
 Mencari S
S=( 1−x ) S liquid + x S uap
S=( 1−0,552 ) 2,61202
S=1,17
J
J
+0,552 .6,2107
gr
gr
J
J
+3,4283
gr
gr
S=4,5983
J
gr
6.26 Sebuah vessel dengan volume 0.15m 3 mengandung uap saturated pada 150 oC
didinginkan menjadi 30 oC. Tentukan volume akhir dan massa air liquid pada
vessel.
SOLUTION :
Berdasarkan persamaan 6.73b,
Vtot = mtot.Vliq + mvap.Vlv
Dari tabel F.1 didapat:

cm3
m3
=0.3924
gr
kg
V vap =392.4
o
Pada suhu 150 C,
3

o
Pada suhu 30 C,
V liq =1.004
3
cm
m
=1.004 . 10−3
gr
kg
3
∆ V lv =32930
3
V tot
0.15 m
m tot =
=
V vap 0.3942m 3 / kg
mvap=
= 0.382 kg
V tot −mtot .V liq
∆ V lv
0.382 kg .1.004 .10−3
¿
3
0.15 m −¿
¿¿
−3
¿ 4.543 ×10 kg
mliq=mtot −mvap
3
m
kg
3
cm
m
=32.93
gr
kg
−3
¿ 0.382 kg−4.543 ×10 kg
= 0.37772 kg
= 377.72 gram
V tot liquid =mliq .V liq
¿ 377.72 gram.1 .004
cm3
gr
= 379.23 cm3
6.27 Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process)
to 101.325 kPa, where its temperature is 378.15 K (105°C). What is the quality
of the steam in its initial state?
SOLUTION :
According from table F.2, 1100 kPa : Hliq
Hvap
H2
= 781.124
J
gm
= 2779.7
J
gm
= 2686.1
J
gm
Interpolate pada 101.325 kPa & 105 degC:
Const. –H throttling : H2
= Hliq + x . (Hvap – Hliq)
=
H 2−H liq
H vap−H liq
x
=
2686.1−781.124
2779.7−781.124
x
= 0.953
x
6.28 Steam at 2,100 kPa and 260 oC expands at constant enthalpy (as in a throttling
process) to 125 kPa. What is the temperature of the steam in its final state and
what is its entropy change? What would be final temperature and entropy
change for an ideal gas?
SOLUTION :
Data, Table F.2 at 2100 kPa and 260 degC, by interpolation :
H1 =
2923.5 J
S1 =
6.5640 J
gm
H2 =
mol wt=18.015 gm
gm.K
mol
2923.5 J
gm
Final state is at this enthalpy and a pressure of 125 KPa
By interpolation at these conditions, the final temperature is 224.80 degC and
S2 =
7.8316 J
∆S = S2 - S1
gm.K
∆S= 1.268 J
gm. K
For steam as an ideal gas, there would be no temperature change and the entropy change
would be given by :
P1 = 2100 KPa
P2 = 125 KPa
∆S = -R
ln P2
molwt
∆S=1.302 J
P1
gm. K
6.29 Steam at 300(psia) and 500(of) expands at constant enthalpy (as in a throttling
process) to 20(psia). What is the temperature of the steam in its final state and
what its entropy change? What would be the final temperature and entropy
change for an ideal gas ?
SOLUTION :
Data pada table F.4, untuk tekanan 300 psia dan 500OF dan enthalpy konstan
P1 = 300 psia
T = 500oF
H1 = 1257.7
BTU
lbm
H2= 1257.7
BTU
lbm
S1 = 1.5703
BTU
lbm rankin
Namun entalpi yang diperlukan pada tekanan akhir yaitu 20 psia
Untuk itu diperlukan interpolasi dan diperoleh hasil interpolasi yaitu :
P2 = 20 psia
T = 438.87
S2 = 1.8606
BTU
lbm rankin
∆S = S2 – S1
=1.8606 – 1.5703
= 0.2903
BTU
lbm rankin
BTU
lbm rankin
Untuk steam pada gas ideal, tidak ada perubahan temperature maka
∆S = −RT ln( P 2 )
P1
mol
=
8.314 x 500 x ln(
¿
−¿
¿
molwt = 18 lbmol
20
)
300
= 0.2903
6.30 Superheated steam at 500 kPa and 300 0C expands insentropically to 50 kPa.
What is its final enthalpy?
SOLUTION :
S2 = S1 = Sliq + x. ( Svap – Sliq)
x=
H
S 1−Sliq
Svap−Sliq
J
gm . K
J
( 7.5947−1 .0912 )
gm . K
( 7.4614−1.0912 )
=
= 0.98
= Hliq + x. ( Hvap – Hliq )
= 340.564
J
gm
= 2599.6
J
gm
+ 0.98. ( 2646.9
J
gm
- x340.564
J
gm
)
6.31 What is the mole fraction of water vapor in air that is saturated with water at
25ºC and 101,33 kPa ? at 50ºC and 101,33 kPa ?
SOLUTION :
At 25ºC
Psat = 3.166 kPa
P = 101.33 kPa
Xwater
= Psat
P
= 3.166 kPa
101.33 kPa
At 50ºC
Psat = 12.34 kPa
= 0.031
Xwater = Psat
P
= 12.34 kPa
= 0.122
101.33 kPa
6.32 A rigid vessel contains 0,014 m3 of saturated-vapor in equilibrium with 0,021 m3
of saturated-liquid water at 1000C. Heat is transferred to the vessel until one
phase just disappears, and single phase remains.Which phase (liquid or vapor)
remains, and what are its temperature and pressure? How much heat is
transferred in the process?
SOLUTION :
Berdasarkan tabel F.1 pada suhu 1000C,maka:
Volume vapor = 0,014 m3
Volume liquid = 0,021 m3
Vliq = 1,044 cm3/gm
Uliq = 41,9 J/gm
Vvapor = 1673,0 cm3/gm
Uvap = 2506,5 J/gm
Vtotal = Vliq + Vvap
= (0,014 + 0,021) m3
= 0,035 m3
Mass = mliq + mvap
=
0,021 m3
1,044 cm3 /gm
+
0,014 m3
1673,0 cm 3 / gm
= 0,02011 gm + 8,3682 x 10-6 gm
= 0,02012 gm
X
=
m. vap
mass
=
8,3682 x 10−6 gm
0,02012 gm
= 4,159 x 10-4
V2
=
Vtotal
mass
=
0,035
0,02012
cm3/gm
= 1,739 cm3/gm
Mencapai saturated liquid pada suhu 349,83 K ( fase liquid)
P = 16,5001 kPa
U2 = 1641,7 J/gm
U1
= Uliq + X(Uvap – Uliq)
= 41,9 J/gm + 4,159 x 10-4 ( 2506,5 – 419,0) J/gm
= 419,0 J/gm + 0,8682 J/gm
= 419,8682 J/gm
Q
= U2 – U1
= 1641,7 J/gm - 419,8682 J/gm
Q
= 1221,832 J/gm
6.33 A vessel of 0,25 m3 capacity is filled with saturated steam at 1500 kPa. If the
vessel is cooled until 25% f the steam has condensed, how much heat is
transferred and what is the final pressure?
SOLUTION :
Of this total mass, 25% condenses making the quality 0,75 since the total volume and mass
don’t change. We have for the final state:
V2=V1=V(l)+X.(V(v)-V(l))
Sementara itu:
X=
V 1−V (l)
V ( v )−V (l)
Find P for which (A) yields the value X=0,75 for wet steam
Since the liquid volume is much smaller than the vapor volume,we make a preliminary
calculation to estimate:
Vvap =
V1
X
= 131,66 cm3/gm/0,75 = 175,547 cm3/gm
This Value Occurs at a pressure a bit above 1100 kPa. Evaluate X at 1100 and 1150 kPa by
(A). Interpolate on X to find P = 1114,5 kPa and
Vliq = 782,41 J/gm
Vvap = 2584,9 J/gm
V2 = Vliq + X (Vvap-Vliq)
= 782,41 J/gm + 0,75 (2584,9 J/gm) -782,41 J/gm)
= 782,41 J/gm + 0,75 (1802,49 J/gm
= 782,41 J/gm + 1351,9 J/gm =2134,31 J/gm
Q = mass (V2-V1)
= 1898,9 gm (2134,3 J/gm -2592,4 J/gm)
= 1898,9 gm ( -458,1 J/gm )
= -869886,09 J = -869,9 KJ
6.34 A vessel of 0,25 m3 capacity is filled with saturated steam at 1500 kPa. If the
vassel is cooled until 25 % of the steam has condensed,how much heat is transferred
and what is the final pressure?
SOLUTION :
Vliq = 1.044 cm3/gr
Vvap = 1673.0 cm3/gr
Uliq = 418.959 J/gr
Uvap = 2506.5 J/gr
m liq =
= 19.157 x 103 gr
=
m vap =
=
= 1.184 x 103 gr
m total = m liq + m vap = 20.341 x 103 gr
x
V1
=
=
= 0.058
= Vliq + x (Vvap - Vliq)
= 1.044 cm3/gr + 0.059 (1673.0 cm3/gr - 1.044 cm3/gr)
= 1.044 cm3 /gr + 0.059 x 1671.956 cm3/gr
= 1.044 cm3 /gr + 98.645 cm3 /gr
= 99.689 cm3 /gr
U1
= Uliq x Uvap x Uliq
= 418.959 J / gr + 0.059 (2506.5 J / gr - 418.959 J / gr )
= 418.959 J / gr + 0.059 x 2087.541 J / gr
= 418.959 J / gr + 123.165 J / gr
= 542.124 J / gr
V1 = V2 = 99.689 cm3 /gr
Berdasarkan tabel F.1 maka didapat T = 212 oC
U liq = 904.5 J/gr
U vap = 2598.0 J/gr
U2
= Uliq x Uvap x Uliq
= 904.5 J/gr + 0.058 (2598.0 J/gr - 904.5 J/gr )
= 904.5 J/gr + 0.058 x 1693.5 J/gr
= 904.5 J/gr + 98.233 J/gr
= 1002.723 J/gr
Q
= m total ( U2 – U1 )
= 20.341 x 103 gr (1002.723 J/gr - 542.124 J / gr)
= 20.341 x 103 gr x 460.599 J / gr
= 9369.044 x 103 J
= 9369.044 kJ
6.35. A rigid vessel of 0,4 m³ volume is filled with steam at 800 kPa and 350 c. How
much heat must be transferred from the steam to bring its temperature to 200c.
SOLUTION :
Appendiks F2 :
P = 800 kPa
T = 350c
V1 = 354,34
Cm ³
gr
T = 2000c
U2 = 2638,7
Q=
=
Vtotal
V1
J
gr
(U₂-U₁)
0,4.10 6
cm ³
cm ³
354,34
gr
U1 = 2878
J
gr
Vtotal = 0,4 m³
6.36 1 Kg of steam is contained in a piston / cylinder device at 800 kPa dan 200oC.
a. If it undergoes a mechanically reversible, isothermal expansion to 150kPa.
How much heat does it absorb?
b. If it undergoes reversible, adiabatic expansion to 150kPa. What is its final
temperature and how much work is done ?
SOLUTION :
Dari Tabel F.2 (Superheted Steam) pada 800 kPa dan 200 oCU1 = 2629.9
S1 = 6.8148
J
gm . K
a. Keadaan Isotermal, Pada Saat 150 kPa dan 200 oC
J
Pada Tabel F.2
U2 = 2656.3 gm
J
gm . K
S2 = 7.6439
T = 473 oK
Q=1 kg .473 K . ( 7.6439−6.8148 )
Q=m .T . ∆ S
Q=473 kg . K ( 0.8291 )
J
gm. K
J
gm . K
Q = 392.1643 KJ
W =(m . ∆ U)−Q
W =26.4
{
W = 1 kg . ( 2656.3−2629.9 )
J
kg−392.1643 KJ
gm
B. Entropy konstant pada saat 150 kPa

Sliq = 1.4336
J
gm . K
Uliq = 466.968
J
gm
}
J
−392.1643 KJ
gm
W =−365.7643 K
J
gm

Svap = 7.2234
J
gm . K
Uvap = 2519.5
J
gm
S vap−¿ S
S −S
X = 1 ¿ liq
liq
J
gm . K
X=
J
( 7.2234−1.4336 )
gm . K
( 6.8148−1.4336 )
X =0.929
U 2=U liq + X ( U vap−U liq )
U 2=2.367 x 103
J
gm
W =m. ( U 2−U 1)
3
W =1 Kg ( 2.367 x 10 −2629.9 )
J
gm
W =−262.527 KJ
6.37
Steam at 2000 kPa containing 6% moisture is heated at constant pressure to
848.15 K. (575°C). How much heat is required per kj ?
SOLUTION :
Data, Table F.2 at 2000 kPa:
H vap
= 2797,2 x
J
gm
Hliq = 908,589 x
J
gm
For superheated vapor at 2000 kPa and 575 degC, by interpolation:
H2 = 3633,4 x
J
gm
Q
= mass H2 H1
= 1 kg . (3633,4 – 2,684X103) J/gm
= 949.52 kJ.
6.38 Steam at 2.700 kPa and with a quality of 0,90 undergoes a reversible.adiabatic
expansion in a nonflow process to 400 kPa.It is than heated at constant volume
until it is saturated vapor. Determine Q and W for the process?
SOLUTION :
Kondisi pertama adiabatik ekspansi
P1=2700kPa
X1=0,90
P2=400 kPa
Kondisi ke dua pemanasan dengan volume konstan
1 Q12 = 0
U12= Q12 + W12,,karena Q12 = 0 . maka:
U12= 0 + W12
U12=W12
U2 –U1=W12
W12 = U2 U1
2 Kondisi ke dua pemanasan dengan volume konstan
W23 = 0
U23= Q23 + W23 karena W23 = 0 , maka:
U23= Q23 + 0
U23= Q23
U3-U2= Q23
Q23 = U3 U2
Untuk proses keseluruhan
Q = U3 U2 dan W = U2 U1
Untuk tekanan P1=2700kPa
Dari tabel F2 halaman 710
J
Uliq=977,968 gm
J
Uvap= 2601,8 gm
J
Sliq=2,5924 gm . K
J
Svap=6,2244 gm . K X1=0,9
U1 Uliq x1Uvap Uliq
J
U1 977,968 gm
J
0,92601,8 gm
J
U1 977,968 gm
J
 gm 
J
977,968 gm 
J
U1= 2,439 103 gm
S1 Sliq x1Svap Sliq
J
S1 2,5924 gm . K 0,96,2244
J
S1 2,5924 gm . K
S1= 5,861
J
 gm . K
J
gm . K
Untuk tekanan P2=400 kPa
Dari tabel F2 halaman 700
J
Sliq = 1,7764 gm . K
J
Svap= 6,8943 gm . K
J
Uliq =604,237 gm
Uvap = 2552,7
J
gm
cm3
Vliq = 1,084 gm
3
Vvap = 462,22
J
gm . K
cm
gm
J
2,5924 gm . K 
Since step 1 is isentropic,
S2 = S1 = Sliq x2Svap Sliq
S 1−Sliq
x2= Svap−Sliq
J
J
−1,7764
gm . K
gm . K
x2=
J
J
6,8943
−1,7764
gm . K
gm . K
5,861
J
gm. K
x2=
J
5,1179
gm . K
4,0846
x2 0.798
U2 Uliq x2Uvap Uliq
J
U2 604,237 gm 0.7982552,7
J
J
604,237
gm
gm 
J
J
U2 604,237 gm  gm
U2= 2,159 10
3
J
gm
V2 Vliq x2Vvap Vliq
3
cm
V2 1,084 gm
3
0.798462,22
3
cm
gm
3
cm
1,084 gm

3
cm
V2 1,084 gm

V2= 369,070528
cm3
gm
cm
gm
V2=V3 (karena pemanasan dengan volume tetap/konstan)
The final state is sat. vapor with this specific volume.Interpolate to find that this V occurs
at T = 509,23 0C
Dari tabel F1 halaman 691
Volume spesifik
U spesifik
373,2(X1)
2560,3 (Y1)
369,070528 (X) / V3
???? (Y)/ U3
355,1 (X2)
2562,1 (Y2)
Interpolasi
X −X 1
Y=Y1 + ( X 2−X 1 ) (Y2-Y1)
369,070528−373,2
Y= 2560,3+ (
) (2562,1 -2560,3)
355,1−373,2
J
Y= 2560,7 gm
Y= U3, maka
J
U3 =2560,7 gm
= 2,5607 103
J
gm
Q U3 U2
J
J
3
2,159
10
gm
gm
Q 2,5607 103
Q=0,4017 10
3
J
gm
Work U2 U1
J
J
W= 2,159 103 gm - 2,439 103 gm
J
W= - 0,28 103 gm
W= - 280
J
gm
Tanda minus menunjukkan bahwa sistem melakukan kerja.
6.39 Four kilogram of steam in a piston/cylinder device at 400 kPa and 175 0C
undergoes a mechanically reversible, isothermal compression to a final pressure
such that stem it just saturated. Determine Q and W for the process.
SOLUTION :
Untuk P = 400 kPa & T = 1750C diperolehlah nilai ( Dari tabel F.2 ),yaitu :
U1 = 2605,8 J/gm
S1 = 7,0548 J/gm.K
Untuk saturated steam pada suhu 1750C diperoleh lah nilai ( Dari tabel F.1 ), yaitu :
U2 = 2578,8 J/gm
S2 = 6,6221 J/gm.K
Jadi, hasilnya ialah : a. Q = m x T x ( S2 - S1 )
= 4 kg x 448,15 K x (6,6221 J/gm.K - 7,0548 J/gm.K )
= - 775,66 kJ
b. W = m x (U2 - U1 ) – Q
= 4 kg x ( 2578,8 J/gm - 2605,8 J/gm ) – (- 775,66 kJ )
= 667,67 kJ
6.40. Steam undergoes a change from an initial state of 450 oC and 3,000kPa to a final
state of 140oC and 235 kPa. Determine ∆H and ∆S:
(a) From steam-table data
(b) By equations for an ideal gas
(c) By appropriate generalized correlations
(a) Table F.2, 3000 kPa and 450 oC
SOLUTION :
H 1=3344.6
J
gm
S 1=7.0854
J
gm. K
Table F.2, interpolate 235 kPa and 140 oC
H 2=2744.5
J
gm
S 2=7.2003
∆ H=H 2−H 1
J
gm . K
∆ H=−600.1
∆ S=S 2−S1
∆ S=0.115
(b) T1 = (450 + 273.15) K
J
gm
J
gm . K
T2 = (140 + 273.15) K
T1 = 723.15 K
T2 = 413.15 K
P1 = 3000 kPa
P2 = 235 kPa
ICPH(723.15, 413.15, 3.470, 1.450×10-3 , 0.0,0.121×105) = -1343.638
ICPS(723.15,413.15, 3.470,1.450×10-3 , 0.0, 0.121×105) = -2.415901
ICPH = -1343.638 K
ICPS = -2.415901
Eqs. (6.86) & (6.87) for an ideal gas:
molwt=18
R .(ICPS−ln
R . ICPH
∆ H ig =
molwt
∆ H ig =−620.6
∆ Sig =
J
gm
∆ Sig =0.0605
(c) Tc = 647.1 K
Pc = 220.55 bar
T r =1.11752
Pr =0.13602
1
gm
mol
1
P2
)
P1
( )
molwt
J
gm . K
w = 0.345
T r =0.63846
Pr =0.01066
2
2
The generalized virial-coefficient correlation is suitable here
HRB(1.11752, 0.13602, 0.345) = -0.13341
HRB1 = -0.13341
SRB(1.11752, 0.13602,0.345) = -0.08779
SRB1 = -0.08779
HRB(0.63846, 0.01066, 0.345) = -0.04422
HRB2 = -0.04422
SRB(0.63846, 0.01066,0.345) = -0.05048
SRB2 = -0.05048
∆ H=∆ H ig +
∆ S=∆ S ig +
R .T c .( HR B2−HR B1 )
molwt
R .(HR B 2−HR B1)
molwt
∆ H=−593,95
∆ S=0.078
J
gm
J
gm. K
6.41 A piston / cylinder device operating in a cycle with steam as the working fluid
executes the following steps:

Steam at 550 kPa and 473.15 K (200°C) is heated at constant volume to a

pressureof 800 kPa.
It then expands, reversibly and adiabatically, to the initial temperature of

473.15 K(200°C).
Finally, the steam is compressed in a mechanically reversible, isothermal
processto the initial pressure of 550 kPa.
SOLUTION :
Data table F.2 Superheat Steam pada 550 kPa dan 200OC:
V1 = 385,19 cm3/gm
U1 = 2640,6 J/gm
S1 = 7,0108 J/gm.K
Pemanasan Volume konstan pada tekanan 800 kPa, pada volume spesifik awal P,
interpolasi yang didapat t = 401,74 oC. V = Konstan
U2 = 2963,1 J/gm
S2 = 7,5782 J/gm.K
Sehingga,
Q12
= U2- U1
= ( 2963,1
-
2640,6 ) J/gm
= 322,4 J/gm
Ekspansi ke T awal, maka:
Q23 = 0
S3 = S2
S3 = 7,5782 J/gm.K
Temperatur konstat dikompresi ke P awal:
T = 473,15 K
Q31 = T ( S1 – S3 )
= 473,15 (7,0108 - 7,5782 ) = -268,465 J/gm
Untuk Siklus perubahan energy dalah adalah = 0
Wsiklus
=
-Qsiklus
n
= - Wsikuls / Q12
n
=
1+
Q 31
Q 12
=
=
- Q12 – Q31
1+
−268,465
322,4
= 0,1672
6.42 A piston/ cylinder device operating in a cycle with steam as the working fluid
executes the following steps :


Saturated – vapor steam at 300(psia) is heated constant preasure to 900ºF
In the expand, reversibly and adiabatically, to the initial temperature of
417,35ºF

Finally, the steam is compressed in a mechanically reversible,isothermal proses
to the initial state
What is the thermal efficiency of the cycle ?
SOLUTION :
Table, F.4, sat. Vapor, 300(psi) :
T1 = (417.35 + 459.67) rankine
H1 = 1202.9
BTU
Lbm. Rankine
T1 = 877.02 rankine
S1 = 1.5105
BTU
Lbm. Rankine
Superheated steam at 300(psi) and 900ºF
H2 = 1473.6 BTU
S2 = 1.7591 BTU
lbm
Q12 = H2-H1
Lbm. Rankine
Q31 = T1. (S1-S3)
Q31 = -218.027 BTU
lbm
6.43 Uap masuk ke dalam turbin pada tekanan 400 kPa dan suhu 400 °C terekspansi
secara reversibel dan adiabatis.
a. Berapakah tekanan yang dikeluarkan dalam bentuk aliran uap jenuh?
b. Berapakan tekanan yang dikeluarkan dalam bentuk aliran uap basah dengan
nilai 0,95?
SOLUTION :
Dari data tabel F.2 superheated steam pada 4000 kPa dan 400 °C:
S1= 6,7733
J
gmK
Dari kedua masalah didapatkan: S2 S1
a. Pertama kita mencari tekanan dimana tekanan uap jenuhnya memiliki entropi.
Hal ini terdapat pada tekanan di bawah 575 kPa, jadi kita mencarinya dengan cara
interpolasi.
P2 = 572,83kPa
b. Untuk uap basa, entropinya di dapatkan:
x = 0,95
S2 = Sliq xSvap Sliq
Jadi kita harus mendapatkan tekanan dimana persamaannya diketahui. Tekanan ini terdapat
jika dibawah tekanan 250 kPa. Pada tekanan 250 kPa nilainya:
J
Sliq = 1,6071 gmK
Svap = 7,0520
J
gmK
S2= Sliq x Svap Sliq
J
S2 = 1,6071 gmK
= 6,7798
+ 0,95 ( 7,0520
J
J
¿
1,6071
gmK
gmK
J
gmK atau > 6,7733
Didapatkan dari interpolasi P2 = 250,16 kPa
6.46 Table F.2 for superheated vapor at the initial conditions, 1300 kPa and 400
degC, and for the final condition of 40 kPa and 100 degC:
SOLUTION :
H1 = 3259.7 kJ/kg
S1 = 7.3404 kJ/kgK
H2 = 2683.8 kJ/kg
If the turbine were to operate isentropically, the final entropy would be
S2 S1
Table F.2 for sat. liquid and vapor at 40 kPa:
Sliq = 1.0261kJ/kgK
Svap = 7.6709 kJ/kgK
Hliq = 317.16 kJ/kg
Hvap = 2636.9 kJ/kg
x2= (S2 Sliq) / (Svap Sliq)
x2= (7.3404 kJ/kgK - 1.0261kJ/kgK) / (7.6709 kJ/kgK - 1.0261kJ/kgK)
x2= 0.95
H' Hliq x2Hvap Hliq
H'= 317.16 kJ/kg + 0.95(2636.9 kJ/kg - 317.16 kJ/kg)
H' =2.522 103 kJ/kg
H2 H1) / (H' H1)
2683.8 kJ/kg - 3259.7 kJ/kg) / (2.522 103 kJ/kg - 3259.7 kJ/kg)
0.78
6.47 From steam table data estimate values for the residual properties V R , HR , SR for
steam at 225 oC and 1600 kPa and compare with values found by a suitable
generalized correlation?
SOLUTION :
P = 1600 kPa = 1.6 x 10 5 Pa = 1.6 atm
V = 132.85 cm3 / gr
H = 2856.3 J / gr
S = 6.5503 J / gr K
Hig = 2928.7 J / gr
Sig = 10.0681 J / gr K
T = 225 oC = 498.15 K
Molwt = 18 gr / mol
R
molwt
R
V =V–
T
P
0.08206 atm
VR = 132.85 cm3 / gr –
18
gr
mol
VR = 132.85 cm3 / gr – 3633 cm3 / gr
VR = - 3500.15 cm3 / gr
HR = H - Hig
HR = 2856.3 J / gm - 2928.7 J / gm
HR = - 72.4 J / gm
Δ Sig =
−R
molwt
ln
P
Po
dm 3
mol K
498.15 K
1.6 atm
−o .08206 atm
=
18
= - 0.0336
dm3
mol K
gr
mol
ln
1600 kPa
1 kPa
atm dm3
gr K
J
gr K
= - 3.36
SR = S – ( Sig + Δ Sig)
J
gr K
= 6.5503
- ( 10.0681
J
gr K
- 3.36
J
gr K
J
gr K
= - 0.2298
Reduced condition
ω = 0.345
Tc = 647.1 K
Pc = 220.55 bar
Tr =
T
Tc
=
498.15 K
647.1 K
Pr =
P
Pr
220.55 ¯¿
¯
1.6 ¿
=
= 0.00725
¿
¿
Bo = 0.083 –
0.422
1.6
Tr
Bi = 0.139 –
0.172
Tr 4.2
= 0.76982
= 0.083 –
0.422
1.6
0.76982
= - 0.558
= 0.139 –
0.172
0.769824.2
= - 0.377
)
Z = 1 + ( Bo + ω Bi )
Pr
Tr
Z = 1 + ( -0.558 + 0.345 x -0.377 )
0.00725
0.76982
Z = 1 – 0.00648
Z = 0.99352
VR =
RT
P molwt
(Z–1)
dm3
498.15 K
mol K
gr
1.6 atm18
mol
0.08206 atm
VR =
( 0.99352 – 1 )
VR =- 9.19755 cm3 / gr
HRB = - 0.17858
SRB = - 0.167101
HR =
R Tc
molwt
HRB
dm 3
647.1 K
molK
gr
18
mol
0.08206 atm
HR =
-0.17858
HR =-52.643 J / gr
SR =
R
molwt
SRB
0.08206 atm
SR =
18
dm 3
molK
gr
mol
SR = - 0.0076 J / gr K
-0.167101
6.48 From data in the steam tables:
(a) Determine values for Gl and Gv for saturated liquid and vapor at 1000 kPa.
Should these be the same?
(b) Determine values for ∆Hlv/ T and ∆Slv at 1000 kPa. Should these be the
same?
(c) Find values for VR, HR , and SR for saturated vapor at 1000 kPa.
(d) Estimate a value for d P
sat
/dT at 1000 kPa and apply the Clapeyron
equation to evaluate ∆Slv at 1000 kPa. Does this result agree with the steamtable value?
Apply appropriate generalized correlations for evaluation of VR, HR , and SR for
saturated vapor at 1000 kPa. Do these results agree with the values found in (c)?
SOLUTION :
So,
∆Vlv = Vv - Vl = 193.8 cm3 gr -1 - 1.128 cm3 gr -1 = 192.672 cm3 gr -1
∆Hlv = Hv - Hl = 2776.3 J gr -1 - 763.1 J gr -1 = 2.0132 J gr -1
∆Slv = Sv - Sl = 6.5819 J gr-1 k-1 - 2.1393 J gr-1 k-1 = 4.4426 J gr-1 k-1
a. Gl = Hl – T. Sl = 763.1 J gr -1 – 453.15 K x 2.1393 J gr-1 k-1 = -206.32 J gr-1
Gv = Hv – T.Sv = 2776.3 J gr -1 - 453.15 K x 6.5819 J gr-1 k-1 = -206.29 J gr-1
b. ∆Hlv = 2.0132 J gr -1 ; T = 453.15 K
r = ∆Hlv / T = 2.0132 J gr -1 / 453.15 K = 4.4427 J gr-1 k-1
∆Slv = Sv - Sl = 6.5819 J gr-1 k-1 - 2.1393 J gr-1 k-1 = 4.4426 J gr-1 k-1
c. VR = Vv – ( R. T / molwt. P)
= 193.8 cm3 gr -1 – ( 8.314 J mol-1 k-1 x 453.15 K / 18.015 gr mol-1 x 1000
= -14.875 cm3 gr-1
For enthalpy and entropy, assume that steam at 179.88 degC and 1 kPa is an ideal gas. By
interpolation in table F.2 at 1 kPa
Hig = 2831.2 J gr-1 ; Sig = 8.7994 J gr-1 k-1 ; P0 = 1 kPa
The enthalpy of an ideal gas is independent of pressure ; the entropy does depend on P :
HR = Hv – Hig = 2776.3 J gr -1 - 2831.2 J gr-1 = -36.9 J gr-1
∆Sig = (-R / molwt ) x ln (P/P0) = ( -8.314 / 18.015) ln (1000/1) = -3.188 J mol-1 k-1
SR = Sv – Sig + ∆Sig = 6.5819 J gr-1 k-1 - 8.7994 J gr-1 k-1 + 3.188 J mol-1 k-1
= -0.1126 J gr-1 k-1
d. Assume ln P vs 1/T linear and fit three data pts @ 975, 1000, 1050 kPa
Data :
pp: ( 975, 1000, 1050 ) kPa ; t: ( 178.79, 179.88, 182.02 ) (degC)
Xi = 1/ ti + 273.15
; Yi = ln (ppi / kPa)
; slope = slope (x,y) slope = -4717
dPdT = (-P / T2) ln slope K = 22.984 kPa k-1
∆Slv = ∆Vlv . dPdT = 192.672 cm3 gr -1 x 22.984 kPa k-1 = 4.4 J gr-1 k-1
Reduced conditions :
ώ = 0.345 ; Tc = 647.1 K ; Pc = 220.55 bar
Tr = T / Tc = 453.15 / 647.1 = 0.7001 K
Pr = P / Pc = 10 / 220.55 = 0.0453 bar
The generalized virial-coefficient correlation is suitable here
B0 = 0.083 – ( 0.422 / Tr 1.6 ) = -0.664
B1 = 0.139 – ( 0.172 / Tr 4.2 ) = -0.63
By Eqs (3.61) + (3.62) and (3.63) along with Eq. (6.40)
Z = 1 + B0 + ώ . B1 . Pr / Tr
= 1 + -0.664 + 0.345 . -0.63 . 0.0453 / 0.7001
= 0.943
VR = ( R.T / P. molwt ) Z-1
= ( 8.314 . 453.15 / 10. 18.015) 0.943 – 1
= -11.93 cm3 gr-1
HR = ( R.Tc / molwt) (HRB. Tr)
i = 1..3
= -43.18 J gr-1
SR = R / molwt ( SRB. Tr)
= -0.069 J gr-1 k-1
6.49 From data in the steam tables :
a)
Determine values for Gl and Gv for saturated liquid and vapor at 150(psia).
Should these be the same?
b) Determine values for ΔHlv /T and ΔSlv at 150(psia). Should these be the same?
c)
Find values for VR, HR, and SR for saturated vapor at 150(psia)
d) Estimate a value for d Psat /dT at 150(psia) and apply the Clapeyron equation to
evaluate ΔSlv at 150(psia). Does this result agree with the steam-table value?
Apply appropriate generalized correlations for evaluation of VR, HR, and SR for
saturated vapor at 150(psia). Do these results agree with the values found in (c)?
T= (358,43+459,67) Rankine = 818,1 Rankine
SOLUTION :
Dari tabel F.4 didapat :
Vl = 0,0181 ft3/lbm
Vv = 3,014 ft3/lbm
Hl = 330,65 btu/lbm
Hv = 1194,1 btu/lbm
Sl = 0,5141 btu/lbm rankine
Sv = 1,5695 btu/lbm rankine
Vlv
= Vv - Vl
= (3,014 – 0,0181) ft3/lbm
Hlv
= Hv - Hl
= (1194,1 – 330,65) btu/lbm = 863,45 btu/lbm
a.
Gl=H l
TSl
-
= 330.65 btu/lbm - (818,1 R . 0,5141 btu/lbm Rankine)
= -89,935 btu/lbm
v
G =H
v
TS
-
v
= (1194,1 btu/lbm) – (818,1 R . 1,5695 btu/lbm Rankine)
= -89,91 btu/lbm
b.
∆ Slv =Sv
-
Sl
= (1,5695 – 0,5141) btu/lbm R
= 1,0554 btu/lbm R
∆ H lv 863,45 btu /lbm
=
T
818,1 R
.
c.
V R =V V −
= 1,0554 btu/lbm R
RT
BM . P
¿ 3.014 ft 3 /lbm−
10.73 ft 3 psia/lbmol R . 818.1 R
18.015lbm /lbmol .150 psia
3
¿−0.2345
ft
lbm
Untuk entalpi dan entropi asumsikan sistem pada 358.43oF dan 1 psi adalah gas
ideal. Dengan interpolasi pada tabel F.4 pada 1 psi, didapat:
T −T b H ig −H b
=
T a−T b H a−H b
T −T b S ig −Sb
=
T a−T b Sa −S b
H ig −1218.7
358.43−350
=
400−350
1241.8−1218.7
Sig −2.1445
358.43−350
=
400−350
2.1722−2.1445
H ig =1222.6
Btu
lbm
S ig =2.1492
Btu
lbm R
Po = 1 psi
Entalpi gas tidak bergantung pada tekanan, tetapi entropi selalu tergantung pada tekanan,
sehingga,
H R =H v −H ig
lbm−1222.6 btu/ ¿ lbm
¿ 1194.1 btu/¿
= -28.5 btu/lbm
∆ Sig =
¿
−R
P
ln
BM
Po
( )
−1.987 btu /lbmol R 150
ln
18.015 lbm/lbmol
1
= - 0.552 btu /lbmol R
S R =Sv −( S ig −∆ Sig )
¿ 1.5695 btu/lbmol R−(2.1492−0.552)btu / lbmol R
= - 0,0277 btu/lbmol R
d. Asumsikan ln P vs
[ ]
145
150 psi
155
Data :
.
X i=
1
T
linear untuk data (145, 150, dan 155 psia).
t=
1
t i+ 459,67
.
X 1=
1
=1,23 .10−3
355,77+ 459,67
.
X 2=
1
=1,22. 10−3
358,43+ 459,67
.
X 3=
1
=1,218 .10−3
361,02+459,67
.
.
y=ln
( Ppi
Psi )
y 1=ln
( 145
150 )
= -0,0339
[ ]
355,77
358,43
361,02
0
F
i=1, 2,3
.
y 2=ln
( 150
150 )
=0
.
y 3=ln
( 155
150 )
= 0,0328
3
Slope = -8,501 . 10
dP dT =
−P
. slope
T2
818,1 R
¿
¿
¿
2
=
¿
−150 psia
¿
= 1,905 psi/R
.
∆ Slv =∆ V lv dPdT
3
2,996
=
ft
psi
. 1,905
. 1,987 btu/lbmol R
lbm
R
ft 3 psi
10,73
lbmol R
= 1,057 btu/lbm R
Kondisi Reduced
W = 0,345
T = 358,43 0F = 454,5 K
Tc = 647,1 K
P = 150 psi = 10,342 bar
Pc = 220,55 bar
Pr =
Tr =
P 10,342
=
=0,0469
P c 220,44
T
Tc
=
454,5
=0,7024
647,1
Korelasi keofisien virial yang sesuai disini :
.
B o=0,083−
=
0,422
Tr 1,6
0,083−
0,422
1,6
(0,7024)
= -0,66
.
B 1=0,139−
=
0,172
4,2
Tr
0,083−
0,172
(0,7024)4 , 2
= -0,62
Dengan persamaan 3.61 , 3.62 , dan 3.63 dan persamaan 6.40 dari interpolasi Tr dan
Pr didapat ω=0,0336 .
Pr
z=1+
Bo+ω
B
(
)
1
.
Tr
−0,62 x 0,0336
0,0469
0,66+(¿)
0,7024
¿ 1+ ¿
= 0,942
.
V R=
=
RT
(z−1)
P BM
ft 3 psia
. 818,1 R
lbmol R
(0,942−1)
150 psia. 18,015 lbm/lbmol
10,73
= -0,1884 ft3/lbm
HR
T r , ω , Br
R.Tc
¿
HRB ¿ )
BM
= -19,024 btu/lbm
T r , ω , Br
R
HR=
SRB¿ )
BM
= -0,0168 btu /lbm rankin
6.49 From data in the steam tables :
e) Determine values for Gl and Gv for saturated liquid and vapor at 150(psia).
Should these be the same?
f) Determine values for ΔHlv /T and ΔSlv at 150(psia). Should these be the same?
g) Find values for VR, HR, and SR for saturated vapor at 150(psia)
h) Estimate a value for d Psat /dT at 150(psia) and apply the Clapeyron equation
to evaluate ΔSlv at 150(psia). Does this result agree with the steam-table
value?
SOLUTION :
Apply appropriate generalized correlations for evaluation of VR, HR, and SR for saturated
vapor at 150(psia). Do these results agree with the values found in (c)?
T= (358,43+459,67) Rankine = 818,1 Rankine
P=150 psi
Molwt= 18,015 gr/mol
Dari tabel F.4 , diketahui :
Vl= 0,0181 ft3/lbm
; Vv=3,014 ft3/lbm
Hl=330.65 BTU/lbm ; Hv=1194,1 BTU/lbm
Sl=0,5141 BTU/lbm.Rankine
ΔVlv=Vv-Vl=2,996 ft3/lbm
ΔHlv=Hv-Hl=863,45 BTU/lbm
; Sv= 1,5695 BTU/lbm.rankine
a. Gl=Hl-T.Sl = 330.65 BTU/lbm-818,1 rankine. 0,5141 BTU/lbm.Rankine
= -89,94 BTU/lbm
Gv= Hv –T. Sv= 1194,1 BTU/lbm-818,1 rankine. 1,5695 BTU/lbm.rankine
= -89,91 BTU/lbm
b. ΔSlv= Sv - Sl= (1,5695 - 0,5141) BTU/lbm.Rankine = 1,055 BTU/lbm.Rankine
r=ΔHlv /T = 863,45 BTU/lbm / 818,1 Rankine = 1,055 BTU/lbm.Rankine
c. VR= Vv-
=
=-0,235 ft3/lbm
Untuk enthalpy dan entropi, asumsikan bahwa uap berda pada 358,43 OF dan 1
psi adalah suatu gas ideal. Dengan interpolasi pada tabel F.4 di 1 psi, maka :
Hig= 1222,6 BTU/lbm
; Sig=2,1492 BTU/lbm.rankinbe pada Po=1 psi
HR=Hv-Hig = 1194,1 BTU/lbm - 1222,6 BTU/lbm = -28,5 BTU/lbm
ΔSig=
SR
=
=Sv-(Sig+
= -0,552 BTU/lbm.Rankine
ΔSig)=1,5695
BTU/lbm.rankine–(2,1492
+0,552)
BTU/lbm.Rankin
= -0,0274 BTU/lbm.rankine
d. Asumsikan ln P VS 1/T linier dan didapatkan tiga data, yaitu :
i
1
Pp (psi)
145
T (F)
355,77
2
150
358,43
3
155
361,02
Slope = -8501
dPdT=(-P.Slope.rankine)/T2 = 1,905 psi/rankine
ΔSlv=
ΔVlv.
dPdT=
2,996
ft3/lbm.
1,905
BTU/lbm.rankine
\
ω=0,345
Tc=647,1 K
Tr= T/Tc= 0,7024
Pc=220,55 bar
Pr= P/Pc = 0,0469
Koefisien virial umum yang cocok disini adalah :
Bo=0,083- (0,422/Tr.1,6) = -0,66
Bl=0,139-(0,172/Tr.4,2) = -0,62
Z=1+ (Bo+ω.B1.Pr/Tr) = 0,942
VR=
= -0,1894 ft3/lbm
HR=
= -19,024 BTU/lbm
SR=
= -0,0168 BTU/lbm.rankine
psi/rankine
=
1,056
6.50 Propane gas at 1 bar and 35 0C is compressed to a final state of 135 bar and
1950C. estimate the molar volume of the propane in the final state and the
enthalpy and entropy changes for the process. In its initial state, propane may be
assumed an ideal gas.
SOLUTION :
Untuk propane :
TC =369.8 K PC =42.28 bar
ω=0.152
T= 195 + 273.15 K = 468.15 K
P=135 bar
Tr=
Pr =
T
TC
P
PC
pO=1bar
=
468.15
369.8
= 1,266
135
42.28
= 3.178
=
Gunakan korelasi antara lee atau kesler dengan interpolasi
Zo=0.6141
Z1=0.1636
Z= ZO + ω .Z1
Z=0.6141 + 0.152 (0.1636)
Z=0.639
ZRT
V= P
V=
0.639 ( 0.08314 ) (468.15)
135
V=184.2
cm3
mol
HRO =-2.496 .R.TC
=-2.496( 0.08314 ) 369.8 K
J
HRO=-7.674 X 103 mol
HR1=-0.586
6.51
Propane at 70 oC and 101.33 kPa is compressed isothermally to 1500
kPa.Estimate ∆H and ∆S for the process by suitable generalized correlations
SOLUTION :
Untuk Propana
Tc = 369.8 K
Pc = 42.48 bar
ω = 0.152
T
= (70+273)K = 343 K
P0
= 101.33 kPa
P
= 1500 kPa
Tr =
Tr=
T
Tc
Pr=
343 K
369.8 K
Tr=0.9275
B 0=0.083−
0.422
T r 1.6
B 0=0.083−
0.422
0.9275
0
B =0.083−0.4549
B 0=−0.3719
1
B =0.139−
1500 kPa
4248 kPa
Pr=0.3531
Diasumsikan propana merupakan gas ideal, maka :
B 1=0.139−
Pr=
0.172
4.2
Tr
0.172
0.9275
P
Pc
1
B =0.139−0.1854
1
B =−0.0464
d B0 0.675
=
d T r T r 2.6
¿
0.675
=−0.2525
0.9275
1
d B 0.722
=
d T r T r 5.2
¿
0.722
0.925
¿−0.203
R
[
0
(
1
H
dB
dB
0
1
=Pr B −Tr
+ω B −Tr
Tc
dTr
dTr
)]
¿ 0.3531 [ −0.3719−0.9275 (−0.2525 )+ 0.152 (−0.0464−0.9275 (−0.203 ) ) ]
¿ 0.3531 [−1.0469+ (−0.178 ) ]
¿ 0.3068
R
(
0
1
S
dB
dB
=−Pr
+ω
R
dTr
dTr
)
SR
=−0.3531 (−0.2525+ 0.152 (−0.203 ) )
R
R
S
=−0.351 (−0.2833 )
R
R
S
=0.0994
R
jadi , dari keterangan rumus diatas , dapat dihitung nilai dari ∆ S dan ∆ H , yaitu:
∆ H=( Cigp ) H ( T 2 −T 1 ) + H 2R−H 1R
∆ S=−1431.3
J
mol
∆ S=( C p ) S ln
T2
P2 R R
−R ln +S 2 −S1
T1
P1
∆ S=−25.287
J
mol . K
ig
6.52 A steam of propane gas is partially liquetied by throttling from 200 bar and 370K
to 1 bar. What fraction of the gas is liquefied in this process? The vapor pressure
of propane is given by Eq. (6.72) with parameters: A=-6,72219, B=1,33236, C=2,13868, D=-1,38551.
SOLUTION :
W=0.152
Tc=369,3 K
Pc=42,48 bar
Zc=0,276
Vc=200 cm3/mol
Jika keadaan akhir adalah campuran dua fase, maka harus ada pada suhu jenuh
pada 1 bar. suhu ini ditemukan dari tekanan uap.
P= 1 bar
A= -6,72219
B= 1,33236
C= -2,13868
D= -1,38551
Panas laten penguapan pada kondisi akhir akan diperlukan untuk energi
keseimbangan. Hal ini ditemukan oleh persamaan Clapeyron. Kami melanjutkan persis
seperti dalam Pb. 6,17.
P=Pc exp [ A . τ (T ) + B .() τ ( T )1,5 + C .()τ ( T )3+ D.( )τ ( T )6 ]
1−τ (T )
T =230,703 K
Panas laten penguapan pada kondisi akhir akan diperlukan untuk energi
keseimbangan. Hal ini ditemukan oleh persamaan Clapeyron. Kami melanjutkan persis
seperti dalam Pb. 6,17.
T = 230,703
P( T )=Pc exp [ A . τ ( T ) + B .() τ ( T )1,5 +C .() τ (T )3 + D .() τ ( T )6 ]
1−τ (T )
d
kPa
P ( T )=4,428
dT
K
dPdT =4,428124
kPa
K
P = 1bar
Pr=
P
Pc
Pr = 0,024
Tr=
T
Tc
Tr = 0,624
B ()=0,083−
0,422
Tr 1,6
B() = -0,815
B 1=0,139−
0,172
4,2
Tr
B1 = -1,109
Vvap=
RT
Pr
[1+() B( )+ W . B 1. ]
P
Tr
Vvap=1,847 x 10 4
cm 3
mol
2
Vliq=Vc . Zc[−Tr 7 ]
Vliq=75,546
cm3
mol
Δ H 1 v=T . () Vvap−Vliq . dPdT
Δ H 1 v=1,879 x 10
4
J
mol
Untuk Langkah (1), menggunakan korelasi umum dari Tabel E.7 & E.8, dan
biarkan Jumlah dari perubahan entalpi untuk langkah-langkah ini ditetapkan sama dengan
nol, dan persamaan yang dihasilkan diselesaikan untuk fraksi dari aliran yang cair.
ENERGY BALANCE: Untuk proses throttling ada entalpi tidak berubah. Jalan
kalkulasional dari keadaan awal ke akhir dibuat lanjut dari langkah-langkah berikut:
(1)
Merubah
gas
awal
menjadi
gas
ideal
di
awal
T
&
P.
(2) Melakukan perubahan suhu dan tekanan untuk T & P pada akhir negara ideal gas.
(3)
Merubah
gas
ideal
menjadi
gas
(4) Sebagian memadatkan gas pada akhir T & P.
R
H
¿1
R . Tc
()
HR
r ( )=
r 1=¿
R . Tc
( )
T1 = 370 K
nyata
pada
akhir
T
&
P.
Tr=
T1
P1
Pr=
Tc
Pc
Tr = 1,001
r() = -3,7733
Pr = 4,708
r1 = -3,568
Δ H 1=−RTc ()r ()+ r 1. ω
Δ H 1=1,27 x 10 4
J
mol
Untuk Langkah (2) perubahan entalpi diberikan oleh Persamaan. (6.86), yang;
ICPH (370, 230.703 , 1.213 , 28.785.10-3 , -8.824.10-6 , 0.0) = -1260.405 K
∆ H 2=R (−1260,405 K ) ∆ H 2=−1,248 x 104
J
mol
Untuk Langkah (3) perubahan entalpi diberikan oleh Persamaan. (6.78), yang;
Tr=
230,703 K
Tr=0,6239
Tc
¯
1 ¿ Pr =0,0235
Pc
Pr =¿
HRB (0.6239 , 0.023 , 0.152 ) = - 0.07555
∆ H 3=R . Tc . HRB ∆ H 3=−232.28
HRB = - 0.07555
J
mol
∆ H 4=−x . ∆ H lv
∆ H 1+ ∆ H 2+ ∆ H 3−x . ∆ H lv =0
x=
∆ H 1+ ∆ H 2+ ∆ H 3
x=0,136
∆ H lv
6.53 Estimate the molar volume,enthalpy and entropy for 1,3-butadiene as a
saturated vapor and as a saturated liquid at 380 K.The enthalpy and entropy
are set equal to zero for the ideal-gas state at 101,33 Kpa and 0 derajat
celcius.The vapor pressure of 1,3-butadiene at 380 K is 1.919,4 kPa.
SOLUTION :
Untuk 1,3-butadiene : ω = 0,190
Tc = 425,2 K
Pc = 42,77 bar
Zc = 0,267
Vc = 220,4 cm3/mol
Tn = 268,7 K
T = 380 K
P = 1919,4 KPa
To = 273,15 K
Po
Tr = T / Tc
Tr = 0,894
Pr = P / Pc
Pr = 0,449
Zo = 0,7442
Z1 = -0,1366
Z = Zo + ω Z1
Z = 0,718
Vvap = Z. R. T
Vvap = 1182,2 cm3 / mol
=
101,33
KPa
P
HRo = -0,689 . R. Tc
HR1 = -0,892 R. Tc
HRo = -2,436 x 103 J/ mol
HR1 = -3,153 x 103 J/mol
SRo = -0,540 R
SR1 = -0,888 R
SRo = -4,49 J/mol.K
SR1 = -7,383 J/mol. K
HR = HRo + ω HR1
SR = SRo + ω SR1
HR = -3,035 x 103 J/mol
SR = -5,892 J/mol. K
Hvap = 6315,9 J/mol
Svap = -1,624 J/mol.K
Untuk saturated vapor,
Vliq = Vc. Zc (1-Tr 2/7)
Vliq. = 109,89 cm3/mol
∆Hn = R.Tn (1,092 ((ln (Pc/bar) – 1,013) / 0,930 – Tn/Tc)
∆Hn = 22449 J/mol
∆Hn = ∆Hn (1- Tr/ 1-Tn/Tc)0,38
∆H = 14003 J/mol
Hliq = Hvap - ∆H
Hliq = -7687,4 J/mol
Sliq = S uap - ∆H/T
Sliq = -38,475 J/mol. K
6.54 Estimate the molar volume, enthalphy,and entropy for n-Butane as a saturated
vapor and as a saturated liquid at 370 K.The enthalpy and entropy are set equal
to zero for the ideal-gas state at 101,33 kPa and 273,15 K .The vapor pressure of
n-Butane at 370 K is 1435 kPa.
SOLUTION :
Untuk n-Butane B.1 halaman 654.
0,200
Tc 425,1K
Pc 37,96bar = 37,96 x105 )Pa =37,96x 105 Pa=3796 kPa.
Zc 0,274
3
Vc = 255
cm
mol
Tn 272,7K
T 370K
P 1435kPa
T0 273,15K
P0 101,33kPa
Tr =
T
Tc
=
370 K
425,1 K
=
1435 kPa
3796 kPa
=0,87
Tr 0.87
Pr=
P
Pc
= 0,378
Pr 0.378
Use Lee/Kesler correlation, however, the values for a saturated vapor lie on the very edge
of the vapor region, and some adjacent numbers are for the liquid phase. These must NOT
be used for interpolation. Rather, extrapolations must be made from the vapor side. There
may be some choice in how this is done, but the following values are as good as any:
Dari tabel E2 halaman 668 didapatkan:
Untuk Z0
0,2 (X1)
0,85 (Y1)
0,8810 (M11)
0,87(Tr) (Y)
0,90 (Y2)
0,9015 (M21)
0,378(Pr) (x)
???? (M)
X 2−X
X −X 1
M = (( X 2−X 1 )M11 + ( X 2−X 1 ) M12)
X −X 1
( X 2−X 1 ) M22)
Y −Y 1
Y 2−Y 1
0,4 (X2)
0,0661 (M12)
0,7800 (M22)
Y 2−Y
Y 2−Y 1
X 2−X
+ (( X 2−X 1 )M21 +
0,4−0,378
M = (( 0,4−0,2 )0,8810 +
0,378−0,2
( 0,4−0,2 ) 0,0661)
0,4−0,378
0,378−0,2
0,4−0,2 )0,9015 + ( 0,4−0,2 ) 0,7800)
0,90−0,87
0,90−0,85
+
((
0,87−0,85
0,90−0,85
M= 0,7692
Z0 0,7692
Dari tabel E2 halaman 669 didapatkan:
Untuk Z1
0,85 (Y1)
0,87(Tr) (Y)
0,90 (Y2)
0,2 (X1)
-0,0715 (M11)
0,378(Pr) (x)
???? (M)
- 0,0142 (M21)
-0,1118 (M22)
X 2−X
X −X 1
M = (( X 2−X 1 )M11 + ( X 2−X 1 ) M12)
X −X 1
( X 2−X 1 ) M22)
M = ((
0,4 (X2)
- 0,0268 (M12)
Y 2−Y
Y 2−Y 1
X 2−X
+ (( X 2−X 1 )M21 +
Y −Y 1
Y 2−Y 1
0,4−0,378
0,4−0,2 ) -0,0715 +
(
0,378−0,2
0,4−0,2 )
0,4−0,378
0,378−0,2
)
-0,0142
+
(
0,4−0,2
0,4−0,2 ) -0,1118)
-0,0268)
0,90−0,87
0,90−0,85
+
((
0,87−0,85
0,90−0,85
M0,1372
Z1 0,1372
Z Z0 Z1 = 0,7692 + 0,200 (0,1372) = 0,7692 - 0,02744= 0,74176=0,742
Z 0.742
j
j
0.742 x 8,314
x 370 K
0.742 x 8,314
x 370 K
Z x RxT
mol K
mol
K
V =
=
=
=
P
1435 kPa
1435 x 103 Pa
3
1590,1
cm
mol
cm3
V = 1590,1 mol
J
mol . K
HR0 0,607RTc 0,6078,314
2,145 10
3
425,1K=
2145,30881
J
mol
J
mol
J
mol
HR0= 2,145 103
HR1 0.831RTc 0.8318,314
J
mol . K 425,1K = - 2936,987843
J
mol
= 2,937
J
103 mol
J
HR1= 2,937 103 mol
SR0 0,485R 0,485 x 8,314
J
mol . K
= -4,032
J
mol . K
J
mol . K = 6,942
J
mol . K
J
SR0= 4,032 mol . K
SR1 0.835R 0.835 x 8,314
SR1= 6,942
=
J
mol . K
HR HR0 HR1
HR 2,145 103
J
mol
HR = 2,733 103
J
mol
SR SR0 SR1
J
SR 4,032 mol . K
J
+ 0,200 x( 2,937 103 mol ) = - 2,7324 103
+ 0,200 x (6,942
J
mol . K )= 5.4204
J
mol . K
J
mol
J
mol . K
SR = 5.421
ICPH273.153701.93536.915103 11.402106 0.0= 1222.048
ICPH 1222,048K
ICPS273.15370 1.93536.915103 11.402106 0.0= 3.80735
ICPS 3,80735
Hvap RICPH HR
J
Hvap 8,314 mol . K
1222,048K 2,733 103
J
mol )
J
mol
Hvap = 7427,1
P
Svap = R (ICPS – ln( Po ))+ SR
J
Svap = 8,314 mol . K
Svap = 4.197
1435 kPa
(3,80735– ln( 101,33 kPa ))+ (5.421
J
mol . K
For saturated vapor, by Eqs. (3.63) & (4.12)
3
Vliq = 255
Vliq =123.85
cm
mol
cm3
mol
(1−0,87)
¿
¿
x
¿
0,274 ¿
J
mol . K )
J
Hn = 8,314 mol . K
Hn = 22514
¯¿
¯¿
37,96 −1,013
¿
¿
¿
x 272,7K(
)
ln ¿
1,092¿
¿
J
mol
6.55 Five moles of calcium carbide is combined with 10 mol of liquid water in a
closed, rigid, high-pressure vessel of 750-cm 3 capacity. Acetylene gas is produced
by the reaction :
CaC2(s) + 2 H2O(l) → C2H2(g) + Ca(OH)2(s)
Initial condition are 250C and 1 bar, and the reaction gas to completion. For a
final temperature of 1250C, determine :
(a) The final pressure ; (b) The heat transferred
At 125 0C, the molar volume of Ca(OH) 2 is 33,0 cm3 mol-1. Ignore the effect of
any gas present in the vessel initially.
SOLUTION :
Vgas = ( 750 cm3 – (5 mol x 33.0 cm3) = 585 cm3
n = 5 mol
0.187
Tc = 308,3 K
T = 398,15 K
T
398,15 K
Tr = Tc = 308.3 K = 1,291
Pc = 61,39 bar
Z = 0, 65
Z nRT
P = V gas =
Pr =
P
Pc
0,65 x 5 mol x 83,14 x 398,15 K
585
61,39 ¯¿
¯¿
= 183,9 ¿
¿
= 2,996
= 183,9 bar
Pressure is clearly high, and requires use of Lee/Kesler correlation. Solution is by
trial, because P is unknown but is required to find Z. Start with reduced pressure from
guess value above. The eventual result for a reduced pressure of 3.07 is:
Z0 0, 6298 ; Z1 0.1948
Z Z0 Z1 = 0,6298 + ( 0,187 x 0,1948)
Z = 0,6298 + 0,00364276
Z 0.666
Z nRT
0,666 x 5 mol x 83,14 x 398,15 K
P = V gas =
= 188,5 bar
585
Pr =
P
Pc
61,39 ¯¿
¯¿
= 188,5 ¿
¿
= 3,07
a). The Final pressure is P = 188,5 bar
b). The heat transferred
By the first law,
Q = nU = nH (PV)
Q = nH VP= nH VgasP
The enthalpy change is evaluated by a three-step process:
(1) Reaction at 298.15 K
(2) Change in T for products in std. states
(3) Transformation to state of real gas
H = H298 HP HR
Step (1): From data of Table C.4
J
∆ H=⌈ −986090+227480−(−59800 )−2(−285830)⌉ x
mol
∆H
298
J
= - 127150 mol
Step (2), Table C.1 data, for acetylene(g):
MCPH298.15398.156.1321.95210-30.01.299105= 5.71731
Table C.2, for calcium hydroxide(s):
MCPH298.15398.159.5975.43510-3 0.00.0= 11.48920
For the products,
MCPH R(5.71731 11.48920)
HP MCPH100K = 143,055 x 100 = 14305,5
J
mol
Step (3), from Tables E.7 & E.8 at the reduced conditions of Part (a) for
acetylene:
HR0 2.340RTc = 2.340 x 8,314 x 308,3 K = 5.998 10
3
J
mol
J
HR1 0.384RTc = 0.384 x 8,314 x 308,3 K = 984.271 mol
HR HR0 HR1 = 5.998 103
J
mol
J
+ 0.187x(984.271 mol )) =6182
J
mol
H H298 HP HR
J
H = 127150 mol + 14305,5
H = 1.19 105
J
mol
+ (6182
J
mol )
J
mol
Q nH Vgas(P 1bar)
Q= 5 mol x ( 1.19 10
5
J
3
mol ) – 585 cm ( 188,5 -1 ) bar
Q = 606101
6.56 Propylene:
0.140
Tc 365.6K
Pc 46.65bar
T 400.15K
P 38bar
P0 1bar
The throttling process, occurring at constant enthalpy, may be split into
two steps:
(1) Transform into an ideal gas at the initial conditions, evaluating property
changes from a generalized correlation.
(2) Change T and P in the ideal-gas state to the
final conditions, evaluating property changes by equations for an ideal gas.
Property changes for the two steps sum to the property change for the
process. For the initial conditions:
SOLUTION :
Tr= T/Tc
Tr 400.15K/ 365.6K
Tr 1.095
Pr= P/Pc
Pr38bar/ 46.65bar
Pr 0.815
Step (1): Use the Lee/Kesler correlation, interpolate.
H0 0.863RTc
H1 0.534RTc
H0 2.623 103J/mol 
H1 = 1.623 103J/mol
HR H0 H1
HR = 2.85 103J/mol
S0 0.565R
S0 =4.697J/molK
SR S0 S1
SR = 5.275J/molK
S1 0.496R
S1 = 4.124J/molK
Step (2): For the heat capacity of propylene,
A 1.637
B= (22.706103)/K 
C=(6.915106)/ K2
Solve energy balance for final T. See Eq. (4.7).
1 (guess) Given
HR= R ((A.T(C/3 T3(
Find
0.908
T T
T 363.27K
ICPS400.15363.271.63722.706103 6.915106 0.0= 0.898338
ICPS 0.898338
Sig = R . (ICPA – ln (p0/p))
Sig =22.774J/molK
S SR Sig
S =28.049J/molK
6.57
Propane gas at 22 bar and 423 K is throttled in a steady state flow process to 1
bar. Estimate the entropy change of the propane caused by this process. In its
final state, propane maybe assumed to be an ideal gas.
SOLUTION :
Dari Appendix B1 pada Tabel B.1. : Properties of Pure Species untuk propane :
ωpropane = 0,152
Tc = 369,8 K
Pc = 42,48 bar
Proses throttling terjadi ketika entalpi konstan, dapat terbagi menjadi dua proses :
1. Transformasi / perubahan bentuk menjadi gas ideal pada keadaan awal, dengan
mengevaluasi perubahan properties dari korelasi umum.
2. Ubah T dan P dari keadaan gas ideal ke kondisi akhir, dengan mengevaluasi
perubahan properties dengan menggunakan persamaan gas ideal. Perubahan
properties untuk dua tahap dijumlahkan ke perubahan properties untuk proses.
Dengan keadaan awal :
Tr =
T
Tc
Tr =
423 K
369,8 K
Pr =
P
Pc
= 1,144
42,48 ¯¿
¯
22 ¿
Pr =
= 0,518
¿
¿
Langkah pertama : Menggunakan rumusan korelasi umum virial.
0
1
R
( HR)
HR (H )
=
+ω
RTc RTc
RTc
Tentukan nilai HR dengan mencari nilai (HR)0 dan (HR)1. Nilai ω dapat dilihat
melalui Appendiks B Tabel B.1. Nilai ω untuk propane adalah 0,152.
1. Nilai (HR)0 diperoleh melalui Appendiks E The Lee/Kesler Generalizedcorrelation Tables Tabel E.5.
Tr/Pr
1,10
1,144
1,15
X = 0,518
X1 = 0,4
X2 = 0,6
Y= 1,144
0,4
-0,367
0,518
0,6
-0,581
?
-0,334
-0,523
Y1= 1,1
Y2= 1,15
M=
[(
]
[(
]
M=
0,518−0,4
1,15−1,144
0,6−0,518
0
(−0,367 )+ (
(−0,581 )
+(
(−0,334 ) + (
([ 0,6−0,518
)
)
)
] 1,15−1,1 [ 0,6−0,4
0,6−0,4
0,6−0,4
X 2− X
X− X 1
Y −Y
X 2− X
X−X 1
Y −Y 1
M 1.1 +
M 1.2 2
+
M 2.1 +
M 2.2
X 2−X 1
X 2−X 1
Y 2−Y 1
X 2−X 1
X 2 −X 1
Y 2−Y 1
) (
)
) (
)
M =−0,45124
( H R)
0
RTc
=−0,45124
(HR)0 = -0,45124 . 8,314 J mol-1 K -1 . 369,8 K = -1387,345 Jmol-1
2. . Nilai (HR)1 diperoleh melalui Appendiks E The Lee/Kesler Generalizedcorrelation Tables Tabel E.6.
Tr/Pr
1,10
1,144
1,15
X = 0,518
0,4
-0,251
0,518
0,6
-0,381
?
-0,199
-0,296
X1 = 0,4
X2 = 0,6
Y= 1,144
Y1= 1,1
Y2= 1,15
M=
[(
]
[(
M=
0,518−0,4
1,15−1,144
0,6−0,518
0
(−0,251 ) + (
(−0,381 )
+(
(−0,199 )+ (
([ 0,6−0,518
)
)
)
] 1,15−1,1 [ 0,6−0,4
0,6−0,4
0,6−0,4
) (
)
) (
M =−0,2648064
( H R)
1
RTc
]
X 2− X
X− X 1
Y −Y
X 2− X
X−X 1
Y −Y 1
M 1.1 +
M 1.2 2
+
M 2.1 +
M 2.2
X 2−X 1
X 2−X 1
Y 2−Y 1
X 2−X 1
X 2 −X 1
Y 2−Y 1
=−0,2648064
(HR)1 = -0,2648064 . 8,314 J mol-1 K -1 . 369,8 K = -814,1518315 Jmol-1
)
0
1
R
( HR)
HR (H )
=
+ω
RTc RTc
RTc
R
H
−1387,345
−814,1518315
=
+0,152
8,314 .369,8 8,314 . 369,8
8,314 . 369,8
HR = -1511,096 J mol -1
3. . Nilai (SR)0 / R diperoleh melalui Appendiks E The Lee/Kesler Generalizedcorrelation Tables Tabel E.9.
Pr/Tr
1,10
1,144
1,15
X = 0,518
0,4
-0,23
0,518
0,6
-0,371
?
-0,201
-0,319
X1 = 0,4
X2 = 0,6
Y= 1,144
Y1= 1,1
Y2= 1,15
M=
[(
X 2− X
X− X 1
Y −Y
X 2− X
X−X 1
Y −Y 1
M 1.1 +
M 1.2 2
+
M 2.1 +
M 2.2
X 2−X 1
X 2−X 1
Y 2−Y 1
X 2−X 1
X 2 −X 1
Y 2−Y 1
M=
[(
0,6−0,518
0,518−0,4
1,15−1,144
0,6−0,518
0,
(−0,23 ) +
(−0,371 )
+
(−0,201 )+
0,6−0,4
0,6−0,4
1,15−1,1
0,6−0,4
0
) (
)
)
(
]
[(
)
) (
]
)
]
[(
M =−0,2757284
(SR)0 / R = -0,2757284
4. . Nilai (SR)1 / R diperoleh melalui Appendiks E The Lee/Kesler Generalizedcorrelation Tables Tabel E.10.
Pr/Tr
1,10
1,144
1,15
X = 0,518
X1 = 0,4
X2 = 0,6
0,4
-0,229
0,518
0,6
-0,35
?
-0,183
-0,275
)
(
Y= 1,144
Y1= 1,1
Y2= 1,15
M=
[(
X 2− X
X− X 1
Y −Y
X 2− X
X−X 1
Y −Y 1
M 1.1 +
M 1.2 2
+
M 2.1 +
M 2.2
X 2−X 1
X 2−X 1
Y 2−Y 1
X 2−X 1
X 2 −X 1
Y 2−Y 1
M=
[(
0,6−0,518
0,518−0,4
1,15−1,144
0,6−0,518
0,
(−0,229 ) +
(−0,35 )
+
(−0,183 ) +
0,6−0,4
0,6−0,4
1,15−1,1
0,6−0,4
0
) (
)
)
]
(
[(
)
) (
]
)
[(
M =−0,2448
(SR)1 / R = -0,2448
R 0
R 1
(S )
SR ( S )
=
+ω
R
R
R
SR
=−0,2757284+0,152(−0,2448)
8,314
S
R
= -2,601766 J mol-1 K-1
Untuk kapasitas panas propane dapat dilihat pada Appendix C Tabel C1
A= 1,213
B=28,785.10-3
C=-8,824.10-6
Cpig / R = A + BT + CT2 + DT-2
Cpig / R = 1,213 + 28,785.10-3T – 8,824.10-6T2
Cpig = ( 1,213 + 28,785.10-3T – 8,824.10-6T2 ) R
T2
ig
ΔS =
∫ Cpig
T1
T2
ig
ΔS =
∫ Cpig
T1
P
dT
−R ln 2
T
P1
P
dT
−R ln 2
T
P1
T2
ΔSig = R/T
P
∫ 1,213+0,028785 T −0,000008824 T 2 dT −R ln P2
T1
ΔSig = 0 – 8,314 J/mol K . ln (1/22) = 25,6998 J/mol K
1
]
)
(
ΔSig – ΔS = -ΔSR
ΔS = ΔSig + SR = 25,6998 J/mol K + -2,601766 J/mol K = 23,098034 J/mol K
6.58. Propane gas at 100oC is compresed isothermally from an initial pressure of 1
bar to a final pressure of 10 bar. Estimate ∆H and ∆S
SOLUTION :
For propane:
Tc = 369.8 K
Pc = 42.48 bar = 0.152
T = (100 + 273.15)K T = 373.15 K
Tr=
T
Tc
Tr = 1.009
P0 = 1 bar
Pr=
P
Pc
P = 10 bar
Pr = 0.235
Assume ideal gas at initial conditions. Use virial correlation at final conditions.
HRB(1.00906, 0.23540, 0.152) = -0.260821
HRB = -0.260821
SRB(1.00906, 0.23540,0.152) = -0.179862
SRB = -0.179862
 ∆ H=−801.9
∆ H = R×T ×HRB
c
P
)
P0
∆ S=R . ¿
SRB−ln (
∆ S=−20.639
J
mol
J
mol . K
6.61 a stream of ethylene gas at 250 0C and 3.800 KPa expands isentropically in a
turbine to 120 KPa. Determine the tenperatur of the expanded gas and the work
produced if the properties of ethylene are calculated by :
a. equations for an ideal gas
b. appropriate generalized correlations
SOLUTION :
6.62
A stream of ethane gas at 220 oC and 30 bar expands isentropically in a turbine
to 2,6 bar. Determine the temperature of the expand gas and the work
produced if the properties of ethane are calculated by :
a. Equation for an ideal gas,b. Appropriate generalized correlation.
SOLUTION :
T0 = 220OC = 493,15 K
P0 = 30 bar
P = 2,6 bar
J
Δ S = 0 mol K
T
ΔS / R ¿∫
T
0
C pig dT
-ln
RT
T
P
P0
1,131+0,019225 T −5,561.10−6 T 2 dT
0 = 493,15
-ln
8,314 . T
∫
2,6
30
T = 367,59 K
a.
Untuk kapasitas panas dari etana dapat dilihat pada Appendix C Tabel C1
A = 1,131
B = 19,225 . 10-3
C = -5,561 . 10-6
Cpig = (1,131 + 0,019225T - 5,561 . 10-6T2 ) R
T2
ΔHig =
∫ C Pig dT
T1
367,59
ig
ΔH =
R
∫
(1,131+0,019225 T −0,000005561T 2) dT
493,15
ΔHig = -8735 J/mol
Ws = ΔHig = -8735 J/mol
b.
Dari Tabel Appendix B Tabel B1 maka diperoleh data untuk ethane adalah :
ω =0,1
Tc = 305,3K
Pc = 48,72 bar
Untuk kondisi awal :
Tr0 = T0/Tc
Tr0 = 493,15 K / 305,3 K = 1,6153
Pr0 = P0/Pc
Pr0 = 30 bar / 48,72 bar = 0,61576
Untuk kondisi akhir :
Tr = T/Tc
Tr = 367,59 K / 305,3 K = 1,20404
Pr0 = P0/Pc
Pr0 = 2,6 bar / 48,72 bar = 0,05337
Menggunakan korelasi koefisien virial :
T = 362,73 K
ΔHig = -9034 J/mol
Ws = -8476 J/mol
6.63 Estimate the final temperature the work required whan 1 mol of n-butane is
comprassed isentropically in a steady-flow process from 1 bar and 50 0C to 7.8
bar.
SOLUTION :
A = 1.935
−3
B=
36.915 . 10
K
−11.402 . 10−6
C=
K2
∆ S=0
J
mol . K
323.15 K
425.1 K
Tro =
¿
Tc
=
Pr0 =
P0
Pc
37.96 ¯¿
¯
1¿
=
= 0.02634
¿
¿
Pr =
P
Pc
37.96 ¯¿
¯¿
= 7.8 ¿
= 0.205
¿
HRB0 = -0,05679

τ
= 1.18
T=
τ . T0
T = 1.18 . 323.15 K
= 381.43 K
Tr =
T
Tc
=
= 0.89726
381.43 K
425.1 K
= 0.76017
∆ Hig = R. 3231,5 K. 381,43 K. 1,935. 36,915 . 10−3 , - 11,402. 10−6
3 J
∆ Hig = 6.551 x 10
mol
Ws = ∆ Hig + (R. Tc. Tr . ω . Pr) – (Tr0. Pr0)
= 6.551 x
= 5680
3
10
J
mol
+ (425,1 K . 0,89726 . 0.200 . 0,205) – (0.76017 . 0.02634)
J
mol
6.65 liquid water at 325 K and 8000 KPa flows into boiler at the rate of 10 kg/s and is
vaporized, producing saturated vapor at 8000 KPa. What is the maximum
fraction of the heat added to the water in the boiler that can be converted into
work in a process whose product is water at initial conditions, if T ð = 300 K ?
what happen to the rest of the heat ? what is the rete of entropy change in the
surroundings as a result of the work producing process? In the system ? total ?
SOLUTION :
T = 325 K
P1 = 8000 K
Pada table :
Psat = 12.87 Kpa
Dari table F1 :
β = 460 x 10-6 /k
Hliq = 217 kj/kg
Sliq = 0.7274 kj/kg K
1.o13 cm3/ gr
H1 = H liq + V liq x ( 1 –β. T) ( P1 – Psat )
H1 = 217 kj/kg + 1.013 ( 1 – 460 x 10-6 x 325 ) ( 8000 – 12.87 ) kj/ kg
1000
H1 = (217 + 6.8813) kj/ kg
H1 = 223.88 kj/kg
S1 = S liq –β. V liq ( P1- Psat )
Vliq
=
S1 = 0.7274 kj/kg – (460 x 10-6 x 1.013 ) ( 8000 – 12.87 ) kj/ kg
1000
S1 = (0.7274 – 3.722 . 10-3 ) Kj/Kg K
S1 = 0.7236 Kj/ Kg K

untuk uap saturated di 8000 KPa
dari table F2 :
H2 = 2759.9 kj/kg
S2= 5.7471 Kj/Kg
Tð = 300 K
PENAMBAHAN PANAS
Q = H2 –H1
Q = 2759.9 – 223.81 ) Kj/ kg
Q = 2536.09 Kj/Kg
Kerja maximum dari sistem
W ideal = ( H1 – H2 ) - Tð ( S1 – S2 )
W ideal = ( 223.88 – 2759.9 ) – 300 ( 0.7236 – 5.7471 )
W ideal = ( 1507.05 – 2536.02 ) kj/ kg
W ideal = - 1028.97 Kj/ kg
Kerja sebagai fraksi sebagai penambahan panas
F frac = I W ideal I / Q
F frac = 1028.97 / 2536.09
F frac = 0.4057
Panas yang tidak converted untuk kerja akhir di surrounding
S surr = Q + W ideal
. 10 kg / sec
Tð
S surr = ( 2536.09 + (- 1028.97 ) / 300 ) 10
S surr = 50.237 kw/K
S systm = ( S1 – S2 ) 10 Kg/ sec
S systm = ( 0.7236 – 5.7471 ) x 10
S systm = 50.235 Kw/ K
Total dari generasi entropi = 0, karena kerja ideal prosesnya reversible.
6.66 Suppose the heat added to the water in the boiler in the preceding problem
comes from a furnace at a temperature of 600°C. What is the total rate of
entropy generation as a result of the heating process? What is Wlost?
SOLUTION :
*untuk menghitung SG dan Wlost , harus di cari dl data dari soal no. 65
>>Dari table F.1 untuk saturated liquid pada 325 K, diperoleh:
Hliq= 217,0 KJ/Kg
Sliq = 0,7274 KJ/Kg.K
Vliq = 1,013 cm3/gr
Psat = 12,87 KPa
P1 = 8000 KPa
T = 325 K
>>untuk compressed liquid pada 325 K dan 8000 KPa, pakai rumus 6.28
dH= Cp.dT+(1-βT). V. dP
β = 460.10-6 K-1
H1= Hliq + Vliq + (1-β.T). (P1-Psat)
= 217,0 KJ/Kg + 1,013 cm3/gr + (1- 460.10-6 K-1 x 325 K) . (8000 KPa - 12,87 KPa)
= 223,881 KJ/Kg
S1= Sliq + β. Vliq. (P1 – Psat)
= 0,7274 KJ/Kg.K + 460.10-6 K-1 x 1,013 cm3/gr . (8000 KPa - 12,87 KPa)
= 0,724 KJ/Kg.K
>>Untuk saturated vapour pada 8000 KPa, dari table F.2 diperolah:
H2= 2759 KJ/Kg
S2= 5,7471 KJ/Kg.K
Panas yang ditambahkan ke boiler:
Q = H2 – H1
= 2759 KJ/Kg – 223,881 KJ/ Kg
= 2536 KJ/Kg
Kerja maksimal dari steam, dengan persamaan 5.27 diperoleh:
Widel = (H2 – H1) - T. (S2 – S1)
= (2759 KJ/Kg – 223,881 KJ/ Kg) - 5,7471 KJ/Kg.K - 0,724 KJ/Kg.K)
= -1029 KJ/Kg
Kerja pada saat panas di tambahkan:
¿ Wideal∨ ¿
Q
Frac =
¿
¿−1029∨ ¿
2536
=
¿
= 0,4058
Panas yg tidk terkonversi menjadi kerja di akhir pada surrounding:
SG.surrounding =
Q+Wideal
T
2536
=
x 10
KJ
KJ
−1029
Kg
Kg
T
x 10
Kg
sec
KW
K
= 50,234
SG system = (S1 – S2) . 10
Kg
sec
= (0,724 - 5,7471)
= - 50,234
Kg
sec
KW
K
KJ
Kg . K
. 10
Kg
sec
Untuk menghitung SG pada soal no. 66:
Q = 2536
KJ
Kg
x 10
Kg
sec
T (600 + 273.15)K = 873.15 K
SG
=
Q
T
+ SG.surrounding
KJ
sec
873.15 K
−25360
=
= 21.19
+50.234
KW
K
KW
K
*dari persamaan (5.34) untuk mencari harga Wlost
T300K
Wlost Tx SG
6356.9 kW
6.67 An ice plant produces 0.5 kg s⁻1 of flake ice at 0°C from water at 20°C (T ) in a
continuous process. If the latent heat of fusion of water is 333.4 kJ kg⁻1 and if
the
thermodynamic efficiency of the process is 32%, what is the power
requirement of the plant?
SOLUTION :
Untuk saturated liquid air pd 20˚ C, table F.1:
H1= 83.86 kJ/ kg
S1= 0.2963 kJ/kg.K
Untuk saturated liquid air pada 0˚C, table F.1:
H0= 0.04 kJ/kg
S0= 0.0000 kJ/kgK
Untuk es pada 0˚C
H2= H0 - 333.4kJ/kg
H2= 333.44 kJ/kg
S2= S0- (333.4/273.15)kJ/kgK
S2= 1.221kJ/kgK
T293.15K
m = 0.5kg/sec
t 0.32
Dengan persamaan (5.26) dan (5.28):
Wideal
= m x [ H2 - H1 -TS2- S1)]
kg/sec x [333.44 kJ/kg - 83.86 kJ/ kg - 293.15K . (1.221kJ/kgK - 0.2963
kJ/kg.K)
13.686 kW
W
=
Wideal
ηt

13.686 kW
0.32
42.77 kW
6.68 An inventor has developed a complicated process for making heat continuously
available at an elevated temperature. Saturated steam at 373.15 K (100°C) is the
only source of energy. Assuming that there is plenty of cooling water available at
273.15 K (O°C), what is the maximum temperature level at which heat in the
amount of 2000 kJ can be made available for each kilogram of steam flowing
through the process?
SOLUTION :
kJ
kg
H1= 2676.0
kJ
S1= 7.3554 kg . K
kJ
kg . K
S2= 0.0
kJ
kg
Q’= -2000.0
kJ
kg
H2= 0.0
Tσ = 273.15 K
SOLUTION:
Wideal = ΔH apparatus.reservoir – Tσ.ΔS apparatus.reservoir
ΔH apparatus.reservoir
= H2-H1-Q’
kJ
= ( 0.0 kg
= - 676.0
ΔS apparatus.reservoir = S2 – S1 -
) - ( 2676.0
kJ
kg ) – (-2000.0
kJ
kg )
kJ
kg
Q'
T'
= (0.0
kJ
−2000.0
kJ
kJ
kg
kg . K ) – (7.3554 kg . K ) - (
T'
kJ
=– (7.3554 kg . K
2000.0
)+ (
kJ
kg
T'
)
)
Jadi ,diketahui W=0
Wideal
= ΔH apparatus.reservoir – Tσ.ΔS apparatus.reservoir
kJ
0.0 kg
676.0
= (- 676.0
kJ
kg
2000.0
kJ
kJ
kg ) – Tσ.[– (7.3554 kg . K
kJ
= (273.15 K)[(7.3554 kg . K
)+(
2000.0
)-(
T'
kJ
kg
T'
)]
kJ
kg
)]
( 676.0)
=(2009.127) -
546300
K
T'
= 1333.127
T’
6.69
546300
K
T'
= 490.78 K
Two boilers,both operating at 200(psia),discharge equal amounts of steam into
the
same steam main. Steam from the first boiler is superheated at 420( 0F)
and steam
from the second is wet with a quality of 96%.Assuming
adiabatic mixing and
negligible changes in
potential and kinetic
energies.What is the equilibirium condition after mixing and what is S G for
each (lbm) of discharge steam ?
SOLUTION :
From Table F.4 at 200 (psia):
At 4200 F
H1 = 1222.6 BTU /lbm
S1= 1.5737 BTU/lbm.rankine
Hliquid= 355.51 BTU/lbm
Hvapour= 1198.3 BTU/lbm
Sliquid= 0.5438 BTU/lbm.rankine
Svapour= 1.5454 BTU/lbm.rankine
x= 0.96
H2= Hliquid + x.Hvapour-Hliquid
S2= Sliquid + x.Svapour-Sliquid
H2 = 355.51 + 0.96 ( 1198.3)-355.51
S2= 0.5438 + 0.96 (1.5454)-0.5438
H2=1.165 x 103 BTU/lbm
S2=1.505 BTU/lbm.rankine
Neglecting kinetic and potential energy changes,on the basis of 1 pound mass of
steam after mixing,Eq. (2.30) yields for the exit stream :
Wet steam
H= 0.5 H1 +0.5 H2
H= 0.5 ( 1222.6) + 0.5 ( 1.165 x103)
H= 1193.6 BTU/lbm
x=_
H - Hliquid____
Hvapour-Hliquid
x = _1193.6-355.51___ = 0.994
1198.3-355.51
S= Sliquid + x.Svapour-Sliquid
S = 0.5438 + 0.994 (1.5454)-0.5438
S = 1.54 BTU/lbm.rankine
By Eq. (5.22 ) on the basis of 1 pound mass of exit steam,
SG = S-0.5 S1-0.5 S2
= 1.54-0.5 (1.5737)-0.5(1.505)
= 2.895 x10-4 BTU/lbm.rankine
6.70 A rigid tank 0f 80 (ft3) capacity contains 4180 ( lbm) of saturated liquid water at
430 (0F). This amount of liquid almost completely fills the tank, the small
remaining volume being occupied by saturated-vapor steam. Since a bit more
vapor space in the tank is wanted, a valve at the top of the tank falls to 420 ( 0F).
Assuming no heat transfer to the contents of the tank, determine the mass of
steam vented.
SOLUTION :
From Table F.3 at 430 degF (sat. liq. and vapor):
V tank = 80 ft3
Mliq = 4180 lbm
Vliq = 0.01909 ft3/lbm
Vvap = 1.3496 ft3/lbm
Uliq = 406.70 Btu/lbm
Uvap = 1118.0 Btu/lbm
VOLliq = mliqx Vliq
= 4180 lbm x 0.01909 ft3/lbm
= 79.796ft3
VOLvap Vtank VOLliq
= 80 ft3 - 79.796ft3
= 0.204ft3
Mvap
=
VOLvap
Vvap
=
0.204 ft 3
1.3496 ft 3 /lbm
= 0,151 lbm
U1
mliq . Uliq+ mvap. Uvap

mliq+ mvap
Btu
Btu
+0,151 lbm x 1118,0
lbm
lbm
4180 lbm+0,151 lbm
4180 lbm x 406,70



Btu
 lbm
By Eq. (2.29) multiplied through by dt, we can write,
dmtUtHdm = 0
(Subscript t denotes the contents of the tank. H and m refer to the exit stream.)
Diintegralkan:
m
m2.U2 – m1.U1+
∫ Hdm
0
=0
From Table F.3 we see that the enthalpy of saturated vapor changes from 1203.9 to
1203.1(Btu/lb) as the temperature drops from 430 to 420 degF. This change is so small that
use of an average value for H of 1203.5(Btu/lb) is fully justified.
Then,
m2U2 m1U1 Havem = 0
Have = 1203,5 Btu/lbm
m2(mass) m1 mass
Property values below
Vliq = 0.01894 ft3/lbm
Vvap = 1.4997 ft3/lbm
are
for
sat.
liq.
and
vap.
at
420
degF
Uliq = 395.81 Btu/lbm
Uvap = 1117,4 Btu/lbm
V2 =
Vtank
m2(mass)
X mass =
V 2 ( mass )−Vliq
Vvap−Vliq
U2(mass) Uliq x(mass)Uvap Uliq
U 1−U 2(mass)
Mass = Have−U 2(mass)
Mass = 55,36 lbm
6.71 A tank of 50 m3 capacity contains steam at 4,500 kPa and 400 0C.Steam is vented
from the tank through a relief valve to the atmosphere until the pressure in the
tank
falls to 3,500 kPa.If the venting process is adiabatic,estimate the final
temperature of the steam in the tank and the mass of steam vented.
SOLUTION :
The steam remaining in the tank is assumed to have expanded isentropically.
from Table F.2 at 4500 kPa and 4000C :
V1 = 64.721 cm3/gr
S1= 6.7093 J/gr.K
Vtank= 50 m3
S2=S1 = 6.7093 J/gr.K
By interpolation in Table F.2 at this entropy and 3500 kPa :
V2 = 78.726 cm3/gr
t2 = 362.460C
m1 = _Vtank__
V1
= _
50 x 106 cm3___
64.721 cm3/gr
m2= _Vtank___
V2
m2= _ 50 x106cm3__
78.726 cm3/gr
Data
= 0.772 x 106 gr
= 0.137 x 106 gr
Δm = m1 – m2
= 0.772 x 106 gr – 0.137x106 gr
= 0.137 x106 gr
= 137 kg
6.72. A tank of 4 m3 capacity contains 1,500 kg of liquid water at 250 oC in
equilibrium with its vapor, which fills the rest of the tank. A quantity of 1,000 kg
of water at 50oC is pumped into the tank. How much heat must be added during
this process if the temperature in the tank is not to change?
SOLUTION :
Q = ∆mt. Ht – H. ∆mt
Here, the symbols with subscript to refer to the contents of the tank, whereas H refers to
the entering stream. We illustrate here development of a simple expression for the first
term on the right. The 1500 kg of liquid initially in the tank is unchanged during the
process. Similarly, the vapor initially in the tank that does not condense is unchanged. The
only two enthalpy changes within the tank result from :
1. addition of 1000 kg of liquid water. This contributes an enthalpy change of : Hliq.∆mt
2. condensation of y kg of sat, vapor to sat. liq. This is contributes an enthalpy change of
Q=∆mt.Ht – Ht.∆mt
y.H liq – H vap = -y. H lv
∆mt. Ht = H liq. ∆mt – y.∆H lv
∆mt.Vt = V liq. ∆mt – y. ∆V lv = 0
Q= H liq. ∆mt – y.∆H lv – H ∆mt
∆mt = 1000kg
Required data from table F.1 are
at 50 degC H = 209.3 kJ
kg
at 250 degC H liq = 1085.8 kJ
V liq = 1251 cm3
kg
gm
∆H lv = 1714 kJ
kg
∆V lv = 48.79 cm3
gm
y = V liq.
y = 25.641 kg
∆mt
∆V lv
Q = ∆mt. H liq – H-y.∆H lv
Q = 832534 kJ
6.74. A well insulated tank of 50 m 3 volume initialy contains 16000 kg of water
distributed between liquid and vapor phases at 25 o C. Saturated steam at 1500
Kpa is atmitted to the tank until the pressure reaches 800 Kpa. What mass of
steam is added ?
Penyelesaian :
m2.( U2 –H ) – m1.(U1 – H) = Q = 0
when m2.(H – U2) = m1.(H – U1 )
Also
U2 = Uliq.1 + x2. Δ U lv.2
V2 = Vliq.2 + x2. Δ V lv.2
V2 = V tank
m2
Eliminating x2 from these equation gives
V tank
- V liq.2
m2
m2. (H – U liq.2 . Δ Ulv.2 ) = m1.( H – U1)
Δ lv.2
Which is later solved for m2
V tank = 50 m3
m1 = 16000 kg
V1 = V tank
V1 = 3.125 x 10 -3 m3 / kg
m1
Data from table F1 : 25 0C
V liq.1 = 1.003 cm3 / gm
ΔV lv.1 = 43400 cm3/ gm
Uliq.1 = 104.8 kJ / kg
Δ Ulv.1 = 2305 kJ / kg
X1 = v1 – v liq..1
U1 = U liq.1 + x1. Δ U lv.1
Δ V lv.1
X1 = 4.889 x 10-5
U1 = 104.913 kJ / kg
Data from Table F2 : 800 kPa
V liq.2 = 1.115 cm3 / gm
U liq.1 = 720.043 kJ / kg
3
Δ Vlv.2 = (240.26 - 1.115) cm / gm
Δ Ulv.2 = (2575.3 - 720.043) kJ / kg
Δ Vlv.2 = 0.239 m3 / kg
Δ Ulv.2 = 1.855 ´ 103 kJ / kg
Data from Table F2 : 1500 kPa
H = 2789.9 kJ / kg
Δ U lv.2
m1. ( H – U1 ) + V tank (
)
Δ U lv.2
m2 =
m2 = 2.086 x 104 kg
Δ U lv.2
H – U liq.2 + V liq.2 . (
)
Δ U lv.2
m steam = m2 – m1
m steam = 4.855 x 103 kg
6.75. An insulated evacuated tank of 1,75m3 volume is attached to a line containing
steam at 400 kPa and 2400C. Steam flows into the tank until the pressure in the
tank reaches 400 kPa. Assuming no heat flow from the steam to the tank, prepare
graphs showing the mass of steam in the tank and its temperature as a function of
pressure in the tank.
Penyelesaian :
Interpolasi dari table F.2 pada buku hal. 696
U = 2943,9 kJ/mol
Mencari temperature dengan cara interpolasi
Pada P = 1 kPa
x−x 1
y= y 1 +
(y −y )
x2 −x1 2 1
2943,9−2889,9 Y −350
=
2969,1−2889,9 400−350
0,6818 =
Y −350
50
Y = 384,09
Pada P = 200 kPa
x−x 1
y= y 1 +
(y −y )

x2 −x1 2 1


2943,9−2887,2 Y −350
=
2966,9−2887,2 400−350
0,7114 =
Y −350
50
 Y = 385,57
Pada P = 400 kPa
x−x 1
y= y 1 +
(y −y )

x2 −x1 2 1

2943,9−2884,5
Y −350
=
2964,6−2884,5 400−350

0, 7415 =
Y −350
50
 Y = 387, 8
Mencari volume dengan cara interpolasi
Pada P = 1 kPa
x−x 1
y= y 1 +
(y −y )
x2 −x1 2 1
2943,9−2889,9
Y −287580
=
2969,1−2889,9 310660−287580
0,6818 =
Y −287580
23080
Y = 303316
Pada P = 200 kPa
x−x 1
y= y 1 +
(y −y )
x2 −x1 2 1
2943,9−2887,2
Y −1432,8
=
2966,9−2887,2 1549,2−1432,8
0,7114 =
Y −1432,8
116,4
Y = 1515,61
Pada P = 400 kPa
x−x 1
y= y 1 +
(y −y )
x2 −x1 2 1
2943,9−2884,5
Y −713,85
=
2964,6−2884,5 772,50−713,85
0, 7415 =
Y −713,85
58,65
Y = 757,23
P ( kPa)
1
T(0C)
384,09
V ( cm3/gm)
303316
200
385,57
1515,61
400
387,8
757,23
V TANK = 1,75 m3
MASSA = VTANK / V INTERPOLASI
Maka :
P ( kPa)
1
V ( m3)
303316 . 10-3
Massa ( kg )
1,75 / 303316 . 10-3 = 5,77 . 10-3
200
1515,61. 10-3
1,75 / 1515,61. 10-3 = 1,155
400
757,23. 10-3
1,75 / 757,23. 10-3 = 2,311
Sumbu Y : Massa ( kg )
2.5
2
1.5
1
0.5
0
1
200
400
Sumbu X : Tekanan ( kPa)
76. A m3 tank contains a mixture of saturated-vapor steam and saturated-liquid
water at 3.000 kPa. Of the total mass, 10% is vapor. Saturated-liquid water is
bled from the tank through a valve until the total mass in the tank is 40% of the
initial total mass. If during the process the temperature of contents of the tank is
kept constant, how much heat is transferred?
Penyelesaian :
Vtank = 2 m3 ;
Data from table 7.2 at 3000 kPa (page 712) :
x1 = 10% vapor
cm3
m3
-3
Vliquid = 1,216 gm = 1,216 x 10
kg ;
cm3
m3
-3
Vvapor=66,626 gm =66,626x10 kg
Hliquid = 1008,4
kJ
kg
kJ
Hvapor=2802,3 kg
Q=
∆ ( mt . H t )
+H.
∆m
tank
V1=Vliquid+x1.(Vvapor-Vliquid)
cm3
=1,216 gm +0,1.
cm3
cm 3
66,626
−1,216
gm
gm
)
cm3
cm3
+
0,1
x
65,41
=1,216 gm
gm
)
(
(
cm3
cm3
+¿
=1,216 gm
6,541 gm
cm3
=7,757 gm
=7,757x10
-3
m3
kg
V tank
m1= V 1
2 m3
3
= 7,757 x 10−3 m
kg
=257,832kg
Q=
∆ ( mt . H t )
+H.
∆m
tank
dimana t menunjukkan kondisi di dalam tangki, dan H adalah entalpi
dari aliran yang mengalir keluar dari tangki. Perubahan yang mempengaruhi
entalpi dari isi tangki adalah:
1.Penguapan(evaporation)
:
y ( H vapor −H liquid )
2.(100%–40%)untuk liquid 
60%=0,6
keluaran :
0,6m1liquid dari tangki :
H liquid
0,6m1.
dengan demikian,
∆ ( mt . H t )= y ( H vapor −H liquid )−¿
Volume tangki konstan, maka:
∆ ( mt . H t )= y ( H vapor −H liquid )−¿
0,6 m1 .
H liquid
0,6m1.
V liquid =0
Dimana:
y=
0,6 m1 .V
( V vapor−V
liquid )
liquid
Maka:
Q =
0,6 m1 .V
.(H
( V vapor−V
liquid )
liquid
dengan H=
H liquid
vapor
−H liquid )
∆m
dan 0,6 m1 =
−¿ 0,6 m . H liquid
1
tank
Persamaannya menjadi :
0,6 m1 .V liquid
Q = V vapor−V liquid . ( H vapor −H liquid )
(
Q=
Q=
Q=
)
3
(
m
0,6 .(257,832 kg). 1,216 x 10
kg
kJ
kJ
. 2802,3 −1008,4
3
3
kg
kg
−3 m
−3 m
66,626 x 10
−1,216 x 10
kg
kg
(
0,1881 m3
kJ
. 1793,9
3
kg
m
65,41 x 10−3
kg
(
337,432 kJ
65,41 x 10−3
−3
)(
Q = 5158,730 kJ
)
)
)(
)
+ H .
∆m
tank
6.77. Stream of water at 85oC, following at the rate of 5 kg-1 is formed by mixing water
at 24oC with saturated steam at 400 Kpa. Assuming adiabatic operation, at what
rates are the steam and water fed to the mixer ?
Penyelesaian :
Data from Table F.1 for sat. Liq
H1 = 100.6 kJ / kg
H3 = 355.9 kJ / kg
Data from Table F2 for sat. Vapor : 400 kPa
H2 = 2737.6 kJ / kg
By Eq. (2.30), neglecting kinetic and potential energies and setting the heat and work terms
equal to zero:
H3mdot3 - H1mdot1 - H2mdot2 = 0
mdot1 = mdot3 - mdot2
mdot3 = 5 kg / sec
Whence
mdot3.( H1 – H3 )
Mdot2 =
H1 – H2
Mdot1 = mdot3 – mdot2
mdot2 = 0.484 kg / sec
mdot1 = 4.516 kg / sec
6.78. In a desuperheater, liquid water at 3,100 kPa and 50 0C is sprayed into a stream
of superheated steam at 3,000kPa and 3750C in an amount such that a single
stream of saturated-vapor steam at 2,900kPa flow from the desuperheater at the
rate of 15 kg/s assuming adiabatic operation, what is the mass flowrate of the
water? What is SG for the process ? What is the irreversible feature of the
process?
Penyelesaian :
P1 = 3,100 kPa dan T1 = 500C
P2 = 3,000kPa dan T2 = 3750C
P3 = 2,900kPa dan m3 = 15 kg/s
a. Dari tabel F.2 saturated vapor untuk tekanan 2,900kPa didapat :
H3 = 2802,2 kJ/kg
S3 = 6,1969 kJ/kg.K M3 = 15 kg/s
Dari tabel F.2 superheated vapor untuk tekanan 3,000kPa dan suhu 3750C didapat :
H2 = 3175,6 kJ/kg
S2 = 6,8385 kJ/kg.K
Dari tabel F.2 saturated liquid dengan suhu 500C didapat :
Vliq = 1,012 cm3/gr Hliq = 209,3 kJ/kg
Sliq = 0,7035 kJ/kg.K
Psat = 12,34kPa
T = 323,15 K
Perubahan volum ekspansi antara suhu 450C dan 550C (pada tabel F.2 sat.liquid )
ΔV = (1,015 – 1,010 )cm3/gr ΔT = 10 K
P = 3,100kPa
1 ΔT
ΔV = 5 x 10-3 cm3/gr
β = V liq Δ V
β
=
1
10 K
1,012 cm 3/gr 5 x 10−3 cm3 /gr
= 4,941 x 10-4/K
Persamaan dengan temperatur konstan :
H1 = Hliq + Vliq (1-β)T ( P-Psat )
H1 = 209,3 kJ/kg + 1,012 cm3/gr (1-4,941 x 10-4/K)T (3,100kPa -12,34kPa )
H1 = 211,926 kJ/kg
S1 = Sliq – β. Vliq ( P-Psat )
S1 = 0,7035 kJ/kg.K – 4,941 x 10-4/K x 1,012 cm3/gr (3,100kPa -12,34kPa)
S1 = 0,702 kJ/kg.K
b. Dengan persamaan pada bab 2 energi kinetik dan potensial diabaikan sehingga
penjumlahan antara panas dan kerja sama dengan nol :
H3. M3 - H2. M2 – H1. M1 = 0
Juga M2 = M3 - M1
M 3 ( H 3−H 2)
Dimana : M1 =
H 1−H 2
M1 =
15 kg /s (2802,2 kJ /kg−3175,6 kJ /kg)
211,926 kJ /kg−3175,6 kJ /kg
= 1,89 kg/s
Maka : M2 = 15 kg /s - 1,89 kg/s
M2 = 13,11 kg/s
Untuk kondisi adibatis persamaa yang digunakan adalah :
Sg = S3. M3 – S1. M1 – S2. M2
Sg = 6,1969 kJ/kg.K x 15 kg /s – 0,702 kJ/kg.K x 1,89 kg/s – 6,8385 kJ/kg.K x
13,11kg/s
Sg = 1,973 kJ/s.K
c. Pencampuran antara 2 aliran yang berbeda temperatur adalah irreversible
6.79 Superheated steam at 700kPa dan 280oC flowing at the rate of 50 kg/s is mixed
with liquid water at 40 oC to produce steam at 700 kPa and 200 oC. Assuming
adiabatic operation, at what rate is water supplied to the mixer? What S G for
process? What is the irreversible feature of the process?
Penyelesaian :
Superheated Steam
- P1 = 700 kPa & T1 = 280oC
- P3 = 700 kPa & T3 = 200oc
- Liquid water T = 40oC (saturated)
- m1 = 50 kg/s
Pada tabel F.2 super heated vapor pada
P3 = 700 kPa & T3 = 200oC, bahwa :
 H3 = 2844,2 KJ/Kg
S3 = 6,8859 KJ/Kg.K
o
P1 = 700 Kpa & T1 = 280 C, Bahwa :
 H1 = 3017,7 KJ/Kg
S1 = 7,2250 Kj/Kg.K
 m1 = 50 Kg/s
Pada tabel F.1 saturated T = 40oC
 Hliq = 167,5 KJ/Kg
Sliq = 0,5721 KJ/Kg.K
Pada persamaan 2.30 dinyatakan bahwa pada keadaan steady state pengurangan
energy kinetik dan energi potensial serta semua usaha dan panas yang bekerja pada
sistem adalah nol, Sehingga :
H2 = Hliq dan S2 = Sliq
H3.m3 – H2.m2 – H1.m1 = 0 sehingga persamaan dapat menjadi
m3 = m2 + m1
m1 ( H 1−H 3 )
m 2=
( H 3−H 2 )
50
m 2=
Kg
KJ
KJ
. 3017,7
−2844,2
s
Kg
Kg
(
)
(3017,7 KJKg −167,5 KJKg )
m2 =3.241
Kg
s
Untuk kondisi adiabatik:
SG = S3.m3 – S1.m1 – S2.m2
SG = 6,8859 KJ/KgK.58.241Kg/s - 7,2250 Kj/KgK. 50 Kg/s - 0,5721KJ/KgK.3.241Kg/s
SG = 3.508 Kj/s.K
80. A stream of air at 12 bar and 900 K is mixed with another stream of air at 2 bar
and 400 K with 2,5 times the mass flowrate. If the process were accomplished
reversibly and adibatically, what would be temperature and pressure of the
resulting
air
stream?
Assume air to be an ideal gas for which CP = (7/2) R.
Penyelesaian :
T1 =900 K
T2 = 400 K
7
R
CP = 2
 CP =
P1 = 12 bar
n1= 1 mol
P2 = 2 bar
n2 = 2,5 mol
7
J
J
8,314
29,099
=
2
mol . K
mol . K
Basis : 1mol air at 12 bar and 900 K + 2,5mol air at 2 bar and 400 K = 3,5
mol at P and T
T −T 1)
T −T 2)
n1. CP (
+ n 2 . CP (
=0
29,099
1 mol .
J
mol . K
J
29,099
( T −400 K )
. ( T −900 K ) + 2,5 mol .
=0
mol . K
(29,099T – 26189,1) K + (72,747T – 29099) K = 0
101,846T = ( 55288,1 ) K
T=
( 55288,1
101,846 )
K
T = 542,859 K
[( ( ( ) ( ))) ( ( ( ) ( )))]
n1 . C P . ln
T
P
−R . ln
T1
P1
+ n2 . C P . ln
T
P
−R . ln
T2
P2
=0
12 ¯¿
P
¿
2 ¯¿
P
¿
(
1 mol . 29,099
)(
(
J
542,859 K
J
J
542,859 K
. ln
−8,314
. ln ( ¿ ) + 2,5 mol . 29,099
. ln
−
mol . K
900 K
mol . K
mol . K
400 K
¿
¿
(
)
=0
P = 4,319 bar
6.81. Hot Nitroge gas at 750(oF) and atmospheric pressure flows into a waste heat
boiler at the rate of 40 (lbm)(S-1), and transfers heat to water boiling at 1 atm.
The water feed to the boiler is saturated liquid at 1 atm and 300( oF). if the
Nitrogen is cooled to 325 (oF) and if heat is lost to the surroundings at rate of 60
BTU for each lbm of steam generated, what is the steam generation rate? If the
(
)
surroundings are at 70 (oF), what is SG for the process? Assume Nitrogen to be an
ideal gas for which Cp =(7/2)R.
Penyelesaian :
Mr N2 =
Cp = 7/2 R = 0.248 BTU/lbm. Rankine
Ms = steam rate in lbm/sec
Mn = nitrogen rate in lbm/sec
Mn =
1. Saturated liquid air pada 212 F masuk
2. Uap steam keluar pada 1 atm dan 300 F
3. Nitrogen masuk pada 750 F
4. Nitrogen Keluar pada 325 F
T3 = 1209.67 Rankine
T4 = 784.67 Rankine
Dari table F.3.
H1 =
S1 =
Dari Tabel F.4.
H2 =
S2 = 1.8158 btu/lbm Rankine
Pada persamaan 2.3. tanpa memperhatikan energy potensial dan energy kinetic dan kerja
sama dengan nol sehingga transfer panas adalah
Ms =
(asumsi)
Q=
Maka persamaan 2.3 menjadi
Persamaan 5.22 menjadi seperti berikut :
Q=
;
Maka nilai akhir dapat kita tentukan dengan memasukkan semua variable
6.82. Hot nitrogen gas at 400oC and aunospheric pressure flows into a waste heat
boiler at the rate of 20 kg.s -1 , and transters heat to water boiling at 101,33
kPa.The water feed to the boiler is saturated liquid at 101,33kPa, and it leaven
the boiler as superheated steam at 1 atm and 300oF.If the nitrogen is cooled to
325oF and if heat is lost to the surroundings at a rate of 8kJ for each kilogram of
steam generated what is the steam generation rate? If the surroundings are at
25oC, What is SG for the process? Assume nitrogen to be an ideal gas for which
Cp=(7/2)R.
Penyelesaian :
Ms = steam rate in kg/sec
Mn= nitrogen rate in kg/sec
(1) = masuk. liq. Water :101.33 kPa masuk
(2) = keluar uap pada 101,33 kPa dan 150 C
(3) = nitrogen dalam : 400 C
(4) = nitrogen keluar pada 170 C
molwt=28,014
Cp=
gm
mol
7 R
J
Cp=1,039
2 molwt
gmK
Mn=20
Kg
, T 3=673,15 K , T 4=443,15 K
sec
Tabel F.2
H 1=419,064
Kj
Kj
H 2=2776,2
Kg
Kg
Tabel F.2
S 1=1,3069
Kj
Kj
S 2=7,6075
Kg
Kg
Dengan Persamaan. (2.30), mengabaikan energi kinetik dan potensial dan
pengaturan istilah work ke nol dan dengan tingkat perpindahan panas yang diberikan oleh;
Ms=1
Kg
Kj
Q=−80
. Ms
sec
Kg
Ms ( H 2−H 1 ) + Mn .Cp ( T 4−T 3 )=−80
Ms=1,961
Kg
sec
Eq (5.22) Here becomes:
sdotG=Ms ( S 2−S 1 )+ Mn ( S 4−S 3 )−
S 4−S 3=Cp . ln (
T4
)
T3
Tσ=298,15 K Q=−80
Cp . ln
Kj
Ms
Kg
( TT 43 )− TσQ
SdotG=Ms ( S 2−S 1 ) + Mn ¿
SdotG=4,194
Kj
sec K
Q
Tσ
Kj
Ms
Kg
6.83 Shows that isobars and iscochorics have positive slopesin the single phase regions
of a TS diagrams. Suppose that Cp=a+bT where a and b positive constants.
Show that the curvature of an isobar is also positive. For specified T and S,
which is sheeper an isobar or an isochore? Why? Note that Cp>Cv
SOLUTION :
Slope isobar dan isokhorik pada diagram TS ditunjukkan pada persammaan 6.17 dan 6.30
( ∂TTS )
T
Cp
P =
dan
( ∂T∂S )
V =
T
Cv
Kedua slope bernilai positif. Dengan Cp > Cv , isokhorik lebih curam
Sebuah persamaan kurva isobarik dari perbedaan persamaan diatas:
2
( )
∂T
2
∂S
P
=
T
2
Cp
1
Cp
P =
[
1−
( ∂TTS )
T ∂ Cp
p
Cp ∂ T
( )
Dwngan Cp = a + bT,
( ∂TTS )
P -
T ∂Cp
Cp 2 ∂ S
( )
P =
T
Cp 2
-
T
Cp 2
∂T
( ∂Cp
)
(
∂T
∂T )
P
]
P= b dan 1 -
T
Cp
( ∂Cp
∂T )
P= 1 -
bT
a+bT
=
a
a+bT
Karena nilainya positif, jadi kurvanya adalah isobar.
6.85 The temperature dependence of the second virial coefficient B is shown for
nitrogen on Fig. 3.11. qualitatively, the shape of B(T) is the same for all gases;
quantitatively, the temperature for which B = 0 corresponds to a reduced
temperature of about Tr = 2,7 for many gases. Use this observations to show by
Eqs. (6.53) trough (6.55) that the residual properties G R, HR, and SR are negative
for most gases at modest pressures and normal temperatures. What can you say
about the signs of VR and C R ?
SOLUTION :
Karena pada saat Tr = 2,7 adalah jauh diatas temperatur gas normal.
Berdasarkan gambar 3.11 bahwa B adalah ( - ) dan dB / dT adalah ( + )
Selain itu d2B / dT2 adalah ( - )
Contoh 6.53 dan 6.55
GR = BP
SR = −P(dB/dT)
Dimana GR dan SR adalah ( - )
Dari definisi GR, HR = GR + T SR, dan HR adalah ( − ).
Contoh 3.37 dan 6.40
VR = B, dan VR adalah ( − ).
Dengan mensubstitusi GR , SR dan HR
HR = P
B – T dB
dT
Dimana,
∂HR
dB
∂T
P
T d2B
dT
PT d2B
dB
dT 2
dT
dT2
Sehingga, CR =
∂HR
∂T
adalah ( + )
P
6.86 An equimolar mixture of methane and propane is discharged froma compressor
at 5,500 kPa and 900 C at the rate of 1,4 kg s-1. If the velocity in the discharged line is
not to exceed 30 m s-1, what is the minimum diameter of the discharged line?
SOLUTION :
Diketahui Data Sebagai Berikut:
P = 5,500 Kpa = 55 bar
T = 900C = 363.15 K
Answer:
Tpr=
T
363.15 K
=
=1.296
Tpc
280.2
44.235 ¯¿=1.243
¯
55 ¿¿
P
Tpr = =¿
Tc
Tpc=Ymetana x Tc metana+Ypropana x Tc propana
Tpc=0.5 x 190.6 K +0.5 x 369.8 K
Tpc=95.3 K +184.9 K=280,2 K
Ppc=Ymetana x Pc metana+Ypropana x Pc propana
¯ x 42.48 ¯¿
Ppc=0.5 x 45.99 +0.5
¯ + 44.235
¯
¯¿
Ppc=22.995 +21.24
By Interpolation In Table E.3 And E.4
Zo = 0.8010
Z1 = 0.1100
w=Y x W ,+Y 2W 2
w=0.5 x 0.012+0.5 x 0.152
w=0.006+0.0160
w=0.082
Z =Zo+W x Z 1
Z =0.8010+0.082 x 0.11
Z =0.8010+0.00902
Z =0.81
Massa Molar Campuran
mol wt =( Y 1 x 16,043,+Y 2 x 44.097 ) gm/mol
mol wt =( 0.5 x 16,043,+0,5 x 44.097 ) gm /mo l
V=
mol wt =30.07 gm/mol
ZRT
0.81 x 0.082 x 363.15 K
3
=
=14.788 cm
P .mol wt 55 bars x 30.07 gm/mol
m ¿˙ 1.4
kg
sec
U=30 m/ sec
V ¿˙ V x m ¿˙
V ¿˙ 14,788
cm 3
kg
x 1,4
gm
sec
V ¿˙ 2.07 x 104
˙
V ¿ x
U
cm3
sec
2.07 x 10
30
4
m
s
cm
s
3
3
cm
2
=690
=6,901 cm
m
A=¿
D=
√ √
√
4 (6.901)cm2
4A
27, 604 cm2
=
=
=√ 8,79108 cm 2 = 2.964 cm
❑
π
3.14
6.87. Estimate VR, HR, and SRfor one of the following by appropriate generalized
correlations:
(a) 1,3-Butadiene at 500 K and 20 bar.
(6) Carbon dioxide at 400 K and 200 bar.
(c) Carbon disulfide at 450 K and 60 bar.
(d) n-Decane at 600 K and 20 bar.
(e) Ethylbenzene at 620 K and 20 bar.
(f) Methane at 250 K and 90 bar.
(g) Oxygen at 150 K and 20 bar.
(h) n-Pentane at 500 K and 10 bar.
(i) Sulfur dioxide at 450 K and 35 bar.
(j) Tetrafluoroethane at 400 K and 15 bar.
SOLUTION :
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
T
P
Tc
Pc
500
400
450
600
620
250
150
500
450
400
20
200
60
20
20
90
20
10
35
15
425.2
304.2
552.0
617.7
617.2
190.6
154.6
469.7
430.8
374.2
42.77
73.83
79.00
21.10
36.06
45.99
50.43
33.70
78.84
40.60
B0
=
0.422
1.6
Tr
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
-0.242
-0.199
-0.512
-0.369
-0.345
-0.200
-0.370
-0.308
-0.320
-0.306
0.083-
1
B = 0.1390.052
0.084
-0.267
-0.055
-0.029
0.084
-0.056
6.718 x 10-3
-4.217 x 10-3
9.009 x 10-3
Tr=
T
Tc
1.176
1.315
0.815
0.971
1.005
1.312
0.970
1.065
1.045
1.069
0.172
4.2
Tr
Pr
P
Pc
0.468
2.709
0.759
0.948
0.555
1.957
0.397
0.297
0.444
0.639
DB0
0.675
2.6
Tr
0.443
0.331
1.148
0.728
0.666
0.333
0.730
0.574
0.603
0.568
= ω
0.190
0.224
0.111
0.492
0.303
0.012
0.022
0.252
0.245
0.327
= DB1
0.722
5.2
Tr
0.311
0.173
2.091
0.841
0.703
0.175
0.845
0.522
0.576
0.510
=
R = 83.14 cm3.bar.mol-1K-1
R
V = [ R.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
Tc
0
1
Pc .(B + ω. B )]
-208.18
-61.723
-314.65
-963.97
-503.44
-68.560
-94.593
-355.907
-146.100
-232.454
HR = [R.Tc.Pr.(B0- ω.Tr.DB0) + (B1–Tr.DB1)
-1.377 x 103
-5.21 x 103
-23.01 x 103
-23.16 x 103
-4.37 x 103
-2.358 x 103
-559.501
-1.226 x 103
-1.746 x 103
-1.251 x 103
SR = [ -R.Pr.(DB0 + ω.DB1 ]
-19.53
-83.27
-87.08
-89.99
-40.55
-54.52
-24.69
-17.40
-27.45
-22.56
Bar.cm3/mol.K