chaptet 6 (1)
Short Description
termodinamika...
Description
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 V ∂V − 2 ∂T P ∂T
( ) ( ) ( ) ( ) ∂Cp ∂V = ∂ P T ∂T
−T
P
P
Sehingga , ∂Cp ∂2 V =−T ∂P T ∂T2
( )
( )
P
Untuk gas ideal : ∂V ∂T
( )
2
=
P
( )
R ∂ V dan =0 2 P ∂T 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 = konstan, diubah menjadi
R C P−C V = =γ−1 , CV CV
maka: d lnT = - ( γ −1 ) d ln(V-b)
⟹
hubungkan dengan persamaan: T
γ−1 d lnT + d ln (V −b) =0
(V −b) γ−1
= 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
U=
Γ (T )
-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: d2 T Cp = -T dT 2
and
d2 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 V=
( ∂G ∂T )
T
;
V =K
;
S=
Dalam S
( ∂∂GT )
S=−
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 )
Cp=−T
P
d2 F d T2
( ∂∂UT )
;
Cv=
;
Cv=−T
V
d2 F dT2
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
−3
β=2.095 x 10 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 −3
β=2.095 x 10 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 V P S
Then at 360 K,
= V1P2P1T1 =
= S
but H = 0
( −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
(P2 – P1) = 0
(P2 – P1)
=
β ĸ
=
4.42 X 10−5 ¿¯−1 36.2 x 10−5 K−1 ¿
(T2 – T1) + P1
= 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 R
ZρRT =
−S VRT ) RT exp ¿ R v−S
ZρRT =
d ln α ( Tr ) −1 qlR d ln Tr VRT RT exp ¿ d ln α ( Tr ) v−z−1+ −1 qlR d ln Tr −z−1+
(
(
)
)
−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
( III.
))
d ln α ( Tr ) v−z−1+ −1 ql ρR d ln Tr
(
exp ¿
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
ZρRT =
−H R 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)
-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
α = [ 1+c.1- Tr0,5]2
0,691 0,606 Pr Pr ¿ β = (. Tr ), maka0,774 β = ( Tr Eq.(3.50) 0,722 0,741 α (Tr ) 0,768 q= E.q.(3.51) Tr 0,715 0,741 Guess: 0,774 0,749 z=1 0,673 Given: 0,769 z−β 0,776 Z = 1+ β – q. β z .(z 0,787 + β) E.q.(3.49)
-6,412 The derivative in the following equations equals; -8,947 -4,795 Tri ( ) 0,5 -7,408 -ci. ∝ i -5,974 -6,02 dimana, -6,246 i=1,.....,14 -6,849 -4,451 βi , qi+ βi -5,098 ¿ βi , qi -7,581 Z (¿) -5,618 Ii = ln Z¿ -4,482 ¿ -6,103 ¿ 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, Z ( βi, qi ) =
HRi=J/mol
SRi = J/mol.K
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 = 0.575
Tr = T / Tc
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 P2 – ln P1 = A – A – ( T 2 T 1 ) sat
sat
Ln
P 2sat P 1sat
Ln
P2 P 1sat
Ln
P 2sat P 1sat
sat
=0+
B B − T 1 T2
1 1 − = B( T 1 T 2 )
= B(
Eleminasi C dari B
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
=0+
Ln
P sat P 1sat
1 1 − = B( T 1 T )
Ln
P sat P 1sat
B B − T1 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 sat P1
: Ln
P 2sat P 1sat
: Ln
P2 sat P1
sat
Ln
P
sat
sat
=
- ln
ln Psat = ln P1sat +
=
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 sat P1
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 sat P1
. Ln
P 2sat P 1sat
…………………(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 D T 0 ( τ +1 ) + T 20 ( τ 2 +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
ig
(C p )s =A+ R
[ ( [
B T 0+ C T 20 +
= 3.470 +
D 2 τ T0 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 P ln −ln 1 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 6
=
0.15 . 10 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 cm3 /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 cm3 −1,209 0,025 gr gr x= 3 cm cm 3 71,45 −1,209 gr gr 3
1−0,030225 cm 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 0.15m3 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
mtot =
V tot 0.15 m3 = V vap 0.3942m3 /kg
mvap=
= 0.382 kg
V tot −mtot .V liq ∆ V lv 0.382 kg .1.004 .10 ¿ 3 0.15 m −¿ ¿¿
¿ 4.543 ×10−3 kg
−3
3
m kg
3
cm m =32.93 gr kg
mliq=mtot −mvap −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
cm gr
3
= 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
= 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
J gm
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
P1
∆S=1.302 J 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
T = 500oF
P1 = 300 psia
BTU lbm
H1 = 1257.7
BTU lbm
H2= 1257.7
BTU lbm rankin
S1 = 1.5703
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 BTU lbm rankin
S2 = 1.8606 ∆S = S2 – S1
BTU lbm rankin
=1.8606 – 1.5703
BTU lbm rankin
= 0.2903
Untuk steam pada gas ideal, tidak ada perubahan temperature maka −RT ln(
P2 ) P1
mol
∆S =
molwt = 18 lbmol
8.314 x 500 x ln (
=
= 0.2903
¿ −¿ ¿
20 ) 300
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
J gm
+ 0.98. ( 2646.9
- 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
= 0.031
101.33 kPa At 50ºC Psat = 12.34 kPa 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
−6
=
8,3682 x 10 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 U vap 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 U vap x U liq
= 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
U1 = 2878
J gr
Vtotal = 0,4 m³
T = 2000c U2 = 2638,7
Q=
Vtotal V1
J gr (U₂-U₁) 6
=
0,4.10 cm ³ cm ³ 354,34 gr
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
Svap = 7.2234
J gm . K
}
J −392.1643 KJ gm
W =−365.7643 K
J gm
J gm
Uvap = 2519.5 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) W =1 Kg ( 2.367 x 103−2629.9 )
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 Q
J gm
= 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 x1Uvap Uliq J U1 977,968 gm
J 0,92601,8 gm
J U1 977,968 gm
J gm
J 977,968 gm
J U1= 2,439 103 gm S1 Sliq x1Svap Sliq J S1 2,5924 gm . K 0,96,2244 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
Since step 1 is isentropic, S2 = S1 = Sliq x2Svap Sliq
J 2,5924 gm . K
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 x2Uvap Uliq J U2 604,237 gm 0.7982552,7
J J gm 604,237 gm
J J U2 604,237 gm gm J U2= 2,159 103 gm V2 Vliq x2Vvap Vliq cm3 V2 1,084 gm
0.798462,22
cm3 V2 1,084 gm
V2= 369,070528
cm gm
cm3 gm
cm3 1,084 gm
cm3 gm
3
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 450oC 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 = 572,83kPa
b. Untuk uap basa, entropinya di dapatkan: x= 0,95
S2 = Sliq xSvap 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/kgK
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/kgK Svap = 7.6709 kJ/kgK Hliq = 317.16 kJ/kg Hvap = 2636.9 kJ/kg x2= (S2 Sliq) / (Svap Sliq) x2= (7.3404 kJ/kgK - 1.0261kJ/kgK) / (7.6709 kJ/kgK - 1.0261kJ/kgK) x2= 0.95 H' Hliq x2Hvap 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 3
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 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.
lv
∆ S =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, R
v
H =H −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 −3 =1,218 .10 361,02+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 2 T
818,1 R ¿ ¿ ¿ 2 = ¿ −150 psia ¿ = 1,905 psi/R
.
∆ Slv =∆ V lv dPdT 2,996 =
ft 3 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 (0,7024)1,6
= -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 3
=
ft 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 V=
ZRT P
0.639 ( 0.08314 ) (468.15) V= 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
0
0
B =0.083−
1500 kPa 4248 kPa
Pr=0.3531
Diasumsikan propana merupakan gas ideal, maka : B =0.083−
Pr=
0.422 T r 1.6 0.422 0.9275
B 0=0.083−0.4549 0
B =−0.3719
B 1=0.139−
0.172 4.2 Tr
B 1=0.139−
0.172 0.9275
B 1=0.139−0.1854
P Pc
1
B =−0.0464
0
d B 0.675 = d T r T r 2.6 ¿
0.675 =−0.2525 0.9275
d B1 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
SR d B0 d B1 =−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 igp ) S ln
T2 P −R ln 2 + S2R−S1R T1 P1
∆ S=−25.287
J mol . K
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 1,6 Tr
B() = -0,815 B 1=0,139− B1 = -1,109
0,172 Tr 4,2
Vvap=
RT Pr [1+() B( )+ W . B 1. ] P Tr
Vvap=1,847 x 10 4
cm 3 mol 2 7
Vliq=Vc . Zc[−Tr ] cm3 Vliq=75,546 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 Tr=
T1 P1 Pr= Tc Pc
Tr = 1,001
Pr = 4,708
nyata
pada
akhir
T
&
P.
r() = -3,7733
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 Pc = 42,77 bar
Zc = 0,267
Tc = 425,2 K Vc = 220,4 cm3/mol
Tn = 268,7 K
T = 380 K
P = 1919,4 KPa
To = 273,15 K
Po
=
101,33
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
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 425,1K Pc 37,96bar = 37,96 x105 )Pa =37,96x 105 Pa=3796 kPa. Zc 0,274
3
Vc = 255
cm mol
Tn 272,7K T 370K P 1435kPa T0 273,15K P0 101,33kPa
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)
0,7800 (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 = ((
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 )0,8810 +
(
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) M= 0,7692
0,4 (X2) 0,0661 (M12)
0,87−0,85 0,90−0,85
0,90−0,87 0,90−0,85
+
((
Z0 0,7692 Dari tabel E2 halaman 669 didapatkan: Untuk Z1 0,2 (X1) -0,0715 (M11)
0,85 (Y1) 0,87(Tr) (Y) 0,90 (Y2)
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)
0,90−0,87 0,90−0,85
-0,0268)
+
((
0,87−0,85 0,90−0,85
M0,1372 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 3 1435 kPa 1435 x 10 Pa 1590,1
cm3 mol
cm3 V = 1590,1 mol
HR0 0,607RTc 0,6078,314 2,145 10
3
J mol
J mol . K
425,1K=
2145,30881
J mol
=
J mol
HR0= 2,145 10
3
HR1 0.831RTc 0.8318,314
J mol . K 425,1K = - 2936,987843
J mol
= 2,937
J 103 mol J HR1= 2,937 103 mol
SR0 0,485R 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 0.835R 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 SR = 5.421
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 . K
ICPH273.153701.93536.915103 11.402106 0.0= 1222.048 ICPH 1222,048K ICPS273.15370 1.93536.915103 11.402106 0.0= 3.80735 ICPS 3,80735
J mol
Hvap RICPH HR J Hvap 8,314 mol . K
1222,048K 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
(1−0,87) ¿ ¿ x ¿ 0,274 ¿
cm3 mol
J Hn = 8,314 mol . K
¯¿ ¯¿ 37,96 −1,013 ¿ ¿ ¿ x 272,7K( ) ln ¿ 1,092¿ ¿
J mol . K )
Hn = 22514
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 ¿ ¿
= 183,9 bar
= 2,996
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
P=
Pr =
Z nRT V gas P Pc
=
0,666 x 5 mol x 83,14 x 398,15 K 585
61,39 ¯¿ ¯¿ = 188,5 ¿ ¿
= 188,5 bar
= 3,07
a). The Final pressure is P = 188,5 bar b). The heat transferred By the first law, Q = nU = nH (PV) Q = nH VP= nH VgasP 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
J 298 = - 127150 mol
Step (2), Table C.1 data, for acetylene(g): MCPH298.15398.156.1321.95210-30.01.299105= 5.71731 Table C.2, for calcium hydroxide(s): MCPH298.15398.159.5975.43510-3 0.00.0= 11.48920 For the products, MCPH R(5.71731 11.48920) HP MCPH100K = 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: J HR0 2.340RTc = 2.340 x 8,314 x 308,3 K = 5.998 103 mol J HR1 0.384RTc = 0.384 x 8,314 x 308,3 K = 984.271 mol
HR HR0 HR1 = 5.998 10
3
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 nH Vgas(P 1bar) Q= 5 mol x ( 1.19 105
J 3 mol ) – 585 cm ( 188,5 -1 ) bar
Q = 606101 6.56 Propylene: 0.140 Tc 365.6K Pc 46.65bar T 400.15K P 38bar P0 1bar 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 400.15K/ 365.6K Tr 1.095 Pr= P/Pc Pr38bar/ 46.65bar Pr 0.815
Step (1): Use the Lee/Kesler correlation, interpolate.
H0 0.863RTc H0 2.623 103J/mol
H1 0.534RTc H1 = 1.623 103J/mol
HR H0 H1 HR = 2.85 103J/mol
S0 0.565R S0 =4.697J/molK
S1 0.496R S1 = 4.124J/molK
SR S0 S1 SR = 5.275J/molK
Step (2): For the heat capacity of propylene, A 1.637 B= (22.706103)/K
C=(6.915106)/ 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
ICPS400.15363.271.63722.706103 6.915106 0.0= 0.898338 ICPS 0.898338 Sig = R . (ICPA – ln (p0/p)) Sig =22.774J/molK S SR Sig S =28.049J/molK 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 Y1= 1,1 Y2= 1,15
0,4 -0,367
0,518
0,6 -0,581
? -0,334
-0,523
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 (−0,367 )+ (−0,581 ) + (−0,334 ) + 0,6−0,4 0,6−0,4 1,15−1,1 0,6−0,4
) ( )
)
]
(
[(
) (
]
)
)
]
[(
)
(
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=
[(
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 (−0,251 ) + (−0,381 ) + (−0,199 )+ 0,6−0,4 0,6−0,4 1,15−1,1 0,6−0,4
) ( )
)
(
]
[(
)
) (
]
)
[(
M =−0,2648064
( H R)
1
RTc
=−0,2648064
(HR)1 = -0,2648064 . 8,314 J mol-1 K -1 . 369,8 K = -814,1518315 Jmol-1
]
)
(
R 0
R 1
R (H ) (H ) H = +ω RTc RTc RTc
HR −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=
[(
]
[(
]
M=
0,518−0,4 1,15−1,144 0,6−0,518 0, (−0,23 ) + ( (−0,371 ) +( (−0,201 )+ ( ([ 0,6−0,518 ) ) ) ] 1,15−1,1 [ 0,6−0,4 0,6−0,4 0,6−0,4 0
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 =−0,2757284 (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 Y= 1,144
0,4 -0,229
0,518
0,6 -0,35
? -0,183
-0,275
Y1= 1,1 Y2= 1,15 M=
[(
]
[(
M=
0,518−0,4 1,15−1,144 0,6−0,518 0, (−0,229 ) + ( (−0,35 ) +( (−0,183 ) +( ([ 0,6−0,518 ) ) ) ] 1,15−1,1 [ 0,6−0,4 0,6−0,4 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
ΔSig =
∫ Cpig T1
T2
ΔSig =
∫ Cpig T1
P2 dT −R ln T P1 P dT −R ln 2 T P1
T2 ig
]
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
ΔS = 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 ΔSig – ΔS = -ΔSR
1
)
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
ig
C p dT ΔS / R ¿∫ -ln RT T 0
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) 3
= 6.551 x
= 5680
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 ) S1 = 0.7274 kj/kg – (460 x 10-6 x 1.013 ) ( 8000 – 12.87 ) kj/ kg
Vliq
=
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
KJ Kg . K
. 10
KW K
Untuk menghitung SG pada soal no. 66:
Kg sec
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 T300K Wlost Tx 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/kgK Untuk es pada 0˚C H2= H0 - 333.4kJ/kg H2= 333.44 kJ/kg S2= S0- (333.4/273.15)kJ/kgK S2= 1.221kJ/kgK T293.15K
m = 0.5kg/sec t 0.32 Dengan persamaan (5.26) dan (5.28): Wideal
= m x [ H2 - H1 -TS2- S1)] kg/sec x [333.44 kJ/kg - 83.86 kJ/ kg - 293.15K . (1.221kJ/kgK - 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 = mliqx 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, dmtUtHdm = 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, m2U2 m1U1 Havem = 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 + x2 −x1 ( y 2− y 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 3 3 cm m Vliquid = 1,216 gm = 1,216 x 10-3 kg ;
3
3
cm m Vvapor=66,626 gm =66,626x10-3 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 3
3
(
cm cm =1,216 gm + 0,1 x 65,41 gm
) )
cm3 cm3 +¿ =1,216 gm 6,541 gm 3
cm =7,757 gm
3
=7,757x10
-3
m 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 )−¿
0,6m1.
H liquid
Volume tangki konstan, maka: ∆ ( mt . H t )= y ( H vapor −H liquid )−¿ V =0 0,6 m1 . liquid Dimana: y=
0,6 m1 .V ( V vapor−V liquid ) liquid
Maka: Q = ∆m
0,6 m1 .V .(H ( V vapor−V liquid ) liquid
vapor
−H liquid )
−¿ 0,6 m . H liquid 1
tank
dengan H=
H liquid
∆m
dan 0,6 m1 =
tank
Persamaannya menjadi : 0,6 m1 .V liquid . ( H vapor −H liquid ) Q= V vapor−V liquid
(
Q=
(
Q=
(
)
m3 kg kJ kJ . 2802,3 −1008,4 3 3 kg kg m m 66,626 x 10−3 −1,216 x 10−3 kg kg 0,6 .(257,832 kg). 1,216 x 10−3
0,1881 m
3
−3
65,41 x 10
3
m kg
)
(
. 1793,9
kJ kg
)
)(
)
+ H .
Q=
337,432 kJ ( 65,41 x 10 ) −3
Q = 5158,730 kJ 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 KJ KJ 3017,7 −167,5 Kg Kg Kg m2 =3.241 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 CP = 2 R 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 . C P ( + n2. CP ( =0 1 mol .
29,099
J mol . K
.
( T −900 K ) + 2,5 mol .
29,099
J ( T −400 K ) mol . K
=0 (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
( TT )−R . ln ( PP )))+( n .( C . ln ( TT )−R . ln ( PP )))] 2
1
P
1
2
2
=0
12 ¯¿ P ¿ 2 ¯¿ P ¿
(
1 mol . 29,099
=0
)(
(
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 ¿ ¿
(
)
(
)
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 300 oF.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 )
P
T Cp
=
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 Cp 2
P
1 Cp
=
[
1−
( ∂TTS )
T ∂ Cp p Cp ∂ T
( )
Dwngan Cp = a + bT,
( ∂TTS )
P
-
T ∂Cp 2 Cp ∂ 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 = =14.788 cm3 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
cm3 sec
V ¿˙ 2.07 x 104
˙ V ¿ x U
2.07 x 104 30
m s
cm 3 s
3
=690
cm =6,901 cm 2 m
A=¿ D=
√ √
√
2 4 (6.901)cm2 4A 27, 604 cm = = =√ 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 : T
P
Tc
Pc
Tr=
T Tc
Pr
= ω
P Pc a. b. c. d. e. f. g. h. i. j.
500 400 450 600 620 250 150 500 450 400
B0
=
20 200 60 20 20 90 20 10 35 15 0.083-
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
1
B = 0.139-
0.422 T r 1.6 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
1.176 1.315 0.815 0.971 1.005 1.312 0.970 1.065 1.045 1.069
0.172 T r 4.2
0.468 2.709 0.759 0.948 0.555 1.957 0.397 0.297 0.444 0.639 DB0 0.675 T r 2.6
0.052 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
0.443 0.331 1.148 0.728 0.666 0.333 0.730 0.574 0.603 0.568
0.190 0.224 0.111 0.492 0.303 0.012 0.022 0.252 0.245 0.327 = DB1 0.722 T r 5.2 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
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