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1. V D 1:
bi:
Cho h C3, C4 pha lng v c thnh phn 30% mol C3 v 70% mol C4. nhit 30 oC th
p sut ca h l bao nhiu. Bit phng trnh trng thi ca kh l l tng.
GII:
Ta s dng phng trnh Antoine tm p sut hi bo ho
Theo c s d liu ta c cc h s ca phng trnh Antoine nh bng sau:
H s
A
B
C
C3H8
6,80398
803,81
246,99
C4H10
6,80896
935,86
238,73
Do :
B
803,31
lg PC3 = A
= 6,80398
= 3,902
T +C
246,99 + 30
2. V D 2:
bi:
Cho h C3 v C4 nh v d 1. Vi p sut lm vic 8 bar hy tm nhit v thnh phn pha
hi ca hn hp.
GII:
Ta gii bi ton theo phng php lp vi s thut ton nh sau:
Nhp A, B, C ca phng
trnh Antoine v sai s
Gi s T
No
P = PC3 + PC4
Yes
Kt qu: T, yC3,yC4
C4
Tnh P C3, P
Tin hnh gii bng phng php lp Solve trn Excel ta thu c kt qu:
PC3
PC4
P C3
P C4
T (oC) LgPC3
P = PC3 +PC4 (bar)
LgP C4
(bar)
(bar)
(bar)
(bar)
47.612 4.076
15.864
4.759
3.541
4.630
3.241
8.000
Vy : T = 47,61 oC
yC3 = PC3 = 4,759 %mol
yC4 = PC4 = 3,24 %mol
3. V D 3 :
bi :
Tnh h s quay w v h s nn ti hn Zc ca nC7.
Bit h s A, B, C ca phng trnh Antoine :
A = 6,89385 . B = 1264,37 . C = 216,636
V cc gi tr ti hn: Pc = 27,4 bar . Tc = 540,1 K
GII:
Ta c: Tr =
T
= 0,7
Tc
B
1264,37
= 6,89385
= 2,964
T +C
105,1 + 216,636
Vy h s quay v h s nn ti hn l:
y1 1
= 1 (*)
x1 x1
y1 P1 P1
=
=
(**)
x1
P
4
T (*) v (**) c:
P
P 4 + P1
1
1
1 = 1 =>
= 1+ 1 =
x1
4
x1
4
4
=> x1 =
4
4 + P1
Do : P1 = P1 * x1 =
4 * P1
4 + P1
(1)
i vi cu t nC4 v nC5:
Lm tng t trn ta c :
P2 = P2 * x 2 =
2 * P2
4 + P2
P3 = P3 * x3 =
2 * P3
(3)
4 + P3
(2)
2 * P3
4 * P1
2 * P2
+
+
4 + P1 4 + P2 4 + P3
C3
6,80398
803,81
246,99
nC4
6,80896
935,86
238,73
Nhp A, B, C ca phng
trnh Antoine v sai s
nC5
6,87632
1075,78
233,205
P C5
Gi s T
No
Tnh PC3,
PC4, PC5
P C3
(bar)
PC3
(bar)
C4
PC4
(bar)
C5
lgP
P C4
(bar)
lgP
P C5
(bar)
PC5
(bar)
10.204
2.874
3.304
2.685
0.803
2.762
0.771
0.323
C3
lgP
3.884
P=
PC3+PC4+PC5
(bar)
4.000
Vy kt qu : T = 28,28 oC
CH :
Nu dng Pro II m phng th ta thu c kt qu : T = 26,806 oC
Nh vy ta c th thy rng s sai khc gia tnh ton bng nhit ng hc ng dng v m
phng bng Pro II l khng ng k.
5. BI TP (T.36 GIO TRNH)
bi: Hy xc nh Pc v Vc ca iC5 v Benzen
GII:
Xt iC5:
Cng thc cu to: CH3 CH CH2 CH3
CH3
Hiu chnh theo cu trc ta thy iC5 c:
3 nhm CH3
1 nhm CH2
1 nhm CH
Do :
vc = 17,5 +
Ni* vi
T
2
= 0,584 + 0,965. N i .Ti ( N i .Ti )
Tc
= 0,584 + 0,965*(3 *0,0141 + 0,0189 + 0,0164) (3*0,0141 + 0,0189 +0,0164)2
= 0,653
=> Tc = T 0,653
Xt Benzen:
Cng thc cu to:
CH
HC
CH
HC
CH
CH
Vc = 17,5 +
Ni*vi
T
2
= 0,584 + 0,965. N i .Ti ( N i .vi )
Tc
= 0,584 + 0.965 * 6 * 0,0122 (6 * 0,0122)2 = 0,649
=> Tc = T 0.649
6. BI TP NG DNG 1 (T.38 GIO TRNH)
bi: Xt qu trnh nn on nhit CH4, co nhit l 25 oC t p sut 40 bars n p sut
50 bars vi cc s liu nh sau:
Cp(T) = A + B*T + C*T2 +D*T3 (J.mol-1.K-1), T [K]
Vi : A = 25,359 ; B = 0,0169 ; C = 7,13*10-5 ; D = -4*10-8
Hy : Xc nh nhit ca qu trnh sau khi nn
Cng sut cn thit cho my nn s dng cho dng kh CH4 c lu lng 100kg/h
Xem kh CH4 trong trng hp ny l kh l tng
GII:
a. Xc nh nhit ca qu trnh sau khi nn:
Ta c: Cp = A + B*T + C*T2 +D*T3
= 25,359 + 0,0169*298 + 7,13*10-5*2982 4*10-8*2983
= 37,7855 (J/mol.K)
V qu trnh on nhit nn :
ds = C P
dT
dP
R
=0
T
P
=> C P
dT
dP
=R
T
P
Ly tch phn 2 v ta c :
T
P
C P . ln 2 = R. ln 2
T1
P1
CP
=> ln T2 = R . ln P2 = ln P2
P
T1 C P
P1
1
Do :
P
T2 = T1 * 2
P1
8, 314
Cp
50 37 ,7855
= 298 *
= 313 K
40
dU = CV .dT + T
P dV = CV dT
T V
=> dA = -CVdT
Ly tch phn hai v ta c :
A = - CV *(T2 T1) = - ( Cp R)*(T2 T1)
= - (37,7855 8,314) * (313 298) = - 442,1 (J/mol)
Vy cng sut nn ca thit b on nhit :
P=
A* n
=
442,1 *
3600
10 5
16 = 767,53 (J/s) = - 767,53 (W)
CH :
Nu thc hin bi ton trn bng m phng Pro II th ta thu c kt qu nh sau :
Nhit : T = 41,86 oC = 314,86 K
Cng sut l thuyt : P = 922,9 W
Vi hiu sut nn on nhit khong 80% th cg thc t nhn c :
P1 = 0,8 * P = 0,8 * 922,9 = 738,32 W
Nhn xt : T kt qu trn y ta c th thy rng s sai khc gia tnh ton v m phng
bng Pro II l khng dng k. y l mt minh chng thy rng Pro II c xy dng trn c s
nhit dng hc ng dng.
7. BI TP NG DNG 2 (T.40 GIO TRNH)
bi : cho bit :
n nhit ho hi ca nC5H12 bng 24,91 kJ/mol
Nhit si ca nC5H12 l 36 oC
Hy xc nh p sut hi bo ho ca nC5H12 50 oC
GII :
Gi P 0 : p sut hi bo ho ca nC5H12 36 oC
Gi P 1 : p sut hi bo ho ca nC5H12 50 oC
Ta tnh p sut hi bo ho ca nC5H12 36 oC bng phng trnh Antoine :
lg P0 = A
B
1075,78
= 6,87632
= 2,88
T +C
36 + 233,205
h
=
R
1 1
.
T1 T0
Vy:
h
P1 = P0 * exp R
1 1
24,91 * 10 3
1
1
. = 1,012 * exp .
= 1,54 (bars)
8,314 273 + 50 273 + 36
T1 T0
Xc nh :
Ta c: Tr =
T
= 0,7
Tc
B
1075,78
= 6,87632
= 3,157
T +C
233,205 + 56
Pr =
P 1,912
=
= 0,0569
Pc
33,6
Xc nh zc:
Zc = 0,291 0,08*w
= 0,291 0,08*0,2449 = 0,2714
Xc inh zRa:
ZRa = 0,29056 0,08775*w
= 0,29056 0,08775*0,2449 = 0,8842
Xc nh nhit si:
P
P
lg c
lg c
3 Pb
P
Ta c: .
1 => Tc
3
1 = . b
7 Tc
1
Tb
7 +1
Tb
P
33,6
lg c
lg
=> Tc 3 Pb
3 1,013
= .
+1 = .
+ 1 = 1,523
Tb 7 + 1
7 0,2449 + 1
=> Tb =
Tc
470
=
= 308,5 K = 35,5 oC
1,523 1,523
Xc nh p sut hi bo ho 50 oC:
Ta c: lg P50 = A
50
B
1075,78
= 6,87632
= 3,078
T +C
50 + 233,205
Ta c:
=>
V L, =
V L , =
2
1+1Tr 7
R.Tc
.z c
Pc
8,314 * 470
* 0,2714
33,6 * 10 5
1+ 1 50 + 273 7
470
= 1,238 * 10 4
(m3) = 123,8 ml
%V
ng TBP
TTBP (oC)
TTBP
ng ASTM D86
TD86 (oC)
TD86
ng FC
TFC (oC)
TFC
40
50
10
10
97
4
111
15
101
7
126
15
160
108
7
141
10
180
95 1 = 94
101
140
98
92
2
120
170
105
95
90
95 0 = 95
20
90
95
20
80
88
90
15
70
87
10
60
85
86
8
50
84
82
80
115
2
150
10
83
70
7
40
76
10
30
57
13
20
70
74
7
10
159
117
2
119
200
160
120
80
40
0
0
10
20
30
TBP
40
50
60
D86
70
80
90
FC
T ng cong FC va c c ta c:
- Nhit im si: 83 oC
- Nhit im sng: 119oC
- Nhit 40 % bc hi (p = 1 atm): 92 oC
Dng th Chart Cox ta c: nhit 40 % bc hi v p = 1,5 atm l 158 oC
Cu 3: Xc nh Kuop ; Mtb ; Tc,m ; Pc,m:
tm Kuop cn c tmav v t trng d
- Tm tmav:
t +t +t
80 + 95 + 140
t v = 20 50 80 =
102 oC
3
3
Hiu chnh tv:
t t
120 57
s = 70 10 =
= 1,05
60
60
Dng bng hiu chnh theo th tch ct HNH P1 ta c:
tmav tv = -11
=> tmav = 102 11 = 91 oC
Mt khc theo bi ra ta c d = 0,75
Do theo HNH P2 ta c: kuop = 11,6
Mtb = 95
Theo HNH P3 ta c: Tc,m = 273 oC
Pc,m = 32,8 bar
Cu 4: Xc nh nht ng lc:
%V
100