CC V D V BI TP CA NHIT NG HC NG DNG

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

=> PC3 = 7980,57 (mmHg) = 10,64 (bar)

B
935,86
lg PC4 = A
= 6,808936
= 3,326
T +C
238,73 + 30

=> P C4 = 2120,465 (mmHg) = 2,83 (bar)

Theo nh lut Raoult ta c:

PC3 = P C3 * x1 = 10,64 . 0,3 = 3,192 (bar)

PC4 = P C4 * x2 = 2,83 . 0,7 = 1,981 (bar)

Vy p sut ca h:
P = PC3 + PC4 = 3,192 +1,981 = 5,173 (bar)

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

Tnh PC3, PC4

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

=> T = 0,7 * Tc = 0,7 * 540,1 = 378,1 K = 105,1 oC

Theo phng trnh Antoine ta c :
LgP = A

B
1264,37
= 6,89385
= 2,964
T +C
105,1 + 216,636

=> P = 920,477 (mmHg) = 1,23 (bar)

P 1,23
=
= 0,045
Do : P =
PC
27,4

Vy h s quay v h s nn ti hn l:

= - lgP r 1 = - lg0,045 1 = 0,347

Zc = 0,291 0,08 * = 0.291 0,08 . 0,347 = 0,263
4. V D 4:
bi:
Xt mt hn hp kh C3, nC4 v nC5 vi thnh phn 50% mol C3, 25% mol nC4 v 25% mol
nC5 (zi). (zi cha xc nh nm pha no).
p sut P = 4 bars hy xc nh nhit ca h hn hp tn ti 50% trng thi hi
GII:
V h tn ti trng thi cn bng hai pha lng hi nn cn bng vt liu ton h:
F=L+V
Ta c cn bng vt liu i vi tng cu t trong hn hp lng hi:

Fzi = Lxi + Vyi

Trong : F l s mol nguyn liu vo thp tch
V l s mol pha kh nhn c t F mol nguyn liu
L l s mol pha lng nhn c t F mol nguyn liu
zi l phn mol cu t I trong dng nguyn liu
xi l phn mol cu t I trong dng lng
yi l phn mol cu t I trong dng hi
Gi s F = 1, ta c: zi = Lxi + Vyi
Theo bi ra thnh phn bay hi l 50% nn ta c:
i vi cu t C3:
z1 = 0,5*x1 + 0,5*y1
0,5 = 0,5x1 + 0,5y1
1 = x1 + y1
=>

y1 1
= 1 (*)
x1 x1

Mt khc theo nh lut Raoult ta c:

k1 =

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)

T (1) ,(2), (3) ta c :

P = P1 + P2 + P3
<=> 4 =

2 * P3
4 * P1
2 * P2
+
+
4 + P1 4 + P2 4 + P3

Dng phng php lp vi s thut ton nh V D 2 vi bng cc h s phng trnh

Antoine nh sau :
H s
A
B
C

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

Tnh P C3, P C4,

P C5

Gi s T

No

Tnh PC3,
PC4, PC5

P = PC3 + PC4 + PC5

Yes
Kt qu: T
Sau khi tnh ton bng phng php lp Solve trn Excel ta c bng kt qu:
T
(oC)
28.278

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

= 17,5 + 3*.65 + 56 + 41 = 309,5 (cm3/mol)

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

Hiu chnh theo cu trc ta thy benzen c 6 nhm = CH trong phn t.

Do :

Vc = 17,5 +

Ni*vi

= 17,5 + 6*38 = 245,5 (cm3/mol)

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

b.Tnh cng sut :

V qu trnh nn on nhit nn : A = -U => dA = - dU
Dng kh CH4 cho vo thit b l kh l tng nn :
P

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

=> Pso = 759,9 mmHg = 1,012 bars

Theo phng trnh Clausius Clapeyron :
P
ln 1
P0

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

8. BI TP NG DNG 3 (T.27 GIO TRNH)

bi:
Hy xc nh w, nhit si, Zc, ZRa, p sut hi v th tch lng bo ho ca nC5 50 oC
GII:
Theo c s d liu ta c:
H s phng trnh Antoine: A = 6,87632 ; B = 1075,78; C = 233,205
Cc gi tr ti hn: Tc = 470 K; Pc = 33,6 bar; Vc = 310,6 cm3/mol

Xc nh :

Ta c: Tr =

T
= 0,7
Tc

=> T = 0,7*Tc = 0,7*470 = 329 K = 56 oC

Theo phng trnh Antoine ta c:
lg P = A

B
1075,78
= 6,87632
= 3,157
T +C
233,205 + 56

P = 1433,96 mmHg = 1,912 bar

Pr =

P 1,912
=
= 0,0569
Pc
33,6

Do : w = - lgP r 1 = - lg(0,0569) 1 = 0,2449

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

= 1196,7 mmHg = 1,596 bar

Xc nh th tch lng bo ho 50 oC:

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

9. BI TP VN DNG CC BNG V CC BIU :

bi: Cho phn on Naphta c t trng d = 0,75 v ng cong chng ct im si
thcTBP.
Cu 1, 2: Chuyn t ng TBP sang ng ASTM D86 v FC
Ta dng th J9 chuyn t TBP sang ASTM D86 v dung th J11 chuyn t TBP
sang FC.
Ta c bng:

%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

th cc ng TBP, ASTM D86, v FC nh sau:

83

70

7
40

76

10
30

57
13

20

70
74

7
10

159

117
2

119

Cc ng TBP, ASTM D86, FC

Nhit

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

Dng HNH P5 vi tmav = 91 oC, Kuop = 11,6 ta c:

nht dng hc: < 1 cSt
=> nht ng lc: = < 1*0,75 = 0,75 cP
Cu 5: Xc nh nhit lng Q bc hi 40 % nguyn liu t 25 oC, p = 1 atm:
Ti im 40% bc hi, dung th J14 ta xc nh li t trng ca hn hp:
d = 0,757
Ta c theo ng cong FC th nhit ti im 40% l 92 oC
Theo th E7 ta c:
Enthalpy bay hi nguyn liu t 25oC: H 25 = 0,978 *12,5 = 12,225 (kCal/kg)
Enthalpy bay hi nguyn liu t 92oC: H 92 = 0,978 * 47,5 = 46,455 (kCal/kg)
Vy: nhit lng bc hi 40% nguyn liu t 25oC, p = 1 atm l
Q = H92 H25 = 46,455 12,255 = 34,23 kCal/kg)