CC V D V BI TP CA NHIT NGHC NG DNG

CC V D V BI TP CA NHIT NG HC NG DNG1. 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 hoTheo c s d liu ta c cc h s ca phng trnh Antoine nh bng sau:

H sC3H8C4H10

A6,803986,80896

B803,81935,86

C246,99238,73

Do :

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

=> PC4 = 2120,465 (mmHg) = 2,83 (bar)Theo nh lut Raoult ta c: PC3 = PC3 * x1 = 10,64 . 0,3 = 3,192 (bar) PC4 = PC4 * 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:

No YesTin hnh gii bng phng php lp Solve trn Excel ta thu c kt qu:

T (oC)LgPC3PC3 (bar)PC3 (bar)LgPC4PC4 (bar)PC4 (bar)P = PC3 +PC4 (bar)

47.6124.07615.8644.7593.5414.6303.2418.000

Vy: T = 47,61 oC

yC3 = PC3 = 4,759 %mol yC4 = PC4 = 3,24 %mol3. 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 = 216,636V cc gi tr ti hn: Pc = 27,4 bar . Tc = 540,1 K

GII:

Ta c:

=> T = 0,7 * Tc = 0,7 * 540,1 = 378,1 K = 105,1 oCTheo phng trnh Antoine ta c:

=> P(mmHg) = 1,23 (bar)Do :

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

= - lgPr 1 = - lg0,045 1 = 0,347 Zc = 0,291 0,08 *

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 + VTa 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=> (*)Mt khc theo nh lut Raoult ta c:

(**)

T (*) v (**) c:

=>

=>

Do : (1) i vi cu t nC4 v nC5:Lm tng t trn ta c:

(2)

(3)T (1) ,(2), (3) ta c: P = P1 + P2 + P3

Dng phng php lp vi s thut ton nh V D 2 vi bng cc h s phng trnh Antoine nh sau:H sC3nC4nC5

A6,803986,808966,87632

B803,81935,861075,78

C246,99238,73233,205

No

Yes

Sau khi tnh ton bng phng php lp Solve trn Excel ta c bng kt qu:T

(oC)lgPC3PC3

(bar)PC3

(bar)lgPC4PC4

(bar)PC4

(bar)lgPC5PC5

(bar)PC5

(bar)P = PC3+PC4+PC5

(bar)

28.2783.88410.2042.8743.3042.6850.8032.7620.7710.3234.000

Vy kt qu: T = 28,28 oCCH :Nu dng Pro II m phng th ta thu c kt qu: T = 26,806 oCNh 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 CH3Hiu 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)

= 0,584 + 0,965*(3 *0,0141 + 0,0189 + 0,0164) (3*0,0141 + 0,0189 +0,0164)2 = 0,653=>

Xt Benzen:

Cng thc cu to: CH HC CH

HC CH

CHHiu 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)

= 0,584 + 0.965 * 6 * 0,0122 (6 * 0,0122)2 = 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*10-8Hy: 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/hXem 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:

=>

Ly tch phn 2 v ta c: =>

Do :

b. Tnh cng sut:

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

=> dA = -CVdTLy 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:

(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 WVi hiu sut nn on nhit khong 80% th cg thc t nhn c:

P1 = 0,8 * P = 0,8 * 922,9 = 738,32 WNhn 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 oCHy xc nh p sut hi bo ho ca nC5H12 50 oC

GII:

Gi P p sut hi bo ho ca nC5H12 36 oCGi P1: p sut hi bo ho ca nC5H12 50 oCTa tnh p sut hi bo ho ca nC5H12 36 oC bng phng trnh Antoine:

=> Pso = 759,9 mmHg = 1,012 barsTheo phng trnh Clausius Clapeyron:

Vy: (bars) 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 oCGII:Theo c s d liu ta c:

H s phng trnh Antoine: A = 6,87632 ; B = 1075,78; C = 233,205Cc gi tr ti hn: Tc = 470 K; Pc = 33,6 bar; Vc = 310,6 cm3/mol Xc nh :Ta c: => T = 0,7*Tc = 0,7*470 = 329 K = 56 oCTheo phng trnh Antoine ta c:

P = 1433,96 mmHg = 1,912 bar

Do : w = - lgPr 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:Ta c: =>

=>

=> K = 35,5 oC

Xc nh p sut hi bo ho 50 oC:Ta c:

P50 = 1196,7 mmHg = 1,596 bar Xc nh th tch lng bo ho 50 oC:Ta c:

=>

EMBED Equation.3 (m3) = 123,8 ml9. 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:

% Vng TBPng ASTM D86ng FC

TTBP (oC)TTBPTD86TD86 (oC)TFCTFC (oC)

24010470183

5507484

721

10577685

1363

20708288

1042

30808690

742

40879092

852

509595 0 = 9595 1 = 94

1063

6010510197

15104

70120111101

20157

80140126108

20157

90160141115

1092

95170150117

1092

98180159119

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

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:

EMBED Equation.3 oC Hiu chnh tv:

Dng bng hiu chnh theo th tch ct HNH P1 ta c:

tmav tv = -11

=> tmav = 102 11 = 91 oCMt 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 barCu 4: Xc nh nht ng lc: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 cPCu 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 oCTheo th E7 ta c:

Enthalpy bay hi nguyn liu t 25oC: (kCal/kg)Enthalpy bay hi nguyn liu t 92oC: (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)Nhp A, B, C ca phng trnh Antoine v sai s

Gi s T

Tnh PC3, PC4

Tnh PC3, PC4

P EMBED Equation.3 = PC3 + PC4

Kt qu: T, yC3,yC4

Tnh PC3, PC4, PC5

Gi s T

Nhp A, B, C ca phng trnh Antoine v sai s

Kt qu: T

Tnh PC3, PC4, PC5

P EMBED Equation.3 = PC3 + PC4 + PC5

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