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90daytona 90daytona
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7 years ago
A liquid mixture of n-pentane and n-hexane containing 40 mol percent n-pentane is fed continuously to a flash separator operating at 250oF and 80 psia.  Determine:

a) the quantity of vapor and liquid obtained from the separator per mol of feed.
b) the composition of both the vapor and the liquid leaving the separator.
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7 years ago
Assume ideal liquid and vapor exists.

Vapor pressure data @ 121oC from Antoine eqs:

a) Note: T(oF) = 1.8T(oC) + 52  ---------> 250oF = 1.8T(oC) + 52 -----------> oC = 121oC.

Antoine eq. logP10 = A BT + C

using table B.4, page 640 for:

n-hexane --------> A = 6.88555,  B = 1175.817,      C= 224.867
n-pentane --------> A = 6.84471,  B = 1060.793,      C= 231.541

n-hexane:

 logP10 = 6.88555 1175.817121 + 224.867

 logP10 =3.48592818P=103.48592818=3061.457114mmHg3061.457114mmHgx(1atm/760mmHg)=4.03atm[b]npentane:[/b][math=inline]logP10 = 6.84471 1060.793121 + 231.541

 logP10 =3.835718138P=103.835718138=6850.434812mmHg6850.434812mmHgx(1atm/760mmHg)=9.07atmP[sub]pentane[/sub]=P[sub]pentane[/sub]x[sup]x[/sup]P=9.07[sup]x[/sup]PP[sub]hexane[/sub]=P[sub]hexane[/sub]x(1[sup]x[/sup]P)=4.03(1[sup]x[/sup]P)P[sub]tot[/sub]=80psiax(1atm/14.696psia)=5.44atm.P[sub]tot[/sub]=5.44=P[sub]p[/sub]+P[sub]H[/sub]=9.07[sup]x[/sup]P+4.03(1[sup]x[/sup]P)5.44=9.07[sup]x[/sup]P+4.03(1[sup]x[/sup]P)5.444.03=9.07[sup]x[/sup]P4.03[sup]x[/sup]P1.41=5.04[sup]x[/sup]P[sup]x[/sup]P=0.280x[sub]hexane[/sub]=1x[sub]pentane[/sub]=1[sup]x[/sup]P=10.280=0.72x[sub]hexane[/sub]=0.72b)[math=inline]ypentane = PpentanePtot= PxPPtot = 9.07 x 0.2805.44

ypentane = yP = 0.467


yhexane = 1 - ypentane = 1 - yP = 1 - 0.467 = 0.533

yhexane = 0.533


we can now answer part a:

Basis: 1 mol of feed

overall balance:

F = L + V -------> 1 = L + V ............................. ....(1)

n-pentane  balance:

F(xF) = L(x) + V (y) --------->  1 (0.40) = 0.280L + 0.467 V ..........................(2)

L + + V = 1 --------------> L = 1 - + V

0.280L + 0.467+ V = 0.4 ----------> 0.280(1 - + V ) + 0.467+ V = 0.4 -----------> 0.467 V - 0.280 V = 0.4 - 0.280

V = 0.642 mol

L = 1 - V = 1 - 0.642 = 0.358

L= 0.358




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