Question

#3 Consider a system that is 30 % n-Butane, 40 % n-pentane and 30 % n-hexane.

a. Estimate the bubble pressure at a temperature of 80 C. What would be the composition of the vapor formed?

b. Estimate the dew pressure at 80 C. What would be the composition of the liquid formed?

c. Estimate the bubble temperature at a pressure of 2 atm? What would be the composition of the vapor formed?

d. If this mixture was at 100 C and 6 atm what fraction of the butane would be in the vapor phase?

Answer 1

Let Butane --1

Pentane --2

Hexane ---3

In order to solve this we will require Antoine equation for all the three component

Antoine coefficients for: log10(Psat[mmHg])=A-B/(T[oC]+C)

	A	B	C
n-butane	6.80776	935.77	238.789
n-pentane	6.85296	1064.84	232.012
n-hexane	6.87601	1171.17	224.408

we will make use of ideal gas law. Writing ideal gas law for all the three components we get

Y1*P =X1*P1sat ............1

Y2*P =X2*P2sat ..........2

y3*P =X3*P3sat ...........3

a)

Now we will evaluate the saturation pressure for all the three components at T = 80 oC

P1sat = 7453 mmHg

P2sat = 2755 mmHg

P3sat = 1068 mmHg

Now we have to find the bubble point pressure.

Since the fluid is at bubble point the feed is liquid and the composition we will be the liquid composition

now we will add equation 1,2, and 3

(Y1 +Y2 +Y3)*P = X1*P1sat + X2*P2sat +X3*P3sat

P =X1*P1sat + X2*P2sat +X3*P3sat =0.3*7453 +0.4*2755 + 0.3*1068 = 3658.3 mmHg

b) Now here we have to find the Dew point pressure. Dew point means all off the feed is now in vapor phase and a slight decrease in temperature will cause all of the feed to condense to liquid. Hence the feed composition for this part will be in the form of Yi's

Now we will re arrange and add equation 1,2 and 3

((Y1/P1sat) + (Y2/P2sat) + (Y3/P3sat) ) *P =X1 +X2 +X3 = 1

((0.3/7453) + (0.4/2755)+(0.3/1068))*P= 1

P = 2144.35 mmHg

c) in this part we are to find the bubble point temperature . Recall part (a)

P =X1*P1sat + X2*P2sat +X3*P3sat

now here we will have all Xi's and all the saturation pressures in terms of temperatures ( in terms of Antoine equation)

We will substitute this and P = 2 atm =1520 mmHg and solve for temperature

On solving we get

T = 68 oC

d) Now we have T and P

Total pressure P = 6 atm = 4560 mmhg

At T = 100 oC, the saturation pressures are as follows

P1sat =11108.5 mmHg, P2sat = 4423.097 mmHg, P3sat = 1844.31 mmHg

Let feed = 1 mol and Vapor be V and liquid be L mol

by overall balance

1 = V + L .......1

by component mass balance of butane

X1f1*F =V*Y1 + L*X1 ......2

X1f = (1-L)*Y1 + L*X1 ....... 3 (From equation 1 and 2)

From Antoine equation

Y1*P =X1*P1sat

Y1*4560 =X1*11108.5  

Y1 =2.436 *X1 .......4

From equation 3 and 4

X1f =0.3= ((1-L)*2.436 + L)*X1 .....5

Similarly we will have to write this for the other 2 components

Y2*4560 =X2*4423.097

Y2 = 0.9699*X2

X2f = 0.4= (1-L)*0.9699*X2 +L*X2  

0.4 = ((1-L)*0.9699 +L)*X2 .........6

Y3*4560 =X3*1844

Y3 =0.404*X3

X3f = 0.3 = ((1-L)*0.404 +L)*X3 ..........7

Rearranging and adding equation 5,6 and 7 we get

0.3/((1-L)*2.436 + L) + 0.4/((1-L)*0.9699 +L) + 0.3/((1-L)*0.404 +L) =X1 +X2 +X3 = 1

Solving the above equation L we get

L = 0.5379 mols

substituting this in equation 5

X1f =0.3= ((1-0.5379)*2.436 +0.5379)*X1 .....5

X1 = 0.18033

SUBSITUTING this in equation 4

Y1 =2.436 *X1 .......4

Y1 = 2.436*0.18033 = 0.43929