A process stream containing a binary mixture is fed into a flash drum where it f

Question

A process stream containing a binary mixture is fed into a flash drum where it flashes into phases as

shown below. The saturation pressures of the two species are given below

ln Psat (bar) ? 9.08? 2572 ; ln Psat (bar) ?10.63? 3504 , 1 T(K) 2 T(K)

y1=0.69 T, P ?

x1=0.17

and the resultant liquid mixture can be described by the following GE ? x1x2 ?1.99x1 ?1.09x2 ?

RT
(a) Use the given data to determine T (K) and P (bar) of the flash drum.

(b) If the flow rates of the equilibrium vapor and liquid phases from the flash drum are in 1:2 ratio, determine the composition of the feed stream.

Explanation / Answer

lnY1= d/dx1( GE/RT)

GE/RT= x1*(1-x1){ 1.99x1+1.09*(1-x1)}

= (x1-x12) { 1.99x1+1.09 -1.09x1) = (x1-x12)(0.9x1+1.09) = 0.9x12+1.09x1-0.9x13-1.09x12 =-0.19x12+1.09x1-0.9x13

lnY1= -0.38x1+1.09 -2.7x12

GE/RT= x2*(1-x2){(1.99(1-x2)+1.09x2)}= (x2-x22){1.99-0.9x2) =1.99x2-0.9x22-1.99x22+0.9x23 =

lnY2= -1.99-1.8x2-3.98x2+2.7x22= -1.99-5.78x2+2.7x22

for x1= 0.17 and x2 =1-0.17= 0.83

lnY1= -0.38*0.17+1.09-2.7*(0.17)2=0.947 and Y1= 2.58

and lnY2= -1.99-5.78*0.83+2.7*(0.83)2 =-4.93 and Y2=0.007246

From y1P= x1Y1P1sat and y2P= x2Y2P2sat

The ratio gives y1/y2= x1p1sat/x2p2sat =0.69/0.31= 0.17*P1sat/0.83P2sat

2.226= 0.2048* P1sat/ P2sat

P1sat/ P2sat= 10.87

P1sat= 10.87*P2sat

Taking ln

lnP1sat= ln10.87+lnP2sat

lnP1sat= 2.386+lnP2sat

9.08-2572/K= 2.386+10.63-3504/T

3504/T-2572/T= 10.63+2.386-9.08=3.396

932/T= 3.396

T= 932/3.396=274.4

At 274.4

Ln P1sat= 9.08-2572/274.4, P1sat= 0.745 bar

And lnP2sat= 10.63-3504/274.4, p2sat= 0.117 bar

P= x1Y1p1sat+x2Y2p2sat= 0.17*2.58*.745+0.83*0.007246*0.117=0.327 bar

Flow rate of vapor = 2 moles and flow rate of liquid =1 moles

Moles of more volatille component = 2*0.69 ( vapor) +1*0.17 ( liquid) = 1.55 moles

Mole of less volatile component= 3-1.55= 1.45

Feed stream will have composition

More volatile compoenet= 1.55/3 = 0.52 and less volatile component=1-0.52=0.48

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