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A 50 mol% mixture of propane (1) and n-butane (2) enters an isothermal flash drum at 37°C. If the flash drum is maintained at 0.6 MPa, what fraction of the feed exits as liquid? What are the compositions of the phases exiting the flash drum?
Work the problem in the following two ways.
a. Use Raoult’s law (use the Peng-Robinson equation to calculate pure component vapor pressures).
b. Assume ideal mixtures of vapor and liquid. (Use the Peng-Robinson equation to obtain fsat for each component.)

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fortuneonyemuwa

Answer:

Explanation:

given data :

xi=0.5 (i.e. mole fraction 50%)

two liquid mixtures are given i.e. propane and n-butane

generally raoults law is applicable to ideal mixtures

the basic equation of raoults law

ki=pi saturated (T)/p...............1

first we take propane liquid

the basic forula to calculate psat is

%log psat=A-B/T(K)+C

for this we need to calculate tv=B-C/A-ln(P)

Substitute the values from antoine data

B=803.997,A=3.92828,C=26.11,P=0.6*10^-3

TV= 44.74K

NOW WE WILL take n-butane and calculate tv

for butane antoine data is as follows

B=2292,A=13.98,C=-27.86

Tv=134.96 k

now calculating psat

%log psat=A-B/T(K)+C

Psat for propane is -11.29 bar

psat for n-butane is -21.27 bar

pure component of vapour pressure is

1. p1=xi*psat=0.5*-11.29=-5.645 bar

2.p2=xi*psat=0.5*-21.27=-10.635 bar

2. now we are calculating the composition of phase

to calculate the composition

first we need to add up all the pure component pressure

p=p1+p2=-5.645+10.635=4.99 bar

i am assuming yi as vapour composition

yi=p1/p=-5.645/4.99=-1.131

y2=p2/p=-10.635/4.99=-2.131

so these are the minimum values of vapour pressure and pure component vapour pressure

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