Separation Process Principles

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Exercise. A hydrocarbon stream in a petroleum refinery is to be separated at 1,500 kPa into two products under the conditions shown below. Using the data given, compute the minimum work of separation, Wmin, in kJ/h for T0 = 298.15 K.

kmol/h kmol/h
Component Feed Product 1
Ethane 30 30
Propane 200 192
n-Butane 370 4
n-Pentane 350 0
n-Hexane 50 0
Feed Product 1Product 2
Phase Condition Liquid Vapor Liquid
Temperature, K 364 313 394
Enthalpy, kJ/kmol 19,480 25,040 25,640
Entropy, kJ/kmol-K 36.64 33.13 54.84

Solution. Mass balances are:

nin = nout1 + nout2

nin = 1000kmol∕hr nout1 = 226kmol∕hr nout2 = 774kmol∕hr

The stream availabilities are:

b = h − T0s bin = hin − T0sin = 8555.78kJ∕kmol bout1 = hout1 − T0sout1 = 15162.29kJ∕kmol bout2 = hout2 − T0sout2 = 9289.45kJ∕kmol

Since,

Wmin = nout1bout1 + nout2bout2 − nin1bin1

then

Wmin = 2061MJ∕hr