# Separation Process Principles

Chapter02 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24

Exercise. This is the solution to exercise 2.1 in the book.

Solution. Mass balances are:

${n}_{in}={n}_{out1}+{n}_{out2}$

$\begin{array}{rcll}{n}_{in}& =& 1000\phantom{\rule{1em}{0ex}}kmol∕hr& \text{}\\ {n}_{out1}& =& 226\phantom{\rule{1em}{0ex}}kmol∕hr& \text{}\\ {n}_{out2}& =& 774\phantom{\rule{1em}{0ex}}kmol∕hr& \text{}\end{array}$

The stream availabilities are:

$\begin{array}{rcll}b& =& h-{T}_{0}s& \text{}\\ {b}_{in}& =& {h}_{in}-{T}_{0}{s}_{in}=8555.78\phantom{\rule{1em}{0ex}}kJ∕kmol& \text{}\\ {b}_{out1}& =& {h}_{out1}-{T}_{0}{s}_{out1}=15162.29\phantom{\rule{1em}{0ex}}kJ∕kmol& \text{}\\ {b}_{out2}& =& {h}_{out2}-{T}_{0}{s}_{out2}=9289.45\phantom{\rule{1em}{0ex}}kJ∕kmol& \text{}\end{array}$

Since,

${W}_{min}={n}_{out1}{b}_{out1}+{n}_{out2}{b}_{out2}-{n}_{in1}{b}_{in1}$

then

${W}_{min}=2061\phantom{\rule{1em}{0ex}}MJ∕hr$