Separation Process Principles

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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:

nin = nout1 + nout2

nin = 1000kmolhr nout1 = 226kmolhr nout2 = 774kmolhr

The stream availabilities are:

b = h T0s bin = hin T0sin = 8555.78kJkmol bout1 = hout1 T0sout1 = 15162.29kJkmol bout2 = hout2 T0sout2 = 9289.45kJkmol

Since,

Wmin = nout1bout1 + nout2bout2 nin1bin1

then

Wmin = 2061MJhr