A butane isomerisation process produces 70 k.mole/hr of pure iso-butane. A purge stream removed continuously, contains 85% n-butane and 15% impurity (mole%). The feed stream is n-butane containing 1% impurity (mole%). The flow rate of the purge stream will be_________________?

Correct answer: C. 5 kmole/hr

  • A. 3 kmole/hr
  • B. 4 kmole/hr
  • C. 5 kmole/hr
  • D. 6 kmole/hr

Explanation

An impurity balance gives 0.01F = 0.15P, so F = 15P. The n-butane balance, 0.99F = 70 + 0.85P, then gives 14P = 70 and P = 5 kmol/h.

Written and checked by , MS Computer ScienceLast updated
Report an error

The more specific you are, the faster it gets fixed. A source beats an opinion.

Prefer email? support@testustad.com

Practise Chemical Process Principles

2,710 free Chemical Process Principles MCQs from Chemical Engineering, each with the correct answer and an explanation. Unlimited attempts, no account needed.

Exams that ask Chemical Engineering questions like this

Chemical Engineering is on this paper prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for it.

More Chemical Process Principles questions