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.
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