Q1: A butane isomerisation process produces 70 kmole/hr of pure isobutane. 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

ANS:C - 5 kmole/hr

To solve this problem, let's first calculate the flow rate of the impurity in the feed stream, then use the information provided to determine the flow rate of the purge stream.

1. Impurity Flow Rate in Feed Stream:
   • The feed stream contains n-butane with 1% impurity.
   • Since we're given the flow rate of pure isobutane produced, we can assume that all the impurities from the feed stream are removed and end up in the purge stream.
   • Therefore, the impurity flow rate in the feed stream is 1% of the flow rate of pure isobutane produced.
2. Impurity Flow Rate in Purge Stream:
   • The purge stream contains 85% n-butane and 15% impurity.
   • Since we're given that the impurity flow rate in the feed stream is the same as the impurity flow rate in the purge stream, we can set up an equation to solve for the flow rate of the purge stream.
   • Let 𝑥x represent the flow rate of the purge stream.
   • The impurity flow rate in the purge stream is 15% of 𝑥x.

Now, let's set up the equation: 1% of 70 kmole/hr=15% of 𝑥1% of 70 kmole/hr=15% of x 1100×70=15100×𝑥1001​×70=10015​×x 0.7=0.15𝑥0.7=0.15x 𝑥=0.70.15x=0.150.7​ 𝑥≈4 kmole/hrx≈4 kmole/hr So, the flow rate of the purge stream is approximately 4 kmole/hr4kmole/hr.