Purge calculation in isomerisation with impurity control:\nA butane isomerisation unit produces 70 kmol/h of pure isobutane. A continuous purge contains 85% n-butane and 15% impurity (mole%). The feed is n-butane containing 1% impurity (mole%). What is the purge flow rate (kmol/h) at steady state?

Introduction / Context:
Purge streams prevent the buildup of inert or contaminant species in recycle or high-conversion processes. Determining the necessary purge rate requires steady-state component balances, particularly on the impurity that must be removed from the system. Here, an isomerisation converts n-butane to isobutane (1:1 basis), and the product is specified as pure isobutane; unreacted n-butane and impurity exit mainly in the purge.

Given Data / Assumptions:

• Product is 70 kmol/h of pure isobutane (no impurity, no n-butane).
• Purge composition: 85% n-butane, 15% impurity (on molar basis).
• Feed: n-butane with 1% impurity → feed composition: 99% n-butane, 1% impurity.
• Steady state; no accumulation of impurity inside the system.
• Isomerisation converts n-butane to isobutane one-for-one; no side reactions producing impurity.

Concept / Approach:
Write component balances. The impurity enters with the feed and must leave entirely in the purge (since the product is pure isobutane). Then relate the n-butane balance to the given production of isobutane, recognizing that consumed n-butane equals isobutane produced.

Step-by-Step Solution:
Let F = feed flow (kmol/h) and P = purge flow (kmol/h).Impurity balance: 0.01 F = 0.15 P ⇒ F = 15 P.n-Butane balance to isobutane: n-butane converted = isobutane out = 70.n-Butane in feed = 0.99 F; n-butane leaving in purge = 0.85 P.Conversion relation: 0.99 F − 0.85 P = 70.Substitute F = 15 P: 0.99(15 P) − 0.85 P = 70 ⇒ (14.85 − 0.85) P = 14.0 P = 70.Solve: P = 70 / 14 = 5 kmol/h.

Verification / Alternative check:
From F = 15 P, the feed is 75 kmol/h. Feed n-butane = 74.25; n-butane purged = 4.25; conversion = 74.25 − 4.25 = 70 → matches product rate, confirming consistency.

Why Other Options Are Wrong:
3, 4, 6 kmol/h do not satisfy both impurity and n-butane balances simultaneously.

Common Pitfalls:
Forgetting to use impurity balance to relate feed and purge; assuming purge composition equals feed composition; neglecting that “pure isobutane” means zero impurity and zero n-butane in the product.

Final Answer:
5 kmol/h