Question

A butane isomerization process produces $70\text{ kmol/h}$ 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:

• A. $3\text{ kmol/h}$
• B. $4\text{ kmol/h}$
• C. $5\text{ kmol/h}$
• D. $6\text{ kmol/h}$

Hint:

When an inert component or impurity does not leave via the main product stream, tracking that single component directly with an independent species balance avoids complex systems of equations.

Answer 1

The Correct Option is C

Concept: In steady-state material balance problems without chemical accumulation, the total mass or molar flow entering a system must exactly balance the total flow exiting the system. This rule holds true both as a global overview and for each individual chemical species involved. Let us define the primary variables for our process network:

• Let $F$ be the molar flow rate of the fresh feed stream entering the system (in kmol/h).

• Let $P$ be the molar flow rate of the product stream leaving the process, which consists of pure iso-butane ($P = 70\text{ kmol/h}$).

• Let $W$ be the molar flow rate of the continuous purge waste stream leaving the loop (in kmol/h).
The compositions of each stream given in mole percent are:

• Feed Stream ($F$): Contains $99\%$ n-butane and $1\%$ impurity. Thus, the mole fraction of impurity is $x_{F} = 0.01$.

• Product Stream ($P$): Consists of $100\%$ pure iso-butane. It contains $0\%$ n-butane and $0\%$ impurity. Thus, $x_{P} = 0.00$.

• Purge Stream ($W$): Contains $85\%$ n-butane and $15\%$ impurity. Thus, the mole fraction of impurity is $x_{W} = 0.15$.

Step 1: Establish the overall total molar balance equation.
The total system input matches the total output streams: \[ F = P + W \] Substituting the known product stream flow rate $P = 70$: \[ F = 70 + W \quad \cdots (1) \]

Step 2: Establish a component balance for the impurity.
Since the impurity enters solely via the feed stream $F$ and exits exclusively through the purge stream $W$ (as the product stream is pure iso-butane), we can equate the input and output rate of the impurity: \[ F \cdot x_{F} = P \cdot x_{P} + W \cdot x_{W} \] Substituting the known mole fractions $x_{F} = 0.01$, $x_{P} = 0$, and $x_{W} = 0.15$: \[ F \cdot (0.01) = 70 \cdot (0) + W \cdot (0.15) \] \[ 0.01F = 0.15W \quad \cdots (2) \]

Step 3: Solve the system of equations to determine the purge rate $W$.
From equation (2), we can express $F$ explicitly in terms of $W$ by multiplying both sides by $100$: \[ F = \frac{0.15}{0.01}W \implies F = 15W \] Now, substitute this expression for $F$ back into the overall balance equation (1): \[ 15W = 70 + W \] Subtract $W$ from both sides of the equation to group like terms: \[ 15W - W = 70 \] \[ 14W = 70 \] Isolate $W$ by dividing both sides by $14$: \[ W = \frac{70}{14} = 5\text{ kmol/h} \] Therefore, the continuous volumetric purge flow rate required is $5\text{ kmol/h}$.