Question

Consider the process in the figure for manufacturing \( B \). The feed to the process is 90 mol\% \( A \) and a close-boiling inert component \( I \). At a particular steady-state: - \( B \) product rate is \( 100 \, \text{kmol/h} \), - Single-pass conversion of \( A \) in the reactor is \( 50\% \), - Recycle-to-purge stream flow ratio is \( 10 \). The flow rate of \( A \) in the purge stream, in \( \text{kmol/h} \), rounded off to 1 decimal place, is: \includegraphics[width=0.5\linewidth]{q54 CE.PNG}

Hint:

In recycle problems, carefully relate recycle-to-purge ratios and overall conversions to set up material balances.

Answer 1

Step 1: Define variables. Let: - \( F \) = feed flow rate of \( A \), - \( P \) = purge flow rate, - \( R \) = recycle flow rate. The relationship between recycle and purge streams is: \[ \frac{R}{P} = 10 \quad \Rightarrow \quad R = 10P. \] Step 2: Material balance on \( A \). The total flow of \( A \) entering the reactor is: \[ F + R = F + 10P. \] Since the single-pass conversion is \( 50\% \): \[ \text{Unreacted } A = 0.5(F + 10P). \] The unreacted \( A \) splits into recycle and purge streams: \[ R + P = 0.5(F + 10P). \] Substitute \( R = 10P \): \[ 10P + P = 0.5(F + 10P). \] \[ 11P = 0.5F + 5P \quad \Rightarrow \quad 6P = 0.5F \quad \Rightarrow \quad F = 12P. \] Step 3: Solve for \( P \). The total product rate \( B = 100 \, \text{kmol/h} \), and the feed is \( 90\% \, A \): \[ F \cdot 0.9 = 100 \quad \Rightarrow \quad F = \frac{100}{0.9} = 111.1 \, \text{kmol/h}. \] Substitute \( F = 12P \): \[ 111.1 = 12P \quad \Rightarrow \quad P = \frac{111.1}{12} \approx 18.2 \, \text{kmol/h}. \] Step 4: Conclusion. The flow rate of \( A \) in the purge stream is \( 18.2 \, \text{kmol/h} \).