Question

An isothermal jacketed continuous stirred tank reactor (CSTR) operating at \(150^{\circ}\mathrm{C}\) is shown. The cold feed at \(30^{\circ}\mathrm{C}\) is first preheated to \(T\) (\(T<150^{\circ}\mathrm{C}\)) in heat exchanger \(HX_1\) (using process-to-process heat recovery) and then heated to \(150^{\circ}\mathrm{C}\) in a utility heater \(HX_2\). Mass flow rate and heat capacity are the same for all process streams and the overall 64ac0493b52af67589bd410c coefficient is independent of temperature. Which action will increase the value of \(T\)?

• A. Increase both 64ac0493b52af67589bd410c area of \(HX_1\) and heat duty of \(HX_2\).
• B. Decrease both 64ac0493b52af67589bd410c area of \(HX_1\) and heat duty of \(HX_2\).
• C. Increase the 64ac0493b52af67589bd410c area of \(HX_1\) and decrease the heat duty of \(HX_2\).
• D. Decrease the 64ac0493b52af67589bd410c area of \(HX_1\) and increase the heat duty of \(HX_2\).

Hint:

- Upstream preheater temperature is governed by \(Q_1=UA\Delta T_{\mathrm{lm}}\); only changes to \(U\), \(A\), or LMTD in that exchanger affect \(T\).
- Any downstream heater cannot raise the upstream preheat temperature; it only supplies the remaining sensible heat to the final setpoint.

Answer 1

The Correct Option is C

Step 1: Write the heat recovered in the preheater \(HX_1\). With equal \(\dot m c_p\) on hot and cold sides and \(U\) independent of \(T\), the recovered heat is \[ Q_1 = U A_1 \,\Delta T_{\text{lm}} = \dot m c_p\,(T - 30^{\circ}\mathrm{C}). \] For fixed inlet temperatures, increasing \(A_1\) (area of \(HX_1\)) \(\;\Rightarrow\) increases \(Q_1\) \(\;\Rightarrow\) increases \(T\).

Step 2: Relate utility duty in \(HX_2\) to \(T\). The utility heater provides the remaining heat to reach \(150^{\circ}\mathrm{C}\): \[ Q_2 = \dot m c_p\,(150^{\circ}\mathrm{C} - T). \] If \(T\) increases (due to larger \(A_1\)), then \(Q_2\) must decrease to keep the reactor inlet at \(150^{\circ}\mathrm{C}\).

Step 3: Eliminate incorrect options.

• Changing the duty of \(HX_2\) alone cannot increase \(T\) because \(HX_2\) is downstream of \(HX_1\).
• Decreasing \(A_1\) would reduce heat recovery, lowering \(T\) and requiring higher \(Q_2\).

Therefore, to increase \(T\): \[ \boxed{(C)\ \text{Increase } A_{HX_1} \ \text{and decrease } Q_{HX_2}} \]