Time to Heat a Stirred Tank and Its Contents: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2013, Q4

Question 4: Time to Heat a Stirred Tank and Its Contents (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A closed, well-stirred system heated at constant rate $\dot Q = 300$ kW.

Body	Mass (kg)	$C_v$ (J/kg·K)
Tank (vessel)	5000	500
Reaction mixture	1500	3765
Temperature change		25 → 250 °C ($\Delta T = 225$ K)

Find. the heating time.

Figure 3 — Closed stirred tank: 300 kW warms both the 5000 kg vessel and the 1500 kg mixture from 25 °C to 250 °C.

Approach. With no reaction, no phase change and negligible stirrer work, the closed-system energy balance is simply the heat added equals the sensible-heat rise of the tank plus contents; divide the required energy by the power.

Lumped heat capacity of tank + contents. Both the vessel and the mixture warm together through the same $\Delta T$: $$C_\text{tot} = m_\text{tank}C_{v,\text{tank}} + m_\text{mix}C_{v,\text{mix}} = 5000(500) + 1500(3765) = 8.148\times10^{6}\ \text{J/K}.$$
Energy required for the 225 K rise. For a closed system with only sensible heating, $Q = C_\text{tot}\,\Delta T$: $$Q = (8.148\times10^{6})(225) = 1.833\times10^{9}\ \text{J} = \boxed{1.833\ \text{GJ}}.$$
Time at constant heating rate. Dividing by the 300 kW input, $$t = \frac{Q}{\dot Q} = \frac{1.833\times10^{9}\ \text{J}}{3.0\times10^{5}\ \text{W}} = 6111\ \text{s} = \boxed{1.70\ \text{h}\;(101.8\ \text{min})}.$$

Quantity	Result
Total heat capacity	8.148 × 10⁶ J/K
Energy required	1.833 GJ
Heating time	6111 s = 1.70 h (101.8 min)

Concept & Theory

Topic. Transient (unsteady) closed-system energy balance with only sensible heat.

Key relations. Closed-system first law $\frac{dU}{dt}=\dot Q - \dot W$; with $\dot W=0$ and constant $C_v$, integration gives $Q = \left(\sum_i m_i C_{v,i}\right)\Delta T$, and $t = Q/\dot Q$ at constant power.

Why this works. Because there is no flow, no reaction and no phase change, all the added heat goes into raising the temperature of the two masses. They share one temperature (well-stirred, in thermal contact), so their heat capacities simply add — the vessel is a "thermal ballast" that the heater must warm along with the batch.

Common pitfall. Forgetting the mass of the tank itself: here it stores $5000\times500\times225 = 0.56$ GJ, about 31 % of the total, so ignoring it under-predicts the heating time badly. Also, use $C_v$ (constant-volume) for a rigid closed batch, not $C_p$.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics §2.11–2.12 (energy balances for closed systems; transient heating).

Question 4 of 7: Time to Heat a Stirred Tank and Its Contents

Question 4: Time to Heat a Stirred Tank and Its Contents (equal value)