Question 10.3.1: Startup of a Batch Reactor A well-stirred batch reactor wrap......

We first note that the conditions of validity of the simplified closed system energy balance equation, Equation 10.3-13, are all satisfied (verify); moreover, since the system has constant volume and the energy input due to the stirrer is presumed negligible, \dot{W} ≈ 0. The equation therefore becomes

Closed System:                  \boxed{MC_{v}\frac{dT}{dt} = \dot{Q}+\dot{W}}                 (10.3-13)

MC_{v}\frac{dT}{dt} = \dot{Q} \\ t=0,   T_{sys} = 25°C

The task is now to integrate this equation from the initial state of the system (t = 0, T = 25°C) to the final state (t = t_{f}, T = 250°C), and to solve the integrated equation for the heating time t_{f}. Rearranging the equation,

MC_{v}dT= \dot{Q}dt \\ \left. \Large{\Downarrow} \right. Integate\\ \int_{25°C}^{250°C}{MC_{v}dT} =\int_{0}^{t_{f}}{\dot{Q}dt} \\ \left. \Large{\Downarrow} \right. \dot{Q}, M,  and  C_{v}  are  constant \\ MC_{v}(250°C  –  25°C) = \dot{Q}t_{f} \\ \left. \Large{\Downarrow} \right. \\t_{f}= \frac{225MC_{v}}{\dot{Q}}

The heat capacity of the system is obtained from Equation 8.3-13 as

(C_{p})_{mix}(T)=\sum\limits_{\begin{matrix}all\\ mixture\\ components \end{matrix} }{y_{i}C_{pi}(T)}               (8.3-13)

C_{v}=\frac{M_{reactants}}{M}(C_{v})_{reactants}+ \frac{M_{reactor}}{M}(C_{v})_{reactor} \\ \left. \Large{\Downarrow} \right.\\ MC_{v}=(1500  g)\left(0.900  \frac{cal}{g\cdot °C} \right) +(3000  g) \left(0.120  \frac{cal}{g\cdot °C} \right)\\=(1710  cal/°C)(4.184  J/cal) \\ = 7150  J/°C

The final result is

t_{f} = \frac{225MC_{v}}{\dot{Q}}\\ \left. \Large{\Downarrow} \right.\begin{matrix} MC_{v}= 7150  J/°C \\ \dot{Q} = 500.0  W=500.0  J/s\end{matrix} \\ t_{f}= \frac{7150  J/°C}{500.0  J/s}(225°C) \\= 3220  s \Longrightarrow \boxed{53.7  min}

Question: Which restriction on the energy balance (Equation 10.3-13) would probably be violated if the reactants were not stirred?