Solution

A species balance for this application starts with the balance equation for moles of A, `n_A`, for two inlet streams and one outlet stream. This is a mixing problem so there is no generation and consumption of species A.

$$\frac{dn_A}{dt} = \dot n_{A1} + \dot n_{A2} - \dot n_{A3}$$

The molar amount, n_A is often measured as a concentration, c_A, and related by the molar flowrate, q.

$$n_A = c_A \, V \quad \quad \dot n_A = c_A \, q$$

$$\frac{dc_A V}{dt} = c_{A1} q_{1} + c_{A2} q_{2} - c_{A3} q_{3}$$

When the tank is well-mixed, the concentration in the tank is equal to the stream leaving the tank. The chain rule is used to separate the differentials of `V` and `c_{A3}`.

$$\frac{dc_{A3} V}{dt} = V\frac{dc_{A3}}{dt} + c_{A3}\frac{dV}{dt} = c_{A1} q_{1} + c_{A2} q_{2} - c_{A3} q_{3}$$

If the volume is constant then `\frac{dV}{dt}=0` and the expression is simplified to a single ordinary differential equation.

$$V\frac{dc_{A3}}{dt} = c_{A1} q_{1} + c_{A2} q_{2} - c_{A3} q_{3}$$

If the volume is not constant then a mass balance must be added.

$$\frac{dm}{dt} = \dot m_{1} + \dot m_{2} - \dot m_{3}$$

With `m = \rho V` and `\dot m = \rho q` the expression becomes:

$$\rho_3 \frac{dV}{dt} = \rho_1 q_1 + \rho_2 q_2 - \rho_3 q_3$$

If the density `\rho` is constant among all streams then the expression is simplified.

$$\frac{dV}{dt} = q_1 + q_2 - q_3$$

Solution

An energy balance for this application starts with the balance equation for enthalpy, h.

$$m\,c_p\frac{dT}{dt} = \dot m_{in} c_p \left( T_{in} - T_{ref} \right) - \dot m_{out} c_p \left( T_{out} - T_{ref} \right) + UA \left(T_\infty-T\right) + W_s + r\,V \, \Delta H_r$$

Enthalpy is related to temperature as m c_p (T-T_ref) where c_p is the heat capacity. With a constant reference temperature (T_ref), constant mass `m=\rho V`, and equal flows `q=q_1=q_2`, this reduces to the following:

$$\rho V \, c_p \frac{dT}{dt} = \rho \, q \, c_p \left( T_{1} - T_{2} \right) + UA_{c} \left( T_c-T_2 \right) + UA_{a} \left( T_a-T_2 \right) + W_s + r\,V \, \Delta H_r$$

If the volume is not constant then a chain rule must be used to account for the non-zero derivative of volume. Also, the inlet flow and outlet flows are no longer equal `q_1 \ne q_2`. There is no reaction so the heat generated by reaction is zero.

$$\rho V \, c_p \frac{dT}{dt} + \rho T \, c_p \frac{dV}{dt} = \rho \, q_1 \, c_p \left( T_{1} - T_{ref} \right) - \rho \, q_2 \, c_p \left( T_{2} - T_{ref} \right) \\+ UA_{c} \left( T_c-T_2 \right) + UA_{a} \left( T_a-T_2 \right) + W_s$$

If the volume is not constant then a mass balance must also be added.

$$\frac{dm}{dt} = \dot m_{1} - \dot m_{2}$$

With `m = \rho V` and `\dot m = \rho q` the expression becomes:

$$\rho_2 \frac{dV}{dt} = \rho_1 q_1 - \rho_2 q_2$$

If the density `\rho` is constant for the inlet and outlet streams then the expression is simplified.

$$\frac{dV}{dt} = q_1 - q_2$$