Energy balance on a reactor

Let us consider a monophasic system where a single chemical reaction occurs in a reactor in continuous mode.

The mass balance for a species \(i\) writes:

\[ \dot{n}_i^{IN}+\nu_i\cdot \dot{\xi} = \dot{n}_i^{OUT} \] Multiplying by the partial molar enthalpy of each species at the reactor temperature, one gets:

\[ \dot{n}_i^{IN}\cdot \bar{h}_i^{R}+\nu_i\cdot \bar{h}_i^{R} \cdot\dot{\xi} = \dot{n}_i^{OUT} \cdot \bar{h}_i^{R} \]

If one sums this equation for all \(i\) species:

\[ \sum_i \dot{n}_i^{IN}\cdot \bar{h}_i^{R}+\Delta_rH\left(T^R\right) \cdot\dot{\xi} = \sum_i \dot{n}_i^{OUT} \cdot \bar{h}_i^{R} \]

The mass balance can be rewritten:

\[ \sum_i \dot{n}_i^{IN}\cdot \left(\bar{h}_i^{R}-\bar{h}_i^{IN}+\bar{h}_i^{IN}\right)+\Delta_rH \left(T^{R}\right) \cdot\dot{\xi} = \sum_i \dot{n}_i^{OUT} \cdot \left(\bar{h}_i^{R}-\bar{h}_i^{OUT}+\bar{h}_i^{OUT}\right) \] That is the same as:

\[ \sum_i \dot{n}_i^{IN}\cdot \left(\bar{h}_i^{R}-\bar{h}_i^{IN}\right) + \sum_i \dot{n}_i^{IN}\cdot \bar{h}_i^{IN}+\Delta_rH \left(T^{R}\right) \cdot\dot{\xi} = \sum_i \dot{n}_i^{OUT} \cdot \left(\bar{h}_i^{R}-\bar{h}_i^{OUT}\right) + \sum_i \dot{n}_i^{OUT} \cdot \bar{h}_i^{OUT} \] Using Euler’s theorem, one can then write:

\[ \sum_i \dot{n}_i^{IN}\cdot \left(\bar{h}_i^{R}-\bar{h}_i^{IN}\right) + \dot{H}^{IN} +\Delta_rH \left(T^{R}\right) \cdot\dot{\xi} = \sum_i \dot{n}_i^{OUT} \cdot \left(\bar{h}_i^{R}-\bar{h}_i^{OUT}\right) + \dot{H}^{OUT} \] Injecting the heat capacity expression:

\[ \sum_i \dot{n}_i^{IN}\cdot \bar{c}_{P,i}^{IN} \cdot \left(T^R-T^{IN}\right) + \dot{H}^{IN} +\Delta_rH \left(T^{R}\right) \cdot\dot{\xi} = \sum_i \dot{n}_i^{OUT} \cdot \bar{c}_{P,i}^{OUT} \cdot \left(T^R-T^{OUT}\right) + \dot{H}^{OUT} \] Applying Euler’s theorem again, one can write:

\[ \dot{n}^{IN}\cdot c_P^{IN} \cdot \left(T^R-T^{IN}\right) + \dot{H}^{IN} +\Delta_rH \left(T^{R}\right) \cdot\dot{\xi} = \dot{n}^{OUT}\cdot c_P^{OUT} \cdot \left(T^R-T^{OUT}\right) + \dot{H}^{OUT} \]

The energy balance writes:

\[ \dot{H}^{IN} + \dot{Q} = \dot{H}^{OUT} \] Thus the mass balance writes:

\[ \dot{n}^{IN}\cdot c_P^{IN} \cdot \left(T^R-T^{IN}\right) +\Delta_rH \left(T^{R}\right) \cdot\dot{\xi} = \dot{n}^{OUT}\cdot c_P^{OUT} \cdot \left(T^R-T^{OUT}\right) + \dot{Q} \]

Multiplying by \(-1\), the “practical” form is obtained:

\[ \dot{n}^{IN}\cdot c_P^{IN} \cdot \left(T^{IN}-T^{R}\right) + \dot{Q} - \Delta_rH \left(T^{R}\right) \cdot\dot{\xi} = \dot{n}^{OUT}\cdot c_P^{OUT} \cdot \left(T^{OUT}-T^{R}\right) \]