# Simple Reactors¶

### Species equation¶

Simple mass balance, where $m_i$ is the mass of species $i$, $V$ is the reactor volume (possibly changing in time), and $\dot{m}_i'''$ is the net species mass production rate per unit volume. $$\frac{dm_i}{dt} = \dot{m}_i'''V.$$ Now, $m_i= my_i$, where $y_i$ is mass fraction and $m$ is the total mass of the reactor, which is constant. Note, if the reactor volume is constant, then so is $\rho$. This gives $$\frac{dmy_i}{dt} = m\frac{dy_i}{dt} = \dot{m}_i'''V,$$ or $$\frac{dy_i}{dt} = \frac{\dot{m}_i'''}{\rho}.$$

This equation holds for any of the four cases listed.

- We can write $\dot{m}_i''' = \dot{\omega}_iM_i$, where $\omega_i$ is a species molar production rate with units of kmol/m$^3$*s. Cantera provides $\omega_i$ via the property
`net_production_rates`

.

### Energy equations¶

- An energy equation is not needed for cases TP, or TV. For cases HP and UV, in a code like Cantera, we can set the gas state as either
`gas.HPY = H, P, Y`

, or`gas.UVY = U, V, Y`

, where $V$ is $1/\rho$.- Temperature is then available simply as
`T = gas.T`

, where inverting $H(T)=H$ to solve for $T$ is done internally by Cantera.

- Temperature is then available simply as
- Otherwise, we can simply write $$\frac{dh}{dt} = 0,$$ or $$\frac{du}{dt} = 0.$$

#### Temperature equation¶

Consider $dh/dt=0$. $$h = h(T, y_i),$$ $$dh = \underbrace{\frac{\partial h}{\partial T}}_{c_p}dT = \sum_i\underbrace{\frac{\partial h}{\partial y_i}}_{h_i}dy_i.$$ Then divide through by $dt$ and solve for $dT/dt$ with $dh/dt = 0$. $$\frac{dT}{dt} = -\frac{1}{\rho c_p}\sum_ih_i\dot{m}_i'''.$$

For $du/dt=0$, we have $$\frac{dT}{dt} = -\frac{1}{\rho c_v}\sum_iu_i\dot{m}_i'''.$$

## CSTR¶

- Here we have an unsteady reactor with an inlet and outlet stream. The reactor is well mixed so the outlet composition is the same as in the reactor. ### Species equation.
- The mass balance in this case gives: $$\frac{dm_i}{dt} = \dot{m}_i^{in} - \dot{m}_i^{out} + \dot{m}_i'''V.$$
- Assume the mass in the reactor is constant in time (this is not true in general). Also assume that $\dot{m}^{in}=\dot{m}^{out}$, which may also not be true.

$$\frac{dy_i}{dt} = \frac{y_i^{in}-y_i}{\tau} + \frac{\dot{m}_i'''}{\rho}.$$

- Here, $\tau=m/\dot{m}$ is a reactor residence time.
- When solved to steady state, the resulting composition is independent of the two assumptions listed above.
- If we set $\frac{dy_i}{dt}=0$, we can solve for species compositions using a nonlinear algebraic solver, rather than time advance the ODEs given using an ODE solver. The ODE solution will be more robust to convergence.

### Energy equations¶

- An enthalpy equation can be written as $$\frac{dh}{dt} = \frac{h^{in}-h}{\tau}.$$
- A similar equation for internal energy can be formulated, as well as a temperature equation using the approach shown above.
- At steady state, we simply have $h-h^{in}$.

### CSTR versus batch¶

- Note, the CSTR equations for species and enthalpy are very similar to the respective equations for batch reactor, but with the additional mixing term involving $\tau$.
- Hence, a single solver can be written to solve both types of reactors.
- It is common to solve a CSTR to steady state, in which case, some measure of the approach to steady state is needed.

### Example code¶

- Solve a Batch reactor HP problem

```
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
from scipy.integrate import odeint
import cantera as ct
```

```
#-------------------- setup
P = 101235 # Pa
T0 = 1500 # K
x0 = "CH4:1, O2:2, N2:7.52" # -
trun = 0.005 # run time (s)
nt = 1000 # number of times to record
#-------------------- define rate function to integrate
def rhsf(y, t):
gas.HPY = h0, P, y
return gas.net_production_rates * gas.molecular_weights / gas.density
#-------------------- initial conditions and constant enthalpy
gas = ct.Solution('gri30.yaml')
gas.TPX = T0, P, x0
y0 = gas.Y
h0 = gas.h
#-------------------- solve problem
t = np.linspace(0, trun, nt)
y = odeint(rhsf, y0, t)
#-------------------- recover the temperature profile corresponding to y(t)
T = np.empty(nt)
for i in range(nt):
gas.HPY = h0, P, y[i,:]
T[i] = gas.T
#-------------------- plot results
plt.rc('font', size=14)
plt.figure(figsize=(8,5))
plt.plot(t, T)
plt.xlabel('t (s)')
plt.ylabel('T (K)');
plt.figure(figsize=(8,5))
plt.plot(t, y[:,gas.species_index("CH4")])
plt.plot(t, y[:,gas.species_index("O2")])
plt.plot(t, y[:,gas.species_index("CO2")])
plt.plot(t, y[:,gas.species_index("CO")])
plt.legend(['CH4', 'O2', 'CO2', 'CO'], frameon=False)
plt.xlabel('t (s)')
plt.ylabel('y')
plt.xlim([0,trun]);
```

```
```