Simple Reactors¶

Batch reactors¶

Cases¶

• TP: constant T, P
• TV: constant T, V
• HP: constant H, P (adiabatic, constant pressure)
• UV: constant U, V (adiabatic, constant volume)

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.
• 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