#### Isothermal Semibatch Reactors: Screencast

Overview of situations where the semibatch reactor has advantages over other types of reactors. The mole balance for the reactor is briefly presented.

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##### Important Equations:

Mass Balances on isothermal, semibatch reactor for the irreversible reactor $$A + B \rightarrow 2C$$ whose rate is $$n^{th}$$ order in $$C_A$$ and $$m^{th}$$ order in $$C_B$$

Accumulation = in – out +rate of generation $\frac{dN_A}{dt} = F_{A_{0}} – F_A -kC_A^n C_B^m V$  $\frac{dN_B}{dt} = F_{B_{0}} – F_B -kC_A^n C_B^m V$  $\frac{dN_C}{dt} = F_{C_{0}} – F_C +2kC_A^n C_B^m V$where $$N_i$$ is the number of moles of component $$i$$ in the reactor time, $$t, F_i$$ are the molar flow rates (mol/s) of component $$i$$ leaving the reactor, $$F_{i0}$$ are the molar flow rates of component $$i$$ entering the reactor, $$V$$ is the volume of the reactor contents(L), $$k$$ is the rate constant, $$C_A$$ is the molar concentration of A (mol/L), and $$C_B$$ is molar concentration of B (mol/L).

Note that the reactor volume changes with time, so an equation for the volume as a function if time is needed:

$\frac{dV}{dt} = v_0 – v$

where $$v_0$$ is the inlet volumetric flow rate (L/s) and $$v$$ is the outlet volumetric flow rate (L/s). The volumetric flow rates can be functions of time.

Note that some of these molar flow rates will be zero for most semibatch reactors. For example, a typical semibatch might have $$F_A, F_B$$ and $$F_C$$ equal to zero (no flow out of the reactor) and $$F_{A0}$$ and $$F_{C0}$$ equal to zero. The reactor starts with A in the reactor and then B is added as a function of time.

These equations use the relations: $F_A = vC_A,\,\,\, F_B = vC_B,\,\,\, F_C = VC_C$ $F_{A0} = v_0C_{A0},\,\,\, F_{B0} = v_0C_{B0},\,\,\, F_{C0} = v_0C_{C0}$ where $$v_0$$ is the inlet volumetric flow rate (L/s), $$v$$ is the outlet volumetric flow rate (L/s) and $$C_{A0}$$, $$C_{B0}$$, and $$C_{C0}$$ are the inlet flow rate of A, B, and C.

These differential equations are solved with initial conditions:

$at \, t = 0, N_A = N_{A0},\,\,\, N_B = N_{B0},\,\,\, N_C = N_{C0},\,\,\, V=V_0$

where $$N_{i0}$$ is the number of moles in the reactor at t = 0, and $$v_0$$ is the initial volume in the reactor.

Conversion needs to be defined carefully for a semibatch reactor and is usually not a good variable to use in solving semibatch reactor problems. If the reactor starts with A in the reactor and only B is added, the fractional conversion of A can be calculated as:

$X = (N_{A0} – N_A)/N_{A0}$

where $$X$$ is the fractional conversion.

Percent conversion = $$100X$$