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#### 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.

We suggest you list the important points in this screencast as a way to increase retention.

##### 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 = $$100X$$