#### 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 = \(100X\)