In a batch reactor maintained at constant volume, the holding time is given by equation (8.1.7):
t=C_{A0}\int_{0}^{0.97}\frac{df_{A}}{-r_{A}}=C_{A0}\int_{0}^{0.97}\frac{df_{A}}{kC_{A0}(1-f_{A})}
=\int_{0}^{0.97}\frac{df_{A}}{k(1-f_{A})} (A)
For isothermal operation at 163^{\circ} C the rate constant is equal to 0.8 h^{-1}. Thus,
\left |t=\frac{1}{0.8}\ln \left ( \frac{1}{1-f_{A}} \right ) \right |_{0}^{0.97}=4.38 hr
The total processing time per batch is the sum of the holding time and the times necessary to fill the reactor, heat the mixture to the reaction temperature, and drain the reaction products from the reactor, or
\tau _{batch}=4.38+\frac{10+14+12}{60}=4.98 h
The total processing time per batch is thus essentially equal to 5 h. While operating 7000 h, one will be able to process 7000/5, or 1400, batches. Each batch must contain 2,000,000/[0.97(1400)], or 1473 lb. Because the liquid has a density of 0.9 g/cm^{3}, the volume required will be 1473/[0.9 × 62.5], or 26.2 ft^{3}. This volume is equivalent to 196 gal.
The maximum heat flux that will have to be maintained to operate isothermally will be that generated at the start of the reaction.
\dot{Q}=\Delta H_{R}(rV_{R})=\Delta H_{R}(kC_{A0}V_{R})
or
\dot{Q}=\Delta H_{R}(kn_{A0})
In consistent units,
\Delta H=-83\frac{cal}{g}\times \frac{1\, Btu}{252\, cal}\times \frac{454\, g}{lb}=-149.5 Btu/lb
Thus,
\dot{Q}=(-149..5)(0.8)(1473)=-176,000 Btu/h
This cooling load is rather large for a system of this size. Kladko found it advantageous to go to a semibatch mode of operation so that incoming cold feed could be used to absorb some of the energy released by reaction and thereby drastically reduce the cooling requirements for the system. If the cold feed enters at room temperature, it acts as a heat sink that is capable of absorbing nearly 80% of the energy released by reaction. This fact enables one to consider operation in the autothermal mode discussed in Section 10.5.
We now consider operation of the batch reactor under adiabatic conditions. We will assume that we need not worry about reaching the boiling point of the liquid and that the rate of energy release by reaction does not become sufficiently great that an explosion ensues.
Equation (10.2.14) is valid for adiabatic operation
T=T_{0}+\frac{\Delta H_{R}f_{A}n_{A0}}{\nu _{A}\sum (n_{i}\bar{C}_{pi})}
In the present case,
T=436+\frac{-83f_{A}n_{A0}}{(-1)n_{A0}(0.5)}=436+166f_{A} (B)
where T is expressed in K. Note that the ultimate temperature rise will be more than 160 K.
From the Arrhenius relation, k=Ae^{-E_{a}/RT} and at 436 K, 0.8 = k=Ae^{-28,960/[1.987(436)]},so
k=2.61\times 10^{14}e^{-14,570/T} (C)
Equations (B) and (C) may now be substituted into design equation (A) to solve for the holding time required using any convenient numerical integration tool. Alternatively, in anticipation of later solving the plug flow analog of this batch reactor problem for different heat transfer conditions, it is convenient to employ the differential form of the material balance to determine the trajectories of conversion and temperature versus time:
\frac{df}{dt}=k(1-f_{A})
This relation may be used with equations (B) and (C) to incorporate the change in temperature (and thus the associated reaction rate) with fractional conversion. The present problem can be solved via machine computation by choosing a reaction time sufficient to obtain f_{A}=0.97. One may employ a differential equation solver such as those in Matlab, MathCad, or other engineering software. With either approach one finds that f_{A}=0.97. at t = 0.117 h or 7.0 min. Evolution of both the fraction conversion and the temperature of the reaction mixture as functions of time is shown in the two panels of Figure I10.1. Plots of the dependent variables as functions of the time for the batch reaction are displayed for operation of the reactor in the isothermal mode and in the adiabatic mode.
The time necessary to accomplish this exothermic reaction under adiabatic operating conditions is only an extremely small fraction of that necessary for isothermal operation. In fact, the times necessary to fill and drain the reactor and to heat it to a temperature where the rate becomes appreciable will be greater than that necessary to accomplish the reaction. Thus,
\tau _{batch, adiabatic}=0.12+\frac{10+12+14}{60}=0.72 h
This time corresponds to an upper limit because the reaction (and release of thermal energy) will begin in the hot feed stream as it approaches the reaction temperature even before entering the adiabatic region of the reactor.
While operating 7000 h, one will be able to process 7000/0.72, or 9722 batches per year. Each batch will then contain 2 × 106/[0.97(9722)], or 212 lb. The requisite reactor volume will then be 28 gal. Under these circumstances the times necessary to perform the nonreactive operations would undoubtedly be somewhat reduced. One should recognize that these steps will be the bottlenecks in this operation. The required reactor size is definitely on the small side, and it would be preferable to operate in a larger reactor and process a smaller number of batches per year with attendant reductions in labor requirements. A flow reactor would be an attractive operating mode that would avoid separate steps for the nonreactive operations, and several alternative approaches of this type are considered in later illustrative examples in this chapter.
