We will carry out an energy balance on the volume ΔV. There is no work done, i.e., , so Equation (11-10) becomes
The heat flow to the reactor, , is given in terms of the overall heat transfer coefficient, U, the heat exchange area, ΔA, and the difference between the ambient temperature Ta and the reactor temperature T.
where a is the heat exchange area per unit volume of reactor. For a tubular reactor
where D is the reactor diameter. Substituting for in Equation (12-1), dividing by ΔV, and then taking the limit as ΔV → 0, we get
Expanding
From a mole balance on species i, we have
Differentiating the enthalpy Equation (11-19) with respect to V
Substituting Equations (12-3) and (12-4) into Equation (12-2), we obtain
Rearranging, we arrive at
and
where
Qg = rAΔHRx
Qr = Ua(T – Ta)
which is Equation (T11-1G) in Table 11-1 on pages 482-483.
For exothermic reactions, we note that when the heat “generated,” Qg, is greater than the heat “removed” Qr (i.e., Qg > Qr), the temperature will increase down the reactor. When Qr > Qg the temperature will decrease down the reactor.
For endothermic reactions Qg will be a negative number and Qr will also be negative because Ta > T. The temperature will decrease if (–Qg) > (–Qr) and increase if (–Qr) > (–Qg).
Equation (12-5) is coupled with the mole balances on each species [Equation (11-33)]. Next, we express rA as a function of either the concentrations for liquid systems or molar flow rates for gas systems, as described in Chapter 4. We will use the molar flow rate form of the energy balance for membrane reactors and also extend this form to multiple reactions.
We could also write Equation (12-5) in terms of conversion by recalling Fi = FA0(Θi + viX) and substituting this expression into the denominator of Equation (12-5).
For a packed-bed reactor dW = ρb dV where ρb is the bulk density,
Equations (12-6) and (12-7) are also given in Table 11-1 as Equations (T11-1E) and (T11-1F). As noted earlier, having gone through the derivation to these equations it will be easier to apply them accurately to CRE problems with heat effects.
If the reaction is in gas phase and pressure drop is included, there are four differential equations that must be solved simultaneously. The differential equation describing the change in temperature with volume (i.e., distance) as we move down the reactor,
must be coupled with the mole balance
and with the pressure drop equation
and solved simultaneously. If the temperature of the heat exchange fluid, Ta, varies down the reactor, we must add the energy balance on the heat exchange fluid. In the next section we will derive the following equation for co-current heat transfer
along with the equation for counter-current heat transfer. A variety of numerical schemes can be used (e.g., Polymath) to solve these coupled differential equations (A), (B), (C), and (D).
For liquid phase reactions the rate is not a function of total pressure, so our mole balance is
Consequently, we need to only solve equations (A), (D), and (E) simultaneously.
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