Heat-Removed Term, R T

Steady-Stat Non isothermal Reactor Design—Flow Reactors with Meat Exchange Chapter 12 

On the other hand, if T Tq, the intercept will move to the right as k increases. 

The heat-generated lerni. Equation (12-29), can be written in terms of conversion. [Recall X = 

To obtain a plot of heat generated. G(7), as a function of temperature, we must solve for Af as a function of T using the CSTR mole balance, the rate law, and stoichiometry. For example, for a first-order liquid-phase reaction, the CSTR mole balance becomes 

Finally, substituting for k in ternis of the Arrhenius equation, we obtain 

---

We search for the solution(s) to this equation. Note that heat removal term R(T) is linear in T, while the heat generation term G(T) contains k(T) and is highly nonlinear. The method we will use to solve this equation is to construct a graph of the left- and right-hand sides and look for intersections that must be the roots of the equation or T solutions to the CSTR. We do this in a plot of R(T) and G(T) versus T. 

The heat removal term Q has to be inserted in the one-dimensional energy balance by implicitly solving the ener balance in the radial direction. The heat flux in the radial direction is = —kf dT/d R, and we replace the radial flux to the wall by assuming that the fluid at any position j is at a constant temperature T while the wall is at temperature to obtain... 

The right-hand side of Equation (8-67) is referred to as the heat-removed term (by flow and heat exchange) R T) ... 

At steady state, this generation term must be exactly balanced by a removal term, which is the sum of the two terms on the right-hand side of Eqn. (8-31). Let s label this sum R(T), the total rate of heat removal. The term UA ST — T fi) is the rate at which heat is... 

If Qr = 0, r = 1,.. . R, so that no heat is added to or removed from the reactor r by direct interchange, we may speak of the sequence of reactors as adiabatic. Then only 2R operating quantities can be chosen, 0, X, and Th+i. This follows from the fact that with Qr = 0 Eqs. (1) and (2) can be solved for c and T, in terms of c +i,Tr+i when only Xr+, and dr are prescribed. The case of an adiabatic sequence with a single reaction admits of an elementary exposition, and this we will now give. 

---