Chapter 12: Steady-State Nonisothermal Reactor Design: Flow Reactors with Heat Exchange

Topics

Overview of User Friendly Energy Balance Equations
Evaluating the Heat Exchanger Term
Multiple Steady States
Multiple Reactions with Heat Effects
Applications of the PFR/PBR User Friendly Energy Balance Equations

User Friendly Energy Balance Equations

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The user friendly forms of the energy balance we will focus on are outlined in the following table.

User friendly equations relating X and T, and F_i and T

1. Adiabatic CSTR, PFR, Batch, PBR achieve this:

(1.A)

(1.B)

2. CSTR with heat exchanger, UA(T_a-T) and large coolant flow rate.

(2)

3 . PFR/PBR with heat exchange

3A. In terms of conversion, X

(3.A)

3B. In terms of molar flow rates, F_i

(3.B)

4. For Multiple Reactions

(4)

5. Coolant Balance

Co-Current Flow

(5)

These equations are derived in the text. These are the equations that we will use to solve reaction engineering problems with heat effects.

Evaluating the Heat Exchanger Term

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Energy transferred between the reactor and the coolant:

Assuming the temperature inside the CSTR, T, is spatially uniform:

At high coolant flow rates the exponential term will be small, so we can expand the exponential term as a Taylor Series, where the terms of second order or greater are neglected, then:

Since the coolant flow rate is high, T_a1T_a2T_a:

Multiple Steady States (MSS)

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CSTR with Heat Effects

	From page 593 we can obtain




where

Finding when R(T) = G(T)

Now we need to find X. We do this by combining the mole balance, rate law, Arrhenius Equation, and stoichiometry.

For the first-order, irreversible reaction A --> B, we have:
where
At steady state:
Substituting for k...

Generating G and R verse T: Single Reaction

Multiple Reactions with Heat Effects

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To account for heat effects in multiple reactions, we simply replace the term (-delta H_RX) (-r_A) in equations (12-35) PFR/PBR and (12-40) CSTR by:

PFR/PBR

CSTR

These equations are coupled with the mole balances and rate law equations discussed in Chapter 6.

Textbook Example (Alternative Solution)

Complex Reactions

Example: Consider the following gas phase reactions

Combined Mole Balance, Rate Law, Stoichiometry, and Energy Balance:

We now substitute the various parameter values (e.g. delta H_RX, E, U) into equations (1)-(13) and solve simultaneously using Polymath.

Applications of the PFR/PBR User Friendly Energy Balance Equations

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NOTE: The PFR and PBR formulas are very similar.



Heat exchange for a PFR:
a = heat exchange area per unit volume of reactor; for a tubular reactor, a = 4/D

Catalyst weight is related to reactor volume by:
Heat exchange for a PBR:
Steady State Energy Balance (with no work):





Final Form of the Differential Equations in Terms of Conversion:
A.
Final Form in terms of Molar Flow Rates
B.




If we include pressure drop:
C. $\frac{dp}{dW} = -\frac{\alpha}{2p}\frac{F_{T}}{F_{T0}}\frac{T}{T_{0}} = \frac{-\alpha(1+\epsilon X)}{2p}(\frac{T}{T_{0}}) = h(X,T)$
Note: the pressure drop will be greater for exothermic adiabatic reactions than it will be for isothermal reactions

Balance on Heat Exchanger Coolant Solve simultaneously using an ODE solver (Polymath/MatLab). If Ta is not constant, then we must add an additional energy balance on the coolant fluid:
Co-Current Flow
Counter-Current Flow	with T_a = T_ao at W = 0

For an exothermic reaction: with counter current heat exchange

A Trial and Error procedure for counter current flow problems is required to find exit conversion and temperature.

Consider an exothermic reaction where the coolant stream enters at the end of the reactor at a temperature T_a0, say 300 K.
Assume a coolant temperature at the entrance (X = 0, V = 0) to the reactor T_a2 =340 K.
Calculate X, T, and T_a as a function of V. We can see that our guess of 340 K for T_a2 at the feed entrance (X = 0) gives a coolant temperature of 310 K, which does not match the actual entering coolant temperature of 300 K.
Now guess a coolant temperature at V = 0 and X = 0 of 330 K. We see that the exit coolant temperature of T_a2 = 330 K will give a coolant temperature at V = V₁ of 300 K.

Finding MSS for an Endothermic Reaction

Multiple Reactions in a PFR with Variable Coolant Temperature

A ↔ B Liquid Phase Adiabatic

A ↔ B Liquid Phase Constant T_a

A ↔ B Liquid Phase Variable T_a, Co-Current

A ↔ B Liquid Phase Variable T_a, Counter Current

Sketch the Ambient Temperature as a function of V.

Elementary Liquid Phase Reaction

Exothermic, Reversible Reaction

Adiabatic Reaction in a PBR.

PBR with heat exchange.

PBR with heat exchange and variable coolant flow rate.

Use this when working with the Variable Coolant Temperature Living Example Problem.

Nonisothermal Reactions.

Objective Assessment of Chapter 12

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Chapter 12: Steady-State Nonisothermal Reactor Design: Flow Reactors with Heat Exchange

Topics

PFR/PBR

CSTR

Complex Reactions

$\frac{dp}{dW} = -\frac{\alpha}{2p}\frac{F_{T}}{F_{T0}}\frac{T}{T_{0}} = \frac{-\alpha(1+\epsilon X)}{2p}(\frac{T}{T_{0}}) = h(X,T)$

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Chapter 12: Steady-State Nonisothermal Reactor Design: Flow Reactors with Heat Exchange

Topics

PFR/PBR

CSTR

Complex Reactions

dpdW=−α2pFTFT0TT0=−α(1+ϵX)2p(TT0)=h(X,T) \frac{dp}{dW} = -\frac{\alpha}{2p}\frac{F_{T}}{F_{T0}}\frac{T}{T_{0}} = \frac{-\alpha(1+\epsilon X)}{2p}(\frac{T}{T_{0}}) = h(X,T)

$\frac{dp}{dW} = -\frac{\alpha}{2p}\frac{F_{T}}{F_{T0}}\frac{T}{T_{0}} = \frac{-\alpha(1+\epsilon X)}{2p}(\frac{T}{T_{0}}) = h(X,T)$