Give step-by-step solution with explanation and final answer:
(give me answer without using bullet-points)An exothermic, reversible, liquid-phase reaction A = B is carried out in a non-isothermal Plu

Question

Give step-by-step solution with explanation and final answer:
(give me answer without using bullet-points)An exothermic, reversible, liquid-phase reaction A = B is carried out in a non-isothermal Plug Flow
Reactor (PFR) surrounded by a cooling jacket. Pure A enters the reactor at a volumetric flow rate of
vo = 10 dm*min, with an inlet concentration Cx = 2 mol/dm? and inlet temperature Ty = 320 K. The
reaction rate law is given by -r = k;(T)Cy - ky(T)Cp, where k; = A exp(—E{/RT) and k, = Azexp(-E/RT).
The heat of reaction is AH, = —50 kJ/mol (assumed constant). The cooling medium flows
countercurrent to the reactant stream with a constant temperature 7c = 290 K. The overall heat transfer
coefficient is U= 500 J/(m?sK) and the heat transfer area per unit reactor volume is a = 20 m™!. The density
of the reaction mixture is p = 1 kg/dm? and its specific heat capacity is C;, =4000 J/(kgK). The reactor has a
length L = 5 m and a diameter D = 0.1 m.
Determine the conversion (X) and temperature (T) profiles along the length of the reactor by formulating and
solving the coupled differential mole and energy balances.
Reactor (PFR) Cooling Jacket
\ i
Feed: vy, Cao, Tg A=B beonm

VIPSolutions · Accepted Answer

Let's break down the solution step by step, focusing on setting up the coupled **mole** and **energy balance** equations for the non-isothermal PFR (Plug Flow Reactor):

---

## **Step 1: Mole Balance for Species A**

For a PFR, the mole balance on A is:
$$
\frac{dF_A}{dV} = r_A
$$
where $ F_A = C_A v_ $, and $ r_A $ is the rate of consumption of A (note: $ r_A <  $).  
Given rate law:
$$
-r_A = k_1(T)C_A - k_2(T)C_B
$$
with
$$
k_1(T) = A_1 \exp\left(-\frac{E_1}{RT}\right)\
k_2(T) = A_2 \exp\left(-\frac{E_2}{RT}\right)
$$

Define **conversion** $ X $ at any position:
$$
X = \frac{C_{A} - C_A}{C_{A}}
$$
So,
$$
C_A = C_{A}(1-X)
$$

For a 1:1 stoichiometry, $ C_B = C_{A}X $.

Substitute into the rate law:
$$
-r_A = k_1(T)C_{A}(1-X) - k_2(T)C_{A}X
$$

Plug into mole balance:

$$
\frac{dF_A}{dV} = r_A
$$
But $ F_A = C_A v_ = v_ C_{A}(1-X) $, so:

$$
\frac{d}{dV}[v_ C_{A}(1-X)] = r_A
$$

Since $ v_ $ is constant (liquid phase, constant density):

$$
-v_ C_{A} \frac{dX}{dV} = -k_1(T)C_{A}(1-X) + k_2(T)C_{A}X
$$
$$
\frac{dX}{dV} = \frac{k_1(T)(1-X) - k_2(T)X}{v_}
$$

---

## **Step 2: Energy Balance**

The general energy balance for a differential element $ dV $ is:

$$
\frac{d}{dV}(\rho v_ C_p T) = -r_A(-\Delta H_{rxn}) - Ua(T - T_C)
$$

Since $ \rho $ is constant and $ v_ $ is constant:

$$
\rho v_ C_p \frac{dT}{dV} = -r_A(-\Delta H_{rxn}) - Ua(T - T_C)
$$

Recall $ r_A $ (already derived above):

$$
r_A = -[k_1(T)C_{A}(1-X) - k_2(T)C_{A}X]
$$

So the energy balance becomes:

$$
\rho v_ C_p \frac{dT}{dV} = [k_1(T)C_{A}(1-X) - k_2(T)C_{A}X](-\Delta H_{rxn}) - Ua(T - T_C)
$$

---

## **Step 3: Substitute Known Values**

Given values:

- $ v_ = 10 $ dm³/min = $ 1.67 \times 10^{-4} $ m³/s (since 1 dm³ = 10⁻³ m³, 1 min = 60 s)
- $ C_{A} = 2 $ mol/dm³ = $ 200 $ mol/m³
- $ T_ = 320 $ K
- $ \Delta H_{rxn} = -50 $ kJ/mol = $-50,000$ J/mol
- $ T_C = 290 $ K
- $ U = 500 $ J/(m²·s·K)
- $ a = 20 $ m⁻¹
- $ \rho = 1 $ kg/dm³ = 100 kg/m³
- $ C_p = 400 $ J/(kg·K)
- $ L = 5 $ m, $ D = .1 $ m

---

## **Step 4: Write Final ODEs**

Define $ V = \text{reactor volume} $, but for a PFR, $ dV = A dx $, with $ x $ the length coordinate. The total reactor volume is $ V_{tot} = \pi(D^2/4)L $.

So the coupled ODEs are:

$$
\frac{dX}{dV} = \frac{k_1(T)(1-X) - k_2(T)X}{v_}
$$

$$
\frac{dT}{dV} = \frac{[k_1(T)(1-X) - k_2(T)X](-\Delta H_{rxn}) - Ua(T - T_C)}{\rho v_ C_p}
$$

Where:
$$
k_1(T) = A_1 \exp\left(-\frac{E_1}{RT}\right)
$$
$$
k_2(T) = A_2 \exp\left(-\frac{E_2}{RT}\right)
$$

Initial conditions at $ V =  $:
$$
X() = , \quad T) = T_ = 320\,K
$$

---

## **Step 5: Final Answer (System to Solve)**

The conversion $ X $ and temperature $ T $ profiles are determined by **numerically integrating** the following coupled ODEs from $ V =  $ to $ V = V_{tot} $:

$$
\boxed{
\begin{align*}
\frac{dX}{dV} &= \frac{k_1(T)(1-X) - k_2(T)X}{v_} \[1.5em]
\frac{dT}{dV} &= \frac{[k_1(T)(1-X) - k_2(T)X](-\Delta H_{rxn}) - Ua(T - T_C)}{\rho v_ C_p}
\end{align*}
}
$$
with initial conditions:
$$
X() = ,\quad T() = 320\, \text{K}
$$

Where each parameter is known or defined above. These equations, along with the rate constants' temperature dependence, are what you solve (typically with a computational tool) to get $ X_{out} $ and $ T_{out} $ at the reactor exit ($ V = V_{tot} $).

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