How do I solve the involved differential and integral equations to find X in both cases so that i can slot in the given values to find the final answer, I dont want to make any mistakes, hence $,$ want to be sure about the steps. Please simplify the steps as much as possible, thank you.

Part $($a$)$: With Pressure Drop

For a packed bed reactor with a pressure drop, the conversion calculation involves using the following equations:

Materlal Balance for the Reactor

The material balance for an elementary reaction $A+BC+D$ in a packed bed reactor considering pressure drop is given by:

$\frac{d{F}_A}{d(W)}=-r_A$

where $F_A$ is the molar flow rate of $A,$ and $r_A$ is the rate of reaction. Since the reaction is elementary:

$r_A=k_A{C}_A{C}_B$

Expression for Conversion

Let $x$ be the conversion of $A.$ Then:

$C_A=C_{A0}(1-x)$

$C_B=C_{B0}-C_{A0}x$

Thus:

$r_A=k_A{C}_{A0}(1-x)(C_{B0}-C_{A0}x)$ Pressure Drop Effect

With a pressure drop, the flow rate decreases due to the pressure drop, and the modified volumetric flow rate is:

$u=u_0(1-*\frac{W}{})$

where $$ is the density of the bed material, and the effective flow rate $u$ affects the reactor performance.

Integration for Conversion

To find $x,$ solve:

$\frac{d{F}_A}{d(W)}=-k_A{C}_A{C}_B$

Substitute $F_A=u_0{C}_A$ :

$\frac{d(u_0{C}_A)}{d(W)}=-k_A{C}_A{C}_B$

Rearrange and integrate to find $x.$

Part $($b$)$: Without Pressure Drop

When there is no pressure drop $(=0),$ the volumetric flow rate remains constant. The conversion calculation simplifies to:

$\frac{d{F}_A}{d(W)}=-k_A{C}_A{C}_B$

Substitute $F_A=u_0{C}_A$ and solve as before:

$\frac{d(u_D{C}_A)}{d(W)}=-k_A{C}_A{C}_B$

Solution Steps

Calculate Converslon with Pressure Drop:

Use the material balance with the pressure drop term included.

Solve the resulting differential equation to find $x.$

Calculate Converslon without Pressure Drop:

Use the material balance with $=0.$

Solve the differential equation to find $x.$