Question

A liquid-phase irreversible reaction $A \rightarrow B$ occurs in a CSTR of volume 1 m$^{3}$ with rate $-r_{A}=164\, x_{A}$ kmol m$^{-3}$ h$^{-1}$. The reactor output is split ideally to obtain pure B in bottoms, while the distillate (free of B) is partly purged (10 kmol h$^{-1}$) and the rest recycled (200 kmol h$^{-1}$). At steady state, pure B product rate is 100 kmol h$^{-1}$. A small amount of inert $I$ is added with the fresh feed. Compute the inert feed rate into the process (kmol h$^{-1}$), rounded to two decimals.

Hint:

For inert species in recycle-purge systems, apply a simple steady-state inert balance: all inert must exit via the purge.

Answer 1

Correct Answer: 0.95 - 1.05

Step 1: Analyze the Splitter

The reactor effluent splits into:

• Product stream (pure B): 100 kmol/h
• Distillate stream (A + I, no B): splits into recycle (200 kmol/h) and purge (10 kmol/h)

Total distillate from splitter: $$F_{distillate} = F_{recycle} + F_{purge} = 200 + 10 = 210 \text{ kmol/h}$$

Step 2: Overall Material Balance Around the Entire System

Input streams: Fresh A feed ($F_A^{fresh}$) + Inert feed ($F_I$) Output streams: Product B (100 kmol/h) + Purge (10 kmol/h)

$$F_A^{fresh} + F_I = 100 + 10 = 110 \text{ kmol/h}$$

Step 3: Component Balance for B

All B is produced in the reactor and leaves as product: $$F_B^{product} = 100 \text{ kmol/h}$$

This equals the rate of reaction of A in the reactor: $$\text{Rate of B production} = (-r_A) \times V = 164 x_A \times 1 = 164 x_A \text{ kmol/h}$$

Therefore: $$164 x_A = 100$$ $$x_A = \frac{100}{164} = 0.6098$$

Step 4: Material Balance Around the Reactor

Let $F_{in}$ be the total molar flow into the reactor.

Input to reactor: Fresh A + Inert + Recycle Output from reactor: Goes to splitter

Since the recycle stream contains A and I (no B), and the reactor produces B:

Reactor inlet flow rate: $$F_{in} = F_A^{fresh} + F_I + F_{recycle}$$

Reactor outlet flow rate: $$F_{out} = F_{in}$$ (liquid phase, constant density)

The reactor outlet splits into:

• Pure B product: 100 kmol/h
• Distillate (A + I): 210 kmol/h

Therefore: $$F_{out} = 100 + 210 = 310 \text{ kmol/h}$$

Step 5: Component Balance for A Around Reactor

A entering reactor: $$F_A^{in} = F_A^{fresh} + F_{recycle} \times x_A^{recycle}$$

The recycle stream has the same composition as the reactor effluent (after removing B):

In the distillate (210 kmol/h), the mole fractions are: $$x_A^{distillate} = \frac{n_A}{n_A + n_I}$$

From reactor outlet (310 kmol/h total):

• Moles of A: $310 \times 0.6098 = 189.04$ kmol/h
• Moles of B: 100 kmol/h (leaves as product)
• Moles of I: $310 - 189.04 - 100 = 20.96$ kmol/h

Wait, let me reconsider. The reactor outlet contains A, B, and I. After the splitter:

• B product: 100 kmol/h (pure B)
• Distillate: 210 kmol/h (contains A and I, no B)

So in the reactor outlet:

• Total flow: 310 kmol/h
• B: 100 kmol/h
• A + I: 210 kmol/h
• Mole fraction of A: $x_A = 0.6098$

Therefore, moles of A in outlet: $310 \times 0.6098 = 189.04$ kmol/h

But this doesn't make sense because A + I = 210 kmol/h in the distillate.

Let me reconsider the mole fraction definition: $x_A$ is the mole fraction of A in the reactor.

Corrected Step 4: Reactor Composition

If reactor has total flow 310 kmol/h and $x_A = 0.6098$:

• Moles of A: $310 \times 0.6098 = 189.04$ kmol/h

This is inconsistent. Let me reconsider: the distillate should contain all the A and I from the reactor.

Actually, in the distillate (210 kmol/h), we have A and I. Moles of A in distillate = Moles of A in reactor outlet Moles of I in distillate = Moles of I in reactor outlet

Since the reactor outlet is 310 kmol/h with B = 100 kmol/h: $$n_A + n_I = 310 - 100 = 210 \text{ kmol/h}$$ ✓

Now, $x_A$ in the reactor (including B): $$x_A = \frac{n_A}{n_A + n_B + n_I} = \frac{n_A}{310} = 0.6098$$

Therefore: $$n_A = 310 \times 0.6098 = 189.04 \text{ kmol/h}$$

But $n_A + n_I = 210$, so: $$n_I = 210 - 189.04 = 20.96 \text{ kmol/h}$$

Step 6: Inert Balance

Inerts don't react, so inert in = inert out: $$F_I = n_I^{purge} = \frac{20.96}{210} \times 10 = 0.998 \text{ kmol/h}$$

Actually, the inert fraction in the distillate is: $$\frac{20.96}{210} = 0.0998$$

Inert in purge: $$F_I^{purge} = 10 \times 0.0998 = 0.998 \text{ kmol/h}$$

Since inerts accumulate but don't react: $$F_I = F_I^{purge} = 0.998 \approx 1.00 \text{ kmol/h}$$

Answer

The inert feed rate into the process is 1.00 kmol/h (rounded to two decimal places).