## Step-by-Step Solution Let's break down the problem: ### **Given Data** - **Reactor Volume, \( V \) = 6 m³** - **Current Conversion, \( X = 55\% = .55 \)** - **Reaction:** \( A \rightarrow R \) (First Order) - **Current reactor is poorly mixed.** - **Pulse tracer suggests the flow is as shown:** - **Inlet flow, \( Q \)** - **Bypass: \( \frac{1}{3} Q \)** - **Rest goes to the tank** ### **Asked:** - **If we replace the stirrer to ensure the tank is perfectly mixed (CSTR), what will the new conversion be?** --- ## **Step 1: Analyze Current System** The flow diagram shows that **1/3 of the flow bypasses the tank** and **2/3 enters the tank**. So, - **Bypass flow:** \( Q_{bypass} = \frac{1}{3}Q \) - **Flow through tank:** \( Q_{tank} = Q - Q_{bypass} = Q - \frac{1}{3}Q = \frac{2}{3}Q \) --- ## **Step 2: Conversion in the Current System** For a first-order reaction in a CSTR: \[ X = \frac{kV}{Q + kV} \] But currently, only **2/3 of the flow passes through the tank**, rest bypasses (**no conversion in bypass**). Let \( X_{tank} \) be the conversion in the tank. The **overall conversion** is: \[ X_{overall} = \text{fraction through tank} \times X_{tank} + \text{fraction bypass} \times \] \[ X_{overall} = \frac{2}{3} X_{tank} + \frac{1}{3}() \] \[ X_{overall} = \frac{2}{3} X_{tank} \] Given: \[ X_{overall} = .55 \] \[ \frac{2}{3} X_{tank} = .55 \] \[ X_{tank} = \frac{.55}{2/3} = .825 \] --- ## **Step 3: Find the Rate Constant \( k \)** From the CSTR formula: \[ X_{tank} = \frac{kV}{Q_{tank} + kV} \] But \( Q_{tank} = \frac{2}{3} Q \): \[ .825 = \frac{k \cdot 6}{\frac{2}{3}Q + k \cdot 6} \] Solve for \( k \) in terms of \( Q \): \[ .825 \left(\frac{2}{3}Q + 6k\right) = 6k \] \[ .825 \cdot \frac{2}{3}Q + .825 \cdot 6k = 6k \] \[ .825 \cdot \frac{2}{3}Q = 6k - .825 \cdot 6k \] \[ .825 \cdot \frac{2}{3}Q = 6k (1 - .825) \] \[ .825 \cdot \frac{2}{3}Q = 6k (.175) \] \[ k = \frac{.825 \cdot \frac{2}{3}Q}{6 \cdot .175} \] --- ## **Step 4: Perfectly Mixed Tank (CSTR, Full Flow)** Now, **all flow \( Q \) enters the tank**. Again, for a CSTR: \[ X_{new} = \frac{kV}{Q + kV} \] Substitute \( k \) from above: First, simplify \( k \): \[ k = \frac{.825 \cdot \frac{2}{3}}{6 \cdot .175} Q \] Calculate the numerator: \[ .825 \cdot \frac{2}{3} = .825 \cdot .6667 \approx .55 \] So, \[ k = \frac{.55}{6 \cdot .175} Q \] \[ 6 \cdot .175 = 1.05 \] \[ k = \frac{.55}{1.05} Q \approx .524 Q \] --- Now, use this in the CSTR formula: \[ X_{new} = \frac{kV}{Q + kV} \] \[ V = 6 \] \[ k = .524 Q \] So, \[ X_{new} = \frac{.524 Q \cdot 6}{Q + .524 Q \cdot 6} \] \[ = \frac{3.144 Q}{Q + 3.144 Q} \] \[ = \frac{3.144 Q}{4.144 Q} \] \[ = \frac{3.144}{4.144} \] \[ = .759 \] --- ## **Step 5: Final Answer** ### **If we replace the stirrer with one powerful enough to ensure mixed flow, the conversion will increase from 55% to 76%** --- ## **Summary Table** | Scenario | Conversion (\(X\)) | |------------------------|-------------------| | Poor mixing | 55% | | Perfect mixing (CSTR) | **76%** | --- ### **Final Answer:** > **If we ensure perfect mixing (CSTR), the conversion will be approximately 76%.**