Absolutely, let’s work through this step-by-step, like we’re at the whiteboard together. **Let’s tackle each part one by one.** --- ### Given: - **Feed:** 40,000 kg/hr, 7.5 wt.% NaOH, at **60°C** - **Product:** 45 wt.% NaOH - **Evaporator steam:** Saturated, **720 kPa** - **Condenser:** **14 kPa** - **Overall U:** 2400 W/(m²·K) --- ## a) **Mass flowrates and temperatures** **Let’s do mass balances first.** ### **1. Mass Balance (NaOH)** Let’s call: - Feed: \( F \) - Product: \( P \) - Vapour: \( V \) \[ \begin{align*} F &= 40,000~\text{kg/hr} \\ w_F &= 0.075 \\ w_P &= 0.45 \end{align*} \] \[ \text{NaOH in feed} = F \times w_F = 40,000 \times 0.075 = 3,000~\text{kg/hr} \] \[ \text{NaOH in product} = P \times w_P \] \[ \text{NaOH in product} = \text{NaOH in feed} \implies P = \frac{3,000}{0.45} = 6,667~\text{kg/hr} \] \[ \text{Water balance:}~F = P + V \implies V = F - P = 40,000 - 6,667 = 33,333~\text{kg/hr} \] ### **2. Convert to kg/s** \[ 1~\text{hr} = 3600~\text{s} \] \[ F = \frac{40,000}{3600} = 11.11~\text{kg/s} \] \[ P = \frac{6,667}{3600} = 1.85~\text{kg/s} \] \[ V = \frac{33,333}{3600} = 9.26~\text{kg/s} \] ### **3. Temperatures** - **Feed:** 60°C - **Vapour:** Boiling point of solution at evaporator pressure. (Assume near atmospheric, for simplicity, say ~100°C for dilute solution. For exact, need boiling point elevation data.) - **Product:** Same as vapour (both leave at boiling T, so ~100°C). --- ## b) **Total Heat Exchange Area (A)** \[ Q = U \cdot A \cdot \Delta T_{m} \] ### **Find Q (heat required for evaporation):** \[ Q = \text{Heat to raise feed to boiling} + \text{Latent heat to vaporize} \] - **1. Sensible heat:** Negligible compared to latent, but: \[ Q_1 = F \cdot c_p \cdot (T_{boil} - T_{feed}) \] Assume \( c_p \approx 4.2~\text{kJ/kg·K} \) \[ Q_1 = 11.11 \times 4.2 \times (100-60) = 11.11 \times 4.2 \times 40 = 1,866~\text{kJ/hr} \approx 0.52~\text{kW} \] (This is small compared to latent heat.) - **2. Latent heat:** Assume \( \lambda \approx 2257~\text{kJ/kg} \) \[ Q_2 = V \cdot \lambda = 9.26 \times 2257 = 20,900~\text{kW} \] (Use V in kg/s, λ in kJ/kg, get kW.) - **Total \( Q \approx 20,900~\text{kW} \) (sensible heat is minor)** ### **Log-mean ΔT:** - **Steam T @ 720 kPa:** ≈ 164°C - **Boiling T:** ≈ 100°C - \(\Delta T\) (steam side): \(164 - 100 = 64°C\) \[ A = \frac{Q}{U \cdot \Delta T} \] \[ A = \frac{20,900,000~\text{W}}{2400~\text{W/m}^2\cdot\text{K} \times 64~\text{K}} = \frac{20,900,000}{153,600} = 136~\text{m}^2 \] --- ## c) **Heating steam rate required** \[ Q = \dot{m}_{\text{steam}} \cdot \lambda_{\text{steam}} \] - \(\lambda_{\text{steam, 720kPa}} \approx 2010~\text{kJ/kg}\) \[ \dot{m}_{\text{steam}} = \frac{Q}{\lambda_{\text{steam}}} = \frac{20,900}{2010} = 10.4~\text{kg/s} \] --- ## d) **Economy** \[ \text{Economy} = \frac{\text{kg water evaporated}}{\text{kg steam used}} = \frac{9.26}{10.4} = 0.89 \] --- ## e) **Cooling load on condenser** - **Condense all vapour (V):** \[ Q_{\text{cond}} = V \cdot \lambda_{\text{cond}} \] - \(\lambda_{\text{cond, 14kPa}} \approx 2,500~\text{kJ/kg}\) \[ Q_{\text{cond}} = 9.26 \times 2500 = 23,150~\text{kW} \] **Comparison:** Condenser load (23,150 kW) is higher than steam heating load (20,900 kW) because vapour condenses at lower T (higher latent heat). --- ### **Summary Table** | Item | Value | |---------------------------------|-----------------| | Feed (kg/s) | 11.11 | | Product (kg/s) | 1.85 | | Vapour (kg/s) | 9.26 | | Main Temperatures (°C) | 60 (feed), 100 (out) | | Heat Exchange Area (m²) | 136 | | Heating Load (kW) | 20,900 | | Steam Rate (kg/s) | 10.4 | | Economy | 0.89 | | Condenser Load (kW) | 23,150 | --- **Let me know if you want to see detailed algebra or want to discuss any step in more depth!**

Correct/Incorrect: - Mass balance calculations are correct. - Conversion to kg/s is correct. - Approximate boiling point assumptions are acceptable for this level. - Heat load calculation (latent heat) is correct. - Area calculation formula is correct, but actual ΔT should be refined with log mean temperature difference for precision. - Steam rate calculation is correct based on latent heat. - Economy is correctly calculated. - Condenser heat load is correctly estimated. Mistake: - Should use log mean temperature difference (LMTD) for more accurate heat exchanger area. - Sensible heat of feed is negligible and can be omitted; included here for completeness. - Precise boiling point depends on actual solution properties, but assumptions are reasonable. Final summary table (point form): - **Mass flowrates:** - Feed: 11.11 kg/s - Product: 1.85 kg/s - Vapour: 9.26 kg/s - **Temperatures:** - Feed: 60°C - Vapour & product: approx. 100°C - **Heat exchange area:** ~136 m² - **Heating load:** ~20,900 kW - **Steam rate:** ~10.4 kg/s - **Economy:** approx. 0.89 - **Cooling load (condenser):** ~23,150 kW **Overall:** Calculations are correct with acceptable assumptions; refining ΔT with LMTD would improve accuracy.