# Condenser Design Problem — Step-by-Step Solution ## Given Data - **Steam mass flow rate (\( \dot{m}_{\text{steam}} \))**: 300 kg/s at 10 kPa, 50°C - **Drain water mass flow rate (\( \dot{m}_{\text{drain}} \))**: 50 kg/s at 10 kPa - **Cooling seawater inlet temperature (\( T_{\text{sw,in}} \))**: 20°C - **Material**: 90-10 copper-nickel - **All processes at steady state** --- ## Assumptions 1. **Steam condenses at constant pressure (10 kPa)**. 2. **All steam leaves as saturated liquid at 10 kPa**. 3. **Complete condensation (no steam leaves as vapor)**. 4. **Neglect heat losses to surroundings**. 5. **Seawater temperature rise limited to 10°C (common practice, unless otherwise specified)**. 6. **Neglect subcooling of condensate**. 7. **Neglect non-condensables and pressure drop on the shell side**. 8. **Drain water mixes with condensate at 10 kPa**. --- ## Step 1: Find Steam Properties at 10 kPa From steam tables: - **At 10 kPa**: - Saturation temperature, \( T_{\text{sat}} \approx 45.8^\circ\text{C} \) - Enthalpy of vaporization, \( h_{fg} \approx 2392 \) kJ/kg - Enthalpy of saturated liquid, \( h_f \approx 191.8 \) kJ/kg - Enthalpy of saturated vapor, \( h_g \approx 2584 \) kJ/kg --- ## Step 2: Calculate Total Heat to be Removed (\( Q \)) ### a) **From Steam Condensation** \[ Q_{\text{steam}} = \dot{m}_{\text{steam}} \times (h_g - h_f) \] \[ Q_{\text{steam}} = 300~\text{kg/s} \times (2584 - 191.8)~\text{kJ/kg} \] \[ Q_{\text{steam}} = 300 \times 2392.2 = 717,660~\text{kJ/s} = 717.7~\text{MW} \] ### b) **From Drain Water Cooling** Assume drain water at 50°C cooled to 45.8°C (condensate temperature): \[ Q_{\text{drain}} = \dot{m}_{\text{drain}} \times c_p \times \Delta T \] Assume \( c_p = 4.18~\text{kJ/kg} \cdot \text{K} \), \( \Delta T = 50 - 45.8 = 4.2~^\circ\text{C} \): \[ Q_{\text{drain}} = 50 \times 4.18 \times 4.2 = 877~\text{kJ/s} \] ### c) **Total Heat Removed** \[ Q_{\text{total}} = Q_{\text{steam}} + Q_{\text{drain}} \approx 717,660 + 877 = 718,537~\text{kJ/s} = 718.5~\text{MW} \] --- ## Step 3: Cooling Water Mass Flow Rate (\( \dot{m}_{\text{cw}} \)) Assume seawater temperature rises by \( \Delta T_{\text{cw}} = 10^\circ\text{C} \) (from 20°C to 30°C): \[ Q_{\text{total}} = \dot{m}_{\text{cw}} \times c_{p,\text{cw}} \times \Delta T_{\text{cw}} \] \[ \dot{m}_{\text{cw}} = \frac{Q_{\text{total}}}{c_{p,\text{cw}} \times \Delta T_{\text{cw}}} \] Assume \( c_{p,\text{cw}} = 4.18~\text{kJ/kg} \cdot \text{K} \): \[ \dot{m}_{\text{cw}} = \frac{718,537}{4.18 \times 10} = \frac{718,537}{41.8} \approx 17,191~\text{kg/s} \] --- ## Step 4: Number of Tubes Assume: - Tube inner diameter (\( d_i \)) = 20 mm = 0.02 m - Tube length (\( L \)) = 8 m ### a) **Velocity of Water in Tubes** Typical design: \( v = 2~\text{m/s} \) \[ \dot{m}_{\text{cw}} = \rho \cdot A_{\text{total}} \cdot v \] \[ A_{\text{total}} = \frac{\dot{m}_{\text{cw}}}{\rho v} \] Seawater \( \rho \approx 1025~\text{kg/m}^3 \): \[ A_{\text{total}} = \frac{17,191}{1025 \times 2} = \frac{17,191}{2050} \approx 8.39~\text{m}^2 \] ### b) **Area per Tube** \[ A_{\text{tube}} = \frac{\pi}{4} d_i^2 = \frac{\pi}{4} (0.02)^2 = 3.142 \times 10^{-4}~\text{m}^2 \] ### c) **Number of Tubes** \[ N_{\text{tubes}} = \frac{A_{\text{total}}}{A_{\text{tube}}} = \frac{8.39}{3.142 \times 10^{-4}} \approx 26,715 \] --- ## Step 5: Pressure Drop in Tubes For turbulent flow in smooth tubes (Darcy-Weisbach): \[ \Delta P = f \frac{L}{d} \frac{\rho v^2}{2} \] Where: - \( f \) = friction factor (for turbulent flow, use Blasius equation: \( f = 0.3164/Re^{0.25} \)) - \( L \) = 8 m - \( d \) = 0.02 m - \( \rho = 1025~\text{kg/m}^3 \) - \( v = 2~\text{m/s} \) ### a) **Reynolds Number** \[ Re = \frac{v d \rho}{\mu} \] Assume seawater \( \mu \approx 1.08 \times 10^{-3}~\text{kg/m}\cdot\text{s} \): \[ Re = \frac{2 \times 0.02 \times 1025}{1.08 \times 10^{-3}} = \frac{41}{1.08 \times 10^{-3}} \approx 37,963 \] ### b) **Friction Factor** \[ f = 0.3164 / (37,963)^{0.25} \approx 0.021 \] ### c) **Calculate \( \Delta P \)** \[ \Delta P = 0.021 \times \frac{8}{0.02} \times \frac{1025 \times (2)^2}{2} \] \[ = 0.021 \times 400 \times \frac{1025 \times 4}{2} = 0.021 \times 400 \times 2050 = 0.021 \times 820,000 = 17,220~\text{Pa} = 17.2~\text{kPa} \] --- ## Step 6: Power Attributed to Water Flow \[ \text{Power} = \dot{m}_{\text{cw}} \frac{\Delta P}{\rho} \] \[ \text{Power} = 17,191~\text{kg/s} \times \frac{17,220~\text{Pa}}{1025~\text{kg/m}^3} \] \[ = 17,191 \times 16.8 = 288,815~\text{W} \approx 289~\text{kW} \] --- ## **Summary Table** | **Quantity** | **Value** | |-------------------------------------|-----------------------| | Number of tubes (\(N_{\text{tubes}}\)) | ~26,700 (rounded) | | Cooling water mass flow (\(\dot{m}_{\text{cw}}\)) | ~17,200 kg/s | | Pressure drop (\(\Delta P\)) | ~17.2 kPa | | Power for water flow | ~289 kW | --- ## **Final Answers** ### a. **Number of tubes of the condenser** **~26,700 tubes** (assuming 20 mm diameter, 8 m length, 2 m/s velocity) ### b. **Cooling seawater mass flow rate** **~17,200 kg/s** ### c. **Pressure drop in the condenser** **~17.2 kPa** ### d. **Power attributed to the water flow** **~289 kW** --- ## **Notes on Assumptions** - If tube diameter/length or seawater temperature rise differ, update calculations accordingly. - Tube material (90-10 Cu-Ni) is suitable for seawater; actual design needs corrosion and fouling factors. - Calculations are first approximations for preliminary design. --- **If you need further breakdown of any specific step or want to adjust assumptions, let me know!**