Let's break down the problem step by step: --- # **Given Data** - **Plate height, \( L \) = 2 m** - **Plate width, \( W \) = 10 m** - **Plate temperature, \( T_s \) = 80°C** - **Saturated steam at 1 atm (100°C)** - **Condensation happens on the vertical plate** --- # **Step 1: Get Steam Properties at 1 atm** From steam tables at 1 atm (100°C): - **Saturation temperature, \( T_{\text{sat}} \) = 100°C** - **Latent heat of vaporization, \( h_{fg} \approx 2257 \) kJ/kg** - **Density of liquid water, \( \rho_f \approx 958 \) kg/m³** - **Dynamic viscosity, \( \mu_f \approx .282 \times 10^{-3} \) kg/m·s** - **Thermal conductivity, \( k_f \approx .679 \) W/m·K** - **Prandtl number, \( Pr_f \approx 1.75 \)** --- # **Step 2: Use Nusselt's Equation for Film Condensation** For a vertical plate (laminar film condensation): \[ q = h \cdot A \cdot (T_{\text{sat}} - T_s) \] where \( h \) is the average heat transfer coefficient, and \( A = W \times L \). Nusselt's correlation for average \( h \): \[ h = .943 \left(\frac{\rho_f (\rho_f - \rho_g) g h_{fg} k_f^3}{\mu_f L (T_{\text{sat}} - T_s)} \right)^{1/4} \] Since \( \rho_g \ll \rho_f \), \( \rho_f - \rho_g \approx \rho_f \). --- # **Step 3: Plug in the Values** - \( g = 9.81 \) m/s² - \( \rho_f = 958 \) kg/m³ - \( h_{fg} = 2,257,000 \) J/kg (convert kJ/kg to J/kg) - \( k_f = .679 \) W/m·K - \( \mu_f = .282 \times 10^{-3} \) kg/m·s - \( L = 2 \) m - \( T_{\text{sat}} - T_s = 100 - 80 = 20 \) K \[ h = .943 \left( \frac{ 958 \times 958 \times 9.81 \times 2,257,000 \times (.679)^3 }{ .282 \times 10^{-3} \times 2 \times 20 } \right)^{1/4} \] Let's calculate step by step: ### Numerator: - \( 958 \times 958 = 917,764 \) - \( 917,764 \times 9.81 = 9,003,209 \) - \( 9,003,209 \times 2,257,000 = 20,322,744,013,000 \) - \( (.679)^3 = .313 \) - \( 20,322,744,013,000 \times .313 = 6,364,221,902,079 \) ### Denominator: - \( .282 \times 10^{-3} = .000282 \) - \( .000282 \times 2 = .000564 \) - \( .000564 \times 20 = .01128 \) ### Full fraction: \[ \frac{6,364,221,902,079}{.01128} \approx 564,454,427,215,425 \] ### Fourth Root: \[ \left(564,454,427,215,425\right)^{1/4} \approx 875.7 \] \[ h = .943 \times 875.7 \approx 825.5 \text{ W/m}^2\cdot\text{K} \] --- # **Step 4: Calculate the Total Heat Transfer Rate** \[ q = h \cdot A \cdot \Delta T \] - \( A = W \times L = 10 \times 2 = 20 \) m² - \( \Delta T = 20 \) K \[ q = 825.5 \times 20 \times 20 = 330,200 \text{ W} = 330.2 \text{ kW} \] --- ## **Final Answers** ### **(a) The rate of heat transfer by condensation to the plate:** \[ \boxed{q = 330,200 \text{ W} = 330.2 \text{ kW}} \] --- ### **(b) The rate at which the condensation drips off the plate at the bottom** Use the heat transfer to find the mass flow rate: \[ q = \dot{m} \cdot h_{fg} \implies \dot{m} = \frac{q}{h_{fg}} \] \[ \dot{m} = \frac{330,200}{2,257,000} \approx .146 \text{ kg/s} \] --- ## **Final Answers (Summary Table)** | Item | Answer | |------|--------| | (a) Rate of heat transfer | **330,200 W (330.2 kW)** | | (b) Condensation rate | **.146 kg/s** | If you need more explanation or steps, let me know!