Saturated steam at 1 atm condenses on a 3-m-high and 8 - \(\mathrm{m}\)-wide vertical plate that is maintained at \(90^{\circ} \mathrm{C}\) by circulating cooling water through the other side. Determine \((a)\) the rate of heat transfer by condensation to the plate, and ( \(b\) ) the rate at which the condensate drips off the plate at the bottom. Assume wavy-laminar flow. Is this a good assumption?

At a pressure of 1 atm, we can look up the saturation temperature \(T_s\), latent heat of vaporization \(h_{fg}\), and the density of the vapor phase \(\rho_v\) and liquid phase \(\rho_l\). From steam tables, we have: Saturation temperature: \(T_s = 100^{\circ}\mathrm{C}\) Density of vapor: \(\rho_v = 0.6 \,\mathrm{kg/m^3}\) Density of liquid: \(\rho_l = 958 \,\mathrm{kg/m^3}\) Latent heat of vaporization: \(h_{fg} = 2.26 \times 10^6 \,\mathrm{J/kg}\)

For wavy-laminar film condensation on a vertical surface, we can use the Nusselt number formula given below: \(Nu = 0.943 \,[g(\rho_l - \rho_v) h_{fg} L/(k_l \Delta T)]^{1/4}\) where: \(g = 9.81\,\mathrm{m/s^2}\), gravitational acceleration \(\rho_l = 958\,\mathrm{kg/m^3}\), density of liquid \(\rho_v = 0.6\,\mathrm{kg/m^3}\), density of vapor \(h_{fg} = 2.26 \times 10^6\,\mathrm{J/kg}\), latent heat of vaporization \(L = 3\,\mathrm{m}\), height of plate \(k_l = 0.68\,\mathrm{W/(mK)}\), thermal conductivity of liquid \(\Delta T = T_s - T_p\), temperature difference between saturation and plate temperature. \(\Delta T = 100^{\circ}\mathrm{C} - 90^{\circ}\mathrm{C} = 10^{\circ}\mathrm{C}\) We can now calculate the Nusselt number: \(Nu = 0.943 \,[9.81(958 - 0.6) (2.26 \times 10^6) 3/(0.68 \times 10)]^{1/4}\) \(Nu = 882.47\)

To find the heat transfer coefficient, we will use the following relation: \(h = Nu \, k_l/L\) \(h = 882.47 \times (0.68/3)\) \(h = 202.92\,\mathrm{W/(m^2K)}\)

Now, we can calculate the rate of heat transfer by condensation using: \(q = hA\Delta T\) where \(A = 3\,\mathrm{m} \times 8\,\mathrm{m} = 24\,\mathrm{m^2}\) is the surface area of the plate. \(q = 202.92\,\mathrm{W/(m^2K)} \times 24\,\mathrm{m^2} \times 10\,\mathrm{K}\) \(q = 48,700.8\,\mathrm{W}\) Therefore, the rate of heat transfer by condensation is about 48,700.8 W.

Using the rate of heat transfer, we can find the rate at which the condensate drips off the plate: \( \dot{m} = q/h_{fg} = (48,700.8\,\mathrm{W})/(2.26 \times 10^6\,\mathrm{J/kg})\) \( \dot{m} = 0.0216\,\mathrm{kg/s}\) The rate at which the condensate drips off the plate at the bottom is 0.0216 kg/s.

We can compare the heat transfer coefficient found using the wavy-laminar flow assumption with that of the turbulent flow regime. For turbulent flow, the Nusselt number relation is different and turbulent films have higher values of heat transfer coefficients. If the heat transfer coefficient of the wavy-laminar flow we found was significantly lower than that of turbulent flow, then the assumption may not hold. However, based on the exercise description and the given flow conditions, the wavy-laminar flow assumption is reasonable.