Saturated steam at \(100^{\circ} \mathrm{C}\) condenses on a 2-m \(\times 2-\mathrm{m}\) plate that is tilted \(40^{\circ}\) from the vertical. The plate is maintained at \(80^{\circ} \mathrm{C}\) by cooling it from the other side. Determine (a) the average heat transfer coefficient over the entire 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 \(100^{\circ} \mathrm{C}\), find the properties of saturated steam using a steam table or online resources. With this information, we can obtain the density (\(\rho\)), dynamic viscosity (\(\mu\)), thermal conductivity (\(k\)), and latent heat of vaporization (\(L\)). For saturated steam at \(100^{\circ} \mathrm{C}\), \(\rho= 958 \mathrm{~kg/m^3}\) \(\mu =2.82 \times 10^{-4} \mathrm{~kg/(m \cdot s)}\) \(k = 0.679 \mathrm{~W/(m\cdot K)}\) \(L=2.257 \times 10^6 \mathrm{~J/kg}\)

To find the average heat transfer coefficient, we need to use the Nusselt equation for condensation on an inclined surface: \(Nu = \frac{hL}{k} = \left( 0.943 \right) \cdot \left(\frac{g \cdot \rho\left(\rho-\rho_{\mathrm{v}}\right) \cdot k^3 \cdot L}{\mu \Delta T \cdot \sin{\theta}} \right)^{\frac{1}{4}}\) Where: \(h\) - average heat transfer coefficient \(Nu\) - Nusselt number \(L\) - plate length (in this case, \(2\mathrm{~m}\)) \(g\) - acceleration due to gravity, \(9.81\mathrm{~m/s^2}\) \(\rho_{\mathrm{v}}\)- vapor density (obtained from the steam table at \(100^{\circ} \mathrm{C}\), \(0.6\mathrm{~kg/m^3}\)) \(\Delta T\) - temperature difference between the saturated steam and the plate (\(100^{\circ} \mathrm{C} - 80^{\circ} \mathrm{C} = 20^{\circ} \mathrm{C}\)) \(\theta\) - angle with the vertical, \(40^{\circ}\)

By substituting our values into the Nusselt equation, we can now calculate \(h\): \(h = \frac{0.679 \cdot \left( 0.943 \right) \cdot \left(\frac{9.81 \cdot 958\left(958-0.6\right) \cdot 0.679^3 \cdot 2.257 \times 10^6}{2.82 \times 10^{-4} \cdot 20 \cdot \sin{40^{\circ}}} \right)^{\frac{1}{4}}}{2}\) \(h \approx 5651 \mathrm{~W/(m^2 \cdot K)}\)

Using Reynolds analogy, we can compute the flow rate of the condensate per unit width: \(Re_wL=\rho (\rho - \rho_{\mathrm{v}}) V_wL^2\sin\theta/\mu\) Where: \(Re_w\) - Reynolds number for wavy-laminar flow (from literature, \(Re_w = 1600\)) \(V_w\) - flow rate per unit width. Now we can solve for \(V_w\): \(V_w = \frac{1600 \times 2.82 \times 10^{-4}}{(958)(958 - 0.6)(2^2)(\sin{40^{\circ}})}\) \(V_w \approx 0.0054\mathrm{~m^3/s}\) The condensate drips off the plate at a rate of approximately \(0.0054\mathrm{~m^3/s}\) per unit width.

To check if the wavy-laminar flow assumption is reasonable, we need to compare the determined flow rate of the condensate (\(V_w\)) with the values from the literature regarding laminar and turbulent flow regimes. The given values for laminar flow are less than 30 cm/s, and turbulent flow is higher than 2 m/s. In this case, \(V_w \approx 5.4\mathrm{~cm/s}\), which lies within the laminar flow regime, indicating that the wavy-laminar flow assumption is reasonable. In conclusion: (a) The average heat transfer coefficient over the entire plate is approximately \(5651\mathrm{~W/(m^2 \cdot K)}\). (b) The rate at which the condensate drips off the plate is approximately \(0.0054\mathrm{~m^3/s}\) per unit width, and the wavy-laminar flow assumption is reasonable.