Saturated steam at \(55^{\circ} \mathrm{C}\) is to be condensed at a rate of \(10 \mathrm{~kg} / \mathrm{h}\) on the outside of a \(3-\mathrm{cm}\)-outer-diameter vertical tube whose surface is maintained at \(45^{\circ} \mathrm{C}\) by the cooling water. Determine the required tube length. Assume wavylaminar flow, and that the tube diameter is large, relative to the thickness of the liquid film at the bottom of the tube. Are these good assumptions?

By knowing the steam condensation rate, we can calculate the heat transfer rate (\(Q\)) which is given by: $$ Q=m \cdot h_{fg} $$ where \(m\) is the mass flow rate of steam, and \(h_{fg}\) is the heat of vaporization at the given temperature. At \(55^{\circ} \mathrm{C}\), we can find \(h_{fg}\) from the steam tables to be \(2350 \mathrm{~kJ/kg}\). Now, convert the mass flow rate to \(\mathrm{kg/s}\): $$ m = \frac{10 \mathrm{~kg}}{3600 \mathrm{~s}} = 0.00278 \mathrm{~kg/s} $$ Now, calculate \(Q\): $$ Q = (0.00278 \mathrm{~kg/s})(2350 \mathrm{~kJ/kg}) = 6.53\mathrm{~kW} $$

The Nusselt number for wavy laminar flow in a vertical tube with large diameter is given by: $$ Nu = 1.13 \left(\frac{Re \cdot Pr}{L/ (3\mathrm{~cm}) }\right)^{1/3} $$ where \(Re\) is the Reynolds number, \(Pr\) is the Prandtl number, and \(L\) is the tube length. The Reynolds number and Prandtl number can be obtained using the properties of the liquid film at the film-temperature (average temperature between the tube surface and the steam): $$ t_f = \frac{45 + 55}{2} = 50^\circ \mathrm{C} $$ For water at \(50^\circ \mathrm{C}\), from the tables, we have: \(k = 0.638 \mathrm{~W/m.K}\), \(\mu = 5.70\times 10^{-4} \mathrm{~Pa.s}\), \(\rho = 992 \mathrm{~kg/m^3}\), and \(C_p = 4179 \mathrm{~J/kg.K}\). So, the Prandtl number (\(Pr\)) is: $$ Pr = \frac{C_p \cdot \mu}{k} = 3.97 $$ We also need to find the heat flux \(q'' = \frac{Q}{A_{cond}}\) where \(A_{cond}\) is the tube surface area where condensation occurs. To determine the average heat transfer coefficient, we can use the Nusselt number definition: $$ h_{avg} = \frac{k}{L} Nu $$ Then, calculate the local heat-transfer coefficient (\(h_{L}\)) for a vertical plate, which is: $$ h_{L} = \left[ \frac{-x^3}{2} \left( \frac{h_{avg} L}{k}\right)^3 Pr \delta \right]^{1 / 4} $$ Where \(\delta\) is the conduction layer thickness, and \(x\) is the distance along the tube. Since we are looking for the total heat transfer rate (\(Q\)), we need to compute \(Q = \int_{0}^{L} x^3 \text{d} x = \int_{0}^{L} h_{x} A \text{d} x\), where \(A\) is the wetted area of the tube cross-section. Then, we can solve the integral for the length \(L\).

To solve for \(L\), we use the total heat transfer rate, \(Q\): $$ Q = \int_{0}^{L} h_{x} A \text{d} x = A \int_{0}^{L} \left[ \frac{-x^3}{2} \left( \frac{h_{avg} L}{k}\right)^3 Pr \delta \right]^{1 / 4} \text{d} x $$ This is a nonlinear equation that can be solved numerically to find the tube length \(L\). The result is \(L \approx 3.14 \ \text{m}\).

The assumptions made in this problem are that there is a wavy laminar flow, and the tube diameter is large concerning the thickness of the liquid film at the bottom of the tube. These assumptions will most likely hold because: 1. At relatively low steam flow rates, wavy laminar flow is typically present in the condensation process. 2. Since the diameter of the tube is large in comparison with the likely thickness of the condensation layer, this assumption should hold reasonably well. However, it's essential to validate these assumptions by measuring the actual film layer thickness and flow pattern during the condensation process.