Inside a condenser, there is a bank of seven copper tubes with cooling water flowing in them. Steam condenses at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\) on the outer surfaces of the tubes that are at a constant temperature of \(68^{\circ} \mathrm{C}\). Each copper tube is \(5-\mathrm{m}\) long and has an inner diameter of \(25 \mathrm{~mm}\). Cooling water enters each tube at \(5^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). Determine the average heat transfer coefficient of the cooling water flowing inside each tube and the cooling water mean velocity needed to achieve the indicated heat transfer rate in the condenser.

#tag_title#Step 2: Calculate the surface area of the tubes#tag_content#We are given the dimensions of the tubes: length \(L = 5 \mathrm{~m}\) and outer diameter \(D_o = 20 \mathrm{~mm}\). There are 1500 tubes, so the total surface area of the tubes can be calculated as: \(A = N_{tubes} \pi D_o L\) where \(N_{tubes}\): Number of tubes \(A = (1500) (\pi) (0.02 \mathrm{~m}) (5 \mathrm{~m}) \approx 471.24 \mathrm{~m^2}\) #tag_title#Step 3: Find the average temperature difference#tag_content#We are given the mean temperature of the cooling water going in and out of the tubes: \(T_i = 20^\circ \mathrm{C}\) and \(T_o = 35^\circ \mathrm{C}\). We can calculate the average temperature difference by finding the difference between the temperature of the tubes (\(T_t = 50^\circ \mathrm{C}\)) and the average temperature of the cooling water (\(\frac{T_i + T_o}{2}\)): \(\Delta T = T_t - \frac{T_i + T_o}{2}\) \(\Delta T = 50^\circ \mathrm{C} - \frac{20^\circ \mathrm{C} + 35^\circ \mathrm{C}}{2} = 22.5^\circ \mathrm{C}\) #tag_title#Step 4: Calculate the average heat transfer coefficient#tag_content#Now that we have the heat transfer rate, surface area of the tubes, and average temperature difference, we can calculate the average heat transfer coefficient: \(h = \frac{Q}{A \Delta T}\) \(h = \frac{1356 \mathrm{~kJ/s}}{(471.24 \mathrm{~m^2})(22.5 \mathrm{~K})} \approx 12.16 \mathrm{~W/(m^2 K)}\) #tag_title#Step 5: Calculate the cooling water flow rate#tag_content#We can now use the heat transfer rate equation to find the cooling water flow rate: \(\dot{m} = \frac{Q}{c_p \Delta T_{water}}\) where \(c_p\) for water is approximately \(4.18 \mathrm{~kJ/(kg K)}\), and \(\Delta T_{water} = 15^\circ \mathrm{C}\): \(\dot{m} = \frac{1356 \mathrm{~kJ/s}}{(4.18 \mathrm{~kJ/(kg K)})(15 \mathrm{~K})} \approx 21.74 \mathrm{~kg/s}\) #tag_title#Step 6: Calculate the cooling water mean velocity#tag_content#Finally, we can calculate the cooling water mean velocity by dividing the flow rate by the total cross-sectional area of the tubes: \(v = \frac{\dot{m}}{N_{tubes} \pi \frac{D_i^2}{4}}\) where \(D_i\) is the inner diameter of the tubes, which can be calculated as \(D_i = D_o - 2t\), where \(t = 2 \mathrm{~mm}\) is the thickness of the tube. This gives us \(D_i = 16 \mathrm{~mm}\). \(v = \frac{21.74 \mathrm{~kg/s}}{(1500) (\pi) (\frac{0.016 \mathrm{~m}}{2})^2} \approx 2.16 \mathrm{~m/s}\) Therefore, the average heat transfer coefficient of the cooling water flowing inside each tube is approximately \(12.16 \mathrm{~W/(m^2 K)}\), and the cooling water mean velocity needed to achieve the given heat transfer rate in the condenser is approximately \(2.16 \mathrm{~m/s}\).