Steam is to be condensed on the shell side of a 1 -shellpass and 8-tube-passes condenser, with 50 tubes in each pass, at \(30^{\circ} \mathrm{C}\left(h_{f g}=2431 \mathrm{~kJ} / \mathrm{kg}\right)\). Cooling water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(15^{\circ} \mathrm{C}\) at a rate of \(1800 \mathrm{~kg} / \mathrm{h}\). The tubes are thin-walled, and have a diameter of \(1.5 \mathrm{~cm}\) and length of \(2 \mathrm{~m}\) per pass. If the overall heat transfer coefficient is \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine \((a)\) the rate of heat transfer and \((b)\) the rate of condensation of steam.

The rate of heat transfer (Q) can be found using the following formula: \[ Q = U \cdot A \cdot \Delta T \] Where: - Q is the rate of heat transfer - U is the overall heat transfer coefficient - A is the surface area of the tubes - ΔT is the temperature difference between the cooling water and the steam First, let's calculate the total surface area of the tubes (A) using the given diameter (D) and length (L) per pass, as well as the total number of tubes (N_tubes): \[ A = N_{tubes} \cdot \pi D \cdot L \] Using the given values, we have: \[ A = 8 \times 50 \times \pi \times 0.015 \mathrm{m} \times 2\mathrm{m} = 3.7699\mathrm{m}^2 \] Next, we need to calculate the temperature difference (ΔT) between the water and the steam. As the initial temperature of the cooling water is given, and we know the condensing temperature of the steam, we can use these directly: \[ \Delta T = 30^{\circ} \mathrm{C} - 15^{\circ} \mathrm{C} = 15 \mathrm{K} \] Now, we can substitute these values into the heat transfer equation to obtain the rate of heat transfer (Q): \[ Q = 3000 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}} \cdot 3.7699 \mathrm{m}^2 \cdot 15\mathrm{K} = 169113.3\mathrm{W} \] So, \((a)\) the rate of heat transfer is 169113.3 W.

The rate of condensation of steam (m) can be found by dividing the rate of heat transfer by the specific enthalpy of vaporization (h_fg): \[ m = \frac{Q}{h_{fg}} \] We are given that the specific enthalpy of vaporization is 2431 kJ/kg, which we can convert to W/kg: \[ h_{fg} = 2431 \frac{\mathrm{kJ}}{\mathrm{kg}} \cdot \frac{1000 \mathrm{W}}{1 \mathrm{kJ}} = 2431000 \frac{\mathrm{W}}{\mathrm{kg}} \] Now, we can substitute these values into the above formula to obtain the rate of condensation of steam (m): \[ m = \frac{169113.3\mathrm{W}}{2431000 \frac{\mathrm{W}}{\mathrm{kg}}} = 0.0695 \frac{\mathrm{kg}}{\mathrm{s}} \] To convert the rate of condensation to kg/h, we multiply by the conversion factor between seconds and hours: \[ m_{\mathrm{h}} = 0.0695 \frac{\mathrm{kg}}{\mathrm{s}} \cdot \frac{3600 \mathrm{s}}{1 \mathrm{h}} = 250.2 \frac{\mathrm{kg}}{\mathrm{h}} \] So, \((b)\) the rate of condensation of steam is 250.2 kg/h.