A shell-and-tube heat exchanger with 1-shell pass and 14-tube passes is used to heat water in the tubes with geothermal steam condensing at \(120^{\circ} \mathrm{C}\left(h_{f g}=2203 \mathrm{~kJ} / \mathrm{kg}\right)\) on the shell side. The tubes are thin-walled and have a diameter of \(2.4 \mathrm{~cm}\) and length of \(3.2 \mathrm{~m}\) per pass. Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(3.9 \mathrm{~kg} / \mathrm{s}\). If the temperature difference between the two fluids at the exit is \(46^{\circ} \mathrm{C}\), determine (a) the rate of heat transfer, \((b)\) the rate of condensation of steam, and \((c)\) the overall heat transfer coefficient.

To determine the rate of heat transfer, we need to know the temperature difference of water as it flows through the heat exchanger. Given the exit temperature difference of 46°C and the initial temperature of the water being 22°C, we can find the exit temperature of the water: 120°C - 46°C = 74°C. Now we can find the temperature change: ΔT = 74°C - 22°C = 52°C.

To calculate the rate of heat transfer, we will use the energy balance equation: Q = mcΔT, where m is the mass flow rate of water, c is the specific heat capacity of water, and ΔT is the temperature change of water. We are given m = 3.9 kg/s, c = 4180 J/(kg·K), and we found ΔT = 52°C in the previous step. Thus, Q = 3.9 kg/s × 4180 J/(kg·K) × 52 K = 848,856 J/s or 848.856 kW.

The rate of heat transfer through the steam can be calculated by applying the energy balance, using the heat of vaporization (h_fg) as follows: Q = m_s×h_fg, where m_s is the mass flow rate of steam and h_fg = 2203 kJ/kg. Since the heat transferred from steam to water is the same, we have: m_s×h_fg = 848.856 kW.

Using the equation from Step 3, we can find the mass flow rate of steam, m_s: m_s = Q/h_fg = 848.856 kW / 2203 kJ/kg = 0.385 kg/s.

To find the overall heat transfer coefficient, we can use the heat transfer equation: Q = UAΔT_m, where U is the overall heat transfer coefficient, A is the heat transfer area, and ΔT_m is the logarithmic mean temperature difference. First, we must calculate the heat transfer area per tube: A_tube = πD×L×n_pass, where D is the diameter of the tube, L is the length per pass, and n_pass is the number of tube passes. We are given D = 0.024 m, L = 3.2 m, and n_pass = 14. Thus, A_tube = π×0.024 m×3.2 m×14 = 3.246 m². Next, we should calculate the logarithmic mean temperature difference, which could be done by assuming that the steam temperature does not change significantly and remain constant at 120°C and that the inlet and outlet temperatures of the water are 22°C and 74°C. We can calculate ΔT_m as follows: ΔT_m = [(120-74)-(120-22)] / ln((120-74)/(120-22)) = 39.09 °C. Finally, we can calculate the overall heat transfer coefficient, U, by rearranging the heat transfer equation, Q = UAΔT_m: U = Q / (A_tube × ΔT_m) = 848.856 kW / (3.246 m² × 39.09 K) = 6.71 kW/(m²·K). #Summary# In this exercise, our calculations yielded: (a) The rate of heat transfer, Q = 848.856 kW (b) The rate of condensation of steam, m_s = 0.385 kg/s (c) The overall heat transfer coefficient, U = 6.71 kW/(m²·K)