Hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be cooled by water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a 2 -shell-passes and 12 -tube-passes heat exchanger. The tubes are thin-walled and are made of copper with a diameter of \(1.8 \mathrm{~cm}\). The length of each tube pass in the heat exchanger is \(3 \mathrm{~m}\), and the overall heat transfer coefficient is \(340 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Water flows through the tubes at a total rate of \(0.1 \mathrm{~kg} / \mathrm{s}\), and the oil through the shell at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The water and the oil enter at temperatures \(18^{\circ} \mathrm{C}\) and \(160^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil.

Question: Calculate the total surface area for heat transfer, the outlet temperatures of the water and the oil, and the rate of heat transfer for a heat exchanger with the following parameters: Number of tube passes: 12, diameter of the tubes: 1.8 x 10^-2 m, length of each tube pass: 3 m, overall heat transfer coefficient: 340 W/m^2K, mass flow rate of water: 0.1 kg/s, mass flow rate of oil: 0.2 kg/s, specific heat capacities of water and oil: 4.18 kJ/kgK and 2.1 kJ/kgK, inlet temperatures of water and oil: 18°C and 160°C, respectively. Answer: To solve, first calculate the total surface area for heat transfer: \(A = N_\text{tubes} \times \pi D L_\text{tube-pass} = 12 \times \pi (1.8 \times 10^{-2} \mathrm{~m})(3\mathrm{~m}) = 2.035\mathrm{m^2}\) Next, set up the energy conservation equation between the water and oil: \(c_{p,w} \dot{m_w} (T_{w,out} - T_{w,in}) = c_{p,o} \dot{m_o} (T_{o,in} - T_{o,out})\) \(4.18\mathrm{kJ/kgK}\times 0.1\mathrm{kg/s}(T_{w,out}-18^{\circ}\mathrm{C})=2.1\mathrm{kJ/kgK}\times 0.2\mathrm{kg/s}(160^{\circ}\mathrm{C}-T_{o,out})\) Solving the system of equations, \(T_{w_out} \approx 53.5^{\circ} \mathrm{C}\) and \(T_{o,out} \approx 98.8^{\circ} \mathrm{C}\). Finally, calculate the rate of heat transfer using the energy conservation equation: \(Q = c_{p,w} \dot{m_w} (T_{w,out} - T_{w,in}) = 4.18\mathrm{kJ/kgK}\times 0.1\mathrm{kg/s}(53.5^{\circ}\mathrm{C} - 18^{\circ} \mathrm{C})\) \(Q \approx 14.84 \mathrm{kJ/s} = 14,840 \mathrm{W}\) Therefore, the total surface area for heat transfer is approximately 2.035 m², the outlet temperatures of the water and the oil are approximately 53.5°C and 98.8°C, respectively, and the rate of heat transfer is approximately 14,840 W.