Question

A shell-and-tube heat exchanger is used for heating $14 \mathrm{~kg} / \mathrm{s}\( of oil \)\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( from \)20^{\circ} \mathrm{C}\( to \)46^{\circ} \mathrm{C}$. The heat exchanger has one shell pass and six tube passes. Water enters the shell side at \(80^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). The overall heat transfer coefficient is estimated to be $1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Calculate the rate of heat transfer and the heat transfer area.

Short answer

Question: Calculate the heat transfer rate and the heat transfer area of a shell-and-tube heat exchanger given the following information: the mass flow rate of oil is 14 kg/s, the specific heat capacity of oil is 2.0 kJ/kg·K, the initial and final temperatures of the oil are 20°C and 46°C, the overall heat transfer coefficient is 1000 W/m²·K, and the initial and final temperatures of the water in the shell side are 80°C and 60°C. Answer: The heat transfer rate of the shell-and-tube heat exchanger is 728,000 W, and the heat transfer area is 15.83 m².

Step-by-step solution

Calculate the rate of heat transfer using the oil data

We know that the mass flow rate of oil is \(14\,\mathrm{kg/s}\), the specific heat capacity is \(2.0\,\mathrm{kJ/kg\cdot K}\) and the initial and final temperatures are \(T_{1} = 20^{\circ}\mathrm{C}\) and \(T_{2} = 46^{\circ}\mathrm{C}\) respectively. We can now calculate the rate of heat transfer using the oil data: \(Q = mc_p(T_2-T_1) = 14\,\mathrm{kg/s} \cdot 2.0\,\mathrm{kJ/kg\cdot K} \cdot (46^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C})\) Convert the specific heat capacity to \(\mathrm{W/kg\cdot K}\): \(2.0\,\mathrm{kJ/kg\cdot K} \cdot \frac{1000\,\mathrm{W}}{1\,\mathrm{kJ}} = 2000\,\mathrm{W/kg\cdot K}\) Now, plug in the converted specific heat capacity: \(Q = 14\,\mathrm{kg/s} \cdot 2000\,\mathrm{W/kg\cdot K} \cdot (46^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C}) = 14\,\mathrm{kg/s} \cdot 2000\,\mathrm{W/kg\cdot K} \cdot 26\,\mathrm{K} = 728,000\,\mathrm{W}\) The rate of heat transfer using the oil data is \(728,000\,\mathrm{W}\).

Calculate the temperature difference across the heat exchanger

We are given the initial and final temperatures of the water entering and leaving the shell side of the heat exchanger as \(80^{\circ}\mathrm{C}\) and \(60^{\circ}\mathrm{C}\). The temperature difference across the heat exchanger (minimum temperature change) can be calculated as follows: \(\Delta T = T_{\mathrm{water}} - T_{\mathrm{oil}} = (80^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C}) - (60^{\circ}\mathrm{C} - 46^{\circ}\mathrm{C}) = 60^{\circ}\mathrm{C} - 14^{\circ}\mathrm{C} = 46\,\mathrm{K}\) The temperature difference across the heat exchanger is \(46\,\mathrm{K}\).

Calculate the heat transfer area using the overall heat transfer coefficient and the rate of heat transfer

We are given the overall heat transfer coefficient as \(1000\,\mathrm{W/m^2\cdot K}\). Now we can calculate the heat transfer area using the previously calculated rate of heat transfer \(Q = 728,000\,\mathrm{W}\) and the temperature difference \(\Delta T = 46\,\mathrm{K}\): \(A = \frac{Q}{U\Delta T} = \frac{728,000\,\mathrm{W}}{1000\,\mathrm{W/m^2\cdot K} \cdot 46\,\mathrm{K}} = 15.83\,\mathrm{m^2}\) The heat transfer area of the shell-and-tube heat exchanger is \(15.83\,\mathrm{m^2}\).