Question

Cold water \(\left(c_{p}=4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) leading to a shower enters a well-insulated, thin-walled, double-pipe, counterflow heat exchanger at \(15^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{kg} / \mathrm{s}\) and is heated to \(45^{\circ} \mathrm{C}\) by hot water \(\left(c_{p}=4.19 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) that enters at \(100^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{kg} / \mathrm{s}\). Determine \((a)\) the rate of heat transfer and \((b)\) the rate of exergy destruction in the heat exchanger. Take \(T_{0}=25^{\circ} \mathrm{C}\)

Short answer

Answer: The rate of heat transfer (a) is 31.35 kW and the rate of exergy destruction (b) is 0.6 kW.

Step-by-step solution

Calculate the Rate of Heat Transfer

Since both the hot and cold water have different heat capacities, we need to first calculate the energy transferred between the two fluids using the following formula: $$Q = \dot{m}_c c_{p_c} (T_{out_c} - T_{in_c}) = \dot{m}_h c_{p_h} (T_{in_h} - T_{out_h})$$ where \(Q\) is the rate of heat transfer, \(\dot{m}_c\) and \(\dot{m}_h\) are the mass flow rates of cold and hot water, respectively, \(c_{p_c}\) and \(c_{p_h}\) are the heat capacities of cold and hot water, and \(T_{in}\) and \(T_{out}\) represent the inlet and outlet temperatures of cold (\(c\)) and hot (\(h\)) fluids, respectively. We are given: $$\dot{m}_c = 0.25~kg/s, ~c_{p_c} = 4.18~kJ/(kg \cdot^{\circ} C), ~T_{in_c} = 15^{\circ} C, ~T_{out_c}= 45^{\circ} C$$ $$\dot{m}_h = 3~kg/s, ~c_{p_h} = 4.19~kJ/(kg \cdot^{\circ} C), ~T_{in_h} = 100^{\circ} C$$ Substituting these values, we can find \(T_{out_h}\): $$Q = 0.25 \cdot 4.18 \cdot (45 - 15) = 3 \cdot 4.19 \cdot (100 - T_{out_h})$$ $$Q = 31.35~kW$$ Solving for \(T_{out_h}\): $$100 - T_{out_h} = \frac{31.35}{12.57}$$ $$T_{out_h} = 75^{\circ} C$$

Calculate the Rate of Exergy Destruction

To find the rate of exergy destruction, we need to first determine the exergy change for both the cold and hot water streams. The exergy change can be calculated using the following formula: $$\Delta Ex = \dot{m} c_p T_0 (T_{out} - T_{in}) \left (1 - \frac{T_0}{(T_{in} + T_{out}) / 2} \right)$$ Calculating the exergy change for the cold water stream: $$\Delta Ex_c = 0.25 \cdot 4.18 \cdot 25 (45 - 15) \left(1 - \frac{25}{(15 + 45) / 2} \right)$$ $$\Delta Ex_c = 9.435~kW$$ Calculating the exergy change for the hot water stream: $$\Delta Ex_h = 3 \cdot 4.19 \cdot 25 (75 - 100) \left(1 - \frac{25}{(75 + 100) / 2} \right)$$ $$\Delta Ex_h = -8.835~kW$$ The rate of exergy destruction can now be found using the second law of thermodynamics: $$\Delta Ex_d = \Delta Ex_c + \Delta Ex_h$$ $$\Delta Ex_d = 9.435 - 8.835$$ $$\Delta Ex_d = 0.6~kW$$ So the rate of heat transfer \((a)\) is \(31.35~kW\), and the rate of exergy destruction \((b)\) is \(0.6~kW\).