Question

Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a crossflow heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of $3 \mathrm{~kg} / \mathrm{s}$, where it is heated by a hot water stream \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(70^{\circ} \mathrm{C}\) at a rate of $1 \mathrm{~kg} / \mathrm{s}$. Determine the maximum heat transfer rate and the outlet temperatures of both fluids for that case.

Short answer

The specific heat capacity of air and water are 1005 J/kg·K and 4190 J/kg·K, respectively. Answer: To find the maximum heat transfer rate and outlet temperatures, follow these steps: 1. Calculate the heat capacities of both fluids: - Heat capacity of air: \(C_{p,air}\cdot m_{air} = 1005 \cdot 3 = 3015 \text{ J/K}\) - Heat capacity of water: \(C_{p,water}\cdot m_{water} = 4190 \cdot 1 = 4190 \text{ J/K}\) 2. Determine the limiting fluid: - In this case, the air has a lower heat capacity (3015 J/K) compared to the water (4190 J/K), so it is the limiting fluid. 3. Calculate the maximum heat transfer rate: - \(Q_{max} = C_{min} \cdot \Delta T_{inlet} = 3015 \cdot (70 - 20) = 3015 \cdot 50 = 150,750 \text{ W}\) 4. Calculate the outlet temperatures: - For air: \(150,750 = 1005 \cdot 3 \cdot (T_{out,air} - 20)\) ⟹ \(T_{out,air} = \frac{150,750}{3015} + 20 = 70°C\) - For water: \(150,750 = 4190 \cdot 1 \cdot (70 - T_{out,water})\) ⟹ \(T_{out,water} = 70 - \frac{150,750}{4190} = 70 - 36 = 34°C\) Therefore, the maximum heat transfer rate is 150,750 W, and the outlet temperatures are 70°C for air and 34°C for water.

Step-by-step solution

Calculate the heat capacities

Begin by calculating the heat capacities of the air and water streams respectively. Heat capacity of air: \(C_{p,\text{air}}m_{\text{air}}\) Heat capacity of water: \(C_{p,\text{water}}m_{\text{water}}\)

Determine the limiting fluid

Next, identify the fluid with the smaller heat capacity since it will dictate the maximum heat transfer rate. This fluid is referred to as the limiting fluid.

Calculate the maximum heat transfer rate

The maximum heat transfer rate, Q_max, is determined by the temperature difference between the inlet streams and the heat capacity of the limiting fluid. \(Q_{\text{max}} = (C_{\text{min}}) \Delta T_\text{inlet}\) Compute the value of \(Q_{\text{max}}\) using the appropriate values.

Calculate the outlet temperatures

Since we now know \(Q_{\text{max}}\), we can use the conservation of energy to find the outlet temperatures of both fluids. For air: \(Q_{\text{max}} = C_{p,\text{air}}m_{\text{air}}(T_{\text{out,air}}-T_{\text{in,air}})\) For water: \(Q_{\text{max}} = C_{p,\text{water}}m_{\text{water}}(T_{\text{in,water}}-T_{\text{out,water}})\) Solve the equations to find the outlet temperatures, \(T_{\text{out,air}}\) and \(T_{\text{out,water}}\).