Question

An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water $\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)$ enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools $5000 \mathrm{~kg} / \mathrm{h}\( of air \)\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( which enters the air handler at \)25^{\circ} \mathrm{C}$. If these streams are in counterflow and the water-stream conditions remain fixed, the minimum temperature at the air outlet is (a) \(5^{\circ} \mathrm{C}\) (b) \(12^{\circ} \mathrm{C}\) (c) \(19^{\circ} \mathrm{C}\) (d) \(22^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

Short answer

Answer: (b) 12°C

Step-by-step solution

Calculate the heat transfer from the water stream

We can determine the heat transfer, Q, from the water stream using the equation: Q = m_water * c_p_water * (T_out_water - T_in_water) where m_water is the flow rate of water, c_p_water is the specific heat capacity of water, T_out_water is the leaving temperature of the water, and T_in_water is the entering temperature of the water. Given values: m_water = 1000 kg/h c_p_water = 4.2 kJ/(kg.K) T_in_water = 5°C T_out_water = 12°C Now, we can plug these values into the equation to obtain the heat transfer: Q = 1000 * 4.2 * (12 - 5) Q = 4200 * 7 Q = 29400 kJ/h

Calculate the heat gained by the air stream

Using the heat transfer, Q, and given information about the air stream, we'll find the final air temperature (T_out_air) using the equation: Q = m_air * c_p_air * (T_out_air - T_in_air) where m_air is the mass flow rate of air, c_p_air is the specific heat capacity of air, T_in_air is the entering temperature of the air, and T_out_air is the leaving temperature of the air. Given values: Q = 29400 kJ/h (from step 1) m_air = 5000 kg/h c_p_air = 1.0 kJ/(kg.K) T_in_air = 25°C Now, we can rearrange the equation to solve for T_out_air: T_out_air = Q / (m_air * c_p_air) + T_in_air T_out_air = 29400 / (5000 * 1) + 25 T_out_air = 5.88 + 25 T_out_air = 30.88°C Since these streams are in counterflow and the water-stream conditions remain fixed, the minimum temperature at the air outlet is the same as the outlet temperature of the water stream, which is 12°C. The correct answer is: (b) \(12^{\circ} \mathrm{C}\)