Consider the psychrometric process denoted by the straight line from state 1 to 2 in the figure. The specific humidity, Dry Bulb Temperature (DBT), and Wet Bulb Temperature (WBT) at the two states are shown in the table. The latent heat of vaporization of water \( h_{fg} = 2440 \, {kJ/kg} \). If the flow rate of air is 1 kg/s, the rate of heat transfer from the air is _________ kW (rounded off to two decimal places).
Show Hint
In psychrometric processes, the heat transferred is closely related to the change in specific humidity and the latent heat of vaporization, especially in cooling or dehumidification processes.
Step 1: Understand the process. The process from state 1 to state 2 is a constant Wet Bulb Temperature (WBT) process. During this process, the air is cooling, and the water vapor is condensing, which leads to heat transfer from the air. Step 2: Determine the heat removed from the air. The heat removed from the air, \( Q \), is related to the change in specific humidity and the latent heat of vaporization: \[ Q = \dot{m} \cdot h_{fg} \cdot (w_1 - w_2), \] where: \( \dot{m} = 1 \, {kg/s} \) is the mass flow rate of air, \( h_{fg} = 2440 \, {kJ/kg} \) is the latent heat of vaporization, \( w_1 = 0.020 \, {kg of water vapor / kg of dry air} \) is the specific humidity at state 1, \( w_2 = 0.015 \, {kg of water vapor / kg of dry air} \) is the specific humidity at state 2. Substituting the values: \[ Q = 1 \cdot 2440 \cdot (0.020 - 0.015). \] \[ Q = 2440 \cdot 0.005 = 12.2 \, {kJ/s}. \] Step 3: Convert to kW. Since \( 1 \, {kJ/s} = 1 \, {kW} \), the rate of heat transfer is: \[ Q = 12.2 \, {kW}. \] Final Answer: The rate of heat transfer from the air is \( \boxed{12.2} \, {kW} \).
