Question

Consider that specific heat (0 to 50°C) of water, water vapour, and air remains constant: 4.48, 1.88, and 1.0 kJ/(kg °C), respectively. Assuming, the heat energy required to convert 1 kg of water to water vapour at 0°C is 2000 kJ, the enthalpy (in kJ/kg dry air) of atmospheric air containing 0.05 kg water vapour per kg dry air at 50°C is ............. (rounded off to 1 decimal place).

Hint:

When calculating enthalpy changes for humid air, consider the individual contributions of both dry air and water vapour, using specific heat values and mass ratios for accurate results.

Answer 1

To calculate the enthalpy, we need to consider the following components:
1. The enthalpy of dry air (\(h_{air}\)) at 50°C, which is given by: \[ h_{air} = c_{air} \cdot T \] Where \( c_{air} = 1.0 \, {kJ/kg°C} \) and \( T = 50°C \). So: \[ h_{air} = 1.0 \times 50 = 50 \, {kJ/kg} \] 2. The enthalpy of water vapour (\(h_{vapor}\)) at 50°C:
The enthalpy of water vapour is calculated as: \[ h_{vapor} = h_{vapor\_0} + c_{vapor} \cdot T \] Where \( h_{vapor\_0} = 2000 \, {kJ/kg} \) at 0°C, \( c_{vapor} = 1.88 \, {kJ/kg°C} \), and \( T = 50°C \). Therefore: \[ h_{vapor} = 2000 + 1.88 \times 50 = 2000 + 94 = 2094 \, {kJ/kg} \] 3. The total enthalpy of the air with water vapour is calculated by considering the mass of water vapour in the air: \[ h_{total} = h_{air} + (h_{vapor} \times {mass of water vapour}) \] The mass of water vapour is \( 0.05 \, {kg water vapour/kg dry air} \), so: \[ h_{total} = 50 + (2094 \times 0.05) = 50 + 104.7 = 154.7 \, {kJ/kg} \] Thus, the enthalpy lies between 152.0 to 156.0 kJ/kg dry air.