Question

Air at 101 kPa, 15°C and 50% relative humidity is first heated to 20°C in a heating coil, and then humidified by spraying water on it. In the final state, the air has temperature of 25°C and relative humidity of 85%. The amount of water sprayed (in gm per kg of dry air) is ........ (rounded off to two decimal places). Use the following data: The saturation pressure of water at 15 °C = 1.7057 kPa The saturation pressure of water at 25 °C = 3.1698 kPa

Hint:

To find the amount of water added to a mixture, calculate the change in humidity ratio before and after heating/humidifying the air.

Answer 1

We are given that:
- Initial pressure of air \( P = 101 \, {kPa} \),
- Initial temperature \( T_1 = 15^\circ C \),
- Relative humidity at 15°C is 50%,
- Final temperature \( T_2 = 25^\circ C \),
- Final relative humidity at 25°C is 85%.
The amount of water vapor in the air can be determined by calculating the humidity ratio at both the initial and final conditions.
Step 1: First, we calculate the saturation pressure of water at 15°C and 25°C: At \( T_1 = 15^\circ C \), the saturation pressure of water is: \[ P_{{sat, 15°C}} = 1.7057 \, {kPa} \] At \( T_2 = 25^\circ C \), the saturation pressure of water is: \[ P_{{sat, 25°C}} = 3.1698 \, {kPa} \] Step 2: The partial pressure of water vapor at the initial condition (15°C) is given by: \[ P_{{water, initial}} = {Relative humidity} \times P_{{sat, 15°C}} = 0.50 \times 1.7057 = 0.85285 \, {kPa} \] Step 3: The humidity ratio \( \omega \) is calculated using the equation: \[ \omega = 0.622 \times \frac{P_{{water}}}{P_{{air}}} \] where \( P_{{air}} = 101 - P_{{water}} \). For the initial condition, this gives: \[ \omega_{{initial}} = 0.622 \times \frac{0.85285}{101 - 0.85285} = 0.622 \times \frac{0.85285}{100.14715} = 0.00525 \, {kg water/kg dry air} \] Step 4: Similarly, for the final condition, where the relative humidity is 85% at 25°C, we get the saturation pressure at 25°C and calculate the water vapor pressure: \[ P_{{water, final}} = 0.85 \times 3.1698 = 2.6953 \, {kPa} \] Step 5: The final humidity ratio is: \[ \omega_{{final}} = 0.622 \times \frac{2.6953}{101 - 2.6953} = 0.622 \times \frac{2.6953}{98.3047} = 0.0168 \, {kg water/kg dry air} \] Step 6: The amount of water vapor added is: \[ {Water added} = \omega_{{final}} - \omega_{{initial}} = 0.0168 - 0.00525 = 0.01155 \, {kg water/kg dry air} \] This value is approximately between 11.5 to 12.0 gm per kg of dry air.