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: - Saturation pressure of water at 15°C = 1.7057 kPa - Saturation pressure of water at 25°C = 3.1698 kPa

Hint:

For humidification problems, always calculate humidity ratio at both states using: \[ \omega = 0.622 \cdot \frac{\phi P_{sat}}{P - \phi P_{sat}} \] Then, subtract to find the amount of water added per kg of dry air.

Answer 1

Correct Answer: 11.5

Step 1: Formula for humidity ratio.
The humidity ratio (kg water vapour / kg dry air) is: \[ \omega = 0.622 \cdot \frac{\phi \cdot P_{sat}}{P - \phi \cdot P_{sat}} \] where \(\phi\) = relative humidity, \(P_{sat}\) = saturation pressure at given \(T\), and \(P\) = total pressure (101 kPa).

Step 2: Initial humidity ratio (at 15°C, 50% RH).
\[ \omega_1 = 0.622 \cdot \frac{0.5 \times 1.7057}{101 - 0.5 \times 1.7057} \] \[ \omega_1 = 0.622 \cdot \frac{0.85285}{101 - 0.85285} = 0.622 \cdot \frac{0.85285}{100.147} = 0.00530 \, \text{kg/kg dry air} \]

Step 3: Final humidity ratio (at 25°C, 85% RH).
\[ \omega_2 = 0.622 \cdot \frac{0.85 \times 3.1698}{101 - 0.85 \times 3.1698} \] \[ \omega_2 = 0.622 \cdot \frac{2.694}{101 - 2.694} = 0.622 \cdot \frac{2.694}{98.306} = 0.01703 \, \text{kg/kg dry air} \]

Step 4: Water added.
\[ \Delta \omega = \omega_2 - \omega_1 = 0.01703 - 0.00530 = 0.01173 \, \text{kg/kg dry air} \] Convert to grams per kg of dry air: \[ \Delta \omega = 0.01173 \times 1000 = 11.73 \, g/kg \, \text{dry air} \]



Step 5: Rounding.
Rounded to two decimal places: \[ \Delta \omega = 11.73 \, g/kg \, dry \, air \] Final Answer:
\[ \boxed{11.73 \, g/kg \, dry \, air} \]