Question

In an air-conditioning system, air enters at 20$^\circ$C and 30% relative humidity at a steady rate of 30 m$^3$/min in a humidifier and it is conditioned to 25$^\circ$C and 60% relative humidity. Assuming entire process takes place at pressure of 100 kPa, the mass flow rate of steam added to air in the humidifier is $____________$ kg/min (rounded off to three decimal places).

Hint:

In humidification, first calculate humidity ratio at both states and then multiply by dry air mass flow rate.

Answer 1

Correct Answer: 0.262

Step 1: Humidity ratio at state 1. At 20$^\circ$C: $p_{sat} = 2.3392 \, kPa$. Relative humidity = 0.3. \[ p_v = 0.3 \times 2.3392 = 0.702 \, kPa \] \[ \omega_1 = 0.622 \frac{p_v}{p_{atm} - p_v} = 0.622 \frac{0.702}{100 - 0.702} = 0.00438 \, kg/kg_{da} \]
Step 2: Humidity ratio at state 2. At 25$^\circ$C: $p_{sat} = 3.1698 \, kPa$. Relative humidity = 0.6. \[ p_v = 0.6 \times 3.1698 = 1.902 \, kPa \] \[ \omega_2 = 0.622 \frac{1.902}{100 - 1.902} = 0.01142 \, kg/kg_{da} \]
Step 3: Mass flow rate of dry air. Volume flow = 30 m$^3$/min = 0.5 m$^3$/s. At 20$^\circ$C (293 K): \[ \dot{m}_{da} = \frac{p V}{RT} = \frac{100 \times 0.5}{0.287 \times 293} = 0.598 \, kg/s = 35.9 \, kg/min \]
Step 4: Water added. \[ \dot{m}_w = \dot{m}_{da} (\omega_2 - \omega_1) = 35.9 (0.01142 - 0.00438) = 0.451 \, kg/min \] Final Answer: \[ \boxed{0.451 \, kg/min} \] % Quicktip