Question 13.16: Heating and Humidification Outdoor winter air enters a heati......
Heating and Humidification
Outdoor winter air enters a heating and humidification system at a rate of 900 cfm (425 L/s) and at 41°F (5°C) dry-bulb and 35°F (1.7°C) wet-bulb temperatures. The air is to be heated to 88°F (31°C) and humidified to a humidity ratio of 0.008 lb_{w}/lb_{a}. Determine the amount of sensible heating that needs to be provided by the heating coil and the amount of saturated steam at 212°F (100°C) that needs to be provided for humidification.
Given: T_{db1} = 41°F, T_{wb1} = 35°F, T_{db3} = 88°F, W_{3} = 0.008 lb_{w}/lb_{a}, and \dot{V}_{a} = 900 ft^{3}/min .
Figure: See Figure 13.18.
Find: \dot{Q}_{h} and \dot{m}_{w}.
Lookup values: Saturated steam enthalpy at 212°F from Table A3 is h_{vap} = 1150.5 Btu/lb_{w} (which is equal to h_{liq,2}). From psychrometric chart, h_{1} = 13 Btu/lb_{a}, v_{1} = 12.67 ft^{3}/lb_{a}, W_{1} = 0.003 lb_{w}/lb_{a}, and h_{3} = 30 Btu/lb_{a}.
The mass flow rate of dry air is
\dot{m}_{a} = \frac{\dot{V}_{a}}{v_{1}} = \frac{900 ft^{3}/min}{12.67 ft^{3}/lb_{a}} = 71.03 lb_{a}/min = 4262 lb_{a}/hThe mass flow rate of steam is
\dot{m}_{w} = \dot{m}_{a} (W_{3} – W_{1}) = 4262 lb_{a}/h \times (0.008 – 0.003) lb_{w}/lb_{a}= 21.31 lb_{w}/h (9.6 kg/h)
An energy balance on the humidifier only is used to determine the air enthalpy at state 2:
h_{2} = h_{3} – \left\lgroup \frac{\dot{m}_{w}}{\dot{m}_{a}} \right\rgroup \times h_{vap}= 30 Btu/lb_{a} – \left\lgroup \frac{21.31}{4262} \right\rgroup \times 1150.5 Btu/lb_{a}
= 24.25 Btu/lb_{a}
From the psychrometric chart, this corresponds approximately to an exit condition of T_{db2} = 87°F, i.e., the humidication occurred essentially at constant T_{db}.
Finally, the heating coil energy is easily found as
= 47,936.8 Btu/h (14.05 kW)