Question 3.6: Moist air at 60 F db and 20 percent relative humidity enters......
Moist air at 60 F db and 20 percent relative humidity enters a heater and humidifier at the rate of 1600 cfm. Heating of the air is followed by adiabatic humidification so that it leaves at 115 F db and a relative humidity of 30 percent. Saturated water vapor at 212 F is injected. Determine the required heat transfer rate and mass flow rate of water vapor.
Figure 3-6 is a schematic of the apparatus. Locate the states as shown in Fig. 3-7 from the given information and Eq. 3-40 using the protractor feature of the psychrometric chart. Process 1 − χ is sensible heating; therefore, a horizontal line to the right of state 1 is constructed. Process χ − 2 is determined from Eq. 3-40 and the protractor:
\frac{i_{2} – i_{1}}{W_{2} – W_{1}} = i_{w} = \frac{Δi}{ΔW} (3-40)
\frac{Δi}{ΔW} = i_{\text{w}} = 1150.4 Btu/lbm
where i_{w} is read from Table A-1a. A parallel line is drawn from state 2 as shown in Fig. 3-7. State χ is determined by the intersection on lines 1 − χ and χ − 2. The heat transfer rate is then given by
\dot{q} = \dot{m}_{a} (i_{x} – i_{1})where
\dot{m}_{a} = \frac{\dot{Q} (60)}{\text{v}_{1}} = \frac{1600}{13.16} 60 = 7296 lbma/hr
and i_{1} and i_{x}, read from Chart 1a, are 16.8 and 29.2 Btu/lbma, respectively. Then
\dot{q} = 7296 (29.2 – 16.8) = 90,500 Btu/hr
The mass flow rate of the water vapor is given by
\dot{m}_{\text{v}} = \dot{m}_{a} (W_{2} – W_{1})
where W_{2} and W_{1} are read from Chart 1a as 0.0193 and 0.0022 lbmv/lbma, respectively. Then
\dot{m}_{\text{v}} = 7296 (0.0193 – 0.0022 = 125 lbmv/hr
