Question 3.4: Find the heat transfer rate required to warm 1500 cfm (ft³/m......
Find the heat transfer rate required to warm 1500 cfm (ft³/min) of air at 60 F and 90 percent relative humidity to 110 F without the addition of moisture.
Equations 3-22 or 3-25 may be used to find the required heat transfer rate. First it is necessary to find the mass flow rate of the dry air:
\dot{m}_{a} i_{2} + \dot{q} = \dot{m}_{a} i_{1} (3-22)
\dot{q}_{s} = \dot{m}_{a} c_{p} (t_{2} – t_{1}) (heating) (3-25a)
\dot{q}_{s} = \dot{m}_{a} c_{p} (t_{2} – t_{1}) (cooling) (3-25b)
\dot{m}_{a} = \frac{\bar{V}_{1} A_{1}}{ν_{1}} = \frac{\dot{Q_{1}}}{ν_{1}} (3-27)
The specific volume is read from Chart 1a at t_{1} = 60 F and \phi = 90 percent as 13.33 ft³/lbma:
\dot{m}_{a} = \frac{1500(60)}{13.33} = 6752 lbma/hr
Also from Chart 1a, i_{1} = 25.1 Btu/lbma and i_{2} = 37.4 Btu/lbma. Then by using Eq. 3-22, we get
\dot{q} = 6752(37.4 – 25.1) = 83,050 Btu/hr
or if we had chosen to use Eq. 3-25,
\dot{q} = 6752(0.244) (110 – 60) = 82,374 Btu/hr
Agreement between the two methods is within 1 percent.