Question 3.5: Moist air at 80 F db and 67 F wb is cooled to 58 F db and 80......
Moist air at 80 F db and 67 F wb is cooled to 58 F db and 80 percent relative humidity. The volume flow rate is 2000 cfm, and the condensate leaves at 60 F. Find the heat transfer rate.
Equation 3-30 applies to this process, which is shown in Fig. 3-5. The following properties are read from Chart 1a: \text{v}_{1} = 13.85 ft^{3} lbma, i_{1} = 31.4 Btu/lbma, W_{1} = 0.0112 lbmv/lbma, i_{2} = 22.8 Btu/lbma, W_{2} = 0.0082 lbmv/lbma. The enthalpy of the condensate is obtained from Table A-1a, i_{\text{w}} = 28.08 Btu/lbmw. The mass flow rate \dot{m}_{a} is obtained from Eq. 3-27:
\dot{q} = \dot{m}_{a}(i_{1} – i_{2}) – \dot{m}_{a}(W_{1} – W_{2}) i_{\text{w}} (3-30)
\dot{m}_{a} = \frac{\bar{V}_{1} A_{1}}{\text{v}_{1}} = \frac{\dot{Q}_{1}}{\text{v}_{1}} (3-27)
\dot{m}_{a} = \frac{2000(60)}{13.88} 60 = 8646 lbma/hr
Then
\dot{q} = 8646[(31.4 – 22.8) – (0.0112 – 0.0082)28.8]
\dot{q} = 8646[(8.6) – (0.084)]
The last term, which represents the energy of the condensate, is seen to be small. Neglecting the condensate term, \dot{q} = 74,356 Btu/hr = 6.2 tons.
