Question 13.14: Cooling of Moist Air Outdoor air at sea level enters an air ......

(a) First, we will find the mass flow rate:

Then, using Equation 13.42, the heat rate can be found since all values on the right-hand side are known:

\dot{Q}_{tot,c} = \dot{m}_{a} [(h_{1}  –  h_{2})  –  h_{liq} (W_{1}  –  W_{2}) ]            (13.42)

= (134,995  –  707.3) Btu/h = 134.3  kBtu/h(39.3  kW)

= 11.2 tons of cooling

In order to determine the contributions of the sensible and latent coil loads, we locate point “a” as the intersection on the psychrometric chart of a horizontal line drawn from point 2 and the vertical line drawn from point 1. The enthalpy is found to be h_{a} = 33.8  Btu/lb_{a}.
From Equation 13.43,

\dot{Q}_{sen} = \dot{m}_{a}(h_{a}  –  h_{2})             (13.43)

= 96,663  Btu/h = 8.06 tons

and from Equation 13.44,

\dot{Q}_{lat} = \dot{m}_{a} (h_{1}  –  h_{a})              (13.44)

= 38,332 Btu/h = 3.19 tons

Finally, the total cooling load on the coil is

It is instructive to compare these values with those from the simplied approach often adopted for hand calculations. The average specic heat of the air water mixture is (Equation 13.25)

Then, from Equation 13.45

\dot{Q}_{sen} = \dot{m}_{a} c_{p,m} (T_{db1}  –  T_{db2})          (13.45)

= 91,871  Btu/h = 7.66 tons

And from Equation 13.46,

\dot{Q}_{lat} = \dot{m}_{a} h_{vap}  (W_{1}  –  W_{2})          (13.46)

= 42,102  Btu/h = 3.51 tons

Finally,

Note that the differences in the sensible, latent, and total loads by both methods are −5%, +10%, and less than 1%, respectively.
(b) The cooling coil sensible heat factor is

We draw a straight line between points 1 and 2 extending till the saturated line (not shown in Figure 13.16). The corresponding dry-bulb temperature T_{3} is the ADP, and this is read off as ADP = T_{db3} = 50°F.
Finally, from Equation 13.48, we find

= \frac{Segment  2 – 3}{Segment  1 – 3} = \frac{T_{db2}  –  T_{db3}}{T_{db1}  –  T_{db3}}            (13.48)

Comments
This cooling effect must be provided by the chiller. Note that the amount of energy leaving the coil in the form of liquid condensate for this problem (707 Btu/h as determined earlier) is very small, about 0.5% of the total coil load.