Chapter 19

Q. 19.1

Building Air-Conditioning

A space is to be maintained at 78°F (25.5°C) drybulb temperature and 50% RH. The total cooling load (heat to be removed from the space to maintain comfort) is 120,000 Btu/h (35 kW) of which 70% is sensible heat. Ventilation air at 1000 ft³/min (472 L/s) is required on the peak day when the outdoor conditions are 95°F (35°C) and 55% RH.
The room supply air temperature is 20°F (11.1°C) below that of the room set point temperature (a typical value for commercial buildings at peak operating conditions). What are the space air supply flow rate and the cooling coil rating (total cooling capacity and SHR)?
Figure: See Figures 19.1 and 19.3.
Assumptions: The location is at sea level. The supply air temperature to the space is 58°F. The duct heat transfer and the fan air temperature rise are ignored for simplicity.
Given: SHR _{space} = 0.7,    \dot{Q}_{space,cool} = 120,000  Btu/h,    \dot{V}_{0} = 1,000  ft^{3}/min = 60,000  ft^{3}/h

T_{db,6} = 78°F, T_{db,o} = 95°F, \phi_{o} = 55\%, T_{db,5} = 58°F
(outdoor and indoor air conditions)
Find: \dot{m}_{a}, \dot{Q}_{cc,tot}, SHR_{cc}
Lookup values: Specific volume v_{o} = 14.4  ft^{3}/lb_{a},  W_{o} = 0.0197  lb_{w}/lb_{a}, W_{6} = 0.0103  lb_{w}/lb_{a}



Verified Solution

The psychrometric chart (Figure 13.6) will be used to illustrate the graphical solution approach.
1. Locate specified points 0 and 6 on the psychometric chart.
2. Locate supply air point 5. The conditioned space inlet and outlet conditions lie on line 3–6 in Figure 19.3. Condition 6 (78°F, 50% humidity) is already known from the problem statement. The slope of line 3–6 is determined from the inner scale of the protractor as shown for SHR = 0.7.
The location of point 3 is the intersection of the line 3–6 and a vertical line drawn
from T_{db,3} = 58°F. This corresponds to a RH value of 80%.
3. Find the space supply airflow rate. The first law for this process is
\dot{Q}_{space,cool} = \dot{m}_{a} (h_{6}  –  h_{5})              (19.6)
The enthalpies can be read from Figure 13.6: h_{5} = h_{3} = 23  Btu/lb_{a}, h_{6} = 30  Btu/lb_{a}
The air mass flow rate is thus
\dot{m}_{a} = \frac{12,000  Btu/h}{(30  –  23)  Btu/lb_{a}}


= 17,140  lb_{a}/h  (3770  ft^{3}/min)
4. Locate mixed air point 1. The end points 0 and 9(6) are known, and so the inverse
relation between mass flows and line segment lengths (Equation 13.32) is used to locate point 1 (see Section 13.6.2). The outdoor air mass flow is

\dot{m}_{a,0} = \frac{\dot{V}_{0}}{v_{0}} = \frac{60,000  ft^{3}/h}{14.4  ft^{3}/lb_{a}} = 4,170  lb_{a}/h
The ratio of air mass ows is (4,170/17,140) = 0.243. Therefore, point 1 is located 24.3% of the distance from point 6 along line segment 6–0. The properties at point
1 can be read from the psychrometric chart. They are T_{db} = 82°F, T_{wb} = 69°F, and moist air enthalpy h_{a} = 33.4  Btu/lb_{a}.
Alternatively, we can use the weighted average rule to calculate the mixed air
humidity ratio:
W_{1} = [0.243 × 0.0197  lb_{w}/lb_{a} + (1  −  0.243) × 0.0103  lb_{w}/lb_{a}] = 0.01216  lb_{w}/lb_{a}
Similarly, assuming constant-specific heats, the mixed air dry-bulb temperature is as follows:
T_{db,1} = [0.243 × 95°F + (1  −  0.243) × 78°F)] = 82°F (consistent with the chart reading).
This allows point 1 to be fixed on the psychrometric chart.
5. Determine cooling coil load. Line 1–3 can now be constructed by connecting points 1 and 3. The slope of the resulting line is transposed to the protractor, and the
coil’s sensible heat ratio SHR_{cc} can be read off as 0.55. The corresponding relative humidity is 80%, and the humidity ratio is 0.0084  lb_{w}/lb_{a}. Finally, the coil heat removal rate (or “coil cooling load”) can be found.

\dot{Q}_{cc,tot} = \dot{m}_{a} (h_{1}  –  h_{3}) = 17,140  lb_{a}/h \times (33.4  –  23)  Btu/lb_{a}

= 178,300  Btu/h       or     14.9  tons (52.2 kW)

Peak-condition coil information (air and water ow rates, SHR, and heat rate) is  supplied to manufacturers for coil selection, costing, and fabrication purposes. Note that the cooling coil load and zone cooling loads are quite different because the coil must remove zone heat and humidity gains as well as the latent and sensible loads imposed by ventilation (or outdoor) air. Hence, in most cases, both the SHR  quantities will be different, i.e., SHR_{Room} ≠ SHR_{cc}.