Question 9.3: For the facility described in Example 9.2, given the followi...

For the facility described in Example 9.2, given the following additional information, what is the cooling load? The front of the facility faces west, and all windows and doors have awnings. The lighting within the facility consists of 575 fluorescent luminaries, each containing two 60 watt lamps. Within the facility, 75 men performing light work and 50 women performing light work will be employed. The outside design temperature is 97°F, and the inside design temperature is 78°F [3].

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Using Equation 9.4 results in the following cooling loads for facility components:

Q_{F} =(25,000ft^{2})\left[0.81Btu/(hr)(ft^{2})(°F)\right](97°F-78°F) Q_{F} =3.85\times 10^{5}Btu/hr

 

Q_{R} =(25,000ft^{2})\left[0.20Btu/(hr)(ft^{2})(°F)\right](97°F-78°F) Q_{R} =0.95\times 10^{5}Btu/hr

 

Q_{G} =(16)(32ft^{2})\left[0.63Btu/(hr)(ft^{2})(°F)\right](97°F-78°F) Q_{G} =0.06\times 10^{2}Btu/hr

 

Q_{D} =(4)(24ft^{2})\left[1.13 Btu/(hr)(ft^{2})(°F)\right](97°F-78°F) Q_{D} =0.02\times 10^{5} Btu/hr

 

Q_{W} =\left[(2)(4500ft^{2})+(2)(1800ft^{2})-(16)(32ft^{2})-(4)(24ft^{2})\right] \left[(0.39 Btu/(hr)(ft^{2})(°F)\right]
(97^{\circ }F-78^{\circ }F)

Q_{W} =0.88\times 10^{5} Btu/hr Q_{V} =\left[(25,000ft^{2})+(12,600ft^{2})\right] \left[(0.20 Btu/(hr)(ft^{2})(°F)\right](70^{\circ }F-0^{\circ }F) Q_{V} =1.43\times 10^{5} Btu/hr

 

The heat absorption factor for the west side of a facility in Chicago, Illinois, is
100 Btu/(hr)(ft^{2}), and for the east side, 75 Btu/(hr)(ft^{2}) [10]. The shade factor awnings are 0.3 [10]. Using Equation 9.5, the cooling load from solar radiation may be calculated as

Q_{S} =\left[(8)(32ft^{2})+(2)(24ft^{2})\right] \left[100Btu/(hr)(ft^{2})(°F)(0.3)\right] +\left[(8)(32ft^{2})+(2)(24ft^2)\right] \left[75Btu/(hr)(ft^{2})(°F)(0.3)\right] Q_{S}=0.16\times 10^{5} Btu/hr

The cooling load from lighting may be calculated via Equation 9.7, as follows:

Q_{L}(fluorescent) = (575)(2)(60 watts) (4.25)\frac{Btu}{(hr)(watt)} Q_{l}=2.93×10^{5} Btu/hr

With the aid of Table 9.2, the cooling load from personnel may be calculated as

Q_{P}= 75(800 Btu/hr) + 50(700 Btu/hr) Q_{P}= 0.95 × 10^{5} Btu/hr

 

Then, using Equation 9.3, the total cooling load may be calculated as follows:
Q_{C}= 3.85 × 10^{5} + 0.95 × 10^{5} + 0.06 × 10^{5} + 0.02 × 10^{5} +0.88 × 10^{5} + 0.14 × 10^{5} + 0.16 × 10^{5} + 2.93 × 10^{5} + 0.95 × 10^{5}

Q_{C}= 9.94 × 10^{5} Btu/hr

Or, utilizing the conversion factor 12,000 Btu/hr= 1 ton, a total cooling load of 83 tons is
required to air condition the total facility.

Table 9.2 Central Equipment Room Area by Use

Use Area of Equipment Rooms
Residential 2%
Office/industrial 5-7%
Public assembly 10-15%
Hospital 25%
Laboratory 25-50%
Animal laboratory 50%

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