2) A building has 3 zones (A, B, and C). The zone design conditions are 75 F DB

Question

2) A building has 3 zones (A, B, and C). The zone design conditions are 75 F DB and 50 % RH. The building zones have the following peak cooling loads. A-660.000 BTU/hr B-400,000 BTU/hr; C-520,000 BTU/hr. The zones all have SHF= 0.8. The ventilation air is 18000 CFM and the outdoor air condition is 95 F DB and 75 F WB. The temperature rise due to the supply fan is 3 F. Ignore temperature difference due to return fan, duct, and heating coil. If the supply air temperature is designed at 55 F DB, For questions a and b, the building is equipped with a CAV Reheat System: a) what is the supply air CFM for each zone and for total building? b) if zones A and B only have half of the peak loads, but zone C has its peak load. Find out the reheaters' heating loads for A and B. Do the CFM requirement and cooling coil load change under this circumstance for Reheat System? c) if the zones A and B only have half of the peak loads, but zone C is having its peak load. Find out the CFM requirement and cooling coil load when the building is equipped with a VAV System;

Explanation / Answer

For people due to the activity they do in office some heat is generated and there is a heat transfer between body and ambient which resisted by clothing.

Assuming the people to be of weight 60 kg and height 1.80 m, and typical office work as typing.

Heat generated per single person = 1.1 met = 1.1 x 58.2 = 64.02 W / m2

Area = 0.202m0.425 h0.725 = 0.202 x (60)0.425 x (1.8)0.725 = 1.7625 m2

Power generated = 64.02 x 1.7625 = 112.83 W

Total power generated = 112.83 x 14 = 1.58 kW

Heat transfer between body and ambient = (36.9 - 23.88) x 1.7625/ 0.155 x 0.6 = 246.56 W

Total heat transfer = 3.45 kW

The resistance is measured in measure of clothing, where 1 clo = 0.155 m2K/W. Assuming a normal clothing of 0.6clo.

Heat generated by lights = 2 x 140 = 280 W

Heat required converting ambient air to zone temperature

Enthalpy of outside air = 43 Btu / lb

Enthalpy of inside air = 27.5 Btu / lb

Heat required = 43 - 27.5 = 15.5 Btu / lb

Ventilation flow rate = 15 ft3/min x 14 = 210 ft3/min

Density of air = 0.0765 lb/ft3

Mass flow rate of air = 0.0765 x 210 = 16.065 lb/min

Power required = 15.5 x 16.065 = 249 Btu/min = 14940 Btu/hr = 4.37 kW

Heat gain due to people = 1.58 + 3.45 = 5.03 kW

Heat gain due to lights = 0.28 kW

Heat gain due to ventilation = 4.37 kW

Total heat gain = 5.03+0.28+4.37 = 9.68 kW

Contribution of heat gain due to people = 5.03/9.68 = 51.96%

Contribution of heat gain due to lights = 0.28/9.68 = 2.89%

Contribution of heat gain due to people = 4.37/9.68 = 45.14%

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