Document Z9opKvgVV209N2kM6wp5xw2L

Heating Ventilating Air Conditioning Guide 1939 this radiation for heating the outside walls and windows. At the same time, it must be controlled to prevent overheating the whole system. ^ few uncontrolled radiators will often change the heat balance and cause excessive overheating of the whole zone, without occupants near the radiators realizing the source of the trouble. All radiation used locally in connection with the control system should be equipped with automatic control to prevent overheating. The apparatus should be placed for minimum piping and duct work but it must be accessible for maintenance, repair, and cleaning. The air conditioning unit should be designed to provide access for cleaning coils drip pans, eliminators and for very easy maintenance on filters. This equipment will be in operation for many years, probably for the life of the building, and a little thought spent in the plan will simplify maintenance and assure successful results. Example 1. With the assumed values as indicated perform the essential calculations, to determine the design loads for heating and cooling and the necessary factors for'designing the distribution system. Solution: Design Conditions from Item 1. Outside air dry-bulb, winter.................... Outside air dry-bulb, summer.... ............ Outside air wet-bulb, summer.... ............. Inside air dry-bulb, winter...... ................ Inside air wet-bulb, winter................ ...... Inside air relative humidity, winter....... Inside air dry-bulb, summer.................... Inside air wet-bulb, summer................... Inside air relative humidity, summer.-- 200 people--4 kw light load. . 0F .95 F .75 F .72 F .56 F .35 per cent .80 F .66.7 F .50 per cent Design Load for Heating from Item S. Sensible heat loss through walls, etc-........................................200,000 Btu per hour Outside air--2000 cfm X 0.075 X 60 X 0.24 X(72 - 0) = 155,500 Btu per hour Humidification- -2-0--0--0---X---0--.-0-7--5- ---X (470qq-q5)--X---6--0---X----1-0--4--0 = 46,9^00 Btu per hour Heat loss through ducts................................................................ 33,200 Btu per hour Heat gain from lights, etc........ ................... ..............Disregard Total heating capacity......... .........................................................435,600 Btu per hour Design Load for Cooling and Dehumidifying from Item 3. \ Sensible Heat gain through walls, etc............................. .... 128,160 Heat gain from occupants (200)....................... .... 44,000 Heat emission from appliances.............................. 2,000 Heat gain from lights (4 kw)....................:........ .... 13,840 Heat gain from solar radiation!............................. 12,000 Latent 36,000 Totals, ...................................................... .... 200,000 36,000 Btu per hour Outside air: 2000 X 15 X 0.075 X 0.024 X 60 -____ 2000 X 0.075 X (110 - 80) X 60 X 1040 7000 , Heat gain through ducts................................... .... 22,200 40,100 Totals. --........................................................... ..:..254,550 428 76,100 Btu per hour 21. Central Systems for Comfort Air Conditioning Total Cooling Capacity = 330,650 Btu per hour Tons'Cooling Effect- . Heat Load Ratio = 27.5 200,000 ,,,,,, 236,000 '85 Design Distribution System for Heating a. Total heating loss in space Using 72 F plus 18 F n =' 200,000 200,000 Btu per hour 90 F entering air 10,300 cfm b Total heat loss in ducts between unit and grilles 33,200 to = 60 X 0.075 X 0.24 X 10;300 c. 90 F plus 3 F Design Distribution System for Cooling 33,200 Btu per hour 3F 93 F temperature leaving coil a. Total sensible heat gain in space Using 80 F minus 18 F 200,000 060 X 0.075 X 0.24 X 18 10,300 cfm 200,000 Btu per hour 62 F entering air 770 lb Latent heat gain -- 36,000 Btu per hour = 600 Btu per minute = 4020grains perminute = 5.2grainsper pound Room condition = 80FDB, 66.7 FWB, 50% RH, 60 FDP, = 77.3 grains Entering air = 62 FDB, 59.5 FWB, 87% RH, 58 FDP, = 72.1 grains b. Duct gain = 22,200 Btu per hour 22 200 to = 60 X. 0.075 X 0.24 X 10,300 = 2 F temperature rise Leaving coil = 60 FDB, 58.7 FWB, 58 FDP These calculations are based on maximum load conditions as set forth in the design. For intermediate loads the calculations may show entirely different relationship. For example, the entering air temperature will approach the space temperature as the sensible heat gain or loss decreases, due to outside temperature change, entrance or exit of people, use of artificial lighting, direction and intensity of sun's rays. The entering dew-point will remain much more uniform as it is affected by changes in room moisture gain only and this does not fluctuate greatly. If intermediate load conditions are im portant the calculations should be repeated for those loads. PROBLEMS IN PRACTICE 1 Consider a central heating system handling 10,000 cfm. The resistance to air flow offered by one coil arrangement is 0.9 in. of water and by another coil arrangement is 0.2 in. of water. The fan operates 4000 hours per year and the combined efficiency of motor and fan is 60 per cent. Determine the annual energy saving if the second coil is used. Difference, in system resistance = 0.9 -- 0.2 = 0.7 in. of water. Reduction in power input = = 1-83 hp. Annual energy saving = 1.83 X 0.764 X 4000 = 5480 kwhr. 2# A group of three drafting rooms, having a total volume of 27,000 cu ft, a transmission loss of 110,100 Btu per hour, and an infiltration loss of 34,200 Btu per 429