Document NEqZNQLLB79bJDRGdNyz67aQb

HEATING VENTILATING AIR CONDITIONING GUIDE 1940 ; 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 36,000 Totals.:..................................... ......--............-200,000 Outside air: 2000 X 0 075 X 60 X 0.24 X (95 - 80) = 32,350 2000 X 0.075 X 60 X (96 - 77.2) X 1060 7000 Heat gain through ducts______________ ____ -- 22,200 36,000 Btu per hour 25,600 Totals.____ ____----, :......................254,550 61,600 Btu per hour Total Cooling Load Tons Cooling Effect Ratio S.H. to T.H. = 316,150 Btu per hour = 26.4 = 200,000 = 0.85 (Room) 236,000 254,550 = .805 (Con316,150 . ditioner) Design Distribution System for Heating a. Total heat loss in space ' Assume grille temperature. 200,000 Q =-60 X 0.075 X 0.24 X (90 - 72) 200,000 Btu per hour 90 F ; , 10,300 cfm b. Total heat loss in ducts between unit and grilles ________ 33,200 A f =-60 X 0.075 X 0.24 X 10,300 c. 90 F plus 3 F 33.200 Btu per hour 3F 93: F temperature leaving coil Design Distribution System for Cooling a. Total sensible heat gain in space Assume grille temperature 200,000 Q ~ 60 X 0.075 X 0.24 X (80 - 62) 200,000 Btu per hour ,,. = ,nonn.r_. .. 10,300 C 62 F e 770 lb per minute' b. Latent heat gain in space = 36,000 Btu per hour = 600 Btu ,per minute 600 X 7000 1060 3970 gr per minute 3970 -5- 770 = 5.2 gr per pound air supplied Room condition = 80 F DB, 50% RH, 60 F DP = 77.2 gr per pound Gain in conditioned space = 5.2 gr per pound Specific humidity leaving conditioner Grille temperature = 62 F DB and 58 F DP 72.0. gr per pound c. Duct gain A t =-60 X 22,200 0.075 X 0.24 X 10,300 Leaving coil = 60 F DB, 58.7 F WB, 58 F DP 22.200 Btu per hour 2 F temperature rise 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 390 CHAPTER 21. CENTRAL SYSTEMS EOR COMFORT AIR CONDITIONING in only and this does not fluctuate greatly. If intermediate load conditions are im portant the calculations should be repeated for those loads. Example 2. Determine the cooling load for a theater if the following design con ditions are assumed. A dehumidifying air washer is to be used with and without the by-pass. , Outside air dry-bulb............. ............,-- ........... ....................... 94 F Outside air wet-bulb.--------------------- ------------------------------------- 75 F Inside air dry-bulb------------------------------------------------- ----------- 80 F Inside air wet-bulb........................... ............................--.............. 67 F Minimum outdoor air..................................................................... 6300 cfm People. --........... ...... ............................... --:-- ..................... ' 600 Lights................ .................................. ............................................. 4 kw Transmission gain--.............................................................. .........110,000 Btu per hour Solution. Without By-Pass. where h = dry-bulb temperature, room or return air, degrees Fahrenheit, ft = dry-bulb temperature at supply grille, degrees Fahrenheit, ft = saturation temperature leaving washer, degrees Fahrenheit, f" = dew-point in room, degrees Fahrenheit, - f" = dew-point at supply grille, degrees Fahrenheit. hi - enthalpy of room air, Btu per pound. he = enthalpy of outside air, Btu per pound. h = enthalpy of air leaving washer, Btu per pound. Design Load for Cooling from Item 3. Sensible Transmission........................................ 110,000 People (600)........ 132,000 Lights (4 kw)......... .............................. 13,800 Latent 720,000 Totals___________ ___ _______.255,800 Btu per hour 720,000- gr. per hour Determine Air Quantity Sensible heat gain in space Assume grille temperature ,, _ 255,800 y 60 X 0.24 (80 - 68) = 255,800 Btu per hour 68 F = 1,480 lb per min = 19,700 cfm ' Minimum outdoor air .= . 470 lb per min = 6,300 cfm Return air = 1,010 lb per min Moisture gain in space = 720,000 gr per hour 12,000 4- 1,480 = 8.11 gr per pound air supplied Room conditions = 80 F DB, 67 F WB, 60 F DP Gain in conditioned space = 13,400 cfm = 12,000 gr per min = 77.21 gr per lb =.. . 8.11 gr per lb Specific humidity leaving conditioner Grille conditions = 68 F DB and 57 F DP = 69.1 gr per lb 391