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AIR CONDITIONING LOAD CALCULATIONS PRESENTATIONS by EVRAJU

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AIR CONDITIONING LOAD CALCULATIONS PRESENTATIONS by EVRAJU

  1. 1. PRESENTATION E VENKATA RAJU E Venkata Raju - MEP Projects&Facilities management Professional
  2. 2. HEATING AND COOING LOADS To estimate the air onditioningload(Both heating and cooling loads) the following factors are to be considered. 1. Building size, shape and orientation. 2. Materials of construction 3. Glass areas 4. People 5. Ventilation 6. Motors, lights and appliances 7. Water, gas and electrical services 8. Equipment location 9. Infiltration 10. Location of doors and windows etc.. E Venkata Raju - MEP Projects&Facilities management Professional
  3. 3. HEATING AND COOING LOADS  Sensible heat gain: When there is a direct addition of heat to the enclosed space, a gain in the sensible heat is said to occur. This sensible heat is to be removed during the process of summer air conditioning. This heat gain may occur due to any one or all of the following sources of heat transfer. 1.The heat flowing into the building by conduction through exterior walls,floors,ceilings,doors and windows due to the temperature difference on their two sides. 2.The heat received from solar radiation .It consists of a) The heat transmitted directly through glass of windows, ventilators or doors, and b)The heat absorbed by walls and roofs exposed to solar radiation and later on transferred to the room by conduction. 3.The heat conducted through interior partition from rooms in the same building which are not conditioned. 4.The heat given off by lights,motors,machinery,cooking operations,industrial process etc. 5.The heat liberated by the occupants. E Venkata Raju - MEP Projects&Facilities management Professional
  4. 4. HEATING AND COOING LOADS  6.The heat carried by outside air which leaks in (infiltration air) through the cracks in doors, windows, and through their frequent openings.  7.The heat gain through the walls of ducts carrying air through unconditioned space in the building.  8.The heat gain by the fan work. Latent heat gain: When there is an addition of water vapour to the air of enclosed space,a gain in latent heat is said to occur. This latent heat is to be removed during the process of summer air conditioning. This heat gain may occur due to any one or all of the following sources of heat transfer. 1.The heat gain due to moisture in the outside air entering by infiltration. 2.The heat gain due to condensation of moisture from occupants. 3. 2.The heat gain due to condensation of moisture from any process such as cooking foods which takes place within the conditioned space. 4.The heat gain due to moisture passing directly into conditioned space through permeable walls or partitions from the outside or from adjoining regions where the water vapour pressure is higher. Total heat load= Sensible heat gain + Latent heat gain E Venkata Raju - MEP Projects&Facilities management Professional
  5. 5. HEATING AND COOING LOADS ROOM TOTAL HEAT LOAD(RTH)=ROOM SENSIBLE HEAT LOAD(RSH) + ROOM LATENT HEAT LOAD(RLH) The ratio of the RSH to the RTH(RSH/RTH) is known as the Room sensible heat factor. E Venkata Raju - MEP Projects&Facilities management Professional
  6. 6. HEATING AND COOING LOADS  By pass air: All the air passing over the cooling coil cannot come in contact with the fins/tube surface of the coil due to the gaps between the fins and tubes. It comes out of the cooling coil at the same condition at which it entered and so it is termed as the by-pass air. The sensible and the latent heats in this by-pass air needs to be removed out.So it is a part of the room load.  By pass factor: It is the ratio of the quantity of by pass air to that of the total air passing through the coil. New terms , Effective room sensible heat Effective room latent heat Effective room total heat E Venkata Raju - MEP Projects&Facilities management Professional
  7. 7. HEATING AND COOING LOADS 1.SHF = (SH)/(SH+LH)=SH/TH Where SHF- Sensible heat factor SH-Sensible heat LH-Latent heat TH-Total heat 2. RSHF= (RSH)/(RSH+RLH)=(RSH/RTH) Where RSHF-Room sensible heat factor RSH-Room sensible heat RLH-Room latent heat RTH-Room total heat 3.EFSHF=(ERSH)/(ERSH+ERLH)=(ERSH)/(ERTH) Where ERSHF- Effective room sensible heat factor ERSH- Effective Room sensible heat ERLH- Effective Room latent heat ERTH- Effective Room total heat E Venkata Raju - MEP Projects&Facilities management Professional
  8. 8. HEATING AND COOING LOADS  Effective room total heat load: It determines the quantity and temperature-humidity condition of the supply-air.  Grand total heat load: It determines the capacity of the refrigeration plant.  Apparatus dew point: It is the effective surface temperature of the cooling coil which determines the condition of (supply) air coming out of the coil. To find out the air quantity required and to select the equipment ERSHF,BF and ADP are required. E Venkata Raju - MEP Projects&Facilities management Professional
  9. 9. HEATING AND COOLING LOADS  A. Effective room sensible heat gain(ERSH) A.1.Solar Gain-glass Heat gain= A X R X MF Where A=Area of glass in sq.m/sq.ft R = Solar gain in kcal/h/sq.m or BTU/h/ sq.ft MF= Multiplying factor/s for the type of glass, shading etc. A.2.Solar transmission gain through the walls and roof The heat gain= A X U X EqTD Where A=Area of the wall or roof in sq.m or sq.ft U=Transmission coefficient in kcal/h/sq.m/ deg.cent EqTD=Corrected equivalent temperature difference in deg.cent E Venkata Raju - MEP Projects&Facilities management Professional
  10. 10. HEATING AND COOING LOADS  A.3.Transmission gain through the walls and partitions (In addition to the solar gain through the glass, there is a heat gain by transmission through the glass because of the temperature difference between the surroundings and the conditioned space) Heat gain= A X U X TD Where A: area of glass/partition in sq.m or sq.ft U:Transmission coefficients for glass/partitions(from tables) in kcal/sq.m/deg. TD: Temperature difference between the surroundings and the conditioned space in deg C E Venkata Raju - MEP Projects&Facilities management Professional
  11. 11. HEATING AND COOING LOADS A.4.Heat gain – (1)Through infiltration (2) (by passed)Fresh Air (1) Design tables give the infiltration air quantity for various types of doors/ windows, for observable cracks and for infiltration due to the opening of the doors. This load is generally ignored as this can be merged with the load due to fresh air intake for ventilation. (1) The room load due to the by passed fresh air(through the cooling) = 1.08 cfm X BF X TD (Deg F) E Venkata Raju - MEP Projects&Facilities management Professional
  12. 12. HEATING AND COOING LOADS A.5.Internal heat gain a) People: Multiply the sensible heat per person by the number of persons to get the sensible heat gain due to occupancy. b) Lights : 1) Fluorescent light=total wattsx1.25x0.86 kcal/hr = total watts x 1.25 x 3.4 btu/hr Incandescent light: total watts x 0.86 kcal/hr total watts x 3.4 BTU/hr c)Fan horsepower in the draw through arrangement: HP X 641 kcal/hr or HP X2545 BTU/hr d)Appliances: E Venkata Raju - MEP Projects&Facilities management Professional
  13. 13. HEATING AND COOING LOADS  A.6. Safety factor: An additional 5% on RSH is taken as a safety factor and this also covers items such as heat gain by the supply duct ,leaks etc Effective room sensible heat Gain=  A.1.Solar Gain-glass+A.2.Solar transmission gain through the walls and roof +A.3.Transmission gain through the walls and partitions+ A.4.Heat gain – (1)Through infiltration (2) (by passed)Fresh Air+ A.5.Internal heat gain+ A.6Safety factor E Venkata Raju - MEP Projects&Facilities management Professional
  14. 14. HEATING AND COOING LOADS B.Effective Room Latent Heat Gain B.1. Room latent heat a)Infiltration : Ignored b)Outside air (by passed air) The heat gain in kcal/h= 0.706cmhXBF(wo -wi) BTU/h=0.68XcfmXBF(wo - wi) wo-wi:difference of moisture content at outside and inside conditions respectively in grams of water vapour/kg dry air or grains of water vapour/lb dry air(lb=7000Gr.) BF=By –pass factor of the cooling coil E Venkata Raju - MEP Projects&Facilities management Professional
  15. 15. HEATING AND COOING LOADS  © People(Occupancy): Heat gain= nXwXh Where n: no. of persons w:moisture released/person in g/person h:Latent heat of condensation of moisture in kcal/g or BTU/gr. In Kcal/ h =n X w X 0.5883 In BTU/h=nXwX0.15 E Venkata Raju - MEP Projects&Facilities management Professional
  16. 16. HEATING AND COOING LOADS  (d) Steam: In industrial or special applications, where steam is used, the corresponding heat gain has to be taken into account.  (e) Appliance : If moisture generating appliances are used in the conditioned space, Heat gain=w X h where w: moisture generated per hour in g or gr. h:latent heat of moisture in kcal/g In Kcal/h:wX0.5883(w= g.moisture generated per hour) In BTU/h:w X 0.15 (w= Gr.moisture generated per hour) E Venkata Raju - MEP Projects&Facilities management Professional
  17. 17. HEATING AND COOING LOADS  B2. Adding up the values of B.1 (a) to (e) and adding up 5% as safety factor, effective room latent heat(ERLH) is obtained.  C.Effective Room Total Heat Gain(ERTH) It is the sum of ERSH + ERLH. D.Outside air heat sensible heat: in Kcal/hr=17.28Xcu.m/min(1-BF)XTD(0C) in BTU/hr= 1.08 X (1-BF) X TD (F) Latent heat: in kcal/hr=0.706xcmhx(1-BF)(wo-wi) In BTU/hr: 0.68 X cfm(1-BF)(wo-wi) E Venkata Raju - MEP Projects&Facilities management Professional
  18. 18. HEATING AND COOING LOADS  (E).Return Duct Heat Gain The gain due to the fan horse power(blow-through system only) In kcal/h: 641 X BHP of fan In BTU/h:2545 X BHP of fan (F) Grand Total Heat (GTH) It is sum of Effective room total heat gain +outside air heat+ Return duct heat gain E Venkata Raju - MEP Projects&Facilities management Professional
  19. 19. HEATING AND COOING LOADS  Determination of air quantity (cu.m/min or cfm) In cu.m/min= (ERSH)/(17.28(trm -tadp)(1-BF)) cfm= (ERSH)/(1.08(trm - tadp)(1-BF) ) where ERSH=Effective room sensible heat trm=Room design temp. tadp=Apparatus dew point temp BF = By pass factor of the coil. E Venkata Raju - MEP Projects&Facilities management Professional
  20. 20. SERVER /DATA CENTER Heat load Calculation 1. Room Area BTU = Length(m) X Width(m)X337 = 50 X 30 X337 =505500 2.Window Size and Position South window BTU=window L(m)XW(m)X870 North Window BTU=window L(m) XW(m)X165 If blinds on the windows multiply by 1.5 If no windows, ignore this. 3.Occupants Total occupants BTU= No.of occupants X 400 = 15 X 400=6000 4.Equipment Add together all the wattages for servers,switches,Routers and multiply by 3.5 Equipment BTU= Total wattage for all equipment X 3.5 = 100000 X 3.5=350000 5.Lighting Take the total wattage of the lighting and multiply by 4.25 Lighting BTU = Total wattage for all lighting X 4.25 =1000 X 4.25=42500 E Venkata Raju - MEP Projects&Facilities management Professional
  21. 21. SERVER /DATA CENTER Heat load Calculation  Total Heat Load=( Room Area BTU + Windows BTU + Total Occupants BTU + Equipment BTU + Lighting BTU) = 505500 + 0 + 6000 + 350000 + 42500 =904000 BTU =(904000)/12000=75.33 TR (1 TR=12000 BTU) Refrigeration plant capacity= 75.33 TR E Venkata Raju - MEP Projects&Facilities management Professional
  22. 22. THANK YOU E. VENKATA RAJU 20+ years experience in MEP (Building services ) and Facilities services i.e. HVAC , ELECTRICAL , FIRE FIGHTING , PLUMBING , SOLAR SYSTEMS ,WATER&WASTE WATER TREATMENT SYSTEMS etc..

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