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AIR CONDITIONING LOAD CALCULATIONS PRESENTATION E VENKATA RAJU E Venkata Raju - MEP Projects&Facilities management Professional

AIR CONDITIONING LOAD CALCULATIONS PRESENTATIONS by EVRAJU

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Page 1: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

AIR CONDITIONING LOAD CALCULATIONS

PRESENTATION

E VENKATA RAJU

E Venkata Raju - MEP Projects&Facilities management Professional

Page 2: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADSTo 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 construction3. Glass areas4. People5. Ventilation6. Motors, lights and appliances7. Water, gas and electrical services8. Equipment location9. Infiltration10. Location of doors and windows etc..

E Venkata Raju - MEP Projects&Facilities management Professional

Page 3: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADSSensible 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.

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Page 4: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADS6.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 gainE Venkata Raju - MEP Projects&Facilities management Professional

Page 5: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

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.

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Page 6: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADSBy 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 heatEffective room latent heatEffective room total heat

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Page 7: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADS1.SHF = (SH)/(SH+LH)=SH/THWhere SHF- Sensible heat factor SH-Sensible heat LH-Latent heat TH-Total heat2. RSHF= (RSH)/(RSH+RLH)=(RSH/RTH)Where RSHF-Room sensible heat factor RSH-Room sensible heat RLH-Room latent heat RTH-Room total heat3.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 heatE Venkata Raju - MEP Projects&Facilities management

Professional

Page 8: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADSEffective 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.

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Page 9: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOLING LOADSA. 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

roofThe heat gain= A X U X EqTDWhere A=Area of the wall or roof in sq.m or sq.ft U=Transmission coefficient in kcal/h/sq.m/ deg.centEqTD=Corrected equivalent temperature difference in

deg.centE Venkata Raju - MEP Projects&Facilities management Professional

Page 10: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADSA.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 TDWhere A: area of glass/partition in sq.m or sq.ftU: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

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Page 11: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

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.

(2)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

Page 12: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADS A.5.Internal heat gaina) 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/hrIncandescent 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/hrd)Appliances:

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Page 13: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADSA.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

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Page 14: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADSB.Effective Room Latent Heat GainB.1. Room latent heata)Infiltration : Ignoredb)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 coilE Venkata Raju - MEP Projects&Facilities management Professional

Page 15: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADS© People(Occupancy): Heat gain= nXwXhWhere 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 0.5883In BTU/h=nXwX0.15

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Page 16: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

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 hwhere w: moisture generated per hour in g or gr. h:latent heat of moisture in kcal/gIn Kcal/h:wX0.5883(w= g.moisture generated per hour)In BTU/h:w X 0.15 (w= Gr.moisture generated per hour)

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Page 17: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADSB2. 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)

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Page 18: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

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 fanIn 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

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Page 19: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

HEATING AND COOING LOADSDetermination 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.

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Page 20: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

SERVER /DATA CENTER Heat loadCalculation

1. Room Area BTU = Length(m) X Width(m)X337 = 50 X 30 X337 =505500 2.Window Size and PositionSouth window BTU=window L(m)XW(m)X870North Window BTU=window L(m) XW(m)X165If blinds on the windows multiply by 1.5If no windows, ignore this.3.OccupantsTotal occupants BTU= No.of occupants X 400 = 15 X 400=60004.EquipmentAdd together all the wattages for servers,switches,Routers and multiply by 3.5Equipment BTU= Total wattage for all equipment X 3.5 = 100000 X 3.5=3500005.LightingTake the total wattage of the lighting and multiply by 4.25Lighting BTU = Total wattage for all lighting X 4.25 =1000 X 4.25=42500

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Page 21: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

SERVER /DATA CENTER Heat loadCalculation

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

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Page 22: AIR CONDITIONING  LOAD CALCULATIONS PRESENTATIONS by EVRAJU

THANK YOU E. VENKATA RAJU20+ years experience in MEP (Building

services ) and Facilities services i.e. HVAC , ELECTRICAL , FIRE FIGHTING , PLUMBING , SOLAR SYSTEMS ,WATER&WASTE WATER TREATMENT SYSTEMS etc..