Cooling Load Calculation

Cooling Load Calculation

Internal Cooling Loads:

A.) Heat Gain from Occupants

QSensible = (SHG)(P)(CLF Sensible)QLatent = (LHG)(P)(CLF Latent)

Where:SHG = Sensible Heat GainLHG = Latent Heat GainP = Number of People/OccupantsCLF = Cooling Load Factor

From Table 1 of ASHRAE Handbook Fundamentals 2005 at moderately active office work,

SHG = 75 WattsLHG = 55 Watts

From Table 4-9, p74 of Refrigeration & Air Conditioning by Stoecker & Jones at both 8 hours after each entry into space and total hours in space,

CLFSensible = 0.84CLFLatent = 1.0

Room 1:

QSensible = (75 watts

Occupant )(2 Occupants)(0.84)

=

QLatent = (55 watts


=QTotal =

Room 2:



=

QLatent = (55 watts


=QTotal =

Room 3:QTotal = 0 , since it has no occupant.

Room 4:



=

QLatent = (55 watts


=QTotal =

Room 5:



=

QLatent = (55 watts


=QTotal =


Room 7:



=

QLatent = (55 watts


=QTotal =

Room 8:



=

QLatent = (55 watts


=QTotal =

Room 9:



=

QLatent = (55 watts


=QTotal =

Room 10:



=

QLatent = (55 watts


=QTotal =

Room 11:

QSensible= (75 watts


=

QLatent = (55 watts


=QTotal =

Room 12:



=

QLatent = (55 watts


=QTotal =

Room 13:



=

QLatent = (55 watts


=QTotal =

Room 14:



=

QLatent = (55 watts


=QTotal =

Room 15:



=

QLatent = (55 watts


=QTotal =

Room 16:



=

QLatent = (55 watts


=QTotal =

Room 17:



=

QLatent = (55 watts


=QTotal =

Room 18:



=

QLatent = (55 watts


=QTotal =


Room 20:



=

QLatent = (55 watts


=QTotal =

Room 21:



=

QLatent = (55 watts


=QTotal =

Room 22:



=

QLatent = (55 watts


=QTotal =

Room 23:



=

QLatent = (55 watts


=QTotal =

Room 24:



=

QLatent = (55 watts


=QTotal =

Room 25:



=

QLatent = (55 watts


=QTotal =

B.)Heat gain from Lightings

Q = (W)(Fu)(Fb)(CLF)

Where:W = Lamp rating in WattsFu = Utilization factor or Fraction of installed lamps in useFb = Ballast factor = 1.2 for Flourescent lamps CLF = Cooling load factor

From Table 4-6, p72 of Refrigeration and Air Conditioning by Stoecker & Jones at Fixture Y, 10 hours max and 8 hours after lights are turned on;

CLF = 0.95

Room 1:

Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 2:Q = ( 40 x 6)(1)(1.2)(0.95)Q =

Room 3:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 4:Q = ( 40 x 1)(1)(1.2)(0.95)Q =

Room 5:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 6:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 7:Q = ( 40 x 1)(1)(1.2)(0.95)Q =

Room 8:Q = ( 40 x 3)(1)(1.2)(0.95)Q =

Room 9:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 10:Q = ( 40 x 8)(1)(1.2)(0.95)Q =

Room 11:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 12:Q = ( 40 x 4)(1)(1.2)(0.95)Q =

Room 13:Q = ( 40 x 6)(1)(1.2)(0.95)Q =

Room 14:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 15:Q = ( 40 x 1)(1)(1.2)(0.95)Q =

Room 16:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 17:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 18:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 19:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 20:Q = ( 40 x 4)(1)(1.2)(0.95)Q =

Room 21:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 22:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 23:Q = ( 40 x 2)(1)(1.2)(0.95)Q =

Room 24:Q = ( 40 x 4)(1)(1.2)(0.95)Q =

Room 25:Q = ( 40 x8)(1)(1.2)(0.95)Q =

C.)Heat gain from Equipments/Appliances

Room 1:Two Computers = 2(200) = 400 Watts

QTotal =

Room 2:Television = 130 WattsSmall radio = 40 WattsTwo computers = 2(200) = 400 WattsPrinter = 110 WattsWater dispenser = 150 Watts

QTotal =

Room 3:Electric stapler = 1250 WattsHydraulic polar motor cutter = 1200 Watts

QTotal =

Room 4:Computer = 200 WattsPhotocopier = 560 Watts

QTotal =

Room 5:Copy printer = 180 Watts

QTotal =Room 6:

Heidelberg letter phrase = 1250 WattsQTotal =

Room 7:Computer = 200 Watts

QTotal =


QTotal =

Room 9:None

Room 10:Two computers = 2(200) = 400 WattsThree printers = 2(110) = 220 WattsRefrigerator = 40 Watts

QTotal =

Room 11:None

Room 12:Computer = 200 WattsPrinter = 110 WattsWater dispenser = 150 Watts

QTotal =

Room 13:Five computers = 5(200) = 1000 Watts

Printer = 110 WattsQTotal =

Room 14:Small radio = 40 Watts

QTotal =

Room 15:Refrigerator = 270 Watts

QTotal =

Room 16:Photocopier = 560 WattsComputer = 200 WattsTwo printers = 2(110) = 220 Watts

QTotal =

Room 17:Laptop = 27.5 WattsTwo printers = 2(110) = 220 Watts

QTotal =

Room 18:Computer = 200 WattsPrinter = 110 Watts

QTotal =

Room 19:Refrigerator = 270 WattsWater dispenser = 150 Watts

QTotal =

Room 20:None

Room 21:None

Room 22:None


QTotal =

Room 24:Four computers = 4(200) = 800 WattsThree printers = 3(110) = 330 WattsTelevision = 130 WattsWater dispenser = 150 Watts

QTotal =

Room 25:Three computers = 3(200) = 600 WattsThree Printers = 3(110) = 330 WattsPhotocopier = 560 Watts

QTotal =

External Cooling Loads:

D.) Heat Gain from Windows

Q=A(SHGFmax)(SC)(CLF)

Where:A = Area of the Window ExposedSHGFmax = Maximum Solar Heat Gain FactorSC = Shading CoefficientCLF = Cooling Load Factor

From Table 4-10, p75 of Refrigeration & Air Conditioning by Stoecker & Jones at 32o North Latitude in March, September,

SHGFmax for:North East/North West: 330 W/m2

South East/South West: 700 W/m2

From Table 4-11, p76 of Refrigeration & Air Conditioning by Stoecker & Jones at Single glass, Regular Sheet with Light Venetian blinds,

SC = 0.55From Table 4-12, p77 of Refrigeration & Air Conditioning by Stoecker & Jones at 9:00 am & 2:00 pm Solar Time,

CLF = 0.58 for Northeast Facing window at 9:00 amCLF = 0.81 for Southeast Facing window at 9:00 amCLF = 0.75 for Southwest Facing window at 2:00 pmCLF = 0.30 for Northwest Facing window at 2:00 pm

Room 1: Southwest Facing window,

Q = 12m2(700 W/m2)(0.55)(0.75)Q =

Southeast Facing window,Q = 6m2(700 W/m2)(0.55)(0.81)Q =

Room 2: Southeast Facing window,

Q = 12m2(700 W/m2)(0.55)(0.81)Q =

Northeast Facing window,Q = 18m2(330 W/m2)(0.55)(0.58)Q =

Northwest Facing window,Q = 12m2(330 W/m2)(0.55)(0.30)Q =

Room 3: Northeast Facing window,

Q = 8m2(330 W/m2)(0.55)(0.58)Q =


Q = 8m2(330 W/m2)(0.55)(0.58)Q =


Room 5: None

Room 6: Northwest Facing window,

Q = 1.25m2(330 W/m2)(0.55)(0.30)Q =


Q = 3m2(330 W/m2)(0.55)(0.30)Q =

Room 8: None

Room 9: None


Q = 6m2(700 W/m2)(0.55)(0.81)Q =

Northeast Facing window,Q = 12m2(330W/m2)(0.55)(0.58)Q =


Q = 6m2(700 W/m2)(0.55)(0.81)Q =


Q = 12m2(700 W/m2)(0.55)(0.75)Q =



Q = 12m2(700 W/m2)(0.55)(0.81)Q =


Q = 6m2(330 W/m2)(0.55)(0.58)Q =


Q = 6m2(330 W/m2)(0.55)(0.58)Q =


Q = 8m2(330 W/m2)(0.55)(0.58)Q =


Q = 8m2(330 W/m2)(0.55)(0.58)Q =



Q = 8m2(330 W/m2)(0.55)(0.30)

Q =

Southwest Facing window,Q = 8m2(700 W/m2)(0.55)(0.75)Q =


Q = 8m2(700 W/m2)(0.55)(0.75)Q =


Q = 16m2(700 W/m2)(0.55)(0.81)Q =


Q = 8m2(700 W/m2)(0.55)(0.81)Q =

Room 22: None


Q = 6m2(330 W/m2)(0.55)(0.30)Q =


Q = 12m2(700 W/m2)(0.55)(0.75)Q =



Q = 12m2(700 W/m2)(0.55)(0.81)Q =

Northeast Facing window,Q = 12m2(330 W/m2)(0.55)(0.58)Q =

E.)Heat Gain from Walls

Q = UA(To - Ti)

Where:U = Overall Heat Transfer Coefficient

= 1/R ; R = Thermal ResistanceA = (Height x Width of the Wall) – Window AreaTo = Outside Design TemperatureTi = Inside Design Temperature

From Table 4-4, p68 of Refrigeration & Air Conditioning by Stoecker & Jones for Concrete Block, sand and gravel aggregate, 200 mm Exterior material,

R = 0.18 m2. K/W

Since,

U = 1/0.18 m2. K/W= 5. 556 W/m2. K

From Philippine Society of Mechanical Engineers (PSME) Code 1898,

To = 35oCTi = 24oC

Room 1:Q = (5. 556 W/m2.K)(17 m2)[(35oC+273) – (24oC+ 273)]KQ =

Room 2:Q = (5. 556 W/m2.K)(29.75m2)[(35oC+274) – (24oC+ 274)]K

Q =

Room 3:Q = (5. 556 W/m2.K)(8.5 m2)[(35oC+274) – (24oC+ 274)]KQ =









Room 12:Q = (5. 556 W/m2.K)(12.75 m2)[(35oC+274) – (24oC+ 274)]K

Q =









Room 21:Q = (5. 556 W/m2.K)(4.5m2)[(35oC+274) – (24oC+ 274)]KQ =




Room 25:Q = (5. 556 W/m2.K)(17 m2)[(35oC+274) – (24oC+ 274)]KQ =

F.) Heat Gain from Roofs

Q = (U)(A)(CLTDadjusted)

Where:

U = Overall Heat Transfer Coefficient= 1/R ; R = Thermal Resistance

A = Area of the RoofCLTDadjusted = Adjusted Cooling Load Temperature DifferenceCLTDadjusted = CLTD + (25-Ti) + (Tave-29)

Ti = inside design dry-bulb temperature, oCTave = Average outdoor dry-bulb temperature, oC

From Table 4-4, p68 of Refrigeration & Air Conditioning by Stoecker & Jones for Built-up, 10 mm Roofing,

R = 0.06 m2. K/W

Since,

U = 1/0.06 m2. K/W= 16.667 W/m2. K

From Table 4-14, p81 of Refrigeration and Air Conditioning by Stoecker & Jones at Roof Type 1 with suspended ceilings and 2:00 pm Solar Time,

CLTD = 43 K

Since,

CLTDadjusted = CLTD + (25-Ti) + (Tave-29)= 43 K + [( 25 + 273) – (24 + 273)]K + [(35 + 273) – (29

+ 273)]KCLTDadjusted = 50 K

Room 1:Q = (16.667 W/m2. K)(17. 2 m2)( 50 K)Q =

Room 2:Q = (16.667 W/m2. K)(110.94 m2)( 50 K)Q =

Room 3:Q = (16.667 W/m2. K)(71.38 m2)( 50 K)Q =

Room 4:Q = (16.667 W/m2. K)(8.6 m2)( 50 K)Q =

Room 5:Q = (16.667 W/m2. K)(8.6 m2)( 50 K)Q =

Room 6:Q = (16.667 W/m2. K)(33. 2 m2)( 50 K)Q =

Room 7:Q = (16.667 W/m2. K)(7. 5 m2)( 50 K)Q =

Room 8:Q = (16.667 W/m2. K)( 24.08 m2)( 50 K)Q =

Room 9:

Q = (16.667 W/m2. K)(8.6 m2)( 50 K)Q =

Room 10:Q = (16.667 W/m2. K)(61.06 m2)( 50 K)Q =

Room 11:Q = (16.667 W/m2. K)(12.9 m2)( 50 K)Q =

Room 12:Q = (16.667 W/m2. K)( 36.98 m2)( 50 K)Q =

Room 13:Q = (16.667 W/m2. K)(45.58 m2)( 50 K)Q =

Room 14:Q = (16.667 W/m2. K)(8.6 m2)( 50 K)Q =

Room 15:Q = (16.667 W/m2. K)(8.6 m2)( 50 K)Q =

Room 16:Q = (16.667 W/m2. K)(16.34 m2)( 50 K)Q =

Room 17:Q = (16.667 W/m2. K)(22.79 m2)( 50 K)Q =

Room 18:Q = (16.667 W/m2. K)(22.79 m2)( 50 K)Q =

Room 19:Q = (16.667 W/m2. K)( 16.34 m2)( 50 K)

Q =

Room 20:Q = (16.667 W/m2. K)(86 m2)( 50 K)Q =

Room 21:Q = (16.667 W/m2. K)(14.19 m2)( 50 K)Q =

Room 22:Q = (16.667 W/m2. K)(17. 2 m2)( 50 K)Q =

Room 23:Q = (16.667 W/m2. K)(12.9 m2)( 50 K)Q =

Room 24:Q = (16.667 W/m2. K)(37.84 m2)( 50 K)Q =

Room 25:Q = (16.667 W/m2. K)(8.97 m2)( 50 K)Q =

G.) Heat Gain due to Infiltration

Q = 1.23Qv(To – Ti)

Where:Qv = Volumetric Flow Rate of Outside Air, L/sTo = Outside Design Temperature, oCTi = Inside Design Temperature, oC

Assumptions:1.) Windows are always close2.) Doors are also always close except when someone enters/exits.

From ASHRAE Handbook Fundamentals:Let Qv = 30 L/s

All Rooms:Q = 1.23(30 L/s)(35 – 24)oCQ =

Calculations:

A.) Mass flow rate of Supply air

QT = maCpΔTWhere:

QT = Total Heat Gain, kWma = mass of air, kg/sCp = Specific heat of the supply air, kJ/kg.K

=1.0 + 1.88WR

W4 = Humidity ratio of air at Temperature 4ΔT = Temperature Difference between Room temperature (Ti)

and Supply temperature (TS), K

From Psychometric Chart at TR = 24oC and 50% Relative Humidity,

WR = 0.00925 kg m/kg da Since,

Cp = 1.0 + 1.88WR

= 1.0 + 1.88(0.00925 kg/kg)Cp = 1.0173 kJ/kg.K

Assume Temperature of Supply air before entering the room is,

TS = 13oCThen,

ΔT = (24 oC + 273) – (13 oC + 273)ΔT = 11 K

Solving for the mass of air (ma),

ma = QT

C p ΔT

We could now get the Volume flow rate of Supply air (Va),

Va = (ma)(Vf)Where:

ma = Mass of airVf = Specific Volume of Supply Air

From Table A – 2 of Psychometric Chart for Moist Air at TS = 13oC,

Vf =

Solving for the Volume of air needed per room:

For Each Rooms:

Room 1:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()=

Room 2:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 3:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 4:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 5:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()=

Room 6:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=Room 7:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 8:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 9:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 10:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 11:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 12:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 13:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()=

Room 14:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 15:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 16:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 17:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()=

Room 18:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 19:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 20:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 21:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()=

Room 22:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 23:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 24:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=

Room 25:

ma = ❑

(1.0173kJkg. K)(11K )

ma =

Va = ()()

=For the Whole Building:

Solving for the Total mass of air in in the Building,

mT = ❑

(1.0173kJkg. K)(11K )

mT =

Solving for the Total Volume of air (VT) needed by the fan or passing through the main duct,

VT = ()()=

It should be equal to the summation of all Volume of air needed in every room,

∑ (Va) = VT

∑ (Va) =

B.)Size of Motor required to drive the fan

Pa = VTρaha (eq. 1)Where:

Pa = Power output or Air power of fanVT = Total Volume of air handled by the fanρa = Density of airha = Total heads

but:Pressure of air = Pressure of water

Pa = Pw

ρaha = ρwhw

From this we get,

ha = ρw hwρa

(eq. 2)

Substituting (eq. 2) in (eq. 1),

Pa = (VT)(ρa¿(ρwhwρ a

)

Pa = (VT)(ρw)(hw)

From Figure 12-38, p437 of Power Plant Engineering by Frederick T. Morse at VT

=

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Cooling Load Calculation