finding load requirements ppt

8/13/2019 finding load requirements ppt

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By

NS MUHAMMAD WAQAS YOUNAS

NS SYED WAJI-UL-HASSAN

NS KAMRAN SIDDIQUE

NS MUNEEB AHMED

PC TALHA NASIR


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What is cooling load ????

Cooling load is the rate at which heatmust be removed from the air to

maintain temperature[1]

There are different methods that can be usedfor the calculation of the cooling load:

1) Transfer function method

2) CLTD/SCL/CLF methodA brief description of these methods is given in

next slides


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Method adopted:

1)Transfer function method:

It is a computer based method which is used tosolve the problems entirely and is more

complicated.2)CLTD/SCL/CLF method:

It is basically a hand made calculation method

which is based on the results of the transfer

function method.

We have adapted CLTD method in our

calculation of the cooling load.


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Factors taken in to account:

The following factors are taken in to account duringthe cooling load calculation:

a) Walls

b) Windows

c) Roofd) People

e) Lights

f) Equipments

g) Infiltration / ventilation

Each of the above factor is considered duringthe cooling load calculation


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Humidity:

Relative humidity outside=0.40

Relative humidity inside=0.30

Using psychometric chart

Outside specific humidity=0.032 Inside specific humidity=0.006


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Walls:Area of walls:

North wall=2280 ft2or 211.812 m2

South wall=3044 ft2 or 282.79 m2 East wall=3106.4 ft2or 288.59 m2

West wall= 801.6 ft2or 74.47m2

Thermal resistances:

Contact resistance(air film):

Rair=0.044

Resistance of cement:

Rcement=L/K=0.0254m/0.72=0.03527

Resistance of brick:

Rbrick=L/K=0.2286m/0.32=0.714

Resistance of air inside:

Rinside air=0.12

Total resistance: Rtot=0.91327


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CLDT calculations:Over all heat transfer co-efficient:

U=1/ Rtot=1/0.91327=1.09496

CLTD for walls:

From tables

For a 100 mm common brick there is C4

For plaster,C4=5At 12:00 noon;

CLTD(north)=5

CLTD(south)=7

CLTD(east)=22

CLTD(west)=5

Corrected CLTD =CLTD+(25.5- Tinternal)+(Tm-39.9)

Where


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So,

Corrected CLTD=CLTD+13.95

Corrected CLTD(north)=5+13.95=18.95

Corrected CLTD(south)=7+13.95=21.95 Corrected CLTD(east)=22+13.95=35.95

Corrected CLTD(west)=5+13.95=18.95

Heat rate:Q=UA (corrected CLTD)

Q north walls=1.09496*211.812*18.95=4394.99

Q south walls=1.09496*282.79 *21.95=6796.68

Q east walls=1.09496*288.59 *35.95=11360

Q west walls=1.09496*74.47*18.95=1545.22


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Windows:

Areas of windows:Area of north windows=456ft2

Area of south windows=456ft2

Total area of both sides windows=912ft2 or 84.73m2

CLTD for glass:

Taking at 12:00 noon = 5

Corrected CLTD=5+13.95=18.95

U=5.9, from tables for single glass for summerQglass windows=UA(corrected CLTDC)

=5.9*84.72*18.95

= 9472.1196


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Athletes

According to ASHRAE fundamentals 1993 page 26.8,

for athletes in a gymnasium

Qathletes=525 watts


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Lights:Factor=0.95

Total bulbs=42

Power of single bulb=250 wattTotal power=42*250

Note: it is case of category C where single story vinyl flooring is

done. Here for 12 hours noonFor other appliances:Factor=0.93

No of exhaust=6

No of ceiling fan=5Qexhaust=30 watts*6=180watts

Qcieling fan=55 watts*5=275watts

Qexhaust + Qcieling fan=180+275=455

Qapplainces=455*0.93=423.15watt


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Ventilation and infiltration air:

Qsensible=1.23Q(TouterTinternal)Where,

No of person = 300

Ventilation per person =7.5 litres/second

Q=total ventilation required

=300*7.5=2250litrs/second

Touter=44oC

Tinternal=25o

CSo,

Qsensible=1.23*2250(44 25)=52583watts


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Qlatent=3010*Q*(wo-wi)

Where

Wo=outside specific humidity=0.032

Wi=in side specific humidity=0.006

Qlatent=3010*2250*(0.032-0.006 )=176085watt

Ventilationandinfiltration air:


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Total cooling load:

Q = Qwalls total+ Qglass windows+ Qpeople+Qathletes +Qlights+Qapplainces+Qsensible+Qroof

=24096.89+9472.1196+36000+525+9975

+423.15+52583+176085+27211.728

soQtotal=336372 watt

or 96 tons


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how much time is required for

cooling:

Q=m*Cp*(T0-Ti) = (air*Vroom)*Cp*(T0-Ti)

= (1.2*11576)*1.005*103*(50-25.5)=342036*103

j

But

Already calculated/recommended

Qtotal=336372 watts or 96 tons

Time=Q /Qtotal= 1017seconds=17minutes.

So system should be turned on a maximum of

17 minutes prior.


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Steps to reduce cooling load:

Since the cooling load is too large, soit is a good decision to use the central

air conditioning unit, instead of so

many small split or window ACs.


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finding load requirements ppt

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