Question

Calculate the following cooling load components at 18:00 hour of August month for the shown hall: (a) Heat transferred through floor, ceiling and doors. (b) Heat gain due to solar effects. (c) Heat gain through the glass windows. (d) Heat gain due to occupants. (e) The heat gain due to lights. (1) Heat transmitted due to infiltration. (g) The total cooling load. The following data are given: The ceiling height is 3.0 m. The hall has eight glass windows, each 3.0 x 1.2m and two wooden doors of 1.5 x 2.1m each. Spaces above the ceiling and below the floor are all at 30 °C. Inside design condition: 24 °C db and 50% RH. Outside design condition: 40 °C db and 28 °C wb temperatures. Outside mean temperature: 35 °C. Overall heat transfer coefficients are: 1.2 W/m2°C for walls (light colour, construction type: A). 1.8 W/m2 °C for ceiling (not suspended, type 3). 2.1 W/m2°C for floor. 6.7 W/m2.°C for glass windows. 3.5 W/m2 °C for wooden doors. Occupancy: 50-persons with total heat gain of 145 W/person and sensible gain is 87 W/ person. Lights: 30 W/m2 of floor area, with ordinary lamps of reccessed fixture. Infiltration: 6.7 L/s per person. Window glass: Clear pattern with 6 mm thickness and have white roller Latitude of location: 24 °N. Assume that the occupants and lights will remain On in the hall for 10 hours after occupant entry at 08:00 hour. shading.
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Answer 1

. Given arorall heat transfer coefficent of wall hu=1.2 w/moc. celling healowinno floor he = 24 lm o Glass window hy = 7 w/moc. Room outline wooden door ind= 3.5 wollen Agters of floor/Celing & lux?o=980m windows = 3X1:2 = 3.6m (thickness=6MA doors = 1.5x2.1 = 3.15m~ Soutside temprature} too out = noma (dht) 2 Inside degen temperature Too in=ruc(dot)s (os.fm temporatwie iwide room Tint Below celing Tc - ( canned with meani tamperature outside * ?5°C

Agreas of! - floor Celing luxro= alom windows = 3x112 = 7.6m (thickness bam doors = 1.5x2.1 = 3.15m Sutside temprature} too out = noc (dit) Inside degin temperatwie Too in = ruc(dot)s (CO-S ROM) temperature inside room Just Below Celing T = c = góc imean temperature outside = 35°C - Dreat Amansferred through flors celings and doos Teo Toro Tool het hete had X one ith e Bestel skaus

o oana = 169.67 w 2) Heat gain to solar effect (es) - Too o Tz A hen he Qalx230 10x280 10 2713.20 0.0368 Scanned with amScanner

- 3) Heat gain throgh a las window a wa hu v Ao NOT o w = on7 x3 :b x lo- Ow= 2016 watt -") Heat gain due to occulants Oo = (508!45) = 7250 watt 3) Heat gain due to lights o = he XARXOT - 30 X 28 OXIo = P400oweit Scanned with .. B . eamScanner

6) Heat transmitted due to in filteration Total volume of room = 280x3 = 84043 - Infile cfraction per person = 6.7|Peron = 67 x 103 m3/pense 671650 - 0335 infletration coi = 7256 x ono 18.(7MT 0.13 2 7) Total cooling load! Qc = Ammound of heat reauires to o remove to feel comfortable Gc169 1674 9.713 Mco + 700 + SUODO es Scanned Qczay:1 32 watt Gainscanner

comment sine the number of occarent are given there fore to focus on the time of lights switching on and off. And the infiltration heat is not accomplished in cooling load, as it is heat released due to infilteration 14 TE