Question

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2. A computer terminal room contains 125 computers with a power consumption of 130W. The room is well-insulated so that only 25% of the energy generated by the computers and their operators is dissipated through the walls. The room is 15 x 40 m with a 2.5m high ceiling and the ventilation system changes 0.9m/s resulting in a net energy flow 20kJ/m. Calculate the temperature of the air in the room at 10am if all of the computers are used continuously by operators with a metabolic output 85W from 9am when the temperature was 18°C. Assume the specific gas constant and specific heat capacity for air are 287 J/kg.K and 1.005 kJ/kg.K respectively and that atmospheric pressure is 101bar.

Answer 1

Solution Sec calculate the work supplied ko - Monputa - 125 x 130 3 x 360o cx 5 kg = 58500 KJ Since the room is insulated, consider Only the beat Out flow from walls and no work come out from the room. Qin 20 Pout = 1250 0.25 (@gen) Powd = 125x0.25(130+85) *3600 x103 Qout = 24187.5 kJ The mass of air is PV=MRT 101 X10%X15X40X2.5 = mx 287x(18+273) 1515000 = mx 83517 m=1814.00194 kg calculate the net Energy DE = Vout Qnet Xtime Scanned with CamScanner

DE = 0.9x20 X3600 DE = 64800 KJ Consider ist law of thermodynamics (Qon - Qout )+(Win-Wow ) + (DE) = AU 0-24187.5 +-58500+ 64800 = MC, AT -17887.5-1814.00194 x 0.718 (OT) AT = -13.73369 1 - 8 : -3 -336 It=4.266

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