Question

A classroom which normally contains 26 people is to be air- conditioned. A person in the classroom typically dissipates heat at a rate of 360 kJ/h. The room is lit with 15 light bulbs which each consume 40 W. A 200 W fan also operates in the room near a door to a computer closet. This fan pulls warm air into the classroom under the door at a rate of 200 kg/h. The room gains 10 kJ in energy (enthalpy) for each kg of warm air which enters. The air comes under the door at 10 m/s and leaves the room with negligible velocity through a large vent 3 m above the floor. Otherwise the room is well sealed. Finally, the room gains 5,000 kJ/h by heat transfer through the walls and windows. Determine the air-conditioner capacity, in kW, that is needed to keep the room at a steady temperature. Are any terms of the energy balance negligible?

Answer 1

Step-by-step explanation:

Heat dissipated by 26 people per second

= 26 x 360 x 1000 / (60 x 60)

= 2600 J

Heat dissipated by bulbs per second

= 15 x 40

= 600 J

Heat dissipated by fans per second

200 J

Heat dissipated by air coming in

= 200 x 10 x 1000 / (60 x 60)

=555.55 J

Heat gain by walls

= 5000 x 1000 / (60 x 60)

= 1388.88 J

Total heat gain per second

= 2600 + 600+200+555.55

= 3955.55 J

Capacity of air-conditioner required

= 3955.55 J/s

= 3.9 kJ/s

= 3.9kW

= 4 kW