26 people is to be airconditioned. 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-conditione.

The capacity of refrigeration =5.35kw

Explanation:

Total people in classroom =26

One person dissipates =360kj/h

So 26 person will dissipates =360 × 26

                                                  =2.6kw

15 light bulb with 40w each

Total heat dissipates     15×40 =600

                                                =0.6kw

Fan will dissipate  =200w

                             =0.2kw

Room gain air at a rate of =200kg/h

Warm air enthalpy =10kj/kg

Total heat coming with warm air = 200×10

                                                         0.555kw

Heat coming from window walls =5000kj/h

                                                       =5000/3600=  1.388kw

Total heat coming in the room =2.6+0.6+0.2+0.55+1.388

                                                   =5.35kw

To maintain room at steady temperature

Heat should be removed at constant rate

Energy balance =energy in/energy out

Capacity of refrigeration =5.35kw

total distance travelled = 60 miles

total time taken =1.5 hours

average speed = total distance / total time

60 / 1.5

= 40

therefore, average speed = 40 m/h

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