Question

A classroom is air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. There are 40 students and 10 bulbs in the class room. Each student emits 360 kJ per hour and the bulb emits 100 W. The walls of the class room transfer 15,000 kJ per hour from the outside. The air conditioning units are of 5kW cooling power. How many air-conditioning units are required to maintain the temperature in the room?

Answer 1

2 air conditioning.

Step-by-step explanation:

Power is the time rate at which energy is used. Its unit is in Watt or J/s.

Mathematically,

Power = Energy/time

Energy rate of 40 students = 40 * 360 kJ/hr

= 14400 kJ/hr

In J/s, 14400 kJ/hr * 1000 J/1 kJ * 1 hr/3600 s

= 4000 J/s

Energy rate of 10 bulbs = 10 * 100 W

= 1000 J/s

Energy rate of the Walls of the classroom = 15000 kJ/hr

In J/s, 15000 kJ/hr * 1000 J/1 kJ * 1 hr/3600 s

= 4166.67 J/s

Total Energy rate added = 4166.67 + 1000 + 4000

= 9166.67 J/s

Energy rate of the air conditioning = 5 kW

= 5000 J/s

To maintain the room temperature at 0 J/s,

= 2 * 5000 J/s

= 10000 J/s

10000 J/s is greater than the total Energy rate added to the room.

Answer 2

unit =1.83

Step-by-step explanation:

The cooling load is due to people, lights, and heat transfer through the walls and the windows so. The total cooling load of the room is determined from

(Equation 1) Q cooling = Q lights + Q people + Q heat gain

there fore

Q cooling = 10 x 100 W = 1 KW

Q people = 40 x 360 KJ/h = 4 KW

Q heat gain = 15.000 kJ/h = 4.17 KW

we replace the values in Equation 1

Q cooling = Q lights + Q people + Q heat gain

Q cooling = 1 + 4 +4.17

Q cooling = 9.17 KW

Thus the number of air — conditioning units required is

9.17 KW /5 KW unit =1.83