Answer :
Answer:
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
To find the heat transfer rate to warm air by 10.0°C, it's necessary to convert the volume of air to mass, then use the formula Q = mcΔT and find the power by dividing by time.
To determine the heat transfer rate needed to warm air by 10.0°C, you need to know the mass of air being warmed, the specific heat capacity of air, and the desired temperature change. First, you convert the volume of air brought into the room to its mass by using the density of air. Then, you can apply the heat transfer equation Q = mcΔT, where Q is the heat transfer, m is the mass of air, c is the specific heat capacity, and ΔT is the temperature change. The power, which is the rate of heat transfer, is then found by dividing the heat transfer by the time over which it occurs.
Given the volume flow rate of the air and the temperature change, one can calculate the necessary power in kilowatts (kW) to achieve the desired heating effect. Note that the specific heat capacity of air at constant pressure is approximately 1.005 kJ/kg·K.
