Question

At winter design conditions, a house is projected to lose heat at a rate of 60,000 Btu/h. The internal heat gairn from people, lights, and appliances is estimated to be 6000 Btuh Ifthis house is to be heated by electric resistance heaters, determine the required rated power of these heaters in kW to maintain the house at constant temperature.

Answer 1

Q = 15.8 kW

Step-by-step explanation:

Given data:

Heat loss rate is 60,000 Btu/h

Heat gain is 6000 Btu/h

Rate of heat required is computed as

Q = (60000 - 6000) Btu/h

Q = 54000 Btu/h

change Btu/h to Kilo Watts

`Q = 54000 Btu/h ((1W)/(3.412142\ Btu/h))`

`Q = 15825.8 W((1 kW)/(1000 W))`

Q = 15.8 kW

Answer 2

15.8529 kW

Step-by-step explanation:

Rate of heat loss = 60000 Btu/h

Internal heat gain = 6000 Btu/h

Rate of heat required to be supplied

`P_(Sup)=\text{Rate of heat loss}-\text{Internal heat gain}\\\Rightarrow P_(Sup)=60000-6000\\\Rightarrow P_(Sup)=54000\ Btu/h`

Converting 54000 Btu/h to kW (kJ/s)

1 Btu = 1.05506 kJ

1 h = 3600 s

`P_(Sup)=54000* (1.05506)/(3600)\\\Rightarrow P_(Sup)=15.8529\ kW`

∴ Required rated power of these heaters is 15.8529 kW