Question

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

Answer 1

15.83 KW

Step-by-step explanation:

Losing heat at a rate = 60000btu/h

Gaining heat at a rate = 6,000btu/h

The difference in heat levels

= 60000btu/h - 6000btu/h

= 54000btu/h

So we need to convert 54000btu/h to KW

1kw = 3412.142btu

? = 54000btu

We cross multiply

54000btu x 1kw = ?3412.142btu

So we divide through by 3412.142 to get the unknown

54000/3412.142

= 15.83kw

So we can conclude that 15.83kw is the required power of this heater to keep the house at a constant temperature