At winter design conditions, a house is projected to lose heat at a rate of 70000 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 required rated power of these heaters in kW to maintain the house at constant temperature.

Question

contestada

Respuesta :

ACCESS MORE ACCESS MORE ACCESS MORE ACCESS MORE

MrRoyal MrRoyal · Answer 1 · 2020-09-18T19:09:21+02:00

Answer:

The power to maintain constant temperature is 18.56kW

Explanation:

Given

[tex]Heat\ Loss = 70000\ Btu/h[/tex]

[tex]Heat\ Gain = 6000\ Btu/h[/tex]

Required

Determine the power to maintain constant temperature

The power to maintain a constant temperature is the difference between heat loss and gain

i.e.

[tex]Power = |Heat\ Loss - Heat\ Gain|[/tex]

This gives:

[tex]Power = |70000\ Btu/h - 6000\ Btu/h|[/tex]

[tex]Power = |64000\ Btu/h|[/tex]

[tex]Power = 64000\ Btu/h[/tex]

Convert Btu/h to kW

[tex]1 Btu/h = 0.00029 kW[/tex]

So:

[tex]Power = 64000\ * 0.00029\ kW[/tex]

[tex]Power = 18.56kW[/tex]

Hence;

The power to maintain constant temperature is 18.56kW