Question

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

Solutions

VerifiedSolution A

Solution B

Step 1

1 of 2(60,000 Btu/h) - (6,000 Btu/h) = 54,000 Btu/h

1 kW = 3412.14 Btu/h

54,000/3412.14 = 15.8 kW

Answered 9 months ago

Step 1

1 of 5$\pmb{\text{Given Data}}:$

- $\dot Q_g=6000\;\frac{\text{Btu}}{\text h}$ - heat gain rate
- $\dot Q_l=60000\;\frac{\text{Btu}}{\text h}$ - heat lost rate

$\pmb{\text{To find:}}$

- $\dot Q_h$ - power of heaters

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