Question

For inside and outside design temperatures of 22°C and 12°C, respectively, a home located in Boise, Idaho, has a design heating load of 38 kW. The house is heated by electric resistance heaters, and the cost of electricity is $0.06/kWh. Determine how much money the home owner will save if she lowers the thermostat from 22°C to 14°C from 11 PM to 7 AM every night in December.

Answer 1

Solution:

We calculate the thermal resistance using,

$\Delta$ T = Q* R

where, $\Delta$ T = change in temperature

Q = heat flow,

R = thermal resistance

In the 1st case,

(22-12) = 38 * R

or, R = 0.263

this runs for 8 hrs from 11 pm to 7am

So total energy consumed = 38 * 8 = 304 kWh

cost = 0.06*304 = $ 18.24

Now we need to calculate the heat load in the 2nd case

(22 -14) = Q * 0.263

Q = 30.418 kW

for 8 hrs, energy consumed = 8 * 30.418 = 243.344 kWh

Cost = 0.06 * 243.344 = $ 14.6

Savings = 18.24 -14.6 = $ 3.64