Answer:
a. 3942 kWh
b. $473.04
c. 1314 kWh
d. $157.68
Explanation:
a.
There are 1000 W in 1 kW, so 450 W = 0.450 kW. The energy used per day is ...
(0.45 kW)(24 h) = 10.8 kWh . . . . energy per day
Then in a 365-day year, the energy used is
(365 da/yr)(10.8 kWh/da) = 3942 kWh/yr
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b.
At the rate of $0.12/kWh, the cost of running the pump is ...
($0.12/kWh)(3942 kWh/yr) = $473.04/yr
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c.
Switching the pump off for 1/3 of the time will save 1/3 of the energy found in part (a):
1/3(3942 kWh) = 1314 kWh . . . . energy saved by switching off the pump
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d.
The savings will be 1/3 of the cost of running the pump full time:
1/3($473.04/yr) = $157.68/yr