what is the power savings?

Question

Determine the resulting % power savings of a typical household in the Philippines if the line voltage drops from 220V to 200V (rms). Assume that 2.5 kW is used by the household for an average of 12.7 hours per 24-hour day and that the equivalent total resistance of the household fixtures and appliances do not change with a decrease in line voltage.

Answer 1

Conservation thru Voltage Reduction (a.k.a CVR)

I will take the long-hand method and walk through the calculations:

Given: 2.5 kW; PF = 1.00; Voltage = 220 V;

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2.5 kW ÷ 1.00 = 2.5 kVA

2.5 kVA ÷ .220 kV = 11.36 amps

220 V ÷ 11.36 amps = 19.4 Ω

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200 V ÷ 19.4 Ω = 10.3 amps

.200 kV x 10.3 amps = 2.06 kVA

2.06 kVA x 1.00 = 2.06 kW

2.5 kW - 2.06 kW = 0.44 kW = Power Savings

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The ENERGY savings is power x time.

0.44 kW x 12.7 hours = 5.6 kWh = Energy savings per day

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Percent Savings

2.06 kW ÷ 2.5 kW = .824

The savings is 1.00 - .824 = 0.176 or 17.6%

Answer 2

If we just need the percentage decrease, we don't need to know the actual power, supposing the effective resistance was 1 Ohm, at 220V we would have 220A and 48400 Watts,
At 200V, 200A, 40000W;
48400 - 40000 = 8400;
8400/48400 * 100 = 17.35%
or (Vhi² - Vlo²) / Vhi² * 100

One problem with this question is that "power" saving does not necessarily mean "energy" saving. Any energy used for heating will not actually decrease unless the duration of use is kept constant, but that is unlikely to be the case in real life, if one is, say, boiling water, one will keep heating for a longer time until the water does boil, so the energy used is the same even though the power is less. that 12.7 hours will tend to increase as the 2.5kW decreases.

[EDIT] OK, 8400/48400 * 100 = 17.35% power savings
I had the percent there, I just didn't label it.

Answer 3

Crappy idea, since the resistance of light bulbs changes with voltage and other things plugged in may not function correctly, but...

You can do this several ways. One is to do it as a proportion, since power (kW) is proportional the the square of the voltage. Hence, the new power consumption will be

(200/220)^2 * 2.5kW = 40000/48400 * 2.5kW =

100/121 * 2.5kW =~2.066kW

Then the percent power saving will be

100(2.5 - 2.066/2.5) = ~17.36%

The energy per day is not needed to solve this, but will be 12.7hr * 2.066 = 26.240kWhr versus 2.5kW * 12.7hr = 31.75kWhr

To check this result, you could determine the equivalent resistance of the load via W = V^2/R

2.5kW = 220^2/R ==

2,500W = 220^2/R ==

R = 220^2/2,500 ==

R = 19.36 ohms

Now, with 200V applied to that load, power will be

200V^2/19.36 = 2,066W = 2.066kW, the same rate found above.

[REPOST] I notice that both answerers above did the math correctly but stopped short of actually answering your question, " % power savings..."

Answer 4

If the voltage goes down by 10% the savings while the electricity is used will be 10%. Total power for 12.7 hours at draw of 2.5kW is 31.75 kWh. If you save 10% that is a savings of
3.175 kWh.

In real life, I am not sure that is really saved because people will probably increase the time they keep appliances like a stove on because the drop in watts means less power and longer cooking times.