Practice test question

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Jeff80

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If you replace a 1/2hp 120v 80% efficient motor with a new 1/2hp 120v 90% efficient motor, what would be the percent difference in amp draw?

I dont even know where too start, I am stumped!!!
 
glene77is said:
If efficiency improved by aprox. 10 percent, then the rpm would increase, and the FLA would drop.
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For a single motor to somehow become more effecient, maybe, but the OP is asking about two different motors, each of which would run at the designed speed.

I.m sure we agree that the more efficient motor would use less current to produce the same output. The question is what would be the numerical difference.

My guess is that the more efficient motor would use 8/9ths the current of the less efficient motor.
 
LarryFine said:
My guess is that the more efficient motor would use 8/9ths the current of the less efficient motor.
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since power is a square of amperage, I would guess that amps woud be a factor of the squares of 8/9 ? (or maybe the square roots ?)
 
Efficiency is based on power consumption. Voltage is fixed therefore amp consumption is proportional to the power consumption.

Aw = Amperage performing work.

Aw/.8 = Amperage at 80% efficiency or 125% of Aw
Aw/.9 = Amperage at 90% efficiency or 111.1% of Aw

Three potential answers based on the reference amperage Ar
%change = (Aw/.8 - Aw/.9)/Ar

Ar @ 100% efficiency %change = 13.89%
Ar @ 90% efficiency %change = 12.50%
Ar @ 80% efficiency %change = 11.11%

Most test questioners presume the best answer is based on reference to the starting conditions (11.11%). Good engineering presumes the answer is based on ideal conditions (13.89%).
 
I would not even bother trying to do calculations on a multiple guess question like this. The efficiency improves by 10%... so wouldnt the current would drop by about 10%?
 
S'mise said:
I would not even bother trying to do calculations on a multiple guess question like this. The efficiency improves by 10%... so wouldnt the current would drop by about 10%?
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Your estimate should be an inverse relationship
10% gain = 90% to 100% = 1/.9 = 11% drop
20% gain = 80% to 100% = 1/.8 = 25% drop
30% gain = 70% to 100% = 1/.7 = 43% drop

Example: 1A needed to drive a circuit. At 70% efficiency: 1A/.7 = 1.43A required to drive the circuit. A 30% gain in efficiency would be a 43% drop in current.

Example: For the OP circuit 1A/.8 = 1.25A improves to 1A/.9 = 1.11A
An engineer says the drop was .14A of the ideal 1A (14%)
Test creators often say the drop was .14A of the starting 1.25A (11.2%)
 
Efficiency = Output ? Input

1 HP = 746W... 1/2 hp = 373W


373W(output) ? .8 = 466.25W(input)

373W(output) ? .9 = 414.44W(input)


466.25W ? 120V = 3.89A

414.44W ? 120V = 3.45A


(3.89A – 3.45A) ? 3.89A ? 100 = 11.3%

That would be the percentage difference in true power amperes.
 
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nakulak said:
since power is a square of amperage, I would guess that amps woud be a factor of the squares of 8/9 ?

(or maybe the square roots ?)
Click to expand...

Most of the roots Ive seen have been round not square !! haha coudnt stand it....sorry

Maybe we could just look at both nameplate FLA's and divide one into the other......:D
 
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