Kilovolt-Ampere per Horsepower

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Issaczion

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A problem I was presented in class. The minimum kilovolt-amperes of a generator that will be required to start simultaneously, two 7.5 horsepower, 3 phase, 230-volt, Code letter F motors.

So I know that to find the minimum kilovolt-amperes you go to Code Letter F for the Locked-Rotor Code letters. Code Letter F=5-5.59

I know you are supposed to then take the lesser of the two numbers(5) and multiply by the horsepower, so the answer would be.
5(Code letter F)*7.5HP=37.5
Then because their are two motors you multiply the result by 2
37.5*2=75

75 kilovolt-amperes.

The question I am uncertain of is.
The minimum kilovolt-amperes of a generator that will be required to start two selectively controlled 7.5 horsepower, 3-phase, 230-volt, Code leter F motors is?

Selectively controlled, does this mean I count the motor as 1 so because the motors are selectively controlled would the calculation just be
5(Code letter F)*7.5HP=37.5 kilovolt-amperes per horsepower
 
Watch the units.
Motor HP is a measurement of output power. The generator needs to supply input power to the motor.
HP is not directly related to kVA. There is a factor of 746W/HP.
A lot of motor data, in the NEC, is provided in Amps.
 
electrofelon said:
What a horrible question. IT is impossible to answer.
Click to expand...
When I was in ninth grade our algebra teacher quit and was replaced by a PE coach who was not the sharpest knife in the drawer. One day on a test he added this "bonus question": How many feet are in a square foot?

When I asked him about it, he answered, "Four." When I pressed him on it he got out a pencil and paper and drew a square, counting out "one, two, three, four" as he drew it, and then he looked at me as though I was the idiot. And yes, he was serious.
 
ggunn said:
When I was in ninth grade our algebra teacher quit and was replaced by a PE coach who was not the sharpest knife in the drawer. One day on a test he added this "bonus question": How many feet are in a square foot?

When I asked him about it, he answered, "Four." When I pressed him on it he got out a pencil and paper and drew a square, counting out "one, two, three, four" as he drew it, and then he looked at me as though I was the idiot. And yes, he was serious.
Click to expand...
actually there are infinitely many 1 foot long line segments within a square with 1 foot sides....
 
Issaczion said:
A problem I was presented in class. The minimum kilovolt-amperes of a generator that will be required to start simultaneously, two 7.5 horsepower, 3 phase, 230-volt, Code letter F motors.

So I know that to find the minimum kilovolt-amperes you go to Code Letter F for the Locked-Rotor Code letters. Code Letter F=5-5.59

I know you are supposed to then take the lesser of the two numbers(5) and multiply by the horsepower, so the answer would be.
5(Code letter F)*7.5HP=37.5
Then because their are two motors you multiply the result by 2
37.5*2=75

75 kilovolt-amperes.

The question I am uncertain of is.
The minimum kilovolt-amperes of a generator that will be required to start two selectively controlled 7.5 horsepower, 3-phase, 230-volt, Code leter F motors is?

Selectively controlled, does this mean I count the motor as 1 so because the motors are selectively controlled would the calculation just be
5(Code letter F)*7.5HP=37.5 kilovolt-amperes per horsepower
Click to expand...
It's much simpler than that.
1 kW = 1.34 HP

FWIW most of the world these uses kW rather than HP..
 
When they say "selectively controlled", I would interpret that to mean they don't both start at once. So you would add the starting kVA of one motor to the running kVA of the other one to get the maximum kVA load.
 
Thanks for the replies, I asked my teacher and the answer he gave was to multiply by 125%. Gonna ask him to walk me through how he would do the problem. I too think the question is not the best. The book the problem is from does not even have an example problem like this, or even talk about selectively controlled.
 
In practice there is not really a way to calculate the answer to something like this. I suppose you could come up with a "worst case" answer by using the locked rotor of the motor(s) and size the generator to supply that, which is probably the approach intended to be used by the question. In practice that will result in an oversized generator. Generators don't strictly have to be able to supply the LRC to start a motor. There is stuff like inertia, source impedance, etc at play. Most generator manufacturers have software and/or will help you figure out what size generator is needed to start a given motor.
 
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