junkhound
Senior Member
- Location
- Renton, WA
- Occupation
- EE, power electronics specialty
Forty-two
Electrical question how to solve?
- Thread starter Danny89
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ptonsparky
Tom
- Occupation
- EC - retired
no, four hundred twenty. you needed to multiply by the sq root of 100. You know that.
JoeStillman
Senior Member
- Location
- West Chester, PA
I think they are looking for an answer like I = HP x K and solve for K. Assuming unity power factor,
Which is totally bogus if you look at table 430.248 where K ranges from 5 to 13.2 for 230V motors. You would need a power factor between 28% and 75% to come close to the table.
Which is totally bogus if you look at table 430.248 where K ranges from 5 to 13.2 for 230V motors. You would need a power factor between 28% and 75% to come close to the table.
- Location
- San Francisco Bay Area, CA, USA
- Occupation
- Electrical Engineer
But…
That “P” is the mechanical power that the motor is CAPABLE of, not the power it is ACTUALLY CONSUMING to do the intended work on the “load”. The original question asked for “amp load”, which to my mind implies the amps pulled under load. We do not know what the actual load on that motor is.
ptonsparky
Tom
- Occupation
- EC - retired
We also don't know if it's a 1 HP or 100 trying to do the work of a 5.
- Location
- San Francisco Bay Area, CA, USA
- Occupation
- Electrical Engineer
True, but if the questioner was just looking for a formula rather than a value, then that is irrelevant. We are left to interpret what they wanted however…
My point is that EVEN IF what they want is the formula, then the best anyone can provide is the MAXIMUM amps that this motor could draw at its rated mechanical power, whatever that rating was. Hv&Lv posted the correct formula, other than the fact that “P(out)” is not a value that is going to be known unless something is actually measured.
So if you want to know A(load), why bother trying to measure P(out), why not just measure A(load)? Using a clamp on ammeter is not rocket surgery…
The real point is that this was, from the very start, a poorly worded question.
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
That's the difference between exam questions and those in the real world, innit? An exam question might be to calculate the minimum wire size that could be used in a certain application under specific conditions, while the real world equivalent might be "Can I use some of the buttload of #10 that I have on the truck?"
kwired
Electron manager
- Location
- NE Nebraska
- Occupation
- EC
Never mind.
bottom line you must convert kW to W then plug that value into the formula. I for some reason was going the wrong way with the conversion.
kwired
Electron manager
- Location
- NE Nebraska
- Occupation
- EC
As a stand alone question as we seen it yes. If on a test and is one question of several based on a previously mentioned situation it may be a valid question. At very least for such stand alone question we would also need to know the voltage rating to be able to come up with any valid amps answer though power factor still is the final detail that gets us more precision in the answer.
JoeStillman
Senior Member
- Location
- West Chester, PA
You're right. This is not an electrical engineering problem, it's an ESP problem.
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