power factor
- Thread starter domnic
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xformer
Senior Member
- Location
- Dallas, Tx
- Occupation
- Master Electrician
Why don't you tell us what you get, and how you got it? then see if we agree or disagree.
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
I must not understand the question, for it seems trivial to me. The meter should read 1978 watts, or 1.978 KW. What am I missing in the question?
xformer
Senior Member
- Location
- Dallas, Tx
- Occupation
- Master Electrician
That is what I was questioning...
We have 3 different variables with the question...
1 Watts
2. power factor
3. Volt amps
Depending on what the op wants.... well thats up to him.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
Motors are never 100% efficient.
The mechanical power out the motor shaft, when converted to watts, will always be less than the electrical input in watts. And that input power will always be less than the input VA.
Tapatalk!
The mechanical power out the motor shaft, when converted to watts, will always be less than the electrical input in watts. And that input power will always be less than the input VA.
Tapatalk!
xformer
Senior Member
- Location
- Dallas, Tx
- Occupation
- Master Electrician
junkhound
Senior Member
- Location
- Renton, WA
- Occupation
- EE, power electronics specialty
Plus, it depends on what type meter you are measuring voltage and current. True rms or an old Fluke 77 (which reads rms only for true sine wave).
Thus distortion factor and harmonics come into play for true watts; also, the PF for the harmonics may not be the same as for the fundamental, e.g. the triplett harmonics cancel themselves out as heat so often have a higher PF than the fundamental, can be lower if a very high leakage inductance in the system.
Thus distortion factor and harmonics come into play for true watts; also, the PF for the harmonics may not be the same as for the fundamental, e.g. the triplett harmonics cancel themselves out as heat so often have a higher PF than the fundamental, can be lower if a very high leakage inductance in the system.
kwired
Electron manager
- Location
- NE Nebraska
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- EC
I have to agree with Charlie, if watts is what is being asked for - it is right there in the question - no calculation necessary. Could be kind of a trick question to see if you are paying attention to details.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
Well, yes, if and only if the meter is in fact a watt-hour meter. I guess you have to assume something to make the question answerable.
Hmm... a watt-hour meter does not indicate "watts kw". :blink:
- Location
- San Francisco Bay Area, CA, USA
- Occupation
- Electrical Engineer
I saw his (eventual) question the same way GoldDigger did; he has a watt meter, he is reading 2.3kW, but he has (mis)calculated the watts as 1.978kW based on amps and voltage alone and he wants to understand why there is a difference, which is what GD explained. The initial question was however poorly worded so as to be ambiguous, he cleared it up later, notably AFTER charlie b's response.
1978W calculated from amps and voltage readings / .86 efficiency = 2300W absorbed power, which is what a watt meter would read.
Motor power DRAW (absorbed power) = A x V x PF x EFF
Although this does appear to work, he would have to be reading a kVA meter, not a kW meter. I think it could just as easily be coincidental and that his motor is 86% efficient, which is right in line with many single phase motors that size.
1978W calculated from amps and voltage readings / .86 efficiency = 2300W absorbed power, which is what a watt meter would read.
Motor power DRAW (absorbed power) = A x V x PF x EFF
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