Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations IDS on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

KWH FOR A 3-PHASE MOTOR 1

Status
Not open for further replies.

durf208

Mechanical
Aug 4, 2004
8
I'm trying to figure out kw used per hour of a 3-phase 480
motor and put a price on the power consumed. the motor uses 10amps per phase and we pay 8 cents a kwh. I'm not sure of the PF of the motor. I cant find anything on this subject so any help would be appreciated.thanks
 
Replies continue below

Recommended for you

Assume 0.8 as p.f and it is good enough for your purpose..
 
kw= 480*10*.8*1.73/1000 or is it kw=480*30*.8*1.73/1000
and will this give me kw per hour? I'm trying to figure out how much it cost per hour to run this motor.
 
It's
kW= (1.732*480*10*0.8)/1000
kwh = kW * hr

where 10 is Line current in any one of the lines, assumed to be identical in all three lines. You need not multiply it with 3. The factor 1.732 takes care of all 3 phases.
 
If the 10 amps is the nameplate full load amps of the motor and the motor is not fully loaded 100% of the time, then the energy usage will be less.
 
Thanks rbulsara, thats what I was looking for.After I read
your first post I realized what I had wrote/asked.
 
Suppose volts, amps, and power factor are measured on each phase. Is the average of each parameter input to the power calculation? The reason asked is because another thread indicated current should be input to the power calculation as the square root of (the sum of the three currents squared) divided by three.
 
If there is slight imbalance you may use average of the 3 line measurements ( no sqaure roots or multiplying by 3).


Or use single phase values and multiply by 3. See my last response in thread238-101450.

Good luck
 
If you have a lot of other motors or electric loads to maintain, you may want to consider a clamp on power meter. These look and work like ordinary multifunction clamp meters, but as the name implies, they also measure power directly. They do this by measuring current (with the clamp part) and voltage (with the plug in test probes) simultaneously. The meter is 'smart'; it measures the phase difference between the voltage and current waveforms, and calculates power directly. (Most of these meters also measure pf itself, as well as kVA, kVAR, and other phase-related measurements. Some also measure harmonics and other fancy things.)
A basic power meter is about 2 to 4 times the cost of a good rms multifunction clamp meter. This may seem expensive for a meter. But, I believe that what it measures - direct power - is much more useful than current. Measuring electrical power is a much more sensitive indicator of motor loads, as well as of power usage.
 
Yeah, I just picked up one of these clamp-on jewels from Harbor Freight for under $20, manufactured by Cen Tec, good three digit resolution too. Let me re-phrase, under what circumstances would the square root of (the sum of the squares) / 3 input to the power calculation apply?
 
The square root of (the sum of the squares) / 3 would apply to power loss calculations because the losses are proportional to the square of the current.
 
Econ makes electronic submeters that you can use for this purpose, particularly if you want to record kilowatthours for different production lines and so forth.

Their meters use current transformers with a built-in 2 volts shunts so that disconnection of the current circuit can be done without having to use special make-before-break current transformer disconnects.

Thier website is
Mike Cole, mc5w@earthlink.net
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor