3 phase power consumption 1

[coley01]

I am trying to calculate the power consumption for a 3 phase motor. The load is varying so I am taking the current readings on each of the 3 phases every second. I know the voltage and power factor. I originally thought I would just calculate Pi=I*V*PF*3600 to get Wh for each phase and then just add the three phases and then add all the Pi's for a day to know the power consumption for one day. Then I was looking at 3 phase calculations and it said you have to multiply Pi by the square root of 3. Is this correct? Would I have to multiply Pi by the square root of 3 and then do the summation. I am confused why you need to multiply by the square root of 3 for 3 phase systems instead of just combining the phases.

 

[desertfox]

Hi coley01

Check this out:-

http://www.informit.com/articles/article.aspx?p=101617&seqNum=8

desertfox

 

[waross]

I know the voltage and power factor.

Really?? How are you measuring the instantaneous power factor.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[wareagle]

coley
There are 2 ways to make the calculation.
1. Voltage phase to phase x phase amps x 1.73 x PF
2. Voltage phase to neutral x phase amps x PF x 3
Assume phase to phase voltage = 208V phase to neutral voltage = 120 v Phase amps = 100 A PF = 0.8

1. watts = 208 x 100 x 0.8 x 1.73 = 28787 W

2. watts = 120 volts x 100 x 0.8 x 3 = 28800 W

Difference is due to rounding sqrt of 3
Take the watts x number of hours/1000 = kwh's

 

[jraef]

But to Bill's point, you cannot assume PF is static, in fact it is just the opposite. It varies with load more than the current does.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
For the best use of Eng-Tips, please click here -> faq731-376

 

[jghrist]

It would be a lot easier to buy/rent a watthour energy meter.

 

[edison123]

Portable power analyzers are dime a dozen now.

Muthu

http://www.edison.co.in

 

[coley01]

Thanks every one for your help. There are sub-electric meters that calculate the power factor, but I don't actually know yet what the sub meters are connected to, so now that you guys brought that up I am thinking the power factor from there is probably not going to be good enough and if the sub-electric meter is connected to what I am testing than that would already give me the info I need any ways. I was trying to just use the tools I already had available because I only have a couple of days to take data.

 

[waross]

If you are not afraid of some calculations, you can determine the values that you need to calculate the consumption.
You have probably found that the currents are close to equal on the three phases. Let's just do one phase and assume that the other two are equal.
You need a power factor capacitor with a rating in KVAR of LESS than 1/5 the motor HP. For a 15HP motor use a 3 KVAR capacitor or smaller. This will be a three phase capacitor. Use caution and never assume that the discharge resistors have discharged the capacitor before working on it.
OK, connect the capacitor and measure the current to the motor, to the capacitor and the incoming line current. (This will be the combined motor and capacitor current.
Use the three current values to construct a scale triangle. The side representing the capacitor current must be verticle. The side representing the line current or the combined currents will extend from the bottom of the verticle side. The side representing the motor current will extend from top of the vertical side.
Note that the side representing the motor current will usually be the longest line except when you have inadvertently applied more than 200% correction and driven the power factopr further leading than it was lagging. From the intersection of the motor current line and the line current line, extend a horizontal line back to the vertical line. If the power factor with the capacitor is leading you will have to construct an extension of the vertical line downwards.
Now, the length of the horizontal line will be the scale value of the in phase current. This will include motor losses. Use this value in your formula to calculate the true instanteneous Watts. (And instanteneous power factor if you wish).
Don't forget to multiply your result by root 3.
Your formula will be the simple Watts formula because the current has been corrected to unity power factor. V x I x 1.73=Watts
There is a formula to solve this. I believe that the law of cosines may be used to calculate the value of the in phase current but it's been a long long time since I had to use this method to do a power factor survey. Google the "Law of Cosines" and see if it works for you.

I was trying to just use the tools I already had available because I only have a couple of days to take data.

This is how you do it with the tools that you have if you're up for it.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Skogsgurra]

Bill's method is certainly interesting. Never used that, but will have a go on it as soon as work situation (and a few other things) permits.

I have used another approach. Pure mechanical, I would say. It has the advantage that mechanical guys almost always accept the measurements. That is not always the case when you do it the 'correct' way.

This is how to do it:
The motor's nominal slip is given on the nameplate. Slip is very nearly linearly proportional to load in most induction motors. Measuring slip is easy these days (electronic tachs and reflective tape) and if your motor's nameplate says 1780 RPM at full load, the slip is 20 RPM, which translates to 1.11 percent.

Now, if you measure .5 % slip, you know that your load is .5/1.11 = 45 percent of full load.

Sometimes, you can not assume exactly 60 (or 50) Hz. So you have to take mains frequency into account. Anyhow, lacking other instruments, this is an easy way to see if a motor is being heavily loaded or not. If voltage deviates from nominal, you have to correct for that. Remember then that available torque is proportional to voltage squared and that slip increases with that factor.

The power measured is power delivered to mechanical load. If all you need to know is current, then use a clamp. Anything in-between: use the necessary tools. They are there for a reason.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[waross]

I was primarily checking power factor. I didn't have to worry about voltage for that. I would take a series of measurements in a plant and then head back to the office. I would lay out the capacitor current vector vertically on the drafting board. The voltage in our area was usually stable enough that I was able to use the same vector for all the solutions. I would construct my line current arcs and motor current arcs from the ends of the vertical vector. Once I had my intersection there was no need to complete the triangle. I would use the square to extend from the intersection of the arcs to the vertical vector or extension thereof.
If I had chosen an appropriate scale, I could read the real current component directly from the scale on the square.
Did it work? Well if I had had a PF meter to check, I would have just used the PF meter and been done.
But, my power factor corrections based on these surveys always left the customer happy and the penalties were gone.
I am looking forward to your experiments Gunnar. It will be interesting to see how my method compares with PF meter reading results.

Bill
--------------------
"Why not the best?"
Jimmy Carter