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Peter_1
24-08-2009, 09:39 PM
I have a 3 phase + neutral system.
Leg 1 pulls 35 A, leg 2 = 41 A, leg 3 = 30 A and neutral = 9A.
What's the actual power on that moment?

I can't find the right answers for this one.

goshen
25-08-2009, 04:37 AM
I have a 3 phase + neutral system.
Leg 1 pulls 35 A, leg 2 = 41 A, leg 3 = 30 A and neutral = 9A.
What's the actual power on that moment?

I can't find the right answers for this one.

hi there need some detailes what are u powering ?
motor? heating element?
Pkw
----------------------=I (AMPERE)
1.73*400*cosphi

U CAN SWING IT AROUAND ANY WAY U LIKE TO GET YOUR ANSWER( this is for 400v systemes):)

Peter_1
25-08-2009, 06:33 AM
Measured on the main supply of a building with mainly lights (rectified) and 4 VRV's running also on this supply.
I know of course the formula but those is only valid in a balanced system.
Complaint of client: new build office, making publicity, +/- 25 x 15 m (75 x 45 ft), 2 floors (2.8 m height/floor), +/- 12 peoples working there is paying 28,000 €/year electricity (+/- 40,000 US$)

Yuri B.
25-08-2009, 07:19 AM
I can't find the right answers for this one.
Peter, install a wattmeter and you will know what is the useful power + power losses.

NoNickName
25-08-2009, 12:44 PM
Ptot=P1+P2+P3 and assuming
V1 = A sin X where A is the peak voltage and X is frequency * time (basicly the phase angle).
V2 = A sin (X+2/3 pi)
V3 = A sin (X+4/3 pi)

P1 = V1 x I1
P2 = V2 x I2
P3 = V3 x I3

Do the remaining easy math.

Please note that In=I1+I2+I3 (vector sum), therefore in your case

In = 35 * 0 + 41 *0.866 - 30 * 0.866 = 9.5A, exactly as you measured.

Your power on the neutral is Ptot = 400 / 0.707 * (0.866x41-0.866x 30)=5.3kW

Rememer the sums are vectorial and shall be split into active power and reactive power, sum of which is apparent power.

RefrigNoob
25-08-2009, 01:17 PM
Taking a stab here...
But don't you get charged by the electrical company for the highest of the phases? +some charge for peak demand.

nike123
25-08-2009, 01:19 PM
http://wapedia.mobi/en/AC_power#4.
http://en.wikipedia.org/wiki/Symmetrical_components

You could calculate vectors here:
http://www.electrician2.com/electa1/electa4htm.htm

goshen
25-08-2009, 04:12 PM
hi there like mentioned before if u are looking to charge tennants then use a meter look up "elnet " sold by control applications OR SATEC i found them very easy and accurate they also calculate peak & low and include a com link to any B.A.S OR

Peter_1
27-08-2009, 06:40 PM
So, like I thought, measuring each phase separately with the same clamp meter and then simple calculating if afterwards is impossible. Right?

nike123
27-08-2009, 06:55 PM
So, like I thought, measuring each phase separately with the same clamp meter and then simple calculating if afterwards is impossible. Right?
It is possible only if you have pure resistive loads. Otherwise, not possible.

Peter_1
27-08-2009, 08:36 PM
I still don't see how you even can do it with resistive loads.
What's the total power in my case?

nike123
27-08-2009, 09:09 PM
I still don't see how you even can do it with resistive loads.
Check my last link!



What's the total power in my case?

Same as above but instead of using 0,120,240 degree for calculation you need to measure power factor of each phase and make correction of angles.
For detailed explanation I need to consult my high school books if I found them.
Or you could buy one of these:
http://www.fluke.co.uk/comx/portal_gen.aspx?locale=uken&page=pqt

Peter_1
27-08-2009, 10:14 PM
This formula and this paper is valid when measuring the currents on exactly the same moment during the 3 phase pass. This is something from my old school days and I did this also with vectors (law of Kirchoff if I remember well)
But I'm measuring the 1st phase, 30 seconds later the 2nd phase and finally, 30 seconds later my 3th phase.
So my measurements are lagged seriously in time.
I'm measuring in each leg max AMP when sinusoid is at it's max (90°) at an RMS voltage.
Or am I seeing this wrongly?

nike123
27-08-2009, 11:09 PM
This formula and this paper is valid when measuring the currents on exactly the same moment during the 3 phase pass. This is something from my old school days and I did this also with vectors (law of Kirchoff if I remember well)
But I'm measuring the 1st phase, 30 seconds later the 2nd phase and finally, 30 seconds later my 3th phase.
So my measurements are lagged seriously in time.
I think that, for your purpose, that doesn't matter.


I'm measuring in each leg max AMP when sinusoid is at it's max (90°) at an RMS voltage.
Or am I seeing this wrongly?What AC Ampere meter measuring and displaying is not momentary value of amperage in any portion of sinusoid. It is max amperage amplitude/1,42 (for sinusoidal current) and it is also called RMS (Root Mean Square) current.

For your calculation it doesn't matter where in sinusoid are currents and voltages.

If your loads during measurement stays same, than there is no (significant) error in measurement because of that (2x30 sec) time difference between measurements.

nike123
28-08-2009, 06:54 AM
Sorry Peter, I was (unintentionally) misleading you! I see now that only concern here is power measured by distributor.
It is simple as NoNickname has alredy stated.

Here in my country, distribution company use power meters (for households and offices) which only measure active power.

So, if that is also case there, you simply measure phase currents and voltages and calculate their product for each phase and than sum all three phases.
Neutral current is not of your concern and all mathematics we discussed is pointless.

Yuri B.
28-08-2009, 05:09 PM
Still, Mr. Wattmeter is calculating better of us all.

shooter
12-09-2009, 07:40 PM
VA is exactly what it says Voltage times amps, so you will have to meaure the amps and the voltage to neutral.
that way you get the VA, however a watt meter also looks at the cos phi. and that you do not measure.
A three phase meter is measuring six values in total, and accounts three cos phi too.
the complete formula is P=V*A*root3*cosphi
root3 is 400/230 because you measure to neutral no root3 is used.
power is 24380 VA
hours/ year is 6000 using 146 MWh costing 28000 euro that is per kWh 19 cents so your reading looks oke.
however it is very much for 12 people.

hook up a recorder for a few days, and check a lot of ampreadings on the fuses or on machines, astonishing how much goes into the ac and light.

are the lights switched off
is airco stopped at 4.
are computers stopped and switched off.