Three Phase Power Measurement

[jimmynora]

Hi

We are measuring 3 Phase Power of an asynchronous Motor-Generator and have equipment installed on Phase 1 (L1) and Phase 3 (L3) to measure the current/voltage/power. I have some questions regarding the results.

Q) Why is the Power of Phase 1 positive and Phase 3 Negative ?

Q) How can I calculate the power of the second phase (L2) from these results?



Thanks
JN

 

[Skogsgurra]

It depends on the voltages being measured phase-phase and thus being + and -30 degrees wrong. The two-wattmeter method requires that there's no load connected to neutral.

Read all about it here:

https://www.ohiosemitronics.com/pdf/tech_papers/Two-wattmeter-method(E).pdf

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[jimmynora]

Thank your for your answer.

The voltages and currents being measured are the line values on Phase 1 and Phase 3.

So we have for Phase 1 the current in that phase and the voltage. Same goes for Phase 3. The power is then also calculated for the two phases.

Using the values how can I calculate the current/voltage/power values for Phase 2 ?

 

[Skogsgurra]

If you have L1-N and L3-N voltage and L1 and L3 current, then you probably do not know anything about L2 power.
The fact that you have negative power in phase 3 speaks for a two-watt meter method and not, as you say, line voltage and currents in phase 1 and 3.
Please check how the voltages are measured/connected.

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[jimmynora]

I have attached the circuit diagram.

What are equipment measures is current of L1 and the voltage of this phase. Same on L3. The current and voltage are in phase to each other.

-> The power graph show Phase 1 shows negative power values and Phase 3 showing positive power values. I need to know what it means that phase 1 power is negative (assuming the measurement contacts have not been connected the wrong way around).

-> I also want to know how I can calculate Voltage of L2 and and the current in the line so I can calculate the power.

 

[Skogsgurra]

That diagram isn't much help. Please try and make a proper diagram. At least one where transformer and meter polarities are indicated. Or, better, make a copy of the installation drawing.

A couple of questions:
1. Are there any loads connected line-neutral or ar all loads connected line-line?
2. Does the meter that shows negative power have the same absolute value as the meter with the positive value - or better, what do the two meters show?

Gunnar Englund

http://www.gke.org

--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.

 

[jimmynora]

just clarified the issue with the supplier of the measuring equipment. He says there should be no negative power and it is due to wrong connections. Will clarify the issue in the morning and update if it solved the problem.

The load connection to the 3 phase supply is a symmetrical load (asynchronous motor).

 

[LionelHutz]

The current should be fairly easy. The 3 phase currents should sum to zero. The voltage is more of a problem. You're not measuring it so the only thing you can do is assume a value.

 

[jimmynora]

The motor is connected to a 6KV AC supply. Because of the wrong connections it shows the phase voltage as 6KV instead of 6KV/sqrt(3)

If the reversal of connections doesnt solve the negative power issue of the phase 1 then we will just take the value of phase 3 and multiply it by 3. I believe it shud be accurate.

 

[rbulsara]

The supplier of the measuring instrument is right.

Rafiq Bulsara

http://www.srengineersct.com

 

[waross]

Use one Watt-meter (or section of a two element meter) to measure the voltage from L1 to L2 and the current in L2. This will meter the power in phase L1-L2 and part of the power in L2-L3.
Use the second Watt-meter (or section of a two element meter) to measure the voltage from L1 to L3 and the current in L2. This will meter the power in phase L1-L3 and the other part of the power in L2-L3.
Check the polarities.
This connection assumes that the three phase to phase voltages are equal. Unequal voltages and phase angle errors will cause errors. A similar connection has been used to meter three phase power. With the addition of a third CT connected in delta, a similar connection may be used to meter unequal currents with the same caution that unequal voltages will cause errors.
We were using two element meters to meter unbalanced currents on three phase circuits until repairs of a submarine cable left us with unequal impedances in our system and unequal voltages. We eventually changed out all the two element meters for three element meters.

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

 

[LionelHutz]

I highly doubt your per phase value is correct since you are trying to measure the line-neutral voltage and getting a line to line voltage. You would have to fix the transformer ratio setting in the power meter or multiply by 3/sqrt(3) before you would get the proper power measurement.

One other thing not pointed out yet. You are showing the trasformers connected line to ground. I hope you really mean line to neutral. A line to ground measurement may not be accurate on a high resistance grounded system and is useless on a floating system.

 

[electricpete]

All the angles of your problem have been covered well by other posters.

If you are confident that your load is balanced and supply is balanced (phase-to-phase and phase-to-ground), you could use single wattmeter connected line to neutral/ground, and multiply it by 3 to get the total.

Otherwise, assuming you have to use more wattmeters.

One approach would be to use three wattmeters, each connected from a different phase to the same arbitrary neutral or ground point. Adding the three readings will give the correct results, regardless of whether the system is balanced and regardless of whether your measurement neutral matches the power supply neutral.

Bill has described the aptly-named 2 wattmeter method, which also gives correct results and saves you hardware and hookup time.

This 2 wattmeter method is surprisingly easy to derive (much easier to derive than to memorize imo):
S = Sum (Vk x Ik*) for k=a, b, c [Equation 1]
(where x in this post means vector multiplication, NOT cross-product, and post-script * means conjugate
P = Real{S}

The Ik in equation 1 are all entering terminal k (a, b, c).
The Vk in Equation 1 all need to be referenced to the same point, but it need not be a neutral!
We will choose phase A for our voltage reference. which results in the following:
Va = 0
Vb = Vba (where Vba = line to line voltage with + meter lead on b and - meter lead on a)
Vc= Vca (where Vca = line to line voltage with +meter lead on c and - meter lead on a)

Plug these values of Vk into equation 1:
S = 0*Ia* + Vba x Ib* + Vca x Ic
S = Vba x Ib* + Vca x Ic

P = Re{S} = Re{Vba x Ib* + Vca x Ic*}
P = Re{Vba x Ib*} + Re {Vca x Ic*} [equation 2]

Equation 2 tells us the total power is the sum of what is shown on two wattmeters:
1st wattmeter measures curent entering phase b with voltage connected b(+) to a(-)
2nd wattmeter measures curent entering phase c with voltage connected c(+) to a(-)

Once you understand the principle of the derivation, you should be able to jump directly to equation 2 without any intermediate steps.

=====================================
(2B)+(2B)' ?

 

[electricpete]

Sorry, I missed the fact that you have some transformers in your setup, presumably because the voltage is too high for your meter. That requires a little bit of attention.

=====================================
(2B)+(2B)' ?

 

[electricpete]

With the setup as drawn, ASSUMING everything is balanced, you should be able to measure 1/3 of total power by hooking a wattmeter to items labeled U1 and I1 (with proper polarity).

If you hooked up to U3 and I3, you should get the same thing (under assumption of balance). That might be a rough double check of the balance conditions (along with comparison of voltage magnitude to voltage magnitude and current mangitude to current magnitude). Don't know how accurate you have to be.

To be able to do the 2 wattmeter method with just two PT's, you would need to reconnect them phase to phase... which may not be an option if your PT rating doesn't support it.

=====================================
(2B)+(2B)' ?

 

[waross]

In support of the single Watt-hour-meter x 3. When I first hooked up a two element meter with delta connected CTs I lacked confidence and did not have anyone to consult with. I hid a single phase Watt-hour-meter as a reality check. I was surprised at how close the indications were. As well as proving my connections, it indicated that the load was more closely balanced than I had realized.

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

 

[wolf39]

I'm sorry to say that an application of the two-wattmeter method (also called "Aron circuit") is not state-of-the-art anymore. But if this method is used one has to consider the terminal input situation of the two wattmeters. In case the wattmeter current inputs are 5 A and the voltage inputs are 100 V, appropriate current transformers with a secondary output of 5 A and 100 V for the voltage transformers have to be used. A diagram is attached which I used for measuring the output power of a large hydro generator about 35 years ago. This circuit does measure the active power accurately regardless of the power factor available or whether the three phase loads are balanced or not.

Nowadays I would suggest to use a digital power meter, like the Yokogawa WT230, for instance. The circuit diagrams shown in the manufacturer's handbook are self-explanatory.

Wolf
www.hydropower-consult.com

 

[LSpark]

If your power meter is a 3 wire 2 element meter then you don't need to calculate the power in L2 (see electricpete's description of the two watt meter method). It's nothing new, most of the 30/40+ year old power stations I look after use this method.

This assumes that the motor/generator is a 3 wire system not 4 wire, if it's a 4 wire system then you need a 4 wire 3 element meter.