Transformer Reactive power requirement 6

[prashious12]

Hello Everyone,

Is there any standard mentioning the reactive Power Requirement for different % loading for power Transformer??
Precisely, If power factor of load is unity and load is 100% of Transformer rating, How much (approximately or maximum) will be the reactive power consumption of transformer?
Is there any Table mentioning loading and reactive power requirement for transformer?
Or even easy method to calculate the reactive power requirement of transformer provided given load details (Load KW and KVAR), Transformer No-load and Full-load Losses, % impedance etc.

Thanks
Prashious

 

[dpc]

The transformer capacity is based on current. The kVA rating on the nameplate applies regardless of the power factor of the load.

It is not the same as a generator.









David Castor

http://www.cvoes.com

 

[jghrist]

Transformers don't care about the load power factor. The losses (resitive and reactive) depend only on the load kVA.

 

[stevenal]

The question was not about load pf, since the OP's load is unity pf.

The %X can be calculated from the %Z and the load loss. Use the %X value to get reactive load loss. You will need the excitation current to get reactive no load loss. At 100% load you will have both the load and no load components.

 

[prashious12]

@dpc: I understand it is different from Generator but is there any standard or guidelines mentioning the reactive power consumption for core excitation for transformer at loading equal to rated load with load power factor equal to 1.

@steve: I have a preliminary data for transformer as follows:

MVA Rating 12/16/20
Voltage 25 KV/4.16KV
% impedance= 7% X/R Ratio= 18.6
No-load losses: 13.56 KW @ 100% voltage
Load loss= 78.99 KW @ 12000 KVA

I wish to estimate the Reactive power consumption of the transformer under consideration that my Load is unity pf = 12 MW.
I was searching for standard (ready to look) but couldn't find any and that's the reason putting it in this forum for expert advice.

Regards,

Prashious

 

[dpc]

As stevenal said, I didn't interpret your question correctly.

Power losses are represented as resistive elements in any equivalent circuit. None of the kW losses stated have any reactive component.

"Reactive Power consumption" can be determined by the transformer equivalent circuit. The leakage reactance %X can be computed from the %Z and the resistive load losses.

The magnetizing current is not a function of load, so the no-load losses and no-load current can be used to determine the power factor of the magnetizing current.

Once the reactances are known the reactive power values can be calculated as a function of load current.

Hope that helps.



David Castor

http://www.cvoes.com

 

[stevenal]

Without knowing the excitation current, you will not know the no-load reactive loss. You have plenty of data to get the load reactive loss.

 

[prashious12]

Thanks to both.
one more que. given X/R ratio is to be considered for excitation branch with respect to Transformer model or with respect to thevenin's equivalent Transformer model referred to primary?

Prashious

 

[dpc]

Transformer model it is a shunt element. It can be referred to either side.

David Castor

http://www.cvoes.com

 

[rcw retired EE]

The MVAR's "lost" in the transformer can be estimated by Qpu= Ipu^2 x Xpu, where Ipu is the per unit current or MVA flowing through the transformer and Xpu is the transformer's per unit reactance.

A good approximation is to take %Z x (MVApu)^2.

For the 12 MVA unit with Z=7%=.07pu, delivering 12 MW @ 1.0 pf = 12 MVA, the per unit MVA and current are 1.0. MVAr's will be = .07 x1.0^2 = 0.07 pu. 0.07 pu x 12 MVA = 0.84 MVAr.

This calculation neglects the no load excitation current and approximates %X with %Z but the result is usually within a few %.

What this means is that to deliver 12 MVA @1.0 pf on the output, you need to supply the transformer with 12.092 MW and 0.84 MVAR, or 12.12 MVA at .9975 pf.

 

[prc]

rcwilson, iam confused!!
To me, when a transformer delivers unity pf load, the voltage drop in trf is IR cos phi + IX sing phi= IR% only as sin phi is zero( IR =load loss as a percentage of MVA )This IRdrop being in phase with voltage, this MW (or rather Kw ) is the only lost component and there is no MVAR loss.In more accurate way, we have to divide the no-load loss by taking no-load pf in to kVAR and kW then "KVA loss"can be found out due to this."Kva loss " due to load current is only kVAR from sin phi X reactance in pu where phi is pf angle of load.

 

[electricpete]

prc - why would you say "this MW (or rather Kw ) is the only lost component and there is no MVAR loss"?
There is "MVAR loss" in reactive components (I^2*X) in the same way as resistive loss in resistive components (I^2*R). If you didn't account for that mvar loss, mvars would not balance.

I think rcwilson is right on track. I notice leakage vars were considered and magnetizing vars neglected as stated. For a motor under full load, the vars lost in magnetizing reactance are roughly same as vars lost in leakage reactance. Transformer is of course different with no airgap. I don't know whether neglecting magnetizing vars is appropriate for transformer or not, but I'll take rwilson's word for it.

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(2B)+(2B)' ?

 

[Skogsgurra]

prc, I wasn't confused at all - until I read your post.

Agree with Pete, rcwilson describes quite well what happens and how to calculate the losses involved.

If you need better than that you probable need to do field measurements on the actual transformer.

Gunnar Englund

http://www.gke.org

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

 

[jghrist]

prc,
You are confusing voltage drop with loss. The voltage drop formula you quote is an approximation of the voltage magnitude decrease. See the attached voltage drop diagram. Your formula calculates VDR. With ø = 0, VDR = I·R, but there is still an I·X component. It just doesn't contribute to the voltage magnitude decrease VDR.

 

[rcw retired EE]

My calculations were developed to explain to power plant operators why the output to the utility was 0.95 power factor when the generator was putting out 0.9 power factor. The missing MVAr's were being used by the Generator Step Up (GSU) Transformer.

Most operators and plant managers understand the watt losses = I squared R, but forgot that MVAr "losses" = I squared X. Invariably we get questions about metering problems during reactive power output tests of our power plants.

Sizing of the GSU's aggravates the situation. Most GSU's are sized so the generator output is somewhere in the second stage of cooling, say 150% of base ONAN transformer rating. So Ipu at full generator load is 1.5 and I squared = 2.25. With typical impedances of 7-10% the transformer MVAr's will be 16%-25% of transformer rating (10-15% of plant output), while the kW losses will be too small to see on the meters.

The calculation is approximate but it yields numbers that match our generator and utilty meters within 1%.

 

[electricpete]

Attached is an example GSU data sheet.

Input data:
Hi side voltage: 345kv nominal.
Transformer rating: 700MVA

Exciting current: 1% at 100% voltage, 2.7% at 110% voltage.

Leakage reactance: 8%

Calculations:

Magnetizing MVARs
For simplicity and without much loss of accuracy, we can move magnetizing branch to the front of the primary leakage branch, directly in parallel with the input. Then we calculate:
Smag = V I* = 1*0.01 = 1% of transformer rating. = 0.01*700 = 7 MVAR

Leakage MVAR:
Sleak = I Z = 1 * 0.08 = 8% of transformer rating 0.08*700 = 56 MVAR

Conclusion:
If I have done my calcs right (please check), the magnetizing mvars for this transformer are about 1/(1+8) = 11% of the total mvars consumed by this particular transformer during normal operation at 100% voltage.

Actually, this transformer normally operates with hi-side voltage of 362kv = 105 of nominal%, so the exciting current would be substantially higher than 1% (gets to 2.7% of 110%). But I'd have to stop and think before I could decide whether to call the harmonic portion of that increase to be vars.

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

 

[electricpete]

Here is attachment to go with previous message.

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(2B)+(2B)' ?

 

[electricpete]

Whoops, I read the guaranteed exciting current as 1%, should have read the actual exciting currrent as 0.59%. So instead of 1% rating = 7MVARs, it would be 0.6% of rrating = 4.2MVARs for the exciting branch, which is still 7% of total MVARs.

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(2B)+(2B)' ?

 

[dpc]

But there is real component to the magnetizing current - it cannot be all reactive. If it was then core losses would be zero.

You can determine the real component from the core losses and then use the total magnetizing current to determine the reactive portion.



David Castor

http://www.cvoes.com

 

[jghrist]

Exciting current is not all reactive. Part is the no-load losses. In your example, total magnetizing branch apparent loss is 0.0059 pu = 4.13 MVA. The resistive loss is 0.384 MVA. The reactive part is sqrt(4.13² - 0.384²) = 4.11 Mvar.