Load Power Factor - Transformer Secondary reflected to Primary

[ron]

A bit theoretical, but maybe practical solution to a troubleshooting issue I'm working on.

If the load of a Delta - Wye transformer is a leading power factor, is it definite that the power factor measured on the primary of that transformer (if not influenced by other distribution on the transformer primary) is leading, or due to the inductive nature of the transformer that it may be lagging on the primary? How much offset toward lagging if any can be expected and what is the calculation?

Any reference material that someone might be able to point to?

 

[gar]

171015-1146 EDT

Make it simple and consider a single phase transformer.

Next consider an equivalent circuit for the transformer. The least complex is an ideal transformer with a series resistor and inductor (leakage inductance) on the ideal transformer output. Clearly primary resistance is reflected to the secondary and added to the secondary resistance to get the equivalent circuit internal series resistance.

Thus, you add the secondary equivalent series impedance to the load impedance, and then reflect this thru the ideal transformer.

If the load is highly capacitive, then it swamps out the leakage inductance, and looking into the primary the load looks capacitive.

If the capacitance and inductance are at resonance, then the input is resistive.

If not much capacitance, then the input looks inductive.

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[Ingenieur]

Assume a xfmr loaded to pu 1 with a pu Z = 5%
so the load is ~ 0.95
the load determines the power factor arctan X/R
so the xfmr only has a very small impact on total ckt X and R, hence on pf

the xfmr config will introduce a ph shift of 30 deg but on both v and i so pf ang stays the same

 

[ron]

Thanks for the replies, but I'm not sure I can make complete heads or tails of them. Maybe too theoretical. Sorry

if I have a DY transformer that is loaded with purely resistive load on the secondary, would I have to expect the power factor on the primary of the transformer to be anything other than 1, since the load is resistive?

 

[gar]

171015-1249 EDT

Slightly inductive from the leakage inductance of the transformer in series with the resistive load, and the internal resistance of the transformer windings adds in series to the load resistance. In addition there is some parallel inductance from the transformer.

You list yourself as Engineer. From that I would expect that means electrical and in turn a familiarity with equivalent circuits.

References for some discussion of transformer equivalent circuit:

https://www.electrical4u.com/equivalent-circuit-of-transformer-referred-to-primary-and-secondary/

http://www.electricaleasy.com/2014/04/equivalent-circuit-of-transformer.html

http://circuitglobe.com/equivalent-circuit-of-a-transformer.html

These are essentially all the same and there are many more.

 

[Ingenieur]

Thanks for the replies, but I'm not sure I can make complete heads or tails of them. Maybe too theoretical. Sorry

if I have a DY transformer that is loaded with purely resistive load on the secondary, would I have to expect the power factor on the primary of the transformer to be anything other than 1, since the load is resistive?

For all intent and purpose, yes, 1

assume 1 mva 480 sec xfmr 100% R load
1200 A so 2.5 Ohm = 2.5 + 0j

Taking xfmr L into consideraton
assume xfmr Z of 5% and x/r of 10:1
xfmr Z = 0.02 Ohm so x ~ 0.002 (r= 0.0002)

new Z total = 2.5002 + 0.002j
pf 0.9999997 w/xfmr L where pf = cos(arctan(x/r))

 

[ron]

You list yourself as Engineer. From that I would expect that means electrical and in turn a familiarity with equivalent circuits.

Thanks for the vote of confidence. ?

Edumacation only gets gets me so far.

 

[Ingenieur]

Math error (I used R = i/v = 1202/480 = 2.5 Ohm, durrrr :dunce: )
load R = 480/1202 = 0.4 Ohm so Z = 0.4 + 0j (pf = 1)

total Z = 0.4002 + 0.002 so pf = 0.999988, pretty close to 1 lol
only an insignificant change in the result

 

[gar]

171915-2341 EDT

ron:

In your first post you had a load with a leading power factor. In other words, a highly capacitive load.

Then in post #4 you specified a purely resistive load.

What is the problem you are trying to troubleshoot?

If you removed all load from the secondary, and maintain full input excitation voltage, then you will have a very distorted current into the primary and this will be a lagging current. Your load is inductive, but inductance varies thru the cycle.

You need to gat a small transformer, 100 to 200 VA, a scope, and other associated measuring , and load components, then play.

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[Julius Right]

Let’s say you have a transformer of 1000 kVA 6% short-circuit impedance 13.8 kV high and 480 V low.PF=0.95 capacitive. The load will be 70% of rated. Then Q[reactive
power of the load]=sqrt(1-0.95^2)*0.7*1000=218.6 kVAR [capacitive]
The reactive power required by the transformer itself is Q1=3*Xtrf*Iload^2
and Q2=sqrt(3)*13.8*Imiu
If no-load current Io=1.55%Irat and no-load losses 1700 W result Io=0.6484731 A and Ife=1.700/sqrt(3)/13.8=0.071122859 A and Imiu=sqrt(0.64873^2-0.0711^)=0.415458937 A
Xtrf=Ztrf=6%*13.8^2/1=11.4 ohm and Iload=0.7*1000/sqrt(3)/13.8=29.3 A
Q1=3*29.3^2*11.4=29.36 kVAR
Q2=SQRT(3)*0.4155*13.8=9.9 kVAR
Then the System will see 218.6-29.3-9.9=179.4 kVAR[capacitive]
The sin(fi)=179.4/1000/.7=0.256 and new power factor =sqrt(1-0.256^2)=0.967.

 

[Besoeker]

171015-1146 EDT

Make it simple.

I agree. The KISS principle.

 

[Besoeker]

A bit theoretical, but maybe practical solution to a troubleshooting issue I'm working on.

If the load of a Delta - Wye transformer is a leading power factor, is it definite that the power factor measured on the primary of that transformer (if not influenced by other distribution on the transformer primary) is leading,

Yes, that is correct.

 

[ron]

171915-2341 EDT

What is the problem you are trying to troubleshoot?

I have power meters on the primary side of step down transformers in a data center. The power meter does not indicate leading or lagging, just a absolute value of power factor.

Typically new server power supplies are capacitive and have a leading power factor. I wanted t know if the power factor read on the primary was shifted relative to the secondary. If the power factor on the primary is not shifted, then I will guess the value shown is leading and derate the upstream UPS accordingly (it is unity rated - kw rating equals the kVA rating).

 

[Ingenieur]

I have power meters on the primary side of step down transformers in a data center. The power meter does not indicate leading or lagging, just a absolute value of power factor.

Typically new server power supplies are capacitive and have a leading power factor. I wanted t know if the power factor read on the primary was shifted relative to the secondary. If the power factor on the primary is not shifted, then I will guess the value shown is leading and derate the upstream UPS accordingly (it is unity rated - kw rating equals the kVA rating).

For an engineer you have simple responses that show you mathematically that
1 it stays the same sign (lead/lag)
2 xfmr reactance has negligable effect and can be ignored

rather than a yes/no answer or long written answer a math based example allows you to see for yourself 'why' and 'how'
in some matters simple is relative lol

 

[gar]

171016-2120 EDT

Photo of AC voltage and currrent.

Is this load resistive, inductive, capacitive, or a mixture?
Is the load leading or lagging?
What is the power factor?
Does this waveform approximately reflect thru a transformer?
Would the input current to a transformer with this current as a load look like this waveform?
.



.

 

[gar]

171017-2401 EDT

In my last post I did not define my load. And I won't yet.

Some new data and plots. All plots use exactly the same scaling. I did not have a suitable capacitor to create a bigger signal. All plots are for a 120 V source. Where a transformer is in the circuit the measurements were made at the input to the transformer. The transformer is an 80 W isolation transformer from the 1940s.






120 V 0.66 A 51.9 W 82.5 VA 0.63 PF
The current pulse is close to being centered at 90 deg, and its base is just over 1 major division.


.
80 W Transformer unloaded.
.

.
120 V 0.12 A 5.6 W 14.8 VA 0.38 PF
The current pulse peaks at 90 deg. It is not a sine wave because of core saturation.



The 80 W transformer with the first load on the transformer secondary. Transformer ratio is about 1 to 1.




120 V 0.64 A 61.5 W 80.6 VA 0.78 PF
The current pulse is slightly beyond 90 deg. The pulse is slanted to the right. close to being centered at 90 deg, and its base is about 1.7 major divisions. The peak current pulse is lower. Power input is greater because of transformer losses, but power factor is better.

You can see the transformer magnetizing current combined with the load current. The pulse is widened resulting from the transformer leakage inductance.



A capacitor alone.




120 V 0.36 A 0.7 W 43.8 VA 0.01 PF
The current is basically a sine wave and leads by about 90 deg.




Transformer loaded with the same capacitor.




120 V 0.30 A 5,6 W 35,7 VA 0.15 PF
The current is basically a sine wave and leads by not quite 90 deg. An oscillation appears on the current waveform, probably from the capacitor in combination with the leakage inductance of the transformer.

Have to leave. I have not proofread.

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[ron]

Gar,

They are computer server power supplies, so switching power supplies.

 

[gar]

171017-1617 EDT

ron:

Why do you think they are a capacitive load?

What proof?

.

 

[ron]

171017-1617 EDT

ron:

Why do you think they are a capacitive load?

What proof?

.

I will try to find a power supply data sheet that reflects lead vs lag, but most don't, just give as absolute value.
such as https://www.google.com/url?sa=t&rct...ta-Sheet.pdf&usg=AOvVaw16RhFBLyfc4e8pG1diowBP


http://www.ecmweb.com/power-quality/power-shift

 

[gar]

171017-2155 EDT

ron:

You apparently have loads (the server power supplies) with "active power factor correction" input circuitry.

For some discussion on this circuitry see
https://www.fairchildsemi.com/application-notes/AN/AN-42047.pdf

I suggest that you start looking at current waveforms into your server power supplies and see what is happening.

Next look at the total facility load on your supply transformer, and then back at the UPS.

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