Transformer Power Factor 1

[Mondy]

Dear experts

How do I go about calculating what the power factor is of a 1500VA transformer off-load and at full load (resistive)

Also how would I go about calculating the values of PFC capacitors connected in parallel with the primary to improve the power factor if it is bad?

Many thanks
Mondy

 

[stevenal]

Need to know what data you have before calcs can be suggested. Do you have test data for the transformer; impedance, excitation current, load and no-load losses? Have you taken any measurements yourself? What is is your target pf for loaded and unloaded conditions?

A fully loaded transformer with unity pf load should show a fairly high pf on the primary. The same transformer unloaded will have a low pf on the primary. Switching the thing off when not in use would be an easy solution for the no-load case.

 

[Mondy]

Dear Stevenal

Thank you very much for your reply. The only data I have is as follows:

Overall VA : 1440VA
Copper Loss: 54.09W
Iron Loss : 6.5W
Regulation : 3.6%
Excitation : 40mA @ 230v
On-load volts: 4 x 24v @ 15A each
S/C primary volts: 8.3v ??

Can you calculate the rest with this info? If so can you show me an example?

Thanks for your time

Ray

 

[stevenal]

This sounds like it may be single phase, so I'll proceed with that assumption. No load is easy: 230*0.04=9.2var loss.
cos(atan(9.2/6.5))=57.7% pf. If three phase then multiply the var losses above by sqrt(3). You still haven't given a target pf. Correcting to unity would require 9.2vars of correction for the single phase case.

When loaded, add the no-load losses to the load losses to the load, keeping vars and watts separate. Then use the cos of the atan of the ratio to find pf. Care to try?

This treatment assumes linear loading.

Is the transformer rated 1500, but loaded to 1440?

Why do you care with such a small load?

 

[Mondy]

Thanks Stevenal

Yes you are correct it is single phase, sorry for not putting that in the question

OK I used this : COS(ATAN(9.2/(1440+54.09+6.5)))= 1

Is this correct? or have I got it wrong (again!) I expected it to be higher but not unity. Also if I need 9.2vars of correction how would I go about calculating the PF capacitor value (to wire in parallel with the primary) to acheive this?

It does seem silly to worry too much at this loading but the customer is always right (right?)

Once again thank you for your time and effort Steve

Ray

 

[stevenal]

Ray,

Pf correction caps are usually bought by the voltage rating and vars. Not sure about ones so small though.

Your treatment left out the load var losses of the transformer. Use percent impedance, transformer rated kVA and load loss watts to determine load var loss using Pythagorean theorem.

% regulation spec bothers me. Usually % impedance is given. I'm not sure there's a direct way to get from regualtion to impedance. Might need an iterative method. Regulation is defined as (|Vno load| -|Vfull load|)/|Vfull load|. Could someone be using the wrong term here?

 

[Mondy]

Thanks Stevenal

OK the reactive losses due to leakage inductance etc is practically zero as the transformer is toroidal. There obviously will be some but will such a small amount make that much difference?

The regulation you defined as the "American" definition the "English" definition is (Vno load-Vfull load)/Vfull load. I am pretty sure that if the transformer just has one secondary winding then the Impedance voltage would be:

%Regulation/100 * Vpri

as the regulation figure expresses all the grouped losses referred to the secondary?

Does this help (probably not!)?

Cheers
Ray

 

[stevenal]

So lets assume the impedance is 3.6%. VA loss would be .036*1500=54. This should exceed the load loss, but maybe as you say load var loss is negligible with the rest of the error due to roundoff. If so your previous calc is correct.

Did you mean to suggest the English definition of regulation is different? Only difference I could detect were the missing magnitude signs. These are necessary if you wish to avoid a complex result.

 

[jbartos]

Comments:
1.
Overall VA=4x24x15=1440 on the transformer low side
2.
S/C Primary Volts=8.3V ?
Assume the Copper Loss is equally divided between the primary side copper loss and secondary side copper loss. 1500VA is the transformer Primary VA. Then:
Vs/c prim=(54.09/2)/(1500/230)~4.14V

 

[stevenal]

After a little playing with Mathcad, I found that X = 5.03% fits pretty good to cause regulation (related to Vfull load) to be 3.6%. This adds 72.4 VA to the full load case.
PF=99.9%. If that's not good enough, you'll need another 72.4 VA in power factor correction to bring it to unity.

 

[jbartos]

Suggestion: The Regulation 3.6% needs to be clarified. Else, the Copper Loss of 54.09Watts is not consistent with (assumed Voltage) Regulation = 3.6% for secondary current 15A for each of 4 voltages of 24V on the secondary side, since
Vreg=3.6%= 100% Vx/24V
or
Vx=24V x 3.6%/100%=0.864V
(VA Loss)x=0.864V x 15A=12.96VA
Total VA Loss=4 x 12.96VA=51.84VA
Therefore, Copper Loss of 54.09W is somewhat excessive since it exceeds Total VA Loss=51.84VA
Either, Voltage Regulation should be higher than 3.6% or Copper Loss smaller than 54.09Watts

 

[Mondy]

Thanks Stevenal/jbartos

I made a mistake when stating the "English" definition of regulation it should be as:

(|Vno load| -|Vfull load|)/|Vno load|

Does this explain the deviation?

Cheers
Ray

 

[stevenal]

jbartos,

Both definitions of regulation use no load and full load voltages. Actual loading (1440VA)is irrelevant to the spec. In this case, I believe the rated full load is the 1500VA from the initial post. I tried to get Mondy to clarify this on Nov. 5, but failed to get an answer. Copper losses are also speced at full load.

Mondy,

This brings up a possible error in your formula of Nov. 6. that I didn't catch before and repeated myself. Copper losses should be adjusted by 1440/1500 (if appropriate) before going into the pf calculation. Load loss vars would also be adjusted by this ratio. Not likely to make mutch difference in the resulting pf.

I like your definition, less trouble getting to impedance. (Anyone else care to verify this difference in definition?)
Please verify the transformer VA rating.

 

[jbartos]

Suggestion to the previous posting: The transformers may be custom-made. Also, they do not have to adhere to industry standards. The transformer nameplate governs the transformer. Some transformer nameplates do not display adherence to any industry standards. Also, the transformer may state the transformer rated secondary voltage and current. In that case, the transformer secondary VA has to be calculated from the given transformer rated secondary voltage and secondary current. Apparently, this requires to be well-rounded around the transformers, transformer manufacturing practices, and transformer industry standards.

 

[jbartos]

Suggestion: Reference:
IEEE Std 100-2000 Dictionary
Definition of Regulation (4) (Transformer-Rectifier System)
Percent Regulation = 100% x (E1-E2)/E2
where
E1 is the no-load voltage
E2 is the voltage at rated load current and the line voltage is held constant at rated value

 

[Mondy]

Hi All

Sorry for the delay in replying. Yes the transformer is custom designed and so does not conform to any standards regarding loading. Here are the actual transformer details

CORE : 160 x 85 x 60 mm. M5 (0.30 mm.)

Fe.Weight : 6.29 kg. Bmax : 1.55 T.
Cu.Weight : 2.48 kg. Excitation : 39.9 mA.
Tot.Weight : 9.03 kg. Power : 1440 VA.
Cu.Loss : 54.09 W. Fill Factor : 21.9%
Fe.Loss : 6.51 W. Temp.Rise : 69 deg.C.
Frequency : 50 Hz. Ave. A/mm2 : 3.3 A/mm2.

Final Dimensions : 174 x 62 x 78 mm. (not including lead out bulges.)

Winding order: Primary first on core.

PRIMARY WINDING DATA.
~~~~~~~~~~~~~~~~~~~~~

Pri 1. Volts, Amps, DCR.*, Cu kg.

115, 6.57, 0.288, 0.577

Pri 2. Volts, Amps, DCR.*, Cu kg.
115, 6.57, 0.303, 0.607


Pri/Sec insulation 3 layers. 13 mm. Polyester, 50% overlap.


SECONDARY WINDING DATA.
~~~~~~~~~~~~~~~~~~~~~~~

Volts. Amps. DCR.* Cu kg. NLV. Reg'n.

Sec 1. 24.2 15 0.03 0.311 25.1 3.6%

Sec 2. 24.2 15 0.031 0.32 25.1 3.6%

Sec 3. 24.2 15 0.032 0.33 25.1 3.6%

Sec 4. 24.1 15 0.033 0.34 25.1 4%

* DCR = 20 deg.C

As this is such a small transformer, the calculations exclude any reactive losses in the windings and consider all Cu losses to be based on I^2R.

Hope this helps. Sorry about the formatting and thanks for all your help

Ray

 

[stevenal]

Okay, your initial posted rating of 1500VA threw me off. Your posted data also doesn't support saying overall regulation is 3.6%. Winding 4 throws the average higher, making room for some load reactive losses. Note that this is test data, so it can't very well exclude the reactive losses when measuring regulation.

reg = 3.69% (your definition), R=3.76%, X=3.70% Z=5.27%. Add 53.2 VArs to the numerator of your formula.

 

[Mondy]

Thanks Stevenal
OK I understand how you came to 3.69% regulation and R=(54.09/1440)% and Z is the quadrature sum of R+X, but how did you come up with X and 53.2VArs?

Will this then evaluate to
COS(ATAN(53.2/(1440+54.09+6.5)))= 0.99937 = 1 ?

Cheers
Ray

 

[jbartos]

Suggestion: It appears that the original posting is more on concept seeking side rather than any computational accuracy side.
Also, there appear to be different definitions of regulations.
VA power is per the transformer secondary
Primary Capacitor capacitance and capacitive reactance -jXc=-j/2pifC has to be equal the transformer input inductive reactance = jXl

 

[stevenal]

I was using a flawed derivation. IEEE C57.12.90 gives the exact formula: REG=sqrt((R+pf)^2+(X+q)^2)-1. q is the reactive factor of the load, 0 in this case. pf of the load is 1. All values in per unit.

Going back to the given data, and recalculating REG per IEEE, then solving for X in the above expression, I now get:
REG=3.83%, R=3.76% and X=3.81%. 1440VA*3.81%=54.9VArs.

Input pf to the fully loaded transformer is then:

pf=cos(atan((54.9+9.2)/(54.09+6.5+1440)))=99.9%

If you wish to correct it to unity for this condition, 54.9+9.2=64.1 VArs of compensation is required.