Is this calculator legit?

[mbrooke]

Are the two calculators on this page legit? Do they at least look accurate enough? I'd like to save the link and use them on the job.



http://voltage-disturbance.com/power-engineering/transformer-voltage-regulation/

 

[Julius Right]

The first voltage regulation calculator use indeed this formula:
voltage regulation=(V_nl-V_load)/V_load
In IEEE 141/1993 [Red Book] chpt.3.11.1 General mathematical formulas
there are two formulas for voltage drop calculation: one approximate and one accurate.
If the voltage drop is divided by V_load the DV% will be 4.048% [approx.] or 4.076%[accurate].
In my opinion, in the second calculator the voltage drop is divided by V_nl and then we get 3.89%.

 

[mbrooke]

The first voltage regulation calculator use indeed this formula:
voltage regulation=(V_nl-V_load)/V_load
In IEEE 141/1993 [Red Book] chpt.3.11.1 General mathematical formulas
there are two formulas for voltage drop calculation: one approximate and one accurate.
If the voltage drop is divided by V_load the DV% will be 4.048% [approx.] or 4.076%[accurate].
In my opinion, in the second calculator the voltage drop is divided by V_nl and then we get 3.89%.

What about the last one? Look to good to be true...

 

[Jraef]

There is some nice info on that site by the way. Thanks for sharing. I can’t figure out who’s site it is though. I’m on my phone and mane it’s optimized for mobile viewing, I’ll check again later.

 

[steve66]

What about the last one? Look to good to be true...

Wasn't a claim made in an earlier thread that the transformer impedance was equal to the voltage regulation? (i.e. for a 5% impedance a fully loaded xformer would have a 5% voltage drop?)

I think that statement would have been too good to be true.

It does hold for a PF = 1 and a low X/R ratio, but as either varies, the voltage drop does change.

BTW, I wish I could just type in the numbers, and not have to try to delicately adjust the sliders on that calculator.

EDIT:

Actually, it seems to be a PF of about 0.7 that gives the same voltage drop that matches the impedance. And as the X/R ratio climbs, the voltage drop seems to decrease.

So in most cases the voltage drop is lower than the transformer Z.

 

[romex jockey]

legit? i wouldn't really know.....my luck it'd tell me i'm wrong shortly after i turned it on.....in a womans voice at that......:dunce:~RJ~

 

[gar]

190416-2118 EDT

Since the author of the calculators uses load voltage at full load as the divisor I get the same answer for a resistive load. But I normally think of open circuit voltage as my divisor.

For the second calculator I get 3.96 instead of 3.89 .

My assumptions were to first calculate the the transformer % impedances from the X/R ratio and assume full load current produces a full load output voltage of 100%, his divisor definition.

Then I assumed output voltage was 100% at the X/R ratio of load defined by PF = 0.8 from this I determined the load X and R % values. These were added to the transformer impedance, and in turn the voltage of the source was determined. Now the regulation can be determined. I did not create one equation, but did partial calculations. Might have made some mistakes, but my answer is close to the author's.

.

 

[mbrooke]

Wasn't a claim made in an earlier thread that the transformer impedance was equal to the voltage regulation? (i.e. for a 5% impedance a fully loaded xformer would have a 5% voltage drop?)

I think that statement would have been too good to be true.

It does hold for a PF = 1 and a low X/R ratio, but as either varies, the voltage drop does change.

BTW, I wish I could just type in the numbers, and not have to try to delicately adjust the sliders on that calculator.

EDIT:

Actually, it seems to be a PF of about 0.7 that gives the same voltage drop that matches the impedance. And as the X/R ratio climbs, the voltage drop seems to decrease.

So in most cases the voltage drop is lower than the transformer Z.

My thoughts word for word

 

[gar]

190416-2359 EDT

For a resistive load the equivalent load circuit is the source jXl internal + R internal + R load. For a low source impedance the load voltage is approximately Voutput = Vsource * ( Rload/(Rload+Rsource) ). The reactive component is small compared to the sum of the resistive components, and it is at 90 deg. Thus, the impedance hypotenuse and resistive side of a right triangle are almost equal.

Where the source reactive component is large compared to its resistive component that makes Z internal not too important compared to just the R internal with a resistive load. Although the reactive component of of the internal impedance is large when only looking at internal impedance it is small when viewed from the total resistive load on the source.

When the load changes to have reactance as well as resistance in the load, then you have a series circuit of R source + R load + jXl source + jXl load with the voltage across Z load being of importance compared to V source. With substantial load reactance we can not ignore the current phase shift that occurs.

Quickly drawing some phasor diagrams for a fixed transformer and loading to full rated current it looks like regulation using source voltage as divisor in the regulation equation that as you go from a resistive load to a more inductive load that the regulation gets worse.

Also it looks like for the same load current that an R load will have a higher resistive magnitude than the Z magnitude of an inductive load.

Is my thinking correct on this?

.

 

[gar]

190417-0844 EDT

Do the following two calculations:

Assume a transformer with an open circuit voltage of 100 V, a full load current rating of 1 A, and an internal impedance of 3 ohms inductive reactance, and 0.3 ohms resistive.

1. Apply a resistive load that produces 1 A of load current. What is the % regulation using loaded load voltage as the divisor?

2. Same as question 1 except that the load is a pure inductance.

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[gar]

190417-1255 EDT

No replies to my two experiments to calculate some values.

I had not compared my results with the results from use of the second calculator. Thus, since I had values that I calculated I put my input data into the second calculator, and found the calculator results close to the results I had determined with my own procedure.

Based on this limited checking I think that both the first and second calculators do provide a useful result.

.

 

[gar]

190418-1956 EDT

mbrooke:

Do you now believe the calculators are "legit" or not?

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