# Transformer With Load (Confusion, need clarification)

Discussion in 'Homework Help' started by jegues, Dec 17, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
For this question I am having some confusion for part 3).

I can't seem to get the same voltage regulation as him and I can't figure out what I'm doing wrong.

I've assumed the load to be on the leftmost portion of the HV side.

I've also assumed since we are at rated load that the current across the transformer is 1000V as per its rating.

Is there anything wrong with how I've drawn the circuit for the HV side? Are all the voltages in the correct places?

If so, I write a KVL,

$-V_{load} -\vec{I_{2}}(R_{eq} + jX_{eq}) + 1000V = 0$

From here,

$V_{load} = 1000 -\vec{I_{2}}(R_{eq} + jX_{eq})$

This gives me,

$V_{load} = 1027$

which is different from his answer.

Note: His answer he uses different names for his voltages. It seems as though he assumes the 1000V is across the load. How does he know this? We've only been given rated current!

Is his answer incorrect? Or is mine?

If it's mine, what am I doing wrong? Is there something wrong with how I've drawn the HV side? Did I place the 1000V in the wrong spot?

I am so terribly confused, I just want to know what I'm mixing up!

Thanks again!

Tnk where are you!?!? I need to clear this up before my exam!

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• ###### TransConfSol.JPG
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Last edited: Dec 17, 2011
2. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
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I think the difference is the note on the 2nd attachment:

Open circuit test is on High side
Short circuit test is on Low side

but the first attachment only shows the high side.

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Yes because as I said in the first part, my confusion resides only in part 3).

I understand everything up until then.

What am I doing wrong in part 3)?

As far as I can tell all we need to concern ourselves with is the HV side.

4. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I think I figured out what I'm confused out.

When someone gives you rated voltages of a transformer, are these the voltages E1, E2 indicated in the figure attached?

The second figure attached is my professors work from class when he had solved this problem.

Here he draws the rated voltage across the load.

Is the rated voltage always across the load, or E1 and E2 where I thought it should be?

Is it just that in the case for voltage regulation, the rated voltage is assumed to be across the load?

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
If the HV equivalent load side voltage were 1027V at a load current magnitude of 100A as you have calculated, then the transformer VA load would be 102.7kVA rather than the nominal rated 100kVA. The implication in part (3) would presumably be that the transformer is passing its rated kVA value with respect to its terminal conditions. It's a moot point and therefore open to conjecture.

To satisfy the rated throughput kVA load requirement you would have to either

(a) back off the 1000V ideal transformer voltage to ~97.37% of 1kV or 973.7V - the latter voltage being nearer to the teacher's solution value. This is equivalent to setting a lower HV side source voltage.

or

(b) back off the (HV side referred) load current draw to ~97.37A at 0.6pf or tweak the load until you have 100kVA throughput at 0.6pf leading. This is equivalent to keeping ~1kV supply side voltage and adjusting the load impedance magnitude. Taking this approach would bring the HV side load voltage down to ~1026V. This would give a regulation of -2.6% - which is about what the teacher gets.

To my way of thinking the convention would be to assume the supply side voltage is 1kV and work off that premise with the supplementary requirement of setting the transformer kVA throughput to exactly 100kVA.

Last edited: Dec 18, 2011
6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Using some simulations with the transformer parameters determined in the analysis and a primary source of 1kV I then tweaked the load to obtain the following terminal conditions:-

Vsec=102.47V
Isec=975.9A

Vprim=1kV
Iprim=97.47A
Input VA = 97.47kVA
Input pf=0.6376
Input power = 62.15kW

Interestingly the input kVA is less than the load kVA.

7. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Thank you for the all the help so far!

But I still have confusion regarding rated voltages.

How do we know when we actually have rated voltages?

I just did a another problem with the following problem statement,

"A 23-kVA, 2300/230-V, 60-Hz, step-down transformer has the following resistance and leakage-reactance values: R1= 4Ω, R2=0.04Ω, X1=12Ω, X2=0.12Ω. (We are neglecting the core for now) The transformer is operating at 75% of its rated load. If the power factor of the load is 0.866 leading, determine the efficiency of the transformer."

I was able to successfully complete this problem with the correct answers, but the only part I mentally stumble on is how I can convince myself that the rated voltage of 230V is indeed at the output terminals of the transformer.

I understand that 75% rated load means 75% of rated current, but what does that mean for the voltage at the output terminals?

How do we know this???

Last edited: Dec 18, 2011
8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
What does one mean by rated voltage?

Normally the rated voltages for the various transformer windings would be written on the nameplate on a real transformer. In practice if the transformer is designed to operate in a particular mode then the 'critical' voltage rating would be the primary excitation voltage. So if the transformer is designed for step-up operation 2300V:230V the 'critical' rating is 2300V. In practice one can exceed this rating with certain caveats set down in the original design specification.

Since one is generally interested in the transformer load conditions then for the purposes of analysis the kVA rating would normally be interpreted in terms of the secondary winding rating - as your comments imply. The interpretation of the rating therefore leads one to a rated secondary current.

I think your misgivings are about a specific issue - What voltage does one assume as the secondary value at rated conditions? I_sec_rated=kVA rating/Vsec(???). In reality, if one has rated current in the load then the secondary voltage can have a range of values subject to the load power factor - notwithstanding the condition that the primary voltage is held constant. As your calculations have shown the secondary value can actually exceed the nominal secondary 'rated' value with a leading power factor load.

It's difficult to correctly estimate the actual loaded secondary voltage. One must do the analysis first by making an assumption that 'something' is fixed - be that the primary voltage or the secondary voltage. So the estimate for secondary current rating is initially based on the nominal primary voltage, the kVA rating and nominal voltage ratio. If the ratio is stated as 2300V:230V then the assumed operating secondary rated voltage would be 230V with 2300V on the primary. By 'fixing' the secondary voltage at the nominal rated value one has then only to set the load current & pf value for the purposes of analysis.

In practice if the load current and secondary voltage are at their rated value the primary voltage must differ from the rated value. The excess (with normally lagging load power factor) is interpreted as the transformer regulation.

When a question is posed stating that there is to be 'such and such' load kVA at certain conditions, then one still needs to select a fixed starting point for a voltage at some point - either primary or secondary terminal voltage. The selection of nominal rated secondary terminal voltage as the load voltage is a convenient means of setting the prescribed load conditions defined in the question.

Last edited: Dec 18, 2011
9. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
So I, the student, must decide what fixed starting point is most convenient for my analysis?

And in the previous case mentioned, this fixed starting point would be to take the voltage across the terminals of the secondary as its rated value, correct?

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Yes - with the proviso that you also satisfy the stated problem conditions.

For instance what does the statement

"The transformer is operating at 75% of its rated load"

actually mean.

Does it simply imply that the load kVA is 75% of the rated 23kVA or 17.25kVA. Does it also imply that 75% of the rated secondary current is flowing - i.e. 75A. If both, then one is constrained to fix the secondary terminal voltage to 230V and the current to 75A. In addition the current phase angle in relation to the secondary voltage must also be fixed to establish the leading 0.866 pf. Given these conditions it would be apparent that the primary supply voltage could not be 2.3kV. Whilst being somewhat 'contrived', this would be the most likely solution scenario and the easiest to implement.

If only the first constraint (load kVA only) applies then one might be tempted to permit the condition that the primary supply voltage is in fact 2.3kV. In one sense this is a more realistic assumption, since the transformer primary voltage is more likely dictated by the supply line conditions than any load side voltage requirements. The problem with using this scenario is that one has to fiddle with load parameters to meet the load 17.25kVA 0.866 pf conditions. Which could be tedious.

In summary choose the solution scenario which is easiest to implement. In the majority of cases I would think the first solution method outlined above is the better choice - from the student's perspective.

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782

Re-stating your problem as posed :

"A 23-kVA, 2300/230-V, 60-Hz, step-down transformer has the following resistance and leakage-reactance values: R1= 4Ω, R2=0.04Ω, X1=12Ω, X2=0.12Ω. (We are neglecting the core for now) The transformer is operating at 75% of its rated load. If the power factor of the load is 0.866 leading, determine the efficiency of the transformer."

Does it make any difference whether I fix the secondary or primary voltage value in determining the efficiency?

Let's assume a secondary load current of 75A flows.

Consider the case of the secondary terminal voltage being fixed as 230V at angle 0°. To constrain a 0.866 pf requires the load current angle to be 30° leading.