# transformer efficiency

Discussion in 'Homework Help' started by power engineer, Sep 24, 2013.

1. ### power engineer Thread Starter New Member

Sep 24, 2013
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the efficiency of single phase 220/35 volt transformer on a certain resistive load is about 66% but as i increase resistive load, efficiency goes decreasing.... why its happening???

2. ### wayneh Expert

Sep 9, 2010
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Impedance mismatch.

Apr 2, 2009
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771

4. ### #12 Expert

Nov 30, 2010
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IR losses in the windings.

5. ### studiot AAC Fanatic!

Nov 9, 2007
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On what basis are you calculating the efficiency, or are you measuring it?

What is the regulation % of the transformer?

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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If the load resistance is increased to finally become an open circuit there will be input power due to magnetization losses but no output power, at which condition the efficiency would be zero percent.

Last edited: Sep 25, 2013
7. ### Ramussons Active Member

May 3, 2013
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I guess Increase of Resistive Load means Reducing the resistance
I agree with #12

Ramesh

8. ### wayneh Expert

Sep 9, 2010
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Alright, now I need to nitpick. I said impedance, implying an AC effect in addition to DC resistance, whereas #12 said IR losses, ie. DC resistance.

There is an optimal load at which the transformer will provide optimal efficiency. Efficiency will fall off at either a lower load (higher resistance) or a higher load (lower resistance).

It's my understanding that in most transformers, the AC component of impedance is relatively larger than the DC resistance of the coils. So while both contribute, the DC resistance is a relatively minor contributor.

Am I all wet?

9. ### power engineer Thread Starter New Member

Sep 24, 2013
2
0
n= VsIs / VpIp

14.4 % regulation
% regulation is too poor and its also increasing as load increases

10. ### studiot AAC Fanatic!

Nov 9, 2007
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So as you increase the load (I assume you mean draw more current?), what happens to Vs and Vp and therefore Vs/Vp?

What happens to Is/Ip?

So what happens to (Vs/Vp)*(Is/Ip)?

My question about regulation was meant to prompt you to think about Vs/Vp.

11. ### #12 Expert

Nov 30, 2010
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Don't start carving this in a brass plaque, but...the AC impedance is the controlling factor for the bulk of the intended energy transfer, but, as I understand it, inductance does not dissipate energy. Energy lost in a transformer is going to show up as heat and that comes from current X resistance in the windings and mis-directed (eddy) current in the iron of the core (times the ohmic resistance of the iron). I think you just got a misplaced factor in your definition between intentional power flow and unintentional power flow.

Anybody else want to shine some light on this idea?

12. ### studiot AAC Fanatic!

Nov 9, 2007
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Do you need more, it is there in the maths?

13. ### wayneh Expert

Sep 9, 2010
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Ah, I see your point. I underestimated your "IR losses" statement as applying only to the coil resistance. My bad.

14. ### #12 Expert

Nov 30, 2010
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You give me too much credit. I was just posting one of the factors as a partial answer. I didn't get serious until post #11.

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
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This depends on what impedance one is talking about.

The impedance of a particular winding of a transformer, with all other windings open-circuited is mostly reactive. The phase angle of the impedance is near 90°; it looks like a not too lossy inductor.

But if we're talking about regulation and efficiency, this isn't the relevant impedance. For this, we need to consider what we might call the "equivalent series impedance" (ESI).

For a two winding transformer, a step down transformer of the sort the OP mentioned, not too big, the ESI is measured by shorting one winding and measuring the impedance seen at the other winding. This impedance appears in series with the windings when the transformer is loaded, and it's mostly resistive.

For example, I took a small concentric wound 32 VA transformer, 120 VAC primary and 24 VAC secondary, shorted the primary and measured the impedance at the secondary. The measured 60 Hz impedance was 2.73 + j .285 ohms. This impedance is mostly resistive, and the I^2*R losses in the resistive part of that impedance accounts for most of the transformer losses at loads near the maximum rated load, and therefore determines the efficiency. For light loads, the core loss becomes dominant. The regulation is determined by both the reactive part and resistive part (the total impedance) of the ESI.

For larger transformers, the reactive part becomes relatively larger. The measured ESI of a 120/12, 2kVA transformer was .00079 + j .00054 ohms.

For transformers which have split bobbin construction rather than being concentric wound, the reactance will be proportionately larger, as with the large transformers.

16. ### wayneh Expert

Sep 9, 2010
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So for a typical wall wart, shorting or overloading the secondary causes over-current and resistive heating in the primary winding?

17. ### #12 Expert

Nov 30, 2010
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I don't know if that's what The Electrician said, but overloading the secondary overloads the magnetic circuit, which lowers the impedance of the primary, which allows more than the rated current...usually. There are also, "impedance protected" transformers and motors (usually tiny shaded pole motors) which have so much ohmic resistance in the primary that you allegedly can't let the magic smoke out.

18. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Doing that causes resistive heating in both primary and secondary.

The transformer impedance, or ESI as I've called it, can be measured at either primary or secondary. It measures the resistance of the other winding as reflected to the measured winding, plus the resistance of the measured winding (plus some reactance, of course). We can use the ESI to calculate what the current drawn and power dissipated will be if we short the secondary with the primary energized.

For example, I pulled a wall wart out of my pile of them. This one has a 120 VAC primary and 6.2 VAC and 1.24 amps secondary rating. If I short the secondary and measure the impedance at the primary, I get 264.7 + j 80.98 ohms.

The DC resistance of the primary is 125 ohms, and the secondary DC resistance is .61 ohms. The turns ratio is 15.43, so the secondary resistance reflected to the primary would be (15.43)^2*.61 = 145.3 ohms. Adding 145.3 ohms (reflected secondary resistance) to the 125 ohms primary DC resistance, we get 270.3 ohms, pretty close to the 264.7 ohms real part of the measured impedance at the primary with the secondary shorted.

Measuring the no load losses of this wall wart with a wattmeter, I get about 1 watt. When I short the secondary with the wattmeter still in the primary circuit, I get about 50 watts absorbed by the wall wart. This power is dissipated in the primary and secondary winding resistances. The core loss is negligible compared to this.

If I have a grid voltage of 120 VAC, and a measured ESI of 264.7 + j 80.98 for the transformer, with the secondary shorted I would expect a primary current of 120/276.8 = .4335 amps (276.8 is the magnitude of the ESI). The real part of the ESI is 264.7 ohms so I would expect a dissipation of I^2*R = (.4335)^2 * 264.7 = 49.75 watts, very close to the measured 50 watt dissipation.

It's this ESI that is effectively in series with the transformer and causes the output voltage to drop when the transformer is loaded.