Transformer losses

mshields · May 16, 2021

What exactly is the relationship between the sum of the load and no load losses and the impedance of the transformer.

I know that the impedance is measured by shorting out the secondary and adding a variable voltage to the primary. When you reach FLA on the secondary, that voltage divided by the nominal voltage is the percent impedance. That, it seems to me is representative of the total losses of the transformer.

So does that mean that if you do the math no-load plus load losses at FLA, you’ll get the same value?

xptpcrewx · May 16, 2021

mshields said:
What exactly is the relationship between the sum of the load and no load losses and the impedance of the transformer.

The no load loss is constant for a given voltage. It has nothing to do with transformer loading. No load losses are measured in real power (watts) and represents energy conversion to/from the electrical domain (in the form of heat or mechanical work)

mshields said:
I know that the impedance is measured by shorting out the secondary and adding a variable voltage to the primary. When you reach FLA on the secondary, that voltage divided by the nominal voltage is the percent impedance. That, it seems to me is representative of the total losses of the transformer.

Yes that’s how you determine %Z but this is not representative of the losses by itself. Total transformer losses are core losses + copper losses. Using the X/R ratio together with %Z you can determine what the full load (or any other loading) copper losses are. The relationship is P=I^2•R. Where R is the effective primary and secondary winding resistance (after reflecting the resistance to primary or secondary side of the transformer).

mshields said:
So does that mean that if you do the math no-load plus load losses at FLA, you’ll get the same value?

Not sure what you’re asking here. Same value as compared to what?

Remember, no load losses only depend on the excitation voltage.

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GoldDigger · May 16, 2021

Xptpcrewx said:
The no load loss is constant for a given voltage. It has nothing to do with transformer loading. No load losses are measured in real power (watts) and represents energy conversion to/from the electrical domain (in the form of heat or mechanical work)

Yes that’s how you determine %Z but this is not representative of the losses by itself. Total transformer losses are core losses + copper losses. Using the X/R ratio together with %Z you can determine what the full load (or any other loading) copper losses are. The relationship is P=I^2•R. Where R is the effective primary and secondary winding resistance (after reflecting the resistance to primary or secondary side of the transformer).

Not sure what you’re asking here. Same value as compared to what?

Remember, no load losses only depend on the excitation voltage.

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If a transformer is designed for maximum efficiency to cost ratio at full load, the core losses are likely to be roughly equal to the copper losses. But the core losses will be roughly independent of load while the copper losses go up as the square of the load current.

electrofelon · May 17, 2021

mshields said:
What exactly is the relationship between the sum of the load and no load losses and the impedance of the transformer.

I know that the impedance is measured by shorting out the secondary and adding a variable voltage to the primary. When you reach FLA on the secondary, that voltage divided by the nominal voltage is the percent impedance. That, it seems to me is representative of the total losses of the transformer.

So does that mean that if you do the math no-load plus load losses at FLA, you’ll get the same value?

I don't know if this is just a coincidence, or an attempt at more discussion, but also see this thread I started a few days ago:

Please discuss transformer losses vs %Z

I know what transformer impedance is and how it is calculated. It is, of course, a somewhat unique use of the word "impedance" How does %Z relate to load and no load losses? I suspect there are other factors at play where we could have transformer A that has a lower %Z yet higher no load losses...

forums.mikeholt.com

Julius Right · May 18, 2021

If transformer short-circuit impedance is Z= R+jX where R it is the winding actual resistance and X is the magnetic field leakage inductance multiplied by angular frequency ω=2.π.f. The main inductance is connected to the magnetic field from the magnetic core. The main magnetic flux passing through the ferromagnetic core produces magnetic losses. The most of leakage magnetic flux passes in the ambient medium as air or oil where no magnetic losses are expected here. A very small part of the leakage magnetic flux passes through the tank or cover indeed, but this is not included in magnetic losses.
The copper losses appear in the load losses, but a small part of no-load losses is also copper losses.
Pcopper=I^2*R. In no-load stage a small current exists-noted Io -due to main reactance of the main flux in series with the leakage flux now and the magnetic losses. Usually the magnetic losses are represented by a virtual resistance Rfe=E^2/Losses [or V^2/Losses].
Magnetic losses are the sum of eddy current losses and hysteresis losses.
Ph = η * Bmax^n * f * V =hysteresis losses V=volume
Pe=Ke*Bmax^2*f^2*V= eddy current losses
Bmax=k*E/nt nt=number of winding turns
E=emf=electromotive force
If E=Volt-I*Z then it is a very low influence of X in magnetic losses. Usually, it may consider E≈Volt and then the magnetic losses depend exclusively on supply voltage.
There are also other losses, less important, as stray losses and losses in other metallic parts [tank, covers]

Besoeker3 · May 18, 2021

This is a typical spreadsheet I have used for calculations of transformer lossesL

Estimated Harmonic Voltage Distortion - Station A

NETWORK HARMONICS All impedance values per phase referred to drive voltage.

	Drive Rating	(kW)	2268			Three running
	Harm Amps at Drive i/p	(A)	447	0,0	0,0	35,8	29,1	0,0	0,0
	Drive Input Voltage	(V)	3300
	Incoming Supply Voltage	(V)	11000
	Supply Frequency	(Hz)	50,0
	Harmonic No		1	5	7	11	13	17	19

Cct No1 - Supply Short Circuit Impedance		0,64%
	Supply Fault Level	(MVA)	40
	Supply resistance	(50Hz Ohms)	0,06534	6,53E-02	6,53E-02	6,53E-02	6,53E-02	6,53E-02	6,53E-02
	Supply reactance	(50Hz ohms)	0,2640825	1,32E+00	1,85E+00	2,90E+00	3,43E+00	4,49E+00	5,02E+00

Cct No2 - Drive Transformer
	Rating	(kVA)	3000
	Impedance	(%)	5,75
	Resistance	(50Hz Ohms)	5,01E-02	5,01E-02	5,01E-02	5,01E-02	5,01E-02	5,01E-02	5,01E-02
	Reactance	(50Hz Ohms)	2,02E-01	1,01E+00	1,42E+00	2,23E+00	2,63E+00	3,44E+00	3,85E+00

Cct No1 + Cct No2
	Resistance	(50Hz Ohms)	1,15E-01	1,15E-01	1,15E-01	1,15E-01	1,15E-01	1,15E-01	1,15E-01
	Reactance	(50Hz Ohms)	4,67E-01	2,33E+00	3,27E+00	5,13E+00	6,07E+00	7,93E+00	8,86E+00

Julius Right · May 18, 2021

Your typical spreadsheet is interesting, indeed. However, I have some remarks -with your permission, of course

-:
It seems to me the source it is a transformer of 40 MVA 9%
If there are actual values it is o.k. If not, then, according to IEC 60076-5 Table 1 – Recognized minimum values of short-circuit impedance for transformers with two separate windings the minimum for 25 up to 40 MVA has to be 10%.
It seems to me, the source impedance is calculated for 11 kV but Z2 impedance is calculated for 3.3 kV.
You have to recalculate the impedance of source for 3.3 kV before the sum.

Besoeker3 · May 18, 2021

Julius Right said:
Your typical spreadsheet is interesting, indeed. However, I have some remarks -with your permission, of course-:
It seems to me the source it is a transformer of 40 MVA 9%
If there are actual values it is o.k. If not, then, according to IEC 60076-5 Table 1 – Recognized minimum values of short-circuit impedance for transformers with two separate windings the minimum for 25 up to 40 MVA has to be 10%.
It seems to me, the source impedance is calculated for 11 kV but Z2 impedance is calculated for 3.3 kV.
You have to recalculate the impedance of source for 3.3 kV before the sum.

It's a Kramer drive (similar to a Static Scherbius Drive. ). It is is a 3.3kV unit stepped down from the site site 11kV supply. All the calcs are based on the 3.3kV supply. The power rating is 2268kW. I don't know what else to add. The units, two of them, have been in service since about 1992..........
It's just what I did for a living. Power electronics.

Julius Right · May 19, 2021

Thank you, Besoeker3, for your information.
I understand you have an induction motor with wounded rotor of 2268 kW supplied-in stator-with 3.3 kV through a transformer of 11/3.3 kV 3 MVA 5.75% short-circuit impedance. The D.C. rotor supply voltage is rectified from the same 3.3 kV of the 11/3.3 kV transformer.
The problem is the 11 kV supply part. Never the less you noted:
Supply Short Circuit Impedance=0.64% and Supply Fault Level=40 MVA
according to the resistance of 0.06534 ohm and reactance of 0.2640825 ohm the impedance then Z1=√(R1^2+X1^2)=0.272 ohm
it is close to 9% short-circuit impedance of 40 MVA transformer and no reference to 0.64% of fault apparent power of 40 MVA.
However, this impedance is referred to 11 kV: Z1=11^2/40*9%=0.2723 ohm

mbrooke · May 19, 2021

Besoeker3 said:
This is a typical spreadsheet I have used for calculations of transformer lossesL

Estimated Harmonic Voltage Distortion - Station A
NETWORK HARMONICS All impedance values per phase referred to drive voltage.
Drive Rating (kW)
2268
Three running
Harm Amps at Drive i/p (A)
447
0,0
0,0
35,8
29,1
0,0
0,0
Drive Input Voltage (V)
3300
Incoming Supply Voltage (V)
11000
Supply Frequency (Hz)
50,0
Harmonic No
1
5
7
11
13
17
19
Cct No1 - Supply Short Circuit Impedance
0,64%
Supply Fault Level (MVA)
40
Supply resistance (50Hz Ohms)
0,06534
6,53E-02
6,53E-02
6,53E-02
6,53E-02
6,53E-02
6,53E-02
Supply reactance (50Hz ohms)
0,2640825
1,32E+00
1,85E+00
2,90E+00
3,43E+00
4,49E+00
5,02E+00
Cct No2 - Drive Transformer
Rating (kVA)
3000
Impedance (%)
5,75
Resistance (50Hz Ohms)
5,01E-02
5,01E-02
5,01E-02
5,01E-02
5,01E-02
5,01E-02
5,01E-02
Reactance (50Hz Ohms)
2,02E-01
1,01E+00
1,42E+00
2,23E+00
2,63E+00
3,44E+00
3,85E+00
Cct No1 + Cct No2
Resistance (50Hz Ohms)
1,15E-01
1,15E-01
1,15E-01
1,15E-01
1,15E-01
1,15E-01
1,15E-01
Reactance (50Hz Ohms)
4,67E-01
2,33E+00
3,27E+00
5,13E+00
6,07E+00
7,93E+00
8,86E+00

Do you, have or can you use this sheet to calculate your Ze and PFC?

Julius Right · May 19, 2021

excuse me please, but what is Ze and PFC?

mbrooke · May 19, 2021

Julius Right said:
excuse me please, but what is Ze and PFC?

Ze is the external earth fault loop impedance (or that of the source), PFC is the perspective fault current.

Besoeker3 · May 19, 2021

mbrooke said:
Do you, have or can you use this sheet to calculate your Ze and PFC?

I don't recall using the Ze. There was a detection circuit, a P&B Golds relay. The fault level is in Cct2.
I don't know if that helps.

mbrooke · May 19, 2021

Besoeker3 said:
I don't recall using the Ze. There was a detection circuit, a P&B Golds relay. The fault level is in Cct2.
I don't know if that helps.

Helps plenty

Julius Right · May 20, 2021

Regarding the impedance of the 11 kV source[see #9], I must admit that if we calculate taking into account only 40 MVA as the apparent short-circuit power and the voltage of 3.3 kV and the ratio X / R = 4 we arrive at the same result.
The 0.64% short-circuit impedance is still an enigma.

Besoeker3 · May 20, 2021

Julius Right said:
Regarding the impedance of the 11 kV source[see #9], I must admit that if we calculate taking into account only 40 MVA as the apparent short-circuit power and the voltage of 3.3 kV and the ratio X / R = 4 we arrive at the same result.
The 0.64% short-circuit impedance is still an enigma.

It's 115 rows further down.
I copied it near the top for convenience.

HARMONIC PHASE VOLTAGES
	At Drive input	(V)	0,0	0,0	5,0	20,9	0,0	0,0
	On Incoming Supply	(V)	0,00	0,00	9,49	39,52	0,03	0,04
	Total HV Distortion	(V)	40,64
	Total HV Distortion	(%)			0,64%
	Total LV Distortion	(V)			21,54
	Total LV Distortion	(%)			1,13%

Transformer losses

mshields

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xptpcrewx

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GoldDigger

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electrofelon

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Please discuss transformer losses vs %Z

Julius Right

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