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Calculation of core losses of a transformer

by clint222
Tags: core losses, losses, transformers
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clint222
#1
Feb4-12, 12:45 PM
P: 36
Hello!

I am a grade 12 student. If you are familar with the IB program, I am writing an extended essay (it's a 4000 word research essay) on how transformer losses are affected by temperature. Through two experiments I was able to determine how the losses of the primary and secondary windings of a transformer change as temperature increases.

In my first experiment I connected the windings in series with each other as well as in series with a 20.0 ohm load. The windings were submerged in hot water. I let the water cool as I collected data. I used the data I collected from this to find how the resistance of the windings increases with temperature and how the increasing resistances decreased the power to the load.

In the second experiment, I set up the transformer and connected the primary windings directly to a constant voltage source. The transformer was submerged in hot water. The water was allowed to cool to collect data for a range of temperatures (between 20 and 70 degrees). I connected the secondary windings directly to the 20 ohm load. I measured to current and voltage in to the primary windings, as well as the voltage and current to the load.

I found the change in primary winding losses with temperature by looking at how the power in decreased with the increasing resistance of the windings.

I was able to find the secondary winding losses and how they changed with temperature by calculating the power the would have been supplied to the load if the windings had no resistance minus the actual power to the load. (I could calculate the expected resistance from the temperature coefficent of the copper wires.)

Now I am trying to figure how to isolate the core losses from the data that I gathered.

I was thinking that I could use:

[Change in core losses from 20 to a given temperature] = [change in power to the load to from 20 to a given temperature] - [change in primary winding losses from 20 to a given temperature] - [change in secondary winding losses from 20 to a given temperature]

The results of this indicated that:
-Primary and secondary winding losses are directly proportional to temperature
-Core losses decrease directly with temperature

I found it strange that the rate of change of core losses with temperature seemed to be the same as the rate of change of primary winding losses with temperature. Might this indicate a flaw in my approach to calculate the core losses?

Does this make sense? Or is it not possible to isolate the core losses? Also, do any of the things I've done sound incorrect?

Any help would greatly be appreciated!
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jim hardy
#2
Feb4-12, 02:35 PM
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did you measure phase angles?

As the core heats up the eddy currents in it decrease.

Eddy currents in the core act like resistive load in parallel with windings that increases in resistance with temperature
but they also tend to cancel primary flux

and i'm not quite sure whether that flux cancelling effect will look like a change of inductance or resistance or combination of the two.

Probably you could tell from careful phase angle measurements.

I'd bet Bassalisk knows.
marcusl
#3
Feb4-12, 04:17 PM
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I am not a transformer expert, but I will venture a couple of comments. First, if you are using a commercial transformer then I suspect the core losses are small compared to ohmic losses in the windings. Second, eddy currents (mentioned in post 2) are not likely to be significant. Power transformers (that is, 60 Hz) are designed with laminated cores (lots of flat plates insulated from each other) precisely to eliminate eddy current losses. The loss mechanism is, instead, magnetic hysteresis as the magnetization of the metal core material cycles through its B-H loop. Third, if your resistance measurement was performed with AC, then the value you measure includes core loss (remember that both windings are always producing field in the core). The resistance measurement should be made instead at DC and with low current, e.g., with a DVM or, for more precision, a four terminal measurement.

Maybe someone else with more transformer knowledge can jump in...

clint222
#4
Feb4-12, 04:44 PM
P: 36
Calculation of core losses of a transformer

Thank you both for your replies!

Quote Quote by marcusl View Post
I am not a transformer expert, but I will venture a couple of comments. First, if you are using a commercial transformer then I suspect the core losses are small compared to ohmic losses in the windings. Second, eddy currents (mentioned in post 2) are not likely to be significant. Power transformers (that is, 60 Hz) are designed with laminated cores (lots of flat plates insulated from each other) precisely to eliminate eddy current losses. The loss mechanism is, instead, magnetic hysteresis as the magnetization of the metal core material cycles through its B-H loop. Third, if your resistance measurement was performed with AC, then the value you measure includes core loss (remember that both windings are always producing field in the core). The resistance measurement should be made instead at DC and with low current, e.g., with a DVM or, for more precision, a four terminal measurement.

Maybe someone else with more transformer knowledge can jump in...
It's actually not a commercial transformer, it's a small educational transformer kit. It consists of two solenoids with an unlaminated iron core that clicks together that you can take in and out. It is very inefficient.

For my resistance measurements I was actually able to take the core out. I did use AC current though, I thought it would be a better idea since it induces core losses wasn't a risk (since I had removed the core), and the measurements would include the AC resistive effects like the skin effect. (I realize these are very insignificant though.)

Quote Quote by jim hardy View Post
did you measure phase angles?

As the core heats up the eddy currents in it decrease.

Eddy currents in the core act like resistive load in parallel with windings that increases in resistance with temperature
but they also tend to cancel primary flux

and i'm not quite sure whether that flux cancelling effect will look like a change of inductance or resistance or combination of the two.

Probably you could tell from careful phase angle measurements.

I'd bet Bassalisk knows.
I don't actually know what phase angles are or how to measure them. Could you briefly explain them?
jim hardy
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Feb4-12, 05:34 PM
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if you look at voltage across transformer coil and voltage across your series resistor you'll find they are not in phase. That's because the transformer has inductance so the voltage across it is not in phase with current through it ,
but voltage across the resistor is in phase with current .
This will show on oscilloscope as the peaks not quite aligned vertically .

test it open secondary at two different temperatures.
If it draws less current warm than cold, well that's combination of eddy currents and warmer copper.
Measure DC resistance change of copper and what isn't accounted for by that change is core losses.
Careful measurement of inductance, by whatever method, will paint a more detailed picture of what's going on.


"...eliminate eddy current losses.." well, at least suppress them to a tolerable level.

Are you equipped to drive the transformer with a triangle wave current source at varying frequencies?
For a triangle wave current you should get a square wave voltage.
Sweep frequency from low to high and you will get a feel for the transformer's frequency response. The edges of the square wave will round off as it departs from ideal.
The effect is described in Thompson's 1901 book "Dynamo Electric Machinery" as "retardation of magnetization".
It's one of the more enlightening experiments i ever did.
The Electrician
#6
Feb5-12, 05:37 AM
P: 761
It would be helpful to know what equipment you have available.

Do you have:

An oscilloscope; if so, how many channels does it have?

A variac; you mentioned a constant voltage source, presumably an AC source?

A wattmeter?

An inductance meter?

How many voltmeter/ammeters? Are they "true RMS" meters? What manufacturer and model number?

You described the transformer as an "educational transformer kit". Who supplied it? Can you provide a link to a web page that has a description and picture of it?

What are the DC resistances of the primary and secondary windings? How many turns are on each winding?
clint222
#7
Feb5-12, 09:37 AM
P: 36
Quote Quote by The Electrician View Post
It would be helpful to know what equipment you have available.

Do you have:

An oscilloscope; if so, how many channels does it have?

A variac; you mentioned a constant voltage source, presumably an AC source?

A wattmeter?

An inductance meter?

How many voltmeter/ammeters? Are they "true RMS" meters? What manufacturer and model number?

You described the transformer as an "educational transformer kit". Who supplied it? Can you provide a link to a web page that has a description and picture of it?

What are the DC resistances of the primary and secondary windings? How many turns are on each winding?
I do not have an oscilloscope, watt-meter (I'm not actually sure what those are), or an inductance meter.

I do have a variable AC power supply from 0 to 16.5V. I have 4 reasonably high precession multi-meters which can measure RMS AC current and voltage up to 10A and 250V or so. They are Elenco M-1750 multi-meters.

I am not sure who provided the kit. I have attached two pictures of it, however.

Honestly, I'm not sure exactly how many turns are in the windings. I think they both have the same number though. If I had to guess, perhaps around 50 turns. The resistance of each winding is around 2.5 ohms at around 24 C.
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jim hardy
#8
Feb5-12, 01:21 PM
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Well that sounds like a pretty interesting setup.

Sorry - i thought you were working in a university lab.

Judging by the pictures, as best i can see your core is not laminated.
In that case core losses should vary strongly with temperature. The reason is eddy currents are free to flow in a non-laminated core. Resistance of iron goes up with temperature which lowers eddy currents, reducing I^2R losses in the iron of the core.

I found the change in primary winding losses with temperature by looking at how the power in decreased with the increasing resistance of the windings.
i dont quite follow. power in should have decreased due to core temperature.

What you can do to separate core losses from winding losses is this: believe in Ohm's law.
Losses in the copper will be I^2 * R where I is current through a winding
and R is resistance of that winding at that temperature.

If you test the transformer with open secondary it can have no loss in secondary copper because current there is zero.
So any power in goes part into primary copper and part into core.
By calculating primary copper loss and subtracting from measured powr in, result is core loss.

Sorry i couldnt follow all your descriptions above - i am a plodder who only takes little steps.

My compliments on your setup and your meticulous tests. Looking forward to more.

old jim
clint222
#9
Feb5-12, 02:11 PM
P: 36
Quote Quote by jim hardy View Post

If you test the transformer with open secondary it can have no loss in secondary copper because current there is zero.
So any power in goes part into primary copper and part into core.
By calculating primary copper loss and subtracting from measured powr in, result is core loss.

old jim
Thanks for the help! That makes sense!

But how would the power out be calculated if there is no current flowing? I realize the voltage could be measured, but how would the power out be found?
jim hardy
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Feb5-12, 02:29 PM
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""But how would the power out be calculated if there is no current flowing? I realize the voltage could be measured, but how would the power out be found?""

there'd be no electrical power out.
All the power that goes in would come out as heat, some from primary copper and some from core. Mostly from core i would wager.

If you allow secondary current to flow,
you'll find copper losses go up because of increased current in both windings.
Core losses should remain pretty constant because flux reemains about constant.
Remember for a sinewave, amplitude of flux is in proportion to voltage.

Have you studied math of sine, cosine and how one is the rate of change(derivative) of the other?
clint222
#11
Feb5-12, 02:48 PM
P: 36
Quote Quote by jim hardy View Post
""But how would the power out be calculated if there is no current flowing? I realize the voltage could be measured, but how would the power out be found?""

there'd be no electrical power out.
All the power that goes in would come out as heat, some from primary copper and some from core. Mostly from core i would wager.

If you allow secondary current to flow,
you'll find copper losses go up because of increased current in both windings.
Core losses should remain pretty constant because flux reemains about constant.
Remember for a sinewave, amplitude of flux is in proportion to voltage.

Have you studied math of sine, cosine and how one is the rate of change(derivative) of the other?
Yes I have. But how would you use that to calculate the core losses then?

Also, if I know the winding resistances (or at least can extrapolate them from the temperature coefficient of copper), would it not be possible to find the core losses by:

Pin - Pout = I2in Rprimary + I2out Rsecondary + Pcore
jim hardy
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Feb5-12, 03:43 PM
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"""would it not be possible to find the core losses by:

Pin - Pout = I2in Rprimary + I2out Rsecondary + Pcore""

exactly. Probably that's what you said in first post i just didnt follow it.

""Yes I have. But how would you use that to calculate the core losses then?""

it wouldn't help with that calculation.
but it can help you understand what's happening inside a transformer core.

Since you're familiar with that concept, you know that derivative of a sine wave is a cosine wave and they have the same shape.They dont look different on an oscilloscope.
Further if flux is sin(wt) then voltage is (some constant) * w * cos(wt) so if flux is sinusoidal so is voltage.

Similarly - the derivative of a triangle shaped wave is a square wave because the triangle wave has constant slope between its peaks.
They look very different on the oscilloscope.

So , since induced voltage is derivative of flux,
if you can produce a triangle wave flux instead of a sinusoidal one in a transformer core,
an open circuited winding on that transformer will produce a square wave voltage.

Since flux is in proportion to current,

a triangle shaped current through a transformer's primary should produce a triangle shaped flux, and a square wave voltage on secondary.
IF you have access to some lab gear you can demonstrate this yourself.

What you will see is almost perfect agreement at low frequency.
As frequency increases the square wave becomes less square - its corners slump.
What is happening is eddy currents in the core are opposing the change in flux, and flux is in proportion to the total current flowing (primary plus eddy) not just primary.

From that experiment you will get a good feel for the useable frequency range of your transformer.

We forget that sine wave is really a mathematic special case.
Other waveshapes can be used to improve our insight.
With sinewave excitation the eddy currents just shift phase a little,
but with triangle excitation they alter the waveshape noticeably.
So we have changed something that's barely noticeable, slight phase shift of sinewave, into something very visible - squareness of squarewave's shoulders.
and that's a good teaching tool.

If your transformer is really unlaminated, as it looks in photos, i suspect that when excited with triangle wave it will give nice looking square waves up to around twenty hertz.

Have you a mentor who could help you try it?
maybe this will interest somebody who's good with that simulation program.


gosh - i dont mean to be obsessive on this point. Just it's good to get a feel for flux-voltage relationship in transformers.

Have fun ! this is the age of technology,so let's enjoy it ...
clint222
#13
Feb5-12, 03:45 PM
P: 36
Quote Quote by jim hardy View Post
"""would it not be possible to find the core losses by:

Pin - Pout = I2in Rprimary + I2out Rsecondary + Pcore""

exactly. Probably that's what you said in first post i just didnt follow it.

""Yes I have. But how would you use that to calculate the core losses then?""

it wouldn't help with that calculation.
but it can help you understand what's happening inside a transformer core.

Since you're familiar with that concept, you know that derivative of a sine wave is a cosine wave and they have the same shape.They dont look different on an oscilloscope.
Further if flux is sin(wt) then voltage is (some constant) * w * cos(wt) so if flux is sinusoidal so is voltage.

Similarly - the derivative of a triangle shaped wave is a square wave because the triangle wave has constant slope between its peaks.

So , since induced voltage is derivative of flux,
if you can produce a triangle wave flux instead of a sinusoidal one in a transformer core,
an open circuited winding on that transformer will produce a square wave voltage.

Since flux is in proportion to current,

a triangle shaped current through a transformer's primary should produce a triangle shaped flux, and a square wave voltage on secondary.
IF you have access to some lab gear you can demonstrate this yourself.

What you will see is almost perfect agreement at low frequency.
As frequency increases the square wave becomes less square - its corners slump.
What is happening is eddy currents in the core are opposing the change in flux, and flux is in proportion to the total current flowing (primary plus eddy) not just primary.

From that experiment you will get a good feel for the useable frequency range of your transformer.

We forget that sine wave is really a mathematic special case.
Other waveshapes can be used to improve our insight.
With sinewave excitation the eddy currents just shift phase a little,
but with triangle excitation they alter the waveshape noticeably.

If your transformer is really unlaminated, as it looks in photos, i suspect that when excited with triangle wave it will give nice looking square waves up to around twenty hertz.

Have you a mentor who could help you try it?
maybe this will interest somebody who's good with that simulation program.


gosh - i dont mean to be obsessive on this point. Just it's good to get a feel for flux-voltage relationship in transformers.

Have fun ! this is the age of technology, let's enjoy it ...
Thank you very much for the help! I was doing a few things wrong in the calculations before I think, but I understand how to do it now!

Transformers are very interesting devices! I'll have to learn more about them!
jim hardy
#14
Feb5-12, 04:00 PM
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Indeed. They are faithful yet subtle .
The Electrician
#15
Feb5-12, 07:20 PM
P: 761
Quote Quote by clint222 View Post
Yes I have. But how would you use that to calculate the core losses then?

Also, if I know the winding resistances (or at least can extrapolate them from the temperature coefficient of copper), would it not be possible to find the core losses by:

Pin - Pout = I2in Rprimary + I2out Rsecondary + Pcore
The problem you face is that you cannot determine Pin with the equipment you have. The input power is not going to be simply the product of input current and input voltage, IF you have the core in place. There will be a phase difference between the primary current and the primary voltage, and you must have a wattmeter (or oscilloscope which can do trace math) to measure the true input power. When you measure the input current and input voltage and multiply them, you are calculating the input volt-amperes, not the input power (watts).

See: http://en.wikipedia.org/wiki/Power_factor

Judging from the picture of your transformer, I suspect that the ratio of the primary reactance at 60 Hz to the primary resistance (the Q) is not very high. The importance of this is the following.

A commercial transformer will have much better material for the core, with higher permeability and a shorter path length, which will give a higher primary inductance, which will then give much smaller no load primary current. With such a transformer there is a standard technique to separate the copper losses from the core losses. To determine the core losses the standard technique is to apply rated voltage to the primary with the secondary unloaded and measure the true power with a wattmeter. If the primary inductance is high, this excitation current is so small compared to the rated current with a load that the I^2*R losses (copper losses) are entirely negligible compared to the core losses (hysteresis and eddy current losses), and the power consumed by the transformer is almost entirely core loss; thus the core loss is measured independent of the copper loss.

But your transformer probably doesn't have a very high primary inductance, so the primary current with rated voltage applied and no load on the secondary may be high enough to cause copper losses comparable to the core loss, and therefore the core losses will not be well separated from the copper losses of your transformer with this measurement.

See: http://en.wikipedia.org/wiki/Open_circuit_test

On the other hand, to measure the copper losses, independent of the core losses, the short circuit test is used. The secondary is shorted with an AC ammeter and the primary is excited with a voltage considerably less that the rated primary voltage, with this excitation passing through a wattmeter. Typically, a variac (variable transformer; you might be able to use your variable AC supply) is used. Start out with the variable AC supply set to zero volts and apply to the primary with the secondary shorted with an ammeter. Slowly turn up the primary voltage until the secondary current (as read on the ammeter which is shorting the secondary) is equal to the rated secondary current. Then the wattmeter which is connected between the variable AC source and the primary indicates the power consumed by the transformer. When the applied voltage is low, as it is in this test, the flux in the core is low and the core losses are so low as to be negligible, and the measured power consumption is almost entirely copper loss. Making the measurement this way includes all the losses (skin effect, proximity losses, etc.) that occur when the transformer is used as a transformer.

See: http://en.wikipedia.org/wiki/Short_circuit_test

It looks like your transformer has equal turns on the primary and secondary, so you could just pass rated (AC) current through both windings in series and multiply the applied voltage by the current and get a good value for the copper losses. This measurement doesn't require a wattmeter because the phase shift between voltage and current is essentially zero when there is no core present.

Another problem with making measurements on your transformer is that because the two windings are so far apart (much farther than on a commercial transformer), the leakage inductance (http://en.wikipedia.org/wiki/Leakage_inductance) is significantly larger than a commercial transformer. This may cause the copper losses measured without a wattmeter (as in the paragraph above) to be noticeably different than calculated just by multiplying primary voltage and current.

These assumptions, namely, that copper loss is negligible during the open circuit test, and that core loss is negligible during the short circuit test, are true for a typical commercial transformer construction, but perhaps less true for your educational transformer.

However, it's very important for you to realize that when the core is in place, and the secondary is unloaded, the losses in the transformer are not just equal to the product of the applied voltage and the primary current; you need a wattmeter (http://en.wikipedia.org/wiki/Wattmeter).
clint222
#16
Feb5-12, 07:42 PM
P: 36
Quote Quote by The Electrician View Post
The problem you face is that you cannot determine Pin with the equipment you have. The input power is not going to be simply the product of input current and input voltage, IF you have the core in place. There will be a phase difference between the primary current and the primary voltage, and you must have a wattmeter (or oscilloscope which can do trace math) to measure the true input power. When you measure the input current and input voltage and multiply them, you are calculating the input volt-amperes, not the input power (watts).

See: http://en.wikipedia.org/wiki/Power_factor

Judging from the picture of your transformer, I suspect that the ratio of the primary reactance at 60 Hz to the primary resistance (the Q) is not very high. The importance of this is the following.

A commercial transformer will have much better material for the core, with higher permeability and a shorter path length, which will give a higher primary inductance, which will then give much smaller no load primary current. With such a transformer there is a standard technique to separate the copper losses from the core losses. To determine the core losses the standard technique is to apply rated voltage to the primary with the secondary unloaded and measure the true power with a wattmeter. If the primary inductance is high, this excitation current is so small compared to the rated current with a load that the I^2*R losses (copper losses) are entirely negligible compared to the core losses (hysteresis and eddy current losses), and the power consumed by the transformer is almost entirely core loss; thus the core loss is measured independent of the copper loss.

But your transformer probably doesn't have a very high primary inductance, so the primary current with rated voltage applied and no load on the secondary may be high enough to cause copper losses comparable to the core loss, and therefore the core losses will not be well separated from the copper losses of your transformer with this measurement.

See: http://en.wikipedia.org/wiki/Open_circuit_test

On the other hand, to measure the copper losses, independent of the core losses, the short circuit test is used. The secondary is shorted with an AC ammeter and the primary is excited with a voltage considerably less that the rated primary voltage, with this excitation passing through a wattmeter. Typically, a variac (variable transformer; you might be able to use your variable AC supply) is used. Start out with the variable AC supply set to zero volts and apply to the primary with the secondary shorted with an ammeter. Slowly turn up the primary voltage until the secondary current (as read on the ammeter which is shorting the secondary) is equal to the rated secondary current. Then the wattmeter which is connected between the variable AC source and the primary indicates the power consumed by the transformer. When the applied voltage is low, as it is in this test, the flux in the core is low and the core losses are so low as to be negligible, and the measured power consumption is almost entirely copper loss. Making the measurement this way includes all the losses (skin effect, proximity losses, etc.) that occur when the transformer is used as a transformer.

See: http://en.wikipedia.org/wiki/Short_circuit_test

It looks like your transformer has equal turns on the primary and secondary, so you could just pass rated (AC) current through both windings in series and multiply the applied voltage by the current and get a good value for the copper losses. This measurement doesn't require a wattmeter because the phase shift between voltage and current is essentially zero when there is no core present.

Another problem with making measurements on your transformer is that because the two windings are so far apart (much farther than on a commercial transformer), the leakage inductance (http://en.wikipedia.org/wiki/Leakage_inductance) is significantly larger than a commercial transformer. This may cause the copper losses measured without a wattmeter (as in the paragraph above) to be noticeably different than calculated just by multiplying primary voltage and current.

These assumptions, namely, that copper loss is negligible during the open circuit test, and that core loss is negligible during the short circuit test, are true for a typical commercial transformer construction, but perhaps less true for your educational transformer.

However, it's very important for you to realize that when the core is in place, and the secondary is unloaded, the losses in the transformer are not just equal to the product of the applied voltage and the primary current; you need a wattmeter (http://en.wikipedia.org/wiki/Wattmeter).

Hmmm, I see what you're talking about. I don't have access to most of those tools though, so I think I'll have to go with the method I'm already using and acknowledge that it isn't fully accurate.
The Electrician
#17
Feb5-12, 11:10 PM
P: 761
How did it happen that you chose to do a study of the variation of transformer losses with temperature for your essay? I'm wondering if it was suggested by a teacher who might not have known that it would require more equipment than is available to you.

As a matter of curiosity, suppose you apply 2 volts AC to one winding without any core in place and measure the current, what is its value? Then, with the core in place, what is the current with the same 2 volts AC applied?

These two values will tell me how much the primary inductance increases due to the core.
es1
#18
Feb7-12, 11:31 AM
P: 324
Quote Quote by The Electrician View Post
The problem you face is that you cannot determine Pin with the equipment you have.
I think Pin can be determined and the ∆P=Conduction+Core loss will work as long as only RMS measurements are used. RMS measurements are easier anyway and don't require all the special tools and apparently Clint222's DMM can just do them.

You can't do the apparent power, etc. but one can still get a meaningful result. It just comes with the caveat that instantaneous power lost could much higher (or lower) then this measured result. As long as the caveat goes into the report all is well, IMHO.


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