Power transformer 60 hz to be tested on 50 hz

[bigjoe5263]

The two power transformer 69/22.9kV and 230/69kV are both rated 60/80/100MVA designed for 60 Hz operation. However it will be tested on test facility were the available supply is only 50 Hz.

During short circuit test at the LV side the 60/80/100MVA rating will be tested for the lowest tap, nominal tap and highest tap (16R, N, 16L) each.

Question:
1. What must be current be to be injected on the HV side for each tap for each MVA rating at 60 Hz?
2. What are the equivalent values of these currents (60Hz) to 50 Hz? is there a formula for that? or standard calculation? (ie 60 to 50 Hz conversion or vice versa)

Thanks in advance

 

[m.s.j]

As you know power supply frequency effect to core losses including eddy current and hysteresis losses, therefore the no-load loss and current should be measured at rated frequency, at a voltage equal to the rated voltage. Other winding is left open-circuited. If the supply frequency is other than the rated frequency, but within +/-3 per cent, the no-load loss may be recorded at that frequency and appropriate correction factors applied to evaluate the loss at rated frequency. The results achieved after such corrections are fairly accurate and may be accepted for all industrial measurements.
One of manufacturer used k= [(f1/f2) + (f1/f2)²]/2 as a Frequency correction factor.

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Creative thinking is breezy, Then think about your surrounding things and other thought products. http://electrical-riddles.com

 

[Bob S]

As you know power supply frequency effect to core losses including eddy current and hysteresis losses, therefore the no-load loss and current should be measured at rated frequency, at a voltage equal to the rated voltage. Other winding is left open-circuited.

Wouldn't running the transformer at rated voltage, but at a lower frequency, over-stress the iron core (volt-seconds)? The iron would be driven into saturation.

∫ V·dt = -N·A∫dB = -N·A·ΔB

Another way of looking at this is using V = L dI/dt:

∫ V·dt = L ∫dI = L ΔI

[added] see page 32 in

http://www.google.com/url?sa=t&sour...ka3lCg&usg=AFQjCNGJe53zK-9F8kJFVn_cAzzKrzwOpA

Bob S

 

[bigjoe5263]

thanks for your replies.. the technical personnel gave this relation V(60)=V(50)x6/5, these is for the voltage, which can be applied to determine the voltage for the 50hz supply during open circuit test. But during short circuit test, the voltage is to be adjusted from zero upto a level where the rated current is injected. But when the transformer is tested on a 50 hz supply what is the percentage of the 50hz rated current to the 60hz rated current? are they just equal?

 

[Studiot]

the technical personnel gave this relation V(60)=V(50)x6/5

I am glad I don't have to rely on your technical department!

Most 'off the shelf' transformers are rated at 50/60 hz and will function equally well on either.

It is only when you are considering expensive special purpose ones that frequency becomes important, and only when they are designed for just the one environment.

Remember that equipment that uses the supply frequency for timing in some way, such as television or rotating machinery will be significantly affected by a frequency change.

 

[Phrak]

thanks for your replies.. the technical personnel gave this relation V(60)=V(50)x6/5, these is for the voltage, which can be applied to determine the voltage for the 50hz supply during open circuit test.

This is the right relationship to use to derate the voltage in order to maintain the same peak flux density. I.e.: For a 60 Hz transformer rated at 120 VAC. Apply 100 VAC for 50 Hz operation.

But during short circuit test, the voltage is to be adjusted from zero upto a level where the rated current is injected.

Inject current? I think you mean that a known load will be applied across the secondary, and the secondary will then be expected to supply the rated current.

But when the transformer is tested on a 50 hz supply what is the percentage of the 50hz rated current to the 60hz rated current? are they just equal?

See the answer above. It depends upon the supplied voltage. If you use the rated voltage you will draw excess magnetization current.

 

[Bob S]

For eddy current and hysteresis core losses, there is no direct way to simulate 60-Hz behaviour with 50-Hz power.

Eddy current losses scale as the square of frequency and the square of the magnetization. Because the primary voltage is reduced at 50 Hz to maintain the same magnetization, the eddy current losses will be less by (50/60)² = 69%.

The hysteresis losses at constant magnetization scale linearly with frequency, so they will also be less at 50 Hz.

Bob S

 

[m.s.j]

Thank you very much Bob S, excellent reply

The eddy loss (Pe) and hysteresis loss (Ph) are thus given by


Pe = k₁ f² t² B²_rms

Ph = k₂ f Bⁿ_mp

where

f is frequency
t is thickness of individual lamination
k₁ and k₂ are constants which depend on material
B_rms is the rated effective flux density corresponding to the actual r.m.s. voltage on the sine wave basis
B_mp is the actual peak value of the flux density
n is the Steinmetz constant having a value of 1.6 to 2.0 for hot rolled laminations and a value of more than 2.0 for cold rolled laminations due to use of higher operating flux density in them.



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Creative thinking is breezy, Then think about your surrounding things and other thought products. http://electrical-riddles.com

 

[bigjoe5263]

during short circuit test, with the LV side short circuited and voltage will be applied on the HV side and it will be adjusted from zero to a level where the rated current at HV or the LV is expected.

The transformer is designed for 60 Hz, however the available supply on the test facility is 50Hz, is this current to be derrated also at 50 Hz? upto what level or is there a relationship/equation between 60 hz and 50 Hz current?

Thank you guys for taking your time..

 

[m.s.j]

As you know the aim of short circuit test is determination of %Uk which related to transformer internal impedance ,%Uk = Z (per unit). On the other hand we can write Z = R + j X, if you obtain amount of R from DC test you can calculate Z (60Hz) easily.
We can write X (60Hz) = X (50Hz). 60/50, because the base of transformer inductance is air leakage flux and independent of magnetizing core circuits ( X = 2пf L) . Consequently we can write:

Z (60 Hz) = R + X (50Hz). 60/50

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Creative thinking is breezy, Then think about your surrounding things and other thought products. http://electrical-riddles.com

 

[Bob S]

during short circuit test, with the LV side short circuited and voltage will be applied on the HV side and it will be adjusted from zero to a level where the rated current at HV or the LV is expected.

The transformer is designed for 60 Hz, however the available supply on the test facility is 50Hz, is this current to be derrated also at 50 Hz? upto what level or is there a relationship/equation between 60 hz and 50 Hz current?

My possibly naive guess is that the short circuit test is to test the I²R winding losses in the primary and secondary windings. This would imply that the measured currents should be the same for both 50 Hz and 60 Hz.

Bob S

 

[bigjoe5263]

thats also what i am thinking Bob S,

and if you neglect the Resistance r of the transformer, I(60) = I(50)

 

[m.s.j]

Load losses are incident to the carrying of a specified load. Load losses include
I ² ⋅ R loss in the current carrying parts (windings, leads, busbars, bushings), eddy losses in conductors due to eddy currents, and stray loss induced by leakage flux in the tank, core clamps, or other parts (IEEE 2002).

The short-circuit test yields the values for the equivalent series impedance
Z_EQ = R_EQ + jX _EQ (referred to the primary side).
These two tests are carried out simultaneously, the two-wattmeter method can be employed for measuring the load (copper) loss of a three-phase transformer, one instrument normally being used, the connections from which are changed over from anyone phase of the transformer to any other by means of a double pole switch. The impedance voltage is given by the voltmeter reading obtained across either phase. The copper loss would be the same if measured on the LV side, but it is more convenient to supply the HV winding.
Not only for impedance voltage test ( refer to my last post) but also the copper-loss test should be carried out at the frequency for which the transformer is designed, because the frequency affects the eddy-current copper-loss component, though not affecting the I²R losses.

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Creative thinking is breezy, Then think about your surrounding things and other thought products. http://electrical-riddles.com

 

[able357]

Everyone is on the right path but remember you have to also consider the power rating of the transformer. If you have a transformer designed at 60 Hertz and now want to apply it into a 50 Hertz system you will use the formula which decreases the voltage to 83.33% while also derating the power to 83%. Let's look at a real example and to keep it simple we will deal with a single phase transformer with the following nameplate data.


Power rating: 2500KVA
HV: 13,800 Volts
LV: 480Volts
Frequency: 60 hertz


Step 1: Let's begin by derating the voltage for 60 Hertz
where the derating factor is 5/6 = .833
HV derating: 13,800 V x .833 = 11500 Volts
LV derating:480 V x .833 = 399 Volts


Step 2: At 60 Hertz your HV and LV currents are
HV I = 2500KVA/13800 = 181.1 Amps
LV I = 2500KVA/480 = 5208 Amps


So we have derived two currents which are used to size the HV and LV conductors. Since this transformer is going into a 50 Hertz application and we are derating the HV and LV to 83% of its rated votlage shown in step 1 we now have to derate the Power rating of 2500KVA or else our current will increase causing higher losses and heating.

If we decide to derate the votlage and not the power rating this is our resultant current

Step 3: At 50 Hertz your HV and LV currents are
HV I = 2500KVA/11500V = 217 Amps (an increase from 181Amps)
LV I = 2500KVA/399V = 6266 Amps (an increases from 5208 Amps)

The conclusion to not derate your power rating is that you have increased the current going through the windings. Therefore you have to derate the power rating:

Step 4:Power Derating
2500KVA x .8333 = 2082KVA

Step 5: After derating everything you get the following

HV derating: 13,800 V x .833 = 11500 Volts
HV I = 2082KVA x 11500V = 181 Amps (back to rated design current)

LV derating:480 V x .833 = 399 Volts
LV I = 2082KVA x 399V = 5218 Amps (back to rated design current)


So when you perform your load loss test you do not simulate using the 60Hertz power rating but rather the derated 50Hertz rating of 2080KVA which maintains the same current rating or with more perspective it maintains the same current density.

Hope this helps