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- Thread starter MineGuy
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- #21
Hi MineGuy;
I'll try to say the same as the others have been telling you, but in a slightly different way.
Transformer windings have resistance and it is measured in ohms.
Transformer windings have reactance and it may also be expressed in ohms.
When the resistive ohms are added vectorily to the reactive ohms, the result is ohms impedance. This doesn't change for a transformer. (Well, it does change somewhat with temperature. The value of impedance at operating temperature is used to express % impedance.)
Now we have the actual ohms impedance.
This is not percent impedance.
Percent impedance is defined as the percentage of the normal
primary voltage that must be impressed upon a transformer to cause rated current to flow in the secondary.
If the rated current is divided by the percent impedance voltage, it yields the symetrical fault current.
If you add forced air cooling to a 20 MVA transformer and push the rating up to 24 MVA, the impedance will remain the same. The rated current will increase 20%. It now takes 20 percent more primary voltage to cause 20% more current to flow in the a short circuited secondary.
As a result, the percent impedance (Which is a figure which is derived for ease of calculation, not a physical property)
will increase by 20%.
If the original percent impedance was 10% it will now be 12%.
The transformer hasn't changed and the impedance of the transformer hasn't changed.
We have re-rated the transformer for a higher output and so the basis on which the percent impedance is calculated has changed.
5% voltage drop. This is a NEC requirement for wiring from the point of common coupling to the end of the circuits. It has to do with voltage drop in the feeders and branch circuit conductors. The voltage drop of the transformer is not part of this requirement.
I am not sure of the European standards, but common motor voltages in North America are less than common system voltages.
For example, if a transformer has a secondary voltage of 600 volts, and a voltage drop of 8.5%, the terminal voltage will be 549 volts. 550 volt motors are commonly used on 600 volt systems.
We still have not considered the voltage drop in the conductors, but the motors will accept a further 10% drop in voltage, although it is not a good idea for heavily loaded motors.
The taps on the transformer may be set 5% high and we will have the 5% we need to address conductor voltage drop.
In your situation. I agree with all of the suggestions that have been made.
Consider forced air cooling to increase the transformer ratings. (Remember to re-calculate your percent impedance voltage.)
Consider one or more synchronous motors, or if that is impractical, capacitor banks or a synchronous condensor to correct the power factor to unity.
Another consideration is to correct some of the power factor with capacitors at the motor terminals and gain the benefit of increased protection against lightning induced surges.
Consider setting the taps higher, or using automatic tap changers or using voltage regulators to maintain a stable voltage.
Many of us have been in plants fed from systems with virtually no voltage compensation and have seen the equipment survive severe voltage swings. Very high voltages when the plant was not operating, and low voltages when the plant was at maximum demand.
If you go with little or no voltage regulation on your power transformers, you may want to consider a regulator for your low voltage transformers. The 600 and 240 volt systems are where you may have the most equipment that is more sensitive to high voltages than large motors.
good luck
respectfully
I'll try to say the same as the others have been telling you, but in a slightly different way.
Transformer windings have resistance and it is measured in ohms.
Transformer windings have reactance and it may also be expressed in ohms.
When the resistive ohms are added vectorily to the reactive ohms, the result is ohms impedance. This doesn't change for a transformer. (Well, it does change somewhat with temperature. The value of impedance at operating temperature is used to express % impedance.)
Now we have the actual ohms impedance.
This is not percent impedance.
Percent impedance is defined as the percentage of the normal
primary voltage that must be impressed upon a transformer to cause rated current to flow in the secondary.
If the rated current is divided by the percent impedance voltage, it yields the symetrical fault current.
If you add forced air cooling to a 20 MVA transformer and push the rating up to 24 MVA, the impedance will remain the same. The rated current will increase 20%. It now takes 20 percent more primary voltage to cause 20% more current to flow in the a short circuited secondary.
As a result, the percent impedance (Which is a figure which is derived for ease of calculation, not a physical property)
will increase by 20%.
If the original percent impedance was 10% it will now be 12%.
The transformer hasn't changed and the impedance of the transformer hasn't changed.
We have re-rated the transformer for a higher output and so the basis on which the percent impedance is calculated has changed.
5% voltage drop. This is a NEC requirement for wiring from the point of common coupling to the end of the circuits. It has to do with voltage drop in the feeders and branch circuit conductors. The voltage drop of the transformer is not part of this requirement.
I am not sure of the European standards, but common motor voltages in North America are less than common system voltages.
For example, if a transformer has a secondary voltage of 600 volts, and a voltage drop of 8.5%, the terminal voltage will be 549 volts. 550 volt motors are commonly used on 600 volt systems.
We still have not considered the voltage drop in the conductors, but the motors will accept a further 10% drop in voltage, although it is not a good idea for heavily loaded motors.
The taps on the transformer may be set 5% high and we will have the 5% we need to address conductor voltage drop.
In your situation. I agree with all of the suggestions that have been made.
Consider forced air cooling to increase the transformer ratings. (Remember to re-calculate your percent impedance voltage.)
Consider one or more synchronous motors, or if that is impractical, capacitor banks or a synchronous condensor to correct the power factor to unity.
Another consideration is to correct some of the power factor with capacitors at the motor terminals and gain the benefit of increased protection against lightning induced surges.
Consider setting the taps higher, or using automatic tap changers or using voltage regulators to maintain a stable voltage.
Many of us have been in plants fed from systems with virtually no voltage compensation and have seen the equipment survive severe voltage swings. Very high voltages when the plant was not operating, and low voltages when the plant was at maximum demand.
If you go with little or no voltage regulation on your power transformers, you may want to consider a regulator for your low voltage transformers. The 600 and 240 volt systems are where you may have the most equipment that is more sensitive to high voltages than large motors.
good luck
respectfully
To all:
In all above discussions is Voltage Regulation is being mixed up with voltage drop during a short circuit or motor starting inrush currents?
Steady state voltage drop is related to Voltage Regulation and has little to do with actual %Z. or at least Voltage regulation does not equal %Z at full load.
A good tranfromers voltage regulation will be less than 3% or 4%.
In all above discussions is Voltage Regulation is being mixed up with voltage drop during a short circuit or motor starting inrush currents?
Steady state voltage drop is related to Voltage Regulation and has little to do with actual %Z. or at least Voltage regulation does not equal %Z at full load.
A good tranfromers voltage regulation will be less than 3% or 4%.
- Thread starter
- #23
Thanks WAROSS and all those who contributed. I am grateful for such an exhaustive explanation. I have still not read it as I want to take my to go through it.
Waross:
I have one more question for you.I would appreciate a word.
I beleive I am looking for a thumb rule. What criterion should I consider for the voltage at a bus.
If the voltage is 99.5 % or 98%. What value is ok.
Or It depends on, if I get the right voltage at the end of my load, which might be a single phase electric panel.
What is the tolerance, that I can allow as a thumb rule may be for example a motor. In case of a 200 HP motor, the volage should 100 % or 98.9% would work. Could you advise me a basic philosophy one should follow while designing a power distribution system
Thanks once again.
Waross:
I have one more question for you.I would appreciate a word.
I beleive I am looking for a thumb rule. What criterion should I consider for the voltage at a bus.
If the voltage is 99.5 % or 98%. What value is ok.
Or It depends on, if I get the right voltage at the end of my load, which might be a single phase electric panel.
What is the tolerance, that I can allow as a thumb rule may be for example a motor. In case of a 200 HP motor, the volage should 100 % or 98.9% would work. Could you advise me a basic philosophy one should follow while designing a power distribution system
Thanks once again.
wareagle
Electrical
- Jul 28, 2003
- 391
MineGuy
I know this may be confusing to you but you need to get some assistance from your work group and have them explain
how the %Z is calculated. You as the engineer must decide how much voltage drop you can permit. We don't know. We do not have all of the parameters to make a decision. I think that the voltage level at the 6.6 kv bus should be 100%. Then you calculate the VD to various points on your system.
The reason for the regulator is to maintain the same voltage level at the bus when the load is low as well as when it is at its maximum. The allowable voltage drop depends on the load. Example - a 480 volt motor will run
properly at 460 volts. Another piece of equipment may not.
Computers and like equipment do not last long when the voltage exceeds the design rating. Get someone to show you
how the %Z is caculated. Maybe then it will become clear.
I know this may be confusing to you but you need to get some assistance from your work group and have them explain
how the %Z is calculated. You as the engineer must decide how much voltage drop you can permit. We don't know. We do not have all of the parameters to make a decision. I think that the voltage level at the 6.6 kv bus should be 100%. Then you calculate the VD to various points on your system.
The reason for the regulator is to maintain the same voltage level at the bus when the load is low as well as when it is at its maximum. The allowable voltage drop depends on the load. Example - a 480 volt motor will run
properly at 460 volts. Another piece of equipment may not.
Computers and like equipment do not last long when the voltage exceeds the design rating. Get someone to show you
how the %Z is caculated. Maybe then it will become clear.
Lets go back to %Z and %Voltage Regulation for a minute (or at least to my understanding of it).
%Z is numerically equal to the percent of primary voltage that you would have to apply to a transformer to get its full rated current to flow when the secondary terminals are shorted together. Maximum fault current is full load amps divided by %Z (assuming infinite bus for the source).
Voltage Regulation of a transformer varies with the power factor of the load.
I just looked at bids that we received for a power transformer last year and selected one as an example. From the 'Performance Data' section of this quote for a 30 MVA transformer: Impedance at 30 MVA = 7.5% H to X, %Regulation at 30 MVA = 0.51 at 1.0pf, 4.85 at 0.80pf, no load loss = 25.4 kW, Load Loss at 75C = 68.8 kW.
The transformer impedance has a much larger X component than R component. To determine voltage drop in the transformer, you have to look at the resistive and reactive components of the load current and apply them to their respective impedance components.
If a 15 MVA unit actually had a voltage drop of 8% as suggested in one of the prior posts, the losses would be insanely high.
%Z is numerically equal to the percent of primary voltage that you would have to apply to a transformer to get its full rated current to flow when the secondary terminals are shorted together. Maximum fault current is full load amps divided by %Z (assuming infinite bus for the source).
Voltage Regulation of a transformer varies with the power factor of the load.
I just looked at bids that we received for a power transformer last year and selected one as an example. From the 'Performance Data' section of this quote for a 30 MVA transformer: Impedance at 30 MVA = 7.5% H to X, %Regulation at 30 MVA = 0.51 at 1.0pf, 4.85 at 0.80pf, no load loss = 25.4 kW, Load Loss at 75C = 68.8 kW.
The transformer impedance has a much larger X component than R component. To determine voltage drop in the transformer, you have to look at the resistive and reactive components of the load current and apply them to their respective impedance components.
If a 15 MVA unit actually had a voltage drop of 8% as suggested in one of the prior posts, the losses would be insanely high.
There has been some confusion in the comments about voltage drop. The percent voltage drop across the transformer at full load is not equal to the transformer percent impedance except in the specific case where the load phase angle equals the transformer impedance angle. That is, if acos(pf) = atan(X/R) and there is no phase angle shift in the voltage.
Regulation is a specific voltage drop, defined as the change in output (secondary) voltage which ocurs when the load is reduced from rated kVA to zero, with the applied (primary) voltage maintained constant and is
expressed as a percentage of the full-load secondary voltage.
Regulation (in pu) = 1 + R·cos(ø) + X·sin(ø) - sqrt((R·cos(ø) + X·sin(ø))² + 1 - X² - R²)
where R and X are in pu and cos(ø) = power factor
Regulation is a specific voltage drop, defined as the change in output (secondary) voltage which ocurs when the load is reduced from rated kVA to zero, with the applied (primary) voltage maintained constant and is
expressed as a percentage of the full-load secondary voltage.
Regulation (in pu) = 1 + R·cos(ø) + X·sin(ø) - sqrt((R·cos(ø) + X·sin(ø))² + 1 - X² - R²)
where R and X are in pu and cos(ø) = power factor
Skogsgurra
Electrical
That is absolutely correct. But as long as the OP hasn't grasped the difference between % tap setting and % voltage drop. I do not think that that will help him. He has to get the idea first.
Gunnar Englund
Gunnar Englund
- Thread starter
- #29
Thanks everybody for this exhaustive explanation.
Here is the thing I was missing. When a transformer size increases, the reactance decreases and the resistance increases. Now, if we have same load but a higher rating transformer as the reactance will decrease but the resistance is still less than reactance, so eventually the voltage regulation will increase.
But I agree %z impedance has nothing to with the voltage drop. I hope I am correct this time.
Thanks Everybody
Here is the thing I was missing. When a transformer size increases, the reactance decreases and the resistance increases. Now, if we have same load but a higher rating transformer as the reactance will decrease but the resistance is still less than reactance, so eventually the voltage regulation will increase.
But I agree %z impedance has nothing to with the voltage drop. I hope I am correct this time.
Thanks Everybody
Typically, percent reactance will be constant or increase when the transformer size increases, and percent resistance will decrease.
Here we might be confusing regulation with voltage drop. By definition, regulation of a transformer is at rated load. Regulation in distribution systems is more generally defined as the percent voltage drop at whatever the load is. The voltage drop for a given load will be less with a larger transformer because the load is a lower percentage of rated load. To look at it another way, the ohmic impedance of a large transformer is less than the ohmic impedance of a smaller transformer with the same percent impedance.
Well, not quite right. Voltage drop is the difference between the magnitude of the sending end voltage and the magnitude of the receiving end voltage. For a transformer, you have to adjust for the transformer ratio, but the sending end voltage is the primary voltage and the receiving end voltage is the secondary voltage. The whole thing gets confused because of phase angles. The sending end voltage equals the receiving end voltage plus the impedance times the current. The voltages, the impedance, and the current all are complex values, so you have to do the math vectorially and then compare the magnitudes.
I hope this doesn't totally confuse you, but I can't think of a way to simplify it.
wareagle
Electrical
- Jul 28, 2003
- 391
LG
You are absolutely correct. I took a simpler method.
Transformer 15 mva at 220 kv FLA = 39 line amps.
Since %Z = 8.5% Primary voltage = 18700 volts with Sec amps = FLA primary amps = FLA
With the primary at 18700 the primary line amps = 39 amps
Phase amps = 39 amps/1.73 = 22.5 amps per phase
Phase Impedance = 220000 volts/22.5 amps =831 ohms per phase
At full load voltage = 220000 volts line amps = 39 amps
Phase amps = 22.5 amps.
Voltage loss in transformer = 22.5 x 831 ohms = 18700 volts.
Is this a proper estimate of the transformer voltage loss?
wareagle,
I couldn't quite follow your reasoning, but you are correct, the voltage loss in the transformer when shorting the secondary and running full load through the transformer is 18700 volts (18700 volts on the primary, zero volts on the secondary). This is how the impedance is measured.
The impedance in ohms can be determined by multiplying the pu impedance times the base impedance which is kV²/MVA = 3227 ohms. Transformer impedance in your example would be 3227·0.085 = 274 ohms.
The current when the secondary is shorted has the same phase angle as the transformer impedance. That's because the transformer is the only impedance in the circuit. This is not generally the case when calculating voltage drops under load. The load current phase angle is usually not the same as the transformer impedance phase angle.
I couldn't quite follow your reasoning, but you are correct, the voltage loss in the transformer when shorting the secondary and running full load through the transformer is 18700 volts (18700 volts on the primary, zero volts on the secondary). This is how the impedance is measured.
The impedance in ohms can be determined by multiplying the pu impedance times the base impedance which is kV²/MVA = 3227 ohms. Transformer impedance in your example would be 3227·0.085 = 274 ohms.
The current when the secondary is shorted has the same phase angle as the transformer impedance. That's because the transformer is the only impedance in the circuit. This is not generally the case when calculating voltage drops under load. The load current phase angle is usually not the same as the transformer impedance phase angle.
wareagle
Electrical
- Jul 28, 2003
- 391
Jgrist
I looked at this from the transmission side as a delta connected load. I made a mistake in the info.I would have thought this method would have given an impedance closer to you answer. I made a typo and should have written:
Transformer 15 mva at 220 kv FLA = 39 line amps.
Since %Z = 8.5% Primary voltage = 18700 volts with Sec amps = FLA and primary amps = FLA
With the primary at 18700 the primary line amps = 39 amps.
Phase amps = 39 amps/1.73 = 22.5 amps per phase delta
Phase Impedance = 18700 volts/22.5 amps =831 ohms per phase
At full load voltage = 220000 volts line amps = 39 amps
Phase amps = 22.5 amps.
Voltage loss in transformer = 22.5 x 831 ohms = 18700 volts.
I looked at this from the transmission side as a delta connected load. I made a mistake in the info.I would have thought this method would have given an impedance closer to you answer. I made a typo and should have written:
Transformer 15 mva at 220 kv FLA = 39 line amps.
Since %Z = 8.5% Primary voltage = 18700 volts with Sec amps = FLA and primary amps = FLA
With the primary at 18700 the primary line amps = 39 amps.
Phase amps = 39 amps/1.73 = 22.5 amps per phase delta
Phase Impedance = 18700 volts/22.5 amps =831 ohms per phase
At full load voltage = 220000 volts line amps = 39 amps
Phase amps = 22.5 amps.
Voltage loss in transformer = 22.5 x 831 ohms = 18700 volts.
wareagle:
You are forgetting phase angles or the power factor. Firstly, you assumed simple arithmetic sum and it is not.
Power factor during the test of %Z measurement is close to zero or very low (secondary is short circuited).
The load power factor is closer to 0.8.
%Z is tested impedance and includes R and X. You really need to know X/R ratio to add the voltage drops correctly.
While voltage regulation can be calculated provided you have all the data, there is a reason published voltage regulation is also a tested value, tested at known load and power factor. The definition of VR is (Vnl-Vfl)/Vfl in per unit. Multiply by 100 for %.
Assume X/R of say 12 (or some reasonable no), calculate X and R, and load pf of 0.8 and plug that in that formula, see what you get. If you get more than 4% I will be surprised.
In fact if VR is more than 4%, the power transformer should be ditched.
You are forgetting phase angles or the power factor. Firstly, you assumed simple arithmetic sum and it is not.
Power factor during the test of %Z measurement is close to zero or very low (secondary is short circuited).
The load power factor is closer to 0.8.
%Z is tested impedance and includes R and X. You really need to know X/R ratio to add the voltage drops correctly.
While voltage regulation can be calculated provided you have all the data, there is a reason published voltage regulation is also a tested value, tested at known load and power factor. The definition of VR is (Vnl-Vfl)/Vfl in per unit. Multiply by 100 for %.
Assume X/R of say 12 (or some reasonable no), calculate X and R, and load pf of 0.8 and plug that in that formula, see what you get. If you get more than 4% I will be surprised.
In fact if VR is more than 4%, the power transformer should be ditched.
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