Per-Unit System
- Thread starter sdilucca
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ron
Senior Member
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- New York, 40.7514,-73.9925
Re: Per-Unit System
The per unit method of calculation is similar to combining fractions with a common denominator. It allows you to convert back and forth between voltages and conditions with ease.
For example, a 500kVA xfmr with a per unit impedance of .0575 lets me determine infinite bus fault current at any voltage I want.
The per unit method of calculation is similar to combining fractions with a common denominator. It allows you to convert back and forth between voltages and conditions with ease.
For example, a 500kVA xfmr with a per unit impedance of .0575 lets me determine infinite bus fault current at any voltage I want.
- Location
- Lockport, IL
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- Semi-Retired Electrical Engineer
Re: Per-Unit System
To give an example of an analysis that uses the per unit system would take up too much space. All I will add to Ron's answer is that the per unit system is a tool that is used in the arena of "Power Systems Analysis." If you have a complicated network of generators, transmission lines, distribution transformers, and loads, the per unit system allows us to reduce the network to a single-line model. If I were to decide that "200 MVA" was going to be represented by "one unit" of power, then a 500 MVA generator would be 2.5 units of power. This simplifies the arithmetic.
The most important feature of the per unit system, and this is hard to explain in words only ? I don't know how to publish images ? is the way it handles voltages. Let me assign "480 volts" as being "one unit of voltage." Let me look at a transformer that has a 480 volt primary (i.e., a per unit voltage of 1.0) and a secondary (line to line) voltage of 208 volts. What is the "per unit voltage" of the secondary? Well the rules for using the per unit system require me to convert the secondary voltage to per unit using the same ratio as the voltage transformation. That is, the secondary voltage is also "one unit on voltage." Thus, when we draw the one-line model, the transformer disappears. All that is left is voltage sources and impedances, and all we need to solve the problem is Ohm's Law.
To give an example of an analysis that uses the per unit system would take up too much space. All I will add to Ron's answer is that the per unit system is a tool that is used in the arena of "Power Systems Analysis." If you have a complicated network of generators, transmission lines, distribution transformers, and loads, the per unit system allows us to reduce the network to a single-line model. If I were to decide that "200 MVA" was going to be represented by "one unit" of power, then a 500 MVA generator would be 2.5 units of power. This simplifies the arithmetic.
The most important feature of the per unit system, and this is hard to explain in words only ? I don't know how to publish images ? is the way it handles voltages. Let me assign "480 volts" as being "one unit of voltage." Let me look at a transformer that has a 480 volt primary (i.e., a per unit voltage of 1.0) and a secondary (line to line) voltage of 208 volts. What is the "per unit voltage" of the secondary? Well the rules for using the per unit system require me to convert the secondary voltage to per unit using the same ratio as the voltage transformation. That is, the secondary voltage is also "one unit on voltage." Thus, when we draw the one-line model, the transformer disappears. All that is left is voltage sources and impedances, and all we need to solve the problem is Ohm's Law.
rcwilson
Senior Member
- Location
- Redmond, WA
Re: Per-Unit System
Calculating in per unit makes the arithmetic a lot easier and less prone to error. With per unit, we think of everything as percentages of a common base.
We don't have to keep converting the voltages and currents at each transformer. In per unit 100% power always draws 100% amps at 100% volts, whether the voltage is 13.8 kV, 480 V or 120V.
If I select 1000 kVA as the base, then 1.0 pu power = 100% power = 1000 kVA = (1.0 pu Volts x 1.0 pu amps x 1.732). Each voltage of the power system will have its own per unit (100%) voltage, amps and impedance, but all voltage levels will use the same kVA base since power doesn?t change when it goes through a transformer.. Once I select the kVA base, all of the rest of the base values at each voltage are calculated using ohms law.
If we are have a 250 kVA load it is 0.25 per unit at 13.8 kV and 120V. It's current will be 0.25 pu at all of the voltages. So when I want to calculate the voltage drop on motor starting for the whole system, I don't have to calculate the ohms and amps at 13.8kv, ohms and amps at 480 V and make sure I do the resistance conversions through the transformer correctly and keep track of which resistance is at which voltage. I just add the per unit impedances at all voltage levels, divide into the per unit voltage at the source and I have the per unit current through the complete system during motor starting. Multiply that per unit current times the motors per unit impedance and I have the voltage during the start. I can then give my voltage drop answer in per unit "Motor voltage at starting = 0.85 pu" or in per cent "15% voltage drop" or actual voltages "0.85 x 480V = 408V".
Most equipment impedance values are expressed in per unit, such as a transformer?s impedance listed as 5.75% (0.0575 per unit). I could convert that impedance to ohms at 13.8 kV or ohms at 480 V, but it is a lot easier to leave it in per unit which is the same value on both sides of the transformer, and then convert to actual values when I am done.
One benefit of per unit is working in percentages and not actual values. If I tell you we have a 100 V drop during motor starting you don't know if that is good or bad until I tell you the voltage. (100 V is only 0.7% at 13.8 kV but 20.8% at 480V). But if I say we have a 0.2 per unit or 20.8% drop, you know the result.
I hope this helps. To fully understand per unit, check out some books on power system analysis and try some sample problems. The IEEE Red Book and books by Lew Blackburn are a good start.
Calculating in per unit makes the arithmetic a lot easier and less prone to error. With per unit, we think of everything as percentages of a common base.
We don't have to keep converting the voltages and currents at each transformer. In per unit 100% power always draws 100% amps at 100% volts, whether the voltage is 13.8 kV, 480 V or 120V.
If I select 1000 kVA as the base, then 1.0 pu power = 100% power = 1000 kVA = (1.0 pu Volts x 1.0 pu amps x 1.732). Each voltage of the power system will have its own per unit (100%) voltage, amps and impedance, but all voltage levels will use the same kVA base since power doesn?t change when it goes through a transformer.. Once I select the kVA base, all of the rest of the base values at each voltage are calculated using ohms law.
If we are have a 250 kVA load it is 0.25 per unit at 13.8 kV and 120V. It's current will be 0.25 pu at all of the voltages. So when I want to calculate the voltage drop on motor starting for the whole system, I don't have to calculate the ohms and amps at 13.8kv, ohms and amps at 480 V and make sure I do the resistance conversions through the transformer correctly and keep track of which resistance is at which voltage. I just add the per unit impedances at all voltage levels, divide into the per unit voltage at the source and I have the per unit current through the complete system during motor starting. Multiply that per unit current times the motors per unit impedance and I have the voltage during the start. I can then give my voltage drop answer in per unit "Motor voltage at starting = 0.85 pu" or in per cent "15% voltage drop" or actual voltages "0.85 x 480V = 408V".
Most equipment impedance values are expressed in per unit, such as a transformer?s impedance listed as 5.75% (0.0575 per unit). I could convert that impedance to ohms at 13.8 kV or ohms at 480 V, but it is a lot easier to leave it in per unit which is the same value on both sides of the transformer, and then convert to actual values when I am done.
One benefit of per unit is working in percentages and not actual values. If I tell you we have a 100 V drop during motor starting you don't know if that is good or bad until I tell you the voltage. (100 V is only 0.7% at 13.8 kV but 20.8% at 480V). But if I say we have a 0.2 per unit or 20.8% drop, you know the result.
I hope this helps. To fully understand per unit, check out some books on power system analysis and try some sample problems. The IEEE Red Book and books by Lew Blackburn are a good start.
rcwilson
Senior Member
- Location
- Redmond, WA
Re: Per-Unit System
I found a link to an older GE publication that has some good data on per unit method and fault calculations. Check the Appendices
http://www.geindustrial.com/products/applications/GET-3550F.pdf
I found a link to an older GE publication that has some good data on per unit method and fault calculations. Check the Appendices
http://www.geindustrial.com/products/applications/GET-3550F.pdf
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