gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
080709-1952 EST
clayton:
mivey has given you a good description.
I want to added to this with a slightly different description, but long and hopefully it gives you an intuitive presentation from basics.
First, consider a resistor. The average steady-state power dissipated in the resistor is V*A for DC or RMS AC values. And in either case VA = watts. This results from the current in the resistor being in phase with the voltage for AC excitation. I won't go into the derviation, but accept it as fact. Note: a true RMS meter will read the same value for DC or AC for voltage or current for a given value of resistance and power dissipation.
A Fluke 87 meter although labeled true RMS really only reads the RMS value of the AC component of a complex signal.
If you view a 3 phase system in its Y configuration, then, except for evaluation of the composite neutral current and any effect it has, you can consider the circuit as three independent single phase circuits.
This means the current in each line is the current in the load of its line-to-neutral phase. With a resistive load the line current times the line-to-neutral voltage is the power to the load and the VA rating for that phase. Since the loads in each of the phases are independent the total power is the sum of the three individual loads. If the loads contain some reactive components, then the VA value will never be less than the power dissipated.
The Y connection is useful for analysis because for a resistive load it is easy and obvious to calculate power from the current in one line and the voltage from that line to neutral.
Now change the load from Y to delta, but have the same total power dissipation. Because the line-to-line voltage is different than the line-to-neutral the resistors must be different for the same power dissipation. This also means the current in a phase (the load) is different than the line current. Using a vector diagram you can easily calculate the line-to-line voltage from the line-to-neutral voltage for a balanced system. Vline-to-line = Vline-to-neutral * (sq-root of 3).
This is obtained from a right triangle with the short triangle leg equal to the line-to-neutral voltage and the long leg equal to 1/2 the line-to-line voltage. The angle adjacent to the short leg is 60 degrees (1/2 of 1/3 of 360). The sin 60 degrees = 1/2 sq-root of 3 (0.8660254). Two times this gets us to the sq-root of 3 or 1.732 .
Thus, in a balanced delta configuration, using the equality of power, the phase current is line current/1.732 . In a resistive load the power dissipated in the total load is 3 times the power dissipated in one phase. In a resistive load P = V*A. = Vline-to-line * Iline/1.732 . Total power is 3*P = 3*Vline-to-line*Iline/1.732 = 1.732*Vline-to-line*Iline .
As mentioned in a previous post the input VA to a transformer equals the output VA assuming no losses. For purposes of understanding consider VA the same as power. From the law of the conservation of energy the requirement is that input equals output. So power doesn't change between input and output, but voltages and currents may change depending upon the turns ratio.
Transformer ratings are based on temperature rise and the materials used and efficiency. A big part of transformer losses at full load are related to I^2*R losses in the transformer windings. Thus, currents in the transformer windings are the determining factor on transformer rating and not transferred power. That is why transformers are rated in VA rather than power transfer. Theoretically you could transfer 0 power to a load but be running the transformer at its full rating.
If you can develop an intuitive concept of how something works and with a knowledge of some basic principles, then you may be able to solve some problems on your own without outside reference.
You can try to remember the sq-root of 3 by correlating this with George Washington's birth year 1732. It does not change the year, but Washington was born under the Julian calendar, and Columbus discovered America under this calendar. We operate under the Gregorian calendar.
From www.dictionary.com
In particular http://dictionary.reference.com/browse/gregorian
At the transition there was a 10 day shift. Thus, relative to the autumnal equinox Columbus did not first see America on 12 October under our calendar. His 12 Oct 1492 in his log was under the Julian calendar. Thus, more like 2 October under our calendar.
.
clayton:
mivey has given you a good description.
I want to added to this with a slightly different description, but long and hopefully it gives you an intuitive presentation from basics.
First, consider a resistor. The average steady-state power dissipated in the resistor is V*A for DC or RMS AC values. And in either case VA = watts. This results from the current in the resistor being in phase with the voltage for AC excitation. I won't go into the derviation, but accept it as fact. Note: a true RMS meter will read the same value for DC or AC for voltage or current for a given value of resistance and power dissipation.
A Fluke 87 meter although labeled true RMS really only reads the RMS value of the AC component of a complex signal.
If you view a 3 phase system in its Y configuration, then, except for evaluation of the composite neutral current and any effect it has, you can consider the circuit as three independent single phase circuits.
This means the current in each line is the current in the load of its line-to-neutral phase. With a resistive load the line current times the line-to-neutral voltage is the power to the load and the VA rating for that phase. Since the loads in each of the phases are independent the total power is the sum of the three individual loads. If the loads contain some reactive components, then the VA value will never be less than the power dissipated.
The Y connection is useful for analysis because for a resistive load it is easy and obvious to calculate power from the current in one line and the voltage from that line to neutral.
Now change the load from Y to delta, but have the same total power dissipation. Because the line-to-line voltage is different than the line-to-neutral the resistors must be different for the same power dissipation. This also means the current in a phase (the load) is different than the line current. Using a vector diagram you can easily calculate the line-to-line voltage from the line-to-neutral voltage for a balanced system. Vline-to-line = Vline-to-neutral * (sq-root of 3).
This is obtained from a right triangle with the short triangle leg equal to the line-to-neutral voltage and the long leg equal to 1/2 the line-to-line voltage. The angle adjacent to the short leg is 60 degrees (1/2 of 1/3 of 360). The sin 60 degrees = 1/2 sq-root of 3 (0.8660254). Two times this gets us to the sq-root of 3 or 1.732 .
Thus, in a balanced delta configuration, using the equality of power, the phase current is line current/1.732 . In a resistive load the power dissipated in the total load is 3 times the power dissipated in one phase. In a resistive load P = V*A. = Vline-to-line * Iline/1.732 . Total power is 3*P = 3*Vline-to-line*Iline/1.732 = 1.732*Vline-to-line*Iline .
As mentioned in a previous post the input VA to a transformer equals the output VA assuming no losses. For purposes of understanding consider VA the same as power. From the law of the conservation of energy the requirement is that input equals output. So power doesn't change between input and output, but voltages and currents may change depending upon the turns ratio.
Transformer ratings are based on temperature rise and the materials used and efficiency. A big part of transformer losses at full load are related to I^2*R losses in the transformer windings. Thus, currents in the transformer windings are the determining factor on transformer rating and not transferred power. That is why transformers are rated in VA rather than power transfer. Theoretically you could transfer 0 power to a load but be running the transformer at its full rating.
If you can develop an intuitive concept of how something works and with a knowledge of some basic principles, then you may be able to solve some problems on your own without outside reference.
You can try to remember the sq-root of 3 by correlating this with George Washington's birth year 1732. It does not change the year, but Washington was born under the Julian calendar, and Columbus discovered America under this calendar. We operate under the Gregorian calendar.
From www.dictionary.com
In particular http://dictionary.reference.com/browse/gregorian
Further from http://dictionary.reference.com/browse/gregorian calendar
This is not quite a complete definition because it fails to say that all other years divisible by 4 are leap years. But further up in one of the other definitions this is covered.
At the transition there was a 10 day shift. Thus, relative to the autumnal equinox Columbus did not first see America on 12 October under our calendar. His 12 Oct 1492 in his log was under the Julian calendar. Thus, more like 2 October under our calendar.
.