2 phase
- Thread starter ddepuy
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- Location
- New Jersey
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- Journeyman Electrician
Where are you using a 2 phase system?
- Location
- New Jersey
- Occupation
- Journeyman Electrician
rattus said:
That's what I was wondering.
- Location
- Lockport, IL
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- Retired Electrical Engineer
For reasons that I cannot begin to explain, I have in mind an answer that I cannot begin to defend. My guess, and it is no more than a guess, is that you take voltage times current times the square root of two. But none of the old college and grad school text books that I keep at my work desk has an answer to this question.
wasasparky
Senior Member
Shooting from the hip, would it not just be 2xVxI?
The coils do not have an interdependence like 3phase, so I don't envision a square root of anything.
Commence flaming!
The coils do not have an interdependence like 3phase, so I don't envision a square root of anything.
Commence flaming!
- Location
- Wisconsin
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- PE (Retired) - Power Systems
From my Ugly's book.
single phase: KW=(E*I*PF)/1000
two phase: KW = (E*I*PF*2)/1000
three phase: KW = (E*I*PF*1.73)/1000
single phase: KW=(E*I*PF)/1000
two phase: KW = (E*I*PF*2)/1000
three phase: KW = (E*I*PF*1.73)/1000
zgozvrm
Member
- Location
- Northern California
Square root of 2
Square root of 2
Volts x Amps X Sqrt(2) is correct!
It is more of a coincidence that the number that is being "square-rooted" matches the number of phases. If you had 2 phase power that had something other than 90 degree separation between phases, this would not be the case. It just so happens that 90 degrees gives us the most efficient use of 2 phase power.
If you look at a vector diagram of the voltage of a Scott-T transformer (which produces 2 voltages separated by 90 degrees), you basically have 2 legs of an isosceles triangle, the 3rd leg, being the resultant vector. Using the Pythagorean Theorem (you remember that, don't you?), you can determine the magnitude of the resulting vector (represented by the hypotenuse of the triangle). This works out to be equal to the square root of 2 times the magnitude of either of the other vectors (i.e. the square root of 2 times the voltage of one of the phases).
Therefore, the power is calculated by multiplying this resultant voltage (vector) by the current, giving you Volts x Amps X Sqrt(2).
Using a similar method, you can see why the square root of 3 is used for 3 phase power.
Hope this helps clarify things for you!
Square root of 2
Volts x Amps X Sqrt(2) is correct!
It is more of a coincidence that the number that is being "square-rooted" matches the number of phases. If you had 2 phase power that had something other than 90 degree separation between phases, this would not be the case. It just so happens that 90 degrees gives us the most efficient use of 2 phase power.
If you look at a vector diagram of the voltage of a Scott-T transformer (which produces 2 voltages separated by 90 degrees), you basically have 2 legs of an isosceles triangle, the 3rd leg, being the resultant vector. Using the Pythagorean Theorem (you remember that, don't you?), you can determine the magnitude of the resulting vector (represented by the hypotenuse of the triangle). This works out to be equal to the square root of 2 times the magnitude of either of the other vectors (i.e. the square root of 2 times the voltage of one of the phases).
Therefore, the power is calculated by multiplying this resultant voltage (vector) by the current, giving you Volts x Amps X Sqrt(2).
Using a similar method, you can see why the square root of 3 is used for 3 phase power.
Hope this helps clarify things for you!
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
zgozvrm said:
Do you have a reference?
According to my Standard Handbook for Electrical Engineers, McGraw-Hill, Tenth edition (1969) page 18-42 which discusses motor circuits:
Locked Rotor current = ((KVA/HP)*HP*1000)/(k*line volts)
where k=1 for single phase, k=2 for 2-phase, and k=1.73 for 3-phase
This agrees with the information in the Ugly's book.
Note, the square root of 2 is used in some 2-phase calculations.
If the 2-phase circuit is only 3-wire, the current in the common conductor will be 1.414 times the current in the other two wires.
zgozvrm said:
And if we had 6 phase power?:grin:
I've never seen a 2-phase system. However all of the pictures I've seen, 2ph is 4-wire - no common. I don't see how sqr(2) would enter into the calc. If we have someone on the forum who has actually seen, laid eyeballs, hands, tools, or meters on a 2ph system - please speak up.
(BTW - I've seen a 6ph alternator)
carl
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
Depending upon how you define your terms and the type of two phase system involved, you may or may not have a root 2 term involved.
If you have a two phase three wire system, with a single common conductor and two phases with 90 degree displacement, and you define voltage as the _phase to phase_ voltage, then power is V * A * sqrt(2) * PF.
If you instead define the V used in your equation as the _phase to common_ voltage, then the answer is V * A * 2 * PF.
You see the same duality in the common three phase calculation. When V is the line to line voltage, and all phases carry the same current, then power delivered is V(LtoL) * A * root(3) * PF. If you instead quoted the line to neutral voltage, then the power delivered is V(LtoN) * A * 3 * PF. The latter equation is much less commonly used, but it is the equivalent of the first equation. Remember that for a three phase wye system, V(LtoL) = root(3) * V(LtoN)
For a six phase system, you need to similarly define what voltage you are looking at, L-N or L-L. Additionally, there are two common six phase systems, one with 60 degree phase displacement (which for motors is the equivalent of common three phase with 120 degree displacement), the other with two 120 degree three phase sets, 30 degrees apart. The latter system has uneven phase displacement, and thus a poorly defined V(LtoL).
-Jon
If you have a two phase three wire system, with a single common conductor and two phases with 90 degree displacement, and you define voltage as the _phase to phase_ voltage, then power is V * A * sqrt(2) * PF.
If you instead define the V used in your equation as the _phase to common_ voltage, then the answer is V * A * 2 * PF.
You see the same duality in the common three phase calculation. When V is the line to line voltage, and all phases carry the same current, then power delivered is V(LtoL) * A * root(3) * PF. If you instead quoted the line to neutral voltage, then the power delivered is V(LtoN) * A * 3 * PF. The latter equation is much less commonly used, but it is the equivalent of the first equation. Remember that for a three phase wye system, V(LtoL) = root(3) * V(LtoN)
For a six phase system, you need to similarly define what voltage you are looking at, L-N or L-L. Additionally, there are two common six phase systems, one with 60 degree phase displacement (which for motors is the equivalent of common three phase with 120 degree displacement), the other with two 120 degree three phase sets, 30 degrees apart. The latter system has uneven phase displacement, and thus a poorly defined V(LtoL).
-Jon
2 phase power
2 phase power
2 phase power has 5 wires. Line to neutral is 120 volts and line to line is 240 volts. The 2 phase transformer is basically an "X" with the neutral out of the center and any phase end being 120 volt. Any 2 phase ends are 240 volt.
I hope this helps to clarify my question of converting 2 phase amps to watts
2 phase power
2 phase power has 5 wires. Line to neutral is 120 volts and line to line is 240 volts. The 2 phase transformer is basically an "X" with the neutral out of the center and any phase end being 120 volt. Any 2 phase ends are 240 volt.
I hope this helps to clarify my question of converting 2 phase amps to watts
2 phase power
2 phase power
The 2 phase power I am refering to is in the old section of phhiladelphia. It is the secondary that is sent to storefronts in the old section of town. It is 2 phase 4 wire with a neutral. The motors in these buildings are 2 phase if original and 3 phase useing a scott t transformer if new.
2 phase power
The 2 phase power I am refering to is in the old section of phhiladelphia. It is the secondary that is sent to storefronts in the old section of town. It is 2 phase 4 wire with a neutral. The motors in these buildings are 2 phase if original and 3 phase useing a scott t transformer if new.
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
coulter said:
I saw this in the literature on high phase order motors, and possibly also in something involving rectification systems.
In a conventional three phase motor, you have _six_ phase angles to work with, because you have two ways that a conductor can go down the slots. So you have the three supply phases and their 'mirrors'. The available conductor current phases (counting direction of the wire) is 0 (A) , 60 (C') , 120 (B), 180 (A'), 240 (C), 300 (B').
In the high phase order motor field, there is a desire to have more separate control of the slot current at the different phase angles. This requires additional inverter phases at different phase displacement. If you make a six phase system with a phase displacement of 60 degrees, then you are simply repeating what you already have, such a system would just give you phase angles 0 (A), 60 (B), 120 (C), 180 (D), 240 (E), 300 (F).
The other approach to a 6 phase system uses the 30 degree offset that I mentioned, giving 0 (A), 30 (B), 120 (C), 150 (D), 240 (E) and 270 (F). With this arrangement, the reverse direction conductors provide non-redundant phase angles.
-Jon
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
ddepuy said:
Again, you have to be quite careful with how you define your variables.
You have 5 conductors. Lets call them A, B, A', B', and N. In this system, the phase angle between A-A' and B-B' is 90 degrees. There are _three_ possible voltages that you could quote: the phase to neutral voltage, the phase to phase voltage, or the voltage across both legs of a single phase.
For example, the phase to neutral voltage could be 120V. If you measure from A to N you get 120V, similarly for B to N, A' to N, and B' to N. If you measure from A to A', you get 240V, similarly for B to B'. Finally, if you measure from A to B, you get 170V, similarly for B to A', A' to B', and B' to A.
If you use the voltage across a single phase (the 240V in the above example), then VA = V * A * 2 and W = V * A * 2 * PF
-Jon
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