mitch cochrell
Member
when calculating a three phase 240v equation do I divide by 240 or is there something I am missing? I know how to calculate Y systems but on the delta I am kind of confused.
240 volt 3 ph high leg
- Thread starter mitch cochrell
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Delta connection is 240 volts. No different from the wye system except the voltage is 208 volts. In the future you do not need to start a new post. Ask your question in your original posting.
Duplicate post.
mitch cochrell
Member
So if I have a load of 45 amps, how do I convert to volt amps? Do a multiply by 240 or 240 times the square root of three? If I could see an example it would help a lot.
- Location
- Wisconsin
- Occupation
- PE (Retired) - Power Systems
It all depends on how the load is connected:
Connections:
A-N or C-N use Amps*120V
A-B or B-C or C-A use Amps*240V
A-B-C use Amps*240V*Sqrt(3)
There have been many posts on this, but the last possibility is:
B-N use Amps*208V
gar
Senior Member
- Location
- Ann Arbor, Michigan
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- EE
100204-1009 EST
mitch cochrell:
Your question and the answer depends upon what purpose your VA rating has.
To a large extent the purpose of a VA rating is to provide a useful tool for determining the temperature rise in a device. In many devices the RMS input current is a measure of the temperature rise in that device. This results from I^2*R losses.
Suppose I consider a single single phase transformer by itself. Under maximum load most of the internal losses are resistive, and temperature rise is approximately proportion to internal dissipated power. For a given physical size transformer the internal temperature rise is approximately the same whether the transformer winding is 120 or 480 if the current in the transformer is based on the VA rating. So a 120 V @ 1 A coil would be equivalent to a 480 @ 0.25 A coil. See if you can reason thru this idea.
Now switch to a three phase system.
The power into a load (work), some label it real power (I see no need for the the real modifier), is never greater than the measured volts * amperes to that load. But the amperes to the load is what is important in terms of transformer heating or supply line losses.
Now consider a balanced Y connected resistive heater connected to a Y source with wires A B C and N. Each leg (hot line) has the same current magnitude, leg to N voltage, and power in each 1/3 of the load. The line current is in phase with the line to N voltage. Thus, the VA rating of each leg Ileg*Vleg to neutral and this equals the power to the load in each leg. Also the power and VA of the total system is 3 times that of one leg.
Note the relationship of Line to Line voltage to Line to Neutral = sq-root of 3 (1.732) for a balanced 3 phase system. 1732 is the year of George Washington's birth.
Next change the resistive load from Y to delta, but keep the same power dissipation in the loads. Intuitively you should be able to see that the hot supply line currents are unchanged and still in phase with the line to neutral voltage. Line currents are obviously out of phase with the Line to Line voltages.
Can you fill in some of the gaps from here on how and why to use certain constants for calculations?
.
mitch cochrell:
Your question and the answer depends upon what purpose your VA rating has.
To a large extent the purpose of a VA rating is to provide a useful tool for determining the temperature rise in a device. In many devices the RMS input current is a measure of the temperature rise in that device. This results from I^2*R losses.
Suppose I consider a single single phase transformer by itself. Under maximum load most of the internal losses are resistive, and temperature rise is approximately proportion to internal dissipated power. For a given physical size transformer the internal temperature rise is approximately the same whether the transformer winding is 120 or 480 if the current in the transformer is based on the VA rating. So a 120 V @ 1 A coil would be equivalent to a 480 @ 0.25 A coil. See if you can reason thru this idea.
Now switch to a three phase system.
The power into a load (work), some label it real power (I see no need for the the real modifier), is never greater than the measured volts * amperes to that load. But the amperes to the load is what is important in terms of transformer heating or supply line losses.
Now consider a balanced Y connected resistive heater connected to a Y source with wires A B C and N. Each leg (hot line) has the same current magnitude, leg to N voltage, and power in each 1/3 of the load. The line current is in phase with the line to N voltage. Thus, the VA rating of each leg Ileg*Vleg to neutral and this equals the power to the load in each leg. Also the power and VA of the total system is 3 times that of one leg.
Note the relationship of Line to Line voltage to Line to Neutral = sq-root of 3 (1.732) for a balanced 3 phase system. 1732 is the year of George Washington's birth.
Next change the resistive load from Y to delta, but keep the same power dissipation in the loads. Intuitively you should be able to see that the hot supply line currents are unchanged and still in phase with the line to neutral voltage. Line currents are obviously out of phase with the Line to Line voltages.
Can you fill in some of the gaps from here on how and why to use certain constants for calculations?
.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
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- Electrical Contractor
Interestingly, the same basic formulae apply with either system. In a Y, you use 1.732 for the voltage, and in a Delta, you apply it to the current, so in the end, you get volt-amps the same way.
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