Transformer KWH

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dnbob

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Rochester, MN
I have an old (1972) 150 KVA 480D to 208Y120 3 phase transformer.

The primary current idling with no secondary load is Phase A 3 amps, Phase B 3 Amps, Phase C 4 amps.

To find out what this is costing to run, with no load, is this the correct formula

Phase Amps (3+3+4) = 10 amps * 480 * 1.73 = 8,304 watts/1000 - 8.304 KW * 24 hours per day * $.11/KWH = 8.304*24*.11 = $21.92 in energy use per day?
 
You have the formula wrong. If you are measuring the line amps, the power will be the sum of the line amps times 480/sqrt(3). Not sum times 480 x sqrt(3).
Unless I screwed up as is often the case...
The phase current going to a single line-to-line load phase will be the line current times sqrt(3)/2.
Then you would also need to take into account the power factor, which for an unloaded transformer may be closer to 0 than to 1.
You need to measure or accurately estimate the PF.

Tapatalk!
 
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dnbob said:
I have an old (1972) 150 KVA 480D to 208Y120 3 phase transformer.

The primary current idling with no secondary load is Phase A 3 amps, Phase B 3 Amps, Phase C 4 amps.

To find out what this is costing to run, with no load, is this the correct formula

Phase Amps (3+3+4) = 10 amps * 480 * 1.73 = 8,304 watts/1000 - 8.304 KW * 24 hours per day * $.11/KWH = 8.304*24*.11 = $21.92 in energy use per day?
Click to expand...

If the PF is 1, then your energy usage would appear to be about 2800 watts. About $7.30 per day.
 
petersonra said:
If the PF is 1, then your energy usage would appear to be about 2800 watts. About $7.30 per day.
Click to expand...
On no load the current is magnetising current and iron losses.
PF will be very much lower than 1.0 .
I usually work on 0.7% of the rated kVA as the off load kW loss.
 
dnbob said:
So I should not add up the amps per phase? I seem to be about 3 times higher than what you are figuring.
Click to expand...
,

Bob
To do it your way, you could average the 3 different line amps to come up with 3 phase current, and then use it in your equation.
The reason the other method is more accurate is that it looks at 3 separate line to neutral, 277 volt, circuits and calculates the va in each. Then add them together and you'll get total va.
The reason va equations are (amps*480v*sqrt3) is that (amps*277v*3)=(amps*480v*sqrt3). Once I understood that, calculating 3 phase power was no longer a mystery.
Jim
 
jtester said:
,

Bob
To do it your way, you could average the 3 different line amps to come up with 3 phase current, and then use it in your equation.
The reason the other method is more accurate is that it looks at 3 separate line to neutral, 277 volt, circuits and calculates the va in each. Then add them together and you'll get total va.
The reason va equations are (amps*480v*sqrt3) is that (amps*277v*3)=(amps*480v*sqrt3). Once I understood that, calculating 3 phase power was no longer a mystery.
Jim
Click to expand...
From that, you can calculate kVA. Not power. You pay for energy which is power time time - kWh. If you don't know the kW you cant calculate energy.
 
jtester said:
,

Bob
To do it your way, you could average the 3 different line amps to come up with 3 phase current, and then use it in your equation.
The reason the other method is more accurate is that it looks at 3 separate line to neutral, 277 volt, circuits and calculates the va in each. Then add them together and you'll get total va.
The reason va equations are (amps*480v*sqrt3) is that (amps*277v*3)=(amps*480v*sqrt3). Once I understood that, calculating 3 phase power was no longer a mystery.
Jim
Click to expand...
The only potential problem with your method of understanding the relationship (and one that bit me the first time) is that if you have an unbalanced load (say only one resistive line to line load instead of three), the the line current at 277V will NOT have a PF of 1, even with a resistive load. And the line currents on the two ends will be equal to the phase current, since there can be no vector cancellation.
Think about it for awhile.....
 
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dnbob said:
So I should not add up the amps per phase? I seem to be about 3 times higher than what you are figuring.
Click to expand...
No you don?t add the ?amps per phase.? In fact, you would benefit by forever forgetting that phrase. It has no meaning, and it tends to confuse. Let me explain it this way:

At my house, I can connect garden hoses in the front, the side, and the back. Suppose I run sprinklers from all three locations. Suppose I can manage to measure the flow rates at each location. I can then add the gallons per minute from the front hose, the gpm from the side hose, and the gpm from the back hose, and I will get the total flow of the three hoses, still in units of gpm.

But three phase currents do not work the same way. The water flowing through the front hose gets thrown onto the front garden, and does not make its way from there to either of the other two hoses. However, the current leaving the transformer on Phase A will return to the transformer on Phase B and Phase C. It is the same current flowing in all three phases. Each of the phases will reach its peak value at different times. At the moment that Phase A is at its peak value in a positive current direction, each of Phase B and Phase C will be at a lower value of current, and they will both be in a negative current direction. The values you measure are really a type of average.

If you have values of 3, 3, and 4 amps on the three phases, I would average them to get a value of 3.3 amps, and I would use that in the calculation. Here again, 3.3 amps is not a ?total? value.


 
charlie b said:

If you have values of 3, 3, and 4 amps on the three phases, I would average them to get a value of 3.3 amps, and I would use that in the calculation. Here again, 3.3 amps is not a ?total? value.
Click to expand...

But, by some fortuitous mathematical circumstance :angel:, the average current which you plug into the balanced three phase formula, is equal to the sum of the currents divided by three. Which allows you to work just as accurately with the sum of the currents and a slightly different three phase formula.
Since that other three phase formula is less likely to be remembered when you need it, the first approach works better in practice. :)
 
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