Because I dont get it...
- Thread starter laketime
- Start date
- Status
Three-phase transformers equal 3 one-phase transformers combined... 25kVA each (75kVA ÷ 3).
What does the kVA stand for?
480V × ?A ÷ 1,000VA/kVA = 25kVA
If you calculate from full kVA rating...
480V × ?A × square-root-of-3 ÷ 1,000VA/kVA = 75kVA
BTW, 25kVA @ 208V is 120A, but the rule holds pretty true: kVA in = kVA out.
Both.
that confused me.....
my take: KVA is KVA excluding transformer losses if you have 100 amps at 208 3 phase you have
36 kva 36 kva at 480 would be 43 amps.
Your 75 KVA would not be the limiting factor, your 100 amp feed would be.
The 75 KVA transformer would have a 112 amp capacity (at 125%) at 480 volts with the proper primary
my take: KVA is KVA excluding transformer losses if you have 100 amps at 208 3 phase you have
36 kva 36 kva at 480 would be 43 amps.
Your 75 KVA would not be the limiting factor, your 100 amp feed would be.
The 75 KVA transformer would have a 112 amp capacity (at 125%) at 480 volts with the proper primary
ptonsparky
Tom
- Occupation
- EC - retired
I'm thinking about 45. FLA 75kva 480 to 208 is about 208 on the secondary with 90 on the primary. Flip it around gives you a bit less than 45, with 100 amp at 208.
Don't take that to the bank though...
ptonsparky
Tom
- Occupation
- EC - retired
HS you guys are fast!!
Ummm... I must not of been in my right mind. Scratch the highlighted red (you have to use the L-N voltage, 277V, if you take one third the kVA rating... 277V × ?A ÷ 1,000VA/kVA = 25kVA)
Anyway, the other formula is correct as is.
75kVA × 1,000VA/kVA ÷ 208V ÷ 1.732 = 208A rated primary current.
If you supply the primary with a 100A rated circuit, you are severely limiting it output potential, which is:
75kVA × 1,000VA/kVA ÷ 480V ÷ 1.732 = 90A rated secondary current.
Feeding it with a 100A limits your output to:
100A in × 208V ÷ 480V = 43A out.
electrofelon
Senior Member
- Location
- Cherry Valley NY, Seattle, WA
Yes 43.33 amps @ 480 volts not counting losses. If the load was continuous than you should be at 80% of that, or increase the primary OCPD size by at least 125% (assuming the conductors remain protected). I wonder why he is using a transformer twice as big as the primary OCPD would allow?
Carultch
Senior Member
- Location
- Massachusetts
Similar question:
If an engine supplies 100 ft-lbs of torque to a shaft rotating at 2080 RPM, connected through a system of gears to an output shaft rotating at 4800 RPM, how much torque can the output shaft supply?
----------------------------
KVA = current * Voltage * a factor that accounts for the 3-phase topology
In this case:
KVA = 100A * 208V * sqrt(3) = 36026 kVA
When selecting a transformer, you usually have to round up to a standard size transformer, unless you get something ultra custom. The next standard size up is 45 kVA. 75kVA would be overkill, but there is nothing wrong about using an oversized transformer.
The fact that you are using an oversized transformer, does not (significantly) affect the transformation of voltage and current. All that means is that you are using a device with more metal in its windings, so that it can withstand the internal heat generated within the unit. And it can withstand more heat, than would be generated for the given load.
To calculate current on the other side of the transformer, assume KVA is conserved. Simply change the voltage for 480V, and let current be an unknown.
KVA = I*V*sqrt(3)
Solve for I:
I = KVA/(V*sqrt(3))
I = 36026 kVA/(480V * sqrt(3))
I = 43.3 A
In reality, KVA is not conserved, because no transformer is 100% efficient. For most of the calculations concerning the NEC, you assume it is conserved to keep track of maximum possible current.
An analogy of transformers is a system of gears. Gears transmit power from an input shaft to an output shaft. It is nearly the same amount of power that gets transferred, however the power is "reformatted" to a different mix of torque and speed. The relative ratio of speeds of each shaft is rigidly set by the number and geometry of the gear teeth. In order for the power to get transmitted, the torque increases from input to output to account for a decrease in speed (reduction gearing), and vice versa (overdrive gearing).
Last edited:
Carultch
Senior Member
- Location
- Massachusetts
Does the transient nature of a load, ever affect how to size a transformer?
Or is it always based on the full load amperes (FLA), or an equivalent measurement for the current you expect in steady-state operation?
electrofelon
Senior Member
- Location
- Cherry Valley NY, Seattle, WA
That's a good question, I'm not sure. Is a transformer ever "upsized" perhaps for a load with a large inrush to lessen the voltage drop?
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
In general the thermal mass of a transformer, especially oil cooled, is so high that the heating effects of transient loads are not really important.
One possible reason to upsize the transformer would be to reduce its contribution to voltage drop during motor starting or other heavy load transients.
But it is an expensive step to take.
To up size or derate the transformer ( to about 85%) may be required to reduce voltage drop during high voltage motor starting.
kwired
Electron manager
- Location
- NE Nebraska
Your transformer has a ratio between primary and secondary voltage. For simplicity lets assume you have 240 and 480 on each side which is a 1:2 ratio and lets also stick to a single phase example. If for any given load the current on low side will be double the current on the high side.
10,000 VA load = ~41.6 amps @ 240
10,000 VA load = ~20.8 amps @ 480
Still keeping it at single phase and 240-480:
If you run a feed capable of handling 100 amps to the 240 volt side then the 480 volt side will be capable of delivering 50 amps - assuming the transformer rating and overcurrent protection devices are high enough to handle those current levels. So if you supplied a 10 kVA transformer to a 100 amp 240 supply, then the supply has more capability then the transfomer can carry without overheating so you are still limited to 41.6 amps input and 20.8 amps output if you want the transformer to last.
208 to 480 is a ratio of 1:2.3, and some of the calculations need to consider number of phases and phase angles with three phase - which brings in the 1.73 multiplier at times - single phase examples are simpler to explain.
kingpb
Senior Member
- Location
- SE USA as far as you can go
- Occupation
- Engineer, Registered
MVA1 = MVA2
So,
I1*V1*sqrt3 = I2*V2*sqrt3
I1*(V1/V2) = I2
In your case, I1 =100A, V1 =208V, and V2 = 480V
Therefore,
100A*(208/480) = 43A
The only place the 75KVA comes into play, would be if your 100A is greater than the rated amps of the transformer, which it is not.
Not sure of your load profile, or the engineer possibly sizing for future load. But either way the 75KVA is quite a bit larger than your current load.
I'm thinking the engineer made a mistake and sized the primary ocpd as though it were connected to the high side (normal primary side when configured in step-down mode). 75kVA@480V is 90A rated high side and 100A would put it close to a 125% max ocpd for primary protected transformer. Actual max' would be 90A × 125% = 112.5A, rounded up to next standard size would put it at 125A ocpd.
Sure it's speculation, but 100A ocpd for a 75kVA, 208V/480V step-up isolation transformer just seems way out of the norm.
- Status