Step Down Transformer from Receptacle - Expected Results?

[MPTESE]

Hello,

I have a 125/250v - 50amp twistlock receptacle in a 3phase building (so, 208v between hots). It's a 4-wire receptacle.

Currently, when needed, I split the phases/hots to attain two lines of 120v, each capable of 50amp current.

If I installed a knock down transformer (208->120) into the receptacle (pigtail a 50amp twist-lock plug), what can I expect to pull current wise after the transformer?

 

[Lucien Nunes]

Unfortunately, not as much as your total 50+50A at the moment. If you use a single-phase step-down from 208 to 120V, and wire it between the two hots that are good for 50A, you'll get 208/120 x 50A = 87A, less a bit for the transformer losses and magnetising current. Call it 80A at 120V.

With your two 50A 120V supplies, the neutral is further away from the hots in voltage, than the hypothetical midpoint between them. Hence within a limit of 50A on each wire, the two 120V supplies give more power in total than the transformed 208V.

 

[MPTESE]

Lucien, Thanks so much.

Can you recommend any visual learning material to understand the theory behind calculating the 87A?

 

[Megawatt]

Hello,

I have a 125/250v - 50amp twistlock receptacle in a 3phase building (so, 208v between hots). It's a 4-wire receptacle.

Currently, when needed, I split the phases/hots to attain two lines of 120v, each capable of 50amp current.

If I installed a knock down transformer (208->120) into the receptacle (pigtail a 50amp twist-lock plug), what can I expect to pull current wise after the transformer?

First off what size of a transformer are you talking about. I can’t understand why you would need a transformer but you have to know the KVA of the transformer and then we can tell you if it will work. Example a 75 KVA transformer is good for 200 amps so if you install a 30 KVA transformer it is good for 66 amps

 

[Lucien Nunes]

what size of a transformer are you talking about.

The information's in the question. He's looking at whether a single-phase transformer, supplied by two hots of a 50A circuit from a 208V Y service, can yield a more useful single-phase source than the alternative of 2 x 120V 50A supplies from the hots to neutral. Hence, the size of transformer under consideration is 208 x 50 = 10.4kVA input.

understand the theory behind calculating the 87A?

It's straightforward and intuitive.

P=VI (power = voltage x current)
Considering the voltage and current at the input and output of the transformer:
Pin = Vin x Iin
Pout = Vout x I out
If the transformer is perfectly efficient i.e. has no losses, the power output would equal the power input. Hence:
Pout = Pin = Vout x Iout = Vin x Iin.

Vout x Iout = Vin x Iin rearranges to:
Iout/Iin = Vin/Vout
I.e. the ratio of output to input current is the inverse of the ratio of output to input voltage. Simplest example: halve the voltage, and double the current will be needed to transfer the same power.

Real transformers dissipate some power as heat in the copper and iron, therefore the power available at the output will be less than the transformer consumes at the input. Efficiencies range from 60-99% or so with the highest efficiencies only in the very largest high voltage transmission transformers. When considering the output current available from a transformer, one must check or estimate the efficiency and reduce the output by that ratio from the 'ideal' transformer's output.

In your case I merely estimated about 8% for losses and excitation as a worst case for a commercial 10kVA transformer, hence 87 x 0.92 = 80A useful output at 50A input.

Beyond this, signifincant maths is required to further understand the characteristics of real transformers.

 

[Megawatt]

Hello,

I have a 125/250v - 50amp twistlock receptacle in a 3phase building (so, 208v between hots). It's a 4-wire receptacle.

Currently, when needed, I split the phases/hots to attain two lines of 120v, each capable of 50amp current.

If I installed a knock down transformer (208->120) into the receptacle (pigtail a 50amp twist-lock plug), what can I expect to pull current wise after the transformer?

In order for me to calculate the size of the transformer is the primary voltage 3 phase 208vac and I’m assuming you want 120vac. If that’s true it is just an isolation transformer, you are still going have 208/120vac on the secondary side
[automerge]1582382966[/automerge]

The information's in the question. He's looking at whether a single-phase transformer, supplied by two hots of a 50A circuit from a 208V Y service, can yield a more useful single-phase source than the alternative of 2 x 120V 50A supplies from the hots to neutral. Hence, the size of transformer under consideration is 208 x 50 = 10.4kVA input.



It's straightforward and intuitive.

P=VI (power = voltage x current)
Considering the voltage and current at the input and output of the transformer:
Pin = Vin x Iin
Pout = Vout x I out
If the transformer is perfectly efficient i.e. has no losses, the power output would equal the power input. Hence:
Pout = Pin = Vout x Iout = Vin x Iin.

Vout x Iout = Vin x Iin rearranges to:
Iout/Iin = Vin/Vout
I.e. the ratio of output to input current is the inverse of the ratio of output to input voltage. Simplest example: halve the voltage, and double the current will be needed to transfer the same power.

Real transformers dissipate some power as heat in the copper and iron, therefore the power available at the output will be less than the transformer consumes at the input. Efficiencies range from 60-99% or so with the highest efficiencies only in the very largest high voltage transmission transformers. When considering the output current available from a transformer, one must check or estimate the efficiency and reduce the output by that ratio from the 'ideal' transformer's output.

In your case I merely estimated about 8% for losses and excitation as a worst case for a commercial 10kVA transformer, hence 87 x 0.92 = 80A useful output at 50A input.

Beyond this, signifincant maths is required to further understand the characteristics of real transformers.

Lucien could you please help me understand your calculations on a 10kva transformer is good for 87 amps. My math just don’t support that much current

 

[Lucien Nunes]

87 x 120 = 10440

 

[Megawatt]

87 x 120 = 10440

Where are you getting 87 from ?
[automerge]1582383919[/automerge]

Where are you getting 87 from ?

If it’s a 4 wire plug then I’m assuming that it will be 3 phase 208 primary.

 

[Lucien Nunes]

See post #2.

He claims his source of power to be a 50A receptacle, from which he can take either 50A from each hot to neutral at 120V, or 50A between hots at 208V. His question is how much current he can get at 120V from a transformer stepping down the 208V. To be fair, he does not explicitly state that he requires the largest single-phase supply that can be achieved from that receptacle, but that is implicit in the use of the transformer as he already knows how much he can get by dividing his single-phase appliances between the hots.

Hence, the largest single-phase transformer that can be powered from that circuit has an input of 50A @ 208V.
50 x 208 = 10.4kVA
Ideal transformer output 50 x 208/120 = 86.6A
Estimating 92% efficiency
0.92 x 86.6 = 80A at 120V.

As I say, I'm inferring what he wants to achieve from the way he asked the question. There may be more to it but the numbers are correct.

 

[Megawatt]

Where are you getting 87 from ?
[automerge]1582383919[/automerge]

If it’s a 4 wire plug then I’m assuming that it will be 3 phase 208 primary.

I just did the math on a 10 KVA single phase transformer calculation with 208 primary and 120vac secondary and actually got 83 amps. I’ve just never seen a transformer that just gives you just 120vac on the secondary
[automerge]1582384487[/automerge]

See post #2.

He claims his source of power to be a 50A receptacle, from which he can take either 50A from each hot to neutral at 120V, or 50A between hots at 208V. His question is how much current he can get at 120V from a transformer stepping down the 208V. To be fair, he does not explicitly state that he requires the largest single-phase supply that can be achieved from that receptacle, but that is implicit in the use of the transformer as he already knows how much he can get by dividing his single-phase appliances between the hots.

Hence, the largest single-phase transformer that can be powered from that circuit has an input of 50A @ 208V.
50 x 208 = 10.4kVA
Ideal transformer output 50 x 208/120 = 86.6A
Estimating 92% efficiency
0.92 x 86.6 = 80A at 120V.

As I say, I'm inferring what he wants to achieve from the way he asked the question. There may be more to it but the numbers are correct.

@Lucian I have the upmost respect for you and your math. I guess I am confused about what he is trying to accomplish

 

[Lucien Nunes]

actually got 83 amps

Probably spot on, I deliberately used a pessimistic value for efficiency.

His question didn't say what he is trying to accomplish, but the figure of 83A was the answer to his original question. We've given him that, maybe it's of interest, maybe not

 

[MPTESE]

Lucien, you assumptions and answer was exactly what I wanted to confirm. Thank you again for your detailed response.

To others who may be wondering, it could be advantageous for me to have a single load that exceeds 50amps at 120. In my scenarios it is for the purpose of cinematography, and powering highwattage tungsten fixtures, which only operate at 120 in the us. Lucien’s answer confirms that I can run a single fixture approaching 8000watts. Alternatively, a lot of HMI ballasts now can indeed operate on 240, so it is also advantageous for me to be prepared for maintaining the 240v operation and attaining the right and safe adapters for these lights.

If using the transformer and with a single high wattage load, I also assume I will need an additional breaker for safety and to avoid burning the transformer - as the current does not transfer back to the original bipole 50amp breakers in the panel?