Calculating Amperage for stepdown transformer

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Hello awesome People,

I am conducting a photovolatic design in which I need to use a step down transformer going from 480V to 240V. I understand that my amperage will roughly double, but how do I calculate exactly what my amperage will be after the step down? I need this in order to correctly size my wires at my inter-connection point so that they may handle the hike in amperage.

Currently have 125A(its a somewhat large system) of backfeed landed in a AC disco w/ production meter and that will lead to my current transformer which after the step down will lead to my interconnection.

Some specs about the system(which may not be necessary but ill just give it anyways) : 91.5KW system


Thanks again for your help!
 

iceworm

Curmudgeon still using printed IEEE Color Books
Location
North of the 65 parallel
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EE (Field - as little design as possible)
Hello awesome People,

I am conducting a photovolatic design in which I need to use a step down transformer going from 480V to 240V. I understand that my amperage will roughly double, but how do I calculate exactly what my amperage will be after the step down? I need this in order to correctly size my wires at my inter-connection point so that they may handle the hike in amperage. ...

No, it will exactly double. A2 = A1 x V1/V2

ice
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
Hello awesome People,

I am conducting a photovolatic design in which I need to use a step down transformer going from 480V to 240V. I understand that my amperage will roughly double, but how do I calculate exactly what my amperage will be after the step down? I need this in order to correctly size my wires at my inter-connection point so that they may handle the hike in amperage.

Currently have 125A(its a somewhat large system) of backfeed landed in a AC disco w/ production meter and that will lead to my current transformer which after the step down will lead to my interconnection.

Some specs about the system(which may not be necessary but ill just give it anyways) : 91.5KW system


Thanks again for your help!
For a 91.5 kW system your current will be about 220A @ 240V 3 phase.
 
how do I calculate exactly what my amperage will be after the step down? I need this in order to correctly size my wires at my inter-connection point so that they may handle the hike in amperage.

Currently have 125A(its a somewhat large system) of backfeed landed in a AC disco w/ production meter : 91.5KW system

Howdy. There might be something in the inverter installation manual about transformer sizing. There's the impedance to factor in, but the efficiency curve matters too.
Sometimes xfmrs run best between 35-65% of rated kVA for instance- if you had a 150kVA xfmr for 91.5 kW, that's 61%, or a 167kVA xfmr would be 55%.
But you might not necessarily want to jump all the way up to 225kVA from 167kVA for 91.5kW just because the xfmr "runs best" at 50%.

Anyhoo, if 125A at 480v, and with a 90% efficient xfmr, you'd get 225A at 240 3-ph. Or like ggunn said, 220A if 88% efficient.

I'm at the same point with a design- it would seem to make sense to use a 98+% efficient xfmr if possible.

The I2R Loss is sometimes referred to as "copper loss." It is power lost in
circulating current in the windings. This represents the greatest loss in the
operation of a transformer. The actual watts of power lost can be determined
(in each winding) by squaring the amperes and multiplying by the resistance in
ohms of the winding.
The intensity of power loss in a transformer determines its efficiency. The
efficiency of a transformer is reflected in power (wattage) loss between the
primary (input) and secondary (output) windings. Here are three formulas for
determining power losses due to efficiency

Efficiency = Secondary Watts (output) / Primary VA (input)
Secondary Watts (output) = Primary VA (input) x Efficiency
Primary VA (input) = Secondary Watts (output) / Efficiency
http://www.edrenzi.com/Code/Trans (1p).pdf
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Disregarding transformer losses, of course.
Any transformer losses will show up as voltage drops rather than current differences.
The difference between primary current and secondary current times turns ratio will be entirely due to the magnetizing current of the core. In a PV interconnect that current will be supplied by the POCO side even though power flows from GTI to POCO.
That current will be largely reactive but will include a component related to hysteresis losses (iron losses.)
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
Any transformer losses will show up as voltage drops rather than current differences.
Well, OK, but the voltage rise that the inverter sees from that voltage drop will result in less current from the inverter and therefore less current from the transformer.

Your move. :D
 

Carultch

Senior Member
Location
Massachusetts
Hello awesome People,

I am conducting a photovolatic design in which I need to use a step down transformer going from 480V to 240V. I understand that my amperage will roughly double, but how do I calculate exactly what my amperage will be after the step down? I need this in order to correctly size my wires at my inter-connection point so that they may handle the hike in amperage.

Currently have 125A(its a somewhat large system) of backfeed landed in a AC disco w/ production meter and that will lead to my current transformer which after the step down will lead to my interconnection.

Some specs about the system(which may not be necessary but ill just give it anyways) : 91.5KW system


Thanks again for your help!

For NEC purposes, you assume conservatively that transformer losses are negligible when sizing circuits on both sides. The ratio of currents equals exactly the reciprocal of the ratio of nominal voltages.
 

Carultch

Senior Member
Location
Massachusetts
Any transformer losses will show up as voltage drops rather than current differences.

So is current ratio rigidly constrained to the actual turns ratio?

I often think of transformers as an "electrical gear system" that trades voltage and current to get leverage for pushing power through high resistance lines due to distance. With mechanical gears, the rotation rate is rigidly constrained to the teeth ratio, while torque is "sacrificed" to the gear inefficiencies. Can one say a similar statement for transformers, with the ratio of currents exactly equaling the actual turns ratio?

I would also anticipate that a transformer designer might build it with slightly extra primary windings to curtail internal voltage losses based on operating at the KVA rating, thus trading them for current losses. This is because voltage tolerance is a power quality issue for certain equipment.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
So is current ratio rigidly constrained to the actual turns ratio?

I often think of transformers as an "electrical gear system" that trades voltage and current to get leverage for pushing power through high resistance lines due to distance. With mechanical gears, the rotation rate is rigidly constrained to the teeth ratio, while torque is "sacrificed" to the gear inefficiencies. Can one say a similar statement for transformers, with the ratio of currents exactly equaling the actual turns ratio?

I would also anticipate that a transformer designer might build it with slightly extra primary windings to curtail internal voltage losses based on operating at the KVA rating, thus trading them for current losses. This is because voltage tolerance is a power quality issue for certain equipment.

Yes, it is rigid as long as the transformer core stays out of saturation. (Which is NOT a function of load).
If the transformer is built with extra turns for voltage drop compensation (as some control transformers are), then the current ratio will be based on the actual turns ratio and not the nominal voltage ratio.

Using your analogy, the ratio of the motion of the input and output gears (~current) is rigidly locked to the tooth ratio. The input torque to output torque ratio (~voltage) will be different depending on the friction losses in the gear train, with the tooth ratio as the ideal starting point.
 

kwired

Electron manager
Location
NE Nebraska
Transformer with no secondary load is open circuit on the secondary - no current can flow, yet there is current in the primary but is mostly reactive current and a little bit is the earlier mentioned iron losses, which is essentially an efficiency loss.
 
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