Transformer current calculation help for In-rush and current consumption

[emiller233]

I am working on a project that is getting a little over my head in knowledge and could use some help!!!

We have a 50KVA dry type transformer feeding a very large 330VDC Power Supply(Ametek SGE 330/91E-01A)

The customer is asking for the In-rush current and current consumption for the supply to the transformer, and I am not positive how calculate this?

Line side = 600VAC, 3-phase

Load side = 480VAC, 3-phase. This feeds the 330VDC Power supply which has a max input current of 73FLA .

Here are links for the power supply
WEBSITE:
http://www.signaltestinc.com/product-p/sga 330-91.htm
DATA SHEET
http://www.signaltestinc.com/v/vspfiles/assets/datasheet/Sorensen_SG_Datasheet.pdf

I found this formula for the breaker size of the line side, but not seeing how to calculate the in-rush current and current consumption
Amps for line side of 3-ph XFMR = Transformer KVA / (Transformer Primary Volts * 1.73)
Amps for line side of 3-ph XFMR = 50KVA / (600VAC * 1.73) = 48.17amps

for the circuit breaker sizing, I got 48.17Amps * 1.25 = 60.21amps

will a 60 amp service be enough for this?

how do i calculate the in-rush and current consumption for the line side of the transformer?

Thanks!
Eric

 

[Ingenieur]

fla
line 50000/(1.732 x 600) = 48 A, call it 50
load 50000/(1.732 x 480) = 60 A

DC supply = 1.732 x 480 x 73 = 60,000 va

reference NEC 450 tables 3(B)

are you having sec protection?


inrush figure 8-12 times fla

 

[emiller233]

yes we will be using circuit breakers on both, line and load side. we have 60A drawn on Line and 100 on load right now

the customer told us they had 60Amps available, so im guessing thats why our salesman quoted a 50KVA originally?

It looks like we need to upsize this to a 60KVA transformer as well instead of a 50KVA?

according to the table , I would need a breaker on the load side of 60Amps and 72 Amps(next availble size up=80? ) on the secondary

FLA
line 60000/(1.732 x 600) = 57.74 A
load 60000/(1.732 x 480) = 72.17 A

inrush on the line side would be 8-12 times FLA of 58 A (462 - 693 Amps)

 

[Ingenieur]

yes we will be using circuit breakers on both, line and load side. we have 60A drawn on Line and 100 on load right now

the customer told us they had 60Amps available, so im guessing thats why our salesman quoted a 50KVA originally?

It looks like we need to upsize this to a 60KVA transformer as well instead of a 50KVA?

according to the table , I would need a breaker on the load side of 60Amps and 72 Amps(next availble size up=80? ) on the secondary

FLA
line 60000/(1.732 x 600) = 57.74 A
load 60000/(1.732 x 480) = 72.17 A

inrush on the line side would be 8-12 times FLA of 58 A (462 - 693 Amps)

load requires 100 A at 480 vac? not sure how a DC power supply rated at 73 draws 100? or is the 100 dc?
if 100 A at 480 ac 3 ph
kva = 1.732 x 100 x 480 / 1000 = 83 kva
if the 100 A is accurate and continuous you need 100 kva (or 112.5 may be the std size)

this makes
primary 96 A, use a 200(code allows up to 250% w/sec protection) this will help ensure no issue with inrush
secondary 120 A, use a 150 (code allows 150%)

 

[petersonra]

the normal maximum xfmr line side current is just 73 * 480/600.

It is hard to say what the inrush might be since it is not specified anywhere that I noticed. You might have to call and ask. Some of these digital power supplies have an amazing amount of inrush for a cycle or two.

 

[Ingenieur]

load requires 100 A at 480 vac? not sure how a DC power supply rated at 73 draws 100? or is the 100 dc?
if 100 A at 480 ac 3 ph
kva = 1.732 x 100 x 480 / 1000 = 83 kva
if the 100 A is accurate and continuous you need 100 kva (or 112.5 may be the std size)

this makes
primary 96 A, use a 200(code allows up to 250% w/sec protection) this will help ensure no issue with inrush
secondary 120 A, use a 150 (code allows 150%)

after rereading the load side CB is 100, not the load current
so yes a 60 kva should do it (iirc next std size is 75)
you want some headroom, the power supply may involve switching so noise/harmonics

75 kva
line 72 A x 2.5 = use a 175 A CB
load 90 A x 1.25 = use a 110 A CB

 

[Jraef]

fla
line 50000/(1.732 x 600) = 48 A, call it 50
load 50000/(1.732 x 480) = 60 A

DC supply = 1.732 x 480 x 73 = 60,000 va

reference NEC 450 tables 3(B)

are you having sec protection?


inrush figure 8-12 times fla

To his point, you can't calculate an exact number on inrush current, but you can estimate it at 8-12x the line current peak for about 1 cycle, decaying after that for a few more cycles.

 

[Ingenieur]

To his point, you can't calculate an exact number on inrush current, but you can estimate it at 8-12x the line current peak for about 1 cycle, decaying after that for a few more cycles.

yep, too many variables
but imho 8-12 x fla or so is a fair approx.

in his case assuming a 75 kva xfmr
fla 72 A, 12 x = 870 A

if primary is 200 A I/t CB instantaneous region should be at least 2000 A, so he should not have nuisance tripping

 

[templdl]

I am working on a project that is getting a little over my head in knowledge and could use some help!!!

We have a 50KVA dry type transformer feeding a very large 330VDC Power Supply(Ametek SGE 330/91E-01A)

The customer is asking for the In-rush current and current consumption for the supply to the transformer, and I am not positive how calculate this?

Line side = 600VAC, 3-phase

Load side = 480VAC, 3-phase. This feeds the 330VDC Power supply which has a max input current of 73FLA .

Here are links for the power supply
WEBSITE:
http://www.signaltestinc.com/product-p/sga 330-91.htm
DATA SHEET
http://www.signaltestinc.com/v/vspfiles/assets/datasheet/Sorensen_SG_Datasheet.pdf

I found this formula for the breaker size of the line side, but not seeing how to calculate the in-rush current and current consumption
Amps for line side of 3-ph XFMR = Transformer KVA / (Transformer Primary Volts * 1.73)
Amps for line side of 3-ph XFMR = 50KVA / (600VAC * 1.73) = 48.17amps

for the circuit breaker sizing, I got 48.17Amps * 1.25 = 60.21amps

will a 60 amp service be enough for this?

how do i calculate the in-rush and current consumption for the line side of the transformer?

Thanks!
Eric

It is more fun, I know, to do it yourself but have you asked the transformer manufacturer? As a former sales and application engineer I have all sorts of no load, 1/4, 1/2, 3/4, and full load losses on the transformers that I sold and supported. There really are no exact numbers available for inrush only ball parks which I am to assume what you may be referring to may be the magnetizing current at the instant that the transformer is energized.
These numbers will be different for 80, 115, 150 deg rise transformers, high efficiency, and the various K-factor transformers and with copper or aluminum windings
The magnetude of the inrush current become an issue as it relates to the magnetic/instantaneous pickup of the breaker which feeds the transformer. Assuring that the breaker will not nuisance trip at the instant such transformer is energized is the objective. Consider that the instantaneous trip of the breaker you select is 10x the rating of the breaker +-20% and is usually toward the +20%.

 

[emiller233]

load requires 100 A at 480 vac? not sure how a DC power supply rated at 73 draws 100? or is the 100 dc?
if 100 A at 480 ac 3 ph
kva = 1.732 x 100 x 480 / 1000 = 83 kva
if the 100 A is accurate and continuous you need 100 kva (or 112.5 may be the std size)

this makes
primary 96 A, use a 200(code allows up to 250% w/sec protection) this will help ensure no issue with inrush
secondary 120 A, use a 150 (code allows 150%)

the breakers they wanted to use did not have anything between 60A and 100A, that's the reason it was so big. The load is a 73Amps and it has an inrush of 150A

I did read that table wrong tho, if I have secondary protection it says 250% protection. is this a maximum? I think I am still good at the 58Amp line side (using the 73A load, not the 100A) which i would need a 75A breaker on the Line side

I think this is irrelevant for the calc's, but we will be using a 75KVA transformer instead of the 50KVA now (was quoted at the same price..?)

 

[Besoeker]

I am working on a project that is getting a little over my head in knowledge and could use some help!!!

We have a 50KVA dry type transformer feeding a very large 330VDC Power Supply(Ametek SGE 330/91E-01A)

The customer is asking for the In-rush current and current consumption for the supply to the transformer, and I am not positive how calculate this?

Line side = 600VAC, 3-phase

Load side = 480VAC, 3-phase. This feeds the 330VDC Power supply which has a max input current of 73FLA .

Something odd about those numbers. Do you have, or can you get, a basic power circuit diagram that you can post?

 

[emiller233]

Something odd about those numbers. Do you have, or can you get, a basic power circuit diagram that you can post?

Here is the website for the Power Supply
http://www.signaltestinc.com/product-p/sga 330-91.htm
http://www.signaltestinc.com/product-p/sga 330-91.htm

Here is the Data Sheet for it
http://www.signaltestinc.com/v/vspfiles/assets/datasheet/Sorensen_SG_Datasheet.pdf
http://www.signaltestinc.com/v/vspfiles/assets/datasheet/Sorensen_SG_Datasheet.pdf

I also attached a working copy of our schematic's containing the XFMR and the power supply. The transformer on our schematics will be changing to a 75KVA WYE secondary (it's a 50KVA D-D on them now).
Transformer is on bottom left, 330VDC power supply is on bottom right


I did just hear back from the MFGR of the 330VDC power supply. They did confirm the 330VDC Power Supply is 73A input current (max) with and inrush of 150Amps to the power supply

 

[Besoeker]

Here is the website for the Power Supply
http://www.signaltestinc.com/product-p/sga 330-91.htm
http://www.signaltestinc.com/product-p/sga 330-91.htm

Here is the Data Sheet for it
http://www.signaltestinc.com/v/vspfiles/assets/datasheet/Sorensen_SG_Datasheet.pdf
http://www.signaltestinc.com/v/vspfiles/assets/datasheet/Sorensen_SG_Datasheet.pdf

I also attached a working copy of our schematic's containing the XFMR and the power supply. The transformer on our schematics will be changing to a 75KVA WYE secondary (it's a 50KVA D-D on them now).
Transformer is on bottom left, 330VDC power supply is on bottom right
View attachment 16405

I did just hear back from the MFGR of the 330VDC power supply. They did confirm the 330VDC Power Supply is 73A input current (max) with and inrush of 150Amps to the power supply

Thank you for your prompt response. Unfortunately, the resolution made it difficult to read the schematics. Specifically, the numbers.

But a couple of points.
The 73Aac line current rectified would give about 90Adc.
The data lists the unit as 30kW 330V 91A.
Those numbers add up. Make sense.

Two things that it doesn't explain.
Why a 60kVA, or now 75kVA transformer, is needed for a 30kW load.

The 73A 3-phase 480V does give about 60kVA.
But that leaves another question. At 480Vac in to the converter you would normally expect much greater than 330Vdc out.

 

[emiller233]

Thank you for your prompt response. Unfortunately, the resolution made it difficult to read the schematics. Specifically, the numbers.

But a couple of points.
The 73Aac line current rectified would give about 90Adc.
The data lists the unit as 30kW 330V 91A.
Those numbers add up. Make sense.

Two things that it doesn't explain.
Why a 60kVA, or now 75kVA transformer, is needed for a 30kW load.

The 73A 3-phase 480V does give about 60kVA.
But that leaves another question. At 480Vac in to the converter you would normally expect much greater than 330Vdc out.

I attached a PDF to the last post that should be a little clearer hopefully (I am at out of the office, so I can't post it again)

I was asking manufacture the same thing, and they told me that the power supply is only about 80% efficient ( I forget the exact number) which also has created another huge problem for me to call this enclosure


Does the in rush of the power supply affect the in rush of the transformer at all in my calculations?

I appreciate all of the help!!!! Thank you!!!

 

[Ingenieur]

eff 0.87
pf 0.69

looks like the 73 A is based on lowest rated input v of 396 vac

i = 30000/(1.732 x 396 x 0.87 x 0.69) = 73 A


in rush does not impact sizing
only selection of cb sizes

 

[GoldDigger]

The inrush of the power supply really is not as important as the magnetic characteristics of the transformer. It might make the difference between 12X FLA and 14X FLA.
And as a general rule it would probably be conservative to have the power power supply turned off when you energize the transformer.

Sent from my XT1585 using Tapatalk

 

[Besoeker]

I attached a PDF to the last post that should be a little clearer hopefully (I am at out of the office, so I can't post it again)

I was asking manufacture the same thing, and they told me that the power supply is only about 80% efficient ( I forget the exact number) which also has created another huge problem for me to call this enclosure


Does the in rush of the power supply affect the in rush of the transformer at all in my calculations?

I appreciate all of the help!!!! Thank you!!!

I would expect the power supply to get powered up after the transformer reaches steady state so I don't think it would impact on transformer inrush.

And I think you are right to flag up the 80% efficiency. At 30kW, that's a lot of heat losses to dissipate. But, just looking at the module dimensions, I find that hard to believe.

I'm sorry that I don't have solid answers fo you but somthing about the numbers doesn't sit right with me.
It might be helpful to know what AC to DC conversion technology is being employed.

 

[emiller233]

eff 0.87
pf 0.69

looks like the 73 A is based on lowest rated input v of 396 vac

i = 30000/(1.732 x 396 x 0.87 x 0.69) = 73 A


in rush does not impact sizing
only selection of cb sizes

Thank you!
i Couldn't figure out how they got the 73A, and they wouldn't help me figure it out!

i = 30000/(1.732 x 480 x 0.87 x 0.69) = 60.11 A

 

[emiller233]

I would expect the power supply to get powered up after the transformer reaches steady state so I don't think it would impact on transformer inrush.

And I think you are right to flag up the 80% efficiency. At 30kW, that's a lot of heat losses to dissipate.

I'm sorry that I don't have solid answers fo you but somthing about the numbers doesn't sit right with me.
It might be helpful to know what AC to DC conversion technology is being employed.

It has a enable input that I could use for that, good idea!

The heat loss is 5KW, I had to get a bigger enclosure just to make room to fit 3 air conditioners to keep it cool

This company refuses to give me any help/info (you would think they would when were buying a $20,000 unit from them) but I will try to find out from them

Any answer is better than no answer! I appreciate any and all help!

 

[Besoeker]

Thank you!
i Couldn't figure out how they got the 73A, and they wouldn't help me figure it out!

i = 30000/(1.732 x 480 x 0.87 x 0.69) = 60.11 A

I'll try.
The rating is given as 91Adc. The factor to calculate the AC current is 0.816.(sqrt(2/3). Hence the 73A. A bit of rounding may have been done.
So that's possibly/probably where the 73A comes from.

Whether that comes directly from the 480Vac for a 330Vdc supply is another matter.