How would I size this boost transformer?

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Hi all,
I have a machine from Canada that will require 575volt 3 phase power. It is rated at 160kw @575volt. This machine does have some motors involved, but the load is mostly resistive (heat strips). The power we have available is 460 volt 3 phase.
My question is how would I size the "boost" transformer and also the feeders, and OCP? Iam being told by the engineer on the project that simple ohms law solves all.
460/160,000=348 amps /1.732= 200.1 amps X 1.25%=251 amps.
That just does not seem right to me. Since the load is mostly resistive would you not recalculate the kw at 460volt for tranformer sizing and wire size/OCP on the primary side ?
Thanks for any help you may offer.
 

augie47

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My first thought was that you needed to look at the heat resistance if you were going to operate them off 480, but if your are going to use a transformer to supply the machine at it's rated voltage of 575 he is correct on the heaters in that kw is kw.
160kw of heat load at 575 volts will still be 160 kw load on your transformer but the primary current will be more in the 192 amp range where as on the 575 volt side it's
closer to 160m amps,
That does not address any "continuous load" multiplier or the motor load.
 

GoldDigger

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First question is what you mean by "boost".
It could be a stepup transformer with isolated windings, in which case both primary and secondary are sized for full nominal load at nameplate wattage.
Or autotransformer or boost windings (simple for wye, not trivial for delta), in which case the boost winding only has to carry the full current times the difference in voltage.
For a step up, their calculation will give the primary current, and you are correct that the secondary current will be lower.
 

templdl

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Wisconsin
First question is what you mean by "boost".
It could be a stepup transformer with isolated windings, in which case both primary and secondary are sized for full nominal load at nameplate wattage.
Or autotransformer or boost windings (simple for wye, not trivial for delta), in which case the boost winding only has to carry the full current times the difference in voltage.
For a step up, their calculation will give the primary current, and you are correct that the secondary current will be lower.
Yes, the problem is stepping up about 100v which is not practical to do with as boost transformer configuration. You may as well use a two winding isolation transformer.
 

GoldDigger

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Yes, the problem is stepping up about 100v which is not practical to do with as boost transformer configuration. You may as well use a two winding isolation transformer.

In terms of boost ratio, 100V out of 500V is not too far out of line (same ratio as 24 out of 120), but I can definitely see a problem finding a 100V secondary with 480V primary to make each boost transformer.
I defer to your experience on that call. :)
 

bob

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Alabama
Hi all,
Iam being told by the engineer on the project that simple ohms law solves all.
460/160,000=348 amps /1.732= 200.1 amps X 1.25%=251 amps.
You need to correct your engineer. When you drop the voltage to 460 volts you no longer have 160000 watts.
The new value is (460?/575?) x 160000 = 0.64 x 160000 w = 102400 watts. Can the machine operate properly like this?
If not you need to transform up from 460 to 575 volts.
 
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GoldDigger

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The engineer is correct in calculating the feeder current to the primary of the transformer (leaving PF and surge out of the calculation.) It also allows sizing of an isolation type transformer.
What it does not do is tell what the secondary current to the unit will be at 575V. But it will nicely overestimate that for sizing the conductors to the unit itself.
 

winnie

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Springfield, MA, USA
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Electric motor research
Is your supply wye (with a neutral) or delta (without)??

You should be able to use a boost transformer (bank of single phase transformers in an autotransformer connection) to get 575V from 460V, (I assume nominal 600V from nominal 480V) using standard components.

-Jon
 

templdl

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Location
Wisconsin
In terms of boost ratio, 100V out of 500V is not too far out of line (same ratio as 24 out of 120), but I can definitely see a problem finding a 100V secondary with 480V primary to make each boost transformer.
I defer to your experience on that call. :)
As I recall while working as a sales and engineer for a transformer manufacture you could design said transformer as a boost but there ended up to be very little cost advantage as one would expect to get over an isolation transformer.
 

topgone

Senior Member
Hi all,
I have a machine from Canada that will require 575volt 3 phase power. It is rated at 160kw @575volt. This machine does have some motors involved, but the load is mostly resistive (heat strips). The power we have available is 460 volt 3 phase.
My question is how would I size the "boost" transformer and also the feeders, and OCP? Iam being told by the engineer on the project that simple ohms law solves all.
460/160,000=348 amps /1.732= 200.1 amps X 1.25%=251 amps.
That just does not seem right to me. Since the load is mostly resistive would you not recalculate the kw at 460volt for tranformer sizing and wire size/OCP on the primary side ?
Thanks for any help you may offer.

Two - 100 kVA, 480/120V single-phase transformers, arranged in an open-delta boost transformer bank to give you 460 x (480+120)/480 = 575V will be more than enough for that 160kW load. You will choose an OCP to protect your transformer and since your open-delta capacity will only be 57% of the full delta capacity, i.e. 3 x 100 x 57% = 173kVA, your primary OCP will be 173,000/(1.732x 460)x 1.25 = 271~ 300A. You can calculate for the secondary protection based on the 173kVA capacity.
 

winnie

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Springfield, MA, USA
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Electric motor research
Two - 100 kVA, 480/120V single-phase transformers, arranged in an open-delta boost transformer bank to give you 460 x (480+120)/480 = 575V will be more than enough for that 160kW load. You will choose an OCP to protect your transformer and since your open-delta capacity will only be 57% of the full delta capacity, i.e. 3 x 100 x 57% = 173kVA, your primary OCP will be 173,000/(1.732x 460)x 1.25 = 271~ 300A. You can calculate for the secondary protection based on the 173kVA capacity.

I am pretty sure that this connection would work with a pair of _25_ (twenty five) kVA transformers in a boost configuration.

See page 9.5 of http://www.jeffersonelectric.com/s3/site/catalog/buckBoost.pdf for example capacities of open delta boost transformer connections.

For a 20% boost going from 240 to 288V, a pair of 5KVA transformers is good for a 51 KVA load. Going from 480 to 576V should have the same proportions (of course with different voltage rated transformers), so for a 160 KVA load you would only need a pair of 16KVA transformers...but if you want off the shelf parts you are probably overloading 15KVA transformers or going to 25KVA each.

Remember that in the boost configuration the transformer is only handling the power for the 'voltage boost'; the bulk of the power is being supplied directly by the line. So you don't need the transformer to be rated for the full capacity of the load.

-Jon
 

bob

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Location
Alabama
For a 20% boost going from 240 to 288V, a pair of 5KVA transformers is good for a 51 KVA load. Going from 480 to 576V should have the same proportions (of course with different voltage rated transformers), so for a 160 KVA load you would only need a pair of 16KVA transformers...but if you want off the shelf parts you are probably overloading 15KVA transformers or going to 25KVA each.
from 460
Remember that in the boost configuration the transformer is only handling the power for the 'voltage boost'; the bulk of the power is being supplied directly by the line. So you don't need the transformer to be rated for the full capacity of the load.

-Jon
Winnie, the customer is going 460 volts to 575 volts or a 25% boost. I think the 25 kva is the correct size.
 

Besoeker

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Location
UK
As I recall while working as a sales and engineer for a transformer manufacture you could design said transformer as a boost but there ended up to be very little cost advantage as one would expect to get over an isolation transformer.
Well, I would expect a sales pitch to to take that line.:p
But 20% of the kVA rating versus 100% of kVA rating?
 
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