Transformer Sizing Help (3 phase)

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Mike32186

Member
Hey everyone,

This is my first time on Mike Holt's forum. I am looking for some help on transformer sizing. I am a little confused about how to handle kva for single phase loads when they are connected to a three phase system. We have 2- 5 ton units that are powered at 208V single phase. The current required to power each unit is about 37 amps. Since the unit is single phase do I calculate the KVA as the following:

37*208V/1000 = 7.7 KVA
2 units @ 7.7 KVA = 15.4 KVA = total connected load.

Will a 20 KVA transformer @ 3 phase be able to take care of this loads?
The maximum full load amps for a 20 KVA transformer is 55.5 amps. But the loads at single phase pull (37*2)= 74 amps, which is more than the 3 phase capacity of the 20 KVA. I am confusing myself thinking about it.

My question is will a 20 KVA be able to handle the load or do I need to calculate the single phase loads as 3 phase loads because they are connected to a 3 phase system? Or do I need to go up to 30 KVA system which is good for 83 amps @ 208/3 Phase.

Your help is greatly appreciated. Thanks guys.:)
 
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petersonra

Senior Member
Location
Northern illinois
Occupation
engineer
Mike32186 said:
Hey everyone,

37*208V/1000 = 7.7 KVA
2 units @ 7.7 KVA = 15.4 KVA = total connected load.
Click to expand...
So far so good.

Will a 20 KVA transformer @ 3 phase be able to take care of this loads?
The maximum full load amps for a 20 KVA transformer is 55.5 amps. But the loads at single phase pull (37*2)= 74 amps, which is more than the 3 phase capacity of the 20 KVA. I am confusing myself thinking about it.
Click to expand...

The loads have to be treated as vectors. The currents flowing through the loads are going to be 120 degrees out of phase with each other. The net current flow into the common phase of the xfmr will end up being less than the current flow in either load.
 
iwire

iwire

Moderator
Staff member
Location
Massachusetts
Mike32186 said:
What do you suggest? How is it too out of balance? There are only two loads.:-?
Click to expand...

I suggest we wait for someone with more expertise in this than myself. :)


As I understand it you have two line to line single phase loads being supplied from a 3 phase transformer.

If that is true you will have to double up on one of the phases.

With one of the loads connected to phase A and B the other load has to be connected to C and A or B.
 
BILLY101

BILLY101

Member
Location
Telford, Pa
The best way to approach this is to work with the KVA rating to match loads with transformer capacities.
In this case the load is 208 volts x 37 amps = 7.7 kva, 1 phase.
The capacity of a 20 kva transformer is 20 kva / 3 phases = 6.67 kva per phase.
The capacity of a 30 kva transformer is 30 kva / 3 phases = 10 kva per phase.
Select the 30 kva transformer for this application.
Better knowledge of the facility may reveal other options such as using (2) 1 phase 10 kva transformers instead of a single 3 phase transformer.
This would depend on how the 480 volt system is distributed and the location of the 1 phase loads.

BILLY
 
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skeshesh

Senior Member
Location
Los Angeles, Ca
Which phases are you connecting the loads to?

check this recent thread: http://forums.mikeholt.com/showthread.php?t=132012

I suggested a very simplified method in that post, but as David commented (correctly), you will need to use phase current/angles (assuming same P.F. or actual angles unless you have means of measuring) and KCL to arrive at exact answers. Perhaps post of an attempt at calculations? If you have no idea let us know and I'll post some numbers but you should really look up the calcs, complex numbers, etc. that you may need in that calc if you dont remember/havent done it before. It may look difficult if you havent tried it before but just crunch the numbers a few times and you'll get it.
 
augie47

augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
skeshesh said:
Which phases are you connecting the loads to?

check this recent thread: http://forums.mikeholt.com/showthread.php?t=132012

I suggested a very simplified method in that post, but as David commented (correctly), you will need to use phase current/angles (assuming same P.F. or actual angles unless you have means of measuring) and KCL to arrive at exact answers. Perhaps post of an attempt at calculations? If you have no idea let us know and I'll post some numbers but you should really look up the calcs, complex numbers, etc. that you may need in that calc if you dont remember/havent done it before. It may look difficult if you havent tried it before but just crunch the numbers a few times and you'll get it.
Click to expand...

From an "in the field (simple)" standpoint, is there any problem with using Billy101's (post #8) method. Since transformers are normally available in specific kva sizes, would that method not work for quick selection ?
 
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jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
augie47 said:
From an "in the field (simple)" standpoint, is there any problem with using Billy101's (post #8) method. Since transformers are normally available in specific kva sizes, would that method not work for quick selection ?
Click to expand...
It is what I do.
 
D

dkarst

Senior Member
Location
Minnesota
I think the original poster specified he had TWO of these units. Need to keep that in mind if the same transformer is supplying both.
 
Smart $

Smart $

Esteemed Member
Location
Ohio
Mike32186 said:
Hey everyone,

This is my first time on Mike Holt's forum. ...
Click to expand...

Welcome :)

If you use a 3? (wye-secondary) xfmr, the common winding will see:
37A ? √3 ? 120V = 7.69kVA​
Multiply that times 3 for your minimum rating...
7.69kVA ? 3 = 23kVA​
Essentially, each winding must be capable of one unit's kVA.
 
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S

skeshesh

Senior Member
Location
Los Angeles, Ca
augie47 said:
From an "in the field (simple)" standpoint, is there any problem with using Billy101's (post #8) method. Since transformers are normally available in specific kva sizes, would that method not work for quick selection ?
Click to expand...

Augie, Peter and Jim: no I don't disagree. I was just pointing him towards solving for values rather than working backwards - my mistake, should've made it clear that I'm not trying to say the suggested approach was wrong.
 
B

BJ Conner

Senior Member
Location
97006
Transformer connections

Transformer connections

IF the loads are hooked up A-B and B-C The currents are A= 37, B = 37 and C= 64.1.
1.) The transformer has to have a KVA > 15.4 (2 x 7.7)
and
2.) AND a kVA of greater than 7.69 KVA ( The size of transformer to have a secondary lead rated at 64.1 amps, 64.1 x120 x 3^2= 7,690va).

Condition 1 rules as all 15.4 KVA transformers are greater than 7.69 KVA.
The next stop is the catalogue where we find standard transformer ratings are 15, 30, 45 etc KVA.
So the answer is 30 KVA
Unless we wanted to buy an 80 Degree C rated 15 KVA transformer and overload it by 2.6%.
 
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suemarkp

Senior Member
Location
Kent, WA
Occupation
Retired Engineer
Um, aren't the currents in A/B/C = 34 / 59 /34 ?

Also curious as to what value the OP was using for amps. FLA, MCA, RLA, something else?
 
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