# Three Phase Transformers - Primary/Secondary Calculations



## JasonCo (Mar 23, 2015)

Hello ET, this week is by far the toughest week of school I've ever had to face. We are now getting into calculations for three phase transformers on the primary and secondary sides of the transformer. I am totally lost though! I think I understand the primary side, its basically just simple ohms law but if its three phase then 1.732 gets added into the equations accordingly. This is correct?

And for the secondary side, this is where I am just incredibly lost at. I have questions that ask me what is the amps on the coil or on the line and etc... I will give a couple examples of some of the questions I have. 

*QUESTION 1:* A 480(Delta) --- 240-volt 3-wire (Delta) transformer is delivering power to a 3phase load and has 75 amps of coil current flowing through each winding. The current flowing in each circuit conductor to the load is ___ amps. 

*QUESTION 2: SOLVED in comments* A 480-volt to 240-volt three-phase transformer is rated for 112.5KVA. What is the maximum available secondary line current on B phase?

*QUESTION 3: SOLVED in comments* Scenario 220.21. A 3phase transformer bank supplies 120/240 volts to a 3phase, 4-wire panelboard. There is a 30kVA 3phase load connected to a 3-pole breaker in spaces 1-3-5. There is a 240-volt 15 kW load connected to a 2-pole breaker in spaces 2-4. There is a 120-volt 10kW load connected to a 1-pole breaker in space 6. 

Under Scenario 220.21, I have questions like finding the amps for the A-phase and C-phase feeder conductors, along with how many amps is on the neutral.

*QUESTION 4: SOLVED in comments* Scenario 220.23. A closed-delta bank of transformers delivers 120/240-volt power to several loads. The 1-phase loads are drawing 120 amps on A-phase and 120 amps on C-phase. The bank is also supplying power to a 3phase load that draws 46 amps. 

Under Scenario 220.23, I have questions like finding the calculated size required for the "power" transformer(s) is ___ KVA, or the calculated size required for the "lighting" transformer(s) has a rating of ___ KVA. 



Our teacher gave us 100 question homework like always. This is just 4 questions out of the 100, so I am really trying to understand this, I just don't at all... I have spent over an hour every day for the last 5 days trying to slowly get this done, but have only done all the questions that don't require calculations which is about 30 or so questions.. ET is my last resort, I really need some help! If you made it this far then I really appreciate your time, thanks and hope to hear from some of you!


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## RePhase277 (Feb 5, 2008)

In a delta, winding current is equal to line current/1.732

And the power out is equal to the power in. In other words,

(Primary volts x primary amps) = (secondary volts x secondary amps)


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## JasonCo (Mar 23, 2015)

InPhase277 said:


> In a delta, winding current is equal to line current/1.732
> 
> And the power out is equal to the power in. In other words:
> 
> kVA/primary volts = kVA/secondary volts


I appreciate that, so the Line current on the secondary side of a transformer is higher than on the coils. By a factor of 1.732 either multiplying or dividing to get the answer. What about for volts and watts?

Edit: Okay so I did some researching, looks like once you have the VA(Watts), if you are trying to find how many watts each coil has, you just divide by 3 because there are 3 different coils in a 3 phase transformer. And for voltage, volts line = volts coil.

Edit #2: Okay what if the transformer is 120/240 volts, the system has coil (or winding) voltages that are ___ volts? Would it be 120 or 240

Edit #3: Found it, answer is 240


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## RePhase277 (Feb 5, 2008)

It would depend on the type of 120/240 transformer. Either single phase, or 3 phase center tapped delta.


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## JasonCo (Mar 23, 2015)

Figured out how to solve *QUESTION 4*. Will Edit this post when I'm done typing it all out in detail

Edit: To find the answer to the "power" KVA of that question. This is what you do
Formula: "power" = [(I_line 3p) / 1.732] + [(1/3) x I_line 1p)] x [240].

So I_line 1phase would be 120amps. Because A phase and C phase have 120 amps each. In a question where you have different amps for each phase, just take the larger of the two. In this case, 120 amps will be plugged into I_line 1p. For I_line 3p, you just take the subtotal of the amps which is 46 amps, because each phase of the 3phase carries 46 amps on it. 

So the formula will look like this:

"power" = *[(46) / 1.732] + [(1/3) x 120)] x [240]*
which gives you an answer of!!! 15974 = *16KVA*.

---------------------------------------------

Now to find the "lighting" transformer KVA. This is what you would do:

Formula you would use is: "lighting" = [(I_line 3p) / 1.732)] + [(2/3) x (I_line 1p)] x [240]

So same principles apply as the last part, just the formula is a bit different.

*"lighting" = [(46) / 1.732)] + [(2/3) x (120)] x [240] =*
The answer comes out to!!! 25574 = *26 KVA*


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## JasonCo (Mar 23, 2015)

Okay found how to find solve the answer to Question 2. Will edit this post with a detailed response

Answer to *QUESTION 2: *

First you must find the maximum available secondary LINE current. Fomula would be: 
*112500 / (240 x 1.732) = 270.6*
Now we are trying to find the B-phase coil current. Which is
*270.6 / 1.732 = 156.3 Amps* running through the line current on B-phase.


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## JasonCo (Mar 23, 2015)

Found how to solve *QUESTION 3*, will update this comment with the detailed answer

Please correct me if I'm wrong but I think this is how it goes...

So there is a 3-phase load that draws 72 amps. so A-phase has 72 amp, B-phase has 72 amp and C-phase has 72 amps. Now there is a 2 pole breaker in slots 2 - 4 (A and B phase) and it draws 62.5 amps. So A-phase gets an additional 62.5 amps added to the 72 amps and B-phase gets the same. So now A phase has 134.7 amps and so does B-phase so far. and C-phase currently has 72 amps. Now the last part of the question says that space 6 (C phase) draws another 83.33 amp load. So when you add all these together. *A phase draws 134.7 amps*. *B phase draws 134.7 amps*. *C phase draws 155.33 amps*. 

Okay, one thing I don't understand is how much the neutral will be carrying? Can someone please help me figure this out. I'd say you add them all up but that is not one of the answer choices on my homework  If each phase carries different amp loads on a panel, what will the neutral amp load be?


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## bkmichael65 (Mar 25, 2013)

JasonCo said:


> Figured out how to solve *QUESTION 4*. Will Edit this post when I'm done typing it all out in detail
> 
> Edit: To find the answer to the "power" KVA of that question. This is what you do
> Formula: "power" = [(I_line 3p) / 1.732] + [(1/3) x I_line 1p)] x [240].
> ...


Let's check one of those to see if you're right. The last one you came up with 25574VA. Divide that by 240 volts and 1.732 and you have an available line current of 61.5 amps on each phase. I don't think that's going to work


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## JasonCo (Mar 23, 2015)

bkmichael65 said:


> Let's check one of those to see if you're right. The last one you came up with 25574VA. Divide that by 240 volts and 1.732 and you have an available line current of 61.5 amps on each phase. I don't think that's going to work


I think its going to have unbalanced-load. Idk at this point my mind is fried! Gotta go to bed anyways, will pick this up tomorrow though! But I believe this is the answer, I feel quite sure


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## RePhase277 (Feb 5, 2008)

The neutral will only carry as much current as the single phase 120 volt equipment connected to it draws. The current in a circuit is the the same going out as it is going in.


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## JasonCo (Mar 23, 2015)

InPhase277 said:


> The neutral will only carry as much current as the single phase 120 volt equipment connected to it draws. The current in a circuit is the the same going out as it is going in.


So the answer would be 83.3 amps? Because in the question, it takes about a single phase load, and that load is carrying 83.3 amps. That single phase load will also be my neutral load in a unbalanced system?


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## RePhase277 (Feb 5, 2008)

JasonCo said:


> So the answer would be 83.3 amps? Because in the question, it takes about a single phase load, and that load is carrying 83.3 amps. That single phase load will also be my neutral load in a unbalanced system?



Without a load attached to it, the neutral would just be a wire dangling in the wind, and therfore could not have a current on it. But since you have an 83.3 amp line to neutral load, the neutral MUST have 83.3 amps on it.


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## JasonCo (Mar 23, 2015)

InPhase277 said:


> Without a load attached to it, the neutral would just be a wire dangling in the wind, and therfore could not have a current on it. But since you have an 83.3 amp line to neutral load, the neutral MUST have 83.3 amps on it.


Well, my A phase load has 134.5 amps. B phase load has 134.5 amps on it. And C phase load has 155.33 amps on it. After working out the problem this is the load answers I came up with. Now it looks like I have an unbalanced load, and to find the current load on my neutral all I do is take my 1-pole breaker load and find the amps on it (which is 83.33 amps) and that is my answer? You are sure about this? Was just about to post a thread asking for help on this specific problem, because it is blowing my mind away. I have searched the internet for it seams like hours and no one yet has a answer it seams o_0


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## telsa (May 22, 2015)

I have to ask...

Do you have your own copy of Ugly's ?

An awful lot of your queries are answered in it's tables.


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## RePhase277 (Feb 5, 2008)

If there is a 10 kW load connected to the neutral, then the neutral current is 10000/120 = 83.33333333333 amps.


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## JasonCo (Mar 23, 2015)

telsa said:


> I have to ask...
> 
> Do you have your own copy of Ugly's ?
> 
> An awful lot of your queries are answered in it's tables.


No I actually don't, I should probably get that book. But I have a feeling Ugly's doesn't tell you how to find the neutral current of an unbalanced 3 phase load on a panel, I have scowered the internet and no one has an answer... The only equation anyone ever talked about was (A^2 + B^2 + C^2) - (AxB + AxC + CxB).. But I tried this equation and it doesn't work for me, none of my answer choices are close to this equation. Plus this equation doesn't apply to loads that have different watts. Literally spent hours on this question, still can't figure it out...


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## JasonCo (Mar 23, 2015)

InPhase277 said:


> If there is a 10 kW load connected to the neutral, then the neutral current is 10000/120 = 83.33333333333 amps.


I understand, but there is also a 30,000 watt and a 15,000 watt loads also connected to the neutral current. Along with the 10,000... So... There are many loads on this neutral with different amps on it. All 3 phases (A,B,C) have different amp loads on it. Unbalanced load


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## RePhase277 (Feb 5, 2008)

JasonCo said:


> I understand, but there is also a 30,000 watt and a 15,000 watt loads also connected to the neutral current. Along with the 10,000... So... There are many loads on this neutral with different amps on it. All 3 phases (A,B,C) have different amp loads on it. Unbalanced load


:001_huh: No there's not. Not according to the question you posted.



> Scenario 220.21. A 3phase transformer bank supplies 120/240 volts to a 3phase, 4-wire panelboard. There is a 30kVA 3phase load connected to a 3-pole breaker in spaces 1-3-5. There is a 240-volt 15 kW load connected to a 2-pole breaker in spaces 2-4. There is a 120-volt 10kW load connected to a 1-pole breaker in space 6.*
> 
> Under Scenario 220.21, I have questions like finding the amps for the A-phase and C-phase feeder conductors, along with how many amps is on the neutral.


There's a three phase 240 V, 30 kVA. There's a single phase 240 V, 15 kW load, and a single phase 120 V, 10 kW load. Only one of them is connected to neutral and therefore only one contributes to neutral current.


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## telsa (May 22, 2015)

http://www.amazon.com/Schaums-Outline-Circuit-Analysis-Outlines/dp/0071756434/ref=pd_sim_14_1?ie=UTF8&dpID=51XIHy7bL2L&dpSrc=sims&preST=_AC_UL160_SR120%2C160_&refRID=16CG6JTWQ4BV3S18TXSD

Schaums has been printing "problems solved" manuals for decades.

MOST of the test questions that you're dealing with are CLASSIC Schaum's problems.

Decades ago, in college, I lent my Schaum's on Physics to the fellas.

Three out of their four take home questions had been directly cribbed from Schaum's.

This is STILL happening.

See your own queries.

Because of the rise of digital circuits, I can't tell which Schaum's is appropriate for power circuits.

You'd just as well buy them all.

Purchase them second-hand via Amazon.

They are paperback// tradeback -- and awfully cheap.


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## JasonCo (Mar 23, 2015)

InPhase277 said:


> :001_huh: No there's not. Not according to the question you posted.
> 
> 
> 
> There's a three phase 240 V, 30 kVA. There's a single phase 240 V, 15 kW load, and a single phase 120 V, 10 kW load. Only one of them is connected to neutral and therefore only one contributes to neutral current.


I thought a 2-pole breaker could have a neutral and ground as well? I've installed plugs before that had 2 hots a neutral and a ground. Last week I did one, it was a plug that charges someones electric car.


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## RePhase277 (Feb 5, 2008)

JasonCo said:


> I thought a 2-pole breaker could have a neutral and ground as well? I've installed plugs before that had 2 hots a neutral and a ground.


You can have a load that uses 2 hots and a neutral, but then the question would have said "120/240 V, 15 kW load" instead of just "240 V".


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## RePhase277 (Feb 5, 2008)

JasonCo said:


> No I actually don't, I should probably get that book. But I have a feeling Ugly's doesn't tell you how to find the neutral current of an unbalanced 3 phase load on a panel, I have scowered the internet and no one has an answer... The only equation anyone ever talked about was (A^2 + B^2 + C^2) - (AxB + AxC + CxB).. But I tried this equation and it doesn't work for me, none of my answer choices are close to this equation. Plus this equation doesn't apply to loads that have different watts. Literally spent hours on this question, still can't figure it out...


That equation is for the neutral current in a WYE connected system. You have a delta connection with a center tap that provides a neutral on one phase. The current on this neutral is equal to the unbalanced load between the ends of that phase.


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## telsa (May 22, 2015)

JasonCo said:


> I thought a 2-pole breaker could have a neutral and ground as well? I've installed plugs before that had 2 hots a neutral and a ground.


The example is *idealized.*

A 240V load -- in idealized test questions ALWAYS means that it has no neutral current.

In idealized test questions you are never to impute alternate possibilities.

The students are stumped enough as it is.

&&&&&&&

Much of the information that you're taught in the first level of electrical circuits is demonstrably false.

BUT, the false part is going to be addressed AFTER you've got the simpler basics down pat.

The false part is either a second or third level effect -- or is a time dependent effect that drops out in seconds or less.

Since the mathematics of these effects is quite demanding -- they are idealized away.

You can design and install practical circuits to the end of your career and never have to adjust for such second or third level effects.

Instead, tables and standard values suffice.

Ugly's is loaded with all of the most common ones.

Plumbers, HVAC, and structural steel is installed likewise: by tables of values.

( Seen in the movie: Bullet. Steve McQueen has to assist his lover with her hydraulic 'look up' tables.)


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## JasonCo (Mar 23, 2015)

InPhase277 said:


> You can have a load that uses 2 hots and a neutral, but then the question would have said "120/240 V, 15 kW load" instead of just "240 V".


Okay awesome, finally I understand this problem thanks to you. Wow... You are a genius, huge thanks! I'm sort of piecing it all together now, really started to get it. Thanks a bunch, really.. Huge help


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## RePhase277 (Feb 5, 2008)

There's more to the neutral current problem. But I'm trying to let you piece it together. One clue is the neutral current is equal to the unbalance of the currents on the phase with the neutral. Another clue is the breaker spaces are numbered. This gives you a clue to the phase relationships of the currents.

Almost there.


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## telsa (May 22, 2015)

A perfect example of a second order effect in our trade is the 'skin effect.'

We never calculate it.

We use ampacity tables ( NEC// Ugly's ) that ALREADY incorporate the 'skin effect.'

Actually calculating it and testing for it -- is a task for PhD EEs at a national testing lab.

It gets complicated very, very, fast.

The same thing is true for heat dissipation in a trench.

And so forth.

&&&&&

In idealized test questions -- the only ones you're going to get -- always assume the simplest case.

Always sketch the circuit. You're never going to get fast results without a sketch.

It's what the pros do. :thumbup:


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## JasonCo (Mar 23, 2015)

InPhase277 said:


> There's more to the neutral current problem. But I'm trying to let you piece it together. One clue is the neutral current is equal to the unbalance of the currents on the phase with the neutral. Another clue is the breaker spaces are numbered. This gives you a clue to the phase relationships of the currents.
> 
> Almost there.


Could I ask one last thing. If I have taps set at position 2 on a 30KVA transformer. Position 2 is 492 Primary Volt. The line-line secondary voltage will be ___ volts if the primary is connected to a 458-volt line?

Really confused about this, unsure if you might know but thought I'd ask!


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## telsa (May 22, 2015)

It's ANOTHER idealized question.

As tapped, 492 V would've generated an idealized 208Y120 Wye output.

IF it were receiving the specified 492 Voltage.

But...

In the real world, the Poco was providing 458 Volts.

Since the taps have not changed.

Then the Wye output HAS TO DROP.

What would that ratio of dropping be ??

Would it be the same ratio as the inputs were to each other ??

Hint, hint. :whistling2:


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## JasonCo (Mar 23, 2015)

telsa said:


> It's ANOTHER idealized question.
> 
> As tapped, 492 V would've generated an idealized 208Y120 Wye output.
> 
> ...


Really appreciate all your help! I got to hit the sack now, will defiantly jump right back onto this thread tomorrow and will really think over what you are saying. Right now though I must sleep! Will answer tomorrow, thanks again


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## hardworkingstiff (Jan 22, 2007)

redundant.


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## RePhase277 (Feb 5, 2008)

JasonCo said:


> Could I ask one last thing. If I have taps set at position 2 on a 30KVA transformer. Position 2 is 492 Primary Volt. The line-line secondary voltage will be ___ volts if the primary is connected to a 458-volt line?
> 
> Really confused about this, unsure if you might know but thought I'd ask!


You know the turns ratio of the transformer, which is just the rated primary voltage divided by the rated secondary voltage. This is a physical quantity and doesn't change. Knowing this, the output voltage will vary in relation to the input voltage by that same ratio. Throwing in the tap connection just adds a layer of difficulty but it is the same principle.

I'll get you started.

If the primary is at 492 volts and the secondary output is 208 volts, the turns ratio is 492/208 = 2.365:1.... So if you put 458 volts in.....


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## JasonCo (Mar 23, 2015)

Ah! Okay so its just simple ratio. Thanks a bunch guys! Answer = 194v
(458 x 208) / 492 = 194v

Edit... Wait how do I know my secondary is 208. If I look at a group H transformer in my school book, it points to diagram 26 in my book. Diagram 26 is Primary: 480 volts Delta and Secondary: 240 Volt Delta/ 120 Volts

I know tap 2 plays a big part in this question, just not sure how that gets you 208

Edit: Also none of my answer choices are close to 194v


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## telsa (May 22, 2015)

I assumed that the test question was for the ever common dry-type delta-wye transform.

If the transform was for delta-delta 480 to 240

Then...

Tap #2 492 V would lead to an idealized 240 V across the hots of the secondary.

So use

458Vx240V/ 492V = 223.4 V 

About 93% of the target value of 240 Volts.


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## JasonCo (Mar 23, 2015)

telsa said:


> I assumed that the test question was for the ever common dry-type delta-wye transform.
> 
> If the transform was for delta-delta 480 to 240
> 
> ...


Yes you are 100% right, just got back from class and went over it. Thanks again, you have been a great help. Will have to go back and Thank your comments, appreciate it


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