208v to 240v Buck Boost Transformer Calculations

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topgone

Senior Member
This link to Acmes catalogue discusses the kVA relation in detail:

http://www.delzer.com/powerproducts/ACM_CAT_005_0412/index.html#87

The 1.5 kVA rating is only accurate when connected as an insulation transformer. When installed as an autotransformer the kVA rating is 5-10 X greater and the 1.5 has no relation to the autotransformer any longer, only for naming purposes as Acme brings out.
The thing is, for an autotransformer, the autotransformer rating is = (load power x [series winding voltage]/[series winding voltage + common winding voltage].

Others call that the "apparent rating", knowing a portion of the power delivered to the load is "conducted" while the remaining power is transferred "magnetically".

In this case the factor will be (32)/(208+32) = 0.1333 or you divide the load power by 7.5 to get the rating of the autotransformer!
 

Sahib

Senior Member
Location
India
And you do that by taking the boost or buck voltage and the load current ....the product of those two is the required VA of the buck boost transformer. You then select a transformer with an equal or higher VA rating.
Based on above, the Tr size is 1.5 KVA in OP case. Now is size of line side O/L protection device for the Tr based on 1.50KVA?
 
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iwire

Moderator
Staff member
Location
Massachusetts
Based on above, the Tr size is 1.5 KVA in OP case. Now is size of line side O/L protection device for the Tr based on 1.50KVA?
Still don't have to do what you are pushing.

In part

Each autotransformer 600 volts, nominal, or less shall be protected by an individual overcurrent device installed in series with each ungrounded input conductor. Such overcurrent device shall be rated or set at not more than 125 percent of the rated full-load input current of the autotransformer.
If I select the transformer as Don has suggested the overcurrent protection will be the same size as the equipment used.
 

Sahib

Senior Member
Location
India
Does line side O/L device, rated at 125% of full load current of Tr of 1.5 KVA, protect load side conductor also?
 

Sahib

Senior Member
Location
India
iwire:You are all wrong. The code clearly states O/L protection for input current to to auto transformer and not just the current taken by it.
 

iwire

Moderator
Staff member
Location
Massachusetts
Does line side O/L device, rated at 125% of full load current of Tr of 1.5 KVA, protect load side conductor also?
Yes

iwire:You are all wrong. The code clearly states O/L protection for input current to to auto transformer and not just the current taken by it.

Thanks for telling me I am doing it wrong, I will let the inspectors know that pass my installations. :D

Typical for me:

  • Expresso machine rated 230 volts, around 40 amps
  • Supply is 208 volts
  • Use a 0.75 kVA buck boost.
  • 50 amp breaker

Sahib, you really need to learn to listen to others, not just me but the engineers and others in this thread.
 

GoldDigger

Moderator
Staff member
If you were required to provide OCP appropriate to the primary winding draw of the autotransformer you would have to put it in the neutral leg or else open the internal connection between winding segments of the autotransformer converting it to an isolation transformer.
Fortunately the OCPD for the load also serves very nicely as secondary winding (segment) protection for the transformer. ;)
 

Ingenieur

Senior Member
Location
Earth
Good illustration. It will help the readers trying to learn. Not sure what you could do for those that refuse to learn.
Some like to be contrary
just the way it is
I'm not exempt but try to temper it

in my example
ocp is sized for 5.76 kva at 208 since it supplies the load directly and the xfmr
no brainier
but the transformer is still 0.77 kva not 5.76
and its prim/sec protection if desired is based on 0.77/208 or 0.77/32
NOT 5.76 kva

I'm guessing some members may not be married
once you are it gets easier to admit when you are wrong lol more practice I guess
 
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winnie

Senior Member
Once you get past the 'I can be unclear so that when you don't understand me I can call you names' I think that Sahib raises an interesting point. ( https://xkcd.com/169/ )

In an AT installation, the OCPD is set by the _total_ load, but clearly the transformer itself is only carrying part of the load. The 'rating multiplication' given by autotransformer use is approximately the voltage ratio; it is completely reasonable to see a 1.5kVA transformer being used to supply a 15kVA load.

The primary of the autotransformer is being protected with OCPD much greater than its rating.

NEC permits this, and these installations are well proven in the field. I am not attempting to argue otherwise, just exploring the way that it works with the _appearance_ of insufficient primary protection.

I see two ways in which this sort of installation is consistent with the rest of the NEC.

1) The primary conductors can be considered very short taps, so an oversized OCPD is considered sufficient for short circuit protection.

2) The secondary conductors _are_ protected against overload, by virtue of being in series with the supply OCPD. This is thus a single phase transformer with overload protection provided on the secondary side, and that secondary protection is sufficient in this case to protect the whole transformer from overload.

As I see it the autotransformer connection is reasonably protected, and providing primary overcurrent protection would be difficult, requiring some sort of overcurrent relay that trips both the primary and secondary OCPD.

-Jon
 

Ingenieur

Senior Member
Location
Earth
I don't think anyone disputes that the ocp is sized for the total load (direct + xfmr)
only the xfmr kva rating does not equal the input or load kva

since it is mostly direct connect it doesn't matter where the fault is
eg 15% boost the fault will be 208 or 240
since the ocp is sized for 208 at 240 the load side fault i will be higher
if buck it will be sized for 240 with potential faults at 240 or 208
at a 208 load side fault i will be 208/240 as much as line side
this typicaly is not a problem, slightly longer trip time

I did see it cause one once
600:300 AT
NGR 15 A sized for input gnd fault of 600 /1.732 /15A = 23 ohm
a fault on the load side only saw 300/1.732/23 = 7.5 A
with voltage drop and droop during a fault <7 A
the relaying was set at 7.5 A and did not trip consistently

this is not an issue with a ph fault on a solidly grounded system
 
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Sahib

Senior Member
Location
India
The autotransformer rating of 1.5 KVA is assigned to it by mistake from the fact that it is derived from a two winding, isolation transformer whose rating is 1.5 KVA actually.

However the code is still not violated by people like iwire because they provide line side O/L device size per the actual rating of the autotransformer (in this case for 11.25 KVA instead of 1.5KVA) and get their installations passed by the inspector.:)

The example 3.2 of slide 10 in the following presentation sheds more light on this issue.

http://r.search.yahoo.com/_ylt=A2oKmM31IN1WtCMA..TnHgx.;_ylu=X3oDMTByMjhxODA5BHNlYwNzcgRwb3MDMwRjb2xvA3NnMwR2dGlkAw--/RV=2/RE=1457361270/RO=10/RU=http://www.faculty.umassd.edu/xtras/catls/resources/binarydoc/3525.ppt/RK=0/RS=qY2t75HsFDqGIkaurnSKFVaR1KY-
 
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mivey

Senior Member
Transformer load from another manufacturer stating that most of the load is passed through (not a load on the transformer):

SolaHD said:
...only the secondary windings are transforming voltage and current. The majority of the kVA load passes directly from the supply to the load.
Sahib keeps trying to twist what Iwire said into something else and simply can't admit he was wrong for emoti-laughing at Iwire's original post. Just quit fooling around with his stubborness.
 

mivey

Senior Member
I think that Sahib raises an interesting point.
With all due respect winnie, the only thing Sahib is trying to do is to twist what Iwire said about the load on the transformer into something about the size load it can manage.

The rest is distraction to cover his original mistake.
 

Sahib

Senior Member
Location
India
mivey: Cool, Cool. This is just a disscusion. By the way you did not comment on example 3.2 of my last post, which shows how to calculate AT apparent power from 2 winding isolation Tr . Study it
 
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