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Joined: Mar 2004
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I have a couple of buckboost transformers that I think are grossly undersized, and I'm hoping someone can check my math:
Two 3kVA, 600V, deltaconnected autotransformers should have a line current of 5 amps, correct?
3000VA * 2 * 0.87 (because it's an open delta) / 1.73 / 600V primary = 5.0A
Thanks a bunch.
John




Joined: Feb 2003
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I have a couple of buckboost transformers that I think are grossly undersized, and I'm hoping someone can check my math:
Two 3kVA, 600V, deltaconnected autotransformers should have a line current of 5 amps, correct?
3000VA * 2 * 0.87 (because it's an open delta) / 1.73 / 600V primary = 5.0A
Thanks a bunch.
John BigJohn., I came up complety diffrent number than your numbers. Here is my verison.,, 3000VA X .87 ÷ 1.73 ÷ 600 = 2.51 amps The 1.73 is allready taken care for triphase { it dont matter if open delta or close delta or wye format } But just wondering why you add X 2 in there so if you try that again like my formaila and also I did used the electrician caluacator to see and the answer and it came pretty close it show 2.8amps on that one. Merci,Marc
Pas de problme,il marche n'estce pas?"(No problem, it works doesn't it?)




Joined: Jul 2002
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Guys, Is there any allowance for the actual impedance ratio of the transformer(s)? BTW, this is a percentage, could be as low as 2.3%, could be as high as 8.0%, per transformer.




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Mike.,
Normally for B/B transfomer I don't useally invoke the impedance ratio not very often unless you are dealing with very large B/B set up.
BigJohn.,
I forgot to add to this the B/B is undersized by design due the secondary part will change the voltage by very small step so you can really load a bit however double check with the manufacters for specs. Merci,Marc
Pas de problme,il marche n'estce pas?"(No problem, it works doesn't it?)




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But just wondering why you add X 2 in there... Marc, I multiplied by two because I was adding the kVA of each transformer... which I now don't think is right. So, these things are supposed to be able to handle 2.5 amps... but somehow they've been running a 10HP 480V pump for days and days. I have to go look and see how they're holding up, but by all rights they should have been smoked a long time ago... right? The more I think about this the less I understand. John




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BigJohn,. If you have time check out this link { it is in PDF file format } Buck/Boost transfomer { PDF } It have the chart that you can able tell if you have correct one for this appaction. Merci,Marc
Pas de problme,il marche n'estce pas?"(No problem, it works doesn't it?)




Joined: Jul 2001
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John, you are correct.
Another way to look at it is: An open delta transformer has 57.7% of the capacity of a closed delta. If you had 3 transformers it would be equal to a 9KVA unit. With an open delta it is effectively a 5.19kVA transformer bank; input amps would be about 5.0A @ 600V
3*57.7% = 2*86.6%




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Alright, this buck boost business is still killing me: So, I have 5220VA available. I'm bucking down 120 volts(600V down to 480V). For some mysterious reason, I only use the bucking voltage when calculating my buck/boost capacity: So, 5220VA/120V = 43.5A of capacity on two opendelta 3kVA buck boost. Now, is that right, or do I still divide 43.5A by 1.73 to get 25.1A? John




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BigJohn;
Please pardon the latency in my delay.
The key factor for sizing a Voltage Buck/Boost Autotransformer arrangement, is the capacity of the _Secondary Winding(s)_  in Amperes.
Two examples:
1: Load requires 5.0 KVA @ 230 Volts, supply Voltage is 208 Volts:
a: 230V  208V = 22 Volts difference. b: Most common Secondary Voltage for small Transformer = 24 Volts. c: Voltage output will be 232 Volts for this arrangement. d: 5,000 VA / 232 Volts = 21.56 Amps. e: Minimum Ampacity for Secondary Winding = 21.56 Amps. f: 21.56 Amps x 24 Volts = 518 VA. g: Minimum Secondary Winding capacity = 518 VA (Transformer rating) h: Closest common size of Transformer = 0.750 KVA (750 VA), which is good for 31.25 Amps through the Secondary Winding(s).

2: Load requires 7.5 KVA @ 208 Volts, supply Voltage is 240 Volts:
a: 240V  208V = 32 Volts difference. b: Most common Secondary Voltage for small Transformer = 32 Volts. c: Voltage output will be 208 Volts for this arrangement. d: 7,500 VA / 208 Volts = 36.06 Amps. e: Minimum Ampacity for Secondary Winding = 36.06 Amps. f: 36.06 Amps x 32 Volts = 1,154 VA. g: Minimum Secondary Winding capacity = 1,154 VA (Transformer rating) h: Closest common size of Transformer = 1.5 KVA (1,500 VA), which is good for 46.88 Amps through the Secondary Winding(s).
 
Now for your scenario.
If the Load requirement is 5220 VA _PER PHASE_, the 3 Phase VoltAmps will be 15,660.
If the _TOTAL 3 PHASE LOAD_ is 5220 VA, the PerPhase Load will be 1740 VA.
Let's use the 5220 VA PerPhase Load requirement for this discussion...
Your Load requires an input Voltage of 480 Volts, 3 Phase 3 Wire. The existing supply Voltage is 600 Volts 3 Phase.
Need to setup the "Autotransformer" connections to "Buck" the Voltage down by 120 Volts, so the output from the Transformer(s) will be capable of supplying no less than 5220 VA @ 480 Volts PerPhase.
5220 VA / 480 Volts = 10.875 Amps. The Transformer's Secondary Winding(s) must be able to carry at least 10.875 Amps.
10.875 Amps x 120 Volts = 1,305 VA.
Minimum Transformer rating = 1,305 VA (1.3 KVA).
A 2.0 KVA 600V x 120V Transformer has a Secondary Ampacity of 16.67 Amps.
A 3.0 KVA 600V x 120V Transformer has a Secondary Ampacity of 25.0 Amps.
Connect Two of the Transformers in an OpenDelta arrangement, with the Windings of each Transformer setup in a SeriesSubtracting connection.
The 2.0 KVA Transformers would be sufficient, as the maximum Amperes that could be drawn through the Secondaries is 16.67 Amps, and the Load's maximum Amps is 10.875 A.
The 3.0 KVA Transformers would be good for 25.0 Amps per Phase, which may be a better choice (if within budget).
Good luck.
Scott
Scott " 35 " Thompson Just Say NO To Green Eggs And Ham!




Joined: Mar 2004
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Well, it's good to know I was on the right track even if I don't really understand why. Thanks for the replies!
Is there some sort of shortcut to determining how much current the secondary of an autotransformer is designed to handle? How do I determine that from the nameplate information on an buck/boost?
John



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