Open Delta Reduced Capacity
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- เผยแพร่เมื่อ 23 ส.ค. 2024
- In this video i prove the reduced capacity of an open Delta transformer and discuss where the two Open Delta percentages come from.
*Edit. Formulas at end should be reversed. Should be SqRt3 /3 =.577 or 57.7%, and SqRt3/2 is .866 or 86.6%
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I’m most impressed by how you’re able to write backwards
Absolute gold nugget on the topic and much needed.
Thank you for the encouraging words!
Red Seal test in a week. Glad I found you Sir!
In some areas in the USA open high leg delta systems are very common in rural areas, especially farms, where the majority of the loads are your typical split phase 120/240 , but you also have some 3 phase motors such as for augers, pumps and grain dryers. In these cases the capacity of a closed delta is not warranted, and the utility companies seem to love that because they only have to add one transformer vs. two and cuts down on costs.
I have heard railyards also commonly use them! Thanks for the info, I was unaware that they were common in rural areas, but will share that with folks as examples!
@@schulerruler I've drove by probably a dozen or so of them, mainly in southeast Michigan I've seen a few in Tennessee, Kentucky, and South Carolina, which can easily be identified by a pole with two transformers, one being much larger than the other. I don't know whether the are common per se, but every single open high leg delta that I've come across was on a farm setting, if they have larger motors to power. But then you'd question yourself whether it would be more economical to install a rotary phase converter for the 3 phase equipment you want to run. I guess it's a situational thing.
I'm in west Texas (retired electrician, but still doing side-work) and run in to this exact scenario A LOT, particularly on larger ranches & farms, and also in light industrial & commercial locations as well.
As you mention, the utility likes to build the open-delta xfmr banks, because it's less expensive (2 vs 3 xfmr's), and only two HV phase-legs need be brought in on the high-side, for additional savings (of time and $$).
3/sqrt (3) = 1.73 and 2/sqrt (3)=1.15. The correct relation can be 1/sqrt (3)=0.58, because the current reduces by sqrt (3).
Ok. Now I see the description. By the way, Good explanation!
@@mstabilito It's sqrt(3)/3=0.577 & sqrt(3)/2=0.866. But yeah, this is very helpful in getting the transformer ratings for 3-phase loads to be connected in v-ph lines.
Excellent video sir. Can you make a tutorial about short circuit analysis sir? Thank you.
God bless you.
Excellent explanation... Thanks a lot , Sir
Thank you for the support!
Awesome video Chris
What if one of the transformers is a different size? Such as one being a 15kva and the other a 10kva?
I don't understand your math at the end. 3/sqrt(3)= sqrt(3) ~1.73 not .577; and 2/sqrt3=1.15 not .866 Could someone please explain these pink equations.
Would the same hold true with a 3phase genset with an open delta? The reason for the question is converting the 3 phases into a single phase/split phase by making a zigzag connection from all 3 phases. In other words, creating an open delta for 1 leg of the split phase and then taking the remaining phase and putting it in series with the open delta for the other leg of the split phase, which would look like this, /\_. L1 /\ N _ L2
When a 3-phase motor is connected to a delta power supply and rotate at full load, will it work if one of power supply legs opens during loaded condition ? What will be the scenario?
Thanks it was helpful
Thanks! How about open wye (star) reduced capacity? How do you get the 50% reduction?
Thank you sir
What will be happen when take current from terminal (B) instead of terminal (A)..? Is line B current increase ?
It will work out the same.
@@schulerruler why current is not increase at terminal B when both phases ( A&B )are healthy ( not disconnected) . As the previous case of balanced delta
@@sheetalsharma1466 Current is flowing BETWEEN B and some other line, no? Either A or C. A and C are both at reduced capacity. This is the circular wizardry of deltas.
Is it just me or you think this guy is awesome at writing oppisite?
Its a video set up (very $$$) that allows one to write on a board (special) with cameras that make it look like they are looking at you and writing properly, its called technology
Hi!!
That is automatic , software managed task !!
Respected sir,
Open Delta is important to supply current to "transformer to 3wire connecting circuit " Method (houshing board circuit device) method
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@Cash Major glad I could help :D
great explanation; but the 3 divide by root 3 is incorrect, so does 2 by root 3
They should both be the other way around for the formula to be correct. I just wrote them down wrong.
how do you write backwards on that glass so well?
Haha trick of the camera! I'm actually writing normally and the camera inverts the image 180°. Hopefully that doesn't ruin the magic.
helpful af, thanks!
If 1000kVA Dyn11 transformer primary (delta side ) is connected to a 33kv distribution line...if a jumper opens (ungrounded) in 1 of the phase, what will be the capacity of the transformer ?..57.7?..pls clarify..thanks in advance.
I’m assuming open jumper in 1phase means it’s inside the delta and all 3 line jumpers are in tact. Then it follows the training in video, 1000Kva x .577= 577Kva. If a line jumper opens this bank goes to single phase and no 3 phase can be delivered.
i hope you get a free 3 floir cheese burger for this, thank you
So good! Very helpful! Thank you!!!!
Thank you and I'm glad they were helpful!
hardly a proof... just a bunch of formulas. should actually explain that the line current on open delta is offset by 30 degrees from voltage across transformer
Greate
Thanks for watching!
You lost me. You seem to skip over or assume too much. Near the end I couldn't follow you reduced capacity calculations. You talked about reduced capacity from 2 coils? What does that mean? The open delta is a closed delta with one transformer removed. So what does it mean to talk about reduced capacity from two coils? I don't get it.