Calculate required KVA for autotransformer in open delta

[haze10]

I have an application where I need to boost 480V 3ph to 600V 3ph. Total KVA required at 600V is 400KVA, or 385A @600V.

I've seen charts on manufacturer's websites to select the correct size KVA of each of the two single phase 480-120V transformers I will need to make an open delta connection, but not formulaes.

Square D rep said (xfrm1 KVA + xfrm KVA) x2 x 0.86 = KVA equivalent - but that doesn't make sense to me.

Does any one know the correct formula?

 

[stevenal]

Assuming the load is at unity pf, you'll find the current through the two transformers is leading by 30 degrees in one transformer and lagging by 30 in the other. .86 is approximately Cos(30) with a very small leeway for less than unity pf loading.

 

[fpietryga]

Divide your load kva (400) by 8.6 and this defines the kva requirement of each single phase transformer. The 385A load current must flow through the 120VAC winding of the single phase transformer being used as the autotransformer. Thus, 385 x 120 = 46,200 VA for each transformer.

 

[waross]

385 Amps x 120 Volts/1000 = 46.2 KVA
Use 50 KVA transformers.
Calculation:
If A phase is common, then B phase and C phase must carry the full load current. There is no other current path for B and C phase than through the transformer winding.
You give the load current as 385 Amps. This is the current that must pass through the 120 Volt winding.

Reality check:
A 50 KVA transformer will carry 416.7 Amps safely. That correlates well with 385 Amps, 120 Volts and 46.2 KVA.
I have used a number of open delta auto-transformers and this calculation has never let me down.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[haze10]

What I am trying to learn is a mathematical method to calculate KVA.
Square D has a chart, but its voltage dependant.
For example:
For 240 to 480 transformation Load KVA = 3.44 (KVA each single phase Xfrm)
For 480 to 600 " " " = 8.6 (KVA each single phase Xfrm)
For 380 to 480 transformation """""""""= 6.88 (KVA each single phase Xfrm)

It doesn't appear to be a simple as just taking the LV rated amps and plugging it into KVA = V x A x 1.73.
I don't know how to calculate it, but with a series winding in the A and C phase, the B phase just passing through.
For a symmetrical load (motor) current is going to be equal in all three legs. So while its easy to understand the current booast in the series windings, the B winding current booast must also be coming from the two series windings. This is why I'm thinking you can't just use rated LV current. But I'm not really sure so that is what I am trying to learn.
Obviously, Square D has different mutlipliers depending on voltage. But how is it calculated.

 

[Parchie]

@haze,
Here's the formula:
S(open-delta) = (load kVA) x (Series winding rating)/(Series winding rating + Common winding rating)

In your case:

Load kVA = 400
Series winding rating = 120V
Common winding rating = 480V​

Then:
S(open-delta) = 400 x (120)/(120+480) = 400 x 0.2 = 80 kVA. Use 100 kVA.
You divide 100 kVA by two, since you will be using two, 1-phase transformers--> 100/2 = 50 kVA!

 

[xnuke]

I used a different approach that doesn't rely on either current or voltage. Since I know that an open-delta configuration will have approx. 57.7% of the kVA rating of a delta configuration (1/√3), I simply took your 400 kVA requirement and multiply it by √3 to find the equivalent delta config rating of 700 kVA. Since that's the three phase total apparent power, I then divided that by 3 to find the per-phase transformer kVA rating, and since it's a boost autotransformer, I divided it by (1+a), where a is the turns ratio, so I divide by 5 (1+ 480/120) in your case. Thus:

S_{auto,per phase}=(S_{auto, three phase}*√3)/[3*(1+a)] which reduces to

S_{auto,per phase}=S_{auto, three phase}/[√3*(1+a)]

S_{auto,per phase}=400 kVA/[√3*(1+480/120)] = 46.2 kVA

Use two 50 kVA transformers as recommended above.

Note that the denominator in the above equation is approximately 8.6, which is the same as fpietryga's first calculation method, though no explanation was provided as to why 8.6 was chosen.


xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[haze10]

Appreciate the input.

The above formulas match the 8.6 constant provided by Square D.
It also works for the 3.44 constant they give for 240 to 480, where you would have a 240 to 240 primary/secondary transformer.
√3*(1+240/240)= 3.46 to the Square D 3.44.

But why doesn't it work on the 380 to 480, where you would have a 380 primary with a 100 secondary.
√3*(1+380/100)= 8.31 which does not match the Square D 6.88 constant.

For bucking applications, Square D has: Load KVA/8.3 = KVA each single pahse xfrm.
With a 575 primary and 95 secondary (575-95=480)
√3*(1+575/95) = 10.48 which doesn't match their 8.3.

I'm just trying to figure out how Square D derived these numbers. I've asked Square D for help, and two weeks ago they said they would, but now they claim with all the manufacturing reorganizations in their transformer group - they don't know who to go to. Scary.

 

[prc]

I think the estimate by waross is the correct one.

The way square D is estimating: When two single phase autos are connected in open delta connection, the net output is 86.6 % of the total line KVA of the single phase units.From line KVA to self kVA conversion in an auto,multiply the line kva by co-ratio (ie HV-LV/HV voltages)

Single phase 2 winding rating= (400/.8666)x1/2 x 600-480/600 = 46.2 KVA

 

[waross]

There are instances when a transformer is used at less than it's rated voltage to develop the required voltage.
Use the rated voltage of the transformer to determine the KVA required.
In open delta configuration each transformer is acting like a single phase transformer.
The boost winding must carry the full load current. The VA divided by the rated voltage of the secondary winding gives the maximum allowable current in the secondary winding. If you are using a 120 Volt winding in an application where it develops only 100 Volts, You must use the rated 120 Volts rather than the 100 actual volts to determine the minimum safe transformer size.
For step down you must differentiate between step-down auto-transformer and a buck connection. The effective ratios are different and the safe working current is different.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Parchie]

@haze,
If you tried Google, you could have found this webpage