Calculate required KVA for autotransformer in open delta
Calculate required KVA for autotransformer in open delta
(OP)
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?
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?






RE: Calculate required KVA for autotransformer in open delta
RE: Calculate required KVA for autotransformer in open delta
RE: Calculate required KVA for autotransformer in open delta
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
RE: Calculate required KVA for autotransformer in open delta
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.
RE: Calculate required KVA for autotransformer in open delta
Here's the formula:
S(open-delta) = (load kVA) x (Series winding rating)/(Series winding rating + Common winding rating)
Series winding rating = 120V
Common winding rating = 480V
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!
RE: Calculate required KVA for autotransformer in open delta
Sauto,per phase=(Sauto, three phase*√3)/[3*(1+a)] which reduces to
Sauto,per phase=Sauto, three phase/[√3*(1+a)]
Sauto,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
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RE: Calculate required KVA for autotransformer in open delta
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.
RE: Calculate required KVA for autotransformer in open delta
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
RE: Calculate required KVA for autotransformer in open delta
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
RE: Calculate required KVA for autotransformer in open delta
If you tried Google, you could have found this webpage