Auto Transformer Problem

[BebeshKing PE]

Hi, 

could anyone explain to me why the answer is 75kVA and not 100kVA?

thanks,

 

[Orchid PE]

Connected in "Buck" means the voltage of the windings is subtractive (180 degrees out of phase). "Boost" is the connection where the voltage is additive.

Normally, we would think the 2-winding transformer connected as an autotransformer would have an input voltage of 480 and an output of 600, but that would be boost. In buck, the input voltage would be 480 and the output would be 480-120=360. 

At rated current, the current on the 120V winding is 25k / 120 = 208.3 A. But with an output voltage of 360, the KVA is 360*208.3= 75KVA.

 

[BebeshKing PE]

Connected in "Buck" means the voltage of the windings is subtractive (180 degrees out of phase). "Boost" is the connection where the voltage is additive.

Normally, we would think the 2-winding transformer connected as an autotransformer would have an input voltage of 480 and an output of 600, but that would be boost. In buck, the input voltage would be 480 and the output would be 480-120=360. 

At rated current, the current on the 120V winding is 25k / 120 = 208.3 A. But with an output voltage of 360, the KVA is 360*208.3= 75KVA.

Why we did not consider the current flowing at the common/combined winding which is:

Ic=25kVA/360V = 69.4A ???

 

[Orchid PE]

Why we did not consider the current flowing at the common/combined winding which is:

Ic=25kVA/360V = 69.4A ???

Ic would actually be Ic = 25k / 480V = 52 A. But since the current through the secondary winding is equal to the current through the load, we really don't need to calculate the Common winding current.

 

[Sparky Bill PE]

I'm way off sadly. I thought power on each side of the transformer is the same (line/load). 

Is the power different here because it was one configuration AND THEN USED AS A BOOST? 

 

[DLD PE]

Someone please correct me if I'm wrong here:

I know these autotransformer problems can be confusing.  The key is in the wording.  It states the autotransformer is in "buck" configuration.  In a typical "step-down" autotransformer circuit, you have two power ratings...the rating of the original transformer (in this case 25kVA), and the rating of the resultant Autotransformer.   The problem presented by the OP is not to be solved using a typical step-down diagram (1st image).  You have to use the buck configuration (2nd image) and we're only concerned with the buck load, which is the secondary current (25kVA/120 = 208.33A) times the load voltage (360V).  

Hope this helps and I hope I've explained this correctly.

 

[Orchid PE]

I'm way off sadly. I thought power on each side of the transformer is the same (line/load). 

Is the power different here because it was one configuration AND THEN USED AS A BOOST? 

The power is the same. Input Power = Output Power.

But that doesn't mean a 25kVA 2-winding transformer can't supply a 75kVA load when connected as an autotransformer.

Since the transformer windings are carrying rated current, that means the connected load is 75kVA. If the transformer is carrying 75kVA of load, the line side (line, not primary winding) must be supplying 75kVA. 

My Ic should be pointing the other direction.

 

[BebeshKing PE]

Someone please correct me if I'm wrong here:

I know these autotransformer problems can be confusing.  The key is in the wording.  It states the autotransformer is in "buck" configuration.  In a typical "step-down" autotransformer circuit, you have two power ratings...the rating of the original transformer (in this case 25kVA), and the rating of the resultant Autotransformer.   The problem presented by the OP is not to be solved using a typical step-down diagram (1st image).  You have to use the buck configuration (2nd image) and we're only concerned with the buck load, which is the secondary current (25kVA/120 = 208.33A) times the load voltage (360V).  

Hope this helps and I hope I've explained this correctly.

View attachment 16541View attachment 16543

I just realized that this book has a formula for buck/boost transformer.  

And I just figured out that the secondary current and load current is equal for both configurations.

Thanks!

 

[Dude99]

Bear in mind it's limited by the rated winding current

Buck: 25 kva / 120 x 360 = 75 kva

if used as boost: 25 kva / 120 x 600 = 125 kva

easy way AT kva = Vout/Vbuck or boost x xfmr kva

360/120 x 25 = 75 kva or 600/120 x 25 = 125 kva

either way windings rated current, V and ratio applies 4:1 in this case, polarity changes

line i changes: buck 156.3 x 480 = 75 kva or boost 260.4 x 600 = 125 kva

156 = 208 - 52 or 260 = 208 + 52