# detailed P.U current and transformer problem



## eng787 (Oct 1, 2010)

Kaplan's problems:

Problem 19 :A power system is like as follows:

Region 1 region 2 region 3 region 4

Generator--------------T1-----------.........T2-----------------T3--------------Load (8 Ohm)

3 transformers ratings are given as:

T1: 240 MVA delta- Y 45 KV:345KV

T2: 210 mva y:y 345kv:138kv

T3: 180 MVA Y: delta 132KV: 13.6 KV

Generator voltage is 45 KV ( Line to Line). A base of 240MVA and 138 KV ( Line to Line) are chosen in region 3. Generator voltage is chosen as reference. The load current is 0.06(-10.35) in polar form. find complex power delivered in generator.

*Here in solution : say since generator, transformers and load are all in series, P.U current is same throughout the system*

But in problem 21:

Problem 21 :

generator----T1-----T2------Load

G: 200 MVA, 22KV, X=0.20 PU

Transformer T1 : 300 MVA, 220/25KV, X=0.10 PU ( Delta- Grounded Y)

T2: 3 single phase transformers each rated at 100 MVA, 130/13 KV, X=0.10 pu (Grounded Y-Delta)

Base values are choosen 200 MVA( 3Phase) and 22KV ( Line to Line) in generator section. if load current is 4375 A, A current in the low winding of the transformer T1 is ?????.

Here they find P.U Load current and then used following :

PU current in windings connected in delta=PU line current/sqrt (3).

Why they divided PU current by sqrt(3) in problem 21. is not PU current same throughout.????

also in problem 21---- is not T1 shall be 25/220KV instead of 220/25 KV.

Can anyone help ??????


----------



## cableguy (Oct 1, 2010)

#19 is a bad problem. They say 45 kV in the problem, but in the solution, suddenly the generator becomes a 25kV generator and they work the problem based on that. There are several Kaplan issues like this. It stinks, I wish they'd spent more time proof reading their stuff. Have you come across the .8 given power factor that becomes .85 in the solution yet? lol.

I worked #19 and came up with a complex power of (14.17 + j2.587) MVA. No guarantees it's right though. 

And for #21, they solved it poorly. They jumped the gun on the delta current conversion. You can solve it as a line current, and then as a very last step convert to delta. My last few lines read:

Ibase=5248

Iline=IbasexIpu=5248x.424=2225 A

and since they're asking for current in the delta windings...

Idelta=2225/sqrt(3) = 1284.7 A


----------



## eng787 (Oct 2, 2010)

I am interested in knowing *is PU current remains same troughout the system *??????


----------



## cableguy (Oct 2, 2010)

Baljit Gill said:


> I am interested in knowing *is PU current remains same troughout the system *??????



Yes, that's the beauty of per unit. Your "actual" current reading will change as you go up/down through transformers, but your Ipu should be the same. To get the actual current, you multiplly Ipu by Ibase, and that will get you the current at any point in the system. (This is assuming it's laid out like these Kaplan problems, with a source on the far left, a load on the far right, and a couple of transformers in between).


----------



## mull982 (Oct 4, 2010)

cableguy said:


> Baljit Gill said:
> 
> 
> > I am interested in knowing *is PU current remains same troughout the system *??????
> ...


It also appears that in problem #19 they did not use the sqrt(3) for calculating the total 3-phase power. They calculated total power as if it was single phase.


----------

