# problem # 513 ncees power



## ros (Mar 16, 2011)

hello,

prob 513..

i need help with this problem...

base (100mva and 12 kv)

i understand how they came up with Ibase. and then how they come up with 1 PU for Transformer... but i am having difficulty in calculating PU for 20 mile long transmission line...

given 0.145/1000ft... so for 20 miles... Z = J0.145*(20000/1000) = 2.9 ohms (IS THIS CORRECT)

SO NOW ZPU FOR TRANSMISSION LINE WILL BE = 2.9 / (Vbase/Ibase) = 2.9/2.49 pu...

can some one help me with this


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## Dolphin P.E. (Mar 16, 2011)

ros said:


> hello,
> prob 513..
> 
> i need help with this problem...
> ...


1 Mile = 5280 ft


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## ros (Mar 16, 2011)

ok ... so Z for transmission line will be (0.145*20*5280) / 1000 = 15.312 ohm

now Zbase is 2.49 ....

so Z pu = 15.312/2.49 is this correct ?

if not .. help me solve this problem...

thanks


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## ros (Mar 16, 2011)

finally i am able to solve this problem.... but still i have confusion in finding short ckt current

i found that Z trans = 1 PU and Z trans line = 10.6 pu

but why SC = I base / ( 1 + 10.6) , where I base = 100mva/(sq 3 * 12 kv)

any help on that

thanks


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## Dolphin P.E. (Mar 16, 2011)

ros said:


> finally i am able to solve this problem.... but still i have confusion in finding short ckt current
> i found that Z trans = 1 PU and Z trans line = 10.6 pu
> 
> but why SC = I base / ( 1 + 10.6) , where I base = 100mva/(sq 3 * 12 kv)
> ...


See Attached file

413.pdf


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## ros (Mar 16, 2011)

dolphin said:


> ros said:
> 
> 
> > finally i am able to solve this problem.... but still i have confusion in finding short ckt current
> ...



thanks a lot ... but i am not able to understand the concept of Isc = Ib/Zpu

any help on that

thanks


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## Dolphin P.E. (Mar 16, 2011)

ros said:


> dolphin said:
> 
> 
> > ros said:
> ...


You are looking for the actual current at the fault. I actual = I base * I pu. = I base / Zpu. since Ipu = Vpu/Zpu note, Vpu=1

same way as S fault = S base / Zpu. I hope I didn't confuse you.


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## ros (Mar 16, 2011)

no u didnt confuse me.. u actually made my concept clear...thank you


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## willsee (Mar 22, 2011)

I solved this using MVA Method as well

z_line=j0.145/1000ft * 5280ft/mi * 20 mi = 15.312 ohms

7.5MVA/.075 = 100MVA

(12kv)^2 / 15.312 ohms = 9.4 MVA

Both are in series

100*9.4 / (100+9.4) = 8.59 MVA

I_line = 8.59MVA / (sqrt(3) * 12kv)) = 413A


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## dianevp (Mar 31, 2011)

willsee said:


> I solved this using MVA Method as well
> z_line=j0.145/1000ft * 5280ft/mi * 20 mi = 15.312 ohms
> 
> 7.5MVA/.075 = 100MVA
> ...


Why are the MVAsc's calculated in parallel? And if following this logic, would parallel MVA's be calculated in as series MVAsc?

Please advice...I'm not very familiar with this method...it does look quicker.

Thanks!


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## Dolphin P.E. (Apr 1, 2011)

dianevp said:


> willsee said:
> 
> 
> > I solved this using MVA Method as well
> ...


Yes, you actually add MVAsc as a capacitors.


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## Dolphin P.E. (Apr 1, 2011)

dolphin said:


> dianevp said:
> 
> 
> > willsee said:
> ...


take a look at this thread. It has the link to the MVA method papers.

http://engineerboards.com/index.php?showtopic=11332


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