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Home > Archive > Electrical Engineering > September 2007 > Distribtuion network design assignment
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Distribtuion network design assignment
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| Warren Thai 2007-09-14, 8:25 pm |
| I'm working on a university assignment where I have to calculate a few
things for an 11kV distribution network. The first thing I have to do is
work out the zone sub transformer impedance. There are 2 x 25MVA subs for
the 11kV network. I have to choose an impedance so that the fault level does
not exceed 10.1kA and also so that the maximum voltage regulation of the
transformer is 10% (assume purely inductive on 0.9 lagging PF). I've done
thing so far but it's not making sense.
to satisfy fault level:
working on 100MVA base
fault level = 10.1kA = (root(3)*11kV*10.1kA)MVA = 1.982pu
with 2 transformers, each will have fault level divide by 2 so 1.982/2 =
0.96pu
Z = 1/0.96 = 1.04pu
so for the fault level to not exceed 10.1kA, the transformer impedance
should be >= 1.04pu
to satisfy voltage regulation (10%):
P= 0.25pu,
with V = 1pu, I = 0.25pu
|V|=|I||Z|
0.1 = 0.25*Z
Z = 0.4pu
so for regulation to be <= 10%, Z should be <=0.4pu
so with the 2 results for Z, there aren't any values for Z that satisfy both
conditions.
What am I doing wrong? (please help, I'm so confused)
-Warren
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| John Rye 2007-09-23, 5:25 pm |
| Hello Warren
In article <46eb1c09$0$31086$afc38c87@news.optusnet.com.au>,
Warren Thai <wthai1@optusnet.com.au> wrote:
> I'm working on a university assignment where I have to calculate a few
> things for an 11kV distribution network. The first thing I have to do is
> work out the zone sub transformer impedance. There are 2 x 25MVA subs for
> the 11kV network. I have to choose an impedance so that the fault level
> does not exceed 10.1kA and also so that the maximum voltage regulation of
> the transformer is 10% (assume purely inductive on 0.9 lagging PF). I've
> done thing so far but it's not making sense.
> to satisfy fault level:
> working on 100MVA base
> fault level = 10.1kA = (root(3)*11kV*10.1kA)MVA = 1.982pu
I make that 1.92 pu
> with 2 transformers, each will have fault level divide by 2 so 1.982/2 =
> 0.96pu
OK,
> Z = 1/0.96 = 1.04pu
OK
> so for the fault level to not exceed 10.1kA, the transformer impedance
> should be >= 1.04pu
OK but they are 25 MVA transformers, so you have to correct back to a 25 MVA
base I make it 26% or 0.26 pu on the 25 MVA rating.
OK
> to satisfy voltage regulation (10%):
> P= 0.25pu,
> with V = 1pu, I = 0.25pu
> |V|=|I||Z|
> 0.1 = 0.25*Z
> Z = 0.4pu
> so for regulation to be <= 10%, Z should be <=0.4pu
I have not checked this bit, but with the correction above the inconsistency
vanishes.
> so with the 2 results for Z, there aren't any values for Z that satisfy
> both conditions. What am I doing wrong? (please help, I'm so confused)
Best Wishes
John
--
John Rye
Hadleigh IPSWICH England
<http://web.ukonline.co.uk/jrye/index.html>
---< On Line using an Acorn StrongArm RiscPC >---
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