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Topic Title: Fault Level Calculation using MVA Method
Topic Summary: MVA equivalent circuit reduction
Created On: 12 August 2009 03:48 PM
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 12 August 2009 03:48 PM
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icoggan

Posts: 27
Joined: 26 November 2002


I have used the MVA Method for verifying some per unit Fault Level Calculations.

Both answers have matched well for my calculations in networks without parallel branches, my question regards a main switchboard fed from a generator, with bus-tie closed, feeding two transformers in parallel both incoming into another switchboard with the bus-tie closed.

The fault level I am calculating is on the switchboard fed from the transformers.

The MVA equivalent circuit has a single branch at the top from the generator, leading to 2 parallel branches with the cables / transformers, these branches recombine and the motor contributing MVA level is at the bottom.

How do I reduce this MVA equivalent circuit?

Regards

Ian Coggan
 13 August 2009 09:45 AM
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hpcompaq

Posts: 78
Joined: 27 September 2007

Reduce the values to the appropriate P.U. or % and then combine the parallel value as you would for resistors in parallel.
 13 August 2009 10:35 AM
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jafarpour

Posts: 198
Joined: 14 August 2005

The simple and reliable solution is to install fault current limiting reactor in series with generator. Of course there are some concern about the power system stability and resulting voltage drop across the reactor and how voltage regulation can be maintained when series reactor are connected.

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 14 June 2010 10:37 PM
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tonysung

Posts: 630
Joined: 14 September 2001

Originally posted by: icoggan

I have used the MVA Method for verifying some per unit Fault Level Calculations.

Both answers have matched well for my calculations in networks without parallel branches, my question regards a main switchboard fed from a generator, with bus-tie closed, feeding two transformers in parallel both incoming into another switchboard with the bus-tie closed.

The fault level I am calculating is on the switchboard fed from the transformers.

The MVA equivalent circuit has a single branch at the top from the generator, leading to 2 parallel branches with the cables / transformers, these branches recombine and the motor contributing MVA level is at the bottom.

How do I reduce this MVA equivalent circuit?

Regards
Ian Coggan


The MVA method can be think of as analogous to dealing with admittances. When the MVA elements are in series, the operation is to sum them as (1/MVA1 + 1/MVA2 +...) and evaluate the inverse of the result. When the MVA elements are in parallel, the result is to sum them up as (MVA1 + MVA2 + ...).

In your case, it will be finding the inverse of ((1/MVA-genny)+(1/(MVA-branch_a+MVA-branch_b))+(1/MVA-motor)).

Hope it will give you the same result as the p.u. method. Alternatively, you can do one branch as a time and sum the results using superposition theorem.

-------------------------
Tony Sung
 30 June 2010 09:45 PM
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Squeeky5

Posts: 1
Joined: 30 June 2010

Your main problem is the transformers in parallel. I am assuming that the bus tie (bus coupler) ties the two transformers together at the LV side. The bus coupler should never be closed with two transformers on. This is like putting two resistors in parallel, the impedance drops and the fault level goes up. The bus coupler should always be open with two transformers on. If one were to fail, the bus coupler can be closed and the plant can run on reduced load. This type of arrangement is normally Castel Key interlocked or done electrically to prevent the two running in parallel with the bus coupler closed..

Aslo if, with both in parallel, one HV side breaker tripped due to a fault in the HV winding, the LV breaker should be interlocked to cause inter tripping or the LV side as it will feed back into the HV fault unless you have directional current flow protection.

BAS
 15 March 2013 09:21 AM
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jonner

Posts: 1
Joined: 15 March 2013

Dear hpcompaq

Can you explain please.... which is the actual Z below I have to choose? for calculating Zero Sequence MVA

of a Synchronous Motor (Y ,netral connected ground), rating 2.219 MVA

if:
a. impedance ( xd"=15.38, x2=15.38, x0=15.38 )
b. dynamic model equivalent ( xd=110%, Xdu=116.93%, xd'=23%, XL=11% )

Locked rotor , LRC= 453.5%
pf =14.8%
grounding connection ( Y. type resistor )

and, are the positive and negative impedance value same or different?

Thanks
jonner
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