secondary fault calculations using utility primary fault current 4

[KVX10]

I am able to calculate secondary fault currents based on an infinite bus, but unable to calculate a sec. fault current with the provided primary fault current. Please lead me in the correct direction.

 

[wareagle]

This ohmic method is given by Bussman
1. Utility X (in ohms) = 1000 x (sec kv)²
Only X is considered since X/R is high
2. Transf X (in ohms) = (10)(%X)(sec kv)²/trans kva
3. Transf R (in ohms) = (10)(%R)(sec kv)²/trans kva
%R may be small and omitted
4. Cable X in ohms
5. Cable R in ohms
6. Z = sqrt((Rt)² + (Xt)²)
7. Isc = sec voltage/(1.73 x Zt)

If you use the per unit method
Utility per unit X = Base kva/utility S.C kva
As with the ohmic method per unit R is small.

 

[wareagle]

Correction
1 should read
1. Utility X (ohms) = 1000 x (sec kv)²/Utility SC KVA
Only X is considered since X/R is high

 

[PMPPEng]

Here is an example:

1 Primary Bus Voltage 132 kV
2 Fault current at Primary Bus 15 kA
3 Fault MVA 3429.36 MVA
4 Transformer Ratio 132/25 kV
5 Transformer imp (on its rating) 15 %
6 Transformer rating 100 MVA
7 Choose a base MVA 100 MVA
8 Fault imp based on base MVA 0.02915996 p.u
(Base MVA/Fault MVA)

9 Imp of Transformer on base MVA 15 p.u
(Since base MVA is equal to transformer rating)

10 Total fault impedance on the secondary side of the transformer 15.02915996 p.u
( sum of p.u impedances of source and transformer)

11 Fault MVA on the transformer secondary side
( Base MVA/ total fault imp) 6.65373183 MVA

12 Fault current on transformer secondary side 0.153665862 kA
( fault MVA/ (1.732* 25)) 1536.66 A

 

[davidbeach]

I'm sure the transformer impedance would be 15% (still high) rather than 15pu. 15% would be 0.15pu. No way your secondary fault current could be even close to correct, you should have gone back and looked for the problem in the calculations.

 

[PMPPEng]

That is true. It was a mistake. Correct the 15% to 0.15 p.u and carry out the calculations.

 

[wareagle]

PMPPEng
using your figures with changes I have the following:
1 Primary Bus Voltage 132 kV
2 Fault current at Primary Bus 15 kA
3 Fault MVA 3429.36 MVA
4 Transformer Ratio 132/25 kV
5 Transformer imp (on its rating) 15 %
6 Transformer rating 100 MVA
7 Choose a base MVA 100 MVA
8 Fault imp based on base MVA 0.02915996 p.u
(Base MVA/Fault MVA)

9 Imp of Transformer on base MVA 0.15 p.u
(Since base MVA is equal to transformer rating)

10 Total fault impedance on the secondary side of the transformer 0.179 p.u
( sum of p.u impedances of source and transformer)

11 Fault MVA on the transformer secondary side
( Base MVA/ total fault imp) 100/.179 = 559 MVA

12 Fault current on transformer secondary side
( fault KVA/ (1.732* 25))
559000/(1.73 x 25) = 12924 amps
Does this look correct?

 

[rbulsara]

Refer to IEEE Buff Book and Red Book. Visit color series at

http://www.ieee.org

 

[jghrist]

wareagle's calculations look correct to me. This method does assume that the source and transformer impedances have the same X/R ratio. Otherwise, they would have to be added as complex numbers.

The X/R ratios will usually be similar. X/R for HV transmission will be high as will that of large HV-MV power transformers. The X/R ratio of MV lines will be lower, but so will the X/R ratio of smaller MV-LV transformers.

Considering that the source impedance is usually much smaller than the transformer impedance, the error introduced by this assumption will be small.

 

[KVX10]

Thanks all....mission accomplished and a lesson learned.

 

[rcw retired EE]

I prefer the MVA method developed by Moon Yuen at Bechtel in the 60’s for quick calculations. It uses the admittance (MVA) of each component expressed as the MVA that could flow through that component from an infinite bus to a short circuit.

1. Draw one line diagram with all major components : Utility Source, Transformer, Cables, Bus & Motors. List MVA rating, voltage, impedance.
2. Calculate the short circuit MVA for each piece of the system:
a. Utility: MVA = short circuit MVA = KV x (short circuit KA) x 1.732
b. Transformer: MVA = (MVA Rating x 100)/%Z
c. Cable: MVA = (KV x KV)/ohms
d. Motors: MVA = (MVA rating) x Locked Rotor Amps/Full Load Amps
3. Write the MVA next to each component on the one line or draw a block diagram one line with the MVA in each box.
4. Reduce the circuit using the rules for adding admittances:
a. If two MVA’s are in parallel: MVA total = MVA 1 + MVA 2
b. If two MVA’s are in series: MVA total = 1/[ (1/MVA1) + (1/MVA2)] = (MVA1 x MVA2 )/ (MVA1 + MVA2)
5. You will end up with one MVA. I short Circuit = MVA x 1000/ (KV x 1.732)

Using the example:

· Utility 132 kV & 15 kA: MVA ut = 3,429 MVA
· Transformer 100 MVA, Z= 15 % . MVA tx = 100/0.15= 667 MVA
· Add the MVA in series: MVAsc = (3429 x 667 ) / (3428 + 667) = 558 MVA
· Calculate the approximate Isc = 558 x 1000/ (25 x 1.732) = 12, 886 A

Same answers, faster, back of envelope, with enough accuracy f or most applications.

I’ll try to find the IEEE Industrial Applications article he published on this method. I always use it to my computer calculations. It has helped catch entry errors.

 

[mykh]

The Moon Yuen article mentioned by rcwilson can be downloaded from

http://www.arcadvisor.com/pdf/ShortCircuitABC.pdf

 

[waross]

The paper at this site may be of some help.

http://www.msha.gov/TECHSUPP/ACC/shortcircuit/ShortCircuitCalculations.pdf#search="gmr acsr"

respectfully