secondary fault calculations using utility primary fault current
secondary fault calculations using utility primary fault current
(OP)
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.






RE: secondary fault calculations using utility primary fault current
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.
RE: secondary fault calculations using utility primary fault current
1 should read
1. Utility X (ohms) = 1000 x (sec kv)²/Utility SC KVA
Only X is considered since X/R is high
RE: secondary fault calculations using utility primary fault current
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
RE: secondary fault calculations using utility primary fault current
RE: secondary fault calculations using utility primary fault current
RE: secondary fault calculations using utility primary fault current
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?
RE: secondary fault calculations using utility primary fault current
RE: secondary fault calculations using utility primary fault current
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.
RE: secondary fault calculations using utility primary fault current
RE: secondary fault calculations using utility primary fault current
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.
RE: secondary fault calculations using utility primary fault current
RE: secondary fault calculations using utility primary fault current
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respectfully