Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
FAULT LEVEL
- Thread starter abrahamJP
- Start date
- Thread starter
- #3
Network impedance:
Zn=11E3^2/250E6=0.484
Transformer impedance in per unit:
Ztpu=0.06
Impedance base on primary with Sbase of 1500kVA:
Zbn=11E3^2/1500E3=80.67
Network impedance in per unit:
Znpu=0.484/80.67=0.006
Total impedance in per unit:
Znpu+Ztpu=0.066
Fault current in per unit at 100% voltage:
Ifpu=1/(0.066)=15.15
Current base on secondary of transformer:
Ib=1500E3/(3^0.5*415)=2087
Actual fault current on secondary:
If=Ifpu*Ib=2087*15.15=31618 amps
Zn=11E3^2/250E6=0.484
Transformer impedance in per unit:
Ztpu=0.06
Impedance base on primary with Sbase of 1500kVA:
Zbn=11E3^2/1500E3=80.67
Network impedance in per unit:
Znpu=0.484/80.67=0.006
Total impedance in per unit:
Znpu+Ztpu=0.066
Fault current in per unit at 100% voltage:
Ifpu=1/(0.066)=15.15
Current base on secondary of transformer:
Ib=1500E3/(3^0.5*415)=2087
Actual fault current on secondary:
If=Ifpu*Ib=2087*15.15=31618 amps
You might want to include a voltage factor in wroggents answer depending on what part of the world you are in. IEC60909 would need 1.05 x 31618=33200A.
If you are working out maximum fault level, you might want to neglect the network contribution in case it is upgraded later and depending on the transformer type, it might be included differently (delta-wye doesn't include zero-sequence of the network contribution). Strictly speaking, impedances should be split into X and R because the network X/R will be different to the transformer X/R (Network X/R may be 10, transformer X/R may be 4), but that is probably going to have a very small impact in this instance.
If you are working out maximum fault level, you might want to neglect the network contribution in case it is upgraded later and depending on the transformer type, it might be included differently (delta-wye doesn't include zero-sequence of the network contribution). Strictly speaking, impedances should be split into X and R because the network X/R will be different to the transformer X/R (Network X/R may be 10, transformer X/R may be 4), but that is probably going to have a very small impact in this instance.
rcw retired EE
Electrical
I prefer the MVA method. Calculate the short circuit MVA of each branch and combine them as admittances.
Utility MVA = 250 = MVAutil
Transformer MVA = MVA rating/Zpu =1.5MVA/.06 = 25 MVA
= Maximum short circuit to the transformer terminals if the primary is connected to an infinte bus.
Combine the MVA's. As admittances, series connected MVA's combine like resistances in parallel. Parallel MVA's add.
1/MVAsc = (1/MVAutil) + (1/MVAtx) = 1/250 +1/25 = 0.044.
MVAsc = 1/0.044 = 22.73 MVA.
At 415V, Isc = 22.73 MVAx 10^6/(415V x 1.732) =31,623 Amps 3 phase, nominal voltage.
For the 105% voltage factor, mulitply by 1.05 = 33,204.
For the fault level with an infinite bus I = 25MVAx10^6/(415V x 1.732)= 34,781A
As stated in the previous post, different X/R ratios can affect the acuracy of this method.
Utility MVA = 250 = MVAutil
Transformer MVA = MVA rating/Zpu =1.5MVA/.06 = 25 MVA
= Maximum short circuit to the transformer terminals if the primary is connected to an infinte bus.
Combine the MVA's. As admittances, series connected MVA's combine like resistances in parallel. Parallel MVA's add.
1/MVAsc = (1/MVAutil) + (1/MVAtx) = 1/250 +1/25 = 0.044.
MVAsc = 1/0.044 = 22.73 MVA.
At 415V, Isc = 22.73 MVAx 10^6/(415V x 1.732) =31,623 Amps 3 phase, nominal voltage.
For the 105% voltage factor, mulitply by 1.05 = 33,204.
For the fault level with an infinite bus I = 25MVAx10^6/(415V x 1.732)= 34,781A
As stated in the previous post, different X/R ratios can affect the acuracy of this method.
- Status
Similar threads
- Question
- Question