Calculate the base MVA maximum fault current using ANSI Method that would appear at the transformer bushing stud terminal. What is the base MVA internal winding fault current inside the tank on the high voltage coil or primary side of the transformer ? What is the 150 MVA internal winding fault current inside the tank? What is the 180 MVA internal winding fault current inside the tank? What is the 216 MVA emergency rating internal winding fault current inside the tank?
If the Digital Fault Recorder (DFR) Line to Ground Fault on Phase A recorded a fault of 3,850 A on the primary or high voltage side, did it exceed the emergency rating fault current? Explain your answer.
1 MVA = 1,000,000 VA ; 1 kV = 1000 V ; √3 = 1.732
Based current _{Line}
Amp
 = 
3 Phase MVA
kV * 1.732

Section 7 of IEEE Std C57.12.002000 specified the general short circuit requirements for liquidimmerse distribution, power, and regulating transformers. For Category I and II transformers, symmetrical short circuit current shall be computed using transformer impedance only. For Category III and IV, system impedance should be included together with transformer impedance to calculate the symmetrical short circuit current. In our example our base MVA is 100 MVA, therefore it is a category IV transformer and should be calculated together with the system impedance. But for now we are neglecting the contribution of system impedance in our symmetrical short circuit calculation. By neglecting the system impedance, you are adding additional safety factor in your transformer short circuit withstand capability. In reality adding the system impedance will lower the calculated available short circuit current but it is difficult to get the real time transmission line impedance value because the network is dynamic due to power flow changes, unplanned outages due to storm and planned outages for scheduled apparatus maintenance.
Category  Single Phase kVA  Three Phase kVA 

I  5 to 500  15 to 500 
II  501 to 1667  501 to 5000 
III  1668 to 10000  5001 to 30000 
IV  Above 10000  Above 30000 
Using the ANSI method the maximum available fault current at the transformer bushing stud terminal will be:
Max. Available Fault Current _{Line}
Amp
 = 
base current
Z %

Now assuming the system impedance Z_{sys} = 5 % is in series connection with transformer impedance. Therefore the total impedance, Z_{total} = Z_{tr} 15% + Z_{sys} 5 % = 20 %. The base short circuit current will be going down to 1255.145 Amp.
Transformer must also withstand the peak asymmetrical short circuit current during the first cycle. ANSI method for calculating the peak asymmetrical short circuit current = K * I_{sc}.
If the x/r ratio of the system is unknown. ANSI method indicates the x/r ratio of the transformer should be used. Assuming the the x/r ratio of the given 100/150/180 MVA power transformer is 25.
Calculate the peak assymetrical short circuit current the given transformer must be able to withstand during the first cycle.
From IEEE C57.12.00 Table 7B1 for x/r = 25 use K =
Therefore the peak asymmetrical line short circuit current during the first cycle (16.67 ms) = Amp.
Keep in mind, this value was based on x/r = 25. You must change your Kfactor value depending on the transformer x/r value that you are analyzing.
For delta connected primary winding, the winding fault current = transformer terminal stud line fault current divided by square root of three or 1.732
Max. Available Fault Current _{Winding}
Amp
 = 
Max. Fault Current

Therefore the peak asymmetrical winding short circuit current = Amp. Keep in mind, this value was based on x/r = 25. You must change your Kfactor value depending on the transformer x/r value that you are analyzing.
HOW TO USE: Enter new number on any white input boxes. Answer is automatic.
For scenario analysis: change the % Z value and the K  Factor value
When you change %Z or Kfactor or both you should click all the white input boxes below to do new calculation base on new changes that you have made.
MVA RATING  LOAD Current  WYEFault Current  WYEAssy. Fault Current  DELTA Fault Current  DELTA Assy. Fault Current 
.  .  .  .  .  . 
.  .  .  .  .  . 
.  .  .  .  .  . 
> ? Is the analysis correct why? and why not?
The above inequality equation is an example of DFR fault current interpretation for delta connected primary coil winding. Here the assumption is 200 A fault current reduction due to transmission lines impedance. Then comparing this value to emergency MVA rating available fault current to make your fault current analysis. One of the questions you may want to know the answer in your fault current analysis, is the actual fault current enough to cause your transformer to fail?
Did your transformer fail at this fault current? Knowing the answer to this question will maybe lead you to update your fault current studies for the substation. Maybe you need to update your % Z of your transformer specifications for the substation to mitigate the problem or add series reactor to limit the fault current.
IEEE C57.12.00 Table 7 B1 Values of K
x/r  K 

200  2.806 
167  2.802 
143  2.798 
125  2.793 
111  2.789 
100  2.785 
50  2.743 
33.3  2.702 
25  2.662 
20  2.624 
16.7  2.588 
14.3  2.552 
12.5  2.518 
11.1  2.484 
10  2.452 
5  2.184 
3.33  1.990 
2.5  1.849 
2  1.746 
1.67  1.669 
1.43  1.611 
1.25  1.568 
1.11  1.534 
1.0  1.509 