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Why is the calculated SLG fault higher than the calculated 3phase fault?

While in most cases it is true that the SLG fault on a bus is lower than the 3Phase fault, there are cases where the SLG fault is higher than 3Phase fault.

This usually occurs when the zero sequence impedance for the fault is less than the positive and negative sequence impedances. This occurs most often when a solidly grounded delta-wye transformer is located nearby.

Figure 1 below is a good example.

 
     
   
  Figure 1  
   
  A sample calculation of the sytem above follows below. See also Figure 2.  
   
  Given:  
   
 
Generator:    
Size: 15625 KVA    
X”pos = .1120pu   X/R = 37.4  Rpos-> = 0.00299pu
X”neg = .1120pu  X/R = 37.4    Rpos-> = 0.00299pu
X”zero = 5.70pu X/R = 37.4  Rpos-> = 0.1524pu
 
   
 
Transformer:    
Size:  1500 KVA 13800V/480V  D/YG  
     
Rpos = .004698 pu Xpos = 0.079862pu    
Rneg = .004698 pu   Xneg = 0.079862pu  
Rzero = .004227pu    Xzero = 0.071876pu  
 
     
  Converting the impedance to 100MVA we get the following:  
     
  Zpunew = Zpuold (Sbasenew/Sbaseold)(Vbaseold/Vbasenew)2  
     
 
Generator:  
X”pos = 0.1120pu(100/15.625)(1.0) = 0.7168pu Rpos = 0.00299pu(100/15.625)(1.0) = 0.0192pu
X”neg = 0.1120pu(100/15.625)(1.0) = 0.7168pu  Rneg = 0.00299pu(100/15.625)(1.0) = 0.0192pu 
X”zero = 5.70pu (100/15.625)(1.0) = 36.48pu  Rzero = 0.1524pu (100/15.625)(1.0) = 0.9754pu 
 
Transformer:  
Rpos = .004698 pu(100/1.5)(1.0) = 0.3132pu Xpos = 0.079862pu(100/1.5)(1.0) = 5.3241pu
Rpos = .004698 pu(100/1.5)(1.0) = 0.3132pu Xpos = 0.079862pu(100/1.5)(1.0) = 5.3241 pu
Rzero = .004227pu(100/1.5)(1.0) = 0.2818pu   Xzero = 0.071876pu(100/1.5)(1.0) = 4.7917 pu
 
   
   
   
  Z+/-thev  = (0.0192, 0.7168)pu + (0.3132, 5.3241)pu= (0.3324, 6.0409)pu = (6.05 ë86.85°)pu  
   
  Z0thev  = (4.7917, 0.2818)pu = (4.799 ë86.63°)pu  
   
  Ibase @ 480V
I = S/VÖ3 = 100MVA / 480 Ö3 = 120281.31 A
 
   
   
   
   
   
  Figure2