CEE Relays Ltd.

87C Whitby Road, Slough, SL1 3DR

Telephone: (01753) 576477        Fax: (01753) 825661

Website: www.ceerelays.co.uk

High Phase-Earth Short-Circuit Currents

In most cases, the 3-phase short-circuit current is higher than the phase-earth (SLG) short-circuit current. However, this is not always the case. This occurs when the zero-sequence impedance for the fault is less than the positive and negative impedances, such as when the fault is close to a solidly-earthed delta-star transformer, as shown below:

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 ∠86.85°)pu

Z0thev            = (4.7917, 0.2818)pu = (4.799 ∠86.63°)pu

 Ibase @ 480V
I = S/V√3 = 100MVA / 480 √3 = 120281.31 A

Calculating the fault current we get:

I3phasepu               = E/Z+
                                               = 1 / (6.05 ∠86.85°) = (0.165∠-86.85°)

I3phaseactual           = (0.165∠-86.85°) x 120281.31 A = (19881.1∠-86.85°) A

Islgpu                                    = 3E / (Z+ + Z- + Z0)
                        = 3E / [(6.05 ∠86.85°)pu + (6.05 ∠86.85°)pu + (4.799 ∠86.63°)pu] =
                        = 3 / (16.9∠86.85°)pu
                        = (0.1775∠86.85°)pu

Islgactual              =  (0.1775∠86.85°)pu x 120281.31 A = (21351.67∠-86.80°) A