FAULT ANALYSIS IN PER UNIT SYSTEM PART -II

Three Phase Fault Example

Figure7: Three phase fault

Per unit analysis can be used to calculate system three phase fault levels and the current distributions. To gain a better understanding, it is worth running through the typical steps required to solve a fault calculation problem.

Given the system single line diagram, construct and simplify the per unit impedance diagram.

The fault level at the point under consideration is given by:

Figure 8: Total impedance

Fault flow through parallel branches is given by the ratio of impedances. As illustrated this can enable fault flows to be found through each branch.

Having calculated the fault flow in each branch, it is then relatively simple to find the current distribution using:

Example:

Calculate Fault current at each stage of following Electrical System SLD having details of.

Main Incoming HT Supply Voltage is 6.6 KV.
Fault Level at HT Incoming Power Supply is 360 MVA.
Transformer Rating is 2.5 MVA.
Transformer Impedance is 6%.

Figure 9: Single line diagram

Calculation:

Let’s first consider Base KVA and KV for HT and LT Side.
Base KVA for HT side (H.T. Breaker and Transformer Primary) is 6 MVA
Base KV for HT side (H.T. Breaker and Transformer Primary) is 6.6 KV
Base KVA for LT side (Transformer Secondary and down Stream) is 2.5 MVA
Base KV for LT side (Transformer Secondary and down Stream) is 415V

Fault Level at HT Side (Up to Sub-station):

(1)Fault Level from HT incoming Line to HT Circuit Breaker

HT Cable used from HT incoming to HT Circuit Breaker is 5 Runs , 50 Meter ,6.6KV 3 Core 400 sq.mm Aluminum Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0990 Ω/Km.
Total Cable Resistance(R) = (Length of Cable X Resistance of Cable) / No of Cable.
Total Cable Resistance=(0.05X0.1230) / 5
Total Cable Resistance=0.001023 Ω
Total Cable Reactance(X) = (Length of Cable X Reactance of Cable) / No of Cable.
Total Cable Reactance=(0.05X0.0990) / 5
Total Cable Reactance =0.00099 Ω
Total Cable Impedance (Zc1)=√(RXR)+(XxX)
Total Cable Impedance (Zc1)=0.0014235 Ω——–(1)
U Reactance at H.T. Breaker Incoming Terminals (X Pu)= Fault Level / Base KVA
U Reactance at H.T. Breaker Incoming Terminals (X Pu)= 360 / 6
Reactance at H.T. Breaker Incoming Terminals(X Pu)= 0.01666 PU——(2)
Total Impedance up to HT Circuit Breaker (Z Pu-a)= (Zc1)+ (X Pu) =(1)+(2)
Total Impedance up to HT Circuit Breaker(Z Pu-a)=0.001435+0.01666
Total Impedance up to HT Circuit Breaker (Z Pu-a)=0.0181 Ω.——(3)
Fault MVA at HT Circuit Breaker= Base MVA / Z Pu-a.
Fault MVA at HT Circuit Breaker= 6 / 0.0181
Fault MVA at HT Circuit Breaker= 332 MVA
Fault Current = Fault MVA / Base KV
Fault Current = 332 / 6.6
Fault Current at HT Circuit Breaker = 50 KA

(2) Fault Level from HT Circuit Breaker to Primary Side of Transformer

HT Cable used from HT Circuit Breaker to Transformer is 3 Runs , 400 Meter ,6.6KV 3 Core 400 sq.mm Aluminum Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0990 Ω/Km.
Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
Total Cable Resistance=(0.4X0.1230) / 3
Total Cable Resistance=0.01364 Ω
Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
Total Cable Reactance=(0.4X0.0990) / 5
Total Cable Reactance =0.01320 Ω
Total Cable Impedance (Zc2)=√(RXR)+(XxX)
Total Cable Impedance (Zc2)=0.01898 Ω——–(4)
U Impedance at Primary side of Transformer (Z Pu)= (Zc2 X Base KVA) / (Base KV x Base KVx1000)
U Impedance at Primary side of Transformer (Z Pu)= (0.01898X6) /(6.6×6.6×1000)
U Impedance at Primary side of Transformer (Z Pu)= 0.0026145 PU——(5)
Total Impedance(Z Pu)=(4) + (5)
Total Impedance(Z Pu)=0.01898+0.0026145
Total Impedance(Z Pu)=0.00261——(6)
Total Impedance up to Primary side of Transformer (Z Pu-b)= (Z Pu)+(Z Pu-a) =(6)+(3)
Total Impedance up to Primary side of Transformer (Z Pu-b)= 0.00261+0.0181
Total Impedance up to Primary side of Transformer (Z Pu-b)=0.02070 Ω.—–(7)
Fault MVA at Primary side of Transformer= Base MVA / Z Pu-b.
Fault MVA at Primary side of Transformer = 6 / 0.02070
Fault MVA at Primary side of Transformer= 290 MVA
Fault Current = Fault MVA / Base KV
Fault Current = 290 / 6.6
Fault Current at Primary side of Transformer= 44 KA

(3) Fault Level from Primary Side of Transformer to Secondary side of Transformer:

Transformer Rating is 2.5 MVA and Transformer Impedance is 6%.
% Reactance at Base KVA = (Base KVA x % impedance at Rated KVA) / Rated KVA
% Reactance at Base KVA = (2.5X6)/2.5
% Reactance at Base KVA =6%
Reactance of the Transformer(Z Pu) =% Reactance /100
Reactance of the Transformer(Z Pu)= 6/100=0.06 Ω—–(8)
Total P.U. impedance up to Transformer Secondary Winding(Z Pu-c)=(Z Pu)+(Z Pu-b)=(7)+(8)
Total P.U. impedance up to Transformer Secondary Winding(Z Pu-c)=0.06+0.02070
Total P.U. impedance up to Transformer Secondary Winding(Z Pu-c)=0.0807 Ω—–(9)
Fault MVA at Transformer Secondary Winding = Base MVA / Z Pu-c
Fault MVA at Transformer Secondary Winding = 2.5/0.0807
Fault MVA at Transformer Secondary Winding =31 MVA
Fault Current = Fault MVA / Base KV
Fault Current = 31 / (1.732×0.415)
Fault Current at Transformer Secondary Winding = 43 KA

Fault Level at LT Side (Sub-station to downstream):

(4) Fault Level from Transformer Secondary to Main LT Panel

LT Cable used from Transformer Secondary to Main LT Panel is 13 Runs , 12 Meter , 1KV, 3.5C x 400 Sq.mm Aluminum Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0618 Ω/Km.
Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
Total Cable Resistance=(0.012X0.1230) / 13
Total Cable Resistance=0.00009 Ω
Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
Total Cable Reactance=(0.012X0.0618) / 13
Total Cable Reactance =0.00006 Ω
Total Cable Impedance (Zc3)=√(RXR)+(XxX)
Total Cable Impedance (Zc3)=0.00011 Ω——–(10)
U Impedance at Main LT Panel (Z Pu)= (Zc3 X Base KVA) / (Base KV x Base KVx1000)
U Impedance at Main LT Panel (Z Pu)= (0.00011X2.5×1000)/(0.415×0.415X1000)
P P.U Impedance at Main LT Panel (Z Pu)= 001601 Ω ——(11)
Total Impedance up to Main LT Panel(Z Pu-d)= (Zc3)+ (Z Pu-c) =(11)+(9)
Total Impedance up to Main LT Panel (Z Pu-d)= 0.001601 +0.0807
Total Impedance up to Main LT Panel (Z Pu-d)=0.082306 Ω.——(12)
Fault MVA at Main LT Panel= Base MVA / Z Pu-a.
Fault MVA at Main LT Panel = 2.5 / 0.082306
Fault MVA at Main LT Panel= 30 MVA
Fault Current = Fault MVA / Base KV
Fault Current = 30 / (1.732X0.415)
Fault Current at Main Lt Panel = 42 KA

(5) Fault Level from Main LT Panel to Sub Panel:

LT Cable used from Main LT Panel to Sub Panel is 2 Runs , 160 Meter , 1KV, 3.5C x 400 Sq.mm Aluminum Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0618 Ω/Km.
Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
Total Cable Resistance=(0.160X0.1230) / 2
Total Cable Resistance=0.008184 Ω
Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
Total Cable Reactance=(0.160X0.0618) / 2
Total Cable Reactance =0.004944 Ω
Total Cable Impedance (Zc4)=√(RXR)+(XxX)
Total Cable Impedance (Zc4)=0.0095614 Ω——–(13)
U Impedance at Sub Panel (Z Pu)= (Zc4 X Base KVA) / (Base KV x Base KVx1000)
U Impedance at Sub Panel (Z Pu)= (0.0095614 X2.5×1000)/(0.415×0.415X1000)
P P.U Impedance at Sub Panel (Z Pu)= 13879 Ω ——(14)
Total Impedance up to Sub Panel (Z Pu-e)= (Zc4)+ (Z Pu-d) =(14)+(12)
Total Impedance up to Sub Panel (Z Pu-e)= 0.13879 +0.082306
Total Impedance up to Sub Panel (Z Pu-e)=0.2211 Ω.——(15)
Fault MVA at Sub Panel = Base MVA / Z Pu-a.
Fault MVA at Sub Panel = 2.5 / 0.2211
Fault MVA at Sub Panel = 11 MVA
Fault Current = Fault MVA / Base KV
Fault Current = 11 / (1.732X0.415)
Fault Current at Sub Panel= 16 KA

(6) Fault Level from Sub Panel to Motor Control Panel:

LT Cable used from Sub Panel to Motor Control Panel is 6 Runs , 150 Meter , 1KV, 3.5C x 400 Sq.mm Aluminum Cable , Resistance of Cable 0.1230 Ω/Km and Reactance of Cable is0.0739 Ω/Km.
Total Cable Resistance(R)= (Length of Cable X Resistance of Cable) / No of Cable.
Total Cable Resistance=(0.150X0.1230) / 6
Total Cable Resistance=0.003075 Ω
Total Cable Reactance(X)= (Length of Cable X Reactance of Cable) / No of Cable.
Total Cable Reactance=(0.150X0.0739) / 6
Total Cable Reactance =0.0018475 Ω
Total Cable Impedance (Zc5)=√(RXR)+(XxX)
Total Cable Impedance (Zc4)=0.003587 Ω——–(16)
U Impedance at Motor Control Panel (Z Pu)= (Zc5 X Base KVA) / (Base KV x Base KVx1000)
U Impedance at Motor Control Panel (Z Pu)= (0.003587 X2.5×1000)/(0.415×0.415X1000)
P P.U Impedance at Motor Control Panel (Z Pu)= 05207 Ω ——(17)
Total Impedance up to Motor Control Panel (Z Pu-f)= (Zc5)+ (Z Pu-e) =(17)+(15)
Total Impedance up to Motor Control Panel (Z Pu-e)= 0.13879 +0.2211
Total Impedance up to Motor Control Panel (Z Pu-e)=0.27317 Ω.——(15)
Fault MVA at Motor Control Panel= Base MVA / Z Pu-a.
Fault MVA at Motor Control Panel = 2.5 / 0.27317
Fault MVA at Motor Control Panel = 9 MVA
Fault Current = Fault MVA / Base KV
Fault Current = 9 / (1.732X0.415)
Fault Current at Motor Control Panel= 13 KA

FAULT ANALYSIS IN PER UNIT SYSTEM PART -II

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