## 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%.

## 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
• U. 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 Aluminium 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%
• U. Reactance of the Transformer(Z Pu) =% Reactance /100
• U. 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 Down stream):

### (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 Aluminium 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 Aluminium 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 Aluminium 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

## Summary of Calculation:

 Sr.No Fault Location Fault MVA Fault Current (KA) 1 At HT Circuit Breaker 332 50 2 At Primary Side of Transformer 290 44 3 At Secondary Side of Transformer 31 43 4 At Main LT Panel 30 42 5 At Sub Main Panel 11 16 6 At Motor Control Panel 9 13

## (1) Service Factor:

• The service factor is a multiplier that indicates the amount of overload a motor can be expected to handle. If a motor with a 1.15 service factor can be expected to safely handle intermittent loads amounting to 15% beyond its nameplate horsepower.
• For example, many motors will have a service factor of 1.15, meaning that the motor can handle a 15% overload. The service factor amperage is the amount of current that the motor will draw under the service factor load condition.

## (2) Slip:

• Slip is used in two forms. One is the slip RPM which is the difference between the synchronous speed and the full load speed. When this slip RPM is expressed as a percentage of the synchronous speed, then it is called percent slip or just “slip”. Most standard motors run with a full load slip of 2% to 5%.

## (3) Synchronous Speed:

• This is the speed at which the magnetic field within the motor is rotating. It is also approximately the speed that the motor will run under no load conditions. For example, a 4 pole motor running on 60 cycles would have a magnetic field speed of 1800 RPM. The no load speed of that motor shaft would be very close to 1800, probably 1798 or 1799 RPM. The full load speed of the same motor might be 1745 RPM. The difference between the synchronous speed and the full load speed is called the slip RPM of the motor.

## (1) Pull Up Torque:

• When the motor starts and begins to accelerate the torque in generally decrease until it reach a low point at a certain speed it called the pull-up torque.
• The Pull-up Torque is the minimum torque developed by the electrical motor when it runs from zero to full-load speed (before it reaches the break-down torque point).
• Pull-up torque is the minimum torque developed during the period of acceleration from locked-rotor to the speed at which breakdown torque occurs.
• Some motor designs do not have a value of pull up torque because the lowest point may occur at the locked rotor point. In this case, pull up torque is the same as locked rotor torque.
• For motors which do not have a definite breakdown torque (such as NEMA design D) pull-up torque is the minimum torque developed up to rated full-load speed. It is usually expressed as a percentage of full-load torque.

## (2) Starting Torque (Locked Rotor Torque):

• The amount of torque the motor produces when it is energized at full voltage and with the shaft locked in place is called starting torque.
• The Locked Rotor Torque or Starting Torque is the torque the electrical motor develop when its starts at rest or zero speed.
• It is the amount of torque available when power is applied to break the load away and start accelerating it up to speed.
• A high Starting Torque is more important for application or machines hard to start – as positive displacement pumps, cranes etc. A lower Starting Torque can be accepted in applications as centrifugal fans or a pump where the start loads is low or close to zero.

• Full load torque is the rated continuous torque that the motor can support without overheating within its time rating.
• In imperial units the Full-load Torque can be expressed as
• T full-load torque (lb ft) = (Rated horsepower of Motor X 5252) / Rated rotational speed (rpm)
• In metric units the rated torque can be expressed as
• Full-load torque (Nm) = (Rated KW of Motor X 9550) / Rated rotational speed (rpm)
• Example :The torque of a 60 hp motor rotating at 1725 rpm can be expressed as
• T full-load torque = 60 X 5,252 / 1725 (rpm)
• T full-load torque = 182.7 lb ft

## (4) Peak Torque:

• Many types of loads such as reciprocating compressors have cycling torques where the amount of torque required varies depending on the position of the machine.
• The actual maximum torque requirement at any point is called the peak torque requirement.
• Peak torques is involved in things such as punch presses and other types of loads where an oscillating torque requirement occurs.

## (5) Pull out Torque (Breakdown Torque):

• Breakdown torque is the maximum torque the motor will develop with rated voltage applied at rated frequency without an abrupt drop in speed. Breakdown torque is usually expressed as a percentage of full-load torque
• The load is then increased until the maximum point is reached.