Engineering Calculation for Power System Analysis

Engineering Calculation for Power
System Analysis

Page 1 of 178


CONTENTS
Modules

Pages

1. Load Flow Calculation.

3-16

2. Short Circuit Analysis.

17-47

Case1-ANSI/IEEE Method
Case2-IEC Method
3. Motor Starting Analysis.

48-66

4. Transient Stability Analysis.

67-96

Case1-Single Machine System
Case2-Multi Machine System

5. Relay Co-ordination.

97-107

6. Harmonic Analysis.

108-125

7. Ground Grid Analysis.

126-141

8. Optimal Capacitor Placement.

142-151

9. Arc Flash Analysis

152-164

10. Underground Raceway System

166-183

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1. LOAD FLOW ANALYSIS

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INTRODUCTION:
Load flow solution analysis is essential for designing a new power system and planning of
the existing one for increased load demand, which determine the steady state operating
condition to calculate,
 Voltage Profile - its magnitude in kV or % of nominal kV
 Current flow throughout the System
 MVA and /or MW plus Mvar power flows throughout each branch of the (i.e.
transformer, cables, line or series reactor etc) electrical system
 Voltage drop and Power factor
 Branch Losses i.e. MW & Mvar losses on each branch

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Input data required for LFA:
Sl.no

Component

1
2
3

Bus
Power Grid
Syn Generator

4

Transformer

5

Motor

6

Syn Motor

7

Static Load

8

Lump Load

9

Cable

8

Transmission Line

9
10

MOV
Protective Devices

11

Capacitor

12

Impedance

Required input
Nominal kv
Voltage rating, MVA sc, X/R ratio
1. Swing (slack) – %v and del
2. Voltage Control (PV) – MW, Mvar limits
3. Mvar Control (PQ) -MW, Mvar and Var limits
4. PF Control-MW and PF
Py kV, Sec kV, MVA, %Z Positive sequence Impedance,
%Tap, Tolerance and LTC settings
1. Status- Continuous , Intermittent or Spare
2. Rating-HP and kV
1. Status- Continuous , Intermittent or Spare
2. Rating-HP and kV
1. Status- Continuous , Intermittent or Spare
2. Rating-kV, MW, Mvar and PF
1. Status- Continuous , Intermittent or Spare
2. Ratings- kV, MW, Mvar and PF
3. Load Type-Motor Load or Static load
1. Length in ft/m/mile/km

2. Cable Type- Size, Insulation, kV and #/Cable
3. Impedance/conductor-Positive sequence
1. Length in ft/m/mile/km
2. Parameter-Phase conductor
3. Impedance per phase-positive sequence
1.
Rating-HP, kV and rated Torque
2. Circuit Breakers-Rated kV
3. Fuses-Rated kV
4. Switches-Rated kV and amps
5. Contactors- Rated kV and amps
1) Status- Continuous , Intermittent or Spare
2) Rating-kV, Max kV, Mvar Bank and No of Banks.
1) Rating-Amps and kV
2) Impedance- Positive sequence Z and X/R

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STEP by STEP Procedure for Load Flow Analysis
Step 1: Build the Single Line Diagram

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Step 2: Input Parameters/data for LFA
Power Grid:Nominal kV
kAsc
X/R

= 11kV
= 40 kA
= 14

Assume it is the base kV=11

Cable:Cable sizes and types are selected from the software library
Insulation
= XLPE
kV
= 15
Conductor type= CU
#/Cable
= 3/c
Size
= 300mm2
Length
= 0.836 km
Transformer:Primary kV = 11
Secondary kV = 0.433
MVA
= 10
%Z and X/R = Typical Value
Z Tolerance = If %Z <10 tolerance is 10
%Z >10 tolerance is 7.5 ( thumb rule)
Tap
= If needed. (To improve the bus voltage)
Lump Load:MW
=4
Mvar

=3
%PF
= 80
Load Type
= 100% Motor Load & 0% Static Load

Step 3: Data needed for Hand Calculation
Power Grid:X/R
MVAsc

= 14
= 762.102

Cable:R
X
Transformer:Z%
X/R

= 0.0801 Ω/km
= 0.1273 Ω/km
= 6.75
= 15.5

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Step 4: ETAP WITH HAND CALCULATION
1. To find Voltage Drop(Vd)
Formula: %Voltage drop= Delta V*100 --------------------------------------- eq1
To find Delta V:

Delta V= (√ (Vr+ (RP+XQ/Vr) ^2+ (XP-RQ/Vr) ^2))-Vr ----------------------- eq2
Where:
Vr = Receiving end voltage
R = Resistance of the cable/transmission in p.u
P = Real Power in MW
Q = Reactive Power in Mvar
X = Reactance of the cable in p.u
S= Apparent power in p.u
We know that,
Base MVA

= 100 MVA

Base kV

= 11 kV

Nominal kV

= 11 kV

Load PF

= 0.8

R Ω/km

= 0.0801 Ω

X Ω/km

= 0.12736 Ω

Length

= 0.836 km

Assume Vr

= 1 p.u

To find Z base:
Zbase

= BasekV^2/Base MVA
= 11^2 / 100
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Z base = 1.21

R

= R*length / Zbase
= 0.0801*0.836 / 1.21
R = 0.0554

X

= X*length / Zbase
= 0.12726*0.836 / 1.21
X = 0.0879

To Find P, Q and S:
We know MW and Mvar of Lump load,
a. To find
MVA =√ (MW^2+Mvar^2)
= √ (4^2+3^2)
MVA=5.033
P

= MW/Base MVA
= 4 / 100
P=0.040

Q

= Mvar/Base MVA
= 3 / 100
Q=0.030

S

= MVA/Base MVA
=5.033 / 100
S=0.0503

The following tabular columns are the results in p.u:
R

0.055

X
0.088

P
0.040

Q
0.030

S
0.0503

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To find % voltage drop:
eq 2=>
Delta V= (√ (Vr+ (RP+XQ/Vr) ^2+ (XP-RQ/Vr) ^2))-Vr
Consider Vs=1 p.u
Assume Vr = 1 p.u
Substitute the above calculated values from the table in the eq2
Delta V= (√ (1+ (0.055*0.040+0.088*0.030/1) ^2+ (0.088*0.040-0.055*0.030) ^2))-1
Delta V = 0.005
Note Vs ≠Vr so we are finding the new Vr new
Vr new=Vr -Delta V-----------------------------------------------eq3
= 1- 0.005
Vr new=0.995
To find Voltage drop:

eq 1=>
Voltage drop= Delta V*100
= 0.005*100
Voltage drop (Vd) = 0.5

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2. To find cable loss:
Formula:
MW =R*S^2*Base MVA*(Base kV/Nominal kV) ^2------------------ eq 4
Mvar = X*S^2*Base MVA*(Base kV/Nominal kV) ^2------------------ eq 5
To find MW:
Eq 4=>
MW

=R*S^2*Base MVA*(Base kV/Nominal kV) ^2
=0.055*0.0503*0.0503*100*(0.433/ 0.433) ^2
MW = 0.014

To Find Mvar:
Eq 5=>
Mvar

= X*S^2*Base MVA*(Base kV/Nominal kV) ^2
=0.088*0.0503*0.0503*100*(0.433/ 0.433) ^2
Mvar = 0.022

Cable losses = MW+jMvar = 0.014+j0.022

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3. To find Transformer Losses

Formula:
MW =R*S^2*Base MVA*(Base kV/Nominal kV) ^2-------------------- eq 6
Mvar = X*S^2*Base MVA*(Base kV/Nominal kV) ^2-------------------- eq 7
We know that
Nominal kV

=11 kV

Source kA

=40 kA

X/R

=14

Base MVA

=100

Z%

= 6.75

X/R

= 15.5

To find transformer new %R, %X and %Z:
%R

=Z%* √ {(1/(X/R) ^2) +1}
=6.75*√ {1 / (15.5 ^2 +1)}
%R=0.43

%X

= √ {%Z^2 - %R^2}
=√ {6.75^2- 0.43^2}
%X=6.74

%Z

= √ {%R^2 + %X^2}
=√ {0.43^2 + 6.74^2}
%Z=6.75

To Find R, X, P, Q and S in p.u:
R

= %R / TFR MVA
Page 12 of 178


= 0.43/10

R=0.0434
X

= %X / TFR MVA
= 6.74/10
X=0.674

P

= PMW / Base MVA
= 4/100
P=0.04

Q

= Q Mvar / Base MVA
= 3/100
Q = 0.03

S

= √ (P^2+Q^2)
= √ (0.04^2+0.03^2)
S=0.05

The following tabular columns are the results in p.u:
R
0.0434

X

0.674

P
0.04

Q
0.03

S
0.05

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To find MW:
eq 6=>
MW

=R*S^2*Base MVA*(Base kV/Nominal kV) ^2
=0.0434*0.05*0.05*100*(11/11) ^2
MW = 0.011

To Find Mvar:
eq 7=>
Mvar

= X*S^2*Base MVA*(Base kV/Nominal kV) ^2
=0.674*0.05*0.05*100*(11/11) ^2
Mvar = 0.199

Transformer losses = MW+jMvar = 0.011+j0.199

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Step 5: Run Load Flow Analysis with ETAP software

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Step 6: Summary Report.

Step 7: Comparison Table.
Hand Calculation
MW
Mvar

ETAP Result
MW
Mvar

Cable Losses

0.014

0.022

0.015

0.02

Transformer Losses

0.011

0.199

0.012

0.188

Cable Voltage drop

0.5

0.5

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2. SHORT CIRCUIT ANALYSIS
(IEEE / ANSI AND IEC METHOD)

Page 17 of 178


CASE 1:-ANSI / IEEE METHOD
INTRODUCTION:
The ETAP Short-Circuit Analysis program analyzes the effect of 3-phase, line-to-ground,
line-to-line, and line-to-line-to-ground faults on electrical distribution systems. The program

calculates the total short circuit currents as well as the contributions of individual motors,
generators, and utility ties in the system. Fault duties are in compliance with the latest
editions of the ANSI/IEEE Standards (C37 series).

Purpose of Short-Circuit Studies:
A Short-Circuit Study can be used to determine any or all of the following:


Verify protective device close and latch capability



Verify protective device Interrupting capability



Protect equipment from large mechanical forces (maximum fault kA)



I2t protection for equipment (thermal stress)



Selecting ratings or settings for relay coordination

Elements that Contribute Current to a Short-Circuit:


Generator



Power Grid



Synchronous Motors



Induction Machines



Lumped Loads
(with some % motor load)



Inverters



I0 from Yg-Delta Connected Transformer

Elements that do Not Contribute Current in PowerStation:


Static Loads



Motor Operated Valves
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

All Shunt Y Connected Branches

Short-Circuit Phenomenon:

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v(t)

i(t)

di
 Vm  Sin(  t   )
dt
equation 1 yields the following

v(t)  Ri  L

RL
-

i(t) 

expression

e

Solving

(1)

t
Vm
Vm
 sin(  t   -  ) 
 sin(  -  ) 
Z
Z
      
       

Steady

State

Transient
(DC Offset)

Copyright 2003 Operation Technology, Inc. – Workshop Notes: Short-Circuit ANSI

Slide 9

Symmetrical Faults:
The ETAP Short-Circuit Analysis program analyzes the effect of 3-phase
Symmetrical faults on electrical distribution systems. The program calculates the total short
circuit currents as well as the contributions of individual motors, generators, and utility ties in
the system. Fault duties are in compliance with the latest editions of the C37 series which
calculates


Momentary symmetrical fault current in kA



Interrupting symmetrical fault current in kA

Unsymmetrical Faults:
The ETAP Short-Circuit Analysis program also analyzes the effect of 3-phase
unsymmetrical Faults like


Line to Ground fault



Double line fault



Double line to ground fault.

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½ Cycle Network
This is the network used to calculate momentary short-circuit current and protective device
duties at the ½ cycle after the fault.
1 ½ to 4 Cycle Network
This network is used to calculate the interrupting short-circuits current and protective device
duties 1.5-4 cycles after the fault.
30-Cycle Network
This is the network used to calculate the steady-state short-circuit current and settings for
over current relays after 30 cycles of the fault.

Device Duty and Usage of Fault Currents from Different Networks:

HV Circuit Breaker
LV Circuit Breaker
Fuse
SWGR / MCC
Relay

½ Cycle Currents
(Sub transient Network)
Closing and Latching
Capability
Interrupting Capability
Interrupting Capability
Bus Bracing
Instantaneous Settings

1 ½ to 4 Cycle Currents
(Transient Network)
Interrupting
Capability
-

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ANSI Calculation Standard Compliance:
ETAP short circuit calculation per ANSI/IEEE Standards fully complies with the latest
ANSI/IEEE and UL Standards, as listed below:
Standard
EEE C37.04

Pub. Year

Title

1979(1988)

Standard Rating Structure for AC High-Voltage Circuit

IEEE C37.04f

1990

Breakers Rated on a Symmetrical Current Basis and

IEEE C37.04g

1986

Supplements

IEEE C37.04h

1990

IEEE C37.04i

1991

IEEE C37.04

1999

IEEE C37.010

1979, 1988,1999

IEEEC37.010b

1985

Circuit Breakers Rated on a Symmetrical Current Basis

IEEE C37.010e

1985

and Supplements

IEEE C37.010

1999

IEEE C37.13

1990

Standard Application Guide for AC High-Voltage

Standard for Low-Voltage AC Power Circuit Breakers
Used in Enclosures

IEEE C37.013

1997

Standard for AC High-Voltage Generator Circuit
Breakers Rated on a Symmetrical Current Basis

IEEE C37.20.1

1993 2002

Standard for Metal Enclosed Low-Voltage Power
Circuit Breaker Switchgear

IEEE Std 399
IEEE Std 141

1990 & 1997
1986, 1993, 2002

Power System Analysis – the Brown Book
Electric Power Distribution for Industrial Plants – the
Red Book

IEEE Std 242

1986 & 2001

IEEE Recommended Practice for Protection and
Coordination of Industrial and Commercial Power
Systems – the Buff Book

UL 489_9

1996, 2000, 2002

Standard for Safety for Molded-Case Circuit Breakers,
Molded-Case Switches, and Circuit-Breaker Enclosures

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Input data required for SCA:
Sl.no

1
2
3

Component
Bus
Power Grid
Syn Generator

4

Transformer

5

Motor

6

Syn Motor

7

Static Load

8

Lump Load

9

Cable

8

Transmission Line

9
10

MOV
Protective Devices

11

Capacitor

12

Impedance

Required input
Nominal kv
Voltage rating, MVA sc, X/R ratio
1. Swing – %v and del
2. Voltage Control – MW, Mvar limits
3. Mvar Control-MW, Mvar and Var limits
4. PF Control-MW and PF
5. Imp/Model-Impedance(Xd’’,Xd’, Xd, Xo, X2 and X/R)
1. Rating-Py kV, Sec kV, MVA

2. Impedance-%Z (+Ve & -Ve sequence Impedance)
3. Tolerance
4. Tap-Fixed Tap and LTC settings
5. Grounding-Py and Sec.
1. Status- Continuous , Intermittent or Spare
2. Rating-HP and kV and select MFR details.
3. Model-%LRC, %PF.
4. Model-Parameters-X0, X/R
1. Status- Continuous , Intermittent or Spare
2. Rating-HP and kV and select MFR details.
3. Model-Impedance(Xd’’,Xo,Xd’,X2, X/R)
1. Status- Continuous , Intermittent or Spare
2. Rating-kV, MW, Mvar and PF
1. Status- Continuous , Intermittent or Spare
2. Ratings- kV, MW, Mvar and PF
3. Load Type-Motor Load or Static load
1. Length in ft/m/mile/km
2. Cable Type- Size, Insulation, kV and #/Cable
3. Impedance/conductor-Positive sequence
1. Length in ft/m/mile/km
2. Parameter-Phase conductor
3. Impedance per phase-positive sequence
1. Rating-HP, kV and rated Torque
1. Circuit Breakers-Rated kV, Amps, Interrupting kA
2. Fuses-Rated kV, Size, Amps, Interrupting kA
3. Switches-Rated kV , BIL ratings and amps
4. Contactors- Rated kV, Interrupting kA and amps
1. Status- Continuous , Intermittent or Spare
2. Rating-kV, Max kV, Mvar Bank and No of Banks.
1. Rating-Amps and kV

2. Impedance- Positive sequence Z and X/R

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Step 1: Build the single line diagram using ANSI / IEEE method

Page 24 of 178


Step 2: Input Parameters/data for short circuit Analysis
Power Grid:Nominal kV

= 33 kV

MVAsc

= 750 kA

X/R

= 14

kAsc

= 13.12

Length

= 4.85km

R

= 0.1ohms/km

X

= 0.1ohms/km

Cable:-

Transformer:Primary kV

= 33 kV

Secondary kV = 6.3 kV
MVA

= 15 MVA

Base MVA

= 1 MVA

Z%

= 14.38

X/R

= 24.94

LRC

= 400%

Input MVA

= 5.676

LRC

= 550%

Input MVA

= 1.282

Motor 1:-

Motor 2:-

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