Power System Simulation Laboratory Manual

DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING
VII SEM - EEE
EE 6711 - POWER SYSTEM SIMULATION LABORATORY
S.NO NAME OF EXPERIMENTS
1. Computation of Parameters and Modeling of Transmission Lines
2. Formation of Bus Admittance and Impedance Matrices and Solution
of Networks
3. Load Flow Analysis - I : Solution of Load Flow And Related
Problems Using Gauss-Seidel Method
4. Load Flow Analysis - II: Solution of Load Flow and Related
Problems Using Newton-Raphson and Fast-Decoupled Methods
5. Fault Analysis
6. Transient and Small Signal Stability Analysis: Single-Machine
Infinite Bus System
7. Transient Stability Analysis of Multimachine Power Systems
8. Electromagnetic Transients in Power Systems
9. Load – Frequency Dynamics of Single- Area and Two-Area Power
Systems
10. Economic Dispatch in Power Systems.

DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING
VII SEM – E.E.E.
EE 6711 - POWER SYSTEM SIMULATION LABORATORY
1. Computation of Parameters and Modeling of Transmission Lines.
2. Formation of Bus Admittance and Impedance Matrices and Solution of Networks.
3. Load Flow Analysis - I : Solution of Load Flow And Related Problems Using
Gauss-Seidel Method.
4. Load Flow Analysis - II: Solution of Load Flow and Related Problems Using
Newton-Raphson and Fast-Decoupled Methods.
5. Fault Analysis.
6. Transient and Small Signal Stability Analysis: Single-Machine Infinite Bus
System.
7. Transient Stability Analysis of Multimachine Power Systems.
8. Electromagnetic Transients in Power Systems.
9. Load – Frequency Dynamics of Single- Area and Two-Area Power Systems.
10. Economic Dispatch in Power Systems.

EXP.NO:1 (A)
DATE:
COMPUTATION OF PARAMETERS
AIM:
To determine the positive sequence line parameters L and C per phase per kilometer of a single
phase, three phase single and double circuit transmission lines for different conductor arrangements.
SOFTWARE REQUIRED: MATLAB
THEORY:
Transmission line has four parameters – resistance, inductance, capacitance and conductance.
The inductance and capacitance are due to the effect of magnetic and electric fields around the
conductor. The resistance of the conductor is best determined from the manufactures data, the
inductances and capacitances can be evaluated using the formula.
ARRANGEMENT INDUCTANCE CAPACITANCE
SINGLE PHASE SYSTEM Lc=2*10-7 ln(D/r’)
r’=0.7788r.
Lloop=2Lc
Cc=2πξ0/ln(D/r)
Cloop=Cc/2.
3PHASE SYMMETRICAL
SYSTEM
Lc=Lph=2*10-7 ln(D/r’) Cc=Cph=2πξ0/ln(D/r)
3PHASE
UNSYMMETRICAL
TRANSPOSED SYSTEM
Lc=Lph=2*10-7 ln(Deq/r’)
Deq=(DAB*DBC*DCA)(1/3)
Cc=Cph=2πξ0/ln(Deq/r)
3PHASE
UNSYMMETRICAL
UNTRANSPOSED
SYSTEM
La=2*10-7[ln√(Dab*Dca)/r’
+j√3*ln√(Dab/Dca)
Lb=2*10-7[ln√(Dbc*Dab)/r’
+j√3*ln√(Dbc/Dab)
Lc=2*10-7[ln√(Dca*Dbc)/r’
+j√3*ln√(Dca/Dbc)
------------

3PHASE SYMMETRICAL
DOUBLECIRCUIT
SYSTEM
Lc=2*10-7 ln(√3D/2r’)
Lph=Lc/2
Cc=2πξ0/ln(√3D/2r)
Cph=Cc*2.
3PHASE
UNSYMMETRICAL
TRANSPOSED SYSTEM
WITH VERTICAL
PROFILE
Lc=2*10-7 ln[2(1/3)(D/r’)(m/n)(2/3)]
Lph=Lc/2
Cc=2πξ0/ln(2(1/3)(D/r)(m/n)(2/3)]
Cph=Cc*2.
3PHASE
UNSYMMETRICAL
TRANSPOSED DOUBLE
CIRCUIT
Lc=2*10-7 ln(Dm/Ds)
Lph=Lc/2
Cc=2πξ0/ln[i2m2jh/r3n3d](1/3)
Cph=Cc*2.
3PHASE LINE WITH
BUNDELED
CONDUCTORS
Lc=2*10-7 ln(Dm/Ds)
Lph=Lc/2
Dm=(DAB*DBC*DCA)(1/3)
Ds=( DSA*DSB*DSC)(1/3)
--------
PROCEDURE:
1. Enter the command window of the MATLAB.
2. Create a new M – file by selecting File - New – M – File
3. Type and save the program in the editor window.
4. Execute the program by either pressing Tools – Run.
5. View the results.

1. (a) Calculate the loop inductance and capacitance of a 1 phase line with two parallel conductors
spaced 3.5m apart. The diameter of each conductor is 1.5 cm.
Manual Calculation:
1(a) CALCULATION OF INDUCTANCE AND CAPACITANCE OF SINGLE PHASE LINE
PROGRAM:
OUTPUT:
1. (b) Calculate the inductance and capacitance of a conductor of a 3 phase system shown which
has 1.2 cm diameter and conductors at the edge of an equilateral triangle of side 1.5m.
Manual Calculation:

1b) CALCULATION OF INDUCTANCE AND CAPACITANCE OF THREE PHASE
SYMMETRIC LINE
PROGRAM:
OUTPUT:

1(c) Calculate the inductance, capacitance and reactance of 3 phase 50 Hz overhead transmission
line which has conductors of 2.5cm diameter. Distance between conductors are
5m between A & B
4m between B & C
3m between C & A Assume conductors are transposed regularly.
Manual Calculation:
1c) CALCULATION OF INDUCTANCE AND CAPACITANCE OF THREE PHASE
UNSYMMETRIC TRANSPOSED LINE
PROGRAM:
OUTPUT:

1. (d) Calculate the inductance and capacitance per phase of a 3 phase transmission line as shown in
figure. Radius of conductor is 0.5 cm. Lines are un transposed.
Manual Calculation:
1d) CALCULATION OF INDUCTANCE AND CAPACITANCE OF THREE PHASE
UNSYMMETRIC UNTRANSPOSED LINE
PROGRAM:
OUTPUT:

1 (e) Calculate the inductance and capacitance of a 3 phase double circuit line as in the figure if the
conductors are spaced 2m apart at the vertices of a hexagon and diameter of conductors is 2cm.
Manual Calculation:
1e) CALCULATION OF INDUCTANCE AND CAPACITANCE OF THREE PHASE
SYMMETRIC DOUBLE CIRCUIT LINE
PROGRAM:
OUTPUT:

Manual Calculation:
1f) CALCULATION OF INDUCTANCE AND CAPACITANCE OF THREE PHASE
UNSYMMETRIC, TRANSPOSED DOUBLE CIRCUIT LINE WITH VERTICAL PROFILE
PROGRAM:
OUTPUT:

1 (g) Calculate inductance and capacitance per phase of a 3 phase double circuit as shown in the figure.
Diameter of each conductor is 2cm. Line is transposed.
Manual Calculation:
1g) CALCULATION OF INDUCTANCE AND CAPACITANCE OF THREE PHASE
UNSYMMETRIC, TRANSPOSED DOUBLE CIRCUIT LINE WITHOUT VERTICAL
PROFILE
PROGRAM:
OUTPUT:

1 (h) A 300 KV, 3 phase bundled conductor with sub conductors per phase has a horizontal
configuration as in the figure. Find inductance per phase and capacitance if the radius of each sub
conductor is 1.2cm.
Manual Calculation:
1h) CALCULATION OF INDUCTANCE OF THREE PHASE LINE WITH BUNDLED
CONDUCTORS
PROGRAM:
OUTPUT:

1(i) A three-phase transposed line composed of one ACSR, 1,43,000 cmil, 47/7 Bobolink
conductor per phase with flat horizontal spacing of 11m between phases a and b and between
phases b and c. The conductors have a diameter of 3.625 cm and a GMR of 1.439 cm. The line
is to be replaced by a three-conductor bundle of ACSR 477,000-cmil, 26/7 Hawk conductors
having the same cross sectional area of aluminum as the single-conductor line. The conductors
have a diameter of 2.1793 cm and a GMR of 0.8839 cm. The new line will also have a flat
horizontal configuration, but it is to be operated at a higher voltage and therefore the phase
spacing is increased to 14m as measured from the center of the bundles. The spacing between
the conductors in the bundle is 45 cm.
Determine the inductance and capacitance per phase per kilometer of the above two lines.
Manual Calculation:
PROGRAM:
OUTPUT:

EXP.NO:1 (B)
DATE:
MODELLING OF TRANSMISSION LINES
AIM:
To understand modeling and performance of short, medium and long transmission lines.
SOFTWARE REQUIRED: MATLAB
FORM ULAE:
Vs=AVR+BIR
Is=CVR+DIR
TYPE METHOD ABCD PARAMETERS
Short ------------------- A=D=1; B=Z; C=0.
Medium Nominal T Method A=D=1+YZ/2;
B=Z(1+YZ/4);
C=Y;
Nominal π Method A=D=1+YZ/2;
B=Z;
C=Y(1+YZ/4);
Long Rigorous Method A=D=cos h(γℓ);
B=Zc sin h(γℓ);
C=1/Zc sin h(γℓ);
γ=√(ZY);
Zc=√(Z/Y);
Equivalent π Method A=D=1+YZ/2;
B=Z;
C=Y(1+YZ/A);
Z=Z sin h(γℓ)/ γℓ;
Y=Y tan h(γℓ/2)/ (γℓ/2);
γ=√(ZY);
Zc=√(Z/Y);
Equivalent T Method A=D=1+YZ/2;
B=Z;
C=Y(1+YZ/A);
Z=Z tan h(γℓ/2)/ (γℓ/2);
Y=Y sin h(γℓ)/ γℓ;
γ=√(ZY);
Zc=√(Z/Y);

PROCEDURE:
3. Type and save the program in the editor window.

1. An overhead 3 phase transmission line delivers 4000KW at 11 KV at 0.8 pf lagging. The
resistance and reactance of each conductor are 1.5Ω and 4Ω per phase. Determine the line
performance.
Manual Calculation:
SHORT TRANSMISSION LINE
PROGRAM:
OUTPUT

2.A balanced 3 phase load of 30 MW is supplied at 132KV, 50Hz and 0.85 pf lag by means
of a line. The series impedance is 20+j52Ω and total admittance is 315*10-6Ʊ. Using
Nominal T method determine A,B,C,D parameters and regulation.
Manual Calculation:
MEDIUM TRANSMISSION LINE
PROGRAM:
OUTPUT:

2.A 50Hz, 3 phase, 100 km transmission line has total impedance of 35Ω, reactance 0f 140Ω and shunt
admittance of 930*10-6 Ʊ. It delivers 40 MW at 220KV, 0.9 pf lag .Using nominal π determine A,B
,C,D Vs, VSA,ISA , pf, Ps Qs,η.
Manual Calculation:
NOMINAL PI
OUTPUT:

3.A 3 phase 50 Hz, 240 KV line is 200m long. The line parameters are R=0.017Ω/ph/km;
L=0.94mH/ph/km; C=0.0111µF/ph/km. Calculate line performance when load is 500MW, 0.9 pf lag at
220KV.
Manual Calculation:
LONG TRANSMISSION LINE (ABCD CONSTANTS)
PROGRAM:
OUTPUT:
EQUIVALENT PI METHOD
EQUIVALENT T METHOD

2. The following data refers to a 3 phase overhead transmission line. The voltage is 220KV.
Total series impedance /ph=200∟30’. Total shunt admittance/ph= 0.0013∟90’Ʊ. Load
delivers is 100MW at 0.8pf lag. Using rigorous method determine line performance.
Manual Calculation:
LONG TRANSMISSION LINE (RIGOROUS METHOD)
PROGRAM:
OUTPUT:
RESULT:Thus the line modeling of different types of transmission lines was done.

EXP NO: 2 (a)
DATE:
FORMATION OF BUS ADMITTANCE AND IMPEDANCE MATRICES AND
SOLUTION OF NETWORKS
AIM:
To develop a program to obtain Ybus matrix for the given networks by the method of inspection.
FORMATION OF Y-BUS MATRIX
Generalized [Y-Bus] = 





jjji
ijii
YY
YY
Each admittance Yii (i =1,2,……n) is called the self admittance or driving point admittance of
bus I and equals the sum of all admittances terminating on the particular bus.
Each off-diagonal term Yij (i,j = 1,2…n; ji) is the transfer admittance between buses I and j,
n=total number of buses. Further, Yij = Yji
SIMULATION
In this exercise matrix, Z-Bus for the system is developed by first forming the Ybus and then
inverting it to get the Z-Bus matrix. The generator and transformer impedances are taken into account.
Ybus is a sparse matrix, Z-Bus is a full matrix, i.e., zero elements of Ybus become non-zero
values in the corresponding Z-Bus elements. The bus impedance matrix is most useful for short circuit
studies.
ALGORITHM
Step (1): Initialize [Y-Bus] matrix, that is replace all entries by zero.
Yij = Yij-yij = Yji = off diagonal element
Step (2): Compute 

n
j
ijii yY
1
= diagonal element.
Step (3) : Modify the Ybus matrix by adding the transformer and the generator admittances

to the respective diagonal elements of Y- bus matrix.
Step (4) : Compute the Z-Bus matrix by inverting the modified Ybus matrix.
START
Readno of buses(NB),Noof
lines(NL) &line Data
InitializeYBusMatrix
Is
l = NL?
ConsiderLine l=1
i = sb(l);j = eb(l)
Y(i,i) =Y(i,i) + yseries(l) +0.5 ysh(l)
Y(j,j) = Y(j,j) +yseries(l) +0.5 ysh(l)
Y(i,j) = - yseries(l)
Y(j,i) = Y(i,j)
l = l + 1
STOP
Modifythe Ybus byaddinggeneratorandtransformer
admittancestothe respective diagonalelements.
Compute the Z-busmatrix byinvertingmodified
Ybus
Printall the Results

1. The [Y-Bus] matrix is formed by inspection method for a four bus system. The line data and is
given below.
LINE DATA
Line
Number
SB EB Series Impedance
(p.u)
Line charging
Admittance (p.u)
1 1 2 0.10 + j0.40 j0.015
2 2 3 0.15 + j0.60 j0.020
3 3 4 0.18 + j0.55 j0.018
4 4 1 0.10 + j0.35 j0.012
5 4 2 0.25 + j0.20 j0.030
Manual Calculation:
FORMATION OF Y-BUS BY THE METHOD OF INSPECTION
PROGRAM:
OUTPUT:
RESULT: Thus the program for the Ybus formation by the method of inspection was executed and
the output is verified with the manual calculation

EXP NO: 2 (b)
DATE:
FORMATION OF BUS ADMITTANCE AND IMPEDANCE MATRICES AND
SOLUTION OF NETWORKS
AIM:
To develop a program to obtain the Z bus matrix for the given network by the method of direct
inspection.
FORMATION OF Z-BUS MATRIX
Z-Bus matrix is an important matrix used in different kinds of power system studies such as
short circuit study, load flow study, etc.
In short circuit analysis, the generator and transformer impedances must be taken into account.
In contingency analysis, the shunt elements are neglected while forming the Z-Bus matrix, which is
used to compute the outage distribution factors.
This can be easily obtained by inverting the Ybus formed by inspection method or by analytical
method.Taking inverse of the Ybus for large systems is time consuming; Moreover, modification in the
system requires the whole process to be repeated to reflect the changes in the system. In such cases, the
Z-Bus is computed by Z-Bus building algorithm.
ALGORITHM
Step (1): Initialize [Y-Bus] matrix, that is replace all entries by zero
Yij = Yij-yij = Yji = off diagonal element
Step (2): Compute 

n
j
ijii yY
1
= diagonal element
Step (3) : Modify the Ybus matrix by adding the transformer and the generator admittances
to the respective diagonal elements of Y- bus matrix.
Step (4) : Compute the Z-Bus matrix by inverting the modified Ybus matrix.

1. Determine Z bus for the given network using direct inspection method.
LINE DATA
Line
Number
SB EB Series Impedance
(p.u)
Line charging
Admittance (p.u)
1 1 2 0.10 + j0.40 j0.015
2 2 3 0.15 + j0.60 j0.020
3 3 4 0.18 + j0.55 j0.018
4 4 1 0.10 + j0.35 j0.012
5 4 2 0.25 + j0.20 j0.030
FORMATION OF Z-BUS BY THE METHOD OF INSPECTION
PROGRAM:
OUTPUT:
RESULT:
Thus the program for the Z bus formation by the method of inspection was executed and
the output is verified with the manual calculation.

EXP NO: 2 (c)
DATE:
FORMATION OF BUS IMPEDANCE MATRICES
AIM:
To develop a program to obtain the Z bus matrix for the given network by the method of bus building
algorithm.
FORMATION OF Z-BUS MATRIX
Z-Bus matrix is an important matrix used in different kinds of power system studies such as
short circuit study, load flow study, etc.
In short circuit analysis, the generator and transformer impedances must be taken into account.
In contingency analysis, the shunt elements are neglected while forming the Z-Bus matrix, which is
used to compute the outage distribution factors.
This can be easily obtained by inverting the Ybus formed by inspection method or by analytical
method.Taking inverse of the Ybus for large systems is time consuming; Moreover, modification in the
system requires the whole process to be repeated to reflect the changes in the system. In such cases, the
Z-Bus is computed by Z-Bus building algorithm.
ALGORITHM:
Step 1: Start the program.
Step 2: Read the number of buses, starting bus and ending bus.
Step 3: Initialize the Z Bus matrix.
Step 4: Form the Z – Bus matrix as follows
Case 1:
When a new bus of impedance Zb is connected to reference bus Zbus,new = [ Zb ]

Case 2: Adding new bus p to existing bus q
Zbus,new = Zorig Z1q
Z2q
..
Zq1 Zq2 …… Zqq+ Zb
Case 3: Adding impedance from an existing bus to reference bus.
Zjk,act = Zjk – Zj(n+1)* Z(n+1)k
Z(n+1)(n+1)
Case 4: Adding Zb between two existing buses h and q
Zbus,new = Zorig Z1h - Z1q
Z2h – Z2q
Zh1 – Zq1 Zh2 – Zq2......Z(n+1)(n+1)
Case 5: Print the Z – bus matrix.

PROBLEM:
1. Find the bus impedance matrix for the given network.
FORMATION OF Z BUS BY BUS BUILDING ALGORITHM
PROGRAM
OUTPUT:
RESULT:
Thus the program for the Z bus formation by the method of inspection was executed and
the output is verified with the manual calculation.

EXP NO: 3
DATE:
LOAD FLOW ANALYSIS BY GAUSS – SEIDAL METHOD
AIM:
To carryout load flow analysis of the given power system by Gauss – Seidal method.
ALGORITHM:
Step 1: Assume a flat voltage profile of 1+j0 for all buses except the slack bus. The
voltage of slack bus is the specified voltage and it is not modified in any iteration.
Step 2: Assume a suitable value of  called convergence criterion. Here  is a specified change in bus
voltage that is used to compare the actual change in bus voltage th
k and th
k )1(  iteration.
Step 3: Set iteration count, k=0 and assumed voltage profile of the buses is denoted as
n
VVVV 0
3
0
2
0
1
0 ,.....,, except slack bus.
Step 4: Set bus count, p=1
Step 5: Check for slack bus. If it is a slack bus then go to step-12,otherwise go to next step.
Step 6: Check for generator bus. If it is a generator bus go to next step, otherwise (i.e., if it is load bus)
go to step=9.
Step 7: Temporarily set
specp
k
p VV  and phase of k
pV as the th
k iteration value if the bus-p is a
generator bus where
specpV is the specified magnitude of voltage for bus – p. Then calculate the
reactive power of the generator bus using the following equation.
  













 




 1
1
1*1
, Im)1(
p
q
n
pq
k
qpq
k
qpq
k
p
k
calp VYVYVQ
The calculated reactive power may be within specified limits or it may violate the limits.

If the calculated reactive power is within the specified limits then consider this bus as generator bus
and set 1
,

 k
calpp QQ for this iteration and go to step-8.
If the calculated reactive power violates the specified limit for reactive power then treat this bus as a
load bus.The magnitude of the reactive power at this bus will correspond to the limit it has violated.
i.e., if min,
1
, p
k
calp QQ 
then min,pp QQ 
(or) min,
1
, p
k
calp QQ 
then max,pp QQ 
Since the bus is treated as load bus, take actual value of k
pV for th
k )1(  iteration.i.e., k
pV need not be
replaced by
specpV when the generator bus is treated as load bus. Go to step9.
Step 8: For generator bus the magnitude of voltage does not change and so for all iteration the
magnitude of bus voltage is the specified value.The phase of the bus voltage can be as shown below.
  











 






 n
pq
k
qpq
p
q
k
qpq
k
p
pp
pp
k
tempp VYVY
V
jQP
Y
V
1
1
1
11
,
1
 
 











1
,
1
,11
Re
Im
tan
k
tempp
k
temppk
p
Val
V

Now that th
k )1(  iteration voltage of the generator bus is given by
11 
 k
pspecp
k
p VV 
After calculating 1k
pV for generator bus go to step -11.
Step 9: For the load bus the th
k )1(  iteration value of load bus –p voltage, 1k
pV can be calculated
using the following equation.
  











 






 n
pq
k
qpq
p
q
k
qpq
k
p
pp
pp
k
p VYVY
V
jQP
Y
V
1
1
1
11 1

Step 10: An acceleration factor,  can be used for faster convergence. If the acceleration factor is
specified then modify the th
k )1(  iteration value of bus-p voltage using the following equation.
)( 11
,
k
p
k
p
k
p
k
accp VVVV  

Then set , 1k
pV = 1
,
k
accpV
Step 11: Calculate the change in the bus-p voltage, using the relation,
k
p
k
p
k
p VVV   11
Step 12: Repeat the steps 5 to 11 until all the bus voltages have been calculated.For this increment the
bus count by1 and go to step-5,until the bus count is n.
Step 13:Find out the largest of the absolute value of the change in voltage.
i.e., Find the largest among 1
1

 k
V , 1
2

 k
V , ….., 1
 k
nV Let this be the largest change maxV
.Check whether this largest change maxV is less than the prescribed tolerance  .If maxV is less
than  then move to the next step. Otherwise increment the iteration count and go to step-4.
Step 14: Calculate the line flows and slack bus power using the bus voltages.

PROBLEM:
The system data for a load flow solution are given below. Determine the voltages by Gauss –
Seidal method
Line admittances Bus Specifications
Bus code P Q V Remarks
1 - - 1.06 Slack
2 0.5 0.2 - PQ
3 0.4 0.3 - PQ
4 0.3 0.1 - PQ
Manual Calculation
PROGRAM:
OUTPUT:
RESULT:
Thus load flow analysis by Gauss – Seidal method was done for the given power system.
Bus code Admittance
1 – 2 2 – j8
1 – 3 1 – j4
2 – 3 0.666 – j2.664
2 – 4 1 – j4
3 – 4 2 – j8

EXP NO: 4(a)
DATE:
LOAD FLOW ANALYSIS BY NEWTON - RAPHSON METHOD
AIM:
To carryout load flow analysis of the given power system by Newton raphson method.
ALGORITHM:
Step-1: Assume a flat voltage profile 1 + j0 for all buses (nodes) except the slack bus. The voltage of
the slack bus is the specified voltage and it is not modified in any iteration.
Step-2: Assume a suitable value of ε called convergence criterion. Hence ε is a specified change in the
residue that is used to compare the critical residues (Δ P and Δ Q or Δ V) at the end of each iteration.
Step-3: Set iteration count k = 0, and assumed voltage profile of the buses are denoted as V10, V20
…Vn0 except slack bus.
Step-4: Set bus count p = 1.
Step-5: Check for slack bus. If it is a slack bus then go to Step 13, otherwise go to next step.
Step-6: Calculate the real and reactive power of bus-p using the following equation.
Step-7: Calculate the change in real power, change in real power, Δ Pk = Pp,spec – Ppk; where Pp,spec =
Specified real power for bus-p.
Step-8: Check for Generator bus. If it is a Generator bus go to next step, otherwise go to Step 12.
Step-9: Check for reactive power limit violation of Generator buses. For this compare the calculated
reactive power Qpk with specified limits. If the limit is violated go to Step 11, otherwise go to next
step.
Step-10: If the calculated reactive power is within the specified limits then consider this bus as
Generator bus. Now calculate the voltage residue (change in voltage) using the following equation.
| Δ Vpk|2 = |Vp|spec
2 - |Vpk|2 where |Vp|spec = specified voltage.
Step-11: If the reactive power limit is violated then treat this bus as a load bus. Now the specified
reactive power for this bus will correspond to the limit violated.

i.e., if Qpk < Qp, min then Qp, spec = Qp, min
(Or) if Qpk > Qp, min then Qp, spec = Qp, max
Step-12: Calculate the change in reactive power for load bus (or for the Generator bus treated as load
bus). Change in reactive power, Δ Qpk = |Qp, spec| - Qpk
Step-13: Repeat steps 5 to12 until all residues (change in P and Q or V) are calculated. For this
increment the bus count by 1 and go to Step 5, until the bus count is n.
Step-14: Determine the largest of the absolute value of the residue (i.e., find the largest among Δ Pk,
Δ Qk or |Δ Vpk|2. Let this largest change be Δ E.
Step-15: Compare Δ E and ε. If Δ E < ε then to Step 20, If Δ E > ε go to next step.
Step-16: Determine the elements of Jacobian matrix (J) by partially differentiating the load flow
equations and evaluating the equation using Kth iteration values.
Step-17: Calculate the increments in real and reactive part of voltages.
Step-18: Calculate the new bus voltage.
Step-19: Advance the iteration count, i.e., k = k + 1 and go to Step 4.
Step-20: Calculate the line flows.

PROBLEM:
Consider the 3 bus system each of the 3 line bus a series impedance of 0.02 + j0.08 p.u
and a total shunt admittance of j0.02 pu.The specified quantities at the buses are given
below : Find the voltages in each bus for the given system using Newton-Raphson Method
Bus Real load
demand,
PD
Reactive Load
demand, QD
Real power
generation,PG
Reactive
Power
Generation,
QG
Voltage
Specified
1 2 1 - - V1=1.04
2 0 0 0.5 1 Unspecified
3 1.5 0.6 0 QG3 =  V3 = 1.04
Manual Calculation:
PROGRAM:

OUTPUT:
RESULT:
Thus the load flow analysis of the given power system by Newton – Raphson method was performed
for the given problem using Matlab-Power Tool Software.
Parameter Calculated Value Simulated Value
Bus voltage after the first
iteration using Newton
Raphson method

EXP NO: 4(b)
DATE:
LOAD FLOW ANALYSIS BY FAST DECOUPLED LOAD FLOW METHOD
AIM:
To carryout load flow analysis of the given power system by Fast decoupled load flow
method.
ALGORITHM:
Step-1: Read system data.
Step-2: Form YBUS matrix
Step-3: For load buses Pischeduled and Qischeduled are specified. Voltage magnitudes and phase angles
are set equal to the slack bus values, or |Vi| = 1.0, |δi| = 0.0 radian. For voltage controlled buses, where
|Vi| and Pischeduled are specified, phase angles are set equal to the slack bus angle, i.e. δi(0) = 0.0
radian.
Step-4: For load buses, Pi(p) and Qi(p) are calculated using equations
And Δ Pi(p) and Δ Qi(p) are calculated from equations
Step-5: For voltage controlled buses, Pi(p) and Δ Pi(p) are computed.
Step-6: Compute elements of J1 and J4 using equations

Step-7: Solve these equations for computing Δ δ and Δ |V|.
Step-8: Compute new voltage magnitudes and phase angles using equations,
Step-9: Check for convergence,
i.e. if max |Δ Pi(p)| ≤ ε and max |Δ Qi(p)| ≤ ε solution has converged go to Step-10, otherwise, go to
step-4.
Step-10: Print output results.

PROBLEM:
Consider the 3 bus system each of the 3 line bus a series impedance of 0.02 + j0.08 p.u
and a total shunt admittance of j0.02 pu.The specified quantities at the buses are given
below : Find the voltages in each bus for the given system using Fast Decoupled method .
Bus Real load
demand,
PD
Reactive Load
demand, QD
Real power
generation,PG
Reactive
Power
Generation,
QG
Voltage
Specified
1 2 1 - - V1=1.04
2 0 0 0.5 1 Unspecified
3 1.5 0.6 0 QG3 =  V3 = 1.04
Manual Calculation:

PROGRAM:
OUTPUT:
RESULT:
Thus the load flow analysis of the given power system by Fast Decoupled method was performed for
the given problem using Matlab-Power Tool Software.
Parameter Calculated Value Simulated Value
Bus voltage after the first
iteration using Newton
Raphson method

EXP NO: 5
DATE:
FAULT ANALYSIS
AIM:
To become familiar with modeling and analysis of power systems under faulted condition and to
compare the fault level, post-fault voltages and currents for different types of faults, both symmetric
and unsymmetric.
OBJECTIVES
To conduct fault analysis on a given system using software available and obtain fault analysis report
with fault level and current at the faulted point and post-fault voltages and currents in the network for
the following faults.
1. Line-to-Ground
2. Line-to-Line
3. Double Line-to-Ground
SINGLE LINE-TO-GROUND FAULT

Sequence Network ofSingle line-to-ground-fault
Fault Current
LINE-TO-LINE FAULT

Sequence Network ofLine-to-Line Fault
DOUBLE LINE-TO-GROUND FAULT

Sequence of Double line-to-ground fault

PROBLEM:
1. The one line diagram of a simple power system is shown in the figure. The
neutral of each generator is grounded through a current limiting reactor of 0.25/3
per unit on a 100MVA base. The system data expressed in per unit on a common
100MVA base is tabulated below. The generators are running on no load at their
rated voltage and rated frequency with their emf’s in phase.
Determine the fault current for the following faults.
a) A balanced three phase fault at bus 3 through a fault impedance Zf=j0.1 per unit.
b) A single line to ground fault at bus 3 through a fault impedance Zf=j0.1 per unit.
c) A line to line fault at bus 3 through a fault impedance Zf=j0.1 per unit.
d) A double line to ground fault at bus 3 through a fault impedance Zf=j0.1 per unit.
Item Base MVA Voltage Rating X1 X2 X0
G1 100 20kV 0.15 0.15 0.05
G2 100 20kV 0.15 0.15 0.05
T1 100 20kV/220kV 0.1 0.1 0.1
T2 100 20kV/220kV 0.1 0.1 0.1
L12 100 220kV 0.125 0.125 0.3
L13 100 220kV 0.15 0.15 0.35
L23 100 220kV 0.25 0.25 0.7125
Manual Calculation:

PROGRAM:
OUTPUT:
RESULT:
Quantity Calculated Value Simulated Value
FAULT CURRENT FOR
1. THREE PHASE FAULT
2. L-G FAULT
3. L-L FAULT
4. DOUBLE LINE FAULT
Thus the modeling and analysis of power system under faulted condition was made familiar and the
fault level, post fault voltage and currents for different types of fault both symmetric and
unsymmetrical was computed.

EXP NO: 6
DATE:
TRANSIENT AND SMALL SIGNAL STABILITY ANALYSIS – SINGLE
MACHINE INFINITE BUS SYSTEM
AIM:
To become familiar with various aspects of the transient and small signal stability analysis of
Single-Machine-Infinite Bus (SMIB) system.
OBJECTIVES
 To understand modeling and analysis of transient and small signal stability of a SMIB power
system.
 To examine the transient stability of a SMIB and determine the critical clearing time of the
system, through stimulation by trial and error method and by direct method.
 To assess the transient stability of a multimachine power system when subjected to a common
disturbance sequence: fault application on a transmission line followed by fault removal and line
opening.
 To determine the critical clearing time.
THEORY :
Stability : Stability problem is concerned with the behaviour of power system when it is subjected to
disturbance and is classified into small signal stability problem if the disturbances are small and
transient stability problem when the disturbances are large.
Transient stability: When a power system is under steady state, the load plus transmission loss equals
to the generation in the system. The generating units run a synchronous speed and system frequency,
voltage, current and power flows are steady. When a large disturbance such as three phase fault, loss of
load, loss of generation etc., occurs the power balance is upset and the generating units rotors
experience either acceleration or deceleration. The system may come back to a steady state condition
maintaining synchronism or it may break into subsystems or one or more machines may pull out of

synchronism. In the former case the system is said to be stable and in the later case it is said to be
unstable.
Small signal stability: When a power system is under steady state, normal operating condition, the
system may be subjected to small disturbances such as variation in load and generation, change in field
voltage, change in mechanical toque etc., The nature of system response to small disturbance depends
on the operating conditions, the transmission system strength, types of controllers etc. Instability that
may result from small disturbance may be of two forms,
(i) Steady increase in rotor angle due to lack of synchronising torque.
(ii) Rotor oscillations of increasing magnitude due to lack of sufficient damping torque.
FORMULA :
Reactive power Qe = sin(cos-1(p.f))
S*
Stator Current It =
Et
*
Pe - jQe
=
Et
*
Voltage behind transient condition
E1 = Et + j Xd
1It
Voltage of infinite bus
EB = Et - j( X3 + Xtr )It
X1 X2
where, X3 =
X1 + X2
Angular separation between E1 and EB
o =  E1 -  EB

Prefault Operation:
X1 X2
X = j Xd
1+ jXtr +
X1 + X2
E1 x EB
Power Pe = sino
X
Pe * X
o = sin-1
E1 * EB
During Fault Condition:
Pe = PEii = 0
Find out X from the equivalent circuit during fault condition
Post fault Condition:
Find out X from the equivalent circuit during post fault condition
E1 x EB
Power Pe = sino
X
max =  - o
Pm
Pe =
sinmax

Critical Clearing Angle:
Pm(max - o ) + P3maxcosmax - P2maxcoso
Coscr =
P3max - P2max
Critical Clearing Time:
2H (cr - o)
tcr =
 fo Pm Secs

PROBLEM:
Transient stability of SMIB system
1. A 60Hz synchronous generator having inertia constant H = 5 MJ/MVA and a direct axis
transient reactance Xd
1 = 0.3 per unit is connected to an infinite bus through a purely
reactive circuit as shown in figure. Reactance are marked on the diagram on a common
system base. The generator is delivering real power Pe = 0.8 per unit and Q = 0.074 per
unit to the infinite bus at a voltage of V = 1 per unit.
a) A temporary three-phase fault occurs at the sending end of the line at point F.When the
fault is cleared, both lines are intact. Determine the critical clearing angle and the
critical fault clearing time.
b) A three phase fault occurs at the middle of one of the lines, the fault is cleared and the
faulted line is isolated. Determine the critical clearing angle.
Manual Calculation:

Small signal stability of SMIB system
2) A 60Hz synchronous generator having inertia constant H = 9.94 MJ/MVA and a direct axis
transient reactance Xd
1 = 0.3 per unit is connected to an infinite bus through a purely reactive circuit as
shown in figure. Reactances are marked on the diagram on a common system base. The generator is
delivering real power Pe = 0.6 per unit and 0.8 power factor lagging to the infinite bus at a voltage of
V = 1 per unit.
Assume the per unit damping power co-efficient is D=0.138. Consider a small disturbance of
∆δ=10°=0.1745 radian. Obtain equations describing the motion of the rotor angle and the generator
frequency.
Manual Calculation:

PROGRAM:
OUTPUT:
RESULT :
Thus the various aspects of transient and small signal stability analysis of single machine infinite bus
system were made familiar.

EXP NO: 7
DATE:
TRANSIENT STABILITY ANALYSIS OF MULTIMACHINE POWER
SYSTEMS
AIM:
(i) To become familiar with modeling aspects of synchronous machines and network for transient
stability analysis of multi-machine power systems.
(ii) To become familiar with the state-of-the-art algorithm for simplified transient stability
simulation involving only classical machine models for synchronous machines.
(iii) To understand system behavior when subjected to large disturbances in the presence of
synchronous machine controllers.
(iv) To become proficient in the usage of the software to tackle real life problems encountered in
the areas of power system planning and operation.
OBJECTIVES
(i) To assess the transient stability of a multi machine power system when subjected to a common
disturbance sequence: fault application on a transmission line followed by fault removal
and line opening.
(ii) To determine the critical clearing time for the above sequence.
(iii)To observe system response and understand its behavior during a full load rejection at a
substation with and without controllers.
(iv)To observe system response and understand its behavior during loss of a major generating
station.
(v) To understand machine and system behavior during loss of excitation.
(vi)To study the effect of load relief provided by under frequency load shedding scheme.

THEORY:
The classical transient stability study is based on the application of a three – phase fault. A
solid three – phase fault at bus k in the network results in Vk = 0. This is simulated by removing the kth
row and column from the pre-fault bus admittance matrix. The new bus admittance matrix is reduced
by eliminating all nodes except the internal generator nodes. The generator excitation voltages during
the fault and post – fault modes are assumed to remain constant.
The electrical power of the ith generator in terms of the new reduced bus admittance matrices
are obtained from
m
Pei = ∑  Ei’   Ej’   Yij  cos ( ij - i + j ) ---------------( 1 ) j =1
The swing equation with damping neglected, for machine i becomes, is given by
Hi d2i m
------ ------- = Pmi - ∑  Ei’   Ej’   Yij  cos ( ij - i + j ) ---------------( 2 )
π f0 dt2 j=1
where Yij are the elements of the faulted reduced bus admittance matrix, and Hi is the inertia constant
of machine i expressed on the common MVA base SB. If HGi is the inertia constant of machine i
expressed on the machine rated MVA SGi, then Hi is given by
Hi = (SGi / SB ) HGi -------------( 3 )
Showing the electrical power of the ith generator by Pe
f and transforming equation ( 2 ) into
state variable model yields
di
------- = ωi , i = 1,2, ……… m ----------------( 4 )
dt
dωi
------- = (π f0 / Hi ) ( Pm - Pe
f ) ------------(5)
dt

For multi-machine transient stability analysis of an interconnected power system, it is
necessary to solve two state equations for each generator, with initial power angles 0 and ω0i = 0.
When the fault is cleared, which may involve the removal of the faulty line, the bus admittance
matrix is recomputed to reflect the change in the network. Next the post- fault reduced bus admittance
matrix is evaluated and the post- fault electrical power of the ith generator shown by Pi
pf is readily
determined from ( 1).
Using the post-fault power Pi
pf, the simulation is continued to determine the system stability,
until the plots reveal a definite trend as to stability or instability.
Usually the slack generator is selected as the reference, and the phase angle difference of all
other generators with respect to the reference machine are plotted.
Usually, the solution is carried out for two swings to show that the second swing is not greater
than the first one. If the angle differences do not increase, the system is stable. If any of the angle
differences increase indefinitely, the system is unstable.

PROBLEM:
1. For bus 1, the voltage is given as V1=1.06∟0 and it is taken as slack bus. The base
value is 100MVA.
GENERATION SCHEDULE
BUS
NO
VOLTAGE
MAG
GENERATION
MW
Mvar LIMITS
Min Max
1 1.06 ------ ----- ------
2 1.04 150 0 140
3 1.03 100 0 90
MACHINE DATA
GEN Ra Xd’ H
1 0 0.20 20
2 0 0.15 4
3 0 0.25 5
A three phase fault occurs on line 5-6 near bus 6 and is cleared by the simultaneous opening of
breakers at both ends of the line. Perform the transient stability analysis and determine the system
stability for a) when the fault is cleared in 0.4 second b) when the fault is cleared in 0.5 second c)
Repeat the simulation to determine the critical clearing angle
Manual Calculation:
LOAD DATA
BUS
NO
LOAD
MW Mvar
1 0 0
2 0 0
3 0 0
4 100 70
5 90 30
6 160 110
LINE DATA
LINE NO
(START)
LINE
NO(END)
R(PU) X(PU) 1/2B(PU)
1 4 0.035 0.225 0.0065
1 5 0.025 0.105 0.0045
1 6 0.040 0.215 0.0055
2 4 0.000 0.035 0.0000
3 5 0.000 0.042 0.0000
4 6 0.028 0.125 0.0035
5 6 0.026 0.175 0.0300

PROGRAM:
OUTPUT:
RESULT:
Thus the multi-machine transient stability analysis is simulated on a given power system network.

EXP NO: 8
DATE:
ELECTROMAGNETIC TRANSIENTS IN POWER SYSTEMS
AIM:
(i) To study and understand the electromagnetic transient phenomena in power systems caused due to
switching and fault by Power system simulation software.
ii) To become proficient in the usage of PSCAD software to address problems in the areas of
overvoltage protection and mitigation and insulation coordination of EHV systems.
OBJECTIVES:
a)To study the transients due to energization of a single-phase and three-phase load from a non-
ideal source with line represented by π model.
b)To study the transients due to energization of a single-phase and three-phase load from a non-
ideal source and line represented by distributed parameters.
c)To study the transient over voltages due to faults for a SLG fault at far end of a line.
d) To study the Transient Recovery Voltage (TRV) associated with a breaker for a
Solution Method for Electromagnetic Transients Analysis
Intentional and inadvertent switching operations in EHV systems initiate over voltages, which might
attain dangerous values resulting in destruction of apparatus. Accurate computation of these over
voltages is essential for proper sizing, coordination of insulation of various equipments and
specification of protective devices. Meaningful design of EHV systems is dependent on modeling
philosophy built into a computer program. The models of equipment’s must be detailed enough to
reproduce actual conditions successfully – an important aspect where a general purpose digital
computer program scores over transient network analyzers. The program employs a direct integration
time-domain technique evolved by Dommel. The essence of this method is discretization of differential
equations associated with network elements using trapezoidal rule of integration and solution of the
resulting difference equations for the unknown voltages. Any network which consists of

interconnections of resistances, inductances, capacitances, single and multiphase π circuits, distributed
parameter lines, and certain other elements can be solved. To keep the explanations simple, however
single phase network elements will be used rather than the more complex multiphase network
elements.
SOLUTION METHOD FOR ELECTROMAGNETIC TRANSIENTS ANALYSIS:
Intentional and inadvertent switching operations in EHV systems initiate over voltages, which might
attain dangerous values resulting in destruction of apparatus. Accurate computation of these voltages is
essential for proper sizing, co-ordination of insulation of various equipments and specification of
protective devices. Meaningful use of EHV is dependent on modeling philosophy built into a computer
program. The models of equipments must be detailed enough to reproduce actual conditions
successfully –an important aspect where a general purpose digital computer program scores over
transient network analyzers.
The program employs a direct integration time-domain technique evolved by Dommel. The essence of
this method is discretization of differential equations associated with network elements using
trapezoidal rule of integration and solution of the resulting equations for the unknown voltages. Any
network which consists of interconnections of resistances, inductances, capacitances, single and
multiphase pi circuits, distributed parameter lines, and certain other elements can be solved. To keep
the explanations simple, however, single phase network elements will be used, rather than the complex
multiphase network elements.

PROBLEM
1. Prepare the data for the network given in the given figure and run PSCAD software .Obtain the
plots of source voltage, load bus voltage and load current following the energization of a
single-phase load. Comment on the results. Double the source inductance and obtain the plots
of the variables mentioned earlier. Comment on the effect of doubling the source inductance.
Energization of a single phase 0.95 pf load from a non ideal source and a more realistic line
representation (lumped R,L,C ) using Power system simulation software.

PSCAD MODEL
RESULT:
Thus the electromagnetic transient phenomena in power systems caused due to switching and
fault by using PSCAD software are analyzed and results were obtained.

EXP NO: 9
DATE:
LOAD – FREQUENCY DYNAMICS OF SINGLE- AREA AND TWO-
AREA POWER SYSTEMS
AIM:
To become familiar with the modeling and analysis of load frequency and tie line flow
dynamics of a power systems with load frequency controller (LFC) under different control modes and
to design improved controllers to obtain the best system response.
OBJECTIVES:
i. To study the time response(both steady state and transient) of area frequency deviation and
transient power output change of regulating generator following a small load change in a
single-area power system with the regulating generator under “free governor action” for
different operating conditions and different system parameters.
ii. To study the time response (both steady state and transient) of area frequency deviation and the
turbine output change of regulating generator following a small load change in a single area
system provided with an integral frequency controller, to study the effect of changing the gain
of the controller and to select the best gain for the controller to obtain the best response.
iii. To analyze the time response of area frequency deviation and net interchange deviation
following a small load change in one of the areas in an inter connected two area power system
under different control modes, to study the effect of changes in controller parameters on the
response and to select the optimal set of parameters for the controllers to obtain the best
response under different operating conditions.
LOAD FREQUENCY CONTROL:
Primary control:
The speed change from the synchronous speed initiates the governor control action resulting in all the
participating generator-turbine units taking up the change in load, and stabilizes the system frequency.
Secondary control:

It adjusts the load reference set points of selected turbine-generator units so as to give nominal value of
frequency. The frequency control is a matter of speed control of the machines in the generating
stations. The frequency of a power system is dependent entirely upon the speed in which the generators
are rotated by their prime movers. All prime movers, whether they are steam or hydraulic turbines, are
equipped with speed governors which are purely mechanical speed sensitive devices, to adjust the gate
or control valve opening for the constant speed.
N = 120 f / P
Therefore N α f
where, N = Speed in rpm
f = Frequency in Hz
P = Number of poles.
ΔXE(s) = [ΔPc (s) – (1/R) ΔF(s) ] x
kG
1+sTG
where R =
k 1kc
𝐤 𝟐
= speed regulation of the governor in Hz/MW
kG =
k3 k1 kc
k4
= Gain of speed governor
TG =
1
k4k5
= Time constant of speed governor
Turbine model:

Tt = Time constant of turbine
k t = Gain constant
ΔPV(s) = per unit change in valve position from nominal value
Generator Load model:
ΔPD(s) = Real load change
k p = 1/B = Power system gain
Tp = 2H/Bfo = Power system time constant
Model of load frequency control with integral control of single area system:

PROBLEM:
1. The load dynamics of a single area system are Pr=2000 MW; NOL=1000MW;H=5s;f=50Hz;
R=4%; TG=0.08s;TT=0.3s; Assume linear characteristics . The area has governor but not
frequency control. It is subjected to an increase of 20MW. Construct simulink diagram and
hence i) determine steady state frequency. ii) If speed governor loop was open, what would be
the frequency drop? iii) Prove frequency is zero if secondary controller is included.
Manual Calculation:
PROGRAM:

OUTPUT:
OUTPUT:
WITH GAIN
Frequency response

WITHOUT GAIN
Frequency response
WITH INTEGRAL CONTROLLER
Frequency response

FOR TWO AREA SYSTEMS
1. A two area system connected by a tie line has the following parameters on a 1000 MVA base.
R1=0.05pu, R2=0.0625pu,D1=0.6, D2=0.9,H1=5,H2=4; Base power1=Base power2=1000MVA,
TG1=0.2s, TG2=0.3s, TT1=0.5s,TT2=0.6s. The units are operating in parallel at the nominal
frequency of 50Hz. The synchronizing power coefficient is 2pu. A load change of 200MW
occurs in area1. Find the new steady state frequency and change in the tie line flow. Construct
simulink block diagram and find deviation in frequency response for the condition mentioned.
Manual Calculation:
PROGRAM:

OUTPUT:

2. A two area system connected by a tie line has the following parameters on a 1000 MVA base.
R1=0.05pu, R2=0.0625pu, D1=0.6, D2=0.9, H1=5, H2=4; Base power1=Base power2=1000MVA,
TG1=0.2s, TG2=0.3s, TT1=0.5s, TT2=0.6s. The units are operating in parallel at the nominal
frequency of 50Hz. The synchronizing power coefficient is 2pu. A load change of 200MW
occurs in area1. Find the new steady state frequency and change in the tie line flow. Construct
simulink block diagram with the inclusion of the ACE’s and find deviation in frequency
response for the condition mentioned.
Manual Calculation:

RESULT:
Thus the modeling and analysis of load frequency and tie line flow dynamics of a power
systems with load frequency controller (LFC) under different control modes and to design improved
controllers to obtain the best system response was done using Matlab simulink.

EXP NO: 10
DATE:
ECONOMIC DISPATCH IN POWER SYSTEMS
AIM :
To understand the fundamentals of economic dispatch and solve the problem using classical
method with and without line losses.
Mathematical Model for Economic Dispatch of Thermal Units
Statement of Economic Dispatch Problem
In a power system, with negligible transmission loss and with N number of spinning thermal
generating units the total system load PD at a particular interval can be met by different sets of
generation schedules
{PG1
(k) , PG2
(k) , ………………PGN
(K) }; k = 1,2,……..NS
Out of these NS set of generation schedules, the system operator has to choose the set of schedules,
which minimize the system operating cost, which is essentially the sum of the production cost of all the
generating units. This economic dispatch problem is mathematically stated as an optimization problem.
Given : The number of available generating units N, their production cost functions, their operating
limits and the system load PD,
To determine : The set of generation schedules,
PGi ; i = 1,2………N (1)
Which minimize the total production cost,
N
Min ; FT =  Fi (PGi ) (2)
i=1
and satisfies the power balance constraint

N
=  PGi –PD = 0 (3)
i=1
and the operating limits
PGi,min  PGi  PGi, ,max (4)
The units production cost function is usually approximated by quadratic function
Fi (PGi) = ai PG2
i + bi PGi + ci ; i = 1,2,…….N (5)
where ai , bi and ci are constants
Necessary conditions for the existence of solution to ED problem
The ED problem given by the equations (1) to (4). By omitting the inequality constraints (4)
tentatively, the reduce ED problem (1),(2) and (3) may be restated as an unconstrained optimization
problem by augmenting the objective function (1) with the constraint  multiplied by LaGrange
multiplier,  to obtained the LaGrange function, L as
N N
Min : L (PG1 ……..PGN , ) =  Fi(PGi) -  [ PGi – PD] (6)
i=1 i=1
The necessary conditions for the existence of solution to (6) are given by
L / PGi = 0 = dFi (PGi) / dPGi -  ; i = 1, 2,……..N (7)
N
L /  = 0 =  PGi – PD (8)
i=1
The solution to ED problem can be obtained by solving simultaneously the necessary conditions (7)
and (8) which state that the economic generation schedules not only satisfy the system power balance
equation (8) but also demand that the incremental cost rates of all the units be equal be equal to 
which can be interpreted as “incremental cost of received power”.

When the inequality constraints(4) are included in the ED problem the necessary condition (7) gets
modified as
dFi (PGi) / dPGi =  for PGi,min  PGi  PGi, ,max
  for PGi = PGi, ,max
`  for PGi = PGi, ,mi ____(9)
Economic Schedule
PGi = ( -bi)/ 2ai ; i=1,2…………….N (10)
Incremental fuel cost






N
i i
N
i i
i
D
a
a
b
P
1
1
2
1
2

ALGORITHM:
1. Read the total number of generating units, power demand, fuel cost and mnB , co-efficient.
2. Find the initial value of lambda by using the given formula






N
i i
N
i i
i
D
a
a
b
P
1
1
2
1
2

3. Calculate the value of
  iiGi abP 2/ 
4. If maxPPGi  ,then maxPPGi 
If minPPGi  , then minPPGi 
5. Calculate the change in power,
  DGi PPp .
6. If 0001.0P , then stop. Otherwise go to next step.
7. Calculate change in lambda.







 

N
i
iaP
1
)2/1(/
8. If DGi PP  ,  
Otherwise,  
9. Read the total number of generating units, power demand, fuel cost and mnB co-efficient.
10. Find the initial value of lambda by using the given formula






N
i i
N
i i
i
D
a
a
b
P
1
1
2
1
2

11. Calculate the value of
        







 

N
j
iiiGiijiGi BaPBbP
1
2/2/2/1 
12. If maxPPGi  ,then maxPPGi 
If minPPGi  , then minPPGi 
13. Calculate the transmission loss
 




N
i
N
j
GjijGiL PBPP
1 1
14. Calculate the change in power,
  DLGi PPPp .
15. If 0001.0P , then stop. Otherwise go to next step.
16. Calculate change in lambda






 

N
i
iaP
1
)2/1(/
17. If DGi PP  ,  
Otherwise,  
18. Stop the program.
PROCEDURE :

3. Type and save the program.
PROBLEM1a: (WITHOUT LOSS AND GENERATING LIMITS)
A power plant has three units with the following cost characteristics:
where GiP ’s are in MW. Find the scheduling for a load of 975 MW.
Manual Calculation:
PROGRAM:
hrRsPPC GG /4005.5006.0 2
2
22 
hrRsPPC GG /5003.5004.0 1
2
11 
hrRsPPC GG /2008.5009.0 3
2
33 

OUTPUT:
PROBLEM1b: (WITHOUT LOSS AND WITH GENERATING LIMITS))
100≤PG1≤450
100≤PG2≤350
100≤PG3≤225
Manual Calculation:
PROGRAM:
OUTPUT
hrRsPPC GG /4005.5006.0 2
2
22 
hrRsPPC GG /5003.5004.0 1
2
11 
hrRsPPC GG /2008.5009.0 3
2
33 

PROBLEM 1c: (WITH LOSS AND GENERATING LIMITS))
10≤PG1≤85
10≤PG2≤80
10≤PG3≤70
PL=0.0218 PG1
2+0.0228 PG2
2+0.0179 PG3
2.
Manual Calculation:
PROGRAM:
OUTPUT:
hrRsPPC GG /1803.6009.0 2
2
22 
hrRsPPC GG /2007008.0 1
2
11 
hrRsPPC GG /1408.6007.0 3
2
33 

RESULT:Thus the economic dispatch problem with and without loss has been written and executed.

Power System Simulation Laboratory Manual

More Related Content

What's hot (20)

Viewers also liked (20)

Similar to Power System Simulation Laboratory Manual (20)

More from Santhosh Kumar (20)

Recently uploaded (20)

Power System Simulation Laboratory Manual