DC Load Flow
Prof Dr. Ashik Ahmed
Islamic University of Technology
Department of EEE
Introduction
• Load flow analysis helps in finding the steady state operating point of
a large power system under normal operating condition having many
generators, transformers, transmission lines, shunt elements, and
loads.
• Normal operating condition implies that,
Vi ,min  Vi  Vi ,max
Pgi ,min  Pgi  Pgi ,max
Sij ,line  Sij ,line ,max
• The steady state operating point implies, the knowledge of
Pg , slack , Qg , slack , Qg , PV ,  PV , VPQ ,  PQ
Ploss , Pline , Qloss , Qline
Introduction
• AC Load flow analysis is accurate but requires the solution of nonlinear set
of equations.
• No analytical solution is possible, requiring adoption of iterative
algorithms: Gauss-Siedel, Newton-Raphson, Fast-decoupled, Backwardforward sweep, etc.
• Being iterative, the time required for AC load flow solution is comparatively
larger.
• When quick and approximate information regarding the network variables
(voltage, current, power flow, power loss) are of interest, DC load flow is
the best way to adopt.
AC Load Flow: Revisit
Power Injected at bus i:
Pi  Pgi  Pdi
Qi  Qgi  Qdi
Current Injected at bus i in terms of complex power: I i  Pi  jQi  Pi  jQi e j i
*
Vi
Injected current and bus voltage relation:
 I bus n
bus x1
Vi
 Ybus n
Vbus n
bus x nbus
bus x1
nbus
nbus
j 1
j 1
Current Injected at bus i in terms of bus admittance: I i   YijV j   Yij  ij V j  j
Yij  Gij2  Bij2
*
Complex power injected at bus i:
nbus
 nbus

*
Si  Vi I i  Vi   Yij  ij V j  j   Vi i  Yij    ij V j    j
j 1
 j 1

nbus
  Vi V j Yij (i   j   ij )
j 1
AC Load Flow: Revisit
nbus
Real power injected at bus i: Pi  real ( Si )   Vi V j Yij cos(i   j   ij )  Pgi  Pdi
j 1
nbus
Reactive power injected at bus i: Qi  imag ( Si )   Vi V j Yij sin(i   j   ij )  Qgi  Qdi
j 1
• The above two equations are nonlinear in nature
• Analytical solution is not possible
• Numerical techniques are adopted to solve the unknowns
AC Load Flow: Revisit
Line Power Flow expressions:
Sij  Vi I ij*
V  V j 
 Vi  i

jx

ij 

*
 Vi   i  V j    j 
 Vi i 


jx


ij
 Vi 2  Vi V j (i   j ) 

 j
xij


 Vi 2  Vi V j cos(i   j )  j Vi V j sin(i   j ) 

 Pij  jQij  j 
xij


Pij 
Vi V j sin(i   j )
xij
Vi  Vi V j cos(i   j )
2
; Qij 
xij
Assumption for DC load flow
• Line resistances are small compared to the line reactances
zij  rij  jxij  jxij ; ij  900
• Bus voltage magnitudes are approximated to 1.0 p.u.
Vi  1.0 p.u
• Deviation of bus voltage angles between the connected buses are very small
i   j is very small. cos i   j   1 ; sin i   j   i   j 
• Shunt admittances are neglected
• Transformer tap settings are neglected
DC load flow Expressions
nbus
• Simplified Pi expression: Pi   Bij (i   j ) 
j 1
nbus

j 1, j  i
Bij (i   j )
(1)
nbus
• Simplified Qi expression: Qi   Bij
(2)
j 1
• Simplified Pij expression: Pij 
(i   j )
xij
• Simplified Qij expression: Qij  0
(3)
(4)
So, the unknowns for DC load flow problem are the bus voltage angles.
 There are no reactive power flowing through the lines.
DC load flow: Observations
• For an Nbus system, the bus voltage angle of the swing bus is known beforehand.
• Thus, the no. of unknowns for the DC load flow becomes Nbus – 1.
• Once the remaining bus voltage angles are known, real power injections at the buses and real
power flows through the lines can be calculated using equations Eq. (1) and Eq. (3), respectively.
• As the line resistances are neglected, there will be no line losses in a DC load flow solution.
• Eq. (1) and Eq. (3) depicts the linear nature of DC load flow.
DC load flow: 3-bus system
• For a 3-bus system, Eq. (1) can be expressed as:
P1  B12 1   2   B13 1  3 
  B12  B13  1  B12 2  B133
P2  B21  2  1   B23  2  3 
Ybus
  B21  B23   2  B211  B233
 B 11 B 12
 j[ B]  j  B 21 B 22
 B 31 B 32
P3  B31 3  1   B32 3   2 
  B31  B32  3  B311  B32 2
• In Matrix form:
 P1   B12  B13
 P    B
21
 2 
 P3    B31
 B12
B21  B23
 B32
 B13  1 
 B23   2 
B31  B32  3 
(5)
B 13 
B 23 
B 33 
DC load flow: 3-bus system
• Considering bus 1 as the swing bus, 𝜃1 = 0.
• So, for a 3-bus system, the only unknowns are 𝜃2 and 𝜃3 .
• Hence, writing Eq. (5) in terms of unknown angles:
• So,
 P2   B21  B23
 P    B
32
 3 
 2   B21  B23
     B
32
 3 
 B23   2 
B31  B32  3 
1
 B23   P2 
B31  B32   P3 
  P ( N
bus
/


B
 
1) X 1
  ( N 1) X 1   B / 
bus
( N bus 1) X ( N bus 1)
1
( Nbus 1) X ( Nbus 1)
 ( N
bus 1) X 1
 P ( N
bus 1) X 1
• It is to be noted that, P1 is still unknown, but can be obtained once P12 and P13 are calculated.
(1  3 )
(1   2 )
P12 
; P13 
x12
x13
P1  P12  P13
DC load flow: 3-bus system
• The line flows can be expressed in Matrix form as:
0
0  1 1 0  1 
 P12  1/ x12
P    0
 0 1 1  
1/
x
0
23
 23  

 2
 P13   0
0
1/ x13  1 0 1 3 
  Pline   bline  A  p.u.
 Pline   MVAbase bline  A 
MW
2
1
3
DC load flow: 3-bus system Example
• Bus data
Bus No.
Bus Type
PD (MW)
QD (MVAr) PG (MW)
1
Slack
0
0
Unknown
2
PV
10
5
63
3
PQ
90
30
0
2
1
• Line Data
3
Line No.
From Bus
To Bus
X (p.u.)
Rating
(MVA)
1
1
2
0.0576
250
2
2
3
0.092
250
3
1
3
0.17
150
Base Apparent Power = 100 MVA
DC load flow: 3-bus system Example
• Bus Admittance Matrix Entries (all in p.u.)
Y11  y12  y13 
1
1

  j17.3611  j 5.8824   j 23.2435
jx12 jx13
Y12  Y21   y12  j17.3611
Y13  Y31   y13  j 5.8824
Y22  y21  y23 
1
1

  j17.3611  j10.8696   j 28.2307
jx21 jx23
Y23  Y32   y23  j10.8696
Y33  y31  y32 
1
1

  j 5.8824  j10.8696   j16.7520
jx31 jx32
DC load flow: 3-bus system Example
Ybus
j 5.8824 
  j 23.2435 j17.3611
  j17.3611  j 28.2307 j10.8696 
 j 5.8824
j10.8696  j16.7520
B12  B21  17.3611; B13  B31  5.8824; B23  B32  10.8696;
 28.2307 10.8696 
 B /   


10.8696
16.7520


1
 2   28.2307 10.8696  (63  10) /100   0.0025
    10.8696 16.7520   (0  90) /100    0.0554  rad
 
 

 3 
DC load flow: 3-bus system Example
0
0  1 1 0   0   4.3403 
 P12 
17.3611
 P   100  0
 0 1 1  0.0025  57.5002 MW
10.8696
0
 23 



 

 P13 
 0
0
5.8824 1 0 1  0.0554 32.5885
Pg1  P12  P13  4.3403  32.5885  36.9288 MW
Total Generation  Pg1  Pg 2  63  36.9288  99.9288 MW
Total demand  Pd 2  Pd 3  10  90  100 MW