Bài giảng Tối ưu hóa vận hành hệ thống điện - Chapter 5: Optimal Power Flow - Võ Ngọc Điều

• Compact expression:
13
Mathematical formulation of the OPF (6)
• Inequality constraints:
– Limits on the control variables: 
– Operating limits on flows: 
– Operating limits on voltages
• Compact expression:
14
2/16/2014
8
Mathematical formulation of the OPF (7)
15
Mathematical formulation of the OPF (8)
16
2/16/2014
9
Mathematical formulation of the OPF (8)
17
Compact form of the OPF problem
18
Subject to:
2/16/2014
10
OPF Challenges 
• Size of the problem
– 1000’s of lines, hundreds of controls
– Which inequality constraints are binding?
• Problem is non-linear
• Problem is non-convex
• Some of the variables are discrete
– Position of transformer and phase shifter taps
– Status of switched capacitors or reactors
19
Solving the OPF
20
2/16/2014
11
Solving the OPF using gradient methods
• Build the Lagrangian function
• The gradient of the Lagrangian indicates the 
direction of steepest ascent: 
• Move in the opposite direction to the point 
with the largest gradient
• Repeat until 
21
Problems with gradient methods
• Slow convergence
• Objective function and constraints must be 
differentiable
• Difficulties in handling inequality constraints
– Binding inequality constraints change as the 
solution progresses
– Difficult to enforce the complementary slackness 
conditions
22
2/16/2014
12
Problems with gradient methods
23
Problems with gradient methods
24
2/16/2014
13
Solving the OPF using interior point method
• Best technique when a full AC solution is 
needed
• Handle inequality constraints using 
barrier functions
• Start from a point in the “interior” of the 
solution space
• Efficient solution engines are available
25
Linearizing the OPF problem
• Use the power of linear programming
• Objective function
– Use linear or piecewise linear cost functions
• Equality constraints
– Use dc power flow instead of ac power flow
• Inequality constraints
– dc power flow provides linear relations between 
injections (control variables) and MW line flows
26
2/16/2014
14
Sequential LP OPF
• Consequence of linear approximation
– The solution may be somewhat sub-optimal
– The constraints may not be respected exactly
• Need to iterate the solution of the linearized problem
• Algorithm:
1. Linearize the problem around an operating point
2. Find the solution to this linearized optimization
3. Perform a full ac power flow at that solution to find 
the new operating point
4. Repeat
27
Advantages and disadvantages
• Advantages of LPOPF method
– Convergence of linear optimization is guaranteed
– Fast
– Reliable optimization engines are available
– Used to calculate nodal prices in electricity 
markets
• Disadvantages
– Need to iterate the linearization
– “Reactive power” aspects (VAr flows, voltages) 
are much harder to linearize than the “active 
power aspects” (MW flows) 
28
2/16/2014
15
DC Power Flow Approximation
Power Flow Equations
• Set of non-linear simultaneous equations
• Need a simple linear relation for fast and 
intuitive analysis
• dc power flow provides such a relation but 
requires a number of approximations
Pk
I
− Vk Vi[Gki cos(θk −θ i) + Bki sin(θk −θ i)]
i=1
N
∑ = 0
QkI − Vk Vi[Gki sin(θ k −θ i ) − Bki cos(θk −θ i)]
i=1
N
∑ = 0
2/16/2014
16
Neglect Reactive Power
Pk
I
− Vk Vi[Gki cos(θk −θ i) + Bki sin(θk −θ i)]
i=1
N
∑ = 0
QkI − Vk Vi[Gki sin(θ k −θ i ) − Bki cos(θk −θ i)]
i=1
N
∑ = 0
Pk
I
− Vk Vi[Gki cos(θk −θ i) + Bki sin(θk −θ i)]
i=1
N
∑ = 0
Neglect Resistance of the Branches
Pk
I
− Vk Vi[Gki cos(θk −θ i) + Bki sin(θk −θ i)]
i=1
N
∑ = 0
Pk
I
− Vk ViBki sin(θ k −θ i)
i=1
N
∑ = 0
2/16/2014
17
Assume All Voltage Magnitudes = 1.0 p.u.
Pk
I
− Vk ViBki sin(θ k −θ i)
i=1
N
∑ = 0
Pk
I
− Bki sin(θ k −θ i )
i=1
N
∑ = 0
Assume all angles are small
Pk
I
− Bki sin(θ k −θ i )
i=1
N
∑ = 0
If α is small: sinα ≈ α (α in radians)
Pk
I
− Bki (θ k −θ i )
i=1
N
∑ = 0 or Pk
I
−
(θk −θ i)
xkii=1
N
∑ = 0
2/16/2014
18
Interpretation
Pk
I
−
(θ k −θ i )
xkii=1
N
∑ = 0
Pk
I
− Pki
i=1
N
∑ = 0
Pki =
(θ k −θ i )
xki
P12 =
(θ1 −θ2 )
x12
; P23 =
(θ 2 −θ3 )
x23
; P31 =
(θ 3 −θ1)
x13
θ1 θ2
θ3
x13
x12
x23
P1
I
P3
I
P2
I
P23P31
P12
Why is it called DC power flow?
• Reactance plays the role of resistance in dc 
circuit
• Voltage angle plays the role of dc voltage
• Power plays the role of dc current
Pki =
(θ k −θ i )
xki
Iki =
(Vk − Vi)
Rki
2/16/2014
19
Example of LPOPF
• Solving the full non-linear OPF problem by 
hand is too difficult, even for small systems
• We will solve linearized 3-bus examples by 
hand
37
Example
1 2
3
BA
450 MW
10 $/MWh
CA
PA
PAMAX=390MW
20$/MWh
CB
PB
PBMAX= 150 
MW
Economic dispatch:
PA = PA
max
= 390 MW
PB = 60 MW
2/16/2014
20
Flows resulting from the economic dispatch
39
1 2
3
BA
450 MW
390MW 60MW
Assume that all the lines have the same reactance
Fmax = 200MW
Fmax = 200MWFmax = 260MW
Do these injection result in acceptable flows?
Calculating the flows using superposition
Because we assume a linear model, superposition is 
applicable
1
60 MW
2
3
450 MW
390 MW
1 2
3
390 MW
390 MW 60 MW
1 2
3
60 MW
2/16/2014
21
Calculating the flows using superposition (1)
1 2
3
390 MW
390 MW
FA FB
FA = 2 x FB because the path 1-2-3 has twice the reactance
of the path 1-3
FA + FB = 390 MW
FA = 260 MW
FB = 130 MW
Calculating the flows using superposition (2)
1 2
3
60 MW
60 MW
FD FC
FC = 2 x FD because the path 2-1-3 has twice the reactance
of the path 2-3
FC + FD = 60 MW
FC = 40 MW
FD = 20 MW
2/16/2014
22
Calculating the flows using superposition (3)
1 2
3
390 MW
390 MW 60 MW
1 2
3
60 MW
260 MW
130 MW 20 MW
40 MW
280 MW
1
60 MW
2
3
450 MW
390 MW
110 MW
170 MW
Fmax = 260 MW
Correcting unacceptable flows
• Must use a combination of reducing the injection at bus 1 and 
increasing the injection at bus 2 to keep the load/generation balance
• Decreasing the injection at 1 by 3 MW reduces F1-3 by 2 MW
• Increasing the injection at 2 by 3 MW increases F1-3 by 1 MW
• A combination of a 3 MW decrease at 1 and 3 MW increase at 2 
decreases F1-3 by 1 MW
• To achieve a 20 MW reduction in F1-3 we need to shift 60 MW of 
injection from bus 1 to bus 2
280 MW1
60 MW
2
3
450 MW
390 MW
Must reduce flow F1-3 by 20 MW
2/16/2014
23
Check the solution using superposition
1 2
3
330 MW
330 MW 120 MW
1 2
3
120 MW
220 MW
110 MW 40 MW
80 MW
1
120 MW
2
3
450 MW
330 MW
260 MW
70 MW
190 MW
Fmax = 260 MW
Comments (1)
• The OPF solution is more expensive than the 
ED solution
– CED = 10 x 390 + 20 x 60 = $5,100
– COPF = 10 x 330 + 20 x 120 = $5,700
• The difference is the cost of security
– Csecurity = COPF - CED = $600
• The constraint on the line flow is satisfied 
exactly
– Reducing the flow below the limit would cost 
more
2/16/2014
24
Comments (2)
• We have used an “ad hoc”method to solve 
this problem
• In practice, there are systematic techniques 
for calculating the sensitivities of line flows to 
injections
• These techniques are used to generate 
constraint equations that are added to the 
optimization problem
Security Constrained OPF (SCOPF)
• Conventional OPF only guarantees that the 
operating constraints are satisfied under 
normal operating conditions 
– All lines in service
• This does not guarantee security
– Must consider N-1 contingencies
2/16/2014
25
Example: base case solution of OPF
1 2
3
BA
450 MW
330MW 120MW
260 MW
70 MW
190 MW
Example: contingency case
1 2
3
BA
450 MW
330MW 120MW
330 MW
0 MW
120 MW
Unacceptable because overload of line 1-3 could lead to
a cascade trip and a system collapse
2/16/2014
26
Formulation of the Security Constrained OPF
Subject to: Power flow equations for the base case
Operating limits for 
the base case
Power flow equations 
for contingency k
Operating limits
for contingency k
∀k
Subscript 0 indicates value of variables in the base case
Subscript k indicates value of variables for contingency k
Preventive security formulation
subject to:
This formulation implements preventive security because
the control variables are not allowed to change after the
contingency has occurred:uk = u0 ∀k
2/16/2014
27
Corrective security formulation
• This formulation implements corrective security because the 
control variables are allowed to change after the contingency 
has occurred
• The last equation limits the changes that can take place to 
what can be achieved in a reasonable amount of time
• The objective function considers only the value of the control
variables in the base case
subject to:
uk − u0 ≤ ∆u
max
Size of the SCOPF problem
• Example - European transmission network:
– 13,000 busses 
13,000 voltage constraints
– 20,000 branches 
 20,000 flow constraints
– N-1 security 
 20,000 contingencies
– In theory, we must consider
20,000 x (13,000 + 20,000) = 660 million 
inequality constraints
• However:
– Not all contingencies create limit violations
– Some contingencies have only a local effect
2/16/2014
28
Limitations of N-1 criterion
• Not all contingencies have the same probability
– Long lines vs. short lines
– Good weather vs. bad weather
• Not all contingencies have the same 
consequences
– Local undervoltage vs. edge of stability limit
• N-2 conditions are not always “not credible”
– Non-independent events
• Does not ensure a consistent level of risk
– Risk = probability x consequences
Probabilistic security analysis
• Goal: operate the system at a given risk level
• Challenges
– Probabilities of non-independent events
• “Electrical” failures compounded by IT failures
– Estimating the consequences
• What portion of the system would be blacked out?
– What preventive measures should be taken?
• Vast number of possibilities