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ELK474E
POWER SYSTEM CONTROL &
COMMUNICATIONS
Chapter 4
LOAD SHEDDING
Chapter 4
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Load Shedding and restoration
Load shedding schemes have began quite important in
present day systems, where there is a lack of adequate
spinning reserve margins , and a shortage of tie-line capacity
to make up for the loss of generations.
It is the function of under frequency load shedding relays, to
detect the onset of the decay in system frequency, and shed
approptiate amounts of system load, so that the generation
and load are again in balanced, ans power system can return
to operating frequency.
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Tie line
AREA 1
AREA 2
Under normal conditions, in this two area system there is a
balance between total lad and generation, and the power
system frequency is within the operation limits.
When there is a load increase in one of the area (say area 1),
The system frequency will decay (before the contro system in
action).
• Firstly if there is enough spinning reserve in area 1, the
generation in area 1 will be increase by the primery
control system, and a balance be provided between the
total load and generation in the system, and frequency will
be come around 50Hz. The supplimentary control brings
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the frequency exactly
50Hz.
•
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• When there is no enough spinning reserve in area 1, the
required power should be taken from area 2 to reach a
balanced, if there is enough tie-line capacity.
• When there is no change to import power from area 2, a
load shedding process will be in action in area 1 to match
the load and generation to bring the frequency to around
50Hz.
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Swing (Motion )Equation of Synchonous Machines
According to the Newton’s law, the torque of
acceleration is given by the moment of inertia J times
the angular acceleration of the rotor. Remember
F=ma for direct motion and T=J(dω/dt) for a rotting
mass).
Where;
J: moment of inertia in kg.
ω(t): the angular velocity in rad/s
Tm: mechanical torque (generator input) N.m
Te: Electrical torque (generator output) N.m
Td: damping torque, Ta: accelerating torque
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Swing (Motion )Equation of Synchonous Machines
Remember:
The angular moment M=Jω (Joule.s/rad)
For directional motion P=F V (W)
For rotational motion P=T ω (W)
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Eq(2)
The Definition of Inertia Constant H
H=Ek/S=Kinetix Energy/ Rated MVA
H=(1/2) MωR/S⇒M=2HS/ωR
Then the swing equation:
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The change in the rotor speed is directionally proportional
to the acceleration power (Pa=Pm-Pe) and inversely
proportional to inertia constant of the machine.
If the power system is modeled by a large generator, the
change in the system frequency, from the motion equation,
will be:
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System Frequency Dynamics
• From the Swing Equation
Where,
dF/dt = Rate of change in frequency
Fo = Base frequency, 50Hz
H = Inertia constant of system
Pgen = Active power generation
Pload = Active power demand
S = Total MVA rating of rotating plant
Pa=P =Pgen-Pload, overload.
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Overload Influence on frequency
• Overload is the difference between the active power
demand and the active power generated, as a
percentage of power generation, i.e.
(Pgen – Pload)
x100
Pgen
• The greater the overload, the greater the rate of
frequency decay
• Small islanded systems: overloads tend to be larger.
• ΔP=Pa=
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Δω =Δ𝐹 =
ΔPa
10
Change in the system load
in pu
With control
Frequency dependent
load (no control)
Frequency independent load (no control)
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Inertia (H) Influence on frequency
50
D=2, P=50%
49
Frequency (Hz)
48
47
46
45
44
H = 7s
H = 4s
43
H = 2s
42
41
40
0
2
4
6
8
10
12
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Time
(seconds)
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Load Self-regulation Influence on frequency
• As system frequency decays, the system demand also
decreases due to the frequency dependent load.
• This effect due primarily to the dynamic behaviour of
motor loads
• Load Reduction Factor :
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Load Self-regulation Influence on frequency
50
49
Frequency (Hz)
48
47
D=d=2
46
45
d = 3.0
44
43
d = 2.0
42
d = 1.2
41
40
0
2
4
6
8
10
12
14
16
18
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Time
(seconds)
20
14
System Response, 2 MW Generation Loss
With Primary Control vs LSD load self response (no
control)
50.2
Frequency
50
49.8
49.6
49.4
49.2
Seconds
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39
36
33
29
26
23
20
16
13
7
3
0
10
49
48.8
15
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Under-frequency Load Shedding (UFLS)
Load shedding is required, when there is lack of
adequate spinning reserve margins, and a
shortage of tie-line power flow capacity to make
up for the loss of generation.
It is the function of under frequency load
shedding relays, to detect the onset of the decay
in frequency, and shed approptiate amount of
system load, so that the generation and loads are
ones again in balance and power system can
return to normal operating frequency.
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•UFLS is a last action, emergency action to balance the
system load and generation.
•In small isolated systems:
-Overloads tend to be larger -System Inertia is smaller, so
more drastic, faster action is required
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Under-frequency Load Shedding
• Design Considerations:
• Spinning Reserves
• Level
• Distribution
• Load Shedding Parameters
• Maximum anticipated overload
• Amount of load to be shed
• Frequency thresholds
• Step size and number of steps
• Time delays
• Priorities and distribution
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Under-frequency Load Shedding
• Generating plant is highly sensitive to
frequency drop
• Low pressure turbine stages
• Motor driven auxiliaries
• Mechanical fatigue is cumulative
• Transient minimum frequency thresholds
• Recovery frequency band
• Governor action - Spinning reserves
• System dispatcher action
• Manual increase of load shedding
• Start-up additional generation
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Under-frequency Load Shedding
53
Maximum Recovery Frequency
52
Frequency (Hz)
51
Example of over-shedding
50
Example of under-shedding
49
48
Minimum Recovery Frequency
47
46
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
Time (seconds)
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Under-frequency Load Shedding
• Number and size of shedding stages:
• Many & Small vs. Few & Large
• Many Small Stages
• Avoids over-shedding
• Too slow for large overloads
• Tends to inhibit system recovery
• Few Large Stages
• Over-shedding may occur
• Faster corrective action for large
overloads
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Under-frequency Load Shedding
H=4s, D=2, P=20%
50
Frequency (Hz)
49
48
47
46
0
1
2
3
4
5
6
7
8
9
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Time
(seconds)
10
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Under-frequency Load Shedding
H=4s, D=2, P=50%
50
Frequency (Hz)
49
48
47
46
0
1
2
3
4
5
6
7
8
9
Time (seconds)
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Example: For a system with 10 000MW connected load, a
generation plant delivering 1500MW is lost due to a
contingency. The system inertia is 5 second.
Determine the settings of under frequency relays which will
accomplish a load shedding plan to drop 1500MW of load
in two steps of 750MW each. Assume that time delay for
tripping is 15 cycle and safety margin is 0.2 Hz.
8500MW
X
X
Power plant
G
1500MW
Initially PL=PG= 10000MW=10 GW and F=50Hz
Load,
10 000MW
Allowable change in frequency ΔF=±1%
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Solution:
Remember: Rate of change of frequency,
∆
=
=
xFo, Hz/s for frequency independent load.
∆
(
)
[
x 𝐹𝑜, 𝐻𝑧/𝑠 for frequency dependent load
]
Immediately after switching off 1500MW of load:
∆𝑃 =
x100=17.65% over loading after load switching
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First load shedding will be happen just after 49.5 Hz.
=
∆
(
)
[
x 𝐹𝑜 =
.
(
.
.
[
]
)
]
x 50 = −0.879
This means frequency decreases by 0.879 Hz in a second.
With a tripping delay of td= 15x20 = 300ms=0.3s
Within 1s
Within 0.3 s
0.879Hz drop
0.264Hz drop,
Then actual load shedding happens at (49.5-0.264)= 49.24Hz
0.879Hz in 1s then 0.76hz in 0.865 s, 0.865s ahter the load switching
operation of 1500MW, the first step of load shedding happens.
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Since the safety margin is 0.2Hz, the next shedding of 750MW will be
at the frequency of (49.5-0.264-0.2)= 49.04 Hz.
The rate of change of frequency at this point will be:
∆𝑃 =
Then
x100=8.8%
=
.
(
[
.
.
.
.
]
)
x 50= -0.434 Hz/s
This means that frequency decrease by 0.434Hz in a second.
The second step of load shedding will be within:
(49.24-49.04)/0.434 = 0.46 second.
Due to the 0.3s time delay, the second shedding occurs at the
frequency of 49.91 Hz.
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The two steps of load setting will be completed
[(50-49.24)/0.879]+[(49.24-48.92)/0.434= 0.865+0.761
=1.626 second at the point of PL=PG,
and the network frequency will begin to return to its
nominal value of 50hz.
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REFERENCE:
Roger Blackman
Barbados Light & Power
REFERENCE:
O Usta
Lecture Notes
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