This document discusses power system stability in three chapters. It defines three types of stability - steady state, transient, and dynamic - and describes each. Steady state stability relates to gradual load increases, dynamic stability involves oscillations from small disturbances, and transient stability concerns large disturbances. The chapter continues by deriving the swing equation that models generator rotor dynamics and stability. It describes how this equation applies to both single and multi-machine systems. The document concludes by discussing power flow under steady state conditions, the equal area criteria for stability, critical clearing angles and times for faults.

1. ASTU
SCHOOL OF ELECTRICAL ENGINEERING AND
COMPUTING
DEPT. OF POWER AND CONTROL ENGINEERING
COMPUTER APPLICATION IN POWER SYSTEM
(PCE5307)
CHAPTER FOUR
POWER SYSTEM STABILITY
BY: MESFIN M.

2. INTRODUCTION
Power system stability refers to the ability of a power
system to move from one steady-state operating point
following a disturbance to another steady-state operating
point, without generators losing synchronism or having
unacceptable voltage magnitude and frequency deviations.
There are three types of power system stability:
Steady-state
Transient
Dynamic.
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3. CONT.…
Steady-state stability relates to the response of
synchronous machine to a gradually increasing load.
It is basically concerned with the determination of the
upper limit of machine loading without losing synchronism,
provided the loading is increased gradually.
Dynamic stability involves the response to small disturbances
that occur on the system, producing oscillations.
The system is said to be dynamically stable if theses oscillations
do not acquire more than certain amplitude and die out quickly.
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4. CONT.…
Transient stability involves the response to large
disturbances, which may cause rather large changes in rotor
speeds, power angles and power transfers.
Transient stability is a fast phenomenon usually evident
within a few second.
Power system stability mainly concerned with rotor
stability analysis.
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5. DYNAMICS OF SYNCHRONOUS MACHINE:
SWING EQUATION
The kinetic energy of rotor in synchronous machine is
given as:
𝐾𝐸 = 1
2 𝐽𝜔𝑠
2
∗ 10−6
𝑀𝑗𝑜𝑢𝑙𝑒
Where
𝐽 = 𝑟𝑜𝑡𝑜𝑟 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑖𝑛 𝑘𝑔 − 𝑚2
𝜔𝑠 = 𝑠𝑦𝑛𝑐ℎ𝑟𝑜𝑛𝑜𝑢𝑠 𝑠𝑝𝑒𝑒𝑑 𝑖𝑛 𝑚𝑒𝑐ℎ. 𝑟𝑎𝑑/𝑠𝑒𝑐
Speed in electrical radian is;
𝜔𝑠𝑒 = 𝑃
2 𝜔𝑠
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6. DYNAMICS OF SYNCHRONOUS MACHINE:
CONT.…
From the above we have:
𝐾𝐸 = 1
2 𝐽
2
𝑝
2
. 𝜔𝑠𝑒 ∗ 10−6
. 𝜔𝑠𝑒 𝑀𝐽
Or
𝐾𝐸 = 1
2 𝑀𝜔𝑠𝑒 𝑀𝐽
Where
𝑀 = 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑖𝑛 𝑀𝐽. 𝑠𝑒𝑐
𝑒𝑙𝑒𝑐 . 𝑟𝑎𝑑
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7. DYNAMICS OF SYNCHRONOUS MACHINE:
CONT.….
The inertia constant H, can be defined as:
𝐺𝐻 = 𝐾𝐸 = 1
2 𝑀𝜔𝑠𝑒 𝑀𝐽
Where
𝐺 = 𝑇ℎ𝑟𝑒𝑒 𝑝ℎ𝑎𝑠𝑒 𝑀𝑉𝐴 𝑟𝑎𝑡𝑖𝑛𝑔 𝑏𝑎𝑠𝑒 𝑜𝑓 𝑚𝑎𝑐ℎ𝑖𝑛𝑒
𝐻 = 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑖𝑛
𝑀𝐽
𝑀𝑉𝐴 𝑜𝑟 𝑀𝑊. 𝑠𝑒𝑐
𝑀𝑉𝐴
From the above equation we have
𝑀 =
2𝐺𝐻
𝜔𝑠𝑒
=
2𝐺𝐻
2𝜋𝑓
=
𝐺𝐻
𝜋𝑓
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8. SWING EQUATION
Consider the following system:
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9. CONT.….
Under steady state conditions (without any disturbance)
we have:
𝑇𝑖 = 𝑇𝑒
Then
𝑇𝑖 𝜔𝑠 = 𝑇𝑒 𝜔𝑠
And
𝑇𝑒 𝜔𝑠 − 𝑇𝑖 𝜔𝑠 = 𝑃𝑖 − 𝑃𝑒 = 0
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10. CONT.….
When a change in load or a fault occurs, then, input power
is not equal with electrical power. Thus there will be an
accelerating torque:
𝑃𝑖 − 𝑃𝑒 = 𝑀.
𝑑2
𝜃𝑒
𝑑𝑡2
+ 𝐷.
𝑑𝜃𝑒
𝑑𝑡
= 𝑃𝑎
Measuring the angular position of the rotor with respect to
a synchronous rotating frame of reference
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11. CONT.….
Thus:
𝛿 = 𝜃𝑒 − 𝜔𝑠 𝑡
So
𝑑2
𝜃𝑒
𝑑𝑡2
=
𝑑2
𝛿
𝑑𝑡2
Neglecting damping (D=0)
𝑃𝑖 − 𝑃𝑒 = 𝑀.
𝑑2
𝛿
𝑑𝑡2
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12. CONT.….
Using earlier equation for 𝑀 we have:
𝐺𝐻
𝜋𝑓
𝑑2
𝛿
𝑑𝑡2
= 𝑃𝑖 − 𝑃𝑒
In Per units
𝐻
𝜋𝑓
𝑑2
𝛿
𝑑𝑡2
= (𝑃𝑖−𝑃𝑒)𝑝𝑢
This is called swing equation of a machine. But for
Dynamic stability analysis damping must be considered.
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13. MULTI MACHINE SYSTEM
In multi machine system a common base must be selected:
𝐺 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 = 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 𝑟𝑎𝑡𝑖𝑛𝑔 𝑏𝑎𝑠𝑒
𝐺𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑠𝑦𝑠𝑡𝑒𝑚 𝑏𝑎𝑠𝑒
Thus the swing equation can be written as
𝐺 𝑚𝑎𝑐ℎ𝑖𝑛𝑒
𝐺𝑠𝑦𝑠𝑡𝑒𝑚
𝐻 𝑚𝑎𝑐ℎ𝑖𝑛𝑒
𝜋𝑓
𝑑2
𝛿
𝑑𝑡2
= (𝑃𝑖−𝑃𝑒).
𝐺 𝑚𝑎𝑐ℎ𝑖𝑛𝑒
𝐺𝑠𝑦𝑠𝑡𝑒𝑚
Thus,
𝐻𝑠𝑦𝑠𝑡𝑒𝑚
𝜋𝑓
𝑑2
𝛿
𝑑𝑡2
= (𝑃𝑖−𝑃𝑒) 𝑝𝑢 𝑜𝑛 𝑠𝑦𝑠𝑡𝑒𝑚 𝑏𝑎𝑠𝑒
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14. CONT.….
Lets consider the swing equation of two machines on common
system base,
𝐻1
𝜋𝑓
𝑑2
𝛿1
𝑑𝑡2
= (𝑃𝑖1−𝑃𝑒1)
𝐻2
𝜋𝑓
𝑑2
𝛿2
𝑑𝑡2
= (𝑃𝑖2−𝑃𝑒2)
Since the rotors swing in unison (coherently)
𝛿1 = 𝛿2 = 𝛿
Therefore for the two generators we have
𝐻𝑒𝑞
𝜋𝑓
𝑑2
𝛿
𝑑𝑡2
= (𝑃𝑖−𝑃𝑒)
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15. POWER FLOW UNDER STEADY STATE
Consider a short transmission line with negligible resistance,
𝑉𝑠 = 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒 𝑠𝑒𝑛𝑑𝑖𝑛𝑔 𝑒𝑛𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
𝑉𝑅 = 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒 𝑟𝑒𝑐𝑒𝑣𝑖𝑛𝑔 𝑒𝑛𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
𝑉𝑠 𝑙𝑒𝑎𝑑𝑠𝑉𝑅 𝑏𝑦 𝑎𝑛𝑔𝑙𝑒 𝛿
𝑋 = 𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡𝑥𝑛 𝑙𝑖𝑛𝑒
𝑃𝑠 =
𝑉𝑠 𝑉𝑅
𝑋
sin 𝛿
𝑄𝑠 =
𝑉𝑠
2
− 𝑉𝑠 𝑉𝑅 cos 𝛿
𝑋
For lossless transmission line
𝑃𝑠 = 𝑃𝑅
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16. EQUAL AREA CRITERIA
Consider a synchronous generating unit connected through a
reactance to infinite bus, plots of the electrical power 𝑝 𝑒 and
mechanical power 𝑝 𝑚 versus power angle 𝛿 is plotted as shown below
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17. CONT.….
For the stability, the area under the graph of accelerating
power versus rotor angle has to be zero.
Or the positive accelerating area has to be equal with the
negative decelerating area.
This criteria is known as equal area criteria for stability.
𝐴𝑟𝑒𝑎 𝐴1 = 𝐴𝑟𝑒𝑎 𝐴2
𝛿0
𝛿1
(𝑃𝑖−𝑃𝑚𝑎𝑥 sin 𝛿)𝑑𝛿 =
𝛿1
𝛿2
(𝑃𝑚𝑎𝑥 sin 𝛿 − 𝑃𝑖)𝑑𝛿
But for steady state 𝑃𝑖 = 𝑃𝑒
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18. CRITICAL CLEARING ANGLE AND TIME
If a fault occurs in a system, δ begins to increase under the
influence of positive accelerating power, and the system will
become unstable if δ becomes very large.
There is a critical angle within which the fault must be cleared
if the system is to remain stable and the equal area criterion is
to be satisfied.
This angle is known as the critical clearing angle.
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19. CONT.….
Expressing 𝐴𝑟𝑒𝑎 𝐴1 = 𝐴𝑟𝑒𝑎 𝐴2 mathematically we have,
𝑃𝑖(𝛿 𝑐 − 𝛿0) =
𝛿1
𝛿2
(𝑃𝑚𝑎𝑥 sin 𝛿 − 𝑃𝑖)𝑑𝛿
cos 𝛿 𝑐 = cos 𝛿1 + ( 𝛿1 − 𝛿0) sin 𝛿0
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20. CONT.….
The critical fault clearing time can be,
𝛿 𝑐 =
𝜋𝑓𝑝𝑖
2𝐻
𝑡 𝑐
2
+ 𝛿0
𝑡 𝑐 =
2𝐻( 𝛿𝑐 − 𝛿0)
𝜋𝑓𝑝𝑖
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21. END OF CHAPTER FOUR
Thank You
END OF THE COURSE
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