ECE 504Law, J.D.

Power System StabilitySpring 2007

Session 33fPage 1/6

Development of the Swing Equation

ANGLES

θm rotor angle in Mechanical degrees

θsm System angle in Mechanical degrees

δm = θm�θsm (1)

dδm

dt=

dθm

dt�

dθsm

dt(2)

ωr =dδm

dt= ωm�ωsm (3)

where

ωm =dθm

dt

ωsm =dθsm

dt

As an aside that will be useful later, solve Eq. 3 for ωm

ωm =dδm

dt+ωsm = ωr +ωsm (4)

Returning to the main flow. Take the derivative of Eq. 3

dωr

dt=

d2δm

dt2 =dωm

dt�

dωsm

dt(5)

Assume ωsm is a constant.

dωr

dt=

d2δm

dt2 =dωm

dt(6)

ECE 504Law, J.D.

Power System StabilitySpring 2007

Session 33fPage 2/6

MECHANICAL ROTATIONAL EQUATIONS

From mechanics

Jdωm

dt+B ωm = τa = τ0

m� τe (7)

where

J is the combined inertia of the turbine, shaft, rotor, and excitor

B is the total damping

τa is the accelerating torque

τ0

m is the prime mover torque (Mechanical)

τe is the electrical counter torque

Equation 7 can be rewritten using Eq. 6 and Eq. 4

Jd2δm

dt2 +B(dδm

dt+ωsm) = τ0

m� τe (8)

Using the distributive property of multiplication.

Jd2δm

dt2 +Bdδm

dt+B ωsm = τ0

m� τe (9)

Multiply by ωm

ωmJd2δm

dt2 +ωmBdδm

dt+ωmB ωsm = ωmτ0

m�ωmτe (10)

Rearrange terms

ωmJd2δm

dt2 +ωmBdδm

dt+ωmτe = ωmτ0

m�ωmB ωsm (11)

ECE 504Law, J.D.

Power System StabilitySpring 2007

Session 33fPage 3/6

Equation 11 can be rewritten in terms of the electrical power, Pe, and mechanical power, Pm.

ωmJd2δm

dt2 +ωmBdδm

dt+Pe = Pm (12)

where

Pe = ωmτe

Pm = ωmτ0

m�ωmBωsm

From Eq. 4 and recognizing that dδmdt �ωsm

ωm =dδm

dt+ωsm

�= ωsm (13)

ωsmJd2δm

dt2 +ωsmBdδm

dt+Pe = Pm (14)

For a round rotor machine

Pe = Pmaxcos(δ) (15)

where

Pmax =VsEg

XT

δ= δm(NP

2)

Multiply the first term by 1 in the form 2ωsm2ωsm

22

ωsm

ωsmωsmJ

d2δm

dt2 +ωsmBdδm

dt+Pmaxcos(δ) = Pm (16)

2ωsm

ω2sm

2J

d2δm

dt2 +ωsmBdδm

dt+Pmaxcos(δ) = Pm (17)

ECE 504Law, J.D.

Power System StabilitySpring 2007

Session 33fPage 4/6

The initial Kinetic Energy stored in the inertia is:

WKE =ω2

sm

2J (18)

2ωsm

WKEd2δm

dt2 +ωsmBdδm

dt+Pmaxcos(δ) = Pm (19)

Multiply the first two terms by 1 in the form NP=2NP=2

2ωe

WKEd2δdt2 +

ωsm

NP=2B

dδdt

+Pmaxcos(δ) = Pm (20)

where

ωe =NP

2ωsm

Dividing Eq. 20 by Sbase yields:

2ωe

WKE

Sbase

d2δdt2 +

ωsm

(NP=2)B

Sbase

dδdt

+Pmaxpucos(δ) = Pmpu (21)

2ωe

Hd2δdt2 +D

dδdt

+Pmaxpucos(δ) = Pmpu (22)

where

H =WKE

Sbase

D =ωsm

(NP=2)B

Sbase

ECE 504Law, J.D.

Power System StabilitySpring 2007

Session 33fPage 5/6

Defining M =2Hωe

finally yields the per unit swing equation.

Md2δdt2 +D

dδdt

+Pmaxpucos(δ) = Pmpu (23)

Page 2 of s25c handout.

With the CB’s open, Pmax = 0.

Md2δdt2 +D

dδdt

= Pmpu (24)

Page 1 of s25c handout.

ECE 504Law, J.D.

Power System StabilitySpring 2007

Session 33fPage 6/6

The Equal Area Criteria (s25b handout) ignores damping; i.e., D = 0. With the CB’s open(Pe = 0),

Md2δdt2 = Pmpu (25)

α =d2δdt2 =

Pmpu

M(26)

α =ωePmpu

2H(27)

α =NP=2ωsmPmpu

2H�= a constant (28)

α =dωr

dt(29)

ωr = αt+ωr0 (30)

ωr0 = 0 (31)

ωr = αt (32)

dδdt

= ωr = αt (33)

Z δ

δ0

dδ=Z t

0αtdt (34)

δ�δ0 =12

αt2 (35)

δ= δ0+12

αt2 (36)