Example Swing Equation

7/23/2019 Example Swing Equation

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Examples: Swing Equation

Example 1

A 3-phase, 60 Hz, 500 MVA, 15 kV, 32-pole hydroelectric generating unit has an H

constant o 2!0 per unit seconds!

a" #eter$ine the synchronous angular %elocity o the rotor,ωsm , in rad&s!

'" #eter$ine the electrical synchronous angular %elocity,ωs , in rad&s!

c" (rite do)n the per unit s)ing e*uation or this unit!d" +he unit is initially operating at e*uili'riu$, i!e! $ p!u! e p!u! 1!0 per

unit, . .s and / 10o )hen a 3-phase to ground 'olted short circuit at the

generator ter$inals causes e p!u to drop to zero or t 0! #eter$ine the

po)er angle 3 cycles ater the short circuit co$$ences! Assu$e $ p!u!

re$ains constant at 1!0 p!u! ,and .p!u.(t) 1!0, )here .p!u.(t) is the per unit

rotor angle gi%en 'y .p!u.(t)

ωm(t )

ωms

Solution

a" or the 60 Hz generator,

N s=120 f

p =¿

120 60&32 225 rp$

s=¿ 2 π

60225❑

ωsm=2π

60 N ¿

23!56 rad&s

ωsm= 2

p ω s

'"

ωs= p

2ωsm =

32

223.56

= 3 rad&s

c"



2 H

ωs

d2

δ

dt 2 = Pm ( pu )− Pe( pu) 'eco$es

2 H

ωs

ω p .u .( t )d

2δ

dt 2 = Pm ( pu )− Pe ( pu) as .p!u.(t)

is taken as 1!0 per unit!

2 x22 π 60

ω p .u .(t ) d2

δ dt

2 = Pm ( pu )− Pe( pu)

4

2 π 60

d2

δ

dt 2 = Pm ( pu )− Pe( pu)

d" At ti$e t 0, the change in tor*ue angle is zero,

d δ (0)

dt =0 ,

and the initial po)er angle& tor*ue angle is /40" 10o 10 &10 0!175 radian

8sing $ p!u! 1!0, e p!u! 0, and .p!u.(t) 1!0, the s)ing e*uation 'eco$es

4

2 π 60

d2

δ

dt 2 =1.0 or t 0!

d2

δ

dt 2 =

2 π 60

41.0

9ntegrating, and using the a'o%e initial condition,

dδ (t )dt

=2π 60

4t +c , at t 0,

d δ (0)

dt =0 , c 0

dδ (t )dt

=2π 60

4t

9ntegrating again,

δ (t )=2π 60

4 x 2t 2+c

, at t 0, /40" 0!175 radian,

δ (t )=2π 60

4 x 2t 2+0.1745



At t 3 cycles 3 cycles&60 cycles&second 0!05 second!

:u'stitute into e*uation;

δ (0.05)=2π 60

4 x 2(0.05)2+0.1745

0!2<23 radian 16!5o

Example 2

A synchronous generator is initially operating in the steady-state condition! At this

steady state condition, /o 23!<5o! A te$porary 3-phase to ground 'olted short

circuit occurs at point )hich is at the generator=s 'us'ar! +hree cycles later theault etinguishes itsel! #eter$ine )hether sta'ility is $aintained or not and

deter$ine the $ai$u$ po)er angle! +he inertia constant o the generating unit is

3!0 per unit seconds! Assu$e $ re$ains constant throughout the distur'ance!

Also assu$e $a 2!763 p!u!

Solution

/o 23!<5o 0!71< radian! $4pu"e4pu" initially e*uals 1!0 per unit! At t 0,

)hen the short circuit occurs, e instantaneously drops to zero and re$ains at zero

during the ault, since po)er cannot 'e transerred past the aulted 'us!

2 H

ωs

d2

δ

dt 2 = Pm ( pu )− Pe ( pu ) , assu$ing .p!u.(t) 1!0

or 3 cycles on a 60 Hz syste$, t 0!05 s )hen ault is cleared, e4pu" 0



2 H

ωs

d2

δ

dt 2 = Pm ( pu )

0≤t ≤0.05 s

d2

δ

dt 2 =

wsPm ( pu)

2 H

9ntegrating t)ice )ith initial conditions /40" 23!<5o anddδ (0)

dt =0 ,

dδ (t )dt

=ws Pm( pu)

2 H t +0

δ (t )=ws Pm( pu)

4 H t

2+¿ /o

At t 3 cycles 0!05 s,

/ 1 δ (0.05)=2π 60(1)4 x 3

0.052+¿ 0!71< 0!7<67 radian 2!77o

+he accelerating area A1 is gi%en 'y

A1 ∫δ o

δ 1

Pm dδ =∫δ o

δ 1

1.0dδ =δ 1−δ 0=

0.4964

−0.4179

=0.0785

At t0!05 s the ault etinguishes and e instantaneously increases ro$ zero to the

sinusoidal cur%e! > continues to increase until the decelerating area A2 e*uals A1!

A2

P

(¿¿ max sinδ − Pm)dδ = ∫0.4964

δ 2

(2.4638 sinδ −1.0)dδ =¿

∫δ 1

δ 2

¿

A1 0.0785

9ntegrating,

2.4638 [(−cosδ ) ]−[δ ]

0!2763?cos40!7<67" @ cos/2 @4 /2 - 0!7<67" 0!05



0!2763cos/2 B /2 2!573

+his nonlinear alge'raic e*uation can 'e sol%ed iterati%ely to o'tain

/2 0!003 radian 70!12o

+hus sta'ility is $aintained!

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Example Swing Equation