practice midterm exam 2 spring 2011 1. swing...

PRACTICE MIDTERM EXAM 2 SPRING 2011

1. Swing Equation

Swing Equationt:

..

60 0.5

3

m e

e

fP P

H

P

Pre-fault:

2 0.2 0.8 2 0.6eq t L

X X X pu

1.1 1.0sin sin sin 1.83 sin

0.6e e

eq eq

E V E VP P

X X

.. 60

0.5 1.83 sin3

0 0

1.1 1.00.5 sin 15.83

0.6

o

During-fault:

0.40.33

0.4 0.8eq

V V

; ' 0.4 0.8

0.2 0.470.4 0.8

eqX

'

'

1.1 0.33sin sin 0.52 sin

0.7

eq

e

eq

E VP

X

.. 60

0.5 0.52 sin3

Post-fault:

''0.2 0.8 1

eq t LX X X pu

''

''

1.1 1.0sin sin 1.1sin

1e

eq

E VP

X

.. 60

0.5 1.1sin3

0

1 11.1sin 0.5 27.04

2. Transient Stability

Pre_fault& Post_fault

Pepre = Pepost =|Ea||V∞|Xeq

sin δ

Xeq = 0.2 + 0.8 = 1pu

Pepre = Pepost =1.15× 1

1sinδ = 1.15sinδ

power angle:

Pepre = Pepost = Pm ⇒ 0.6 = 1.15× sinδ0 ⇒ δ0 = 0.55rad = 31.45◦

δmax = π − δ0 = π − 0.55 = 2.59rad

P_fault

Pe = 0

ˆ δcc

δ0

(Pm − Pefault)dδ =ˆ δmax

δcc

(Pepost − Pm)dδ

ˆ δcc

0.55

(0.6− 0)dδ =

ˆ 2.59

δcc

(1.15sinδ − Pm)dδ

0.6× (δcc − 0.55) = −1.15cosδ|2.59δcc − 0.6× (2.59− δcc)

0.6× (2.59− 0.55) = −1.15cos2.59 + 1.15cosδcc

0.24 = 1.15cosδcc ⇒ δcc = 1.36rad = 77.95◦

δ̈ =πf

H(Pm − Pefault)⇒ δ̈ =

πf

H(Pm − 0)

δ(t) =1

2

πf

HPmt

2 + δ0 ⇒ δcc =1

2

πf

HPmt

2cc + δ0

tcc =

√(δcc − δ0)× 2H

πfPm⇒ tcc =

√(1.36− 0.55)× 2× 2

π × 60× 0.6

= 0.169

1

3. Synchronous Machine Modeling

a) Classical machine model

b) Exciter Model

Exciter provides excitation to the field (Efd)

c) PSS Model

PSS counters the effect of the excitation response to the disturbance. It uses the frequency deviation to modify the excitation signal

4. Synchronous Machine- Algebraic equations

5. Large Signal Stability

Pre_fault

Pepre =|Ea||V∞|Xeq

sin δ

Xeq = 0.2 + 0.8/2 = 0.6pu

Pepre =1.2× 1

0.6sinδ = 2sinδ

power angle:

Pepre = Pm ⇒ 0.6 = 2× sinδ0 ⇒ δ0 = 0.30rad = 17.46◦

P_fault

Pe = 0

Post_fault

Pepost =|Ea||V∞|X ′′eq

sinδ

X′′

eq = 0.2 + 0.8 = 1pu

Pepost =1.2× 1

1sinδ = 1.2sinδ

Pepost = Pm ⇒ 0.6 = 1.2× sinδ1 ⇒ δ1 = 0.52rad = 30◦

δmax = π − δ1 = π − 0.52 = 2.62rad

part a)

ˆ δcc

δ0

(Pm − Pefault)dδ =ˆ δmax

δcc

(Pepost − Pm)dδ

ˆ δcc

0.30

(0.6− 0)dδ =

ˆ 2.62

δcc

(1.2sinδ − 0.6)dδ

0.6× (δcc − 0.30) = −1.2cosδ|2.62δcc − 0.6× (2.62− δcc)

0.6× (2.62− 0.30) = −1.2cos2.62 + 1.2cosδcc

0.35 = 1.2cosδcc ⇒ δcc = 1.01rad = 57.99◦

part b)

ˆ δcc

δ0

(Pm − Pefault)dδ =ˆ δmax

δcc

(Pepost − Pm + PR)dδ

ˆ δcc

0.30

(0.6− 0)dδ =

ˆ 2.62

δcc

(1.2sinδ − 0.6 + 0.4)dδ

0.6× (δcc − 0.30) = −1.2cosδ|2.62δcc − 0.2× (2.62− δcc)

−0.6× 0.3 + 0.2× 2.62 + 1.2cos2.62 = 1.2cosδcc − 0.4δcc

−0.7 = 1.2cosδcc − 0.4δcc

δcc w 1.62rad = 92.82◦

=> Critical clearing angle has increased

6. Synchronous Machine Equations:

E′d + jE′

q = (ed + jeq) + Ra(id + jiq) + jX ′did −X ′

qiq

E′d = ed + Raid −X ′

qiq

E′q = eq + Raiq + X ′

did

Model2:

Xq = X ′q,

dΨd

dt= 0, i1q=0

from equation sheet:

ed =dΨd

dt− ωsΨq −Raid

Ψq = −Lqiq + La1qi1q ⇒ Ψq = −Lqiq ⇒ ωsΨq = −Xqiq ⇒ ωsΨq = −X ′qiq

⇒ ed = X ′qiq −Raid

⇒ E′d = X ′

qiq −Raid + Raid −X ′qiq

⇒ E′d = 0

jE′q = Et + RaIt + jX ′

did −Xqiq

This model is valid for transient period which is 30ms-10s

6. Short Answer (20 points):

a) What is (are) the input(s) to the power system stabilizer? What is the function of the stabilizerin generator control? (Be specific.)

€nb.rh +o P55 * f ic ' t ' tcnuTd'vvio'h'ar" '

: rt" ;a."pro,,rideb *aa;l)^u], 4*F,T io,r'uc vo wdr^r-e,cs'illahor$'

b) Generator excitation control serves to maintain the voltage at the generator terminals. What isthe drawback of this feedback control?

--+ ZV resul*l in rtaucxnL do"mpir'6 larcyu tea'd"'rg lo nTah ''s' 6'ompt".( +aYe''

n\ flnnciJo. o o'.oll c\/cfaffi rrri+lr frrra arcaq qea/ Area A "t1d

afga B. Afea A iS heing a frf.l<.1<{v / ' v v u o a s v e J r l v . r r v v r u r l v r v s w * J , J - j u v 4 . i o i g 6 ! - u . j _ l Q - ^ i - u i x i S a 6 u u u

citizen on the frequency control with a Bias setting larger than the ideal setting (i.e., above thenatural governor response). Area B has the Bias setting exactly at the natural governor response.If Area B experiences an increase in load, then the initial Area Control Error (ACE) for area Awill be:

greater than zero (i.e., AGC sends lower pulses to the generators in area A)zetoless than zero (i.e., AGC sends raise pulses to the generators in area A)

d) Conceptually one can consider two types of torque that act on a generator and are needed tomaintain operation following a disturbance. What are those two types of torque?

-+ { hsso +r,oo +11* "1"*3"t ^*-'O)S.Xrvohronrztn6

ioropte;

e) Da.mfr ( io"qro

e) Consider a generator modeled by the swing equation tied to an infinite bus. Explain in wordswhy if the steady-state rotor angle is 45 degrees the rotor angle must not exceed 145 degreesdurins a disturbance. 4 . .

t vs w u r t s q s l D l u u q r v v . J a ^ ^ _ l ^ t

ihr, q,+or "^g" --t: "",:l .I:'*;^, i-T#:il"

7t-

l}"f; "ffi;,^-1*. (g^*,,'"":::,t:tr*,5 * "#,'; J; ::i""?ffH ***;:;::#r* s

,'zPrrv a.neL '.h:*TtL 'oo,..t"l- fF-e ro.l-or ,? ̂ O:{7

,1*, o"g,, rv :'^,

O"l,:ilJ".i'rtlo , w for c,ry lu ^r t t uior+,aq, o,qatrw so iKett tl ie' bP(ec;[-"': for crnqlu ^ritL

I V

,5i4n'c hrdneu., spoel- ' Et c n't'rbe' :k l T:

' uJ*ro .

t

\ J ' L J - . . 1 * r . - - . ' l - . J i r l * : t A D l H ' 7Jnvrx,, t

o+ e

b , { n c h f r n 9 ! ' t , o f 6 ' v v " " t - . r ' . r . L . r O c i l A f b ,

,$ ;* ; - i t to lF L" ' ' '1 - 'k r rb}ou; t^ t '4 uw\ ' <do--159