MIDTERM EXAM 2 SPRING 2014
ECE 422/522
1. (25 points) Consider three interconnected areas. The connected load at 60Hz is 20,000MW in Area 1, 30,000MW in Area 2, and 40,000MW in
Area 3. Respectively in Area 1, Area 2 and Area 3, the load varies 1%, 1.5% and 2% for every 1% change in
frequency. Area 1 is exporting 1,200MW, Area 2 is importing 1,500MW, and Area 3 is exporting 300MW. The
speed regulation, R, is 4% for all units. If the load in Area 3 decreases by 1000MW, and there are no
supplementary load frequency controls, determine:
a. the new steady-state system frequency
b. the new generation and load of each area
c. the new MW export or import of each area
a. 12 points
PG1=20000+1200=21200MW
PG2=30000-1500=28500MW
PG3=40000+300=40300MW
1/R1=1/0.04 x 21200/60 = 8833.33MW/Hz
1/R2=1/0.04 x 28500/60 =11875MW/Hz
1/R3=1/0.04 x 40300/60 = 16791.67MW/Hz
1/R = 1/R1 + 1/R2 + 1/R3 = 37500MW/Hz
3
4
D1=1 x 2000/100 x 100/60 = 333.33MW/Hz
D2=1.5 x 3000/100 x 100/60 = 750MW/Hz
D3=2 x (4000-1000)/100 x 100/60 = 1300MW/Hz
D = D1+D2+D3 = 2383.33 MW/Hz
4
f = -PL/ (1/R +D ) = -(-1000) / (37500 + 2383.33 )
=0.0251Hz
So, fnew = 60.0251Hz
1
b. 10 points
PD1=D1f= 8.36MW
PD2=D2f=18.80MW
PD3=D3f=32.60MW
PG1= -f/R1= -221.49MW
PG2= -f/R2= -297.74MW
PG3= -f/R3= -421.02MW
3
3
PL1new=PL1+PL1+PD1=20000+0+8.367=2008.35MW
PL2new=30000+0+18.80 =30018.80MW
PL3new=40000-1000+32.60
=39032.60MW
PG1new=PG1+PG1=21200-221.49=20978.52MW
PG2new=28500-297.74
=28202.26MW
PG3new=40300-421.02
=39878.98MW
4
c. 3 points
Area 1:
Area 2:
Area 3:
1
PG1new-PL1new =970.16MW (exported)
PG2new-PL2new =-1816.55MW (imported)
PG3new-PL3new =846.39MW(exported)
3
2. (25 points) Consider two interconnected areas. The connected load at 60Hz is 19,000MW in area 1 and 32,000MW in area 2. The load in each
area varies 1.2% for every 1% change in frequency. Area 1 is exporting 2,000MW to Area 2. The speed regulation, R, is 5% for all units. Area 1
is operating with a spinning reserve of 2,000MW spread uniformly over a generation of 5,000MW capacity, and Area 2 is operating with a
spinning reserve of 2,000MW spread uniformly over a generation of 10,000MW. If the generation carrying spinning reserve in each area is on
supplementary control with frequency bias factor settings of 800MW/0.1Hz for area 1 and 500MW/0.1Hz for area 2.
Assume Area 1 loses 1000MW generation, which is not carrying any spinning reserve, and at the same time, Area 2 loses 2,000MW generation,
which is part of the generation carrying spinning reserve. Determine the new steady-state frequency, generation and load, final ACE and remaining
spinning reserve of each area, and the new tie-line power flow.
Area 1:
P12=2000MW ->
PL1= 19000MW
PG1=19000+2000=21000MW
D1=1.2 x 19000/100 x 100/60=380MW/Hz
B1=8000MW/Hz
Area 2:
PL2= 32000MW
PG2=32000-2000=30000MW
D2=1.2 x 32000/100 x 100/60=640MW/Hz
B2=5000MW/Hz
In Area 1:
Generation loss: 1000MW,
It causes spinning reserve loss: 0MW
So available spinning reserve: 2000MW
Since 1000MW < 2000MW, the spinning reserve may be sufficient to eliminate ACE1 to 0.
Assume ACE1=0:
ACE1=B1 x f+P12=0
P12= -B1x f= -8000 x f
5
In Area 2:
Generation loss: 2000MW
It causes spinning reserve loss: 2000 x 2000/(10000-2000) =500MW
So available spinning reserve = 1500MW
Since 2000MW > 1500MW, spinning reserve is not sufficient. So ACE2 is not 0
2
The 2000MW generation loss is made up by the remaining spinning reserve, load change due to frequency drop and change of tie-line flow,
i.e.
2000 = 1500
- D2 x f
+P12
= 1500 - 640 x f -8000 x f
Gen. loss Spinning reserve
Load change in area 2 Increase of tie-line flow
5
f =(-2000+1500)/8640= -0.0579Hz
fnew=59.942Hz
1
P12 = -8000x-0.0579=463MW
P12new = P12+P12=2000+463=2463MW
2
PL1=D1 x f =380 x (-0.0579)= -22.00MW
PL2=D2 x f=640 x (-0.0579)= -37.06MW
PL1new=19000-22 =18978MW
PL2new=32000-37.06=31963.4MW
4
PG1=PL1+P12 = -22+463 =441MW
PG1new= PG1+PG1=21441MW Or PG1new= PL1new+P12new=21441MW
Remaining reserve in Area 1: 2000-1000-PG1=559MW>0
That confirms ACE1=0
2
2
Remaining reserve in Area 2: 0MW
PG2new= PL2new -P12new=31963.4-2463=29499.4MW
2
ACE2=B2 x f - P12=5000 x (-0.0579) – 463 = -752.5MW
2
3
3. (25 points) A generating unit has a simplified linearized AVR system
as shown in the figure. If A=0.07 sec, E=0.4 sec, G=0.8 sec, R=0.03 sec,
KR=1, KE=1 and KG=0.9,
a.
Use the Routh-Hurwitz array to find the range of KA for
control system stability (15 points)
b.
If KA is set to 2/3 of the upper limit determined from 3.a, then
estimate the steady-state step response (10 points)
a. 15 points
Characteristic equation:
5
1
658
1487+1339
51
1934
0
620
1487+1339
0
1810-111
0
0
1487+1339
0
0
5
1810-111
>0
5
b. 10 points
(
)
( )
10
4
4. Short Answer (25 points).
a. As frequency is a common factor throughout an interconnected power system, a change in real power demand at one single point is reflected
through the entire system by a change in frequency. True or false, and why?
True
3 points
Reason: See Kundur’s Page 582
2 points
c. The performances of SVC and STATCOM are tested near the same induction
model load. The plot gives the voltage performances following a temporary
short-circuit event for three tests: (1) without SVC or STATCOM, (2) with SVC
only and (3) with STATCOM only. Based on the plot, comment on two
statements. Are they true or false, and why?
i) SVC and STATCOM perform equally well in reducing steady-state
voltage deviation for this event.
ii) SVC and STATCOM must have different maximum Mvar output limits.
i)
ii)
Correct
1.5 points
Reason: Same steady state value
1 point
Incorrect.
1.5 points
Reason: SVC and STATCOM may have the same maximum Mvar
output limit. Why they perform differently during the post-contingency
transient period may be because they have Mvar outputs proportional to
|V|2 and |V| respectively after they hit the limit.
1 point
Induction motor terminal voltage (pu)
b. For a power system with multiple generators operating in parallel to support the system load. Generator speeds are controlled only by their
governing systems (i.e. governor droop control without AGC). When there is a load increase, the generator with the largest speed regulation R
will pick up the largest amount of load increase. True or false, and why?
False
3 points
P1/P2=R2/R1 (Slide 28)
2 points
1.20
0.96
0.
0.72
With STATCOM
With SVC
0.48
0.24
Without VAr
support
0.00
0.00
2.00
4.00
6.00
Time (sec)
d. For a NERC Balancing Authority in the US, what are considered to be its generation resources and load requirements?
The generation resources may be internal or purchased from other BAs and transferred over tie-lines.
2.5 points
5
8.00
10.00
Load requirements may include internal customer loads, losses or scheduled sales to other BAs.
2.5 points
e. Consider a small system with two areas, say area 1 and area 2, whose area control errors are ACE1=P12+B1 and ACE2= -P12+B2. If
B1 for area 1 is set to be smaller than its 1, i.e. frequency bias factor. If area 2 experiences a 100MW increase in load, which of the five
statements is (are) correct, and why?
i) Following that load increase, ACE1 becomes positive
True 1 points
ii) AGC will decrease generation in area 1
True 1 point
iii) ACE1 will always go back to zero
False 1 point
iv) Following that load increase, ACE1 becomes negative
v) AGC will increase generation in area 1
Reasons
2 points
Without using spinning reserve
Let B1=k1
(
PL 2
1
D1 ) (
R2
1
R1
ACE1
P12
k
1
PL 2
D2 )
1
1
P12
(
2
k
1
(k 1)
PL 2 (1 k )
1
D1 )
1
PL 2
1
2
1
1
1
1
100 (1 k )
1
R1
2
1
2
If B1<1, i.e. k<1, then ACE1>0. So, AGC of area 1 will decrease its generation, and AGC of area 2 will increase generation to
balance the load increase and the generation decrease in area 1. ACE1 will go back to zero only if area 1 and area 2 have sufficient
capabilities to increase or decrease their generations.
i) and ii) are correct.
The AGC of an area with Bi<i tends to react AGAINST the other area in which load/generation changes, or in other words, its AGC
may even decrease generation if the other area increases load. Thus, the other area has to overreact to its load increase to increase
more generation in order to eliminate ACE.
6