MIDTERM EXAM 1 SPRING 2015
ECE 422
1. Load modeling (20 points): A bus in a power system has this steady-state operating condition:
Frequency f0=60Hz
Real power load
P0=10MW
Voltage magnitude V0=10kV Reactive power load Q0=5MVAR.
̅ 0. The following estimates are from measurement data at
P and Q are the actual real and reactive power loads. Let 𝑃̅=P/P0, 𝑄̅ =Q/Q0, 𝑉̅ =V/V0 and 𝑓=f/f
̅ 1.0 𝑄̅ /𝑓 =
̅ 2.0.
the bus 𝑃̅/𝑉̅ = 1.6 𝑄̅ /𝑉̅ = 3.0 𝑃̅/𝑓 =
If P is modeled as a frequency-dependent constant-impedance-constant-power (IP) load and Q is modeled as a frequency-dependent constantimpedance-constant-current (ZI) load,
a. Represent P and Q as functions of actual voltage V and frequency f with all coefficients calculated.
b. At frequency f=59.4Hz, draw the characteristics of functions P(V) and Q(V) for V=0~15kV
b
a
̅ = (𝑷𝟏 𝑽
̅ 𝟐 + 𝑷𝟐 )[𝟏 + (𝒇̅ −
𝑷
̅
𝝏𝑷
𝟏) 𝝏𝒇̅ ]
̅ = (𝑸𝟏 𝑽
̅ 𝟐 + 𝑸𝟐 𝑽
̅ )[𝟏 + (𝒇̅ −
𝑸
f=59.4
̅
𝝏𝑸
𝟏) ̅ ]
𝝏𝒇
4’
̅ 𝟎 = 𝑷𝟏 + 𝑷𝟐 = 𝟏
𝑷
̅ 𝟎 = 𝑸𝟏 + 𝑸𝟐 = 𝟏
𝑸
= 𝟐𝑷𝟏 = 𝟏. 𝟔
𝑸 = 𝟓. 𝟏(𝟎. 𝟎𝟐𝑽𝟐 − 𝟎. 𝟏𝑽)
2’
15
4’
= 𝟐𝑸𝟏 + 𝑸𝟐 = 𝟑. 𝟎
10
𝑷𝟏 = 𝟎. 𝟖
𝑷 = 𝟎. 𝟐
{ 𝟐
𝑸𝟏 = 𝟐. 𝟎
𝑸𝟐 = −𝟏. 𝟎
P/Q (MVA)
̅
𝝏𝑸
{𝝏𝑽̅
̅
𝝏𝑷
̅
𝝏𝑽
𝑷 = 𝟗. 𝟗(𝟎. 𝟎𝟎𝟖𝑽𝟐 + 𝟎. 𝟐)
4’
5
0
𝑷=
𝑽
𝟏𝟎[𝟎. 𝟖(𝟏𝟎)𝟐
𝑽
+ 𝟎. 𝟐](𝟏 +
𝑽
𝑸 = 𝟓[𝟐. 𝟎(𝟏𝟎)𝟐 − 𝟏𝟎](𝟏 +
𝒇−𝟔𝟎
𝟔𝟎
𝒇−𝟔𝟎
𝟔𝟎
)
(−𝟐. 𝟎))
2’
2’
-5
0
5
10
V (kV)
15
2’
2. Short Answer (30 points).
a. (4 points) The figure plots all bus voltages of a power system responding
to a temporary short-circuit fault lasting 0.1 second. Are these statements
regarding system dynamics following the fault true or false, and why?
i. The system must be stable
T
1pts
ii. The system must be secure
F
1pts
Reason: criteria are unknown 1pts
iii. The system must be oscillatory
T
1pts
b. (4 points) About a synchronous machine, these two figures show how
stator mutual inductance lbc and stator-to-rotor mutual inductance lbF
change with rotor position . Is it more likely a round rotor or salient
pole machine, and why?
Round-rotor machine 2 pts
No 2nd order harmonic in lbc 2 pts
c. (4 points) List at least two conditions, under which the d-axis current id and q-axis stator currents iq of a synchronous generator are not direct
currents.
1) Unbalanced conditions 2pts
2) rs
2pts
d. (4 points) Briefly explain why T’d0>T’d>>T”d and Ld>L’d>L”d for a typical synchronous machine (use their formulas about inductances and
resistances)
T’d0>T’d
1pt
-> T’d>>T”d
Ld>L’d>L”d
1pt
2pts
e. (4 points) True or false? Briefly explain why.
i. Any thermal heater load is more likely a constant impedance load than a constant current or power load.
F
2 pts
Heater loads with thermostat controllers are more like constant power loads over a long period 2 pts
ii. The 2nd order classic model of a synchronous generator has no state variables that are currents, voltages or flux linkages of the equivalent daxis and q-axis circuits.
T
2pts
Only state variables are angle and speed
f. (4 points) For a power system with two generators operating together to support one load. Assume that speeds of two generators are controlled
only by their governors with speed regulation factors R1=0.04 and R2=0.06, and there is no supplementary control by AGC. The load is increased
by 100MW, how much load increase is picked up by each generator?
Generators 1 and 2 will pick up 60MW and 40MW, respectively
4 pts
g. (6 points) Consider a two-area power system with area 1 and area 2. Their area control errors are ACE1=P12+B1 and ACE2= -P12+B2.
For area 1, assume B1 > 1, i.e. frequency bias factor. If area 2 has a sudden load decrease of 100MW, are these statements true or false, and why?
i. Right after that load decrease, ACE1 becomes positive
T
ACE1
1 pt
P12
k
1
1
k
PL 2 (1 k )
100 (1 k )
1
1
1
2
1
1
0
2
ii. AGC will decrease generation in area 2
F
1pt
Since B2 is unknown, PG2 may or may not decrease
1pt
iii. ACE1 will always go back to zero
F
1pt
It’s true only when Area 1 has enough spinning reserve (Prefmax – Pref)
1pt
1 pt
3. (20 points) Consider the following equivalent circuits for a 60Hz, 3-phase synchronous generator, which has the following inductance and
resistance parameters in per unit values in the Lad–Laq base per unit system and open-circuit time constants in seconds:
Ld=1.700
Lq=1.600
Ll=0.15
Ra=0.0025
L’d=0.300
L’q=0.500
L”d=0.200
L”q=0.200
T’d0=4.0 s
T’q0=0.50 s T”d0=0.030 s T”q0=0.05 s
The transient and subtransient parameters are based on the classical expression and unsaturated values of Lad and Laq.
a. Determine per unit values of Lad, Laq, Lfd, Rfd, L1d, R1d, L1q, R1q, L2q and R2q.
b. Determine transfer function Ld(s)
c. Assume that the generator is operated with the armature terminal voltage at rated value and its steady-state outputs are Pt=0.9pu and Qt= 0.4pu. Neglect saliency, i.e. let Xs=Xq=Xd, X’q=X’d and X”q=X”d,
i Calculate air-gap torque Te in per unit.
ii Calculate Eq for the simplified steady-state model, E” for the Voltage behind X” model and E’ for the classic model.
iii Draw a phasor diagram showing phasors about Et, It, jXsIt, RaIt, Eq, E’, jX’dIt, E” and jX”dIt
𝑳𝒂𝒅 = 𝑳𝒅 − 𝑳𝒍 = 𝟏. 𝟕 − 𝟎. 𝟏𝟓 = 𝟏. 𝟓𝟓
𝟏
𝑳′𝒅 = 𝑳𝒍 +
𝟏
𝟏
+
= 𝟎. 𝟏𝟕 +
𝑳𝒂𝒅 𝑳𝒇𝒅
𝑳′′𝒅 = 𝑳𝒍 +
𝟏
𝟏
+
𝟏
+
𝟏
= 𝟎. 𝟏𝟕 +
𝑳𝒂𝒅 𝑳𝒇𝒅 𝑳𝟏𝒅
𝑳
𝑹𝒇𝒅 = 𝑻′𝒂𝒅
𝒅𝟎
+𝑳𝒇𝒅
⁄𝒕
𝒃𝒂𝒔𝒆
=
𝟏
𝟏
𝟏
+
𝟏.𝟓𝟓 𝑳𝒇𝒅
𝟏.𝟓𝟓+𝟎.𝟏𝟔𝟔
𝟒∗𝟑𝟕𝟕
1’
= 𝟎. 𝟑 𝑳𝒇𝒅 = 𝟎. 𝟏𝟔𝟔
𝟏
𝟏
𝟏
𝟏
+
+
𝟏.𝟓𝟓 𝟎.𝟏𝟔𝟔 𝑳𝟏𝒅
= 𝟎. 𝟎𝟎𝟏𝟏𝟒
= 𝟎. 𝟐
1’
𝑳𝟏𝒅 = 𝟎. 𝟎𝟕𝟓
1’
1’
𝑹𝟏𝒅 =
𝟏
+𝑳𝟏𝒅
𝟏⁄
𝟏
𝑳𝒂𝒅 + ⁄𝑳𝒇𝒅
𝑻′′𝒅𝟎
⁄𝒕
𝒃𝒂𝒔𝒆
=
𝟏
+𝟎.𝟎𝟕𝟓
𝟏⁄
𝟏
𝟏.𝟓𝟓+ ⁄𝟎.𝟏𝟔𝟔
𝟎.𝟎𝟑∗𝟑𝟕𝟕
= 𝟎. 𝟎𝟏𝟗𝟗
1’
𝑳𝒂𝒒 = 𝑳𝒒 − 𝑳𝒍 = 𝟏. 𝟔 − 𝟎. 𝟏𝟓 = 𝟏. 𝟒𝟓
𝟏
𝑳′𝒒 = 𝑳𝒍 +
𝟏
+
= 𝟎. 𝟏𝟓 +
𝟏
𝑳𝒂𝒒 𝑳𝟏𝒒
𝟏
𝑳′′𝒒 = 𝑳𝒍 +
𝟏
+
𝟏
𝟏
+
𝟏
𝟏
𝟏
+
𝟏.𝟒𝟓 𝑳𝟏𝒒
= 𝟎. 𝟏𝟓 +
𝑳𝒂𝒒 𝑳𝟏𝒒 𝑳𝟐𝒒
𝑳𝒂𝒒 +𝑳𝟏𝒒
𝑹𝟏𝒒 = 𝑻′𝒒𝟎
⁄𝒕
𝒃𝒂𝒔𝒆
𝑹𝟐𝒒 =
=
𝟏.𝟒𝟓+𝟎.𝟒𝟔𝟏
𝟎.𝟓∗𝟑𝟕𝟕
𝟏
+𝑳𝟐𝒒
𝟏⁄
𝟏
+
𝑳𝒂𝒒 ⁄𝑳𝟏𝒒
𝑻′′𝒒𝟎
⁄𝒕
𝒃𝒂𝒔𝒆
=
1’
= 𝟎. 𝟓
𝟏
𝟏
𝟏
𝟏
+
+
𝟏.𝟒𝟓 𝟎.𝟒𝟔𝟏 𝑳𝟐𝒒
𝑳𝟏𝒒 = 𝟎. 𝟒𝟔𝟏
= 𝟎. 𝟐
1’
𝑳𝟐𝒒 = 𝟎. 𝟎𝟓𝟖𝟑
= 𝟎. 𝟎𝟏𝟎𝟏
𝟏
+𝟎.𝟎𝟓𝟖𝟑
𝟏⁄
𝟏
𝟏.𝟒𝟓+ ⁄𝟎.𝟒
𝟎.𝟎𝟓∗𝟑𝟕𝟕
1’
1’
= 𝟎. 𝟎𝟐𝟏𝟕
1’
(𝟏+𝒔𝑻′ )(𝟏+𝒔𝑻′′ )
𝑳𝒅 (𝒔) = 𝑳𝒅 (𝟏+𝒔𝑻′ 𝒅 )(𝟏+𝒔𝑻𝒅′′ )
𝒅𝟎
𝟏
𝟏⁄
𝟏 +𝑳𝒇𝒅
𝑳𝒂𝒅 + ⁄𝑳𝒍
𝑻′𝒅 =
𝑹𝒇𝒅
𝑻′′𝒅 =
𝒅𝟎
∗ 𝒕𝒃𝒂𝒔𝒆 = 𝟎. 𝟕𝟎𝟔 𝒔
𝟏
𝟏⁄
𝟏
𝟏 +𝑳𝟏𝒅
+
⁄
𝑳𝒂𝒅
𝑳𝒇𝒅 + ⁄𝑳𝒍
𝑹𝟏𝒅
𝑳𝒅 (𝒔) = 𝟏. 𝟕
∗ 𝒕𝒃𝒂𝒔𝒆 = 𝟎. 𝟎𝟐 𝒔
1’
1’
(𝟏+𝟎.𝟕𝟎𝟔𝒔)(𝟏+𝟎.𝟎𝟐𝒔)
1’
(𝟏+𝟒𝒔)(𝟏+𝟎.𝟎𝟑𝒔)
̅ 𝒕 = −𝟎. 𝟒
̅𝒕 = 𝟏 𝑷
̅ 𝒕 = 𝟎. 𝟗 𝑸
𝑬
̅
̅
𝑷 +𝒋𝑸
𝑰̃𝒕 = ( 𝒕 𝑬̅ 𝒕)∗ = 𝟎. 𝟗𝟖𝟓∠𝟐𝟑. 𝟗𝟔°
𝒕
cos=0.9138
1’
̅𝒆 = 𝑷
̅𝒕 + 𝑹
̅ 𝒂 𝑰̅𝟐𝒕 = 𝟎. 𝟗𝟎𝟐𝟒
𝑻
1’
b
̃𝒒 = 𝑬
̃ 𝒕 + (𝑹𝒂 + 𝒋𝑿𝒅 )𝑰̃𝒕 = 𝟎. 𝟑𝟐𝟐 + 𝒋𝟏. 𝟓𝟑𝟏 = 𝟏. 𝟓𝟔𝟓∠𝟕𝟖. 𝟏𝟏𝟒°
𝑬
1’
̃ =𝑬
̃ 𝒕 + (𝑹𝒂 + 𝒋𝑿′𝒅 )𝑰̃𝒕 = 𝟎. 𝟖𝟖𝟐 + 𝒋𝟎. 𝟐𝟕𝟏 = 𝟎. 𝟗𝟐𝟑∠𝟏𝟕. 𝟎𝟕𝟓°
𝑬′
1’
̃ =𝑬
̃ 𝒕 + (𝑹𝒂 + 𝒋𝑿′′𝒅 )𝑰̃𝒕 = 𝟎. 𝟗𝟐𝟐 + 𝒋𝟎. 𝟏𝟖𝟏 = 𝟎. 𝟗𝟒∠𝟏𝟏. 𝟏𝟎𝟒°
𝑬′′
1’
Phasor diagram
Eq
2’
jXsIt
It
E`
E``
δ
φ
jXd`It
jXd``It
Et
RaIt
d
4. (15 points) An isolated power station operating at 60Hz has the LFC system
as shown in the figure with
Turbine time constant T=0.5 sec
Governor time constant g=0.25 sec
Generator inertia constant H=10 sec
Governor speed regulation = R per unit
The load varies by 1% for 1% change in frequency, i.e. D=1.
a. Use the Routh-Hurwitz array to find the range of R for control system stability. (10’)
b. If R=0.05 per unit and the reference power Pref has a step increase of Pref to raise the steady-state frequency to 60.3Hz. Find Pref /Pref (5’)
T  0.5
g  0.25
T( s ) 
H  10
D  1
K
T( s ) 
( 1  g  s )  ( 1  T  s )  ( D  2 H s )
expand
ch  (0.25s
  1) (0.5s
  1)(20s  1)  K
a
a
n 1 n 2
1
a
b a
c 
b
b
1
1
2
c
b 
b
a
n
n 1
b 0  a
2
c 
2
1
1
b
0
n 1
n 1 2
1
R 
3
2
 2.5 s  15.125s
  20.75s
 K1
1
124.5375
c
1
Pref =(D+1/R) f
2’
f = (60.3-60)/60=0.005pu
1’
Pref = (1+20) x 0.005 =0.105 pu
1
c
0 solve K  124.5375
K  124.5375
0  a 0
n 1
2
n 1
c b  a
1
a
n n 3
n 1
( 0.25s
  1)  ( 0.5s
  1)  ( 20 s  1)
Under the steady-state condition, (Pref - f/R)/D=f
(all in p.u.)
 a a
a
1 n 3
1
d 
collect s
K  1


20.75 

a 
 15.125
 2.5 


n  3
b 
ch
K
0 solve K  1
3
R  8.03 10
2’
5. (15 points): Consider a two-area power system. The connected load at 60Hz is 20,000MW in Area 1
and 30,000MW in Area 2. Area 1 is exporting 1,000MW. Ignore transmission loss. Area 1 has speed
regulation R1=4% for all units based on its total generation, and Area 2 has speed regulation R2=6% for
all units based on its generation. The load in Area 2 is increased by 1000MW, and there is no
supplementary frequency control by AGC. If respectively in Areas 1 and 2, the load varies 1% and 2%
for every 1% change in frequency. Determine:
a. the new steady-state system frequency
b. the new generation, load and MW export or import of each area
a. 12 points
PG1=20000+1000=21000MW
PG2=30000-1000=29000MW
1/R1=1/0.04 x 21000/60 = 8750MW/Hz
1/R2=1/0.06 x 29000/60 =8056MW/Hz
1/R = 1/R1 + 1/R2 = 16806MW/Hz
1’
1’
2’
D1=1 x 20000/100x100/60 =333.33MW/Hz
D2=2 x (30000+1000)/100 x100/60=1033.33MW/Hz
(If the student uses D2=2 x 30000/100
x100/60=1000MW/Hz, it is also correct. Then answers
below should be updated accordingly)
D = D1+D2 = 1366.67 MW/Hz
2’
f = -PL/ (1/R +D )
= -(1000) / (168060 + 1366.67 ) =-0.055Hz
So, fnew = 59.45Hz
1’
b. 10 points
PD1=D1f= -18.34MW
PG1= -f/R1= 481.5MW
1’
PD2=D2f=-56.86MW 1’
PG2= -f/R2=443.3MW
PL1new=PL1+PL1+PD1=20000+0-18.34=19981.66MW
1’
PL2new=30000+1000-56.86=30943.14MW
1’
PG1new=PG1+PG1=21000+481.5=21481.5MW
PG2new=29000+443.3 =29443.3MW
1’
1’
c. 3 points
Area 1: PG1new-PL1new =1500MW (exported) 1’
Area 2: PG2new-PL2new =-1500MW (imported)
1’