MIDTERM EXAM 1 SPRING 2014
ECE 422/522
NAME:
BE SURE TO SHOW YOUR WORK CLEARLY AND FULLY. SHOWING YOUR THINKIING HELPS
YOU GET MORE POINTS.
All questions are required for both ECE422 and ECE522 students. ECE422 students will receive 10% more
points.
1 and 2 should be finished in class. 3 and 4 are take-home questions due by 12AM midnight today.
1. Load modeling (25 points): Under a steady-state operating condition, a bus has real power load P0=100MW,
reactive power load Q0=50MVAr, voltage magnitude V0=220kV and frequency f0=60Hz. If its real and reactive
power loads, P and Q, under this condition can both be modeled as frequency-dependent ZI (constant
̅ 0.8, 𝑄̅ /𝑉̅ = 2.7 and 𝑄̅ /𝑓=
̅ 2.3, where 𝑃̅=P/P0,
impedance and current) loads with 𝑃̅ /𝑉̅ = 1.3, 𝑃̅/𝑓=
̅ 0.
𝑄̅ =Q/Q0, 𝑉̅ =V/V0 and 𝑓=f/f
a. Represent P and Q as functions of V and f and calculate all coefficients.
b. When f=59.7Hz, draw the characteristics of functions P(V) and Q(V)
a
̅ = (𝑷𝟏 𝑽
̅ 𝟐 + 𝑷𝟐 𝑽
̅ )[𝟏 + (𝒇̅ − 𝟏)𝑲𝒑𝒇 ]
𝑷
̅ = (𝑸𝟏 𝑽
̅ 𝟐 + 𝑸𝟐 𝑽
̅ )[𝟏 + (𝒇̅ − 𝟏)𝑲𝒒𝒇 ]
𝑸
8’
𝑽 = 𝑽𝟎 , 𝒇 = 𝒇𝟎
̅
𝝏𝑷
= 𝑲𝒑𝒇 = 𝟎. 𝟖
𝝏𝒇̅
̅
𝝏𝑸
= 𝑲𝒒𝒇 = −𝟐. 𝟑
𝝏𝒇̅
̅ 𝟎 = 𝑷𝟏 + 𝑷𝟐 = 𝟏
𝑷
̅ 𝟎 = 𝑸𝟏 + 𝑸𝟐 = 𝟏
𝑸
̅
𝝏𝑷
= 𝟐𝑷𝟏 + 𝑷𝟐 = 𝟏. 𝟑
̅
𝝏𝑽
̅
𝝏𝑸
= 𝟐𝑸𝟏 + 𝑸𝟐 = 𝟐. 𝟕
{ 𝝏𝑽
̅
𝑷𝟏 = 𝟎. 𝟑
𝑷 = 𝟎. 𝟕
{ 𝟐
𝑸𝟏 = 𝟏. 𝟕
𝑸𝟐 = −𝟎. 𝟕
4’
𝑷 = 𝟏𝟎𝟎[𝟎. 𝟑(
𝑸 = 𝟓𝟎[𝟏. 𝟕(
𝑽 𝟐
𝑽
𝒇 − 𝟔𝟎
) + 𝟎. 𝟕
](𝟏 +
𝟎. 𝟖)
𝟐𝟐𝟎
𝟐𝟐𝟎
𝟔𝟎
𝑽 𝟐
𝑽
𝒇 − 𝟔𝟎
) − 𝟎. 𝟕
](𝟏 +
(−𝟐. 𝟑))
𝟐𝟐𝟎
𝟐𝟐𝟎
𝟔𝟎
4’
b
f=59.7
𝑷 = 𝟗𝟗. 𝟔(𝟎. 𝟑(
𝑽 𝟐
𝑽
) + 𝟎. 𝟕
)
𝟐𝟐𝟎
𝟐𝟐𝟎
𝑸 = 𝟓𝟎. 𝟓𝟕(𝟏. 𝟕(
𝑽 𝟐
𝑽
) − 𝟎. 𝟕
)
𝟐𝟐𝟎
𝟐𝟐𝟎
4’
5’
(Using the exponential model receives up to 10’)
2. True or false? If it is false, explain why briefly (25 points).
a. In NERC’s reliability standards, power system events resulting in the loss of a single element are
referred to the Category A contingencies and events resulting in the loss of two elements belong to the
Category B contingencies.
False. Loss of a single element are referred to the Category B. Loss of two elements belong to the
Category C
(Background - Slides #24~25)
b. Under a steady-state operating condition, all the operating quantities that characterize the condition are
constant.
False. Actually, not constant but treated as constant only for purpose of analysis. (Background – Slide
#29)
c. A bulk power system that has adequate energy resources to support the electrical demands of its
customers does not have reliability problems.
False. Reliability includes adequacy and security. A system without adequacy issue could still have
security issue. (Background – Slide #23)
d. Power system stability is essentially a single problem; Because of high dimensionality and complexity
of stability problems, it helps to make simplifying assumptions to analyze specific types of problems
using an appropriate degree of detail of system representation and appropriate analytical techniques.
Ture. (Background – slide #37)
e. A dynamic load model enables modeling dynamic characteristics of a load with changes of the bus
voltage magnitude and/or frequency at any instant of time. It can be described by algebraic functions of
the bus voltage magnitude and frequency at that instant.
False. Described by differential equations. (Load modeling – slide #10)
NAME:
3. Equivalent Circuits (35 points): Consider the following equivalent circuits for a 483 MVA (3-phase), 24kV
(line-to-line RMS), 0.9 power factor, 60Hz, 3 phase, 2 pole 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=1.800
Lq=1.720
Ll=0.17
Ra=0.0027
L’d=0.285
L’q=0.490
L”d=0.220
L”q=0.220
T’d0=3.7 s
T’q0=0.48 s T”d0=0.032 s T”q0=0.06 s
The transient and subtransient parameters are based on the classical definitions and unsaturated values of Lad
and Laq.
a. Determine per unit values of fundamental parameters, i.e. all inductances and resistances in the d- and qaxis equivalent circuits
b. Explain why inequalities Ld>L’d>L”d and T’d0 >> T”d0 hold
c. If ifd base (iFbase)=1875A. Determine the following parameters in mH or .
Ll, Ld, Lq, Lad, Laq, MF (i.e. Lafd in Kundur’s book), LF (i.e. Lffd in Kundur’s book), Lfd, Rfd and
Ra
d. Determine transfer function Ld(s)
a (Kundur’s Example 4.1 in Page 153)
𝑳𝒂𝒅 = 𝑳𝒅 − 𝑳𝒍 = 𝟏. 𝟖 − 𝟎. 𝟏𝟕 = 𝟏. 𝟔𝟑
𝑳′𝒅 = 𝑳𝒍 +
𝟏
𝟏
𝟏
𝑳𝒂𝒅 + 𝑳𝒇𝒅
= 𝟎. 𝟏𝟕 +
𝟏
𝟏
𝟏
𝟏. 𝟔𝟑 + 𝑳𝒇𝒅
= 𝟎. 𝟐𝟖𝟓
𝑳𝒇𝒅 = 𝟎. 𝟏𝟐𝟑𝟕
𝑳′′𝒅 = 𝑳𝒍 +
𝟏
𝟏
= 𝟎. 𝟏𝟕 +
= 𝟎. 𝟐𝟐𝟎
𝟏
𝟏
𝟏
𝟏
𝟏
𝟏
𝑳𝒂𝒅 + 𝑳𝒇𝒅 + 𝑳𝟏𝒅
𝟏. 𝟔𝟑 + 𝟎. 𝟏𝟐𝟑𝟕 + 𝑳𝟏𝒅
𝑳𝟏𝒅 = 𝟎. 𝟎𝟖𝟖𝟓
𝑹𝒇𝒅 =
𝑹𝟏𝒅
𝑳𝒂𝒅 + 𝑳𝒇𝒅 𝟏. 𝟔𝟑 + 𝟎. 𝟏𝟐𝟑𝟕
=
= 𝟎. 𝟎𝟎𝟏𝟐𝟔
𝑻′𝒅𝟎⁄
𝟑. 𝟕 ∗ 𝟑𝟕𝟕
𝒕𝒃𝒂𝒔𝒆
𝟏
+ 𝟏⁄𝑳
𝟏
+ 𝑳𝟏𝒅
+ 𝟎. 𝟎𝟖𝟖𝟓
𝟏⁄
𝟏
𝟏⁄
⁄
+
𝑳𝒂𝒅
𝒇𝒅
𝟎. 𝟏𝟐𝟑𝟕
=
= 𝟏. 𝟔𝟑
= 𝟎. 𝟎𝟏𝟔𝟗
𝑻′′𝒅𝟎⁄
𝟎. 𝟎𝟑𝟐 ∗ 𝟑𝟕𝟕
𝒕𝒃𝒂𝒔𝒆
𝑳𝒂𝒒 = 𝑳𝒒 − 𝑳𝒍 = 𝟏. 𝟕𝟐𝟎 − 𝟎. 𝟏𝟕 = 𝟏. 𝟓𝟓
𝑳′𝒒 = 𝑳𝒍 +
𝟏
𝟏
𝟏
𝑳𝒂𝒒 + 𝑳𝟏𝒒
= 𝟎. 𝟏𝟕 +
𝟏
𝟏
𝟏
+
𝟏. 𝟓𝟓 𝑳𝟏𝒒
= 𝟎. 𝟒𝟗𝟎
𝑳𝟏𝒒 = 𝟎. 𝟒𝟎𝟑𝟑
𝑳′′𝒒 = 𝑳𝒍 +
𝟏
𝟏
= 𝟎. 𝟏𝟕 +
= 𝟎. 𝟐𝟐𝟎
𝟏
𝟏
𝟏
𝟏
𝟏
𝟏
+
+
+
+
𝑳𝒂𝒒 𝑳𝟏𝒒 𝑳𝟐𝒒
𝟏. 𝟓𝟓 𝟎. 𝟒𝟎𝟑𝟑 𝑳𝟐𝒒
𝑳𝟐𝒒 = 𝟎. 𝟎𝟓𝟗𝟑
𝑹𝟏𝒒 =
𝑹𝟐𝒒
𝑳𝒂𝒒 + 𝑳𝟏𝒒 𝟏. 𝟓𝟓 + 𝟎. 𝟒𝟎𝟑𝟑
=
= 𝟎. 𝟎𝟏𝟎𝟖
𝑻′𝒒𝟎
𝟎. 𝟒𝟖 ∗ 𝟑𝟕𝟕
⁄𝒕
𝒃𝒂𝒔𝒆
𝟏
𝟏
+ 𝑳𝟐𝒒
+ 𝟎. 𝟎𝟓𝟗𝟑
𝟏⁄
𝟏⁄
𝟏
+
⁄𝟏. 𝟓𝟓 + 𝟏⁄𝟎. 𝟒𝟎𝟑𝟑
𝑳𝒂𝒒
𝑳𝟏𝒒
=
=
= 𝟎. 𝟎𝟏𝟔𝟖
𝑻′′𝒒𝟎
𝟎. 𝟎𝟔 ∗ 𝟑𝟕𝟕
⁄𝒕
𝒃𝒂𝒔𝒆
10’
b
𝑳𝒅 = 𝑳𝒍 +
𝑳′𝒅 = 𝑳𝒍 +
𝑳′′𝒅 = 𝑳𝒍 +
𝟏
𝟏⁄
𝑳𝒂𝒅
𝟏
𝟏⁄
𝟏
𝑳𝒇𝒅 + ⁄𝑳𝒂𝒅
𝟏
𝟏⁄
𝟏
𝟏
𝑳𝒇𝒅 + ⁄𝑳𝒂𝒅 + ⁄𝑳𝟏𝒅
𝑳𝒅 > 𝑳′𝒅 > 𝑳′′𝒅
𝑻′𝒅𝟎 =
𝑻′′𝒅𝟎 =
𝑳𝒂𝒅 + 𝑳𝒇𝒅
𝑹𝒇𝒅
𝑳𝟏𝒅 + 𝑳𝒂𝒅 //𝑳𝒇𝒅
𝑹𝟏𝒅
𝑹𝒇𝒅 ≪ 𝑹𝟏𝒅 ,
𝑳𝒂𝒅 + 𝑳𝒇𝒅 ≫ 𝑳𝟏𝒅 + 𝑳𝒂𝒅 //𝑳𝒇𝒅
𝑻′𝒅𝟎 ≫ 𝑻′′𝒅𝟎
2’
c (Kundur’s Example 3.1 in Page 90, slide #48)
𝑬𝑺𝒃𝒂𝒔𝒆 =
𝟐𝟒
√𝟑
= 𝟏𝟑. 𝟖𝟔 𝒌𝑽
𝒆𝑺𝒃𝒂𝒔𝒆 = √𝟐𝑬𝑺𝒃𝒂𝒔𝒆 = 𝟏𝟗. 𝟔𝟎 𝒌𝑽
𝑰𝑺𝒃𝒂𝒔𝒆 =
𝑽𝑨𝒃𝒂𝒔𝒆
𝟒𝟖𝟑 ∗ 𝟏𝟎𝟔
=
= 𝟏𝟏. 𝟔𝟐 𝒌𝑨
𝟑𝑬𝑺𝒃𝒂𝒔𝒆 𝟑 ∗ 𝟏𝟑. 𝟖𝟓𝟔 ∗ 𝟏𝟎𝟑
𝒊𝑺𝒃𝒂𝒔𝒆 = √𝟐𝑰𝑺𝒃𝒂𝒔𝒆 = 𝟏𝟔. 𝟒𝟑 𝒌𝑨
𝒆𝑺𝒃𝒂𝒔𝒆
= 𝟏. 𝟏𝟗𝟐𝟓 𝜴
𝒊𝑺𝒃𝒂𝒔𝒆
𝒁𝑺𝒃𝒂𝒔𝒆 =
𝑳𝑺𝒃𝒂𝒔𝒆 =
𝑳𝒂𝒇𝒅𝒃𝒂𝒔𝒆 (= 𝑴𝑭𝒃𝒂𝒔𝒆 ) =
𝒁𝑺𝒃𝒂𝒔𝒆
= 𝟎. 𝟎𝟎𝟑𝟏𝟔 𝑯
𝝎𝒃𝒂𝒔𝒆
𝑳𝑺𝒃𝒂𝒔𝒆
𝟎. 𝟎𝟎𝟑𝟏𝟔
𝒊𝑺𝒃𝒂𝒔𝒆 =
∗ 𝟏𝟔. 𝟒𝟑 ∗ 𝟏𝟎𝟑 = 𝟎. 𝟎𝟐𝟕𝟕 𝑯
𝒊𝒇𝒅𝒃𝒂𝒔𝒆
𝟏𝟖𝟕𝟓
(Generator modeling – slide #36)
𝒆𝒇𝒅𝒃𝒂𝒔𝒆 =
𝑽𝑨𝒃𝒂𝒔𝒆
= 𝟐𝟓𝟕. 𝟔 𝒌𝑽
𝒊𝒇𝒅𝒃𝒂𝒔𝒆
𝒁𝒇𝒅𝒃𝒂𝒔𝒆 =
𝒆𝒇𝒅𝒃𝒂𝒔𝒆
= 𝟏𝟑𝟕. 𝟑𝟗 𝜴
𝒊𝒇𝒅𝒃𝒂𝒔𝒆
𝑳𝒇𝒅𝒃𝒂𝒔𝒆 =
𝒁𝒇𝒅𝒃𝒂𝒔𝒆
= 𝟑𝟔𝟒. 𝟒 𝒎𝑯
𝝎𝒃𝒂𝒔𝒆
𝑳𝒍 = 𝑳̅𝒍 𝑳𝑺𝒃𝒂𝒔𝒆 = 𝟎. 𝟓𝟑𝟖 𝒎𝑯
𝑳𝒂𝒅 = 𝑳̅𝒂𝒅 𝑳𝑺𝒃𝒂𝒔𝒆 = 𝟓. 𝟏𝟓𝟔 𝒎𝑯
𝑳𝒂𝒒 = 𝑳̅𝒂𝒒 𝑳𝑺𝒃𝒂𝒔𝒆 = 𝟒. 𝟗𝟎𝟑 𝒎𝑯
𝑳𝒅 = 𝑳𝒂𝒅 + 𝑳𝒍 = 𝟓. 𝟔𝟗𝟒 𝒎𝑯
𝑳𝒒 = 𝑳𝒂𝒒 + 𝑳𝒍 = 𝟓. 𝟒𝟒𝟏 𝒎𝑯
̅ 𝑭 = 𝑳̅𝒂𝒅
𝑴
𝑳𝒂𝒇𝒅 (= 𝑴𝑭 ) = 𝑳̅𝒂𝒅 𝑳𝒂𝒇𝒅𝒃𝒂𝒔𝒆 = 𝟒𝟓. 𝟏𝟖𝟖 𝒎𝑯
̅ 𝑹 − 𝑳̅𝒂𝒅 = 𝟎
𝑴
̅ 𝑹 = 𝟏. 𝟕𝟓𝟑𝟕 𝒑𝒖
𝑳̅𝒇𝒇𝒅 (= 𝑳̅𝑭 ) = 𝑳̅𝒇𝒅 + 𝑴
(Generator modeling – Slide #44)
𝑳𝒇𝒇𝒅 (= 𝑳𝑭 ) = 𝑳̅𝒇𝒇𝒅 𝑳𝒇𝒅𝒃𝒂𝒔𝒆 = 𝟔𝟑𝟗. 𝟏𝟏𝟏 𝒎𝑯
𝑳𝒇𝒅 = 𝑳̅𝒇𝒅 𝑳𝒇𝒅𝒃𝒂𝒔𝒆 = 𝟒𝟓. 𝟎𝟗𝟏 𝒎𝑯
̅ 𝒂 𝒁𝑺𝑩 = 𝟎. 𝟎𝟎𝟑𝟐𝟐 𝜴
𝑹𝒂 = 𝑹
̅ 𝒇𝒅 𝒁𝒇𝒅𝒃𝒂𝒔𝒆 = 𝟎. 𝟏𝟕𝟑 𝜴
𝑹𝒇𝒅 = 𝑹
19’
d
𝑳𝒅 (𝒔) = 𝑳𝒅
(𝟏 + 𝒔𝑻′𝒅 )(𝟏 + 𝒔𝑻′′
𝒅)
′
(𝟏 + 𝒔𝑻𝒅𝟎 )(𝟏 + 𝒔𝑻′′
𝒅𝟎 )
𝟏
𝟏⁄
𝟏⁄ + 𝑳𝒇𝒅
+
𝑳𝒂𝒅
𝑳𝒍
𝑻′𝒅 =
∗ 𝒕𝒃𝒂𝒔𝒆 = 𝟎. 𝟓𝟖𝟔 𝒔
𝑹𝒇𝒅
𝟏
𝟏⁄
𝟏
𝟏⁄ + 𝑳𝟏𝒅
+
+
⁄
𝑳𝒂𝒅
𝑳𝒇𝒅
𝑳𝒍
𝑻′′𝒅 =
∗ 𝒕𝒃𝒂𝒔𝒆 = 𝟎. 𝟎𝟐𝟒𝟕 𝒔
𝑹𝟏𝒅
𝑳𝒅 (𝒔) = 𝟏. 𝟖
(𝟏 + 𝟎. 𝟓𝟖𝟔𝒔)(𝟏 + 𝟎. 𝟎𝟐𝟒𝟕𝒔)
(𝟏 + 𝟑. 𝟕𝒔)(𝟏 + 𝟎. 𝟎𝟑𝟐𝒔)
4’
4. Simplified synchronous machine models (15 points): Continuing with Question 1, assume that the
generator is operated with the armature terminal voltage at rated value and its steady-state outputs are
Pt=250MW and Qt=115MVAr. Neglect saturation and saliency (i.e. let Xq=Xd, X’q=X’d and X”q=X”d),
a. Calculate air-gap torque Te in per unit and in Nm.
b. Calculate Eq for the simplified steady-state model, E” for the Voltage behind X” model and E’ for the
classic model.
c. Draw a phasor diagram showing phasors about these quantities: Et, It, jXsIt, RaIt, Eq, E’, jX’dIt, E” and
jX”dIt
a (slide #52, Kundur’s Page 101)
̅𝒕 = 𝟏
𝑬
̅𝒕 =
𝑷
𝟐𝟓𝟎
= 𝟎. 𝟓𝟏𝟕𝟔
𝟒𝟖𝟑
̅𝒕 =
𝑸
𝑰̃𝒕 = (
𝟏𝟏𝟓
= 𝟎. 𝟐𝟑𝟖
𝟒𝟖𝟑
̅𝒕
̅ 𝒕 + 𝒋𝑸
𝑷
)∗ = 𝟎. 𝟓𝟕𝟎∠ − 𝟐𝟒. 𝟕𝟎°
̅𝒕
𝑬
cos=0.9085
̅𝒆 = 𝑷
̅𝒕 + 𝑹
̅ 𝒂 𝑰̅𝟐𝒕 = 𝟎. 𝟓𝟏𝟖𝟓
𝑻
𝑻𝒃𝒂𝒔𝒆 =
𝟒𝟖𝟑 ∗ 𝟏𝟎𝟔
= 𝟏. 𝟐𝟖𝟏 ∗ 𝟏𝟎𝟔 𝑵𝒎
𝟑𝟕𝟕
̅𝒆 𝑻𝒃𝒂𝒔𝒆 = 𝟎. 𝟔𝟒𝟑 ∗ 𝟏𝟎𝟔 𝑵𝒎
𝑻𝒆 = 𝑻
6’
b
̃𝒒 = 𝑬
̃ 𝒕 + (𝑹𝒂 + 𝒋𝑿𝒅 )𝑰̃𝒕 = 𝟏. 𝟒𝟑𝟎 + 𝒋𝟎. 𝟗𝟑𝟏 = 𝟏. 𝟕𝟎𝟔∠𝟑𝟑. 𝟎𝟔𝟖°
𝑬
̃ =𝑬
̃ 𝒕 + (𝑹𝒂 + 𝒋𝑿′𝒅 )𝑰̃𝒕 = 𝟏. 𝟎𝟔𝟗 + 𝒋𝟎. 𝟏𝟒𝟕 = 𝟏. 𝟎𝟕𝟗𝟑∠𝟕. 𝟖𝟐𝟏°
𝑬′
̃ =𝑬
̃ 𝒕 + (𝑹𝒂 + 𝒋𝑿′′𝒅 )𝑰̃𝒕 = 𝟏. 𝟎𝟓𝟒 + 𝒋𝟎. 𝟏𝟏𝟑 = 𝟏. 𝟎𝟓𝟗𝟖∠𝟔. 𝟏𝟑𝟐°
𝑬′′
6’
c
3’