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ECE473 Exam 3, Spring 2008: Electrical Engineering Problems, Exams of Electrical and Electronics Engineering

The instructions and problems for exam 3 of the ece473 course in electrical engineering, held in spring 2008. The exam covers topics such as nonlinear equations, gauss algorithm, power system components, and load flow equations. Students are required to perform calculations and find solutions using the provided information.

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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ECE473 EXAM #3 SPRING 2008
Instructions:
1. Closed-book, closed-notes, open-mind exam.
2. Work each problem on the exam booklet in the space provided.
3. Write neatly and clearly for partial credit. Cross out any material you do not want graded.
Name:
Problem 1: /20
Problem 2: /20
Problem 3: /30
Problem 4: /30
Total: /100
+
jX
P1
Q1
V1 1
θ
+
V2 2
θ
+
P1
Q1
V1 1
θ
R
+
V2 2
θ
P1=V1V2
Xsin(θ1θ2)P1=V2
1
RV1V2
Rcos(θ1θ2)
Q1=V2
1
XV1V2
Xcos(θ1θ2)Q1=V1V2
Rsin(θ1θ2)
Per-Unit System:
Zb=Vb
Ib
=V2
b
Sb
=V2
LN
S1φ
=V2
LL
S3φ
Zu=Z
Zb
=Z×Sb
V2
b
Znew
u=Zold
u×ÃVold
b
Vnew
b!2
×Snew
b
Sold
b
1
pf3
pf4
pf5

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ECE473 EXAM #3 SPRING 2008

Instructions:

  1. Closed-book, closed-notes, open-mind exam.
  2. Work each problem on the exam booklet in the space provided.
  3. Write neatly and clearly for partial credit. Cross out any material you do not want graded.

Name:

Problem 1: /

Problem 2: /

Problem 3: /

Problem 4: /

Total: /

jX

P 1 Q 1

V 1 θ 1

V 2 θ 2

P 1 Q 1

V 1 θ 1

R

V 2 θ 2

P 1 =

V 1 V 2

X

sin(θ 1 − θ 2 ) P 1 =

V 12

R

V 1 V 2

R

cos(θ 1 − θ 2 )

Q 1 =

V 12

X

V 1 V 2

X

cos(θ 1 − θ 2 ) Q 1 = −

V 1 V 2

R

sin(θ 1 − θ 2 )

Per-Unit System:

Zb = Vb Ib

V (^) b^2 Sb

V LN^2

S 1 φ

V LL^2

S 3 φ

Zu =

Z

Zb

= Z ×

Sb V (^) b^2

Znewu = Zuold ×

( V (^) bold V (^) bnew

) 2 × Sbnew Soldb

The following nonlinear equation,

F (x) = sin

( 5 π 4

− x

)

e−x √ 2

can be recast under the following Gauss formulation:

x =

5 π 4

  • sin−^1

( e−x √ 2

)

Perform five iterations using the Gauss algorithm.

Iteration k x(k) 0 π 1 2 3 4 5

T1 T

G M

The ratings of the power system components in the above one-line diagram are:

Generator G: 60 MVA 22 kV XG = 15% Transformer T1: 50 MVA 20-kV/200-kV XT 1 = 10% Line 2-3 200-kV XL = 100 Ω Transformer T2: 60 MVA 200-kV/10-kV XT 2 = 12% Motor M: 50 MVA 11 kV XM = 15%.

a) Using bases of 100 MVA and 200 kV as system bases in the high-voltage (transmission) zone, find the system bases in all other zones of this power system.

Base MVA (3φ) Base kV (LL) Zb (Ω) Generator Side Transmission Line 100 200 Load Side

(b) Compute all of the impedances in the per-unit impedance diagram below.

M

E

G E

(c) Assuming that the motor is drawing 45 MVA at a 0.8 lagging power factor and at a line-to-line voltage of 9 kV at Bus 4, determine the terminal voltage magnitude | V˜ 1 | in per-unit and in kV line-to-line.

G

G

Q = 0

P = 0

L

L

G

G

L

L

G

G

Q = 0

P = 2. Q =? P = 0

0 o θ 2

L

P 3 Q 3 P 2 Q 2 P 1 Q 1

θ 3

P =? Q =?

Q =?

P = 0

Q = 0

P = 0

L

1 1 1 j0.2 0.

PV PV

(a) Write two (2) load-flow equations needed to solve for the two basic unknown variables (θ 2 and θ 3 ). (Derive these equations using the fundamental power transfer equations across inductive or resistive transmission lines.)

(b) Verify that the solutions θ 2 = 60 o^ and θ 3 = 90 o^ satisfy the two load equations in Part (a).

(c) Find P 1 and Q 1.