ECE 307 – Exam 3 – Spring 2003 – Date – Name (print):
This test is closed book and closed notes. SHOW ALL WORK. Partial credit may be given. Work the problems according to the method specified. Be sure to specify whether your answers are in degrees or radians.
1 Power Calculation (25 points)
You are a non-electrical engineering employee working for an electronics manufacturer that has been producing defective DVD players. Your boss decides that, although EE is not your focus area, since you have obtained a quality engineering education from South Carolina’s most prestigious university, that you should lead the R&D team that is investigating the defects in the DVD player’s design. You are presented with a capacitor ( C = 2 mF), resistor
( R = 2 Ω), and an inductor ( L = 20 mH) that are connected in series. This series connection of elements is connected across a voltage source with v ( t ) = 10 2 sin (120 πt +
π
) One of your
EE colleagues has determined this voltage phasor to be 10
π
2
, but you know that this is not
2 the case. Find his mistake and finish the job that he doesn’t seem capable of doing, and you’ll be up for promotion by the end of the month.
(A) Write the voltage source as an RMS voltage phasor.
V v ( t ) = 10
= 10
√
√
2 sin (120 πt +
2 cos (120 πt + 0
π
2
◦
)
)
RM S
= 10 0
◦
(B) Find the current that flows through the series connection of elements. Express this current as an RMS current phasor and as a function of time.
Since V = IZ , we need to calculate the series equivalent impedance, Z
S
.
Z
S
= Z
C
+ Z
R
+ Z
L
=
=
1
+ R 0 + jwL jwC
1 j ∗ 120 π ∗ 2 ∗ 10 − 3
+ 2 0 + j ∗ 120 π ∗ 20 ∗ 10
− 3
≈ 2 + j 6 .
214
= 6 .
53 72 .
2
◦
= 6 .
53 1 .
26
I =
V
Z
=
10 0
6 .
53 72 .
2 ◦
= 1 .
53 ( − 72 .
2
◦
) = 1 .
53
√
2 cos (120 πt − 72 .
2
◦
)
(C) Now calculate the maximum instantaneous and average power in each element.
= (1 .
53
√
2) 2 ∗ 2 = 9 .
364 P
R
M
P
R
AV G
P
C
M
P
C
AV G
P
L
M
P
L
AV G
=
= (1 .
53) 2
I 2
M
R
∗ 2 = 4 .
682
=
=
=
=
I
2
RM S
X
C
0
I 2
RM S
X
L
0
=
= (1 .
53) 2
(1 .
53)
2
120 π ∗ 2 ∗ 10 − 3
∗ 120 π ∗ 20 ∗ 10
− 3
= 3 .
= 17
105
.
65
1

2 Power Factor Correction (25 points)
A 900 MWatt (average power) load with a power factor (pf) of .65 lagging is costing your company too much money. Duke power indicates that you could save substantial money if you correct your load’s pf to .9 lagging, thereby making your company eligible for much lower rates per KWH. Unfortunately, your resident electrical engineer has fallen ill with some strange venereal disease and the responsibility has become yours to correct the load’s pf, and save your company from going under.
(A) What size capacitor bank (expressed in KVARs) will be required?
We know that pf bef ore
P
A
= P + jP
X
= cos ( θ ) = .
65, then
= 900 ∗ 10 6 + jP
X
. This means that tan ( θ ) = P
X
/P . Since,
P
X bef ore
= P tan ( θ ) = 900 ∗ 10 6 tan ( cos
− 1 ( .
65)) ≈ 1 .
05 ∗ 10 9
Since the pf after the correction needs to be .9, this means that:
P
X af ter
= P tan ( θ ) = 900 ∗ 10 6 tan ( cos
− 1 ( .
9)) ≈ 435 .
9 ∗ 10 6
This means that in order to correct the pf, a capacitor bank rated at
435 .
9 ∗ 10
6 − 1 .
05 ∗ 10
9
= − 616 .
32 ∗ 10
6
VARs will be required. (= − 616 .
32 ∗ 10 3 KVARs). The sign can be negelected.
(B) How much capacitance (expressed in Farads) will this capacitor bank have? (Assume that the load is supplied power from a 2400 Volt RMS (at 60Hz) transmission line)
P
X
C
=
V
2
RM S
X
C
=
V
2
RM S
− 1 wC
C =
V
P
XC
2
RM S
∗ w
≈ 616 .
32 ∗ 10
6
2400
2 ∗ 377
= 283 .
82 ∗ 10
− 3
Farads. (= 283 .
83 mF)
(C) The load described above can be represented as a resistance in parallel with an inductance. What would this resistance R (expressed in ohms) be?
Since we know that P
AV G
= P
REAL
=
V
2
RM S
R
,
R =
V
2
RM S
P
AV G
=
2400
2
900 ∗ 10
6
≈ 6 .
4 ∗ 10
− 3 ohms. (= 6 .
4 m Ω).
(D) What would the inductive reactance X
L for the load be BEFORE correcting it’s pf.
In part (A) we found that P
X bef ore
= 1 .
05 ∗ 10 9 . Therefore,
P
X
L
=
V
2
RM S
X
L
X
L
=
V
2
RM S
P
XL
=
2400
2
1 .
05 ∗ 10
9
= 5 .
49 ∗ 10
− 3
2

3 3 Phase Question (25 points)
A balanced Wye (Y) 3 phase load is connected to a 3 phase voltage source where the effective voltage is 440 Volts (RMS).
V
AB is the reference phase with an angle of 0
◦
. Each leg of the load has an impedance of Z = 10 + j 10.
(A) Find the line-to-line voltages, phase voltages, line currents, phase currents. (give answers in RMS phasor form)
V
AB
= 440 (0
◦
)
V
V
AB
AB
= 440 ( − 120
◦
= 440 ( − 240
◦
)
)
Since V
L − L
=
√
3 V phase
30
◦
, V phase
=
V
√ L
3
( − 30
◦
).
V
V
V
AN
BN
CN
=
=
=
√
3
√
3
√
3
(0
◦ −
(( − 120)
◦
30
−
◦
)
30
◦
≈ 254 .
03 ( − 30
◦
)
) ≈ 254 .
03 ( − 150
◦
)
(( − 240)
◦ − 30
◦
) ≈ 254 .
03 ( − 270)
◦
)
Since V = ZI and Z = 10 + j 10 = 10
√
2 45
◦
,
I
A
I
B
= I
= I
AN
BN
=
=
I
C
= I
CN
=
V
AN
Z
V
BN
Z
V
CN
Z
=
=
=
√
6
√
6
√
6
( − 75
◦
) = 17 .
97 ( − 75
◦
)
( − 195
◦
) = 17 .
97 ( − 195
◦
)
( − 315
◦
) = 17 .
97 ( − 315
◦
)
(B) Determine the apparent power delivered to the entire 3 phase load.
P
A
= 3 ∗ V phase
∗ I
= 3 ∗ √
3
( − 30
≈ 13 .
689 ∗ 10 3
∗ phase
◦
)
45
∗
◦
√
6
75
◦
=
58 , 080
18
45
◦
3

4 Transformer Questions (25 points)
An ideal transformer that has 100 turns on the primary side and 10 turns on the secondary side. The secondary is connected to a 40 ohm resistor. The primary is connected to 200
Volts RMS.
(A) Determine the average power dissapated in the resistor.
To determine the average power dissipated by the resistor, we could find the voltage on the secondary side and use the fact that P
AV G know that
V p
= V
2
RM S
/R . Since this in an ideal transformer we
=
N p
V s
N s
This implies that
V s
=
V p
N s
N p
=
200 ∗ 10
100
= 20 volts.
Therefore the power dissipated in the resistor is
P = 20
2
/ 40 = 10 watts.
(B) Determine the current that flows on the primary side.
There are several ways to calculate this quantity, but since we already know the power in the load and since this is an ideal transformer, we know that the power supplied to the primary must be the same as the power supplied by the secondary. Since P = V I , we have that
I p
=
P p
V p
=
10
200
=
1
20 amps.
4

5 Extra Credit (+5 points)
What would be required to correct the power factor of the 3 phase load in Problem 3 to
.9 lagging? Draw a diagram illustrating the what you will need to do assuming that you cannot place any capacitor banks directly in parallel with impedances in the Y load. (do not attempt to solve here, just make sure you draw what is needed)
You would simply connect the correction that you would normally place parallel to each impedance in the 3 phase load across the lines serving the load in a delta shaped configuration.
6 Extra Credit (+5 points)
Solve for the total KVARS required in capacitor banks to achieve the above.
From Problem 3, we know that the total apparent power is
P
A
=
58 , 080
18
45
◦ which means that:
P
X bef ore
P
AV G
= P
A bef ore sin ( θ ) =
= P
A bef ore cos ( θ ) =
58 , 080
58 ,
18
080
18 sin (45
◦
) = cos (45
◦
) =
58 , 080
6
58 , 080
6
= 9680
= 9680
To correct to .9 lagging, we know that after correction
P
X af ter
= P
AV G tan ( cos
− 1
( .
9)) ≈ 4688
That means the correction will be a total of about
9680 − 4688 = − 4992
VARS. The sign can be neglected. The total would about 5 KVAR’s.
5