Test2 Solution:
Problem#1
a)
when one cell is shaded the rest cells provide
When ISC = 5A, it has the maximum power if
Note the Rp = 3. So the load should be
b)
Pmax = 10*20 = 200 W
c)
When ,
Problem#2
a)
b)
c)
d)
ECE420 Spring 2012
TEST 1 Solution
Problem#1
(a)
Caution: this is not current, so no  sign at
(b)
(c)
()
(
)
(
)
For those who were asked to solve full load efficiency and VR:
()
(
)
(
Problem#2
(a)
For ZL2, because they are balanced:


Because of 0.9
ECE 420 Midterm #3
Spring 2012
ECE Department
Clemson University
Name: _
CUID: _
Instructions:
Answer all questions completely and show all work.
This is a closed book/closed notes test. You may use a single side of an 8.5x11 piece of
paper with formula
Problem#1
A 60Hz, 4 pole, 11kV, Y connected synchronous generator has stator impedance
per phase. A 5MVA, 0.8 p.f lagging load is connected to the generator.
Rotational losses: (
)
Core loss:
Rotor copper loss:
Calculate:
a)
b)
c)
d)
The rotor speed in rp
ECE 420 Midterm #1
Spring 2012
ECE Department
Clemson University
Name: _
CUID: _
Instructions:
Answer all questions completely and show all work.
This is a closed book/closed notes test. You may use a single side of an 8.5x11 piece of
paper with formula
ECE 420 Midterm #1
Spring 2012
ECE Department
Clemson University
Name: _
CUID: _
Instructions:
Answer all questions completely and show all work.
This is a closed book/closed notes test. You may use a single side of an 8.5x11 piece of
paper with formula
ECE 420 Spring 2012
HW12 Solution
Problem#1:
N
a)
P VK I K cos( K K ) =2195.29+5.49+2.54+0.76=2204 W
K 1
b) Q
~
V
K 1
c)
K
I K sin( K K ) =1274 VAR
S= VRMS I RMS
Vrms (
I rms (
280.1
)2 (
7.4
2
18.1
1.75
)2 (
2.2
2
2
S 2589 A
V
)2 (
)2 (
2
) 2 (1.6) 2 19
ECE420
Spring 2012
HW # 11
(Due 09:30 am, Apr/17/2012)
Problem #1
Consider the design of a small PV powered lightemittingdiode (LED) light. The PV array consists of 8
series cells, each with rated current 0.3 A @ 0.6 V at 1kW/M2 full sun. Storage is pr
ECE 420 Spring 2012
HW10 Solution
Problem #1
a)
Full sun:
Voc 0.0257ln(
I sc
1)
Io
I sc 1A, I o 109 A
Voc 0.53258
V
Half sun
ISC = 0.5 A
Voc 0.0257ln(
I sc
1)
Io
= 0.514774 V
b)
Full sun
I Isc Io (e38.9V 1) 1 109 (e38.9( 0.5) 1) 0.7200 A
Half sun
I I sc
ECE420
Spring 2012
HW # 10
(Due 09:30 am, Apr/10/2012)
Problem #1
For the simple equivalent circuit for 0.005 m2 photovoltaic cell shown below, the reverse
saturation current is I0 = 109 A, the short circuit current is ISC = 1A. At 25oC, find the follow
ECE420
Spring 2012
HW # 9
(Due 09:30 am, Apr/3/2012)
Problem#1
A double fed induction generator (DFIG) has a rotor current of 10A and stator current of 8A. The
stator side resistance is 3 and the rotor side resistance is 2. The slip is 0.1. The friction
ECE 420 Spring 2012
HW08 Solution
Problem#1
a)
b)
This question is not accurate with stator copper loss. Your solution is still considered as correct if you did
not count the stator copper loss.
c)
(
)
(
Problem#2
(
)
(
)
Problem#3
a)
(
)
(
)
)
b)
(
(
)
)
ECE420
Spring 2012
HW # 8
(Due 09:30 am, Mar/13/2012)
Problem#1
A 480V 60Hz Y connected 4pole synchronous generator has synchronous impedance
0.1+j0.015 ohm per phase. This generator supplies 1200A at 0.8pf lagging. The friction and
windage losses are 4
ECE 420 Spring 2012
HW07 Solution
Problem#1
a)
b)
Problem#2
a)
Pout 3V I 1 cos 3(220 / 3 )(77)(0.88) 25820.03W
PAG Pout Pcore PSCL 25820.03 485 1033 27338.03W
b)
s PRCL / PAG 0.0475
c)
Pconv PAG PRCL 27338.03 1299 28637.03W
Pm Pconv Pf & w 29677.03W 39.78
ECE420
Spring 2012
HW # 7
(Due 09:30 am, Mar/6/2012)
Problem#1
A 40 KW, 60Hz, 6 poles, 3phase single fed induction generator is running at 1230
rpm. Find:
a) Slip (s)
b) Rotor frequency in hertz
Problem#2
A 3phase, 4 poles, 60Hz, Yconnected single fed
Spring 2012
HW06 Solution
Problem#1
a)
Phase voltage, V =
11000
6351
3
Load, S = 2 MVA
Armature current, I A
2000
3 *11
104.98 A
RA 1.2
X s 25
p.f is given to be 0.8 lagging
VnL = (V cos IA Re )2 (V sin IA X s )2 = (6351* 0.8 104.98 *1.2) 2 (6351* 0.6
ECE420
Spring 2012
HW # 6
(Due 09:30 am, Feb/28/2012)
Problem#1
A three phase wye connected synchronous generator is rated at 2.5 MVA, 11 kV.
The armature effective resistance and synchronous reactance are 1.2 and 25
respectively per phase. Calculate the
ECE420 Spring 2012
HW #05
(Due 09:30 am, Feb/21/2012)
Problem#1
Develop a table showing the speed of magnetic field rotating in AC machines of 2, 4, 6, 8, 10, 12
and 14 poles operating at frequencies of 50Hz
Problem#2
Using the parameters given in Proble
Spring 2012
HW04 Solution
Problem #1
(a)
(b )
(c)
(d )
= 91.19%
(
)
Problem #2
Vca
Vab
n
36.87
36.87
0
30
Ibc
Iab
Vbc
Ia
a)
Vab 2400, Vbc 240 120, Vca 240120
240
3
10
240 103
3
I a ,3
577.35 A
240
3 240
3
I a,3 577.35 66.87 A
I b,3 577.35 186.87 A
I
ECE420 Spring 2012
HW #04
(Due 09:30 am, Feb/14/2012)
Problem#1
The perunit quantities of a threephase 150 kVA, 34.64 kV/5kV, 60 Hz, wyedelta transformer
are:
RP = RS = 0.003 p.u
XP = XS = 0.001 p.u
RC = 40 p.u
Xm = 50 p.u
Based on the three phase rat
Spring 2012
HW03 Solution
Problem#1
(a)
(b)
Problem#2
(a)
 ohms law
(b)
(
(
)
(
)
(
)
)  Because the difference between voltage angle and current angle equals the
angle of Z or if this still confuse you please see the equation:
This may easier than you
ECE420 Spring 2012
HW #03
(Due 09:30 am, Feb/7/2012)
Problem #1
a) A transmission line which impedance is 45 82 per phase. Sbase = 100 MVA and
Vbase = 124.7 KV, calculate the perunit (p.u) value of the line impedance.
b) An 80 MVA, 12.47Y/124.7Y kV tran
Spring 2012
ECE 420 HW02 Solution
Problem#1
(
)
(
)
p.f = cos(
) =0.8 lagging
Problem#2
a)
For Y connection load
Convert load to Y connection:
Total impedance is:
Phase current:
Power supplied to load
(
(
(
)
)
) =77754.84
P = 62203.79 W, Q = 46653.02 Var
ECE420 Spring 2012
HW #02
(Due 09:30 am, Jan/31/2012)
Problem#1
Three impedances of 4+j3 are connected and tied to a threephase 208V power line, find
and the power factor of this load.
Problem#2
Following figure shows a threephase power system with two
Spring 2012
HW01 Solution
Problem #1
The polar form of the source voltage is
A
Therefore the instantaneous current is
(
)
)[
(
(
(
(
)
)
)]
(
)[
(
)] VA
The instantaneous value of P and Q are:
P=
(
)[
Q=
(
)[
)]
(
(
)]
Problem #2
a)
(
A
)
V
(
)
lagging
b)