DG Systems
Problem #1
EL6663
A solar cell array is required to provide a peak power of 3 kW at 24 Vdc to a private
home. The array is constructed of series and parallel connected solar cells.
Each solar cell is characterized by the relationship of its ter

DG Systems
Solution #1
EL6663
At open circuit, the solar cell terminal current is i = 0 . Therefore, the open circuit voltage
can be calculated from: 0 = 0.56 1.5657 10 8 e38.65VOC 1 , which results in: VOC = 0.45 V .
(
)
e38.65v 1
From the above, the te

DR systems
Problem 8
EL6663
A bank of fuel-cells generates 180 Vdc at 20 Adc.
Use a single-phase line commutated inverter ( q = 2 ), and a single-phase transformer,
having a leakage reactance of XC = 0.05 , to connect the unit to the utility line at 120
V

DG Systems
Prob. 12 - solution
1. The reactive power supplied by the capacitor is: QC =
The load power is: P =
EL6663
6000 2
= 240 kVAr
150
6000 2
= 800 kW
45
800 2 + 240 2
= 139.2 A
6
3. The reactive power absorbed by the line is: Qline = 139.22 15 = 290

DG Syst.
The 60 Hz, 3 distribution network on
the right supplies a non-linear load.
The line impedance Zline is 10% of the
transformer impedance.
Problem 11
EL6663
45*62
!"#$%&'($')(%*+#,-$
*./000*123
F%(=(%G
D)E
The line current of the load itself is
giv

DG systems
Problem 2
EL6663
For a given AC motor: 3, 625 kVA, 480 V, 670 HP, nn = 886 rpm (n0 = 900
rpm), 91% Efficiency. Also, R1 = 0.01 , X1 = 0.06 , R2 = 0.017 , X2 = 0.066 ,
and Xm = 2.23 . Assume Rm .
The motor operates as a generator driven by wind

DG systems
Solution 3
EL666
940
The frequency at 940 rpm is: f2 = 60
= 62.7 1/sec
900
Using the simplified equivalent circuit for the machine, knowing that the
load is directly connected to the generator terminals (no line or internal
impedances), the no-

DG systems
Solution 6
The per-phase electrical diagram of generator and load is shown below.
&'$($)*$+
+
&7
8962
7:62
!"#$%
The generator internal voltage is: P =
',-./$01$($*2#*#$3-44564
V0 Va
sin
V0 =
X
The phasor diagram for the generator is shown on

DG Systems
Prob.9 solution
EL6663
VB2
230 2
The base impedance on the transformer secondary is: ZB =
=
= 1.76
SB 30 103
The leakage reactance of the transformer is: X = X pu ZB = 0.164 1.76 = 0.289
6
sin = 310.6 V
6
qX
I
(a) The delayed firing angle wou

DG Systems
Problem 4
EL6663
An induction generator 13.8 kV , 4.494 MVA , R1 = 0.31 , R2 = 0.153 ,
X1 = 4.958 , X2 = 4.195 is connected to the utility line through a
feeder/transformer having a combined impedance of Rx = 0.44 and X x = 2.96
(see figure bel

DG Systems
G(!%$(H)%-E
I-:-$F%#$
;*<*=>*0?
@*<*A
Problem 10
EL6663
D):-E*'F+F'(%#$
B*<*A
7%(8(%9*8(:*3*01
C*<*A
34/5./6*01
B)!%#J-$
=>*0?
K5L>*012$
!"#$%&'($')(%*+#,-$
*./*012
M%"-$
')!%#J-$!
A 3-phase distributed generator is connected to a customer bus

DG Systems
Prob. 8 solution
EL6663
The DC voltage on the load side is:
q
Vd = Vdo cos
Xc I d
2
1q
12
1
sin = 2 120
sin = 108.04 V
a
q
a
2
a
Where Vph is the phase voltage on the primary side, and a is the transformer ratio.
Substituting Vdo in the above

DG systems
Problem 5
EL6663
A wind-turbine rotates the shaft of a doubly-fed slip-ring induction generator at
n = 1080rpm , and supplies a torque of T = 8, 950Nm (to the generator shaft). The
generator has the following data: 3- , 2400 V, 6-poles, 60 Hz,

DG Systems
Problem 12
EL6663
The transmission line in the figure below possesses a reactance of X = 15 .
The equivalent load resistance is R = 45 , the capacitive reactance in parallel with the
load is X C = 150 , and the output voltage is VR = 6000 V .
!

DG systems
Problem 3
EL6663
A standalone induction generator is connected to a load in parallel with a
capacitor C = 530.5 F per-phase (for the generator, you may assume a simplified
equivalent circuit that includes only R2 S in parallel with jXm ).
The s

DG Systems
Problem 6
EL6663
A 3, Y connected symmetrical load is connected to the utility network and takes 50 A
current at 0.707 lagging power factor. The utility voltage is 220 V line-to-line.
A distributed synchronous generator is added in parallel wit

DG systems
Solution 5
a) The current supplied to the generator slip-rings in is:
EL6663
*
463 10 3 150
I 2 =
= 356.1 j205.6 = 411.3 150A
3 650
2400
= 1385.6V :
From the electrical equivalent circuit, using KCL Vph =
3
E 1385.6
E
+
+ 411.3 150 = 0E = 1

DG Systems
Problem 10 Solution
The short-circuit current at the customer bus is: I sh =
The Thevenins reactance is: XTh
200 10 3
= 555.14 A
3 208
208
= 0.216 / ph
3 555.14
Series resonance at the fifth harmonic occurs at: 2 60 5 =
The parallel resonance

DG systems
Problem 7
1,234,
0,
./
,+*-
0%
EL6663
0,
1%234%
0%
61778
5
5
!"#$
1!234!
%&'('&)
('*+
The synchronous generator in the figure above supplies power to the load and to
the utility line. The internal generator-voltage is VG, and the voltage at the

DR systems
Problem 9
EL6663
A 3 transformer rated 30 kVA, V1/V2 = 460/230 V, / Y , 16.4% per-unit impedance,
supplies power to a 6-pulse phase controlled rectifier. The rectifier is connected to a
fuel-cell bank that provides a voltage E = 120 Vdc at a cu

DG Systems
Prob. 7 solution
EL6663
From the power equations (the generator QG = 0) :
208
VG
7200
3 sin
=
3
5
208
208
VG
3
3
0=
cos
5
5
From the above:
and
= 39.8
The generator current is: I =
2
VG = 156.2 V
7200
= 20 A
3 208
and the phasor diagram is