Prove that the three-phase system of voltages on the secondary of the Y-∆ transformer shown in Figure 2-38b lags the three-phase system of voltages on the primary of the transformer by 3
Trang 155
The resulting equivalent circuit referred to the primary at 50 Hz is shown below:
+
-VS
VP
IS
+
-IP
R C jX M
R EQ jX EQ
260 Ω j917 Ω
2-20 Prove that the three-phase system of voltages on the secondary of the Y-∆ transformer shown in Figure
2-38b lags the three-phase system of voltages on the primary of the transformer by 30°
SOLUTION The figure is reproduced below:
VC
VC '
VA '
VB '
+
+
+
+
-
-+ +
+ +
VA '
VB '
VC '
VA
VB
VC
Trang 256
Assume that the phase voltages on the primary side are given by
°
∠
A Vφ
V VB =VφP∠−120° VC =VφP∠120°
Then the phase voltages on the secondary side are given by
°
∠
=
S
A Vφ
V VB′ = VφS∠ − 120 ° VC′ = VφS∠ 120 °
where VφS =VφP /a Since this is a Y-∆ transformer bank, the line voltage Vab on the primary side is
°
∠
=
°
−
∠
−
°
∠
=
−
V
and the voltage V b′ = VA′ = VφS∠ 0 ° Note that the line voltage on the secondary side lags the line
voltage on the primary side by 30°
2-21 Prove that the three-phase system of voltages on the secondary of the ∆-Y transformer shown in Figure
2-38c lags the three-phase system of voltages on the primary of the transformer by 30°
SOLUTION The figure is reproduced below:
+ +
-VA
VA '
Assume that the phase voltages on the primary side are given by
°
∠
A Vφ
V VB =VφP∠−120° VC =VφP∠120°
Trang 357
Then the phase voltages on the secondary side are given by
°
∠
=
S
A Vφ
V VB′ = VφS∠ − 120 ° VC′ = VφS∠ 120 °
where VφS =VφP/a Since this is a ∆-Y transformer bank, the line voltage Vab on the primary side is just
equal to VA =VφP∠0° The line voltage on the secondary side is given by
°
−
∠
=
°
∠
−
°
∠
=
−
=
V
Note that the line voltage on the secondary side lags the line voltage on the primary side by 30°
2-22 A single-phase 10-kVA 480/120-V transformer is to be used as an autotransformer tying a 600-V
distribution line to a 480-V load When it is tested as a conventional transformer, the following values are measured on the primary (480-V) side of the transformer:
Open-circuit test Short-circuit test
V OC = 480 V V SC = 10.0 V
I OC = 0.41 A I SC = 10.6 A
V OC = 38 W P SC = 26 W
(a) Find the per-unit equivalent circuit of this transformer when it is connected in the conventional manner
What is the efficiency of the transformer at rated conditions and unity power factor? What is the voltage regulation at those conditions?
(b) Sketch the transformer connections when it is used as a 600/480-V step-down autotransformer
(c) What is the kilovoltampere rating of this transformer when it is used in the autotransformer connection? (d) Answer the questions in (a) for the autotransformer connection
SOLUTION
(a) The base impedance of this transformer referred to the primary side is
( ) (2 )2 base,
480 V
23.04
10 kVA
P P
V Z
S
The open circuit test yields the values for the excitation branch (referred to the primary side):
, ,
0.41 A
0.000854 S
480 V
OC EX
OC
I Y
V
φ φ
OC
OC OC
P
0.000854 78.87 0.000165 0.000838
1/ 6063
1/ 1193
The excitation branch elements can be expressed in per-unit as
6063
263 pu 23.04
C
Ω
1193
51.8 pu 23.04
M
Ω
The short circuit test yields the values for the series impedances (referred to the primary side):
10.0 V
0.943 10.6 A
SC EQ SC
V Z
I
Trang 458
SC
SC SC
P
V I
0.943 75.8 0.231 0.915
The resulting per-unit impedances are
0.231
0.010 pu 23.04
EQ
0.915
0.0397 pu 23.04
EQ
Ω The per-unit equivalent circuit is
+
-VS
VP
IS
+
-IP
R C jX M
R EQ jX EQ
0.010 j0.0397
263 j51.8
At rated conditions and unity power factor, the input power to this transformer would be PIN = 1.0 pu The core losses (in resistor RC) would be
( )2 2
core
1.0 0.00380 pu 263
C
V P
R
The copper losses (in resistor REQ) would be
( ) (2 )
2
CU EQ 1.0 0.010 0.010 pu
The output power of the transformer would be
OUT OUT CU core 1.0 0.010 0.0038 0.986
and the transformer efficiency would be
OUT IN
0.986
1.0
P P
The output voltage of this transformer is
OUT= IN − ZEQ=1.0− 1.0 0∠ ° 0.01+ j0.0397 =0.991∠ −2.3°
The voltage regulation of the transformer is
1.0 0.991
0.991
−
(b) The autotransformer connection for 600/480 V stepdown operation is
Trang 559
+
-+
-480 V
600 V
N C
N SE VSE
VC
+
+
-(c) When used as an autotransformer, the kVA rating of this transformer becomes:
SE IO
SE
4 1
10 kVA 50 kVA 1
C
W
N
(d) As an autotransformer, the per-unit series impedance ZEQ is decreased by the reciprocal of the power advantage, so the series impedance becomes
0.010 0.002 pu 5
EQ
0.0397
0.00794 pu 5
EQ
while the magnetization branch elements are basically unchanged At rated conditions and unity power factor, the input power to this transformer would be PIN = 1.0 pu The core losses (in resistor R C) would
be
( )2 2
core
1.0 0.00380 pu 263
C
V P
R
The copper losses (in resistor REQ) would be
( ) (2 )
2
CU EQ 1.0 0.002 0.002 pu
The output power of the transformer would be
OUT OUT CU core 1.0 0.002 0.0038 0.994
and the transformer efficiency would be
OUT IN
0.994
1.0
P P
The output voltage of this transformer is
OUT = IN− ZEQ=1.0− 1.0 0∠ ° 0.002+ j0.00794 =0.998∠ −0.5°
The voltage regulation of the transformer is
1.0 0.998
0.998
−
2-23 Figure P2-4 shows a power system consisting of a three-phase 480-V 60-Hz generator supplying two loads
through a transmission line with a pair of transformers at either end
(a) Sketch the per-phase equivalent circuit of this power system
Trang 660
(b) With the switch opened, find the real power P, reactive power Q, and apparent power S supplied by the
generator What is the power factor of the generator?
(c) With the switch closed, find the real power P, reactive power Q, and apparent power S supplied by the
generator What is the power factor of the generator?
(d) What are the transmission losses (transformer plus transmission line losses) in this system with the
switch open? With the switch closed? What is the effect of adding Load 2 to the system?
base1
S = 1000 kVA Sbase2 = 1000 kVA Sbase3= 1000 kVA base2
,
L
V = 480 V V L,base2 = 14,400 V V L,base3 = 480 V
SOLUTION This problem can best be solved using the per-unit system of measurements The power system can be divided into three regions by the two transformers If the per-unit base quantities in Region 1 are chosen to be Sbase1 = 1000 kVA and V L,base1 = 480 V, then the base quantities in Regions 2 and 3 will be as shown above The base impedances of each region will be:
( )2 2
1 base1
base1
0.238
1000 kVA
V Z
S
φ
( )2 2
2 base2
base2
3 3 8314 V
207.4
1000 kVA
V Z
S
φ
( )2 2
3 base3
base3
3 3 277 V
0.238
1000 kVA
V Z
S
φ
(a) To get the per-unit, per-phase equivalent circuit, we must convert each impedance in the system to
per-unit on the base of the region in which it is located The impedance of transformer T1 is already in per-unit to the proper base, so we don’t have to do anything to it:
010 0
pu ,
R
040 0 pu ,
1 =
X
The impedance of transformer T2 is already in per-unit, but it is per-unit to the base of transformer T2, so
it must be converted to the base of the power system
2 base 1 base 2
pu on base 2 pu on base 1 2
base 2 base 1
( , , )R X Z ( , , )R X Z V S
2
8314 V 1000 kVA
8314 V 500 kVA
2
8314 V 1000 kVA
8314 V 500 kVA
Trang 761
The per-unit impedance of the transmission line is
line line,pu
base2
1.5 10
0.00723 0.0482 207.4
Z
Ω The per-unit impedance of Load 1 is
load1 load1,pu
base3
0.45 36.87
1.513 1.134 0.238
Z
Z
Ω The per-unit impedance of Load 2 is
load2 load2,pu
base3
0.8
3.36 0.238
Z
Ω The resulting per-unit, per-phase equivalent circuit is shown below:
+
-1∠0°
T1 Line T2
L1 L2
0.010 j0.040 0.00723 j0.0482 0.040 j0.170
1.513
j1.134
-j3.36
(b) With the switch opened, the equivalent impedance of this circuit is
EQ 0.010 0.040 0.00723 0.0482 0.040 0.170 1.513 1.134
EQ 1.5702 1.3922 2.099 41.6
The resulting current is
1 0
0.4765 41.6 2.099 41.6
∠ °
I
The load voltage under these conditions would be
Load,pu = 0.4765ZLoad = ∠ −41.6° 1.513+ j1.134 =0.901∠ −4.7°
( )( )
Load Load,pu base3 0.901 480 V 432 V
The power supplied to the load is
( ) (2 )
2 Load,pu Load 0.4765 1.513 0.344
Load Load,pu base 0.344 1000 kVA 344 kW
The power supplied by the generator is
( )( )
,pu cos 1 0.4765 cos 41.6 0.356
G
( )( )
,pu sin 1 0.4765 sin 41.6 0.316
G
( )( )
,pu 1 0.4765 0.4765
G
,pu base 0.356 1000 kVA 356 kW
,pu base 0.316 1000 kVA 316 kVAR
,pu base 0.4765 1000 kVA 476.5 kVA
The power factor of the generator is
Trang 862
PF=cos 41.6° =0.748 lagging
(c) With the switch closed, the equivalent impedance of this circuit is
EQ
1.513 1.134 3.36 0.010 0.040 0.00723 0.0482 0.040 0.170
1.513 1.134 3.36
EQ 0.010 0.040 0.00788 0.0525 0.040 0.170 (2.358 0.109)
EQ 2.415 0.367 2.443 8.65
The resulting current is
1 0
0.409 8.65 2.443 8.65
∠ °
I
The load voltage under these conditions would be
Load,pu = 0.409ZLoad = ∠ −8.65° 2.358+ j0.109 =0.966∠ −6.0°
( )( )
Load Load,pu base3 0.966 480 V 464 V
The power supplied to the two loads is the power supplied to the resistive component of the parallel combination of the two loads: 2.358 pu
( ) (2 )
2 Load,pu Load 0.409 2.358 0.394
Load Load,pu base 0.394 1000 kVA 394 kW
The power supplied by the generator is
( )( )
,pu cos 1 0.409 cos 6.0 0.407
G
( )( )
,pu sin 1 0.409 sin 6.0 0.0428
G
( )( )
,pu 1 0.409 0.409
G
,pu base 0.407 1000 kVA 407 kW
,pu base 0.0428 1000 kVA 42.8 kVAR
,pu base 0.409 1000 kVA 409 kVA
The power factor of the generator is
PF=cos 6.0° =0.995 lagging
(d) The transmission losses with the switch open are:
( ) (2 )
2 line,pu line 0.4765 0.00723 0.00164
line line,pu base 0.00164 1000 kVA 1.64 kW
The transmission losses with the switch closed are:
( ) (2 )
2 line,pu line 0.409 0.00723 0.00121
line line,pu base 0.00121 1000 kVA 1.21 kW
Load 2 improved the power factor of the system, increasing the load voltage and the total power supplied to the loads, while simultaneously decreasing the current in the transmission line and the transmission line losses This problem is a good example of the advantages of power factor correction in power systems
Trang 963
Chapter 3: Introduction to Power Electronics
3-1 Calculate the ripple factor of a three-phase half-wave rectifier circuit, both analytically and using
MATLAB
SOLUTION A three-phase half-wave rectifier and its output voltage are shown below
π/6 5 π/6
2π/3
( ) sin
v t =V ωt
( ) sin( 2 / 3)
v t =V ωt− π ( ) sin( 2 / 3)
v t =V ωt+ π
SOLUTION If we find the average and rms values over the interval from π/6 to 5π/6 (one period), these values will be the same as the average and rms values of the entire waveform, and they can be used to calculate the ripple factor The average voltage is
( )
5 / 6
/ 6
( ) sin
2
T
π π
π
5 6
6
π π ω
The rms voltage is
( )
5 / 6
rms
/ 6
( ) sin
T
π π
π
5 / 6 2
rms
/ 6
sin 2
M
V
π π
π
Trang 1064
2 rms
sin sin
M V
π
rms
sin sin
rms
0.8407
M
The resulting ripple factor is
rms
DC
0.8407
0.8270
M M
r
The ripple can be calculated with MATLAB using the ripple function developed in the text We must right a new function halfwave3 to simulate the output of a three-phase half-wave rectifier This output
is just the largest voltage of v A( )t , v B( )t , and v C( )t at any particular time The function is shown below: function volts = halfwave3(wt)
% Function to simulate the output of a three-phase
% half-wave rectifier
% wt = Phase in radians (=omega x time)
% Convert input to the range 0 <= wt < 2*pi
while wt >= 2*pi
wt = wt - 2*pi;
end
while wt < 0
wt = wt + 2*pi;
end
% Simulate the output of the rectifier
a = sin(wt);
b = sin(wt - 2*pi/3);
c = sin(wt + 2*pi/3);
volts = max( [ a b c ] );
The function ripple is reproduced below It is identical to the one in the textbook
function r = ripple(waveform)
% Function to calculate the ripple on an input waveform
% Calculate the average value of the waveform
nvals = size(waveform,2);
temp = 0;
for ii = 1:nvals
temp = temp + waveform(ii);
end
average = temp/nvals;
% Calculate rms value of waveform
Trang 1165
temp = 0;
for ii = 1:nvals
temp = temp + waveform(ii)^2;
end
rms = sqrt(temp/nvals);
% Calculate ripple factor
r = sqrt((rms / average)^2 - 1) * 100;
Finally, the test driver program is shown below
% M-file: test_halfwave3.m
% M-file to calculate the ripple on the output of a
% three phase half-wave rectifier
% First, generate the output of a three-phase half-wave
% rectifier
waveform = zeros(1,128);
for ii = 1:128
waveform(ii) = halfwave3(ii*pi/64);
end
% Now calculate the ripple factor
r = ripple(waveform);
% Print out the result
string = ['The ripple is ' num2str(r) '%.'];
disp(string);
When this program is executed, the results are
» test_halfwave3
The ripple is 18.2759%
This answer agrees with the analytical solution above
3-2 Calculate the ripple factor of a three-phase full-wave rectifier circuit, both analytically and using
MATLAB
SOLUTION A three-phase half-wave rectifier and its output voltage are shown below
Trang 1266
T/12
( ) sin
v t =V ωt
( ) sin( 2 / 3)
v t =V ωt− π ( ) sin( 2 / 3)
v t =V ωt+ π
SOLUTION By symmetry, the rms voltage over the interval from 0 to T/12 will be the same as the rms voltage over the whole interval Over that interval, the output voltage is:
v t =v t −v t =V ωt+ π −V ωt− π
2 cos sin 3 cos
3
M
Note that the period of the waveform is T=2 /π ω, so T/12 is / 6π ω The average voltage over the
interval from 0 to T/12 is
/ 6
/ 6 0 0
T
π ω
π ω
3 3
1.6540
π
The rms voltage is
/ 6
rms
0
T
π ω
π
Trang 1367
/ 6 2
rms
0
sin 2
M
π ω
rms
The resulting ripple factor is
rms DC
1.6554
1.6540
M M
r
The ripple can be calculated with MATLAB using the ripple function developed in the text We must right a new function fullwave3 to simulate the output of a three-phase half-wave rectifier This output
is just the largest voltage of vA( ) t , vB( ) t , and v C( )t at any particular time The function is shown below: function volts = fullwave3(wt)
% Function to simulate the output of a three-phase
% full-wave rectifier
% wt = Phase in radians (=omega x time)
% Convert input to the range 0 <= wt < 2*pi
while wt >= 2*pi
wt = wt - 2*pi;
end
while wt < 0
wt = wt + 2*pi;
end
% Simulate the output of the rectifier
a = sin(wt);
b = sin(wt - 2*pi/3);
c = sin(wt + 2*pi/3);
volts = max( [ a b c ] ) - min( [ a b c ] );
The test driver program is shown below
% M-file: test_fullwave3.m
% M-file to calculate the ripple on the output of a
% three phase full-wave rectifier
% First, generate the output of a three-phase full-wave
% rectifier
waveform = zeros(1,128);
for ii = 1:128
waveform(ii) = fullwave3(ii*pi/64);
end
% Now calculate the ripple factor
r = ripple(waveform);
% Print out the result
string = ['The ripple is ' num2str(r) '%.'];
disp(string);
Trang 1468
When this program is executed, the results are
» test_fullwave3
The ripple is 4.2017%
This answer agrees with the analytical solution above
3-3 Explain the operation of the circuit shown in Figure P3-1 What would happen in this circuit if switch S1
were closed?
SOLUTION Diode D1 and D2 together with the transformer form a full-wave rectifier Therefore, a voltage oriented positive-to-negative as shown will be applied to the SCR and the control circuit on each half cycle
(1) Initially, the SCR is an open circuit, since v1 < VBO for the SCR Therefore, no current flows to the
load and vLOAD = 0
(2) Voltage v1 is applied to the control circuit, charging capacitor C1 with time constant RC1
(3) When vC > VBO for the DIAC, it conducts, supplying a gate current to the SCR
(4) The gate current in the SCR lowers its breakover voltage, and the SCR fires When the SCR fires, current flows through the SCR and the load
(5) The current flow continues until i D falls below I H for the SCR (at the end of the half cycle) The process starts over in the next half cycle