Upload
ananya-pal
View
217
Download
0
Embed Size (px)
Citation preview
8/8/2019 EE1101 Circ N 5
1/43
3-Phase Circuits
Balanced 3-phase
8/8/2019 EE1101 Circ N 5
2/43
Balanced Complex Numbers
Consider the
complex roots of
unity:
rootstheofsumthefindThen
j
ofrootstheallFind
3
3 1
:
8/8/2019 EE1101 Circ N 5
3/43
Balanced Complex Numbers
...3,2,1,0;120112011
3601
31
3
!!!@
!@!
nnn
mxZzxZz
n
o
o
mm
o
8/8/2019 EE1101 Circ N 5
4/43
Balanced Complex Numbers
!@!
!
3
1
3
3
00
2
3
2
1
,2
3
2
1
,011
2401,1201,011
:
:
jroots
jjj
or
arerootshe
thatshowcanWe
ooo
8/8/2019 EE1101 Circ N 5
5/43
Balanced Complex Numbers
o01
o1201
o2401
8/8/2019 EE1101 Circ N 5
6/43
Balanced Complex Numbers
!@
!
!
!@
!
!!
3
1
3
00
0,
2
1
2
3,
2
1
2
3
2701,1501,301
301201
903601
9010
jroots
jjj
nj
n
jj
ooo
oo
oo
o
8/8/2019 EE1101 Circ N 5
7/43
Balanced Complex Numbers
It is evident from above that the sum of
any three roots of the cube-root of any
complex number is absolutely zero.
This is the essence of a balanced three
phase system and indeed the most
powerful tool in power transmission.
The results below are self-evident.
8/8/2019 EE1101 Circ N 5
8/43
Balanced Complex Numbers
Consider a general complex number z
!@
!
!
!
!!
3
1
3
1
3
3
1
2
3
1
1
3
1
3
00
)23
21(
)2
3
2
1(
)01(
120.....0
jroots
jZz
jZz
jZz
nZzZz oo
8/8/2019 EE1101 Circ N 5
9/43
Balances 3-phase Currents
Let us use the above discovery and
KCL:
Suppose we have three currents suchthat
:
int
;,,sup
240,120,0
321
321
sconclusion
erestingakewebelowshownas
loadatoiiiplytodecideweIf
IiIiIi ooo
@
!!!
8/8/2019 EE1101 Circ N 5
10/43
Balances 3-phase Currents
nI
1i
2i
3-phase
supply
3-phase
load
3i
8/8/2019 EE1101 Circ N 5
11/43
Balances 3-phase Currents
Clearly,
systemphasebalancedain
ireneutralNOal aysisThere
ireneutralhavetoneedlessI
I
iii
ButiiiI
n
n
n
@
!
!@
!
!
3
;0
0
0
,
321
321
8/8/2019 EE1101 Circ N 5
12/43
Advantages of balanced 3-phase
supply
1.The elimination of a neutral wire is clearly
the biggest saving.
2. If we wanted to supply the same amount of
power using single phases, we would haveused 6 wires! But we have managed with just
3 of them.
3. The fact that line voltages are higher means
that the line currents are lower hence reduced
losses.
RI2
8/8/2019 EE1101 Circ N 5
13/43
Line & Phase Values
In practice, when a 3-phase system is
used, we do have both the currents &
voltages as 3-phase. Their values may
be given as phase or line values.
Let us consider a balanced 3-phase
voltage system.
Just as we did in the case of currents,
the voltage is given by:
8/8/2019 EE1101 Circ N 5
14/43
Line & Phase Values
.&
;
)2
3
2
1(240
)2
3
2
1
(120
)01(0
3
2
1
earthirephasebet een
measuredvaluephaseVhere
jVVv
jVVv
jVVv
o
o
o
!
!!
!!
!!
8/8/2019 EE1101 Circ N 5
15/43
Line & Phase Values
!@
!!!
!!!
!!!
00
1503)2
3
2
3(
903)30(
303)
2
3
2
3(
1331
3223
2112
jvoltagesline
jvvv
jvvv
jvvv
o
o
o
8/8/2019 EE1101 Circ N 5
16/43
Line & Phase Values
It is however usual to measure voltage
between one phase & another phase. So
the value is that of one phase with
respect to the other. Since it is measured
between one phase & another it is
referred as the line voltage.
The line voltages must the phasordifferences as shown below:
8/8/2019 EE1101 Circ N 5
17/43
Phasors for Phase & Line Voltages
1v
2v
3v 12v
23v
31v
8/8/2019 EE1101 Circ N 5
18/43
Phase & Line Voltages
The following conclusions are made:
1. The line voltages also constitute a balanced 3-
phase system.
2.
351,63
000,11
3
000,11
41632403
240
:
.3
}!!@
!!
}!!@
!!
!
voltagephase
voltageline
voltageline
voltagephase
Exa ples
voltagephasevoltageline
8/8/2019 EE1101 Circ N 5
19/43
Line & Phase Values
Please note that:
1. In a 3-phase system; the line values
(voltage & current) are the ones specified.
2. It is very dangerous to touch any two line
wire (because of much higher voltages).
3. All values are rms.
4. Red, Yellow, Blue represent . 321 ,, vvv
8/8/2019 EE1101 Circ N 5
20/43
Line & Phase Values
While it is easy to identify phase & line
voltages; caution is taken on phase &
line currents.
It is easier to assume that phase & line
currents are the same.
8/8/2019 EE1101 Circ N 5
21/43
Star & Delta 3-ph representation
There are two distinct ways of
representing 3-phase voltages/currents;
Star or Delta. [Star is sometimes calledWye].
Let us re-draw the phasors for 3-phase
voltages.
8/8/2019 EE1101 Circ N 5
22/43
Star & Delta 3-ph representation
Phasors
1v
2v
3v
1v
2v
3v
Star
Delta
8/8/2019 EE1101 Circ N 5
23/43
Star & Delta 3-ph connections
Power connections
1v
2v
3v
1v
2v
3v
Star
Delta
R
Y
B
B
R
Y
n
8/8/2019 EE1101 Circ N 5
24/43
It is noted that:
A Star connection has a neutral point (n)and can be accessed when need arises.But the Delta connection does not haveone.
Hence in a Delta connection only the line
voltages can be measured. But in theStar connection both line & phasevoltages are measurable.
8/8/2019 EE1101 Circ N 5
25/43
Star-Star connection
Supply-load connections
1i
supply Load
n n
2i
3i
8/8/2019 EE1101 Circ N 5
26/43
Star-Delta connection
Supply-load connections
1i
supply
Loadn
2i
3i
1I
2I
3I
8/8/2019 EE1101 Circ N 5
27/43
Line & Phase currents
In the star connection; the line current=phase
current.
But in Delta connection; the line current DOES
NOT equal to phase current e.g.
But we may use KCL to find the relationshipsbetween line and phase currents.
11 Ii {
0311 ! IIi
8/8/2019 EE1101 Circ N 5
28/43
Power in 3-phase
We know that is 1-phase;
)(
,
cos3cos3
3
.,
;cos3
;3
cos
L
LL
LLLL
IIassu e
valueslineareIwhere
IIpowerotal
valuesphaseareIwhere
I
powerotal
phaseinut
IPower
!
!!@
!
!
UU
U
U
8/8/2019 EE1101 Circ N 5
29/43
8/8/2019 EE1101 Circ N 5
30/43
3-phase exercises
Exercise 3:
A 0.6MW, 416V, 3-phase, 50Hz load
has a power factor of 0.5. Determine the necessary capacitors and
show their connections so as to improve
the power factor to 0.9.
8/8/2019 EE1101 Circ N 5
31/43
Solution 1
The 3-phase voltages differ by angle only:
)120sin()240sin()(
)120sin()(
)sin()(
3
2
1
oo
o
wtwttv
wttv
wttv
!!
!@
!
8/8/2019 EE1101 Circ N 5
32/43
Solution 2
connectionstarif
currentphaselineincurrent
I
I
ILoad
WLoadWply
L
L
LL
!!
}!@
!@
!
!!@!
@
0.168311
3200
5.0**10*11*310*6.1
cos3
6.12*8.02sup
36
U
8/8/2019 EE1101 Circ N 5
33/43
Solution 3
The power factor improvement is done
by connecting a capacitor between @
phase & the neutral point.
[Though it is possible to connect a
capacitor between phases but then the
capacitors are more expensive because
of the higher (line) voltages].
8/8/2019 EE1101 Circ N 5
34/43
Star-Star connection
Supply-load connections
1i
Supply Load
n n
2i
3i
C
C
C
8/8/2019 EE1101 Circ N 5
35/43
Solution 3
Since @ capacitor is connected across a
phase & neutral; phase voltages must be
used.
We are at liberty to use the phasor
diagram studied earlier.
8/8/2019 EE1101 Circ N 5
36/43
Solution 3
1
--
U Jc- V
8/8/2019 EE1101 Circ N 5
37/43
Solution 3
)tancos(sin
cos
cos
coscos
sinsin
9.0cos
5.0cos
1
1
1
JUU
J
U
JU
JU
J
U
!@
!@
!
!
!
!
II
I
I
II
III
C
C
8/8/2019 EE1101 Circ N 5
38/43
Solution 3
388,13416
10
5.0*416*3
10*6.0
cos3
cos3
6
6
$
!!@
!
!
U
U
L
L
LL
V
I
IIhere
IV
But
8/8/2019 EE1101 Circ N 5
39/43
Solution 3
fIZIV
capacitoracrossVoltage
I
T
JU
2
1*
9.865
]10
19
9
5
2
3[3416
10
]9.0
sin5.0[sin3416
10
6
6
!!!
$
!
!@
8/8/2019 EE1101 Circ N 5
40/43
Solution 3
.
18390@3
1839001839.0
3416*100100
100
marketonavailablevaluable
highernearestusepraticeIn
Fofcap
FF
IV
I
I
@
!$
!!@
!
Q
Q
TT
T
8/8/2019 EE1101 Circ N 5
41/43
Question 4
A 440V, 3-ph, Y-connected source has
two loads connected as:
One load is balanced in @ phase withZ=10+j5.
The second load is balanced in @ phase
with Z=15+j0.
(i) Find the average power to @ load.
(ii) Find the total power delivered.
8/8/2019 EE1101 Circ N 5
42/43
kWVI
phaseino erZ
V
Z
Vi
5.1510*5.15cos3
3
0
3
1
$!
!
!
!
U
U
U
8/8/2019 EE1101 Circ N 5
43/43
Total load =(10+j5)+(15+j0)=25+j5=Z
kW
Z
iV po erTotal
ZZi
ZjZ
o
phasephase
phase
45.7
3.11cos650
440cos*
3
440*3
440*3
cos**3
3
4400
3440
25
5arctan;525
2
}
!!
!@
!
!@
!!!
U
U
UU
UU