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EE212 Passive AC Circuits
Lecture Notes 5
Three Phase Systems
1
Balanced 3-Phase Systems
Example:
A balanced 3-phase, line-to-line voltage of 220 V is applied to a
balanced 3-phase load. The 3-phase load consists of 3 identical
loads (each with an impedance of 6+j8 )
(a) connected in Y
(b) connected in D
Calculate the line current and the total power consumed for
both cases. Draw the complete phasor diagrams for each case.
2
Balanced 3-Phase Systems
A 3-phase load in D-connection consumes 3 times more power
than in Y- connection.
A 3-phase load in D-connection draws 3 times more current than
in Y- connection. Y-D motor starting method to reduce starting
current
3
Delta-Wye (D-Y) Transformation
A
A
Zab
B
B
Za
Zb
Zca
Zbc
Zc
C
C
ZY =
Za = (Zab. Zca) / (Zab + Zbc + Zca)
Zb = (Zab. Zbc) / (Zab + Zbc + Zca)
Zc = (Zca. Zbc) / (Zab + Zbc + Zca)
(product of adjacent ZD)
sum of all 3 ZD
For balanced load,
Zab = Zbc = Zca = ZD
Therefore, ZY = ZD / 3
4
Wye-Delta (Y- D) Transformation
A
B
Za
A
Zab
B
Zb
Zbc
Zca
Zc
C
C
(product of ZY taken in pairs)
ZD =
Zab = (Za.Zb + Zb.Zc + Zc. Za) / Zc
the opposite ZY
Zbc = (Za.Zb + Zb.Zc + Zc. Za) / Za
For balanced load,
Za = Zb = Zc = ZY
Zca = (Za.Zb + Zb.Zc + Zc. Za) / Zb
Therefore, ZD = 3.ZY
5
Balanced 3-f Source and Load
ZL
Y-Y System:
B
b
a
~
~
A
ZB
ZL
n
ZA
N
ZC
~
c
Balanced Y
source
Per phase equivalent circuit:
~
n
Line
impedance
ZL
a
Vp
ZL
a
C
Balanced Y
load
Ip = IL
A
ZA
N
6
Balanced 3-f Source and Load
Y-D System:
ZL
Ia
IAB
b
a
~
~
n
a
~
ca
ZL
B
Z
Z
a
ZL
A
Z
C
7
ZL
Y-D System:
Ia
IAB
b
a
~
ZL
~
n
a
B
Z
Z
a
~
ZL
ca
A
Z
C
Convert D load to Y load,
ZL
Ia
b
a
~
~
n
a
ZL
N
a
c
Z/3
IAN
Z/3
~
a
B
A
Z/3
ZL
C
Per phase equivalent circuit:
8
Example:
In a balanced Y-D 3-f circuit, the Y-connected source
has a Vp = 110 V. Line impedance between the source
and the load is ZL = (10 + j5)  per phase. Per phase
impedance of the D-connected load is ZD = (75 + j225)
. Determine the phase currents in the D-connected
load.
10+j5 
Ia
IAB
b
a
~
Vp = 110 V
~
B
n
75+j225 
~
ca
A
C
9
10+j5 
b
a
~
Ia
IAB
~
n
Vp = 110 V
A
B
75+j225 
~
ca
C
Convert D load to Y load,
10+j5 
Ia
b
a
~
~
Vp = 110 V
n
~
ca
A
IAN
B
N
(75+j225)/3 
C
Per phase equivalent circuit:
10
Multiple Balanced Loads in 3-f Systems
a
b
A1
B1 A2
Z1
N
Z1
c
B2
Z2
Z1
Z2
Z2
C2
C1
A number of balanced 3-f loads are connected in parallel.
Such systems can be evaluated using different methods:
1.Convert all loads to Y, and parallel the impedances
2.Convert all loads to D, and parallel the Z’s
3.Combine the per phase powers
11
Multiple Balanced Loads in 3-f Systems
Example:
A 3-f, Y-connected motor takes 10 kVA at 0.6 p.f. lagging from
a source of 220 volts line-to-line. The motor is in parallel with a
balanced D load having 16- resistance and 12- capacitive
reactance in series in each phase. Find the total P, line current
and p.f. of the combined load.
12
Example: Unbalanced 3-f Load
A 3-phase unbalanced load shown below is energized by a
balanced 3-phase voltage with 208 volts line-to-line, and
phase sequence a-b-c. Determine the total real power
delivered to the load.
a
+
10 
208 V
_
-j10 
10 
b
+
208 V
_
c
13
Example: Unbalanced 3-f Load
a
+
Loop Analysis:
Vab =
208/00
V
10 
I1
10 
-j10 
_
b
+
Vbc = 208/-1200 V
_
c
I2
14
Example: Unbalanced 3-f Load
Ia
→
a
+
10 
Vab = 208/00 V
-j10 
n
Nodal Analysis at node n:
10 
_ Ib →
b
+
Vbc = 208/-1200 V
_
c
Ic →
The voltage at node n is taken with
respect to the neutral node at the source.
That is, VnN
15
Watt-meter Connection in Single Phase System
Most watt-meters have CC and VC tap
settings such that they can be operated
close to the rated values (or at high loads)
Watt-meter
+
cc
+
vc
I
V
+
Single phase load
cc
I
Power factor cos 
vc
+
V
CC: 0 impedance
VC: ∞ impedance
Power measured by a watt-meter is
16
Watt-meter Connection in 3-f (3-wire) System
A
Ia
+
P1
+
b
a
Zb
B
n
Zc
+
P2
C
+
Ic
Q3f = 3 (P2 – P1)
P3f = P1 + P2
P1 = |Vab|.|Ia|. cos f1 where
Za
f1 = angle between Vab and Ia = 300 - 
P2 = |Vcb|.|Ic|. cos f2 where
c
f2 = angle between Vcb and Ic = -300 - 
2-Wattmeter Method (connected line – line)
Vcb
Ic
p.f. angle,  = angle between Vph and Iph
Vca
Vcn
Vab
2
= angle between Van and Ia (or Vcn and Ic)
300

1
Van
Ia
P1 + P 2
= VL.IL [cos (300 - ) + cos (-300 - )]
= VL.IL [2 cos 300 cos ] = 3.|VL|.|IL|. cos 
Ib
Vbn
Vbc
17
Watt-meter Connection in 3-f Balanced System
a
Pa
Ia
+
b
A
B
+
Ib
N
n
c
Ic
C
When neutral point is accessible (3-f, 3-wire system)
Three phase power = 3 x Pa
18
Example
A balanced Y load having per phase impedance of 10 /450 , is supplied by a 3-f source
with a line-to-line voltage of 220 V. Two-wattmeter method is used to measure the power
delivered to the load. Determine the reading on each wattmeter, and find the real &
reactive power from the wattmeter readings.
A
Ia
+
P1
+
VL = 220 V
b
a
Z
B
n
Z = 10/450 
Z
C
+
P2
+
Ic
c
19

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