Chapter1 - 3 phase system 17feb13 - Universiti Malaysia...
Transcript of Chapter1 - 3 phase system 17feb13 - Universiti Malaysia...
COURSE OUTCOME (CO)
CO1: Ability to define and explain the concept of single-phase and three-phase system.
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• A sinusoid is a signal that has the form of the sine or cosine function.
• A general expression for the sinusoid,
where
Vm = the amplitude of the sinusoidω = the angular frequency in radians/s
Ф = the phase
Revision
)sin()( φω += tVtv m
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Revision
A periodic function is one that satisfies v(t) = v(t + nT), for all t and for all integers n.
ω
π2=T
HzT
f1
= fπω 2=
• Only two sinusoidal values with the same frequency can be compared by their amplitude and phase difference.
• If phase difference is zero, they are in phase; if phase difference is not zero, they are out of phase.
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Revision
Example 1
Given a sinusoid, , calculate its amplitude, phase, angular frequency, period, and frequency.
)604sin(5 ot −π
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Revision
Example 1
Given a sinusoid, , calculate its amplitude, phase, angular frequency, period, and frequency.
Solution:
Amplitude = 5, phase = –60o, angular frequency = 4π rad/s, Period = 0.5 s, frequency = 2 Hz.
)604sin(5 ot −π
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Revision
Example 2
Find the phase angle between and , does i1 lead or lag i2?
)25377sin(41
oti +−=
)40377cos(52
oti −=
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Revision
Example 2
Find the phase angle between and , does i1 lead or lag i2?
)25377sin(41
oti +−=
)40377cos(52
oti −=
Solution:
Since sin(ωt+90o) = cos ωt
therefore, i1 leads i2 155o.
)50377sin(5)9040377sin(52
oootti +=+−=
)205377sin(4)25180377sin(4)25377sin(41
oooottti +=++=+−=
SingleSingle--Phase CircuitPhase Circuit
Three wired system
• same magnitude
• same phase
A single phase circuit consists of a generator connected through a pair of wires
to a load
Two wire system
Aa
TwoTwo--Phase CircuitPhase Circuit
Three wired
system
Second source
with 90° out
of phase
Three wired system
• same magnitude
• different phase
• It is a system produced by a generator consisting of three sources having the same amplitude and frequency but out of phase with each other by 120°.
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What is a ThreeWhat is a Three--Phase Circuit?Phase Circuit?
Three sources
with 120° out
of phaseFour wired
system
• A three-phase generator consists of a rotating magnet (rotor) surrounded by a stationary winding (stator).
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Balance ThreeBalance Three--Phase Voltages Phase Voltages
A three-phase generator The generated voltages
• Two possible configurations:
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Balance ThreeBalance Three--Phase Voltages Phase Voltages
Three-phase voltage sources: (a) Y-connected ; (b) ∆-connected
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Phase sequences
a) abc or positive sequence b) acb or negative sequence
Balance ThreeBalance Three--Phase Voltages Phase Voltages
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If the voltage source have the same amplitude and frequency ω and are out of phase with each
other by 120o, the voltage are said to be balanced.
0VVV cnbnan =++
cnbnan VVV ==
Balanced phase voltages are equal in magnitude and out of phase with each other by
120o
Balance ThreeBalance Three--Phase Voltages Phase Voltages
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0
p
0
pcn
0
pbn
0
pan
120V240VV
120VV
0VV
+∠=−∠=
−∠=
∠=
abc sequence or positive sequence:
acb sequence or negative sequence:
0
p
0
pbn
0
pcn
0
pan
120V240VV
120VV
0VV
+∠=−∠=
−∠=
∠=
pV is the
effective or
rms value
Balance ThreeBalance Three--Phase Voltages Phase Voltages
Example 1
Determine the phase sequence of the set of
voltages.
)110cos(200
)230cos(200
)10cos(200
°−=
°−=
°+=
tv
tv
tv
cn
bn
an
ω
ω
ω
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Balance ThreeBalance Three--Phase VoltagesPhase Voltages
Solution:
The voltages can be expressed in phasor form as
We notice that Van leads Vcn by 120°and Vcn in turn
leads Vbn by 120°.
Hence, we have an acb sequence.
V 110200V
V 230200V
V 10200V
°−∠=
°−∠=
°∠=
cn
bn
an
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Balance ThreeBalance Three--Phase VoltagesPhase Voltages
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Two possible three-phase load configurations:
a) a wye-connected load b) a delta-connected load
Balance ThreeBalance Three--Phase Voltages Phase Voltages
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A balanced load is one in which the phase impedances are equal in magnitude and in phase.
For a balanced wye connected load:
Y321 ZZZZ ===
∆=== ZZZZ cba
For a balanced delta connected load:
YZ3Z =∆ ∆= Z3
1ZY
Balance ThreeBalance Three--Phase Voltages Phase Voltages
• Four possible connections
1. Y-Y connection (Y-connected source with
a Y-connected load)
2. Y-∆ connection (Y-connected source with
a ∆-connected load)
3. ∆-∆ connection
4. ∆-Y connection
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Balance ThreeBalance Three--Phase Connection Phase Connection
Balance YBalance Y--Y Connection Y Connection
• A balanced Y-Y system is a three-phase system with a balanced y-connected source and a balanced y-connected
load.
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=
=
=
=
Y
L
s
Z
Z
Z
Z
l
Source impedance
Line impedance
Load impedance
Total impedance per phase
LY ZZ =
Balance YBalance Y--Y Connection Y Connection
LsY ZZZZ ++=l
Since all impedance are in series,
Thus
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Applying KVL to each phase:
Y
ana
Z
VI =
0
a
Y
0
an
Y
bnb 120I
Z
120V
Z
VI −∠=
−∠==
0
a
Y
0
an
Y
cnc 240I
Z
240V
Z
VI −∠=
−∠==
0IIII ncba =−=++
0IZV nnnN ==
Balance YBalance Y--Y Connection Y Connection
Balance YBalance Y--Y Connection Y Connection
0
pca
0
pbc
0
pab
210V3V
90V3V
30V3V
−∠=
−∠=
∠=
Line to line voltages or line voltages:
Magnitude of line voltages:
pL V3V =
cabcabL VVVV ===
cnbnanp VVVV ===
Example 2
Calculate the line currents in the three-wire Y-Y system
shown below:
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Balance YBalance Y--Y Connection Y Connection
Example 2
Calculate the line currents in the three-wire Y-Y system
shown below:
A 2.9881.6I
A 8.14181.6I
A 8.2181.6I
Ans
°∠=
°−∠=
°−∠=
c
b
a
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Balance YBalance Y--Y Connection Y Connection
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BalancBalance e Y-∆ CConnection onnection
• A balanced Y-∆ system is a three-phase system with a balanced y-connected source and a balanced ∆-connected
load.
BalancBalance e Y-∆ CConnection onnection
A single phase equivalent circuit
3
ZZY
∆=
3/∆
==Z
V
Z
VI an
Y
an
a
BalancBalance e Y-∆ CConnection onnection
A single phase equivalent circuit
CA
0
pca
BC
0
pbc
AB
0
pab
V210V3V
V90V3V
V30V3V
=−∠=
=−∠=
=∠=
Line voltages:
BalancBalance e Y-∆ CConnection onnection A single-phase equivalent circuit of a balanced Y-∆∆∆∆ circuit
°∠=−=
°−∠=−=
°−∠=−=
903
1503
303
ABBCCAc
ABABBCb
ABCAABa
IIII
IIII
IIII
Line currents:
∆
∆
∆
=
=
=
Z
VI
Z
VI
Z
VI
CA
CA
BC
BC
AB
ABPhase currents:
BalancBalance e Y-∆ CConnection onnection A single-phase equivalent circuit of a balanced Y-∆∆∆∆ circuit
)24011(IIII
240II
0
ABCAABa
0
ABCA
−∠−=−=
−∠=
0
ABa 303II −∠=
pL II 3=
cbaL IIII ===
CABCABp IIII ===
Magnitude line currents:
Example 3
A balanced abc-sequence Y-connected source with ( )
is connected to a ∆-connected load (8+j4)Ω per phase. Calculate the phase and line currents.
Solution
Using single-phase analysis,
Other line currents are obtained using the abc phase sequence
°∠= 10100Van
A 57.1654.3357.26981.2
10100
3/Z
VI an °−∠=
°∠
°∠==
∆
a
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Balance Balance Y-∆ ConnectionConnection
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Balance Balance ∆-∆ CConnectiononnection
• A balanced ∆-∆ system is a three-phase system with a balanced ∆ -connected source and a balanced ∆ -connected
load.
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∆
∆
∆
=
=
=
Z
VI
Z
VI
Z
VI
CA
CA
BC
BC
AB
AB
CAca
BCbc
ABab
VV
VV
VV
=
=
=
Line voltages: Line currents:
3II pL =
Magnitude line currents:
3
ZZY
∆=
Total impedance:
Phase currents:
°∠=−=
°−∠=−=
°−∠=−=
903
1503
303
ABBCCAc
ABABBCb
ABCAABa
IIII
IIII
IIII
Balance Balance ∆-∆ CConnectiononnection
Example 4
A balanced ∆-connected load having an impedance 20-j15 Ωisconnected to a ∆-connected positive-sequence generator having ( ). Calculate the phase currents of the load and the line currents.
Ans:
The phase currents
The line currents
V 0330Vab °∠=
A 87.1562.13IA; 13.812.13IA; 87.362.13I °∠=°−∠=°∠= ABBCAB
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A 87.12686.22IA; 13.11386.22IA; 87.686.22I °∠=°−∠=°∠= cba
Balance Balance ∆-∆ CConnectiononnection
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Balance Balance ∆-Y ConnectionConnection
• A balanced ∆-Y system is a three-phase system with a balanced y-connected source and a balanced y-connected
load.
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Balance Balance ∆-Y ConnectionConnection
Applying KVL to loop aANBba:
Y
p
baZ
VII
00∠=−
From:
0
ab 120II −∠=
0
aba 303III ∠=−
Y
p
aZ
V
I
0303
−∠
=
Line currents:
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Replace ∆ connected source to equivalent Y
connected source.Phase voltages:
0p
cn
0p
bn
0p
an
903
VV
1503
VV
303
VV
+∠=
−∠=
−∠=
Balance Balance ∆-Y ConnectionConnection
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A single phase equivalent circuit
Y
p
Y
an
aZ
V
Z
VI
0303
−∠
==
Balance Balance ∆-Y ConnectionConnection
Example 5
A balanced Y-connected load with a phase impedance 40+j25 Ωis supplied by a balanced, positive-sequence ∆-connected source with a line voltage of 210V. Calculate the phase currents. Use Vab as reference.
Answer
The phase currents
A; 5857.2I
A; 17857.2I
A; 6257.2I
°∠=
°−∠=
°−∠=
CN
BN
AN
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Balance Balance ∆-Y ConnectionConnection
phase-single ,2'2
2
L
Lloss
V
PRP =
( ) phase- three,'3
'3''3'2
2
2
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L
L
L
LLloss
V
PR
V
PRRIP ===
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Power in a Balanced SystemPower in a Balanced System
• Comparing the power loss in (a) a single-phase system, and (b) a three-phase system
• If same power loss is tolerated in both system, three-phase system use only 75% of materials of a single-phase system
( ) phase-single ,222
22
L
LLloss
V
PRRIP ==
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)120cos(2
)120cos(2
cos2
0
0
+=
−=
=
tVv
tVv
tVv
pCN
pBN
pAN
ω
ω
ω
For Y connected load, the phase voltage:
Power in a Balanced SystemPower in a Balanced System
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Phase current lag phase voltage by θ.
θ∠= ZZY
If
)120cos(2
)120cos(2
)cos(2
0
0
+−=
−−=
−=
θω
θω
θω
tIi
tIi
tIi
pc
pb
pa
The phase current:
Power in a Balanced SystemPower in a Balanced System
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Total instantaneous power:
cCNbBNaANcbaivivivpppp ++=++=
θ= cosIV3ppp
Average power per phase:
Apparent power per phase:
Reactive power per phase:
Complex power per phase:
θ= sinIVQppp
θ= cosIVPppp
*
pppppIVjQPS =+=ppp
IVS =
Power in a Balanced SystemPower in a Balanced System
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Total average power:
θ=θ== cosIV3cosIV3P3PLLppp
Total reactive power:
θ=θ== sinIV3sinIV3Q3QLLppp
Total complex power:
*
p
2
p
p
2
p
*
pppZ
V3ZI3IV3S3S ====
θ∠=+=LL
IV3jQPS
Power in a Balanced SystemPower in a Balanced System
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Power loss in two wires:
2
L
2
L2
LlossV
PR2RI2P ==
Power loss in three wires:
2
2
2
22
333
L
L
L
LLloss
V
PR
V
PRRIP === PL : power absorbed by the load
IL : magnitude of line current
VL : line voltage
R : line resistance
Power in a Balanced SystemPower in a Balanced System
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Example 6
A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the
line voltage is 220 V and the line current is 18.2 A.
Determine the power factor of the motor.
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Example 6
A three-phase motor can be regarded as a balanced Y-load. A three-phase motor draws 5.6 kW when the
line voltage is 220 V and the line current is 18.2 A.
Determine the power factor of the motor.
The apparent power is
VAIVS LL 13.6935)2.18)(220(33 ===
The real power is WSP 5600cos == θ
The power factor is8075.0cos ===
S
Ppf θ
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Exercise 6
Calculate the line current required for a 30-kW three-phase motor having a power factor of 0.85
lagging if it is connected to a balanced source
with a line voltage of 440 V.