ee207 electrical power - lecture 3
TRANSCRIPT
EE207 Electrical Power
Lecture 3
3 phase circuits
Rajparthiban Kumar EE207 Electrical Power 2
Introduction• Three phase power is preferred to single phase
power for the following reasons:-
a) Three phase motor, generators and transformers are simpler, cheaper and more efficient.
b) Three phase transmission line can deliver more power for a given weight and cost
c) The voltage regulation of 3 phase transmission line is inherently better.
Rajparthiban Kumar EE207 Electrical Power 3
Three Phase System
• As the magnet turns it sweeps across the conductors, inducing in them a voltage given as:
• Induced voltage will be maximum when the poles are in the position shown by the Fig. because the flux density is greatest at the centre of the pole.
• Induced voltage will be zero when the poles have rotated 90o.
S
N
ωm
Single Phase
Generator
BlvEa ====1
1
a
R
Rajparthiban Kumar EE207 Electrical Power 4
Three Phase System
• The voltage and current are
in phase as load is resistive.
• The Power supplied is
always positive and at twice
the voltage and current
frequency and also has a
peak:
Pp=VpIp
• The average power supplied:
90 180 270 360 450 θθθθ
0
Vp
0
-Vp
Ip
90 180 270 360 450 θθθθ
Pp
rmsrmsppavg IVIVP ========2
11φφφφ
Rajparthiban Kumar EE207 Electrical Power 5
• In a two phase generator
another set of windings
(b,2) is placed 90o from
the first set (a,1). Thus at
90o voltage will also be
induced and power will be
generated.
• In this case power will be
supplied almost in
continuous manner and
thus vibration no longer
exists.
Two Phase
Generator
SN
ωm
1
a
R
b2
R
Three Phase System
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• The voltage in phase a and
b are given respectively
as:
• The total power Ptotal has
average value equal to the
peak power
)cos(Vv
cosVv
pb
pa
2
ππππθθθθ
θθθθ
++++====
====0 1.57 3.14 4.71 6.28 7.85 9.42 10.99-1
0
1
0 1.57 3.14 4.71 6.28 7.85 9.42 10.990
0.5
1
0 1.57 3.14 4.71 6.28 7.85 9.42 10.990
0.5
1
va vbIa Ib
Pa Pb
Pa+Pb=Ptotal
avgppavg PIVP φφφφφφφφ 12 2========
Three Phase System
Rajparthiban Kumar EE207 Electrical Power 7
Three Phase System
• Three phase generator is
similar to two phase generator
except that it has three coils
placed 120o from each other.
• The three phase voltages are
given as:
SN
ωm
1
a
R
b
2
R
c
3
R
Three Phase
Generator
)cos(Vv
)cos(Vv
cosVv
pc
pb
pa
3
4
3
2
ππππθθθθ
ππππθθθθ
θθθθ
−−−−====
−−−−====
====
Rajparthiban Kumar EE207 Electrical Power 8
Three Phase System
• Note that the voltage signals are phase shifted by 120o
• The total power supplied to a three phase load is:
• Note also that the sum of three phase voltages is Zeroat any instant.
• Thus if these voltages are applied across three equalimpedances, the sum of the resultant three phase currents is also Zero
0 60 120 180 240 300 360 420 480 540 600 660-1
-0.5
0
0.5
1
va vb vc
ppavgavg IV.PP 513 13 ======== φφφφφφφφ
0====++++++++ cba vvv
Rajparthiban Kumar EE207 Electrical Power 9
Three Phase System
• The windings of a three phase system are connected to three resistive loads using 6 wires.
• However the return wires form all the three phases can be joined in one conductor called the Neutral.
• Incase the loads in three phase system are equal (balanced) , the sum of the three currents at any instant will be Zero and thus the Neutral wire can be removed.
a
c b
1
2
3
R
R
R
va1
vb2
vc3
Ia
Ib
Ic
a
c b
12
3
R
R
R
va1
vb2
vc3
Ia
Ib
Ic
Ia +Ib+Ic=0
Rajparthiban Kumar EE207 Electrical Power 10
Wye (Y) Connection
• The voltages (van, vbn and
vcn) between line and neutral
are known as phase voltages
(vφφφφ)))).
• The voltage difference
between two phase voltages
(vab, vbc, vca) is known as
line voltage (vL). Line
voltages are given as:
a
cb
R
R
R
van
v bn
vcn
Ia
Ib
Ic
n vab
vca
vbc
ancnca
cnbnbc
bnanab
vvv
vvv
vvv
−−−−====
−−−−====
−−−−====
• Currents in the lines (Ia, Ib and Ic
are known as line currents.
• In aY connected system line
currents are equal to those currents
flow in the load phases.
Y System
Rajparthiban Kumar EE207 Electrical Power 11
van
-vnb
vbn
vcn
-vna
-vnc
bnbc vv 3====
γ=30γ=30γ=30γ=30οοοο
120o
120o
120o
)cos(Vv
)sincos(V
)sincos(V
)sincos(cosV
)cos(VcosVv
opab
p
p
p
ppab
303
2
1
2
33
2
3
2
3
2
3
2
1
3
2
++++====
−−−−====
−−−−====
−−−−++++====
−−−−−−−−====
θθθθ
θθθθθθθθ
θθθθθθθθ
θθθθθθθθθθθθ
ππππθθθθθθθθ
In a Y connected system The line
voltage is √√√√3 times the phase voltage
and it leads by 30o.
Wye (Y) Connection
Y system Phasor Diagram
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• Example
• A 3-phase, 60 Hz, wye connected generator has a line voltage of 23900 V
Calculate
• the line-to-neutral voltage
• voltage induced in the individual windings
• the time interval between the positive peak voltage of phases A and B
• the peak value of the line voltage
Rajparthiban Kumar EE207 Electrical Power 13
VV
V LLLN 13800
3
23900
3===
5.55ms s 1/180 s) (1/60 1/3 T
120 is waveformsltagebetween vo angle The
s 1/60 takescyle One
===
°
VVV rmsLp 338002, ==
Rajparthiban Kumar EE207 Electrical Power 14
Delta Connection
• Three phase load can also be
connected to form what is
known as Delta System.
• In a ∆ the line voltages
appear across the load
phases. However, the
currents in the load phases
are not the same as the line
currents.
• The line current in ∆∆∆∆system is √√√√3 times the
phase current and it
lags by 30o.
R
R
R
Ia
Ib
Ic
vab
vbc
vca
a
b
c
Iφ1φ1φ1φ1
Iφ2φ2φ2φ2
Iφ3φ3φ3φ3
∆∆∆∆ System
23
12
31
φφφφφφφφ
φφφφφφφφ
φφφφφφφφ
III
III
III
c
b
a
−−−−====
−−−−====
−−−−====
Rajparthiban Kumar EE207 Electrical Power 15
• Note from the phasor
diagram that the phase
currents are in phase with the
line voltages because the load
is resistive.
• The line currents Ia, Ib and Ic
are phase shifted from each
other by 120o.
Iφ1φ1φ1φ1
γ=30γ=30γ=30γ=30οοοο
120o
120o
120o
Iφ2φ2φ2φ2
Iφ3φ3φ3φ3
-Iφ3φ3φ3φ3
vab
vca
vbc
-Iφ1φ1φ1φ1
Ib
Ic
θθθθIIa 3====
∆∆∆∆ system Phasor Diagram
Delta Connection
Rajparthiban Kumar EE207 Electrical Power 16
Example
Three identical impedances are connected in
delta across a 550 V line. The current drawn
is 10 A. Calculate
• the current in each impedance
• the impedance values
Rajparthiban Kumar EE207 Electrical Power 17
Ω===
==
9577.5
550
I
VZ
550V isresistor across Voltage
77.53
AI
I aφ
Rajparthiban Kumar EE207 Electrical Power 18
Power In Balanced Three Phase System
• If the magnitude of the phase voltages for a Y connected
system is :
• And if the magnitude of the phase currents for a delta
system is:
• Then the total three phase power is:
Where θ is the power factor angle (angle between the voltage and the current)
cnbnan vvvv ============φφφφ
cnbnan IIII ============φφφφ
θθθθφφφφφφφφφφφφ cosIvP 33 ====
Rajparthiban Kumar EE207 Electrical Power 19
• For a Y system is the magnitudes of the line voltage and
line current are |VL| and |IL| respectively. Then:
• The three phase power can be expressed in terms of the
line quantities as:
• Also the total Reactive power (Q) in a three phase system
is given as
Power In Balanced Three Phase System
φφφφφφφφ IIandV
v LL ========3
θθθθφφφφ cosIVP LL33 ====
θθθθ
θθθθφφφφφφφφφφφφ
sinIV
sinIVQ
LL3
33
====
====
Rajparthiban Kumar EE207 Electrical Power 20
Power In Balanced Three Phase System
• The total apparent Power (S) in a three phase system is
also given as:
• In a Delta system all the above equations are valid.
• For ∆ system, the total three phase power (P) is:
• However for ∆, the phase voltages and currents can be
express as line quantities as:
LL IVS 33 ====φφφφ
θθθθφφφφφφφφφφφφ cosIvP 33 ====
3
φφφφφφφφ
IIandVv LL ========
Rajparthiban Kumar EE207 Electrical Power 21
Power In Balanced Three Phase System
• Then the total power in a ∆ system is given as:
LL
LL
LL
IVS
and
sinIVQ
,similarly
cosIVP
3
3
3
3
3
3
====
====
====
φφφφ
φφφφ
φφφφ
θθθθ
θθθθ
Rajparthiban Kumar EE207 Electrical Power 22
Examples
• Example 1
For the circuit in the figure below calculate:
– the current in each line.
– The voltage across the inductor terminals
– The Power absorbed by the load
– The Reactive power
– The apparent Power
– The power factorR
R
R
j4Ω
j4Ω
j4Ω
3Ω
3Ω 3Ω
a
b
c
440V
3 phase line
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Ω=+= 534 22Z
VV
V LLLN 254
3==
VIXV
AZ
VI
L
LN
2.203)4(8.50
inductor theacross Voltage
8.50
===
==
254V
3+4j
Rajparthiban Kumar EE207 Electrical Power 24
LL
LL
LL
L
IVS
IVIVQ
IVIVP
AI
3
sin3sin3
cos3cos3
87.368.5087.365
0254
=
==
==
−∠=∠
∠=
θθ
θθ
φφ
φφ
Rajparthiban Kumar EE207 Electrical Power 25
Examples
• Example 2
A manufacturer plant draws a total of 415kVA from 2400V 3
phase line. If the plant power factor is 87.5% lagging,
calculate:
• The impedance of the plant
• The phase angle between the line to neutral voltage and the
line current.
• The complete phasor diagram of the plant.
Rajparthiban Kumar EE207 Electrical Power 26
AI
IVS
VVV
L
LL
LN
100)2400(3
415000
3
13863
2400
==
=
=== φ
Ω===∴ 9.13100
1386branch)per (
I
VZ
φ
Rajparthiban Kumar EE207 Electrical Power 27
°=
=
29
875.0_
θ
factorpower
Rajparthiban Kumar EE207 Electrical Power 28
Example
Motor
3594 HP
93% efficient
Pf = 0.9
Capacitor
Bank
1800 kvar
4000V
3phase
• A 5000 HP wye connected motor is connected to a 4000V,
3phase, 60 Hz line.
• A delta connected capacitor rated at 1800 kvar is also
connected to the line.
• If the motor produces an output of 3594 hp at 93%
efficiency and power factor of 90% lagging, calculate:-
Rajparthiban Kumar EE207 Electrical Power 29
a) The active power absorbed by the motor
kWP
P
kWP
m 288393.0
2681
2681)746(3594
===
==
η
Rajparthiban Kumar EE207 Electrical Power 30
kVarPSQ
kVAPS
S
P
1395
32039.0/2883cos/
cos
22 =−=
===
=
θ
θ
b) The reactive power absorbed by the motor
Rajparthiban Kumar EE207 Electrical Power 31
• c) The reactive power supplied by
transmission line
var40513951800
var1395
var1800
kQ
kQ
kQ
net
m
c
−=+−=
=
−=
Rajparthiban Kumar EE207 Electrical Power 32
d) The apparent power supplied by the line
kVAQPS
kQ
kWP
2911
var405
2883
22 =+=
−=
=
Rajparthiban Kumar EE207 Electrical Power 33
e) The line current f) Motor line current
Ak
I
IVS
L
LL
420)4000(3
2911
33
==
=φA
kIm 462
)4000(3
3203==
Rajparthiban Kumar EE207 Electrical Power 34
• Phasor diagram
°=
=
8.25
9.0cos
mθθ
°=
===
8
99.02911
2883cos
θ
θLS
P
25.8°
8°2309V
IL = 420A
Im = 462 A
Ic = 260A
Rajparthiban Kumar EE207 Electrical Power 35
• Capacitor current
• Q =1800kvar
• V = 4000V
• Q = 3 VI sin θ
• Ic =260 A
Rajparthiban Kumar EE207 Electrical Power 36
Phase Sequence
3-phase systems have a property called phase sequence
– it determines the direction of
• rotation of 3-phase motors
– it means the order in which the three line
• voltages become successively positive
– two sequences: ABC & ACB
• positive sequence: ABC
• •negative sequence: ACB
Rajparthiban Kumar EE207 Electrical Power 37
• Phase sequence is determined by a simple
• circuit
• – one lamp always burns brighter
• – the sequence order is bright lamp, dim
• lamp, capacitor
Rajparthiban Kumar EE207 Electrical Power 38
• Wattmeters measure active power
• single-phase wattmeters have two pairs of terminals: voltage & current
• each terminal pair has polarity markings polarity markings determine the direction of power flow
• 3-phase, 3-wire circuits
• – power is measured by two single-phase wattmeters
• – total power is the sum of the two meters
• 3-phase, 4-wire circuits
• – power is measured by three single-phase wattmeters
• – total power is the sum of the three meters
Rajparthiban Kumar EE207 Electrical Power 39
Rajparthiban Kumar EE207 Electrical Power 40
• Questions
• 8.5
• 8.11
• 8.15
• 8.26