Lecture1b-Three Phase Systems
Transcript of Lecture1b-Three Phase Systems
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School of Electrical Engineering & telecommunications
ELEC4612 – Power System Analysis
Lecture 1b: Three Phase Systems
Dr Jayashri Ravishankar
Properties of 3-phase Systems
• Three-phase systems have either three or four conductors.
• There are three-phase conductors identified as A, B, and C.
• There is sometimes a fourth conductor, whichis the neutral.
• The three phases are 120 degrees out of phase with each other (360 divided by 3).
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Balanced 3-phase systems
1. All 3 sources are represented by a set of balanced 3-phase variables
2. All loads are 3-phase with equal impedances
3. Line impedances are equal in all 3 phases
Using “phase a” as a reference, the voltage in
three phases can be described as follows:
0°
120° 240° 120°
Advantages of 3-phase
• The rating of 3-phase machines are 150%
greater than 1-phase machines of similar frame
size.
• The power delivered by a 1-phase system
pulsates and falls to zero. The 3-phase power
never falls to zero and approximately remains
the same at any instant.
• A 3-phase system needs three conductors;however, each conductor is only 75% the size
of the equivalent kVA rated 1-phase conductors.
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1-phase Vs 3-phase Power
Wye/Star – Connection (Y)
30||3
2
3
2
1||||
)120sin120(cos1||
120||0||
ph
ph ph
ph
ph ph
bnanab
V
jV V
jV
V V
V V V
210||3
90||3
phca
phbc
V V
V V
ph L ph L I I V V &3
Reference
Phase voltageLine voltage
Phase current
Line current
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Delta - Connection ( )
ph L
ph L
I I
V V
3
Phase voltage
Line voltage
Phase current
Line current
30||3 ph L I I
Line & Phase Values
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-Y Transformation
3-phase Power
Total instantaneous power absorbed into load =
instantaneous power in each phase
Under balanced conditions the average power,
Total average power,
balanced
3
source
balanced
3
load
a
b
c
abv
bcvbi
ai
ci
+
+
_
_
balanced
3
source
balanced
3
load
a
b
c
abv
bcvbi
ai
ci
+
+
_
_
3 an a bn b cn c p t v t i t v t i t v t i t
∅ 3 √3
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Example 1.1
Assume a -connected load is supplied from a 313.8 kV (L-L) source with Z = 10020Calculate the power consumed.
© 2012 Cengage Learning Engineering. All Rights Reserved.
13.8 0
13.8 0
13.8 0
ab
bc
ca
V kV
V kV
V kV
Method 1
© 2012 Cengage Learning Engineering. All Rights Reserved.
*
138 20 138 0
239 50 amps
239 170 amps 239 0 amps
3 3 13.8 0 kV 138 amps
5.7 MVA5.37 1.95 MVA
pf cos20 lagging
a ab ca
b c
ab ab
I I I
I I
S V I
j
13.8 0138 20
138 140 138 0
ab
bc ca
kV I amps
I amps I amps
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Method 2
13.80°10020°
138 20°
13.8
138 20°x 3 239 20°
3 3 x13800 x 239 5.71 MW
OR 3 3 x13800 x 138 5.71 MW
Example 1.2
480
8 60
60 x 3 104
240
3 138.57
138.57
6 23
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Example 1.3
6 x 6
18 2
4||2 1.333
?
440
3 254.03
254.03
1.333 .
4 4
4 x 4
4 12 12||6 4
440 ;
440
4 110 110 x 3 .
Example 1.4
A 3-phase line has an impedance of 2+j4 as shown in the figure. The linefeeds two balanced loads that are connected in parallel. The first load is Y-
connected and has an impedance of 30+j40 per phase. The second load is-connected and has an impedance of 60-j45 . The line is energised at thesending end from a 3-phase balanced supply of line voltage 207.85 V. Taking
phase voltage Va as reference, determine:
(a) The current, real power and reactive power drawn from the supply.
(b) The line voltage at the combined loads.
(c) The current per phase in each load.
(d) The total real and reactive powers in each load and the line.
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Summary
• The voltages of a three-phase system are 120°out of phase with each other.
• The two types of three-phase connections arewye and delta.
• In a wye connection, the phase voltage is lessthan the line voltage by a factor of 1.732. Thephase current and the line current are thesame.
• In a delta connection, the phase voltage is thesame as the line voltage. The phase current isless than the line current by a factor of 1.732.
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Exercise 1.1
1. A single-phase voltage source with V = 100130 voltsdelivers a current I = 1010 A, which leaves the positiveterminal of the source. Calculate the source real and
reactive power, and state whether the source delivers or
absorbs each of these. What is the type of the source?
Ans: - 500 W, + 866 VAR
2. An industrial plant consisting primarily of induction motor
loads absorbs 500 kW at 0.6 pf lag. (a) Compute the
required kVA rating of a shunt capacitor to improve the
power factor to 0.9 lag. (b) Calculate the resulting pf if a
synchronous motor rated 500 HP with 90% efficiency
operating at rated load and at upf is added to the plantinstead of the capacitor. Assume constant voltage. 1 HP =
0.746 kW. Ans: (a) 424.5kVA (b) 0.808 lag
Exercise 1.1
3. Three loads are connected in parallel across a single-
phase source voltage of 240 V (RMS). Load 1 absorbs
12 kW and 6.667 kVAR; Load 2 absorbs 4 kVA at 0.96
pf lead; Load 3 absorbs 15 kW at upf. Calculate the
values of R and X, for the three parallel loads, for two
cases: (a) Series combination of R and X, and (b)
parallel combination of R and X.
Ans: (a) R = 1.809 , X = 0.325
(b) R = 1.868 , X = 10.384
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Exercise 1.2
?
1. Find the current from the alternator.
Ans: 61.45 A
2. Two balanced Y-connected loads, one drawing 10 kW at 0.8
pf lag and the other 15 kW at 0.9 pf leading, are connected
in parallel and supplied by a balanced three-phase Y-
connected, 480-V source. (a) Determine the source current.
(b) If the load neutrals are connected to the source neutral
by a zero-ohm neutral wire through an ammeter, what will
the ammeter read? Ans: (a) 30.07 A (b) 0
Exercise 1.2
3. A three-phase line with an impedance of (0.2+j0.1) /phfeeds three balanced three-phase loads connected in
parallel. Load 1: Absorbs a total of 150 kW and 120
kVAR; Load 2: Delta connected with an impedance of
(150-j48) /ph; Load 3: 120 kVA at 0.6 pf leading. If theline-to-neutral voltage at the load end of the line is
2000V (RMS), determine the magnitude of the line-to-
line voltage at the source end of the line.
Ans: 3478.62V