Bài giảng Power Circuits and Electromechanics - Lecture 2 - Nguyễn Quang Nam
ECE430
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Fall 2009
Lecture 2
1
Three-phase systems
Voltage in each phase differ from the other phases by 1200. In
positive (also a-b-c) phase sequence, the three voltages are given by
t 1200
vcc' Vm cos
t 1200
vaa' Vm cost
vbb' Vm cos
Three-phase connections: wye connection and delta connection
In wye connection, terminals a’, b’, and c’ are joined and labeled as
neutral terminal n.
a
ia
+
ia, ib, and ic are line currents, which are
also equal to the phase currents. in is
neutral line current.
n
in
ib
c
b
ic
Lecture 2
2
1
Three-phase systems (cont.)
In delta connection, terminals a’ is connected to b, and b’ to c. Because
vac’ = vaa’(t) + vbb’(t) + vcc’(t) = 0, as can be verified mathematically, c’ is
finally connected to a.
ia
c’ a
Line and phase quantities
Since both supply and load can be wye or
delta connected, there are 4 possible
combinations: wye-wye, wye-delta, delta-
wye, and delta-delta (supply-load).
ib
ic
a’
c
b’
b
• wye-wye connection, balanced condition:
Van V 00
Vbn V 1200
Vcn V 1200
Lecture 2
3
Three-phase systems (cont.)
where V is the rms value of phase-to-neutral voltage.
The line-to-line voltages are given by
Vca Vcn Van
Vab Van Vbn
Vbc Vbn Vcn
V
For example, magnitude of ab can be calculated as
Vab 2V cos
300
3V
Vcn
Vab
Vca
From the phasor diagram, it can be seen
Van
Vab 3V 300
Vca 3V 1500
Vbc 3V 900
Vbn
Vbc
Under balanced condition, in = 0 (no neutral current)
Lecture 2
4
2
Three-phase systems (cont.)
• wye-delta connection, balanced condition:
Without losing generality, assuming line-to-line voltages are
Vab VL00
Vbc VL1200
Vca VL1200
Phase currents I1, I2, and I3 in the three legs of
delta-connected load lag the respective
Vca
I3
voltages by , and have the same magnitude of
I. It can be seen from the phasor diagram
Vab
I2
I1
Ia
3I 300
Ib 3I 1500
Ic 3I900
Ia
Vbc
Wye connection: V 3V and I I , delta connection:
VL V
L
L
and
IL 3I
Lecture 2
5
Power in balanced three-phase circuits
Balanced Y-connected load
In a balanced system, the magnitudes of voltages in all phases are the
same, and so also the current magnitudes. Let these be V and I. The
power per phase is then
P V I cos
Total power is
P 3P 3V I cos 3VL IL cos
T
S V I* V I
Complex power per phase is
And total complex power is
ST 3S 3V I 3VL IL
Note that is the phase angle between the phase voltage and phase current
Lecture 2
6
3
Power in balanced three-phase circuits (cont.)
Balanced -connected load
Similar to the case of balanced Y-connected load, per phase and total
power can be calculated using the same formulae.
It can be seen that for a balanced load, the expression for total complex
power is the same both for wye and delta connections, provided line-to-
line voltages and line currents are used.
Hence, calculations can be done on a three-phase or per-phase basis.
Ex. 2.12 and 2.13: see text book
Lecture 2
7
Per-phase equivalents
-Y conversion
Given a delta-connected load where impedance of each phase is Z, the
equivalent wye circuit has a phase impedance of ZY = Z/3. This can be
proved by equating the impedance across 2 arbitrary lines in both cases.
Instead of analyzing the delta circuit, the per-phase equivalent circuit can be
used after doing the -Y conversion.
Ex. 2.14: Draw the per-phase equivalent circuit for a given circuit.
Replace -connected capacitor bank by a Y-connected bank with phase
impedance of –j15/3 = -j5 . The resulting wye-connected circuit can then be
simplified to give the per-phase equivalent circuit.
Lecture 2
8
4
Class examples
Ex. 2.15: 10 induction motors in parallel, find three-phase kVAR
rating of a capacitor bank to improve overall PF to unity?
Per-phase real power is 30 x 10 / 3 = 100 kW, at lagging PF = 0.6. Per-
phase kVA is therefore 100/0.6. Hence,
100103
S S cos1
0.6
0.6 j0.8 VA 100 j133.33 kVA
0.6
A capacitor bank can be connected in parallel to the load for improving
overall PF. The capacitor bank needs to supply all the reactive power to
bring PF to unity. This means per-phase Qcap = 133.33 kVAR, and three-
phase kVAR required will be 3(133.33) = 400 kVAR.
Lecture 2
9
Class examples
Ex. 2.16: Suppose in Ex. 2.15, the new PF is to be 0.9 lagging, what
is the kVAR needed?
S 100 j133.33 kVA
New PF is 0.9 lagging, therefore new per-phase reactive power is
2
2
Qnew P 1 PF 1 100 1 0.9 1 48.43kVAR
133.33
kVAR
The capacitor bank therefore needs to supply
133.33 + 48.43 = 84.9 kVAR, and three-phase
kVAR required will be 3(84.9) = 254.7 kVAR.
48.43
kVAR
Ex. 2.17: see text book
100 kW
Lecture 2
10
5
In class quiz
Problem 2.21: A three-phase load of 15 kVA with a PF of 0.8
lagging is connected in parallel with a three-phase load of 36
kW at 0.6 PF leading. The line-to-line voltage is 2000 V.
a) Find the total complex power and power factor
b) How much kVAR is needed to make the PF unity?
Special question: A three-phase wye-connected load with a
unity PF is being supplied from a three-phase power system.
How does the load power change if the load is now rewired in
delta configuration?
Lecture 2
11
6
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