# Giao trinh bai tap ece 110 hw3

ECE430
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Fall 2009

http://www4.hcmut.edu.vn/~nqnam/lecture.php

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

v aa'  Vm cost 

vbb '  Vm cos t  120 0

v cc '  V m cos t  120 0

 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, ib, and ic are line currents, which are

n

+

ib

+

c

in

also equal to the phase currents. in is
neutral line current.

ia

+

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’ = v aa’(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-

b’

a’

+

c

ib

b
ic

• wye-wye connection, balanced condition:

Van  V 0 0

Vbn  V   120 0

Vcn  V 120 0

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

V ab  V an  Vbn

Vca  Vcn  V an

Vbc  Vbn  V cn

For example, magnitude of V ab can be calculated as

 

Vab  2V cos 30 0  3V
From the phasor diagram, it can be seen

Vab  3V 30 0

V cn

V ab

V ca
V an

Vbc  3V   90 0
Vbn

Vca  3V 150 0

Under balanced condition, in = 0 (no neutral current)
Lecture 2

Vbc
4

2

Three-phase systems (cont.)
• wye-delta connection, balanced condition:
Without losing generality, assuming line-to-line voltages are
Vab  VL 00

Vbc  VL   1200

Vca  VL 1200

Phase currents I1, I2, and I3 in the three legs of

V ca

I3

voltages by , and have the same magnitude of
I. It can be seen from the phasor diagram
I a  3I    30 0  

I b  3I    150 0  

I c  3I  90 0  

 Wye connection: V L 
and I L 

I2

V ab

I1
Vbc

Ia

3V and I L  I  , delta connection: V L  V

3I 
Lecture 2

5

Power in balanced three-phase circuits
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 PT  3P  3V I cos  

3VL I L cos 

Complex power per phase is S   V I *  V I  
And total complex power is S T  3S   3V I   

3VL I L 

Note that  is the phase angle between the phase voltage and phase current

Lecture 2

6

3

Power in balanced three-phase circuits (cont.)
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-toline 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. Perphase kVA is therefore 100/0.6. Hence,

S   S  cos 1 0.6  

100  10 3
0.6  j 0.8 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 threephase 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

Qnew  P 1 PF   1  100 1 0.9   1  48.43 kVAR
2

2

133.33
kVAR

133.33 + 48.43 = 84.9 kVAR, and three-phase

kVAR required will be 3(84.9) = 254.7 kVAR.
 Ex. 2.17: see text book

ol
d

The capacitor bank therefore needs to supply

new

48.43
kVAR

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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