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 cost

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, and delta-delta (supply-load).

• 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

delta-connected load lag the respective

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

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 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.)

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-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

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 cost

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, and delta-delta (supply-load).

• 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

delta-connected load lag the respective

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

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 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.)

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-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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