EE2022 Electrical Energy Systems Lecture 6: Three‐Phase Circuit Analysis

EE2022 Electrical Energy Systems

1/29/2015 EE2022: Electrical Energy Systems ‐ Three phase Circuit Analysis 1

Lecture 6: Three‐Phase Circuit Analysis

Learning Outcomes

• To calculate the complex power, voltages and currents in single phase and balanced three‐phase AC circuits and able to describe their relationships using Phasor diagrams.– Be able to solve balanced three‐phase circuit problems.

– Be able to calculate three‐phase power.


Outline

• Three‐Phase Circuit Analysis– Generation, Transmission, and Distribution.– Three‐phase balanced systems.– Advantages of three‐phase balanced systems.

• Three‐Phase voltage and current– Line‐to‐neutral voltage– Line‐to‐Line voltage– Line current.– Delta/Wye configuration.


Generation, Transmission and Distribution

1/29/2015 EE2022: Electrical Energy Systems ‐ Three phase Circuit Analysis 4Source: http://venturebeat.com/2010/10/29/super‐grid‐introduction/

Three lines

Three lines

Three lines

Three lines

A Three‐Phase Circuit System


3‐Phase Generation system.

3‐Phase Transmission system.

Generation (11 – 36 KV) Transmission (110 – 765 KV) Industrial customer (23 – 138 KV)Commercial customer (4.16 – 34.5 KV)Residential customer (120 – 240 V)

Generation Transmission and Distribution LoadThree‐phase voltage source Three transmission lines Three‐phase load

Three‐Phase Voltage Sources


Va

Vb

Vc120° All three voltage sources

have the same voltage magnitude,with 120 degrees apart.

Phase sequence: abc(positive)

Three Single‐Phase Circuits


Generation Transmission Load

These lines can be combined.

Three Phase Circuit


Generation Transmission Load

This line is called ‘neutral’ line. Later on, we’ll see why this line can be removed.

Balanced Three‐Phase Circuit


Va

Vb

VcThree‐phase circuit is said to be balanced when the impedances in the 3 phases are identical.

Identical impedances in 3 phases!

Three‐phase voltage source with identical magnitude and 120 phase differences

Balanced Three‐Phase Circuit


Va

Vb

Vc

Ia

Ic

Ib

Ia

Va

0 n a b cI I I I

IaIc

Ib

0

Neutral line can be removed

Advantages of Balanced 3‐Phase Systems

• When compared to three single‐phase circuits, three‐phase circuits have better use of equipment and materials– More power can be transmitted

per conductor– Lesser power losses in the

conductors• This implies reduced capital

and operating costs of transmission and distribution.

• We can calculate voltage and current for only one phase and refer to other phases easily.


Example 1

• Consider the following three‐phase system shown below. Find the current ia when z’ = 10 Ω.


Z’V 0100 aV

V 120100 bV

V 120100 cV

= 100 Ω

= 100 Ω = 100 Ω

Ans: A 1201 A, 1201 A, 01 cba iii

THREE‐PHASE CURRENT AND VOLTAGE

Line‐to‐Neutral VoltageLine‐to‐line voltageLine currentWye‐Delta connection


Line‐To‐Neutral (Phase) Voltage


, , are called line‐to‐neutral voltage or phase voltage.

anV bnV cnV

Line‐To‐Line Voltage

• Voltage is given as line‐to‐line voltage by convention.

• KVL:


anV

bnV

cnVbnV

abV

n

303 anbnanab VVVV

neutral-LineLine-Line 3VV 30°

Line Current


, , are called line currents.ai bi ci

3‐Phase Circuit ConnectionWye Connection Delta Connection


n

a

b c

+

‐

+

‐

aI

bIcI

abVanVa

bc

aI

bI

cI

abI

bcI

caI

These two types of connections apply to both three‐phase voltage sources and three‐phase loads.

Line Current VS Phase CurrentWye Connection Delta Connection


n

A

B C

aI

bI

cI

A

BC

abI

bcI

caI

0

Currents through the three‐phase conductor lines are called ‘Line currents’.Currents carried by the load impedance are called ‘Phase currents’ or ‘Load Current.

aI

bI

cI

Delta‐Connected Load

• , , are called Phase currents.

• We can find relationship between line currents and phase currents using KCL,


A

B C

aI

bI

cI

abI

bcI

caI

n

aI

bI

cI

abI

bcI

caI

caI

303 abcaaba IIII

abI bcI caI

PhaseLine 3 II

Delta‐Wye Load Transformation


303 anab VV

Z Z

Z

aI

bI

cI

A

B C

+

‐

abVYZ

YZ YZ

aI

bI

cI

A

B C

+

‐

abV

303 aba II

Y3ZZ

Example 3• For a balanced Y‐connected

three phase voltage source and Y‐connected load system with a line voltage of 440 V and three equal resistive loads of 100 Ω per phase, assume positive sequence, what will be the magnitudes of – (a) the line‐to‐neutral voltage,– (b) the phase current, – (c) the line current?


Example 4

• For a balanced Y‐connected three phase generator with the line‐to‐neutral voltage of 80 V, ∆‐connected load of 120 Ω, assume positive sequence, find– (a) the line‐to‐line voltage,– (b) the voltage across a

resistor, – (c) the current through a

resistor?


Summary• Three‐phase voltage sources

– Positive and negative sequences• Balanced three‐phase circuit

– Conditions– Advantages

• Balanced three‐phase circuit– Line‐to‐neutral (phase) voltage– Line‐to‐line (line) voltage

– Line current• Wye‐Delta connection• Delta‐Wye load transformation


neutral-LineLine-Line 3VV

Y3ZZ

• Three‐Phase Circuit Analysis– Three‐phase complex power– Per Phase analysis


Three Phase Power Calculation

• Three phase power is found from summation of each phase power.


***3 ccnbbnaan IVIVIVS

Balanced Three‐Phase Power

• From three phase power,

• When the system is balanced, (assume positive sequence) we can write,


*3 3 aan IVS

***3 120120120120 aanaanaan IVIVIVS

***3 ccnbbnaan IVIVIVS

01anV

1201bnV

1201cnVPositive sequence,

abc

Balanced Three‐Phase Load

• Three‐phase load can be connected in either Wye or Delta connection.

• 3‐phase load parameter is given as total apparent power ( ) with power factor.

• The voltage given is Line‐to‐line voltage.• We can find three‐phase real and reactive power as follows.


3S

p.f.3 313 SPP

sinp.f.cossin3 31

313

SSQQ

Delta/Wye Connected 3‐Phase Load


LineLineToLine3 3 IVS

Z Z

Z

aI

bI

cI

a

b c

+

‐

abV YZ

YZ YZ

aI

bI

cI

a

b c

+

‐

abV

aababab IVIVS 33 *3

303 anab VV 303 aba II

aabaan IVIVS 33 *3

Example 1

• A 3Ф load of 300 MVA, 100 kV at 0.85 p.f. lagging, find– The magnitude of line current– Three‐phase (real) power


LineI

LoadP

Ans: 1732 A, 255 MW.

3Φ, 300 MVA, 0.85 pf lagging

?line I+

‐kV 100line-to-line V

Phase a

Phase b

Phase c

Instantaneous Three‐Phase Power

• Given by,

• Recall that single phase instantaneous power,


titvtitvtitvtp ccbbaa 3

iv

iv

tIV

IVtitvtp

2cos

cos

-1/120 0 1/120 1/60

-20

0

20

40

60

80

100

t

vtVtv cos itIti cos

Instantaneous Three‐Phase Power

• For a balanced three‐phase system,

• We can find three phase instantaneous power as,

• Constant power transfer to load.


PIVtp aa 3cos33

4802coscos

2402coscos

2coscos3

ivccivcc

ivbbivbb

ivaaivaa

tIVIV

tIVIV

tIVIVtp

=0

An Additional Advantage of Balanced 3‐Phase Circuit

• Constant power transfer to load.– This also implies constant mechanical power input for a generator.

– When mechanical power input is constant, mechanical shaft torque is also constant.

– This helps to reduce shaft vibration and noise, extending the machine’s lifetime.


PER PHASE ANALYSIS

AssumptionSingle‐line diagramExample example example…


Per Phase Analysis: Assumption


Va

Ia

Vb

Ib

120º

Vc Ic

120º

It must be balanced three‐phase circuit.

0 n a b cI I I I

Steps of Per Phase Analysis

• Make sure that the three‐phase system is balanced.– The three‐phase sources need to have the same magnitude with 120 degree phase difference.

– The three‐phase impedances must be of the same value (both phase and magnitude).

• Convert all Delta‐connected sources/loads toWye‐connected sources/loads.

• Per phase analysis reduce three‐phase circuit to single‐phase circuit. We can apply the same concept used in single‐phase.


Single‐Line Diagram• Show the interconnections of a transmission system– Generator– Load– Transmission line– Transformer

• This is a representation of a 3Ф circuit. Each line represents three conductors in three‐phase system.


G1 G2

Load L1

TL1

TL2 TL3

T1 T2

T3

Example 2

• Given a one‐line diagram,

If the voltage source is . Find,1. Current magnitude supplied by source, , and,2. Current magnitude through a capacitor, .


V 3Line-Line V

Source,Δ‐Connected

Positive sequence.Load 1, Δ‐Connected Capacitor: ZC = ‐j15

Load 2, Y‐Connected Inductive: ZL = j10

j5j2.5

12 3SI

CI

SICI

Example 2: 3Ф Circuit Diagram


‐+

‐ +

+‐

j2.5

j2.5

j2.5

‐j15

‐j15

‐j15

j5

j5

j5

j10

j10

j10abV

bcV

caV1a

1b1c 2c

2a

2b

3a

3b3c

SI

CI

Example 2: Convert from Δ → Y

5315

3jjZZY


j2.5

j2.5

j2.5

j5

j5

j5

j10

j10

j10

anV

bnVcnV

1a

1b1c2c

2a

2b

3a

3b3c

SI

+‐

+ ‐ ‐ + ‐j5

‐j5

‐j5

3013030330

3ab

anVV

Use voltage source as angle reference

Example 2: 1‐Phase diagram

305.15.22 Sana IjVV1/29/2015 EE2022: Electrical Energy Systems ‐ Three phase Circuit Analysis 41

5

551055105.2 jjjj

jjjjZeq

j2.5 j5

j10 301anV

1a 2a 3a

SI

+‐ ‐j5

We will use this to find CI

A 602.0905301

5301

jZVI

eq

anS

Example 2: Final Calculation


12022 ab VV

A 9010

39015

120305.1305.115

22

jVVI ba

C

‐+

‐ +

+‐

j2.5

j2.5

j2.5

‐j15

‐j15

‐j15

j5

j5

j5

j10

j10

j10abV

bcV

caV1a

1b1c 2c

2a

2b

3a

3b3c

SI

CI

Ans: 901732.0,602.0 CS II

Practice Problem 1

• (Final EE2022 AY2011/12 semester 1) Find the voltage V1 and current I2.– Hint: remove the impedances that are in neutral line (why can we do this?) and transform delta to wye connection.


j0.1 Ω j0.1 Ω

j1.0 Ω

j1.0 Ω+‐

+ ‐ ‐ +

‐j2.4 Ωj1.0 Ω

‐j2.4 Ω

‐j2.4 Ω

j0.1 Ω

j0.1 Ω

j0.1 Ω

j0.1 Ω

0.01+ j0.05 ΩHz 50 ,0120 AV

C’

A’

B’

A”

B”C”C

A

B

+

‐1V

2I

Ans: 12041.103,037.125 21 IV

Practice Problem 2• (Final EE2022 AY2011/12 semester 1) A balanced three‐phase

voltage source feeds a balanced three‐phase load shown below, find .– Hint: Assume that line‐to‐line voltage at the load has reference angle

of 0 degree. Then, find line current magnitude and angle at the load. Calculate line‐to‐neutral voltage from source using V_source(line‐to‐neutral) = I_line×j1.0 + V_load(line‐to‐neutral). Then, use relationship between line‐to‐line voltage and line‐to‐neutral voltage to find V_source(line‐to‐line).


linetoline, SV

Ans: 601.61 V

j1.0 Ω

j1.0 Ω

j1.0 Ω

+

‐

V(rms) 230linetoline, LV3Φ Load,

Y‐connected,100 kVA at 0.8 p.f. lagging

3Φ Source

+

‐

linetoline, SV

lineI

Summary

• Three‐phase complex power,

• Per phase analysis– Only applied to balanced three‐phase circuit.– Need to convert all Delta‐connected load/sources to Wye‐connected load/sources.

– Same analysis as single‐phase circuit.


LineLineToLine3 3 IVS

EE2022 Electrical Energy Systems Lecture 6: Three‐Phase Circuit Analysis

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