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BALANCED THREE-PHASE AC CIRCUIT • Balanced Three-Phase Voltage Sources Delta Connection Star Connection • Balanced 3-phase Load Delta Connection Star Connection • Power in a Balanced Phase Circuit Introduction Three Phase System Balanced Three Phase Voltages Three-phase voltage sources a) wye-connec...

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BALANCED THREE-PHASE AC CIRCUIT •

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Balanced Three-Phase Voltage Sources Delta Connection Star Connection Balanced 3-phase Load Delta Connection Star Connection Power in a Balanced Phase Circuit

Introduction Three Phase System

Balanced Three Phase Voltages Three-phase voltage sources

a) wye-connected source

b) delta-connected source

If the voltage source have the same amplitude and frequency ω and are out of phase with each other by 120o, the voltage are said to be balanced.

Van = Vbn = Vcn

Van + Vbn + Vcn = 0

Balanced phase voltages are equal in magnitude and out of phase with each other by 120o

Balanced Three Phase Voltages Vm

Va

Vb

Vc

ωt V

abc sequence or positive sequence:

Van = Vp ∠00 Vbn = Vp ∠ − 1200 Vcn = Vp ∠ − 2400 = Vp ∠ + 1200 acb sequence or negative sequence:

Van = Vp ∠00 Vcn = Vp ∠ − 1200 Vbn = Vp ∠ − 2400 = Vp ∠ + 1200 Vp

is the effective or rms value

Balanced Three Phase Loads Two possible three-phase load configurations:

a) a Star or Y-connected load For a balanced wye connected load:

Z1 = Z 2 = Z3 = Z Y 1 ZY = Z∆ 3

b) a delta-connected load For a balanced delta connected load:

Za = Z b = Zc = Z ∆

Z ∆ = 3Z Y

Example 1 Determine the phase sequence of the set of voltages van = √2 200 cos(ωt + 10◦) vbn = √2 200 cos(ωt − 230◦), vcn = √2 200 cos(ωt − 110◦) Solution: The voltages can be expressed in phasor form as We notice that Van leads Vcn by 120◦ and Vcn in turn leads Vbn by 120◦. Van = 200∠10°V Vbn = 200∠ − 230°V Vcn = 200∠ − 110°V

Hence, we have an acb sequence. Given that Vbn = 110∠30°V

, find Van and Vcn, assuming a

positive (abc) sequence. Answer:

Van = 110∠150°V

Vcn = 110∠ − 90°V

Balanced Y-Y Connection A balanced Y-Y system is a three phase system with a balanced Y connected source and balanced Y connected load.

Zs =

Source impedance

Z =

Line impedance

ZL =

Load impedance

ZY = Z s + Z + Z L

ZY = ZL

Balanced Wye-Wye Connection Zs =

Source impedance

Z =

Line impedance

ZL =

Load impedance

Z Y = Total impedance per phase Z Y = Zs + Z + Z L ZY = ZL

Balanced Y-Y Connection Line to line voltages or line voltages given that phase a voltage is reference Vcn can be shown to be:

Vab

Vab = 3Vp ∠300 Vbc = 3Vp ∠ − 90

0

o

120

30o

Vca = 3Vp ∠ − 2100 Vbn

VL = Vab = Vbc = Vca Vp = Van = Vbn = Vcn

VL = 3V p IL = I p

Van

Balanced Y-Y Connection Given the phase voltages, the line current can be calculated as: Applying KVL to each phase: Van Ia = ZY Vbn Van ∠ − 1200 Ib = = = I a ∠ − 1200 ZY ZY Vcn Van ∠ − 2400 Ic = = = I a ∠ − 2400 ZY ZY

I a + I b + I c = −I n = 0

VnN = Z n I n = 0

Thus, the per-phase equivalent circuit can be expressed as:

Ia =

Van ZY

Y-Y configuration Example:1 • A balanced positive-sequence Y-connected 60 Hz three-phase source has phase voltage Va=1000V. Each phase of the load consists of a 0.1-H inductance in series with a 50-Ω Ω resistance. • Find the line currents, the line voltages, the power and the reactive power delivered to the load. Draw a phasor diagram showing line voltages, phase voltages and the line currents. Assuming that the phase angle of Van is zero.

Z = R + jωL = 50 + j 37.7 = 62.62∠37 ∴θ = 37 0

0

Van = 15.97∠ − 370 Z ∴ I bB = 15.97∠ − 157 0 , I cC = 15.97∠830 I aA =

Vab = Van × 3∠300 = 1732∠300 ∴Vbc = 1732∠ − 900 ,Vca = 1732∠1500

Example 2 1- Calculate the line currents in the three wire Y-Y system of figure below.

2- A Y-connected balanced three-phase generator with an impedance of 0.4+j0.3 per phase is connected to a Y-connected balanced load with an impedance of 24 + j19 per phase. The line joining the generator and the load has an impedance of 0.6 + j0.7 per phase. Assuming a positive sequence for the source voltages and that

V an = 120 ∠ 30 0 V Find: (a) the line voltages

(b) the line currents

Balanced Y-Delta Connection A balanced Y- system consists of balanced Y connected source feeding a balanced connected load. Line voltages:

Vab = 3Vp ∠300 = VAB Vbc = 3Vp ∠ − 900 = VBC Vca = 3Vp ∠ − 2100 = VCA Phase currents: I = V AB ; I = VBC ; I = VCA AB BC CA Z∆ Z∆ Z∆ Line currents:

I a = I AB − I CA = 3I AB ∠ − 30° I b = I BC − I AB = 3I AB ∠ − 150° I c = I CA − I BC = 3I AB ∠90°

I CA = I AB

Balanced Y-Delta Connection ∠ − 240 0

I a = I AB − I CA = I AB (1 − 1∠ − 2400 )

I a = I AB 3∠ − 300 Magnitude line currents:

IL = Ia = Ib = Ic

IL = Ip 3

I p = I AB = I BC = I CA

A single phase equivalent circuit ZY =

Ia =

Z∆ 3

Van V = an ZY Z∆ / 3

Y-Delta configuration: Example 3 1- A balanced abc sequence Y-connected source with Van = 100∠10 0 V is connected to a -connected balanced load (8+j4) Calculate the phase and line currents.

per phase.

2-One line voltage of a balanced Y-connected source is If the source is connected to a the phase and line currents. Assume the abc sequence.

V AB = 180∠ − 20 0 V

-connected load of 20 ∠ 40 0 Ω , find

Balanced Delta-Delta Connection A balanced - system is one in which both balanced source and balanced load are connected.

Balanced Delta-Delta Connection A balanced - system is the one in which both balanced source and balanced load are connected. Line voltages:

Vab = VAB Vbc = VBC Vca = VCA Line currents: I a = I AB − I CA = 3I AB ∠ − 30°

Phase currents:

I b = I BC − I AB = 3I AB ∠ − 150° I c = I CA − I BC = 3I AB ∠90°

Magnitude line currents:

IL = Ip 3

Total impedance:

Z∆ ZY = 3

I AB =

V AB Z∆

I BC =

V BC Z∆

I CA =

V CA Z∆

Example4:

A delta-connected source supplies a delta-connected load through wires having impedances of Zline=0.3+j0.4 , the load impedance are Z =30+j6 , the balanced source ab voltage is Vab=1000...