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Pharos UniversityEE-272

Electrical Power Engineering 1“Electrical Engineering Dep”

Prepared By:Dr. Sahar Abd El Moneim Moussa

Dr. Sahar Abd El Moneim Moussa

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SINGLE PHASE SYSTEM

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Review on AC CircuitBasic Principles: Sinusoidal voltage source: it is a source that

produces voltage that varies with time as sine wave

Waveform:

Where:

T: periodic time: the time of one complete cycle.

f: number of cycles per second = 1/T Hz

ω: angular frequency of the sine wave = 2πf rad/sec

T

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Equation: V(instantaneous)= Vm sin ωt

Where: Vm

: The maximum voltage value & it is knows as the amplitude

Vrms : Root mean square of the voltage =

Symbol:

~ V(t)= Vm sin ωt

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Resistive Circuit Circuit diagram:

Equation: V= Vm sin ωt , I= Im sin ωt Waveform: “in terms of the time domain”

Phasor diagram:

∴ The Resistive current is in phase with the voltage

VmIm

V(t)= Vm sin ωt

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Capacitive Circuit Circuit diagram:

Equation: V= Vm sin ωt , I= Im sin (ωt+90) Waveform:

Phasor Diagram:

Where: θ is the angle between the voltage and the current

(cos θ is called the power factor )

∴ The Capacitive current leads the voltage by 90o

Vm

Im

Xc = 1/ωc Ω

θ

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Inductive Circuit Circuit diagram:

Equation: V= Vm sin ωt , I= Im sin (ωt-90) Waveform:

Phasor Diagram:

Where: θ is the angle between the voltage and the current

( cos θ is called the power factor )

∴ The Inductive current lags its voltage by 90o

XL = ωL Ω

VmI

m

θ

Dr. Sahar Abd El Moneim Moussa

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THREE-PHASE SYSTEM

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Balanced Three Phase System

Balanced three-phase voltage consists of three sinusoidal voltage having the same amplitude & frequency but are out of phase with each other exactly by 120o

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3 Phase Voltages in Time Domain

Va = Vm Sin ωt Vb = Vm Sin (ωt-120) Vc = Vm Sin (ωt-240)

Phase

(a)

Phaseb

Phase (c)

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3-Phase Voltages in Terms of Phasors

Va = Vm ∠0

Vb = Vm ∠-120

Vc = Vm ∠-240 = Vm ∠120

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Types of Connections in 3-phase system

Wye”Y”

Delta”∆”

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Wye Connection “Y”

Wye Connection: “Y”

Iline = Iphase

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Delta Connection “∆”

For Delta Circuit:

Eline = Ephase

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Example 1:

A balanced three-phase Y-connected generator with positive sequence has an impedance of 0.2 +j0.5 / and internal voltage 120V/ feeds a -connected load through a distribution line having an impedance of 0.3 +j0.9 /. The load impedance is 118.5+ j85.8 /. Use the a phase internal voltage of the generator as a reference.

A. Construct the single-phase equivalent circuit of the 3- system.

B. Calculate the line currents IaA , IbB and IcC.

C. Calculate the phase voltages at the load terminals.

D. Calculate the phase currents of the load.

E. Calculate the line voltages at the source terminals.

F. Calculate the complex power delivered to the -connected load.

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Solution:

A. The load impedance of the Y equivalent is

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B. The a-phase line current is

A.

Therefore,

IbB=204-156 A.

IcC= 2.483.13 A.

C. because the load is - connected, the phase voltages are the same as the line voltages. To calculate the line voltages,

VA=(39.5 + j28.6)(2.4-36.87) = 117.04-0.96

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The line voltage VAB is

= 202.72 -0.96 V

Therefore,

VBC=202.72 -90.96 V

VCA= 202.72 149.04 V

D. The phase currents of the load will be,

= 1.39 -6.87 A.

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Therefore,

IBC=1.39-126.87 A

ICA=1.39113.13 A

E. The line voltage at the source terminals will be,

Va=(39.8 + j29.5) (2.4-36.87)

=118.9 -0.32 V.

The line voltage will be

= 205.9429.68 V.

Therefore ,

Vbc=205.94 -90.32 V.

Vca= 205.94149.68 V.

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F. The total complex power delivered to the load will be,

V=VAB= 202.72 29.04 V.

I=iAB=1.39-6.87 A.

Therefore,

ST= 3 (202.72 29.04) (1.396.87)

= 682.56 +j 494.21 VA