Pharos University EE-272 Electrical Power Engineering 1 “Electrical Engineering Dep” Prepared By: Dr. Sahar Abd El Moneim Moussa Dr. Sahar Abd El Moneim Moussa

Single Phase system Dr. Sahar Abd El Moneim Moussa

Review on AC Circuit Basic 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 Dr. Sahar Abd El Moneim Moussa

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 Dr. Sahar Abd El Moneim Moussa

Waveform: “in terms of the time domain” 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 V(t)= Vm sin ωt Im Vm Dr. Sahar Abd El Moneim Moussa

V= Vm sin ωt , I= Im sin (ωt+90) Waveform: 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 Xc = 1/ωc Ω Im θ Vm Dr. Sahar Abd El Moneim Moussa

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 Vm θ Im Dr. Sahar Abd El Moneim Moussa

Three-Phase system Dr. Sahar Abd El Moneim Moussa

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 Dr. Sahar Abd El Moneim Moussa

3 Phase Voltages in Time Domain Va = Vm Sin ωt Vb = Vm Sin (ωt-120) Vc = Vm Sin (ωt-240) Phase (a) Phase (c) Phase b Dr. Sahar Abd El Moneim Moussa

3-Phase Voltages in Terms of Phasors Va = Vm ∠0 Vb = Vm ∠-120 Vc = Vm ∠-240 = Vm ∠120 Dr. Sahar Abd El Moneim Moussa

Types of Connections in 3-phase system Wye”Y” Delta”∆” Dr. Sahar Abd El Moneim Moussa

Wye Connection “Y” Wye Connection: “Y” Iline = Iphase Dr. Sahar Abd El Moneim Moussa

Delta Connection “∆” For Delta Circuit: Eline = Ephase Dr. Sahar Abd El Moneim Moussa

Construct the single-phase equivalent circuit of the 3- system. 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. Construct the single-phase equivalent circuit of the 3- system. Calculate the line currents IaA , IbB and IcC. Calculate the phase voltages at the load terminals. Calculate the phase currents of the load. Calculate the line voltages at the source terminals. Calculate the complex power delivered to the -connected load. Dr. Sahar Abd El Moneim Moussa

Solution: A. The load impedance of the Y equivalent is 118.5+𝑗85.8 3 =39.5+𝑗28.6 / Dr. Sahar Abd El Moneim Moussa

B. The a-phase line current is 𝐼 𝑎𝐴 = 1200 0. 2+0. 3+39. 5 +𝑗(0. 5+0 B. The a-phase line current is 𝐼 𝑎𝐴 = 1200 0.2+0.3+39.5 +𝑗(0.5+0.9+28.6) = 120<0 40+𝑗 30 =2.4 −36.87 A. Therefore, IbB=204-156 A. IcC= 2.483.13 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 Dr. Sahar Abd El Moneim Moussa

D. The phase currents of the load will be, 𝐼 𝐴𝐵 = 1 3 𝐼 𝑎𝐴 +30 The line voltage VAB is 𝑉 𝐴𝐵 = 3 𝑉 𝐴 +30 = 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, 𝐼 𝐴𝐵 = 1 3 𝐼 𝑎𝐴 +30 = 1.39 -6.87 A. Dr. Sahar Abd El Moneim Moussa

Therefore, IBC=1. 39-126. 87 A ICA=1. 39113. 13 A E Therefore, IBC=1.39-126.87 A ICA=1.39113.13 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 𝑉 𝑎𝑏 = 3 𝑉 𝑎 +30 = 205.9429.68 V. Therefore , Vbc=205.94 -90.32 V. Vca= 205.94149.68 V. Dr. Sahar Abd El Moneim Moussa

F. The total complex power delivered to the load will be, V=VAB= 202 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) (1.396.87) = 682.56 +j 494.21 VA Dr. Sahar Abd El Moneim Moussa