1. 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

2. Single Phase system
Dr. Sahar Abd El Moneim Moussa

3. 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

4. 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

5. 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

6. 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

7. 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

8. Three-Phase system
Dr. Sahar Abd El Moneim Moussa

9. 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

10. 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

11. 3-Phase Voltages in Terms of Phasors Va = Vm ∠0 Vb = Vm ∠-120
Vc = Vm ∠-240 = Vm ∠120
Dr. Sahar Abd El Moneim Moussa

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

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

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

15. 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 j0.5 / and internal voltage 120V/ feeds a -connected load through a distribution line having an impedance of j0.9 /. The load impedance is 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

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

17. 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 𝑗( ) = 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=( j28.6)(2.4-36.87) = -0.96
Dr. Sahar Abd El Moneim Moussa

18. 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
= -0.96 V
Therefore,
VBC=  V
VCA=  V
D. The phase currents of the load will be,
𝐼 𝐴𝐵 = 𝐼 𝑎𝐴 +30
= 1.39 -6.87 A.
Dr. Sahar Abd El Moneim Moussa

19. Therefore, IBC=1. 39-126. 87 A ICA=1. 39113. 13 A E
Therefore, IBC=1.39 A ICA=1.39 A E. The line voltage at the source terminals will be, Va=( j29.5) (2.4-36.87) =118.9 -0.32 V. The line voltage will be 𝑉 𝑎𝑏 = 3 𝑉 𝑎 +30 = 29.68 V. Therefore , Vbc=  V. Vca=  V.
Dr. Sahar Abd El Moneim Moussa

20. 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= 29.04 V. I=iAB=1.39-6.87 A. Therefore, ST= 3 ( 29.04) (1.396.87) = j VA
Dr. Sahar Abd El Moneim Moussa