*FUNDAMENTAL OF ELECTRICAL POWER SYSTEMS (EE 270)Chapter 3Basic Principles

*ObjectivesReview basic concepts and establish terminology and notation encountered in electric circuit theory.Review phasors, instantaneous power, complex power, network equations and elementary aspects of balance three-phase circuits.

*Power System Network

*Three-Phase Power Transformer

*Sub-station

*Distribution Transformer

*PhasorsSinusoidal voltage or current at constant frequency characterized by:Phase angleMaximum value

*PhasorsThe root-mean-square effective valueLet the rms value ofVoltage: Current:

And let

*Power in Single Phase AC CircuitAssume a single phase sinusoidal source supplying a load.v(t) = instantaneous voltagei(t) = instantaneous currentFind the instantaneous power p(t)

*Power in Single Phase AC CircuitUse trigonometric identity

*Power in Single Phase AC Circuit

*Real PowerThe average power, P is also referred as the active power or real power.The power absorbed by the resistive component of the load.Standard unit: Watt

*Reactive PowerThe power absorbed by the reactive component of the load.Standard unit: var (volt-ampere reactive)

*Complex PowerThe complex power, S is the product of voltage and the conjugate of the current.Standard unit: VA (volt-ampere)

*Complex Power

*Phasor Diagram

Purely Resistive LoadQ=0; S=PPurely Inductive LoadP=0

Purely Capacitive LoadP=0

*Phasor Diagram

*Power Triangle

Purely Inductive LoadP=0, Q=+ve

Purely Capacitive Load

P=0, Q=-ve

*Power Triangle

*ImpedanceImpedance of complex power is given by:

*The Complex Power BalanceThe sum of real and reactive power supplied by the source is equal to the sum of real and reactive powers transferred to the load.Law of energy conservation

*The Complex Power Balance

DC

V

*The Complex Power BalanceExample:

DC

V

*SolutionCurrent at each load:

*SolutionComplex power absorbed at each load:Total load complex power:

*ExerciseTwo loads connected in parallel are supplied from a single-phase 240Vrmssource. The two loads draw a total realpower of 400kW at a power factor of 0.8lagging. One of the loads draws 120kW at apower factor of 0.96 leading. Find the complex power of the other load.

*ExerciseTwo impedances, Z1=0.8+j5.6 and Z2=8-j16, and a single phase motor are connected in parallel across a 200Vrms, 60Hz supply. The motor draws 5kVA at 0.8pf lagging. Find S1, S2 and S3 for the motor.

*Power Factor CorrectionPF = 1 } unity power factorIf PFreal power PCurrent increase, cost of utility increase.Major loads of the system should be near to unity power factor.

*Power Factor CorrectionInductive load : lagging pfCapacitive load : leading pfHow to fix PF? Capacitor is added to the system (inductive load).PF is mostly considered in industrial consumers (using inductive load) and not in residential and small commercial since the power factor is near unity.

*Power Factor Correction

*Power Factor Correction

*ExerciseFind the total real and reactive power, the power factor at the load, and the total current without C.Find the capacitance of the capacitor connected across the loads to improve the overall power factor to 0.8 lagging.Two loads Z1=100+j0 and Z2=10+j20 are connected across a 200Vrms, 50Hz source.

DC

200V

*Solution (a)Current at each load:Power at each load:

*Solution (a)Total real and reactive power: Total current without C: Power factor at the load:

*Solution (b)Power at the capacitor:

*Capacitance of the capacitor:Solution (b)

*ExerciseThree loads are connected in parallel across a1400Vrms, 50Hz single-phase supply.Load 1: Inductive load, 125kVA at 0.28 power factorLoad 2: Capacitive load, 10kW and 40kvarLoad 3: Resistive load of 15kWFind the total kW, kvar, kVA and the supply power factor.A capacitor of negligible resistance is connected in parallel with the above loads to improve the power factor to 0.8 lagging. Determine the kvar rating of this capacitor and the capacitance in F.

*ExerciseTwo loads are connected in parallel across a 200Vrms, 50Hz single-phase supply.Load 1: 0.8 + j5.6Load 2: 8 - j16Find the total kW, kvar, kVA and the supply power factor.A capacitor is connected in parallel with the loads. Find the kvar and the capacitance in F to improve the overall power factor to unity.What is the new line current?

*Complex Power FlowNeed to consider two way current (i.e. From V1 to V2 and from V2 to V1) and two way S (i.e. From V1 to V2 and from V2 to V1).If P is negative than the P in which the source is associated to receives/absorbs the P. If P is positive than the P in which the source is associated to generates/delivers the P.If Q is negative than the Q in which the source is associated to receives/absorbs the Q. If Q is positive than the Q in which the source is associated to generates/delivers the Q.

*Generator & Load Convention

ConventionP/Q Characteristic Generator Convention+Delivered/Generated Absorbed/Received Load Convention+Absorbed/Received Delivered/Generated

Vsource

+

I

Vload

+

I

*ExerciseConsider two voltage sources V1=1205V and V2=1000V are connected by a short line of impedance Z=1+j7. Determined the real and reactive power supplied or received by each source and the power loss in the line.

DC

V1

V2

Zshort line

*SolutionFind current which flows from V1 to V2 i.e. I12

Find current which flows from V2 to V1 i.e. I21

Find S from V1 to V2 i.e. S12

Find S from V2 to V1 i.e. S21

*SolutionEvaluate the source from the previous calculated S.Based on S1:

Source 1 receive 97.5W and delivers 363.3var.

Based on S2

Source 2 generates 107.3W and receives 294.5var.

*SolutionPower loss in the line

Check!

Thus, the real power loss in the line is 9.8W and the reactive power loss in the line is 68.6var.

*ExerciseTwo single-phase ideal voltage sources areconnected by a line of impedance of 0.7+j2.4. V1=50016.26V and V2=5850V. Find the complex power for each source and determine whether they are delivering or receiving real and reactive power. Also, find the real and the reactive power loss in the line.

*Review..Power in single-phase ACS, P, Q, p(t)Phasor analysis/diagramPower triangleComplex power balancePower factor correctionComplex power flowGenerator / Load Convention

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