BASIC PRINCIPLES Power In Single-Phase AC Circuit Let instantaneous voltage be v(t)=v m cos(ωt+θ v ) Let instantaneous current be i(t)=i m cos(ωt+θ i ) The instantaneous p(t) delivered to the load is p(t)=v(t)i(t)=v m i m cos(ωt+θ v ) cos(ωt+θ i ) p(t) = V I cosθ[1+cos2(ωt+θ v ) ] + V I sinθsin2(ωt+θ v ), where V = Vm / 2, I = Im / 2 The first term is the energy flow into the circuit The second term is the energy borrowed and returned by the circuit
BASIC PRINCIPLES Power In Single-Phase AC Circuit p(t) = V I cosθ[1+cos2(ωt+θ v ) ] = V I cosθ+ V I cosθ cos2(ωt+θ v ) ] The first term is the average power delivered to the load The second term is the sinusoidal variation with twice of source frequency in the absorption of power by the resistive load power absorbed by the resistive component is active power or real power V I is called apparent power in VA P= V I cosθ is real power in watts, cosθ is called power factor when current lags voltage, cosθ is lagging, and vice versa
BASIC PRINCIPLES Power In Single-Phase AC Circuit p X (t) = V I sinθsin2(ωt+θ v ), This term is the power oscillating into and out of load of reactive element (capacitive or inductive) Q= V I sinθ is the amplitude of reactive power in var (voltage-ampere reactive) For inductive load, current lagging voltage, θ = (θ v -θ i ) > 0, Q is positive For capacitive load, current leading voltage, θ = (θ v -θ i ) < 0, Q is negative
BASIC PRINCIPLES The Instantaneous Power Has Some Characteristics For a pure resistor, cosθ=1, apparent power V I = real power In a pure inductive or capacitive circuit, cosθ=0, no energy change but the instantaneous power oscillate between circuit and source When p X (t) > 0, energy is stored in the magnetic field with the inductive elements, when p X (t) < 0, energy is extracted in the magnetic field with the inductive elements If load is pure capacitive, the current leads voltage by 90 o, average power is 0, vise versa for pure inductive load
BASIC PRINCIPLES Example 2.1(PSA-Saadat) supply voltage v(t)=100cosωt, inductive load Z=1.25 60 o Ω determine i(t), p(t), p R (t), p x (t)
COMPLEX POWER The rms voltage phasor and rms current phasor V = V θ v and I = I θ i The term VI* = V I (θ v -θ i ) = V I θ = V I cosθ + j V I sinθ θ θ i θ v Define a complex power quantity S = VI* = P + jq, where V I 2 S = P + apparent power The units of complex power S are kva or MVA apparent power S is used as a rating unit of power equipment apparent power S is used for an utility to supply power to consumers S θ P Q 2, Q
Define a reactive power quantity Q COMPLEX POWER Q is (+) when phase angle θ is (+), i.e. load is inductive. Q is (-) when load is capacitive I capacitive P θ θ v θ i V θ S Q (-) Given a load impedance Z V=ZI S=VI*=ZII*= (R+jX)II* = (R+jX) I 2 S=VI*=V(V/Z)*=V(V*/Z*)=(VV*)/Z*= V 2 /Z* Z= V 2 /S* Complex power balance real power supplied by source = real power absorbed by load see Example 2.2
POWER FACTOR CORRECTION AND COMPLEX POWER FLOW Apparent Power S > P if power factor (p.f.) < 1 current supplied in p.f. < 1 will be larger than current supplied in p.f. = 1 a larger current will cost more! Power factor inductive load causes lagging current and result in a lagging power factor (p.f. < 1), Z=jX, I=V/(jX)=(V/X) -90 o (lagging) capacitive load causes leading current and result in a leading power factor (p.f. < 1), Z=-jX, I=V/(-jX)=(V/X) +90 o (leading) resistive load has a unity power factor (p.f. = 1) To keep p.f. 1, we need to install capacitor banks through the network for inductive load industry probably operate at low lagging p.f. since most of them are motors (inductive load) to correct low power factor, capacitor banks are connected in parallel to the load for p.f. correction, see Ex. 2.3 in pp.23, Ex. 2.4 in pp.24 (PSA- Saadat)
COMPLEX POWER FLOW Complex power flow (PSA-Saadat) see Figure 2.9 in pp. 26 Z γ complex power S 12 S real power at the sending end P reactive power at the sending end Q 2 V2 12 = γ γ + δ1 δ2 Z Z 2 12 = δ Z V2 cosγ cos( γ + δ1 2) Z 2 12 = δ Z V2 sinγ sin( γ + δ1 2) Z V 1 δ 1 V 2 δ 2
COMPLEX POWER FLOW OBSERVATION Observation on complex power flow assume transmission lines have small resistance compared to the reactance (Z=X 90 o, cosγ=0), P 12 and Q 12 becomes Z γ V2 P12 = sin( δ1 δ2) X V V 1 δ 1 1 V Q12 = ( V2 cos( δ1 δ )) 2 δ 2 2 X maximum power occurs at δ = 90 o, where maximum power transfer: P max = V 1 V 2 /X real power flow P 12 could be changed by small change in δ 1 or δ 2 reactive power Q 12 is determined by terminal voltage difference, (i.e., Q V 1 - V 2 ) See Ex. 2.5 for line loss (PSA-Saadat)
COMPLEX POWER FLOW OBSERVATION Ex. 2.6 direction of power flow between two voltage sources voltage source 1 can change phase from ±30 o in step of 5 o phase of voltage source 2 is constant
PROJECT 2 Two voltage of sources V 1 =120-5 V rms and V 2 =100 0 V rms are connected by a short line of impedance Z=1+j7 Ω. The figure is shown below Write a Matlab program such that the voltage magnitude of the source 1 is changed from 75% to 100% of the give in steps of 1V. The voltage magnitude of source 2 and phase angle are kept constant. Compute the complex power for each source and the line loss. Tabulate the reactive powers and plot Q 1, Q 2, and Q L versus voltage magnitude V 1. From the results, show that the flow of reactive power along the interconnection is determined by the magnitude difference of the terminal voltages. Z γ V 1 δ 1 V 2 δ 2