1. Lecture 27Electro Mechanical System1  The fact that power is always positive reveals that it always flows from the generator to the resistor.  This is one of the basic properties of active power: although it pulsates between zero and max. it never changes direction.  The average power is clearly midway between 2P and zero, and so its value is P = 2EI/2 = EI watts.  That is precisely the power indicated by the wattmeter.  The generator is an active source and the resistor is an active load.  The symbol for active power is P and the unit is the watt (W). The kilowatt (kW) and megawatt (MW) are frequently used. Active power

2. Lecture 27Electro Mechanical System2  The resistor is replaced by a reactor X. As a result, current I lags 90° behind the voltage E  we can draw the waveforms of E and I & by multiplying their instantaneous values, we obtain curve of instantaneous power  Positive waves correspond to instantaneous power delivered by the generator to the reactor and the negative waves represent power delivered from the reactor to the generator  Duration of each wave corresponds to one quarter of a cycle  Frequency of power wave is therefore twice the line frequency  Power that surges back and forth is called reactive power (Q)  To distinguish this power from active power, another unit is used the var. Its multiples are kilovar (kvar) and megavar (Mvar) Reactive power

3. Lecture 27Electro Mechanical System3  Reactive power involves real power that oscillates back and forth between two devices over a transmission line.  Consequently, it is impossible to say whether the power originates at one end of the line or the other.  It is useful to assume that some devices generate reactive power while others absorb it.  In other words, some devices behave like reactive sources and others like reactive loads.  A reactor is considered to be a reactive load that absorbs reactive power. Reactive load & reactive source

4. Lecture 27Electro Mechanical System4 A reactor having an inductive reactance of 4 Ω is connected to the terminals of a 120 V ac generator. a)Calculate the value of the current in the reactor I L = E / X L = 120V / 4 Ω = 30 A b)Calculate the power associated with the reactor Q = EI = 120 X 30 = 3600 var = 3.6 kvar c)Calculate the power associated with the ac generator Because the reactor absorbs 3.6 kvar of reactance power, the ac generator must be supplying it. Consequently, the generator is a source of reactive power: it delivers 3.6 kvar. d)Draw the phasor diagram for the circuit The phasor diagram show that current I, lags 90° behind voltage E Example

5. Lecture 27Electro Mechanical System5  If we add capacitor having a reactance of 4 Ω to the circuit.  capacitor current I C = 120 V/4Ω = 30A & leads the voltage by 90°  The vector sum of I L and I C is zero and so the ac generator is no longer supplying any power at all to the circuit.  The current in the reactor has not changed; consequently, it continues to absorb 30 A X 120 V = 3.6 kvar of reactive power.  Where is this reactive power coming from? It can only come from the capacitor, which acts as a source of reactive power.  Q = El c = 120 V X 30 A = 3600 var = 3.6 kvar  It means a capacitor is a source of reactive power. It acts as a reactive power source whenever it is part of a sine-wave-based, steady-state circuit. Capacitor and reactive power

6. Lecture 27Electro Mechanical System6  Remove the reactor from the circuit. Capacitor is now alone.  It still carries a current of 30 A, leading the voltage E by 90°  Capacitor acts as a source of reactive power, delivering 3.6 kvar  Where does this power go? Capacitor delivers it to the generator  How, can a passive device(capacitor) produce any power?  Reactive power really represents energy, like a pendulum, swings back and forth without ever doing any useful work  The capacitor acts as a temporary energy-storing device  Instead of storing magnetic energy, it stores electrostatic energy  If we connect a varmeter, it will give a negative reading of EI = – 3600 var, showing that reactive power is indeed flowing from the capacitor to the generator. Capacitor and reactive power

7. Lecture 27Electro Mechanical System7 An ac generator is connected to a group of R, L, C circuit elements carrying, the currents shown. Calculate the active and reactive power associated with the generator. The two resistors absorb active power given by P = I 2 R =(16.12 2 X 2) + (14 2 X 4) = 784 + 520 = 1304 W The 3 Ω reactor absorbs reactive power: Q L = I 2 X L = 14 2 X 3 = 588 var The 3.5 Ω capacitor generates reactive power: Q C =I 2 X c = 20 2 X 3.5 = 1400 var R, L, C circuit generates a net reactive power of 1400 – 588 = 812 var In conclusion, the ac generator is a source of active power (1304 W) and a receiver of reactive power (812 var). Example

8. Lecture 27Electro Mechanical System8  Active and reactive power, are separate quantities and one cannot be converted into the other.  Both place a burden on the transmission line.  Active power eventually produces a tangible result (heat, mechanical power, light, etc.)  Reactive power represents power that oscillates back and forth.  All ac inductive devices such as magnets, transformers, ballasts, and induction motors, absorb reactive power because one component of the current they draw lags 90° behind the voltage.  The reactive power plays a very important role because it produces the ac magnetic field in these devices.  A building, shopping center, or city may be considered to be a huge active/reactive load connected to an electric utility system.  Load centers contain thousands of electromagnetic devices that draw both reactive power (to sustain their magnetic fields) and active power (to do the useful work). Distinction between active and reactive power

9. Lecture 27Electro Mechanical System9  Loads that absorb active power P and reactive power Q is made up of a resistance and an inductive reactance.  Resistor draws a current I p, while the reactor draws a current I q  Consequently, I p is in phase with E while I q lags 90° behind.  Resultant line current I lags behind E by an angle .  If we connect a wattmeter and a varmeter, the readings will both be positive, indicating P = EI p watts and Q = EI q vars  An ammeter will indicate a current of I amperes, so the power supplied to the load should be equal to EI watts.  This is incorrect because the power is composed of an active component (watts) and a reactive component (vars). For this reason the product EI is called apparent power, symbol is S.  Apparent power is expressed in voltamperes. (kVA) & (MVA). Combined active & reactive loads: apparent power