1. Lecture 28Electro Mechanical System1  Consider a circuit with a source, a load, & appropriate meters.  P and Q are positive, so the load absorbs both active and reactive power. The line current I lags behind E ab by an angle .  Current I has two components I p & I q. Value I p = P/E & I q = Q/E  The apparent power S = EI, from which I = S/E  The phasor diagram shows S =apparent power [VA|:P =active power [W]:Q =reactive power [var]  The angle  = arctan I q / I p = arctan Q/P Relationship between P, Q, & S

2. Lecture 28Electro Mechanical System2  The power factor of an ac device or circuit is the ratio of the active power P to the apparent power S. given by the equation: power factor = P / S = EI P / EI = I P / I = cos  Where: P = active power delivered or absorbed by circuit or device [W] S = apparent power of the circuit or device [VA]  Power factor is expressed as a simple number, or a percentage.  The power factor can never be greater than unity (or 100 percent).  The power factor of a resistor is 100 percent because the apparent power it draws is equal to the active power.  The power factor of an ideal coil having no resistance is zero, because it does not consume any active power.  The power factor of a circuit or device is simply a way of stating what fraction of its apparent power is real, or active, power.  In a single-phase circuit the power factor is also a measure of the phase angle  between the voltage and current  If we know the power factor, we can calculate the angle. The power factor is said to be lagging or leading. Power factor

3. Lecture 28Electro Mechanical System3  Relationship S 2 = P 2 + Q 2 is right-angle triangle, known as a power triangle. Following rules apply:  Active power P absorbed by a circuit or device is considered to be positive and is drawn horizontally to the right  Active power P that is delivered by a circuit or device is considered to be negative and is drawn horizontally to left  Reactive power Q absorbed by a circuit or device is considered to be positive and is drawn vertically upwards Power triangle  Reactive power Q that is delivered by a circuit or device is considered to be negative and is drawn vertically downwards  The concept of the power triangle is useful when solving ac circuits that comprise several active and reactive power components.

4. Lecture 28Electro Mechanical System4  A resistor and capacitor are connected to the source.  Capacitor acts as a reactive source.  A wattmeter connected into the circuit will give a positive reading P = El p watts, but a varmeter will give a negative reading Q = EI q.  Source G delivers active power P but receives reactive power Q.  Two powers flowing in opposite directions over the same line.  An electrical outlet can act not only as an active or reactive source, but it may also behave as an active or reactive load.  It depends upon the type of device connected to the receptacle. Sources and loads

5. Lecture 28Electro Mechanical System5  Consider, a group of loads connected in an unusual way to a 380 V source.  We wish to calculate the apparent power absorbed & current supplied by source.  We simply draw a block diagram of the individual loads, indicating the direction of active and reactive power flow  Add all the active powers in a circuit to obtain the total active power P.  We can add the reactive powers to obtain the total reactive power Q.  Total apparent power S is then found by:  Adding reactive powers, we assign a +ve value to those that are absorbed by the system and a –ve value to those that are generated (such as by a capacitor) System with several loads

6. Lecture 28Electro Mechanical System6  Similarly we assign a +ve value to active powers that are absorbed and –ve value to those that are generated (by an alternator)  we cannot add the apparent powers to obtain the total apparent power S. Add them only if their power factors are same  Let us now solve the circuit : 1. Active power absorbed by the system: P = (2 + 8 + 14) = +24 kW 2.Reactive power absorbed by the system: Q 1 = (5 + 7 + 8) = +20 kvar 3.Reactive power supplied by the capacitors: Q 2 = (–9 – 16) = – 25 kvar 4.Net reactive power Q absorbed by the system: Q = ( + 20 – 25) = – 5 kvar System with several loads

7. Lecture 28Electro Mechanical System7 5.Apparent power of the system: 6.Because the 380 V source furnishes the appaent power, the line current is I = S/E = 24 500/380 = 64.5 A 7.The power factor of the system is cos  L = P/S = 24/24.5 = 0.979 (leading)  The 380 V source delivers 24 kW of active power, but it receives 5 kvar of reactive power.  This reactive power flows into the system of the electrical company, where it is available to create magnetic fields.  The magnetic fields may be associated with distribution transformers, transmission lines, relays etc. System with several loads

8. Lecture 28Electro Mechanical System8  Let us make power triangle for the system.  It is the graphical solution to our problem.  Starting with the 5 kvar load, we move from one device to the next around the system.  We draw the magnitude and direction (up, down, left, right) of each power vector.  At the end, we can draw a power vector from the starting point to the end point. System with several loads  Vector having a value of 24.5 kVA.  The horizontal component of this vector has a value of 24 kW and, because it is directed to the right, we know that it represents power absorbed by the system.  The vertical component of 5 kvar is directed downward, so it represents reactive power generated by the system.

9. Lecture 28Electro Mechanical System9  Loads sometime absorb reactive power without creating any magnetic field at all.  This can happen in electronic power circuits when the current flow is delayed by means of a rapid switching device.  Consider, a circuit in which a 100 V, 60 Hz source is connected to a 10 Ω resistive load by means of a mechanical switch.  The switch opens and closes its contacts so that current only flows during the latter part of each half cycle.  This forced delay causes the current to lag behind the voltage.  If we connect a wattmeter and varmeter, they would read +500 W and + 318 var. Reactive power without magnetic fields  This corresponds to a lagging power factor (sometimes called displacement power factor) of 84.4 percent.  Reactive power is associated with the rapidly operating switch rather than with the resistor itself.