Lecture 28 Electro Mechanical System 1

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 Eab by an angle . Current I has two components Ip & Iq. Value Ip = P/E & Iq = 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 Iq / Ip = arctan Q/P

Relationship between P, Q, & S

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222 S so QPE

Q

E

P

E

SIII qp

Lecture 28 Electro Mechanical System 2

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 = EIP / EI = IP / 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

Lecture 28 Electro Mechanical System 3

Relationship S2 = P2 + Q2 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.

Lecture 28 Electro Mechanical System 4

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 = Elp watts, but a varmeter will give a negative reading Q = EIq . 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

Lecture 28 Electro Mechanical System 5

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

22 QPS

Lecture 28 Electro Mechanical System 6

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 kW2. Reactive power absorbed by the system:

Q1 = (5 + 7 + 8) = +20 kvar3. Reactive power supplied by the

capacitors:Q2 = (–9 – 16) = – 25 kvar

4. Net reactive power Q absorbed by the system:

Q = ( + 20 – 25) = – 5 kvar

System with several loads

Lecture 28 Electro Mechanical System 7

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 A7. 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

kVAQPS 5.24)5(24 2222

Lecture 28 Electro Mechanical System 8

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.

Lecture 28 Electro Mechanical System 9

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.