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Introduction to Power

Engineering

Lecture # 01


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Definition of Electric Power

Power is a measure of energy per unit time.

Power therefore gives the rate of energy consumption or

production.

The units for power are generally watts (W).

For example, the watt rating of an appliance gives the rate atwhich it uses energy.


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The total amount of energy consumed by an appliance is the

wattage multiplied by the amount of time during which it was

used.

This energy can be expressed in units of watt-hours or, more

commonly, kilowatt-hours.


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Power dissipated by a circuit element is given by the product of itsresistance and the square of the current through it

P=I2R

The term dissipated indicates that the electric energy is beingconverted to heat.

This heat may be part of the appliances intended function, as in

any electric heating device.

It may also be considered a loss, as in the resistive heating oftransmission lines.


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Another, more general way of calculating ower is as the product of current

and voltage.

P=IV For a resistive element, we can apply Ohms law (V = IR) to see that the

formulas P = I2R and P = IV amount to the same thing:

P = IV = I(IR) = I2R


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Example:

Consider an incandescent light bulb, rated at 60 W.

This means that the filament dissipates energy at the rate of 60 W when

presented with a given voltage.

This assume to be the normal household voltage, 120 V.

The power equals the voltage applied to the light bulb times the current

through it.


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The current is 0.5 A.

60 W = 0.5 A. 120 V

The power rating is specific to a given voltage but one property of the lightbulb that is always the same is its resistance, in this case, 240 ohms.

We could determine the resistance from Ohms law:

120 V = 0.5 A* 240 ohm.

And also verify that the power also corresponds to I2R:

60 W = (0.5 A)2 *240 ohm


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Difference between the power dissipated

and the power transmitted by the line

The dissipated power is simply given by P = I2R.

We could also write this as P = IV.

But that would be less convenient for two reasons:

First, although it is tempting to think of a power line as just a resistivewire, it actually has a significant reactance.

Because of the phase shift involved, the presence of reactance means

that multiplying I and V together will be more complicated.

On the other hand, taking the square of the current magnitude isalways easy, regardless of phase shifts.


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Second, if we tried to calculate dissipated power by using P = IV, we

would have to be very careful about which V to use.

Since Ohms law refers to the voltage drop across a resistor, then V

must be the voltage difference between the two ends of the line,otherwise known as the Line Drop.

This line drop is distinct from the Line Voltage.

Line voltage specifies the voltage with respect to ground.


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Typically, the line drop would be a few percent of the line voltage, but

it is usually not known precisely.

If transmission lines had no resistance at all, there would be zero line

drop.

For these reasons, Thermal Losses are better calculated using

P = I2R, and are often referred to as I2R losses.


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When we ask about the power transmitted by the line, we can think of the

line as extended terminals, like battery terminals.

The power that is available to a load connected to this line can be

calculated with the formula P = IV.

But now V refers to the line voltage, which is that seen by the load

between the two terminals.

We say that the power has been transmitted by the line at the voltage V.


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Intuitive understanding of P = IV

We can have an intuitive understanding of P = IV.

Voltage is a measure of energy per unit charge.

Current is the flow rate of charge.

The product of voltage and current therefore tells us how many electrons

are passing through, multiplied by the amount of energy each electron

carries.

[Energy/Charge] x [Charge/Time] = [Energy/Time]


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Complex Power

Applying the simple formula P = IV becomes more problematic when

voltage and current are changing over time, as they do in a.c. systems.

Let us begin by considering power, voltage, and current as real quantities

that vary in time.

The fundamental and correct way to interpret the statement P = IV when I

and V vary in time is as a statement ofinstantaneous conditions.

Instantaneous Power is equal to the instantaneous product of current andvoltage.


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Complex Power

In other words, at any instant, the power equals the voltage times the

current at that instant.

This is expressed by writing each variable as a function of time.

P(t) = I(t) x V(t)

t is the same throughout the equation (i.e., the same instant).

However, instantaneous power as such is usually not very interesting for

us.


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Complex Power

In power systems, we generally need to know about power transmitted or

consumed on a time scale much greater than 1/60 of a second.

Therefore, we need an expression for power as averaged over entire cycles

of alternating current and voltage.

Consider first the case of a purely resistive load.

Voltage and current are in phase.

They are oscillating simultaneously.


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Complex Power

The Average Power, that is the average product of voltage and current,

can be obtained by taking the averages, rms values, of each and then

multiplying them together.

Pave = IrmsVrms (resistive case)

Power for the resistive case is illustrated in Figure on the next slide.


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Complex Power

Power as the product of voltage and current, with voltage and current in phase


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Complex Power

But now consider a load with reactance.

The relative timing of voltage and current has been shifted.

Their maxima no longer coincide.

In fact, one quantity is sometimes negative when the other ispositive.

As a result, the instantaneous power , the product of voltage andcurrent, is sometimes negative.

This is shown in Figure on next slide.


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Complex Power

Power as the product of voltage and current, with current lagging behind voltage by a

phase angle f.


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Complex Power

We can interpret the negative instantaneous power.

It is the power that flows backwards along the transmission line, or out

of the load and back into the generator.

Since instantaneous power is sometimes negative, the average power is

clearly less than it was in the resistive case.

But just how much less?

Fortunately, this is very easy to determine.


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Complex Power

The average power is directly related to the amount of phase shift

between voltage and current.

Here we skip the mathematical derivation and simply state that:

The reduction in average power due to the phase shift is given

by the cosine of the angle of the shift.

Pave = IrmsVrms cos


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Complex Power

The factor of cos is called the POWER FACTOR, often abbreviated p.f.

This same equation can also be written as:

Pave = (ImaxVmax cos )/2

Each rms value is related to the maximum value by a factor of 1/2.


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Complex Power

This equation is true for any kind of load.

In the special case where there is only resistance and no phase shift, we

have =0 and cos= 1.

So there is no need to write down the cos, and we get the formula fromthe previous page.

Pave = IrmsVrms (resistive case)


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Complex Power

Consider another special case where the load is purely reactive, having no

resistance at all.

In this case, the phase shift would be =90o and cos=0.

It means that power only oscillates back and forth, but is not dissipated

the average power is zero.


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Complex Power

The Average Power corresponds to the power actually transmitted or

consumed by the load.

It is also called Real Power, Active Power or True Power, and is measured

in watts.


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Complex Power

There are other aspects of the transmitted power that we wish to specify.

The product of current and voltage, regardless of their phase shift, is called theApparent Power, denoted by the symbol S.

Its magnitude is given by:

S = IrmsVrms

Although apparent and real power have the same units physically, they are

expressed differently to maintain an obvious distinction.

Thus, the units of apparent power are called volt-amperes (VA).


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Complex Power

Apparent power is important in the context of equipment capacity.

Actually, the crucial quantity with respect to thermal capacity limits is only

the current.

In practice, though, the current is often inconvenient to specify.

Since the operating voltage of a given piece of equipment is usually quite

constant, apparent power is a fair way of indicating the current.

The point is that apparent power is a much better measure of the current

than real power, because it does not depend on the power factor. Thus,

utility equipment ratings are typically given in kVA or MVA.


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Complex Power

So what is the difference between Apparent and Real Power.

It is the Reactive Power.

Reactive power is the component of power that oscillates back and forththrough the lines, being exchanged between electric and magnetic fields

and not getting dissipated.

It is denoted by the symbol Q, and its magnitude is given by:

Q= IrmsVrms sin


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Complex Power

Again, note how the equation converges for the resistive case.

When =0, then sin=0, and there will be no reactive power at all.

Reactive power is measured in VAR, for Volt-ampere Reactive.

We can represent power as a vector in the complex plane.


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Complex Power

The complex power S, with real power P in the real and reactive power Q in the

imaginary direction.


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Complex Power

The angle is the same as the phase difference between voltage andcurrent.

The projection of the apparent power vector onto the real axis has length

P and corresponds to the real power.

The projection of apparent power onto the imaginary axis has length Q

and corresponds to reactive power.

This agrees with the factors of cos and sin in the formulas for P and Q,respectively.


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Complex Power

In mathematical terms, S is the vector sum of P and Q.

In this sense, it is completely analogous to the complex impedance Z,

which is composed of the resistance R in the real and the reactance X in

the imaginary direction.


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Complex Power

Note that when Q and S are pointing upward in Figur, is positive and thepower factor is said to be lagging

It is like the current lagging behind the voltage.

For a leading power factor, would be negative, and Q and S would pointdownward.


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Complex Power

Example: Consider a vacuum cleaner that draws 750W of real power, at a

voltage of 120V a.c. and a power factor of 0.75 lagging. How much current

does it draw?

Since the real power is given by the apparent power times the power

factor, the apparent power equals 750 / 0.75 = 1000 VA = 1 kVA.

The rms current is the apparent power divided by the rms voltage: 1000

VA / 120 V = 8.33 A.


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Complex Power

When we say that a load draws power, it means that whenpresented with a given voltage, a certain amount of current willflow through this device.

Accordingly a certain amount of power will be dissipated or

exchanged.

Just as a load draws real power in relation to its resistance, it drawsreactive power in relation to its reactance.

In fact, the ratio of resistance to reactance determines the ratio ofreal to reactive power drawn by a load.

In other words, the angle in Z is the same as the angle f in S.


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Complex Power

Specifically, inductive loads are said to consume reactive power,whereas capacitive loads are said to supply reactive power.

This is merely a terminological convention, and a rather misleadingone.

Recall that inductors and capacitors produce opposite phase shifts.

Either type of shift causes reactive power to oscillate through thecircuit.

But because of the difference in timing, the contributions ofinductance and capacitance to reactive power are opposite.


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Complex Power

At the instant that the inductor magnetic field absorbs energy, the

capacitor electric field in the same circuit releases energy.

Conversely, at the instant that the magnetic field releases energy, the

electric field absorbs it.

Although on average neither inductor nor capacitor gains or loses energy,

their effects are complementary.

Following the law of energy conservation, the amount of energy going intothe circuit must equal the energy coming out of the circuit at every

instant.


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Complex Power

In principle, therefore, inductance and capacitance in a circuit must

always be matched.

A circuit will behave in such a way as to provide equal absorption and

release of reactive power at any instant.

The preferable way to satisfy the reactive power balance is by adjusting

the a.c. power source to compensate for the loads circulation of reactive

power.

Thus, in operational terms, the problem of managing reactive power is

analogous to that of managing real power.


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Complex Power

Just like the utility must supply the precise amount of real power that is

demanded at any instant, the utility must compensate for the precise

amount of reactive power that is being circulated at any instant.

In practice, electric loads are dominated by inductance, not capacitance.

Utilities therefore associate supplying real power with compensating for

inductive reactance, a lagging current.

This operational perspective explains the use of the physically improperterminology of consuming and supplying reactive power.

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