power engineering full class notes

8/10/2019 Power Engineering Full Class Notes

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ECSE - 361 Prof. F.D. Galiana 5

Text Book

A.E. Fitzgerald, C. Kingsley, S.D.Umans, 6 thEd., Electric Machinery, McGraw Hill


Main Contents of CourseElements of power systems (test 1);

Magnetic devices and circuits, Amperes law,electromagnetic inductance, Faradays law,energy and co-energy (test 2);

Force & torque in electro-mechanical systems,steady torque in rotating machines, fielddistributions, rotating magnetic fields,characteristics of various machine types (test 3);

Basic synchronous, induction and dc machines(test 4).


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Objectives of Course

To introduce the field of power engineering; To unify the knowledge acquired in the

fields of electric circuits, electro-magnetics,mechanics, control systems, and economics;

To introduce student to interdisciplinarythinking and problem-solving.


Sub-disciplines of PowerEngineering

Energy conversion (304-361, 462); Power electronics (304-565); Power systems (304-464, 563).


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Energy Conversion

Electro-mechanical; conversion ofmechanical to electrical energy and vice-versa,

Direct conversion; e.g. solar cells, Energy storage (one of the most

fundamental engineering problems still to be solved in an efficient, large-scalemanner).


Electro-mechanical EnergyConversion

Generation: Primary fuel: moving water, nuclear,hydro-carbons, wind, tides, solar rays, oceantemperature gradients; Intermediate fuel: high

pressure steam; Rotating turbine interacts with electro-magnetic

field in generator air gap and produces anelectrical voltage; current and power then flowthrough an electric circuit.


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Electro-mechanical System

Primary/SecondaryFuel

Boiler-turbineor Hydro-turbine

Pmech = Mechanical power of rotating mass,Pelec = Electrical power of 3-phase AC

voltages and currents.

Pmech

Generator Pelec


Pmech

Generator Pelec

Pmech

Motor P

elec

Generator Mode

Motor Mode

Input

Output

Output

Input

An electro-mechanical energy conversion devicecan be used as either a motor or as a generator.


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Not all electro-mechanical devices require rotational motion orthree-phase power.

There exist many sorts of electro-mechanical relays or linear displacement devices. Magnetically levitated electric trains isone dramatic example of a large object where the mechanical

power is obtained from a linear force rather than a torque.

Similarly, many small electro-mechanical devices operate on asingle, rather that on three-phases. However, as we shall see,3-phase AC power becomes essential when large amountsof power are generated, consumed or transmitted.


Comparative Power Levels

Light bulb 100W, Hair dryer 1 kW, Household breaker box 20 kW, 200 A,

Small town 100 MW, Montreal 15,000 MW=15 GW Quebec 30,000 MW Canada 100,000 MW Check this


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Rates of Electric Energy

Quebec about 5 c/kWh, Ontario about 8 c/kWh, N.Y. about 20 c/kWh, California unknown wholesale rates,

market forces decide.


Costs of Power System Elements

Generation from 1000 to 4000 $/kW, E.g., 30 GW HQ system costs 60 B$, HV Transmission from 50 to 1000 k$/km,

E.g., a 1000 km, 735 kV line from JamesBay to Montreal costs about 1 B$. Thereexist 6 such lines.


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Revenue of Power Utility

E.g., 30 GW capacity at 60% load factorcollecting 5 c/kWh yields a revenue of (30GW)*(0.6)*(5 c/kWh)*(8760 h/y)=7.88 B$/y,

Large revenues are needed to pay for largeinvestments of about 70 B$ plus maintenance.

Fuel is free in hydro systems but otherwise is amajor cost component.


Peak Load and Load Factor Pd (MW)

Time of day (h)

peak d P

aved

_ ave

d peak

d

P Load Factor

P =


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Comments on Load Factor

It is desirable for the load factor of a utility orgenerator to be as high as possible. A low loadfactor means that expensive generation has to beinstalled and used sparingly, only during the short

periods of peak demand. Of course, the shape ofthe demand curve depends on uncontrollableconsumption habits and on weather effects. Oneway to control the load factor is by introducingtime-of-day rates that stimulate consumptionoutside periods of peak demand.


Role of Power Electronics(Profs. Ooi and Joos)

Power supplies that convert (a) AC to AC atdifferent frequencies; (b) AC to DC; (c) DC toAC; (d) DC to DC at different voltages.

High power or voltage devices that control powerflows or voltage magnitude levels in transmissionnetworks.

Devices that control the speed of motors or the position of robotic actuators.


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Electric Power Systems A system consisting of energy storage or conversion facilities,

power electronics, transmission and distribution systems, andenergy utilization devices or loads,

These basic systems must be protected and operated to minimizethe impact of failures (short or open circuits, lightning, humanerror, failure of sub-components) and attain a very highminimum level of reliability (e.g. 3 hours/year),

To achieve this aim requires, on-line sensors, data collection, processing and transmission, distributed and centralized control,detailed forecasting and planning. Economic considerations are

paramount after meeting the minimum reliability norms. The costof a major power outage to industry and society is so high that itsoccurrence is practically unacceptable.


Examples of Power Systems Orbiting space station, consisting of solar

panels, storage batteries, power electronicconversion, protection equipment,distribution, controls, and loads.

Uninterrupted power supplies in a hospitalor computer facility. Interconnected power utility.


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Interconnected power utilities sell or buy power among each other.

A

DC

B

G

G

G

G

150 MW

1005030

75


Monopolies and Electricity Markets Traditional power utilities were monopolies, both owning

and operating all generation, transmission and distributionfacilities.

Recent legislation allows for competition in the electricityindustry, primarily at the wholesale level, in bothgeneration and distribution.

Producers and consumers bid to sell or buy in an electricityexchange, the price being set by the highest accepted bid. Competing generation companies can own but cannot

operate transmission and distribution networks, whichmust be open to all competitors.


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Voltage Types and Levels inPower Systems

Mostly 3-phase AC transmission; Some DC, usuallyfor long overhead distances or for underwatercrossings.

AC power is easy to convert through transformersfrom low (generation and distribution) to high voltages(transmission).

AC is generally cheaper. Typical standard phase-to-phase AC voltages: 25, 40,

69, 138, 230, 345, 400, 500, 735, 765, 1100 kV.


Why HV Transmission?

The higher the voltage, the lower the current. Smaller, cheaper conductors can then be used. Transmission losses which vary as the current

magnitude squared are significantly reduced. Doubling the voltage, quadruples the power that

can be transmitted through a transmission line.


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Why 3-phase AC?

Although, the instantaneous power carried by

each phase is pulsating, the total 3-phase power transmitted is constant over time.Generating and consuming devices thereforeoperate at a steady, non-pulsating power level.

Under balanced conditions, the sum of the 3- phase currents is zero. No return wires are

therefore needed.

Review of Time Domain Sinusoids and Phasors

Power systems operate primarily in sinusoidal steady state. This meansthat all currents and voltages inside a power network are time varyingsinusoids with a constant frequency (50 or 60 Hz), magnitude and phaseangle.

Let ( )v t be a standard sinusoid in the time domain defined by,

( ) 2 sin( )v t V t = + (1.1)

where is the phase angle in radians, is the frequency in radians persecond, t is time in seconds, and V is the root mean square (rms) of the

sinusoid, defined as,

2 2

0

1( ) ;

2

T

t

V v t dt T

T

=

=

=

(1.2)

Associated with the sinusoid, ( )v t , we define the phasor, jV Ve = , that is,

2 sin( ) jV t Ve + (1.3)


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Examples of Phasors

(1) 2(100)sin 100 0 100100

(2) 100sin( / 3) / 32

100 100(3) 100sin( / 3) / 3 2 / 3

2 2

(4) 2(100) cos 2(100)sin( / 2 )

2(100)sin( / 2)

2(100) sin( / 2)

100 / 2

t

t

t

t t

t

t

=

+

+ =

=

=

= +

Phasor Diagrams

Consider two phasors 0; ( )V V I I = = . These can be depictedgraphically by,

V

I


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Example of phasor diagram,

0V V =

( / 2) I I =

/ 2

Advantages of Phasors RLC circuits driven by sinusoidal sources

can be analyzed in steady-state through a setof algebraic instead of differentialequations;

The calculation of power injections, flowsand losses is rendered considerably simpler.


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------KVL

Capacitor, C

------KCL

Inductor, L

Resistor, R

AdmittanceImpedanceComplexdomainrelation

Timedomainrelation

( ) ( )v t Ri t =

V RI = Z R=

1/

Y G

R

=

=

( )( )

di t v t L

dt = V j LI =

/ Z V I = /Y I V =

Z j L = /Y j L =

/ Z j C = Y j C = I

V j C

=( )( ) dv t

i t C dt

=

( ) 0imesh

v t = 0k mesh

V =

( ) 0k node

i t = 0k node

I =

Time vs Phasor Domain Behaviour of RLC Circuits

Properties of impedances and admittances:

(1) Impedances in series add, that is,1

n

ser k k

Z Z =

= ;

(2) Admittances in parallel add, that is,1

n

par k k

Y Y =

= ;

(3) The effect of an inductor ( j L ) can be neutralized by adding a capacitor in series so that,

0 j j LC

= , or by adding a capacitor in parallel

so that, 0 j

j C L

= .


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Lecture 3

RLC Circuit and instantaneous real/reactive powerSep. 8 th, 2003

RLC Circuits

RLC circuits driven by sinusoidal sources can be analyzed in steady-state through a set ofalgebraic instead of differential equations;

v(t)R, L, C

Transformer

( ) ( ) ( ) p t v t i t =

where ( ) p t is the total instantaneous power consumed.

1 1 2 2( ) ( ) ( ) ( ) ( ) ... p t v t i t v t i t = + +

1

( ) ( )n

k k k

v t i t =

= .

1( )v t

2( )v t

3( )v t

4()v t

( )nv t

1( )i t

2 ( )i t

3( )i t

4 ( )i t

( )ni t


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cosVI

Consuming

Generatingt

p(t)

INSTANTANEOUS, REAL AND REACTIVE POWER

The instantaneous power consumed by a load with voltage v(t) and current i(t) ,

( ) ( ) ( ) p t v t i t = (1.1)

If the load is R,L,C and the input voltage is sinusoidal with

( ) 2 sinv t V t = (1.2)

then, the output current is also sinusoidal,

( ) 2 sin( )i t I t = (1.3)

In terms of phasors, these two sinusoids become,

0V V = (1.4)and,

I I = (1.5)

For sinusoidal inputs the instantaneous power can be expressed as,

( )( ) 2 sin sin( )

2 cos( ) cos(2 ) / 2

cos( ) cos(2 )

p t VI t t

VI t

VI VI t

= = =

(1.6)

P = average power per cycle =0

1( ) ( )

T p t d t

T ;Also P = real power = active power.


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We now define the average power consumed per cycle ( 2 /T = ) as P, where from(1.6),

cos( ) P VI = (1.7)

The average power, P , is also called the active or real power. Note also that, moregenerally, the phase angle should be interpreted as,

V I = (1.8)

The instantaneous power can also be expressed as,

{ }

{ }

( ) cos(2 )

cos2 cos sin 2 sin

1 cos2 sin 2

p t P VI t

P VI t t

P t Q t

= = +

=

(1.9)

where we define the reactive or imaginary power as,

sinQ VI = (1.10)

The reactive power Q can be interpreted as the peak value of the component of theinstantaneous power that on the average does zero useful work. On the other hand, thereal power P , is the average power of the component of the instantaneous power thatalways produces non-negative useful work.

There are practical reasons for studying the notions of real and reactive power. Real power, measured in watts, defines real work done and is assigned a price by theregulatory bodies which is passed on to the general consumer. Reactive power, measuredin volt-amperes reactive or vars, is a measure of the reactive current drawn by the device,

sin I . Although reactive power does no useful work, the reactive current drawnrequires wires to flow through and can certainly create losses.

To compute real and reactive power, we first define the notion of complex power,

S P jQ= + (1.11)

We can show in general that,

*

*

cos( ) sin( )

( )( ) j jS P jQ VI jVI

VIe V Ie

VI

= + = += ==

(1.12)


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IMPEDANCE OR ADMITTANCE LOADS

If the load behaves as an impedance, then,

V ZI = (1.14)

where Z R jX = + . One can also express this relation in terms of the admittance

Y G jB= + so that, I YV = (1.15)

In this case, it readily follows that,2 2

2 2

P I R V G

Q I X V B

= == =

(1.16)

E.g. if Y j C = as for a capacitor, then,

2 2

0

1 1( ) ( )

P

Q I V C C

=

= = (1.17)

Prove that if V V = , a constant voltage source, that the identity (1.17) is true.

Capacitive loads generate vars, while inductive loads consume vars. From the aboveexample, we see that a capacitor consumes a negative amount of vars, in other words, itgenerates vars.


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Power Factor CompensationElectric loads, whether pure admittances, PVpf loads, or PQV loads, may have powerfactors that differ from 1. A unity power factor is however desirable for a number ofreasons. One is that at a power factor of one, the device consumes only real power, theimaginary power being zero. The other is that with pf=1 the total current drawn by the

device is there uniquely for the purpose of doing useful work. A device that consumes (orgenerates) reactive power requires a current flow, current that the supplying utility mustsend through its transmission system wires. Even a device that consumes zero watts but anon-zero amount of vars requires some current. This purely reactive current flow,although resulting in zero watt load consumption, nonetheless produces some real lossesin the transmission lines leading to the load, in addition to causing network voltagestability problems. The transmission utility therefore has to supply the resultingtransmission losses and must provide voltage stability resources, all of which cost money.Consequently, large electricity consumers who use up or generate vars are charged a ratefor both reactive and real power, one in c/kVAr-h and the other in c/kWh. It is thereforeeconomically important for consumers to adjust their loads to reduce their var

requirements as much as possible, or equivalently to bring their power factor close tounity.

Power Factor Compensation in Admittance or Impedance LoadsIf the load behaves as an impedance with Z R jX = + , then the consumer can add acompensating device in series whose impedance is c c Z jX = . The combined impedanceis then ( )c c Z Z R j X X + = + + , which if c X X = , becomes purely resistive andtherefore has unity power factor.

The phasor diagram of the above load compensated to obtain a pf=1, could be denoted by,

Z R jX = + c c Z jX =

V

+

I

'V

+

I V

'V

c jX I


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Note that in the above phasor diagram, the angle between the voltage, V , and the current, I , is zero, which corresponds to a pf of unity. Note also that an alternative phasordiagram when the original load is inductive is,

If the desired pf is not unity, but something less than unity, then the phasor diagramwould be as shown below, first for a case of final lagging power factor, then, for the caseof a leading power factor. In both cases, the original power factor is assumed to beleading.

To calculate the value of the level of series compensation, c X , needed to achieve a non-

unity power factor, recall that in general for a load with Z R jX = + or with Y G jB= + ,the power factor can be shown to be given by,

2 2 2 2

R G pf

X R B G

= =

+ +

(1.1)

For the series compensated load, therefore,

2 2( )c

R pf

X X R=

+ + (1.2)

I

V

'V

c jX I

I V

'V

c jX I

I V

'V c jX I


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or a given desired value of pf , therefore we can solve (1.2) for c X ,21

c

R pf X X

pf

= (1.3)

In (1.3), one solution yields the leading pf, while the other yields the lagging pf.

If the load is an admittance with Y G jB= + , then the consumer can add a compensatingdevice in parallel with admittance c cY jB= . The resulting admittance is then

( )c cY Y G j B B+ = + + , which if c B B= , becomes purely resistive and therefore hasunity power factor.

In both of the above alternatives, if a load is capacitive then the compensating elementmust be inductive and vice-versa. A similar analysis to that developed above for theseries compensation case applies to when the desired power factor is less than unity.

Power Factor Compensation in PVpf or PQV LoadsIf the device is a PQV load, then its real and reactive power consumptions, P and Q, areknown constants. To carry out pf-compensation, the consumer can install a compensatingdevice in parallel or in series whose var consumption plus that of the load leads to acombined var level that corresponds to the desired power factor. Thus, for a given

cos pf = , with either series or shunt compensation, we must satisfy the condition,

sintan

coscQ Q

P

+= = (1.4)

This means that the compensating element has to consume cQ vars where from (1.4),

tancQ P Q = (1.5)

Moreover, since the total complex power consumed with compensation is*( )cS P j Q Q V I = + + = , the following is also true,

Y G jB= +

c cY jB= V +

I

c I

' I


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2 2( )cS S VI P Q Q= = = + + (1.6)

Let the load be compensated by a parallel element,

The compensating device placed in parallel has an admittance of c cY jB= . Then, the

complex power consumed by this device is** 2( )c c c cS V I V V jB jV B = = = . Thus, the

compensating device consumes 2c cQ V B= vars, where the input line voltage magnitude,V , is known. From (1.5), we then have,

2tanc cQ P Q V B = = (1.7)

which defines the required compensating susceptance, c B ,

2tanc Q P B V

= (1.8)

If the device is inductive, then Q will be positive and, generally, so will c B . This meansthat the compensating element must be a capacitor. The opposite is true when the deviceis capacitive.

For series compensation,

Here, equation (1.5) still applies, but now the power consumed by the compensatingdevice is given by, 2c cQ I X = . We then have to find the current magnitude I in order to

S P jQ= +

c cY jB= V

+

I

c I

S G jB= + c c Z jX =

V

+

I

'V

+


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be able to find the required series compensating reactance, c X . This current is obtainedfrom (1.6),

2 2( )cP Q Q I V

+ += (1.9)

From (1.9), 2

2 2 2( )c c

cc

Q Q V X

I P Q Q= =

+ + (1.10)

If the device is a PVpf load, then the compensating procedure is identical to the onedescribed above for a PQV load, with one variation. This being that the reactive powerconsumption of the PVpf load must be first found from sinQ VI = , where the current

magnitude I is found according to,

( )

P I

V pf

= , where the power factor is defined by

cos pf = . Recall also that0

0

if pf is lagging

if pf is leading

>< .

Finally, it must be noted that the installation of a compensating device or devices at theterminals of a large load (e.g. a manufacturing plant) is costly and this cost must beweighed against the savings obtained by not paying the utility rate for reactive powerconsumption. As indicated above, a consumer may therefore choose to compensate theload only partially, that is, to aim for a pf near unity but not necessarily exactly one.

Sample problems Problem 1 : An impedance load with 1 2 Z j= + ohms is to be compensated to achieve aunity power factor with both series and parallel devices. Calculate the impedance of thesecompensating devices and indicate the reactive power consumed by them when the linevoltage is 1 volt, indicating whether the compensating device is inductive or capacitive.

Solution : For series compensation, add 2c Z j= . Since this compensating device has anegative reactance it is capacitive. Its reactive power consumption is 2 2 ( 2)c cQ I X I = = .

To find the current magnitude, I , we solve,1 0

1 0(1 2 ) ( 2 )c

V I

Z Z j j

= = = + + + . Note

that the current phasor I is in phase with the current, as required by the unity powerfactor requirement. Thus, 2cQ = vars, that is, it consumes 2 vars, or it generates 2vars, which is exactly what the load consumes. The parallel compensation is left to thestudent as an exercise.

Problem 2 : A PQV load with 2P = watts and 1Q = var is to be compensated to achievea power factor of 0.95 lagging using both series and parallel devices. Calculate theimpedance of these compensating devices and indicate the reactive power consumed by


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tan z pv c

z pv

Q Q Q

P P

+ +=

+ (1.11)

where 1tan tan(cos (0.99)) 0.1425 = = , the minus sign accounting for the leadingnature of the desired power factor. The real power consumed by the PVpf load is 1 pvP =

watt, while that consumed by the impedance load is, 2 2 22 1.65 z

P V G= = = watts.

Similarly, as shown above, 0.75 pvQ = vars, and 0.8 zQ = vars. From (1.11) then,

0.050.1425

2.6cQ+ = (1.12)

or 0.4205cQ = vars. Note that to make the overall power factor leading, it is necessaryfor the compensating device to generate 0.4205 vars. Since for a shunt device,

2c cQ V B= , it follows that 0.4205/ 4 0.1051c B = = S.

The series compensation is considerably more difficult since the power consumed by theimpedance load depends on the voltage across it and this voltage is itself dependent onthe compensation level. In the interest of brevity, we leave this exercise to thoseinterested in a good mathematical challenge.


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Lecture 5

Power factor compensation (Lectures 2002 in the webpage). Power system dynamics.

Load

I+

-

Z R jX = +cos( )

cos

Power Factor pf V I

=

=

V

I The goal of pf compensation is to modify the load, so that the compensated load

has a new pf ;' 1 pf ;

We can have shunt or series compensation.

Z R jX = +c c Z jX =

+

-

+

-

V Series


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( )

cos sin

V V V Z V I

I I I V V

j I I R jX

= = = =

= +

= +

V R Z pf

I = = another way to write pf.

As an exercise : ShowG

pf Y

= ;

Similarly, if the load consumes a given P & Q.

P pf S

= ;

With a series compensation,

Where :2 2( )c Z R X X = + +

Z R jX = +c c Z jX =

+

-

+

-

V

' ( )

''

c Z R j X X

R pf

Z

= + +

=

Y G jB= +

c cY jB=

+

-

V

parallel

I


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New pf : 2 2'

( )c

c

R pf

R X X

We need to make X X

=+ +

=

Example:

jXc= -2j

1+2jV

More generally, pf = 0.95 ;

So in that case we have to solve the equation:

2 2'

( )c

R pf

R X X =

+ +;

And the other form of compensation :

Y G jB= +

25c

Y j=

+

-

V 1 1 2

1 2 5 j

Y j

= =

+

I


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Power System Dynamics :

Consider the rotating mass of a large generator.

Wm

Pe(t)Pm(t) Turbine

The rotating mass has a large amount of K.E. :

212 m

KE I W = i ;

In steady state ( ) ( )m e P t P t = ;

However, if there is an imbalance ( ) ( )m e P t P t ;And this, it takes while to be corrected.

( )e P t - Fast changing in milliseconds to seconds.( )m P t - Slow changing in seconds to minutes.

So if ( )e P t changes instantaneously, ( )m P t cannot react.

The energy comes from the rotating mass. i.e. if ( ) ( )e m P t P t > , the rotating mass slow

down i.e. ( )mW t decreases.

Dynamic Balances

( )( ) ( )m e

d Ke P t P t

dt = - Dynamic balance equation.

In steady state, Kinetic energy is constant.

( )0

d Kedt

= ; 212 m

Ke I W const = =i


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Power Transmission

Power is transmitted through transmission lines at different voltages. Usually, the longerthe distance over which the power must flow, the higher is the transmission voltage. Atthe local distribution level, the three-phase, phase-to-phase voltage employed ranges from25 to 40 kV. Of course, at the residential level the 3-phase voltage is transformed down to

120 volts, single phase. For very long distances and large power levels, 400, 500 and 735kV are standards followed in North America and Europe.

Generally, high voltage levels lower the current magnitude and therefore reduce thetransmission losses, increasing the overall efficiency of getting energy from the sourcesto the consumers. At the generating level, the voltage is relatively low, of the order of 10kV, given that insulation inside the closed space of a power plant is difficult for highervoltage levels. In addition, the flux densities inside the iron rotor and stator of thegenerating units are limited by the magnetic saturation properties of ferromagneticmaterials. This limitation then restricts the level of the voltages induced in the machinewindings.

To examine how power flows through a line consider a simple transmission line model,

The complex power sent (or injected at the sending end) is,

*

* * *

* * 2 *

s s

s r s r s s

s s s r s s r

S V I

V V V V V V

R jX R jX

V V V V V V V R jX R jX

=

= = + = =

(1.1)

If we denote the sending and receiving voltage phasors in polar form, s js sV V e = and

r j

r r V V e =

, then, if the sending and receiving voltage magnitudes are equal, s r V V V = =

,the complex power sent becomes,

R+jX

r jr r V V e

+=

s j

s sV V e

+=

I I


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Problem 5 : Prove that, if the line is lossless, the maximum real power that can betransmitted is,

2max V P

X = (1.8)

Thus, the maximum real power that can be transmitted across a line with reactance X is proportional to the square of the voltage . This is one of the fundamental reasons for usinghigh voltages to transmit power over long distances.

Problem 6 : Prove that at the condition of maximum power transfer, the angle between thesending and receiving voltages is 90 degrees.

Problem 7 : It is undesirable to operate a transmission line close to its maximum powertransfer capability as this can result in unstable operation possibly, that is, large powerswings that can result in the disconnection of some generators or loads, in other words, a

so-called black-out. Generally, lines are operated at angle differences of less thanmax 30 = degrees from sending to receiving end. Suppose then, that, for reasons of

stability, the angle difference should be less than max , prove that the power transfer islimited by,

2maxsin

V P

X (1.9)

Problem 8 : Explain which design alternative would result in a higher power transfercapability: Doubling the transmission voltage magnitude or doubling the number of lines.

Problem 9 : Find the maximum real power that can be transmitted through a lossy line.Determine the angle difference at the maximum power transfer condition.

Problem 10 : For a lossless line, find the reactive power sent and received at maximumreal power transfer. Find the power factor at the sending and receiving ends under thesame maximum transfer condition.


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Typically R < X, 20% R X . The equation for the sending complex power is

( )( )

* * **

2

2 2

s r s s s r ss s s

js s r

s

V V V V V V S V I V

R jX R jX

V V V e R jX S R X

= = =

+

+=

+

Substituting 2 2 j R jX R X e + = + , 1tan X

R

=

, then

( )2 22 2 2 2

j j j js s r s s r

s

V V V e e V e V V eS

R X R X

+ = =

+ +

Then the real power is { }Res sP S =

( ) ( )2

2 2 2 2cos coss s r s

V V V P

R X R X = +

+ +

And substituting ( )2 2

cos R

R X =

+, ( )

2 2sin

X

R X =

+, and using the trigonometric

identity ( ) ( ) ( ) ( ) ( )cos cos cos sin sin x y x y x y+ = , we can express the last equation as,

( ) ( )( )2

2 2 2 2 cos sins s r

sV R V V P R X

R X R X =

+ +

If we define the conductance and the negative of the susceptance as

2 2 2 2, R X

g b R X R X

= =+ +

respectively, then

( ) ( )( )2 cos sins s s r P gV V V g b =

The receiving power will be now less than the sending power because there are losses inthe line. P loss = P s P r > 0, i.e. some energy is lost to the atmosphere in the form of heat.

To calculate P r we take the real part of the receiving complex power as follows,

{ } { }*

*Re Re Re s r r r r r r

V V P S V I V

R jX

= = =

+


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Following a similar procedure as for the sending end one obtains,

( ) ( )( )2

2 2 2 2 cos sins r r

r

V V V RP R X

R X R X = + +

+ +

( ) ( )( )2 cos sinr r s r P gV V V g b = + +

Then the loss in the line is P loss = P s P r

( )( )2 2 2 cosloss s r s r P g V V V V = +

Note that if R = 0 then P loss = 0. Typically in a 735 kV line sending 2000 MW, lossescould be as much as 5% or 100 MW. To generate 100 MW of losses one needs to build a

power plant costing around 200 M$. To reduce losses normally they increase the crosssection of the transmission line, but that makes the line costly and heavier, requiring also

new supporting towers. Another method to reduce losses is to increase the voltage inorder to reduce the current.

Consider now the transmission line with shunt elements as illustrated in Figure 3 .

Figure 3 - Transmission line with shunt elements

In this equivalent Y 1 and Y 2 account for the capacitance of the line to ground and otherlines, and the resistance of the insulators and corona effect losses. The sending complex

power is then

( )**

1 0ss s s s s r S V I V V Y V V Y = = +

If Y1 = Y 2 = j B sh, i.e. a pure capacitance, then P s, P r , and P loss are exactly the same as thecase without shunt elements.The shunt capacitors tend to inject vars into the line. This might be problematic withlightly loaded lines (e.g. at night) because the injected vars tend to increase the linevoltage and this leads to voltage control problems. Generally an increase in Q causes anincrease in the voltage magnitude and a decrease in Q generates a decrease in voltagemagnitude.


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Lecture 8

Effect of power imbalance on system frequency

Consider the analogy of a mechanical system, where a car is traveling with a velocity v,there is mechanical power, and friction, as depicted in Figure 1 .

Figure 1 - Mechanical analogy

The kinetic energy of the car with mass m is

212e

k mv=

If P mech < P friction then the car slows down. Alternatively if we push the pedal then the carspeeds up, i.e. v decreases, as well as the kinetic energy.

A similar situation occurs in rotating machines. A generator is a rotating deviceconnected to a turbine with a moment of inertia I .

Figure 2 - Power Conversion

If P mech = P elec then the speed of rotation ( m) is constant and the electric frequency ( )as well.This is called the steady state mode of operation. Hence if P elec increases, analog toPfriction , then m decreases and the input of steam or water should be increased, but thisresponse is slow (couple of minutes).


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Therefore if P mech cannot keep up with the change on P elec there will be a frequency dropor increase (depending on the sign of the change).The kinetic energy of rotation is given by,

21

2e mk I =

Then the change ratio of the kinetic energy with respect to time is equal to the powermisbalance.

e mm mech elec

dk d I P P

dt dt

= =

Hence if dk e/dt = 0 then k e is constant ( 0ek ).The demand of power in MW versus time varies during the day as exemplified in Figure3

Figure 3 - Demand vs. Time

The mismatch between the electrical and the mechanical power causes a frequencyvariation over time.

Figure 4 - Frequency vs. Time


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Generally the frequency does not vary more than 0.5 Hz .

The inertia constant for a generator is defined as

ek H rated power

= (seconds)

Typically H = 3 s. Therefore the kinetic energy of a machine rated at 100 MVA is k e =300 MJ, i.e. 300, 000, 000 joules of energy stored in the rotating mass.

If there is an imbalance between P mech and P elec of -10 MW, then the kinetic energy willchange, dk e/dt = -10 MW. Integrating with respect to time at both sides yields

( )0 10e ek k t =

If we did nothing to correct the power imbalance, then in ten seconds the kinetic energywould have dropped to 200 MJ. This means the load is extracting energy from therotating mass instead of taking it from the input P mech .

The new frequency is calculated as follows.

( ) ( )

( ) ( )

2

2

121

0 2 602

e m

e

k t I t

k I

=

=

Assuming m = , since m then,

( )( )

( )( )

( ) ( )

2

2

2

0 2 60

2002 60

300

e

e

k t t

k

t

=

=

Therefore ( ) 60 2 / 3 48.98 f t = = . This change is not permissible since it is too big.

The power consumed has to be generated instantaneously. There are ways of storingenergy in small scale, e.g. in batteries, or even in large scale, e.g. dams, but the lattercannot be built everywhere. Hence a way to store energy is in rotating masses.

Power systems are provided with primary frequency regulation. Basically this is a closedloop control signal related to the frequency deviation from the normal standardfrequency.


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The problem with the primary frequency regulation is that the new frequency is stable butdifferent to 0. This problem is solved with the secondary frequency regulation. This typeof regulation acts in a time frame of 5 to 10 minutes to readjust ( )t back to 0. Itdoes this by modifying the set point 0mechP .


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( ) ( )

( ) ( )( ) ( )

2 sin

2 sin 120

2 sin 120

a

b

c

V t V t

V t V t

V t V t

=

=

= +

Similarly a balanced set of currents

120

120 240

a

b

c

I I

I I

I I I

=

=

= + =

Where a b ca b cV I V I V I = = = .

Figure 3 - Phase diagram with currents

Some important properties are:

1. 0a b c I I I + + = or in time domain, ( ) ( ) ( ) 0a b ci t i t i t + + = . Therefore the neutral orground current is i g = i a + i b + i c = 0. Hence there is no need for returning wires.

2. The 3 instantaneous power is ( ) ( ) ( ) ( )3 a b cP t P t P t P t

= + + = constant.

Where ( ) ( ) ( ) ( )( ) ( )1 cos 2 sin 2a a aP t V t i t P t Q t = = , and cos , sinP VI Q VI = = .


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0 0.01 0.02 0.03 0.04 0.05 0.06-2000

-1000

0

1000

2000

3000

4000

5000

6000

t

P a

( t )

Figure 4 - Instantaneous power of phase a

Therefore individual phases deliver pulsating instantaneous power. P b(t) and P c(t) are

similar in form but displaced by -120 and 120 degrees respectively. (Check as anexercise).

( )3 3 3 cosP t P VI = =

And it is independent of time.

0 0.01 0.02 0.03 0.04 0.05 0.062437

2437.5

2438

2438.5

2439

2439.5

t

P 3 p

h a s e

Figure 5 - P 3 with respect to time

It is important that P 3 be constant because, e.g. if we take a 3 motor, we want it to haveconstant torque so that it works in a steady form.

Since each phase still carries a current or power component characterized by Q that doesno useful work then vars are relevant. Therefore we still define vars under 3 systems.


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( )3 3 3 sinQ t Q VI = =

Note that V is the rms value of the phase to ground voltages, and I is the rms value of the phase or line current.

However, in 3 systems when we speak of voltage magnitude (or rms voltage) we usuallymean phase to phase voltage, and not phase to ground. For example if in the Power

jargon we say A three phase 735 kV line, that means there are 735kV rms line to linevoltage. So in the diagram of Figure 1 and Figure 2 V a, V b, and V c are referred toground. The line to line voltages are given by,

ab a bV V V =

Where 3ab aV V =

The reason why in the power industry they rate devices considering the line to linevoltage is because it is higher than the line to ground voltage. Therefore if 3V V =

then

3

3

3 cos 3 cos

3 sin 3 sin

P VI V I

Q VI V I

= =

= =

Similarly in a 3 system when we refer to a real and a reactive power consumed or

generated we mean 3 power. For example,

A 3 balanced load operates at a voltage of 1000 V rms and consumes 9 kW at a lagging power factor of 0.8. Therefore

( ) ( )3 9 3 1000 0.8P kW I = =

Solving for I,

( )

90006.49

0.8 3 1000

I = =

This is the current magnitude (rms) that flows through each wire.And ( )1cos 0.8 36.86 = = , then

( )( ) ( )3 3 1000 6.49 sin 36.86 6750Q Vars = =


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Lecture 10

3 loads. Per unit system (p.u.)

Consider the three phase loads of Figure 1 and Figure 2 .

Figure 1 - Wye (Y) load

Figure 2 - Delta ( ) load

A load is balanced if the input voltages are balanced and the currents are also balanced.Alternatively if we apply balanced currents then the voltages are balanced.Mathematically, a balanced set of voltages are,


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2

1a

babc

c

V

V V a V

V a

= =

Where 120 ja e

= .

Figure 3 - Phasor diagram

Therefore a balanced set of currents would be of the form

2

1a

babc

c

I

I I a I

I a

= =

Some useful identities for a are:

3

2 1

2

1

1 0

aa a

a a

==

+ + =

To show that the load is balanced we apply balanced voltages and calculate thecurrents.


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( ) ( )( ) ( )( ) ( )

2

2

1

1

ab a b

ac a c

bc b c

I V V Y a V Y

I V V Y a V Y

I V V Y a a VY

= =

= =

= =

Then from Kirchhoffs current law (KCL),

( ) ( )22 2

1 1

2 3 1

3

a ab ac I I I a a V Y

a a V Y a a V Y

V Y

= + = = =

=

Similarly (check as an exercise),

23

3

b

c

I a VY

I a V Y

= =

Thus

2

1

3abc I a V Y

a

=

To prove that a Y load is balance it is easier to apply balanced currents and verify thatthe voltages are balanced. The current through Y g will be zero since the currents are

balance and they add up to zero. Therefore the voltages will be

2

1a Y a

b Y Y babc

c Y c

V I Y

V V I Y a IY

V I Y a

= = =

If the 3 load is very complex it can be represented by an admittance matrix.


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Figure 4 - Load admittance matrix

abc abc abc I Y V =

The load will be balanced if

2 2

1 1

abcY a V a I

a a

=

Per Unit System

Consider the following statements as examples.

1. Motor A operates at 1 kW while motor B operates at 900 kW.

2. Device A operates at 1 kV while device B operates at 2 V.

It is difficult to compare their mode of operation because such statements do not tell uswhether one device is more overloaded than another.

One must also add a statement concerning the relative loading with respect to a base or anorm or a rated level.

The p.u. defines all variables (voltages, currents, powers, impedances, and admittances)not just in S.I. units but with respect to a rated level.

For the first example given, it is better to state the following: Motor A is rated at 950 VAand operates at 1.05 pu. This latter value tells you not only at what power it is operating


51/155

but also that it is overloaded. The motor hence operates at P = 1.05 * P base = 1.05 (950) =1000 W. And the motor is operating at a P 5% above the rated or normal level.

For the second example, a better statement is: Motor B is rated at 1000 kVA and operatesat a power of 0.9 pu. Then P = 0.9 (1000k) = 900 kW it operates at 90% of its rated value.

Power devices are assigned two rated values: the base power (S B) in VA and the basevoltage (V B) in Volts.

The base current is defined as B B B

S I

V

= , and the impedance base is defined

as2

B B B

B B

V V Z

I S = = .

Then since V I Z = is in SI units, to put in per unit each quantity must be referred to its

base.

pu

B

pu

B

pu

B

V V

V

I I

I

Z Z

Z

=

=

=

So that Ohms law is stated now as pu pu puV I Z = .


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Similarly,

2

_ sin pu

transmitted pu pu

V P

X =

Where is in radians.

Conversion from one pu base to another.

Suppose you are given two devices with impedance Y 1 and Y 2.

Figure 1 - Change of base

Y1 is known in pu on a base V B1 and S B1 , Y2 is known in pu on a base V B2 and S B2 . SinceY1 and Y 2 are given in different bases it is not possible to solve the circuit in a straight

forward fashion. To be able to solve this circuit, put both admittances on a common base,lets say V B1 and S B1, then to change Y 2 into the same base,

2_ 2 _

2

SI pu

B

Y Y

Y =

Where SI stand for international units and 22 22

B B

B

S Y

V =

( )22 _ 22 _

2

SI B pu

B

Y V Y S =

( )2 _ 22 _ 2

2

pu BSI

B

Y S Y

V =

So the new pu value will be


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( ) 22 _ _ 2 _ 2 12 _ _ 2

1 2 1

pu old SI B B pu new

B B B

Y S V Y Y

Y V S = =

More generally, if X is any variable in pu on a base X B1 , to change to a new base X B2 ,

2 1

1 _ _

2 B B

B pu X pu X

B

X X X

X =

Normally we do not write X pu , instead use only X, where it is understood that thequantities are on a common base in pu.

Effect of p.u. on transformers

Consider the ideal transformer of Figure 2 .

Figure 2 - Ideal transformer

The voltage and current relations are

11 1 2

22 2 1

, N N V I N N V I

= =

To put everything in pu, since the voltage and current are different in each side, we needto modify our definition of base quantities. So the primary side will have a base V B1 , S B and the secondary side will have a base V B2 , SB. Note that the power base is the same,since power is transferred unaltered.

* *1 21 2S V I V I = =


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Putting the transformer relations in pu,

1

1

2

2

1 2

2 1

1_ 1 2

2 _ 2 1

B

B

V V BV

BV

pu B

pu B

N V N V

N V V N V V

=

=

Which is the turn ratio in pu.

For the current,

1_ 2 1

2 _ 1 2

pu B

pu B

N I I N I I

=

If we let 1 12 2

B

B

N V N V

= by selecting the base voltages, then the per unit turns ratio is 1, which

means that

1_ 2 _

1_ 2 _

pu pu

pu pu

V V

I I

=

=

Figure 3 - Transformer in pu

Therefore under this condition transformers are left out of the circuit.

Example. Consider the circuit of Figure 4 .


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Where

11

22

33

B B

B

B B

B

B B

B

S I

V S

I V

S I

V

=

=

=

To find the complex power consumed by the load,

( )* 2

2 _ 2 _ 2 _

2 2 _

pu pu pu pu pu pu

pu B

S I Z I I Z S S S

= ==

Another advantage is that the impedances in pu (Z s_pu , Z1_pu , Z2_pu ) have similarmagnitudes, whereas in SI units the magnitudes are much different.


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Lecture 12

Per unit in 3 systems

In 1 systems we defined 2 base quantities for each 1 device: V B, S B. For example, I =1.1 pu implies that it is 10% above the rated current. In addition we have seen that thetransformers can be replaced by solid wires yielding the following relations,

1_ 2 _

1_ 2 _

pu pu

pu pu

V V

I I

=

=

Hence the analysis is simpler without the transformers. The rationale behind thedefinition of the base quantities was to maintain KCL, KVL, and Ohms laws invariant tothe new set of functions. Therefore if V I Z = in SI units and pu pu puV I Z = .

In 3 systems the same rationale is applied. Consider the 3 power consumed by a 3 balanced load. We have seen that 3 3 cosP VI = .

Figure 1 - 3f load

Recall that,0, 120, 120, , 120, 120a b ca b cV V V V V V I I I I I I = = = = = = + .

Note that V is the rms value of the phase to ground voltage. Be aware that in 3 systemsvoltages are rated phase to phase 3V V = .

Example. A 3 motor is rated at 1000 V implies that V = 1000V, therefore1000 / 3V = Volts.

Example. A 3 motor is rated at 3000 KVA implies that the normal 3 power is 3000KVA, but the single phase power is 3000/3 KVA = 1000 KVA.


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But if using per unit system, this factors (3 and 3 ) are not present in the equations. The per unit system for 3 devices tries to render all relations derived from KVL, KCL andOhms laws invariant with respect to the 1 relations.

In a 3 device the two base quantities are: V B is the phase to phase base voltage and S B is the 3 base power.

If we consider 3 3 cosP VI = in SI units, then to put it in pu we have to divided the 3 power by the base power,

33 _

3 cos pu

B B

P VI P

S S

= =

To define a per unit voltage in 3 , since V B is to , the voltage that appears in theequation must also be phase to phase / 3V V

= .

3 _

3 _ _

3 cos 3 cos

3

3 cos

B pu

B B B

B pu pu

B

V I V I V P

S V S

V P V I

S

= =

=

Where _ pu B

V V

V

= . So if we define3

B B

B

S I

V

= then

3 _ _ cos pu pu puP V I =

Dropping the pu and subscripts,

cosP VI =

Usually, when dealing with 3 systems it is understood that V is phase to phase and P is3.

To calculate the base impedance consider the definition for the 1 system,

( )2

2 _

_1

3

3

B to g B B

B B

V V Z

S S

= =


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And this is equivalent to2 _

_1

B to g B

B

V Z

S

= . Hence the definition is the same either for the 1

system or 3 system taking into account the respective bases.

Example. A 1kV, 3 MVA 3

load operates at rated voltage and consumes 0.95 real powerat a leading power factor of 0.8.

Then V B = 1kV, S B = 3MVA, using 3 bases( )21

0.333 B

kV Z

MVA= = . Or using 1 bases

1 / 3, 1 B BV kV S MVA= = , then( )

21 / 3

0.331 B

kV Z

MVA= = .

In pu cosP VI = , since we operate at rated voltage V = 1pu, P = 0.95 pu, and

( )( )0.95

1.18751 0.8 I pu pu= =

Then ( )( )( )sin 1 1.1875 0.6 0.7125Q VI pu = = = . Note that the minus sign is becausethe pf is leading, i.e. the load generates vars.

To put the results in SI units,

( )( )

( )

0.95 0.95 3

0.7125 0.7125 3

31.1875 1.1875 2.056

3 1

B

B

B

P S MW

Q S MVar

MVA I I kA

kV

= =

= =

= = =

Consider now the power transfer equation for Figure 2 .

Figure 2 - Transmission Line


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Considering V s = V r = V, then2

sinV

P X

= , where V is phase to ground. For 3 , if one

phase carries2

sinV

P X

= , then the 3 carry2

3 3 sinV

P X

= , and in per unit (show as an

exercise)2

_ 3 _ sin pu pu

pu

V P X

= , which is the same relation as for the 1 . Removing again

the subscripts,

2

sinV

P X

= in pu, 3 .


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Lecture 13

Introduction to electromechanical energy conversion

The main idea is to take electrical energy, pass it through an electromechanical converterand produce a force or torque, or vice versa.

Consider the following experiment: an iron rod sitting in a piece of wood and a windingaround the rod as in Figure 1 .

Figure 1 - Experiment

When the current flows through the coil and the flux through the rod, the space aroundthe rod changes magnetically. If we place a metal ring into the rod and if the current iszero the ring will fall into the bottom and hit the wooden surface. If we increase thevoltage, and hence the current, the ring will levitate due to a magnetic force that willcounteract the gravity force.

Figure 2 - Forces acting on the ring

The metal ring will also increase its temperature due to a current flowing along it.

Figure 3 - Cross-section of the ring and current flowing


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The temperature will be proportional to the power lost in the ring given by ( )2 R R RP i t R= ,where R R is the resistance around the ring.

If we keep V S constant and we force the ring to move up and down, then we wouldobserve that the current i(t) will vary in order to compensate the force.

What has been observed in time is that when a current flows through a conductor thespace around it changes due to a magnetic field set up in the neighborhood of the currentelement.

Figure 4 - Magnetic field

A magnetic field is characterized by its density B in Wb/m 2 or Teslas (T) and by the fieldintensity H in A/m.The field is a vector in 3 dimensions and therefore the notation will be from now on withan upper bar B .

Current i generates the field B and this is described by Amperes Law. Magnetic fieldsinteract with current to produce force and this is described by Lorenzs Law.

Amperes law in differential form.

From the following vector diagram,

3

id l r d H

r

Note that is the cross product. Using the right hand rule, with the thumb pointing in thedirection of the current, the rest of the fingers will point in the direction of the field.


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Where indicates the current is going out perpendicular to the page and indicates thecurrent is going into the page perpendicularly.

Example. Given a long wire carrying i(t), from electromagnetic theory we know that H

around an infinite long wire isi

H r

. So to get a magnetic field we only need to inject a

current!

Figure 5 Amperes law

Lorentzs law states that when there is a magnetic field due to a conductor carrying acurrent, a force is produced.

d F id l B

Using the right hand rule, with the fingers pointing in the direction of the current and bending towards the direction of B, the thumb will point to the direction of the force.

The presence of an existing field, a current carrying element experiences a force.


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Figure 6 - Lorentz's law

Both i 1 and i 2 generate a magnetic field in the neighborhood of the two wires. Conductor2 will be attracted to conductor 1 due to the force induced by B 1.

Amperes law in integral form. This law helps find the relation between the currents andmagnetic fields for complicated structures.

C S

H dl J d S = i i = Total current flowing perpendicular to the surface S.

Note that i indicates the dot product, i.e. the projection of one vector to the other one. J isthe current density in A / m 2.

Figure 7 - Ampere's integral form

Example. Consider an iron core with a coil. The magnetic field will flow along the iron because it has less reluctance than the air. So we assume that all the magnetic flux flowsthrough the iron. We assume also that H is tangential to the path inside the center of the

iron core, and H is constant throughout the contour.


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Figure 8 - Example

Then by Amperes law,

cC

H dl Hl= i

Where l c is the length of the contour and H H = .

And,

S

J d S Ni= i

Where Ni is the total current inside the contour. Thus,

c

Ni H

l=

This is an instantaneous magnitude, i.e. ( ) ( )c

Ni t H t

l= as long as the frequency is low

(60Hz).


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Lecture 14

Magnetic Circuits

An electromagnetic-mechanical system is formed with the following blocks.

We know that the integral form of Amperes Law isC S

H dl J d S = i i . Considering thenext figure, an iron core with cross-sectional area A and with mean length of perimeter l,several assumptions were made:

AmperesLaw

Material properties

FaradaysLaw

OhmsLaw, KCL,and KVL

LorenzsLaw andenergy

conservation

Newtonsand

LagrangesLaws

Speed,displacement

e

B, Hi


68/155

1. H is tangential to the path.2. H H is constant throughout the path.

3. There is no field outside the iron.

Since there are N turns in the coil then Hl Ni= .

Consider now an iron core with an air gap.

Figure 1 - Iron core with air gap

Let l c be the mean length of the iron core and g the length of the air gap, then by

Amperes Law,

c c g

C

H dl H l H g Ni= + = i

We need to relate the field in the core and in the gap with a common variable in order tosolve the equation. The magnetic field density B is related to the magnetic field intensityH in magnetically linear materials by,

B H =

Where is the permeability of the material. A typical curve is shown in Figure 2 . The permeability of the material is defined as,

0r =

Where 70 4 10 /Wb A turn m = or in henrys per meter and r is the relative

permeability with ranges from 2000 to 80,000 for materials used in transformers androtating machines.


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Figure 2 - B-H curve

The magnetic flux is defined as,

S

B d S = i

If B c = cHc is uniformly distributed in the core then = B cAc.

In the air gap the flux tends to fringe outwards.

Figure 3 - Fringing fields

The cross-section in the air gap is therefore bigger than in the core A g > A c. This is calledthe fringing effect. But there is continuity of flux,

core gap =

Hence,


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0

c c c c c

g g g g

B A H A

B A H A

= = = =

Then we can express Amperes Law as,

0

cc c g

c c g

l g Ni H l H g

A A = + = +

And

0

c

c c g

Nil g A A

=+

Lets consider now an iron core with two coils as shown in Figure 4 .

Figure 4 - Iron core with two coils

Now current i1 is entering the surface created by the contour C and i2 is going out,therefore,

1 1 2 2S

J d S N i N i= i


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It follows that,

C

H dl Hl l A

= = i

Then,

1 1 2 2

l N i N i

A

=

For complicated magnetic circuits we need a more systematic approach. To do this let uslook an analogy between electric and magnetic circuits.

Electric Circuit Magnetic Circuiti

Vs = voltage (emf) Ni = mmf

Resistance along path ( R) Reluctance along path ( ) Ohms law (V=iR) Mmf drop across path

KCL 0k k

i = Flux continuity 0k k

=

KVL 0k k

V = Amperes law 0k k

mmf =


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Lecture 15

Equivalent Circuits and Inductance Matrix

There is a perfect equivalence between magnetic and electric circuits. Recalling figuresfrom previous lectures, let Ac be the cross-sectional area of the iron core, Ag the cross-sectional area of the air gap so that Ag>Ac. The steps to convert the magnetic circuit toits electric equivalent are as follows:

1. Identify sources of magneto motive force (mmf), which are the coils. Theequivalents for these are voltage sources with magnitude Ni.

Figure 1 - mmf to voltage source

Note the polarity of the voltage source; it has to be consistent with the direction of theflux and the current I.

2. Identify and label fluxes in branches and equivalent currents I.3. Identify branch reluctances and their equivalent resistances.

0

,cc c g g c c g

l R R

A A = = = =

Thus the equivalent circuit for an iron core with an air gap and a single coil would be,

Es

Rc

Rg

Figure 2 - Equivalent circuit


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4. Solve the circuit by the usual circuit analysis methods.

For the example above, we can solve the circuit by using KVL.

( ) ( ) s c g c g E R R I Ni= + = +

Ag is found empirically, it can be considered as ( )1 g c A g A+ .

Consider the following example.

Figure 3 - 3 branches magnetic circuit

Note that the directions of the currents of the coils were arbitrarily set. There are 3 branches in the circuit. The lengths are defined by the points a, b, c, and d such that l ac,ladc , and l abc are the lengths of the branches. The reluctances are hence,

, ,ac adc abcac adc abcl l l A A A

= = =

The cross-sectional area A is the same through the entire core. The equivalent electriccircuit is,

Radc

Rac

Rabc

N3i3N1i1

N2i2

I1 I3

I2

a

Figure 4 - Equivalent circuit


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By KCL at node a, since this yields a single equation,

1 2 3

1 2 3

0

0

I I I + =

+ =

Let F a be the equivalent voltage to ground at node a. Then,

1 1 2 2 3 31 2 3, ,

a a a

adc ac abc

i F F N i N i F R R R

+ = = =

Substituting,

3 31 1 2 2 1 1 1 0aadc ac abc adc ac abc

N i N i N i F

R R R R R R

+ + =

And solving for F a,

3 31 1 2 2

1 1 1adc ac abc

a

adc ac abc

i N i N i R R R

F

R R R

+

=

+ +

Note that F a and therefore 1, 2, and 3 are linear functions of the mmf sources andhence linear with respect to the coil currents.

1 1

2 2

3 3

i

M i M i

i

= = =

Where for our example,

( )( )

( )

1 2 3

1 2 3

1 2 3

1 abc ac abc ac

abc abc adc adcac abc adc abc adc ac

ac adc ac adc

N R R N R N R

M N R N R R N R R R R R R R

N R N R N R R

+ = + + + +

The flux linkage i of coil i is defined as,


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i i i N

Therefore,

1 1 1

2 2 2

3 2 2

N Li

N

= = =

And L is called the inductance matrix.

( )( )

diag N

diag N Mi

=

=

Then,

( ) L diag N M =

For our example,

( )( )

( )

21 1 2 1 3

21 2 2 2 3

21 3 2 3 3

1 abc ac abc ac

abc abc adc adcac abc adc abc adc ac

ac adc ac adc

N R R N N R N N R

L N N R N R R N N R R R R R R R

N N R N N R N R R

+

= + + + +

In general, when there are n mutually coupled coils, they are characterized by aninductance matrix L of dimension n x n.


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Lecture 16

Nonlinear Magnetic Circuits

Consider an iron core with a single coil as shown in Figure 1 .

Figure 1 - Iron Core

If we inject dc current to the coil there will be a flux flowing around the core. If wekeep increasing the current, at some point the flux will lose its linearity and the flux in thecore will saturate, i.e., there is a limit in the total flux that can flow in a magneticconductor. This occurs because the magnetic moments of the conductor are all alignedwith the magnetic field. Figure 2 shows this effect.

Figure 2 - Flux saturation of magnetic core.

If instead we apply an ac current to the coil, then the flux will also tend to be sinusoidal, but then another effect takes place called remanent magnetization . Take for instance asinusoidal current in its positive part. When the current is increasing, the magneticmoments of the core will start aligning with the magnetic field following a sinusoidal


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pattern. When the current reaches its maximum, then if the core saturated the flux will nolonger be sinusoidal, if this is not the case then the flux will remain sinusoidal. When thecurrent starts decreasing, but still positive, the magnetic field will decrease as well, andthe magnetic moments will lose their orientation until the current goes back to zero. Atthis point, since the iron core is a good magnetic conductor, the magnetic moments will

no longer be what they were at the beginning, i.e., the final orientation will not coincidewith the initial orientation. This gives rise to the hysteresis loop shown in Figure 3 .

Figure 3 - Hysteresis loop.

Taking into account the nonlinearities of the magnetic circuit, Amperes Law remainsinvariant.

H dl Hl Ni= = i

But the relation B H = is no longer valid and hence the reluctance. Typically a B-Hrelation will have the shape as in Figure 2 changing for B and i for H.

Therefore, to find B it is necessary to go through the material specifications, i.e., eitherthe curve or a table. A table would be of the following form,

Table 1 - B-H Table

Hc X

1 X

2 X

3 X

4

Bc Y1 Y2 Y3 Y4 = B cAc Y1Ac Y2Ac Y3Ac Y4Ac

Hclc X1lc X2lc X3lc X4lc

Once B is found, can be calculated as,


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BA =

For more complicated system, e.g. a core with an air gap, then two problems might arise:

Problem A: Given find i.

a. From Amperes Law c c g Ni H l H g= +

b. Since is given then ,g cg c

B B A A

= = .

c. Air is magnetically linear hence0

gg

B H

= , and from the B-H curve or table for the

core we can obtain ( )c c H f B= .

d. Finally, c c g H l H g

i N

+= .

Problem B: Given i calculate . This is a more complicated problem and requires aniterative procedure that will be explained briefly.

a. Since the mmf drop in the core is much smaller than in the air gap, neglect H c,i.e., H clc = 0.

b. From Amperes Law, c cg Ni H l

H g

= therefore c c

g

Ni H l =

, where0

gg

g A

=

is the reluctance of the air gap.

c. Then cc

B A

= is the flux density in the core and from the B-H curve or table we

can calculate H c.

d. Let ' c c g H l H g

i N

+= , then if 'i i < stop, otherwise go to step b.

Problem B Alternative (Graphic): This method consists on overlapping two curves andfinding the intersection point, which in turn will be * for the given i.

a. From the B-H table graph H clc vs. . b. From Amperes Law we can solve for H

clc so that,

c c g H l Ni= , then plot H

clc

vs. .c. Find the intersection point and thus *.

Example. Consider the specifications for M19 steel given in Table 2. The cross sectionalarea of the core is 0.0018 m 2 and the mean length is 0.6 m. The air gap length is 0.0023 m


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Lecture 17

Faradays Law

Recalling the iron rod experiment and the levitating ring, we saw that the field interactswith the ring producing an induced current flowingthrough it. The current opposes the change in flux.Faradays Law can be stated hence as follows:

A voltage is induced in the coil equal in magnitude to d

dt

, with a

polarity such that it opposes the change of flux. In Figure 1 e is theinduced voltage. The current is clockwise so that it would generatea flux (by the right hand rule) that opposes (t). The inducedvoltage source and the resistance of the ring are distributed along

the coil as depicted in the picture. The sum of all of these fictitiousdistributed V-sources adds to

d dt

.

Figure 1 - Faraday's Law

If we connect several coils instead of one (as if the ring were made of several loops) theflux (t) links all N turns and the induced voltage across the N turns is

d e N dt

=

This is depicted in Figure 2 .


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Figure 2 - Induced voltage and current.

Note that e and I must satisfy KVL and Ohms law, thus Ri = e, and d

N Ridt

= . i will

generate a flux that opposes (t). It is assumed that there exists a source of mmfgenerating (t). Faradays law works also for a loop that is open circuited. In this casethe induced current will be zero since R = . The polarity of the induced voltage is still

such that if tries to create a current that would oppose the change in flux. If 0d dt

> , i.e.

(t) is increasing, then i > 0 and hence by the right hand rule i(t) will induce a flux

component that goes down. If 0d dt

< , i.e. (t) is decreasing, then i


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Then by KVL and Ohms law,

s

d e Ri

dt

=

If instead of a voltage source we have an independent current source as shown in Figure4,

Figure 4 Loop with an independent current source

Then,

s

s

i id

Ri edt

=

=

i andd dt

are known in this case and we can solve for e s.

Consider now Figure 5 . If e s = E s is a constant, then in steady state i = Es / R since

0di

e L dt = = because the coil can be treated as an inductor.

To analyze this devie we have the following,

1. Amperes Law: Ni = , where ( )1 2 2c c g x = + + .

2. Faradays Law: d e N dt

=


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3. KVL and Ohms Law: se Ri e= +

Figure 5 - Variable reluctance case

Combining 1 and 2,

2

di d N

dt dt N di

e dt

=

=

And if2 N

L =

thendi

e Ldt

= , therefore substituting in 3,

s

die Ri L

dt = +

Now let e s = E s, then e = 0 and is constant. Therefore there is no induced or back emf.However, if x varies with time,

0dx vdt

= >

Where v is the speed then ( ) x varies with time and hence L(x) as well. Then,


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( ) ( )( )d L x i t e

dt =

Applying the derivative of a product,

( ) di dLe L x idt dt

= +

The second termdL

idt

is called the generated voltage, and since ( )( )

2 N L x

x=

then

applying the chain rule again we get,

dL dL dx dLv

dt dx dt dx= =

Where( )

2

2

dL N d dx x dx

=

And since ( )1 2 2c c g x = + + , and the reluctance of air is0

gg

x A

= , then

0

2

g

d dx A

=

And finally,

( )

2

20

2

g

dL N dx x A

=


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Lecture 18

Examples of Faradays Law.

So far we have seen that magnetic circuits can be either linear or nonlinear. They can also be fixed or have moving parts. Both can be characterized by an inductance matrix L,which in the latter case will be dependant of the variable x (position). It is in generaldifficult to obtain matrix L by inspection and therefore easier to do it systematically bysolving the magnetic circuit.

Example 1 . Consider the following non-moving magnetic circuit.

This has the following equivalent circuit,

1 = 2. Then by loop analysis,

( )( )

1 1 2 2 1 1 2 1 3

3 3 3 3 2 3 1

N i N i

N i

+ = +

= +

By simple algebraic manipulation and substitution one can find that,


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( ) ( )( ) ( )

( )

1 1 2 3 2 2 3 3 2 1

2 1 2 3 2 2 3 3 2 21 2 1 3 2 3

3 1 2 2 2 3 1 2 3

1 N N N i

N N N i

N N N i

+ + = = + +

+ + +

And since N Li = = then ( )diag N Li = = , therefore

( ) ( )( ) ( )

( )

21 2 3 1 2 2 3 1 3 2

21 2 2 3 2 2 3 2 3 2

1 2 1 3 2 3 21 3 2 2 3 2 3 1 2

1 N N N N N

L N N N N N

N N N N N

+ +

= + + + + +

Example 2 . Consider the following moving magnetic circuit.

Note that the direction of the currents and polarity of the voltages have been arbitrarilychosen and therefore special care must be taken when assigning signs in Faradays law.The equivalent circuit is the following,

1 = - 2. Then by loop analysis,

( )

( )1 1 2 2 1 2 1

1 1 2 2 1

2 g N i N i x

N i N i x

= + + =


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Where L is the matrix of linkages, LI = , thus

E j LI =

Example 3 . Consider the following circuit with a given L.

Then,

1 1 211 12

2 1 221 22

2 2

s E E j L I j L I

E j L I j L I

E RI

= = +

= +

=

Where L ij is the ij-element of the L matrix.

If there were moving parts, then L (x), where x is the relative position of the moving part.Therefore,

( )d d i d L dxe L x idt dt dx dt

= = +

If i = I is constant (i.e., dc) then di/dt = 0,

d L dxe i

dx dt

=

So if 0dxdt

there is an induced voltage. So if ( )0 sindx

x t dt

= then e will also be

sinusoidal.


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Lecture 19

Energy and co-energy

Consider a ferromagnetic, electromagnetic device without moving parts or stationarymoving parts (device with movable parts but that are kept fixed), either linear = Li or

nonlinear = f(i). By Faradays law we knowd

edt

= . The instantaneous power

consumed by the device is

( ) ( ) ( ) ( ) ( )d t

p t e t i t i t dt

= =

Therefore in the time interval dt, the device absorbs

( )dW p t dt id = = joules

Since this device does not produce heat, nor does it produce mechanical energy, thendW must be stored in the device as a magnetic field. If we consider a device operating at

a current i then

( )0

W dW p t dt id

= = =

Figure 1 - Nonlinear characteristic curve of a device


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The co-energy stored at ( 0, i0) is defined as

0

'

0

i

W di =

And corresponds to the area under the curve (i).

One identity relating W and W is

'0 0W W i + =

This means the sum of the areas on both sides of the curve is equal to the total area of therectangle.

If the device is magnetically linear ( = Li) then

0 0

20

0 0

12

W id d L L

= = =

And since 0 = Li 0 then 2012

W Li = and is equal to the co-energy'W since

0 0

' 20

0 0

12

i i

W di Lid Li = = =

In a magnetically linear device the characteristic curve has the following shape,

Figure 2 - Magnetically linear device

Therefore ' 2 21 1 12 2 2

W W Li i L

= = = =


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In a magnetically nonlinear device this is not true. Generally 'W W > .

For example, if ( )0/0 1 i ie =

0

W id

=

This could be solved by solving for i and then using integral tables. Alternatively,

( ) ( )0 0/ /' 0 0 00 0

1 1i i

i i i iW di e di i i e = = =

And, ( )0/' 0 0 1 i iW i W i i i e = =

Let us consider now devices with multiple coils. If the device is linear then

1

n

Li

d d i d Le L i

dt dt dt

= =

= = +

And if there are no moving parts then dL / dt = 0, hencedi

e L dt =

Therefore the total power absorbed by the device is

( ) ( ) ( ) ( )1 1

n n

k k k k k

p t p t e t i t = =

= =

In vector form

( ) ( ) ( ) ( ) ( )T T p t e t i t i t e t = =

Substituting e = d / dt,

( ) ( ) ( )T d t

p t i t dt

=

If we consider an infinitesimal interval of time then the energy is


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( ) ( ) ( )T dW p t dt i t d t = =

Integrating,

12

T T W i d i Li = =

Show as an exercise.

This equation is valid even if the device has moving parts that are free to move, since W was derived independently of time.

( ) ( ) ( )12

T W t i t Li t =

If the device is linear 1'1 1 12 2 2

T T T W W i Li i L = = = =

If we take into account moving parts, the fundamental law is energy conservation.

elec mechdW dW dW = +

Figure 3 - Energy conservation


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( ) ( ) ( )12

T W i t L t i t =

So, for example, if we consider again the relay magnetic circuit

Then we know by Amperes Law ( )c g Ni x = + , where ( )0

2g

g

x x A

= and

therefore the inductance is ( )2

0

2c

g

N L x

x A

= +

Consider now the changes in the energy stored in the magnetic field of n-mutuallycoupled coils. Suppose you are given L(x), and that initially 1i i= and 1 x x= . Then aftera change in the device, e.g., a displacement of the moving part, 2i i= and 2 x x= . As a

result the change in the energy stored in the field W is,

final initialW W W =

( ) ( )2 2 1 12 11 12 2

T T W i L x i i L x i =

This change depends only in the initial values ( )1 1,i x and final values ( )2 2,i x and not onthe path followed by the variables i and x.

Example. Let ( )2 1

1 3

x L x

=

and 1 2

1

2i i

= =

. Changing x from 0 to 1 would yield a

change in the energy as follows,


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[ ] [ ]2 1 1 1 2 0 1 11

1 2 1 21 3 2 1 3 22

W

=

[ ]1 117 182 2

W Joules = =

From the first law of thermodynamics, which says that energy can be transferred fromone system to another in many forms and that it cannot be created nor destroyed, thechange in the electrical energy at one side must be reflected as a change in mechanicalenergy at the other side of the system.

Figure 1 - First law of thermodynamics (energy conservation)

elec mechW W W = +

Where

W elec: energy supplied by electric sources.W: change of energy stored in the magnetic field.Wmech : mechanical energy or work

Therefore mech elecW W W =

We already know how to find W, hence we need to find W elec so as to obtain the workdone. Let the flux linkage be a function of a variable inductance,

( ) L x i =

Suppose that x changes from x 1 to x 2 at constant current si i= , then

( ) ( )2 11 12 2

T T s s s sW i L x L x i i Li = =

And the electric energy delivered can be obtained as


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( ) ( )1

n

elec ii

W p t dt p t dt =

= =

( ) ( ) ( )1 1

n n

elec i i i ii i

W e t i t dt i t d = =

= =

Since the change in x occurred at constant currents then,

_ _ 1 1

n n

elec s i i s i ii i

W i d i = =

= =

Where( )

( )

( ) ( )2

1

2 1

i

i

x

i i i i x

d x x

= = . But Li = so ( ) ( )k k x L x i = and therefore

( ) ( ) ( ) ( )2 1 2 1 x x L x L x =

Thus the electric energy is

( ) ( )

( ) ( )

[ ]

_ 2 11

2 1

nT selec s i i

i

T s selec

T s selec

W i i x x

W i L x L x i

W i L i

=

= =

=

=

And finally, solving for the mechanical energy,

[ ] [ ]

[ ]

12

12

T T s s s smech elec

T s smech

W W W i L i i L i

W i L i

= =

=

If the currents are not constant then this derivation is no longer the same. Note that W elec is path dependent whereas W is path independent.


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Lecture 21

Force and Torque

Given a system of initially coupled coils with a moving part ( ) L x i = , where x is the position of the moving part, then by the first law of thermodynamics,

elec mechW W W = +

If x changes by x at constant current i then,

12

T W i Li =

where ( ) ( ) L L x x L x = + . In addition the electric energy delivered by the n externalsources during the change can be obtained as,

T T elecW i d i = =

where ( ) ( ) L x x i L x i = + . Then

12

T mech elecW W W i Li = =

Let us consider an infinitesimal change x dx , and then the differential of the energywould be

mech elecdW dW dW =

If additionally we consider a differential of the current (no longer constant) then thediffe

power engineering full class notes

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