PPT Ch.2 AC Networks

7/26/2019 PPT Ch.2 AC Networks

1/70

AC NETWORKS


2/70

Definition of Alternating

Quantity An alternating quantity changes continuously

in magnitude and alternates in direction

at regular intervals of time.


3/70

Advantages of AC System OverDC System

1. AC voltages can be eciently stepped up/down

using transformer.2. AC motors are cheaper and simpler in

construction than C motors.

!. "witchgear for AC system is simpler than C

system.


4/70

Generation of Single Phase EMF

Consider a rectangular coil of # turns placed in a uniform magnetic $eldas shown in the $gure. %he coil is rotating in the anticloc&wise directionat an uniform angular velocity of ' rad/sec.


5/70

%he ma(imum )u( lin&ing the coil is in the downward direction as

shown in the $gure. %his )u( can be divided into two components* onecomponent acting along the plane of the coil +ma( sin't and another

component acting perpendicular to the plane of the coil +ma( cos't.


6/70

%he component of )u( acting along the plane of the coil does notinduce any )u( in the coil. ,nly the component acting perpendicularto the plane of the coil i.e. +ma( cos't induces an emf in the coil

Hence the emf induced in the coil is a sinusoidal emf. This will induce a

sinusoidal current in the circuit given by


7/70

Angular -requency ' Angular frequency is de$ned as the number of radians covered

in one second ie the angle covered by the rotating coil. %heunit of angular frequency is rad/sec.

Problem 1An alternating current i is given by i = 141.4 sin 314t

Find i) The maimum value ii) Fre!uency iii) Time "eriod iv) The

instantaneous value when t=3ms.

#ollution $i = 141.4 sin 314t i) %aimum value m!1"1#" $

ii) &= 314 rad'sec

f =&'( = %& '(

iii) T=1'f = &) se*

iv) i=141.4 sin*314+.++3) = 11"#+%A


8/70

%he arithmetic average of all the values of an alternating quantity overone cycle is called its average value.

Average $alue

For #ymmetrical waveforms, the average value calculated over one cycle

becomes e!ual to -ero because the ositive area cancels the negative

area. Hence for symmetrical waveforms, the average value is calculated for

half cycle.


9/70

Average /alue of #inusoidal 0urrent$


10/70

,MS or Effe*tive $alue%he e0ective or " value of an alternating quantity is that steady

current dc which when )owing through a given resistance for a giventime produces the same amount of heat produced by the alternatingcurrent )owing through the same resistance for the same time.


11/70

2%# /alue of #inusoidal 0urrent$


12/70

Form Factor

%he ratio of " value to the average value of analternating quantity is &nown as -orm -actor

Pea- Fa*tor or Crest Fa*torThe ratio of maimum value to the 2%# value of an alternating !uantity is

nown as the ea factor.


13/70

,esistive Cir*uit0onsider an A0 circuit with a ure resistance 2 as shown in the figure. Thealternating voltage v is given by$

sing ohms law, we can write the following

relations

Fig $ 2esistive 0ircuit


14/70

-rom equation 1 and 2 we conclude that in a pure resistive

circuit* the voltage and current are in phase. 3ence the voltageand current phasor and waveform can be drawn as below.

Fig $ "hasor 5iagram and waveforms


15/70

6nstantaneous "ower of 2esistive 0ircuit$ %he instantaneous power in the above circuit can be derived asfollows 4

The instantaneous ower consists of two terms. The first term is called as the

constant ower term and the second term is called as the fluctuating ower term.


16/70

Average "ower of 2esistive 0ircuit4 -rom the instantaneous power we can $nd the average power

over one cycle as follows4

As seen above the average ower is the roduct of the rms voltage and the rms

current.


17/70

As seen from the waveform, the instantaneous ower is always ositive meaning

that the ower always flows from the source to the load.

The voltage, current and ower waveforms of a urely resistive circuit is as

shown in the figure.


18/70

Problem 2An ac circuit consists of a pure resistance of 156 and is connected

to an ac supply of 2!5 7* 85 39.Calculate the i current ii power consumed and iii equations forvoltage and current.

"ollution4


19/70


20/70

-rom equation 1 and 2 we observe that in a pure inductive

circuit* the current lags behind the voltage by :5;. 3ence thevoltage and current waveforms and phasors can be drawn as below.



21/70

is given as

6t is e!uivalent to resistance in a resistive circuit. The unit is ohms *8)

6nstantaneous owerThe instantaneous ower in the above circuit can be derived as follows

As seen from the above e!uation, the instantaneous ower is fluctuating in nature.


22/70

Average powerFrom the instantaneous ower we can find the average ower over onecycle as follows

The average ower in a ure inductive circuit is -ero. 9r in other words, the

ower consumed by a ure inductance is -ero.


23/70

%he voltage* current and power waveforms of a purelyinductive circuit is as shown in the $gure.

7hen the ower is ositive, the ower flows from the source to the inductor

and when the ower in negative, the ower flows from the inductor to the

source. The ositive ower is e!ual to the negative ower and hence the

average ower in the circuit is e!ual to -ero. The ower :ust flows between the

source and the inductor, but the inductor does not consume any ower.


24/70

?roblem !A pure inductive coil allows a current of 15A to )ow from a 2!57* 85 39 supply.

-ind i inductance of the coil ii power absorbed and iii equations for voltage

and current.

"olution4


25/70

Pure Ca.a*itive Cir*uit Consider an AC circuit with a pure resistance as shown in the

$gure. %he alternating voltage v is given by4

7e can find the current through the caacitor

as follows $

Fig $ 0aacitive 0ircuit


26/70

-rom equation 1 and 2 we observe that in a pure capacitive

circuit* the current leads the voltage by :5;. 3ence the voltageand current waveforms and phasors can be drawn as below.



27/70

Capacitive reactance%he capacitive reactance =C is given as

6t is e!uivalent to resistance in a resistive circuit. The unit is ohms *8)

6nstantaneous owerThe instantaneous ower in the above circuit can be derived as follows


28/70

Average power

-rom the instantaneous power we can $nd the average power

over one cycle as follows 4

The average ower in a ure caacitive circuit is -ero. 9r in other words, the

ower consumed by a ure caacitance is -ero.


29/70

%he voltage* current and power waveforms of a purely capacitivecircuit is as shown in the $gure4

As seen from the ower waveform, the instantaneous ower is alternatelyositive and negative. 7hen the ower is ositive, the ower flows from the

source to the caacitor and when the ower in negative, the ower flows from the

caacitor to the source. The ositive ower is e!ual to the negative ower and

hence the average ower in the circuit is e!ual to -ero. The ower :ust flows

between the source and the caacitor, but the caacitor does not consume any

ower.


30/70


31/70

,/ Series Cir*uit

0onsider an A0 circuit with a resistance 2 and an inductance ; connected in series as shownin the figure. The alternating voltage v is given by

Fig $ 2; #eries 0ircuit

The current flowing in the circuit is i. The voltage

across the resistor is /2 and that across the inductoris /;.

/2=62 is in hase with 6

/;=6


32/70

%he current < is ta&en as the reference phasor. %he voltage 7 is in phase with < andthe voltage 7> leads the current by :5;. %he resultant voltage 7 can be drawn as shownin the $gure. -rom the phasor diagram we observe that the voltage leads the current

by an angle + or in other words the current lags behind the voltage by an angle +. Thewaveform and equaton!for an > series circuit can be drawn as below.

From the hasor diagram, the eressions for the resultant voltage / and the angle > can be derived as

follows.

The imedance in an A0 circuit is similar to a

resistance in a 50 circuit. The unit for imedance

is ohms *8).


33/70

Pha!e An"le

nstantaneous .o0erThe instantaneous ower in an 2; series circuit can be derived as follows

he instantaneous .o0er *onsists oft0o terms# The first term is called as the

constant ower term and the second

term is called as the fluctuating ower

term.


34/70

Avera"e Power-rom the instantaneous power we can $nd the average power over one cycle as follows

The voltage, current and ower waveforms of a 2; series circuit is as shown in the figure.

The ositive ower is not e!ual to the negative

ower and hence the average ower in the circuit isnot e!ual to -ero.

Hence the ower in an 20 series circuit is consumed only in the

resistance. The caacitance does not consume any ower.


35/70

Power Factor%he power factor in an AC circuit is defined as the cosine of the angle between voltage and current i.e cos+.

?7< cos+

%he power in an AC circuit is equal to the product of voltage* current and power factor.

#m$edance Tran"leDe can derive a triangle called the impedance triangle from the phasor diagram of an > series circuit as shown


36/70

The imedance triangle is right angled triangle with 2 and . The imedance

triangle enables us to calculate the following things.


37/70

Power6n an A0 circuit, the various owers can be classified as

1. Real or Active power2. Reactive power

3. Apparent power

2eal or active ower in an A0 circuit is the ower that does useful wor in the cicuit.2eactive ower flows in an A0 circuit but does not do any useful wor. Aarent ower

is the total ower in an A0 circuit.

From the hasor diagram of an 2; series circuit,

the current can be divided into two comonents.

9ne comonent along the voltage 6cos>, that iscalled as the active comonent of current and

another comonent erendicular to the voltage

6sin> that is called as the reactive comonent of

current.


38/70

Real Power%he power due to the active component of current is called as the active

power or real power.


39/70

Power Tran"le-rom the impedance triangle* another triangle called the power triangle

can be derived as shown.

The ower triangle is right angled triangle with " and ? as two sides and # as the

hyotenuse. The angle between the base and hyotenuse is >. The ower triangle enables us

to calculate the following things.


40/70

Problem )

A coil having a resistance of @8 and an inductance of 31.mH is connected to (3+/, B+H-suly. 0alculate *i) the circuit current *ii) hase angle *iii) ower factor *iv) owerconsumed

Solution2


41/70

Problem *

A (++ /, B+ H-, inductive circuit taes a current of 1+A, lagging 3+ degree. Find*i) the resistance *ii) reactance *iii) inductance of the coil

Solution2


42/70

,C Series Cir*uit

0onsider an A0 circuit with a resistance 2 and an caacitance ; connected in series as shownin the figure. The alternating voltage v is given by

Fig $ 2; #eries 0ircuit


across the resistor is and that across the inductoris .

/r =62 is in hase with 6

/c =6


43/70

%he current < is ta&en as the reference phasor . %he voltage 7 is in phase with can

be derived as follows.

The imedance in an A0 circuit is similar to

a resistance in a 50 circuit. The unit for

imedance is ohms *8).


44/70

Pha!e An"le

Average .o0er

Hence the ower in an 20 series circuit is consumed only in the resistance. The caacitance does not

consume any ower.


45/70

#m$edance Tran"le

7e can derive a triangle called the imedance triangle from the hasor diagram of an 20series circuit as shown

Po0er riangle

From the imedance triangle, another triangle called the ower triangle can be derived as

shown.


46/70

Problem +

A Capacitor of capacitance F:.8B- is connected in series with a noninductive resistance of !56 across a 1557* 8539 supply. -ind iimpedance ii current iii phase angle iv Gquation for theinstantaneous value of current.

Soluton,


47/70

,/C Series Cir*uit

Fig $ 2;0 #eries 0ircuit


across the resistor is and that across the caacitoris .

0onsider an A0 circuit with a resistance 2, an inductance ; and a caacitance 0 connected in

series as shown in the figure. The alternating voltage v is given by


48/70


49/70


50/70


51/70

Problem -

A 2!5 7* 85 39 ac supply is applied to a coil of 5.5H 3 inductance and 2.8 6resistance connected in series with a H. B- capacitor. Calculate i


52/70

Series ,esonan*e Cir*uit 6n a series 2;0 circuit there becomes a fre!uency oint were the inductive reactance of the

inductor becomes e!ual in value to the caacitive reactance of the caacitor. 6n otherwords,


53/70


54/70

"eries esonance -requency

where$ Cris in Hert-, ; is in Henries and 0 is in

Farads.

Dlectrical resonance occurs in an

A0 circuit when the two

reactances which are oosite and

e!ual cancel each other out

as


55/70


56/70

"eries >C Circuit at esonance


57/70


58/70

Parallel ,/ Cir*uit


59/70

?hasor iagram of ?arallel > circuit

6n comle form the currents are written as,

#o the total current 6T,


60/70

#m$edance of Parallel R0 Crcut

;et, G = total imedance of the circuit in ohms.2= resistance of circuit in ohms.

; = inductor of circuit in Henry.


61/70

',EE P'ASE AC C,C4S

A three hase suly is a set of three alternating !uantities dislaced from each other by an

angle of1(+ . A three hase voltage is shown in the figure. 6t consists of three hases hase A,

hase J and hase 0. "hase A waveform starts at + . "hase J waveform stars at 1(+ and hase

0 waveform at(4+ .


62/70

The three hase voltage can be reresented by a set of three e!uations as shown below.

The sum of the three hase voltages at any instant is e!ual to -ero.

The hasor reresentation of three hase voltages is as shown.

The hase A voltage is taen as the reference

and is drawn along the ais. The hase J

voltage lags behind the hase A voltage by

1(+ . The hase 0 voltage lags behind the

hase A voltage by(4+ and hase J voltage by

1(+ .


63/70

Generation of hree Phase $oltage

Three "hase voltage can be generated by lacing three rectangular coils dislaced in

sace by 1(+ in a uniform magnetic field. 7hen these coils rotate with a uniformangular velocity of rad'sec, a sinusoidal emf dislaced by 1(+ is induced in these coils.


64/70

5e*essity and advantages of three .hase

systems26

3 ower has a constant magnitude whereas 1 ower ulsates from -ero to ea value

at twice the suly fre!uency

A 3 system can set u a rotating magnetic field in stationary windings. This is not

ossible with a 1 suly. For the same rating 3 machines are smaller, simler in construction and have better

oerating characteristics than 1 machines

To transmit the same amount of ower over a fied distance at a given voltage, the

3 system re!uires only 3'4th the weight of coer that is re!uired by the 1 system

The voltage regulation of a 3 transmission line is better than that of 1 line


65/70


66/70

sing KirchhoffLs voltage law, the line voltages can be written in terms of the hase voltages

as shown below.

The hasor diagram shows the three hase voltages and the line voltage DAJdrawn from DAand DJ hasors. The hasor for current 6Ais also shown. 6t is assumed that the load is

inductive.


67/70

From the hasor diagram we see that the line voltage DAJleads the hase voltage DAby 3+ .

The magnitude of the two voltages can be related as follows.

Hence for a balanced star connected load we can mae the following conclusions.

;ine voltage leads hase voltage by 3+


68/70


69/70

sing KirchhoffLs current law, the line currents can be written in terms of the hase currents asshown below.

From the hasor diagram we see that the linecurrent 6A lags behind the hase hase current 6AJ

by 3+ . The magnitude of the two currents can be

related as follows.

Hence for a balanced delta connected load we can mae the following conclusions.

;ine current lags behind hase current by 3+


70/70

THANK YOUTHANK YOU

PPT Ch.2 AC Networks

Documents

Transcript of PPT Ch.2 AC Networks