ch1 harmonics

7/26/2019 CH1 Harmonics

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7. Harmonics in Single-phase A.C. Circuits

(b) Pure inductance Let the inductance of the circuit be L henry whose reactance

varies directly as the frequency of the applied voltage.

Its reactance for the fundamental would be X1 = L; for the

second harmonic,X2= 2L, for the third harmonic,X3= 3Land

for thenth harmonicXn= n L.

However for ever harmonic term the current will la behind the

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voltage by 90.

1 m1

Ei (t) sin( t )

L 2

=

2 m2

Ei (t) sin( 2 t )L 2

=

nm

n

Ei (t) sin( n t )

nL 2

=

Instantaneous current due to

fundamental :


2nd harmonic:


nth harmonic:


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(b) Pure inductanceTotal current: 1 2 n

1 m 2 m nm

1 m 2 m nm

i ( t ) i ( t ) i ( t ) ... i ( t )

E E Ei ( t ) sin( t ) sin( 2 t ) ... sin( n t )

L 2 2 L 2 nL 2

i ( t ) I sin( t ) I sin( 2 t ) I sin( n t )2 2

=

=

=

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(i)the waveform of the current differs from that of the applied

voltage.

(ii)for the nth harmonic, the percentage harmonic content in the

current-wave is (1/n) of the corresponding harmonic content in thevoltage wave. It means that in an inductive circuit, the current

waveform shows less distortion that the voltage waveform. In this

case, current more nearly approaches a sine wave than it does in a

circuit containing resistance.


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(c) Pure Capacitance Let the capacitance of the circuit be C (Farad) whose reactance

varies directly as the frequency of the applied voltage.

Its reactance for the fundamental would beX1= (1/C); for the

second harmonic,X2= (1/2C), for the third harmonic,X3= (1/3C)

and for thenth harmonicXn= (1/nC).

However for ever harmonic term the current will lead behind

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the voltage by 90.

1 1 mi (t) E C sin( t )

2 2 mi (t) E 2C sin( 2 t )

n nm

i (t) E nC sin( n t )

Instantaneous current due

to fundamental :


to2nd harmonic:


tonth harmonic:


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(c) Pure Capacitance Total current:

1 2 n

1 m 2 m

nm

i ( t ) i ( t ) i ( t ) ... i ( t )

i ( t ) E C sin( t ) E 2C sin( 2 t )

2

... E nC sin( n t )2

=

=

=

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This equation shows that

(i) the current and voltage waveforms are dissimilar.

(ii) percentage harmonic content of the current is larger than that of

the applied voltage wave. For example, fornthharmonic, it would be

n timelarger.

(iii) the current wave is more distorted than the voltage wave.

(iv) Capacitor effect on distortion is the reverse of that of inductance.

1 m 2 m nms n s n s n n


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Exercise .5. A complex wave of r.m.s. value 240 V has 20% 3rd harmonic content, 5% 5th harmonic content and 2% 7th harmonic

content. Find the r.m.s. value of the fundamental and of each harmonic.

Solution Exercise .5.

Let V1, V3, V5 and V 7 be the r.m.s. values of the fundamental and

harmonic volta es.


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Then V3= 0.2 V1; V5= 0.05 V1and V7= 0.02 V1

2 2 2 2

1 3 5 7

2 2 22 21 1 1 1 1

1 3

5 3

240 V V V V

40 V 0 ,2V 0 ,05V 0 ,02V 1 ,0429V

240V 235V ; V 0 ,2 235 47V

1 ,0429

V 0 ,05 235 11 ,75V ; V 0 ,02 235 4 ,7V

=

= =

= = = =

= = = =


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Exercise .6.An alternating voltage of:

is applied to a series circuit of resistance R = 20 and inductance L = 0,05

Henry with fundamental angular frequency = 314 rad/s. Derive:

1) The instantaneous expression for the current

2) The r.m.s. value of the current and for the voltage

3) The total power supplied


e( t ) 250 sin t 50 sin 3 t 60 2 sin 5 t 150

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Exercise .7.

An alternating voltage of :

is applied across a capacitor which can be represented by a capacitance ofC = 0,5 F shunted by a resistance of 4000 . Determine:

1) The r.m.s. value of the current I

2) The r.m.s. value of the applied voltage V

3) The power factor of the circuit PF.

v ( t ) sin 500t 0 ,5 sin 1500t


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Generally, with the help of ammeter and voltmeter

readings, the value of impedance, inductance and capacitance

of a circuit can be calculated. But while dealing with complex voltages, the use of

instrument readings does not, in general, give correct values of

8. Effect of Harmonics on Measurement of Inductance and

Capacitance

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inductance and capacitance except in the case of a circuitcontaining only pure resistance.

It is so because, in the case of resistance, the voltage and

current waveforms are similar and hence the values of r.m.s.volts and r.m.s amperes (as read by the voltmeter and

ammeter respectively) would be the same whether they ware

sinusoidal or non-sinusoidal (i.e. complex).


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Effect on Inductances

Let L be the inductance of a circuit and E and I the r.m.s.

values of the applied voltage and current as read by the

instruments connected in the circuit. For a complex voltage:

Hence:


Capacitance

1 m 3 m 5 m

2 2 2E 0 ,707 E E E ....

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3 m 5 m

1 m

3 m 5 m

1 m

2 223 m 5 m1 m

2 2

2

2 2

2

E EEI 0 ,707 ....

L 3 L 5 L

E E0 ,707I E ....L 9 25

E E0 ,707L E ....

I 9 25

=

=

=


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For calculating the value ofLfrom the above expression, it is

necessary to known the absolute value of the amplitudes of

several harmonic voltages. But, in practices, it is moreconvenient to deal with relative values than with absolute

values.


Capacitance

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or s purpose, e us mu p y an v e e r g - an

side of the above expression by E but write the E in the

denominator in its form:

Hence:1 m 3 m 5 m

2 2 2E 0 ,707 E E E ....

3 m 5 m

1 m

1 m 3 m 5 m

2 2

2

2 2 2

E E0 ,707 EL E ....

I 9 25 0 ,707 E E E ....

=


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Capacitance

3 m 5 m

1 m

1 m 3 m 5 m

2 22 2

3 m 5 m2

1 m 1 m2 22 2 2

3 m 5 m

1 m 1 m

E E1 1E E 1 ....E ....

9 E 25 EE E9 25LI E E E .... I E E

1 ....E E

= =

Or:

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e e ec o armon cs were o e neg ec e , en e va ue o

the inductance would appear to be(E /I) but the true or actual

value is less than this.

The apparent value has to be multipliedby the quantity under the

radical to get the true value of inductance when harmonics arepresent. The quantity under the radical is called the correction

factor.

True Inductance Apparent Inductance correction

L L factor

=


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Effect on Capacitances

Let the capacitance of the circuit beC FaradsandEandIthe

instrument readings for voltage and current. Since the

instruments read r.m.s. values, hence, as before:


Capacitance

1 m 3 m 5 m

2 2 2E 0 ,707 E E E ....

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2 2 2

3 m 5 m1 m

2 22

1 m 3 m 5 m

2 22

1 m 3 m 5 m

2 2 2

1 m 3 m 5 m

E EEI 0 ,707 ....

1 C 1 3C 1 5C

I 0 ,707 C E 3C E 5C E ....

I 0 ,707 C E 3 E 5 E ....

I 0 ,707 C E 9 E 25 E ....

=

=

=

=


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Capacitance

Again, we will multiply and divide the right-hand side Ebut

2 2 21 m 3 m 5 m

IC

0 ,707 E 9 E 25 E ....

=

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1 m 3 m 5 m

2 2 2E 0 ,707 E E E ....

in this case, we wil write inthenumerator in its orm:

1 m 3 m 5 m

2 2 2

2 2 2

1 m 3 m 5 m

0 ,707 E E E ....IC

E0 ,707 E 9 E 25 E ....

=


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Capacitance

1 m 3 m 5 m

2 2

3 m 5 m2 2 2

1 m 1 m2 22 2 2

1 m 3 m 5 m3 m 5 m

1 m 1 m

E E1 ....

E E E .... E EI ILE E 9 E 25 E .... E E E

1 9 9 ....E E

= =

Or:

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If the effects of harmonics were neglected, the value of capacitancewould appear to be(I/E)but its true value is less than this.

For getting the true value, this apparent value will have to be

multiplied by the quantity under the radical (which, therefore, is

referred to as correction factor)

True Apparentcorrection

capacitance capacitancefactor

C C

=


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Exercise 7. A current of 50-Hz containing first, third and

fifth harmonics of maximum values 100, 15 and 12 A

respectively, is sent through an ammeter and an inductive coil

of negligibly small resistance. A voltmeter connected to theterminals shows 75 V.

hat would be the current indicated by the ammeter and what


Capacitance

58

SolutionExercise 7.

The r.m.s. current is:

The r.m.s. voltage across the terminals of the coil is :

1 m 3 m 5 m

2 2 2 2 2 2I 0 ,707 I I I 0 ,707 100 15 12 72 A = =

1 m 3 m 5 m

2 2

E 0 ,707 E E E


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Capacitance

1 m 2 m 3 mE 100 ; E 20 ; E 10

= =

Solution Exercise 8.

True value,C = 20F

Apparent value:C = value read by the instruments

Now,C = C correction factor.Let us find the value of correction factor.

Here:

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Correction factor2 2 2

1 m 2 m 3 m

2 2 2

1 m 2 m 3 m

2 2 2

2 2 2

'

E E ECorrection factor

E 4 E 9 E

100 20 10Correction factor 0 ,966

100 4 20 9 10

C 20C 21 ,82 F

Correction factor 0 ,966

=

= =

= = =


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In three-phase systems, harmonics may be produced in the

same way as in single-phase systems.

Hence, for all calculation they are treated in the same

9. Harmonics in Different Three-phase Systems

62

Usually, even harmonics are absent in such systems. But care

must be exercised when dealing with odd, especially, third

harmonics and all multiples of 3rd harmonic (also called the

triple-n harmonics).


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(a) Expressions for Phase E.M.Fs.

Let us consider a 3-phase alternator having identical phase

windings (R, Y and B) in whichharmonics are produced. The

three phase e.m.fs. would be represented in their proper phase

sequence by the equation.


R 1 m 1 3 m 3 5 m 5e ( t ) E sin t E sin 3 t E sin 5 t

2 2 2e t E sin t E sin 3 t E sin 5 t

=

=

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On simplification, these become

m m m

B 1 m 1 3 m 3 5 m 5

3 3 3

4 4 4e ( t ) E sin t E sin 3 t E sin 5 t

3 3 3

=

=

R 1 m 1 3 m 3 5 m 5

Y 1 m 1 3 m 3 5 m 5

B 1 m 1 3 m 3 5 m 5

e ( t ) E sin t E sin 3 t E sin 5 t

2 4e ( t ) E sin t E sin 3 t E sin 5 t

3 3

4 2e ( t ) E sin t E sin 3 t E sin 5 t

3 3

=

=

=


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(c)Line Voltage for a-connected System

If the winding of the alternator aredelta-connected (), then

the resultant e.m.f. acting round the closed mesh would be the

sum of the phase e.m.fs.

The sum of these e.m.fs. is zero for fundamental,5th,7th,11th

etc. harmonics.


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Since the third harmonics are in phase, there will be a

resultant third harmonic e.m.f. of three times the phase value

acting round the closed mesh.

It will produce a circulating current whose value will depend

on the impedance of the windings at the third harmonic

frequency.


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(c)Line Voltage for a-connected System

It means that the third harmonic e.m.f. would be short-

circuited by the windings with the result that there will be no

third harmonic voltage across the lines. The same is applicable

to all triple-n harmonic voltages.

Obviously, the line voltage will be the phase voltage but


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- .

Exercise 9.

A 3- phases generator has a generated e.m.f. of 230 V with 15

per cent third harmonic and 10 per cent fifth harmonic

content. Calculate

(i) the r.m.s. value of line voltage for Y-connection.

(ii) the r.m.s. value of line voltage for-connection.


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(d)Line Voltage for a-connected Alternator

Let the three symmetrical phase e.m.fs. of the alternator be

represented by the equations:


R 1 m 1 3 m 3 5 m 5

Y 1 m 1 3 m 3 5 m 5

e ( t ) E sin t E sin 3 t E sin 5 t

2 2 2e ( t ) E sin t E sin 3 t E sin 5 t

3 3 3

4 4 4

=

=

=

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The resultant e.m.f. acting round the -connected windings

of the armature is the sum of these e.m.fs. Hence it is given by:

B 1 m 1 3 m 3 5 m 5e t s n t s n t s n t

3 3 3

=

R Y B

3 m 3 9 m 9 15 m 15

e( t ) e ( t ) e ( t ) e ( t )

e( t ) 3 E sin 3 t 3E sin 9 t 3 E sin 15 t

=

=


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(e)Line-phase four-wire System

In this case, there will be no third harmonic component in

line voltage.

For the 4-wire system, each phase voltage (i.e. line to neutral)may contain a third harmonic component. If it is actually

present, then current will flow in theY-connected load.


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In case of balanced load , the resulting thirdharmonic linecurrents will all be in phase so that neutral wire will have to

carry three times the third harmonic line current.

There will be no current in the neutral wire either at

fundamental frequency, or any harmonic frequency other than

the triple-n frequency.


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ch1 harmonics

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