Effect of Harmonic Components on an Induced E.M.F.

Effect of Harmonic Components on an Induced E.M.F.

The flux density distribution around the air gap in all well designed alternators is symmetrical with respect to abscissa and also to polar axes. Thus it can be expressed with the help of a Fourier series which do not contain any even harmonics.

So flux density at any angle 0 from the interpolar axis is given by,

B = B_ml sin θ + B_m3 sin 3 θ + ... + B_mx sin x θ + ...

where          x = Order of harmonic component which is odd

B_ml = Amplitude of fundamental component of flux density

B_m3 = Amplitude of 3rd harmonic component of flux density

B_mx = Amplitude of X^th (odd) harmonic component of flux density

The e.m.f. generated in a conductor on the armature of a rotating machine is given by,

e_c = B l v

Substituting value of B,

e_c = [ B_ml sin θ + B_m3 sin 3 θ + ... + B_mx sin x θ + ...] l v  

l = Active length of conductor in metre

d = Diameter of the armature at the air gap

v = Linear velocity = π d n_s

Where n_s = Synchronous speed in r.p.s.

Now N_S = 120f/P

n_s = 120f/60P = 2f/P

v = π d 2f/P

Substituting in the expression for e_c,

It can be observed that the magnitude of harmonic e.m.f.s are directly proportional to their corresponding flux densities.

The R.M.S. value of resultant e.m.f. of a conductor is,

 

1. Effect of Harmonic Components on Pitch Factor

We know that,

ɑ = Angle of short pitch for fundamental flux wave

then it changes for various harmonic component of flux as,

3 ɑ = For 3rd harmonic component

5 ɑ = For 5th harmonic component

xɑ = For Xth harmonic component

Hence the pitch factor is expressed as,

K

where

K_cx = cos (xɑ/2)

where x = Order of harmonic component

 

2. Effect of Harmonic Components on Distribution Factor

Similar to the pitch factor, the distribution factor is also different for various harmonic components.

The general expression to obtain distribution factor is,

 

3. Total e.m.f. Generated due to Harmonic Components

Considering the windings to short pitch and distributed, the e.m.f. of a fundamental frequency is given by,

E_lph = 4.44K_cl K_dl  ϕ₁ f T_ph V

where T_ph = Turns per phase in series

ϕ₁ = Fundamental flux component

While the phase e.m.f. of  xth order harmonic component of frequency is given by,

E_xph = 4.44K_cx K_dx  ϕ_x f T_ph V

The total phase e.m.f. is given by,

Line e.m.f. : For star connected, the line or terminal induced e.m.f. is √3 times the total phase e.m.f. But it should be noted that with star connection, the 3rd harmonic voltages do not appear across line terminals though present in phase voltage

In delta connection also, 3rd, 9th, 15th ... harmonic voltages do not appear at the line terminals.

Taking ratio of fundamental frequency e.m.f. and xth order harmonic frequency e.m.f. we can write,

 

Example 1.14.1 An 8 pole, 3 phase, 60^o spread, double layer winding has 72 coils 72 slots. The coils are short pitched by two slots. Calculate the winding factor for the fundamental and third harmonic.

Solution :

 

Example 1.14.2 A 3 phase, 4 pole, 50 Hz, star connected alternator has an armature with a diameter of 0.25 m and length 0.3 m. The equation for flux density distribution is, B = 0.15 sin θ + 0.03 sin 3 θ + 0.02 sin 5 θ V/b/m² The armature has 60 coils with 10 turns per coil arranged in double layer. The phase spread is 60°.

a) Determine voltage generated per coil.

b) Determine voltage generated per phase.

c) Determine terminal voltage available.

Assume coil span 13/15 of a pole pitch.

Solution : The given values are , d = 0.25 m, l = 0.3 m, P = 4

A₁ = Area of each fundamental pole = πdl / P

= π × 0.25 × 0.3 / 4 =0.0589 m²

From the given equation of B we can write,

B_ml = 0.15, B_m3 = 0.03, B_m5 = 0.02

ϕ₁= Average value of fundamental flux per pole

= 2 /π B_m1 A₁ = 2/ π × 0.15 × 0.589 = 5.62 × 10^-3 Wb

 

a) To find voltage generated per coil

The R.M.S. value of the fundamental frequency e.m.f. generated in a single conductor is,

E_cl = 1.11 × 2 f ϕ₁ = 0.6238 V

Now winding coil span is 13/15 of pole pitch.

So Coil span = 13/15 × 180° = 156° i.e ɑ = 180° - 156° = 24^o

K_cl = cos ɑ/2 = 0.9781, K_cl = cos 3ɑ / 2 = 0.809, K_c5 = cos 5ɑ/2 = 0.5

The R.M.S. value of the fundamental frequency e.m.f. generated in one turn is,

E_t1 = K_cl × (2 E_cl) as two conductors constitute 1 turn

= 0.9781 × 2 × 0.6238 = 1.22 V

Note : Remember that for a single turn, K_d is unity for all harmonic components. 

There are 10 turns in each coil.

Voltage generated per coil = Number of turns per coil × Total e.m.f. per turn

= 10 × 1.24 = 12.4 V

 

b) To find voltage generated per phase

E_1ph = 4.44 K_cl K_dl ϕ₁ f T_ph

There are 60 coils with 10 turns per coil.

• Total turns = 60 × 10 = 600

T_ph = 600/3 = 200

The winding is double layer i.e. each slot has 2 coil sides.

In each coil there are 10 turns i.e. in each coil side there are 10 turns i.e. in each coil side there are 10 conductors. Hence total conductors per slot = 20 as shown in the Fig. 1.14.1.

Now total turns = 600

2 conductors constitute 1 turn

Total conductors = 2 × 600 = 1200

No . of slots = Total conductors / conductors per slot = 1200 / 20 = 60

Remember that for double layer winding, number of slots is equal to number of coils of armature. 

 

c) To determine terminal voltage

The 3rd harmonic e.m.f. does not appear at the terminals.

 

Examples for Practice

Example 1.14.3 A 3 phase 10 pole 600 rpm star connected alternator has 12 slots per pole with 8 conductors per slot and the winding is short chorded by 2 slots. The flux per pole contains a fundamental of 0.09 Wb, a third harmonic of 20 % and fifth harmonic of 10 % of the fundamental. Determine the rms values of phase and line voltages.  

[Ans.: 2971.625 V, 5121.528 V]

Example 1.14.4 A 3 phase, 6 pole, 1000 r.p.m., star connected alternator has an air gap diameter of 28 cms and a core length of 23 cms. A double layer winding is employed with 4 slots per pole per phase and 8 conductors per slot. The coils are short pitched by one slot. The flux density wave in the air gap has an amplitude of 0.87 Wb/m2 for the fundamental and 0.24 Wb/m2 for the 3rd harmonic. Estimate the r.m.s. value of resultant voltage per phase and the line voltage.

[Ans.: 383.509 V, 654.276 V] 

Example 1.14.5 The flux density distribution curve of a smooth core, 50 Hz alternator is B = sin θ + 0.2 sin 3θ + 0.2 sin 5θ + 0.2 sin 7θ Wb/m² where 0 is the angle measured from the neutral axis. The pole pitch is 35 cm and core length is 32 cm. The stator coil span is 45th the pole pitch. Find the equation for e.m.f. induced in one turn and its r.m.s. value.

[Ans.: 21.288 sin 0 + 2.6304 sin 3 0 + 2.6304 sin 7 0 V, 15.2811 V]

Example 1.14.6 A three phase, 6 pole, 1000 r.p.m. alternator has an air gap diameter of 28 cm and a core length of 23 cm. A two layer winding with 60° phase spread is used and the winding is accommodated in 4 slotsfpole/phase with 8 conductors/slot. The coils are short pitched by one slot. The flux density wave consists of a fundamental of 0.87 Wb/m2, a 3rd harmonic of 0.24 Wb/m1 and a 5th harmonic of 0.14 Wb/m1. Estimate the resultant phase and line voltages for star and delta connections.

[Ans.: E_ph = 383.654 V, E_line (star) = 654.52 V, E_line (Delta) = 377.89 V]

Review Question

1. Explain the effect of harmonics on induced e.m.f. and winding factors of an alternator. AU : Dec.-11, Marks 8