Thursday, February 7, 2008
jntu eee electrical machines unit-2 synchronous generator characteristics
UNIT – 2
SYNCHRONOUS GENERATOR CHARACTERISTICS
2.1 HARMONICS IN GENERATED EMF:
The main reason for the production of harmonics in generated e.m.f is the non-sinusoidal field flux waveform in air gap.
When the field flux waveform is sinusoidal, there would be no harmonics. The harmonics in the generated e.m.f are of two types
(1) Space harmonics
(2) Slot harmonics.
(1) Space harmonics:
The order of space harmonics are (6k±1) where k=1, 2,3…..n integers
The space harmonics are 5, 7, 11, 13….
The speed of harmonics of order 6k±1 is 1/ (6k±1) Ns =120f/p.
Therefore the speed of harmonics of order 6k+1 i.e. 7, 13, 19……. is (1/7, 1/13, 1/19…..) Ns respectively which rotate in same direction as that of fundamental
The speed of harmonics of order (6k-1) i.e. (5, 11, 17…..) is (1/5, 1/11, 1/17…) Ns which is in reverse opposite direction to that of fundamental.
(2) Slot harmonics:
Slot harmonics is caused due to variable reluctance caused by slotting the stator, rotor
Slot harmonics are of order (2s/p±1)
Example:
Spp=4, for 3-d s/p=4*3=12
The order of slot harmonics can be eliminated by skewing the armature slots with an angle of skew = 1 slot pitch.
2.2 SUPPRESSION OF HARMONICS:
As mentioned above if the field flux wave is sinusoidal then there would be no harmonics in the generated e.m.f so the field flux wave can be made sinusoidal by the following methods.
(i) By using chamfered salient poles.
(ii) At the pole center air gap should be small in at pole ends the air gap should be large in salient pole synchronous machines.
(iii) By skewing salient poles
(iv) By using short pitch armature winding, the dominant harmonics can be eliminated.
(v) By distributing the armature winding, the effect of higher order harmonics can be eliminated.
(vi) The triple harmonics can be eliminated by connecting the armature winding is Y or Δ
(vii) The slot harmonics can be removed, by using fractional slot armature winding.
(viii) By skewing the armature slots, the slot harmonics or tooth harmonics can be eliminated.
NOTE:
Ø The e.m.f equation is given by
Em=4.44Kpn Kdn Фfn7.
Ø Therefore to eliminate nth harmonic
Case (i) Kpn= 0
Cos n α/2 = 0
n α/2=90°
α=180/n
Therefore coil should be short pitched by (180/n)°
Coil span =180-α
= 180-180/n
= 180 (n-1/n)
Ex: - (i) to eliminate 5th harmonic α= 180/5= 36°
β= 180-36=144°
Case (ii) Kdn =0
Sin (m n γ/2)/m sin (nγ/2) =0
m n γ/2 =180
m γ =360/n
= phase spread.
Questions:
1) What are the two types of harmonics present in generated e.m.f?
2) What are order of space harmonics and slot harmonics?
3) What should be the short pitch angle in order to eliminate both 5th and 7th harmonics as much as possible?
2.4 PHASE DIAGRAM OF CYLINDRICAL ROTOR ALTERNATOR:
To draw the phase diagram, it is assumed that the per phase quantities Vt (terminal voltage), Ia (armature leakage reactance), ra (armature resistance), Xal (armature leakage reactance) cosθ (load power factor) are known. Here θ is the angle between Vt and Ia.
Ø Armature current (Ia) lags terminal voltage (Vt) by θ.
Ø Ia ra drop is in phase with Ia and Iaxar is perpendicular to Ia. when these quantities are added to terminal voltage behind leakage impedance Ef.
Vt +Iara+IaXal = Er
Ø This air gap voltage Er is induced by flux Фr which leads it by 90°
Ø In uniform air-gap machine, the flux Фr is in phase with resultant m.m.f Fr which is the phasor sum of field m.m.f Fa
· Fr =Ff + Fa
Ø The armature reaction m.m.f Fa is in phase with armature current Ia. at no load Ia =0. Then Fa =0 which implies Fr=Ff. At this condition, the flux phase Фr align along Ff. the e.m.f generated at no load lags Ff by 90° which is indicated by Ef.
Ø Ef is also called no-load voltage or excitation voltage’s Ψ is the angle between Ef and Ia which is called internal p.f angle
Ø Ψ is the angle between Ef and Ia which is called internal p.f angle.
Ø δ is the angle between Ef and Vt which is called load angle or power angle.
Ø The angle between Ef and Er is same as the angle between Ff and Fr.
2.5 Open circuit and short circuit characteristics
2.5.1 Open circuit characteristics (o.c.c)
To obtain the o.c.c, the alternators open circuit terminal voltage is noted as the field current is varied gradually from zero, driving the alternator at rated speed. The circuit diagram for obtain o.c.c is draw below
Ø Ef is noted as If is increased gradually from zero driving the alternator at rated speed. Therefore o.c.c is graph which is plotted taking If on x-axis and Ef on y-axis
Ø The final value of Ef should be about 125% of the rated voltage.
Ø The o.c.c is straight line for low values of If, but at higher values, it bends because of saturation in the iron part of the magnetic circuit.
Ø At small values of Ff, the air gap requires whole m.m.f and m.m.f required by iron is zero.
Ø But at large values of Ff for equal to ‘oa’, m.m.f requires by iron is ‘bc’, as shown in figure 6.2(b) where ‘ab’ is the m.m.f required by air gap.
Ø When m.m.f required by iron is assumed to be zero, then the graph between Ef and If is called “air gap line”
2.5.2 Short circuit characteristics (s.c.c)
To obtain the s.c.c, the alternator is driven at rated speed and the armature terminals are short circuited through an ammeter.
If is increased gradually from zero till Isc reaches 125%-150% of rated current. The connection for obtaining s.c.c is shown below.
Ø When terminals are short circuited Vt =0
o Er = Iara+jIaXal
Ø ra is very small compared to Xal. Therefore Er leads Ia by nearly 90°
Ø Fa is almost opposite to Ff and therefore it is entirely demagnetizing in nature.
Ø Saturation doesn’t occur under short circuit conditions because (Ff-Fa) is almost equal to Fr and hence the resultant air gap flux is decreased.
TOPICS 2.4, 2.5
1) Draw the phasor diagram of cylindrical rotor alternator
2) Draw the circuit connections to draw O.C.C and S.C.C
3) Why O.C.C bends at higher values of field current but S.C.C does not bend?
2.6 Zero power factor characteristics
Ø Z.p.f characteristics is plotted between armature terminal voltage and its field current, when the speed and armature current of alternator is kept constant.
Ø The alternator is driven at rated and a purely inductive load is connected across the armature terminals. Then If is increased till full load armature current flows.
Ø Then the load is varied in steps such that always rated armature current flows by varying the field current.
Ø The z.p.f characteristics and phasor diagram under zero-power-factor over excited conditions
Terminal voltage VT=Er-IaXal and
Field m.m.f Ff =Fr+ Fa.
For field excitation equal to op, the no-load voltage is Pk (from o.c.c) when a purely inductive load is connected at the armature terminals, then the net excitation is of, which is less than the net excitation is of, which is less than the net excitation is of which is less than op by Fa (armature m.m.f) . The air gap voltage for field m.m.f of is Fc. When drop due to leakage reactance is subtracted from f we get terminal voltage which is equal to Fb. Since z.p.f characteristics is a plot between terminal voltage Vt and field m.m.f Ff it is not changed from no-load value of op. therefore point aliens on z.p.f. The triangle ABC is called potier triangle.
Where CB=IaXar
BA=Fa
Since the armature current at which point a is obtained can be known, the value of leakage reactance xal can be calculated from the potier triangle
Ø Draw DB parallel to f’o such that AD=f’o. Now draw DC parallel to air gap line.
Ø Therefore z.p.f characteristics are nothing but the o.c.c dropped by leakage reactance drop IaXal and shifted horizontal right by armature m.m.f Fa.
2.7 Concept of synchronous reactance and synchronous impedance
For a cylindrical rot synchronous machine as we studied earlier
Fr = Ff + Fa--------------> (1)
Er = Ef + Ear--------------> (2)
The m.m.f can be replaced by their corresponding e.m.f.
Since the corresponding e.m.f. lags their field m.m.f by 90(as they can expressed as
Ef = -jkFf -----> (3)
Er = -jkFr--------> (4) (where k=Ef/Ff)
Ear = -jkFa------> (5)
Here k is the slope of air-gap line.
But Fa = C Ia
Ear = -j Ck Ia --- à(6)
Substituting equation (6) in (7)
Er= Ef-j Ck Za
Ef =Er +j Ck Za -- à(7)
From the phasor
Er = Vt +Ia (ra+ jXal)
From (7) Ef = Er+j CkZa
Ef =Er-Vt+Ia ra+j Ia (xal+Ck) ----------- à(8)
Ck has dimensions of reactance due to Fa.
There fore Ear = -j K Fa
= -j k (C Ia)
= -j Xar Ia.
The total equivalent reactance Xal +Xar = Xs
Where Xs is called synchronous reactance.
The term ra = j Xs = Zs
Where Zs is called the synchronous impedance of cylindrical rotor synchronous machine.
From equation (8)
Ef = Vt +Iara +j Za Xs
= Vt + za (ra +j xs)
Ef = Vt + Ia Zs------- à(9)
Therefore the equivalent circuit for cylindrical rotor synchronous machine is drawn as shown below
Experimental determination of Xs and Zs
Ø O.C.C and SCC are required for the experimental determination of Zs.
In open circuit test Ia=0 which implies VT=Ef.
In short circuit test, entire e.m.f is wasted in circulating the short circuit current Is through the synchronous impedance
Zs = Ef \Isc = open circuit terminal voltage \short circuit current
The equivalent circuit under short circuit test and its phase or diagram is drawn below.
For a field current of OC amperes
Zs (ohms) = CD = AC (volts) \BC (amperes)
We can see that Zs curve is not a straight line. It decreases for high field current due to the onset of saturation in O. C. C.
From the formula Zs = √ra2+xs2
Xs = √zs2 –ra2
From the ammeter voltmeter method, d.c resistance r d.c can be calculated
rd.c = ½ voltmeter reading /ammeter reading (when armature winding is star connected)
And rd.c = 3/2 voltmeter reading / ammeter reading (when armature winding is delta connected)
ra = 1.3 rd.c
So by knowing Zs and ra, Xs can be calculated from the formula
Xs = √ (zs) 2 – (ra) 2
TOPICS 2.6, 2.7
1) What is potier triangle?
2) How leakage reactance can be determined using potier method?
3) How synchronous reactance can be determined experimentally?
4) Draw the equivalent circuit for cylindrical rotor synchronous machine shows synchronous reactance, leakage reactance
2.8 Load Characteristics
The load characteristics of an alternator is plot between armature terminal voltage VT and field current If for a constant armature current Ia.
The load characteristic of an alternator which is driven at constant speed is drawn below.
Figure 2.9
2.9 EXTERNAL LOAD CHARACTERISTICS
The external load characteristics of an alternator are a plot of armature terminal voltage Vt with armature current when field current, p.f and speed are kept constant.
TOPICS 2.8, 2.9
1) Draw the load characteristics of an alternator
2) Draw the circuit diagram to determine external load characteristics
LONG ANSWER QUESTIONS
1) Explain how2 harmonics are generated and how it can be suppressed?
Ans (refer section 2.1, 2.2)
2) Draw and explain the components of phasor diagram of cylindrical rotor alternator
Ans (refer section 2.4)
3) Write a short notes about the following
A) O.C.C
B) S.C.C
C) Z.p.f characteristics
Ans (refer section 2.5, 2.6)
4) Explain the concept of synchronous reactance and synchronous impedance and how they can be determined experimentally
Ans (refer section 2.7)
Multiple choice questions
1) The order of space harmonic is
a. (4k±1)
b. (6k±1)
c. (8k±1)
d. (12k±1) Ans: -(b)
2) Which of the following is not a space harmonic
a. 5th
b. 13th
c. 9th
d. 17th Ans: -(c)
3) The speed of 5th harmonic is
a. 5Ns
b. 25Ns
c. 1/5Ns
d. 1/25Ns Ans: -(c)
4) For a 3Φ,2-pole,12 slot alternator, the order slot harmonics are
a. 11,13
b. 5,7
c. 17,19
d. 23,25 Ans: -(a)
5) What adjustment is made in the field side of an alternator for the suppression of harmonics
a. Chamfering salient poles
b. Skewing armature slots
c. Distributed winding
d. Short pitch winding Ans: -(a)
6) _________harmonic can be eliminated by connecting the armature winding in star (or) delta.
a. Dominant
b. Higher order
c. Triple
d. Slot Ans: -(c)
7) Higher order harmonics are removed from generated e.m.f of alternator by using
a. Distributed winding
b. Chording winding
c. Fractional winding
d. Skewing armature slots. Ans: -(a)
8) The short pitch angle to eliminate both 5th and 7th harmonics is
a. 36°
b. 25°
c. 30°
d. 45° Ans: -(c)
9) The phase spread of distributed winding to eliminate 9th harmonic is
a. 20°
b. 40°
c. 60°
d. 90° Ans: -(b)
10) What is the coil span when the armature is short pitched to eliminate 5th harmonic
a. 36°
b. 72°
c. 216°
d. 144° Ans: -(d)
11) Which of the following about cylindrical rotor alternator is true
a. Vt = Er + Iara + jZaXal
b. Ef = Vt + Iara
c. Er = Vt + Iara +j ZaXal
d. Vt = Ef + Iara + jIaXal Ans: -(c)
12) Er____________ Fr by ___________ in a cylindrical rotor alternator
a. Lags, >90°
b. Leads, <90°>20A
c. <20a>30A
d. 40A Ans: -(c)
20) In the above question if the load is pure capacitive then the load current is
a. 30A
b. <30a>30A
d. 40A Ans: -(b)
21) While plotting z.p.f characteristics of alternator is not kept constant
a. If (field current)
b. Ia (armature current)
c. P.f (power factor)
d. N (speed) Ans: -(a)
22) From potier triangle we can find
a. Armature resistance
b. Armature leakage resistance
c. Synchronous reactance
d. Synchronous impedance Ans: -(b)
23) Z.p.f is plotted for
a. Ef Vs If
b. Vt Vs If
c. Ef Vs Ia
d. Vt VS Ia Ans: -(b)
24) Which of the following pertaining to cylindrical rotor alternator is true
a. Er = Vt+jIa Xar
b. Ef = Vt+Ia(ra+j Xs)
c. Xal + Xar = Xs
d. All the above Ans: -(d)
25) The z.p.f characteristics for the potier diagram can be obtained by loading the alternator with
a. Induction motor
b. Synchronous motor
c. Dc motor
d. Reluctance motor Ans: -(a)
26) OC test is conducted on an alternator at
a. Nominal flux
b. Reduced flux
c. Zero armature current
d. Nominal current Ans: -(a)
27) In a cylindrical motor alternator
a. Air-gap voltage lags the field by 90°
b. Armature reactance due to lagging p.f is purely magnetizing
c. Air-gap voltage leads terminal voltage
d. Ef lags Ff by an angle greater than 90° Ans:- (c)
28) Potier reactance of an alternator is same as that armature
a. Leakage reactance
b. Field winding leakage reactance
c. Synchronous reactance
d. Armature resistance Ans: -(b)
29) Load characteristics of an alternator is a plot between
a. Vt VS Ia
b. Vt VS If
c. Ef VS Ia
d. Ef VS If Ans:-(b)
30) Which of the following is not kept constant for plotting external load characteristics of alternator
a. Field current
b. Armature current
c. Power factor
d. Speed Ans: -(b)
31) The effect of armature reaction on in main field flux under U.p.f load is
a. Purely magnetization
b. Purely cross magnetization
c. Purely demagnetization
d. Partly magnetization + partly cross magnetization Ans:- (b)
32) The effect of armature reaction on main field flux under z.p.f lead for an alternator is
a. Partly magnetization + partly cross magnetization
b. Partly demagnetization + partly cross magnetization
c. Purely magnetization
d. Purely demagnetization Ans: -(c)
33) If the effect of armature reaction is purely demagnetization then the load connected is
a. U.p.f load
b. Purely inductive load
c. Purely capacitive load
d. 0.8 p.f lag Ans: -(b)
SYNCHRONOUS GENERATOR CHARACTERISTICS
2.1 HARMONICS IN GENERATED EMF:
The main reason for the production of harmonics in generated e.m.f is the non-sinusoidal field flux waveform in air gap.
When the field flux waveform is sinusoidal, there would be no harmonics. The harmonics in the generated e.m.f are of two types
(1) Space harmonics
(2) Slot harmonics.
(1) Space harmonics:
The order of space harmonics are (6k±1) where k=1, 2,3…..n integers
The space harmonics are 5, 7, 11, 13….
The speed of harmonics of order 6k±1 is 1/ (6k±1) Ns =120f/p.
Therefore the speed of harmonics of order 6k+1 i.e. 7, 13, 19……. is (1/7, 1/13, 1/19…..) Ns respectively which rotate in same direction as that of fundamental
The speed of harmonics of order (6k-1) i.e. (5, 11, 17…..) is (1/5, 1/11, 1/17…) Ns which is in reverse opposite direction to that of fundamental.
(2) Slot harmonics:
Slot harmonics is caused due to variable reluctance caused by slotting the stator, rotor
Slot harmonics are of order (2s/p±1)
Example:
Spp=4, for 3-d s/p=4*3=12
The order of slot harmonics can be eliminated by skewing the armature slots with an angle of skew = 1 slot pitch.
2.2 SUPPRESSION OF HARMONICS:
As mentioned above if the field flux wave is sinusoidal then there would be no harmonics in the generated e.m.f so the field flux wave can be made sinusoidal by the following methods.
(i) By using chamfered salient poles.
(ii) At the pole center air gap should be small in at pole ends the air gap should be large in salient pole synchronous machines.
(iii) By skewing salient poles
(iv) By using short pitch armature winding, the dominant harmonics can be eliminated.
(v) By distributing the armature winding, the effect of higher order harmonics can be eliminated.
(vi) The triple harmonics can be eliminated by connecting the armature winding is Y or Δ
(vii) The slot harmonics can be removed, by using fractional slot armature winding.
(viii) By skewing the armature slots, the slot harmonics or tooth harmonics can be eliminated.
NOTE:
Ø The e.m.f equation is given by
Em=4.44Kpn Kdn Фfn7.
Ø Therefore to eliminate nth harmonic
Case (i) Kpn= 0
Cos n α/2 = 0
n α/2=90°
α=180/n
Therefore coil should be short pitched by (180/n)°
Coil span =180-α
= 180-180/n
= 180 (n-1/n)
Ex: - (i) to eliminate 5th harmonic α= 180/5= 36°
β= 180-36=144°
Case (ii) Kdn =0
Sin (m n γ/2)/m sin (nγ/2) =0
m n γ/2 =180
m γ =360/n
= phase spread.
Questions:
1) What are the two types of harmonics present in generated e.m.f?
2) What are order of space harmonics and slot harmonics?
3) What should be the short pitch angle in order to eliminate both 5th and 7th harmonics as much as possible?
2.4 PHASE DIAGRAM OF CYLINDRICAL ROTOR ALTERNATOR:
To draw the phase diagram, it is assumed that the per phase quantities Vt (terminal voltage), Ia (armature leakage reactance), ra (armature resistance), Xal (armature leakage reactance) cosθ (load power factor) are known. Here θ is the angle between Vt and Ia.
Ø Armature current (Ia) lags terminal voltage (Vt) by θ.
Ø Ia ra drop is in phase with Ia and Iaxar is perpendicular to Ia. when these quantities are added to terminal voltage behind leakage impedance Ef.
Vt +Iara+IaXal = Er
Ø This air gap voltage Er is induced by flux Фr which leads it by 90°
Ø In uniform air-gap machine, the flux Фr is in phase with resultant m.m.f Fr which is the phasor sum of field m.m.f Fa
· Fr =Ff + Fa
Ø The armature reaction m.m.f Fa is in phase with armature current Ia. at no load Ia =0. Then Fa =0 which implies Fr=Ff. At this condition, the flux phase Фr align along Ff. the e.m.f generated at no load lags Ff by 90° which is indicated by Ef.
Ø Ef is also called no-load voltage or excitation voltage’s Ψ is the angle between Ef and Ia which is called internal p.f angle
Ø Ψ is the angle between Ef and Ia which is called internal p.f angle.
Ø δ is the angle between Ef and Vt which is called load angle or power angle.
Ø The angle between Ef and Er is same as the angle between Ff and Fr.
2.5 Open circuit and short circuit characteristics
2.5.1 Open circuit characteristics (o.c.c)
To obtain the o.c.c, the alternators open circuit terminal voltage is noted as the field current is varied gradually from zero, driving the alternator at rated speed. The circuit diagram for obtain o.c.c is draw below
Ø Ef is noted as If is increased gradually from zero driving the alternator at rated speed. Therefore o.c.c is graph which is plotted taking If on x-axis and Ef on y-axis
Ø The final value of Ef should be about 125% of the rated voltage.
Ø The o.c.c is straight line for low values of If, but at higher values, it bends because of saturation in the iron part of the magnetic circuit.
Ø At small values of Ff, the air gap requires whole m.m.f and m.m.f required by iron is zero.
Ø But at large values of Ff for equal to ‘oa’, m.m.f requires by iron is ‘bc’, as shown in figure 6.2(b) where ‘ab’ is the m.m.f required by air gap.
Ø When m.m.f required by iron is assumed to be zero, then the graph between Ef and If is called “air gap line”
2.5.2 Short circuit characteristics (s.c.c)
To obtain the s.c.c, the alternator is driven at rated speed and the armature terminals are short circuited through an ammeter.
If is increased gradually from zero till Isc reaches 125%-150% of rated current. The connection for obtaining s.c.c is shown below.
Ø When terminals are short circuited Vt =0
o Er = Iara+jIaXal
Ø ra is very small compared to Xal. Therefore Er leads Ia by nearly 90°
Ø Fa is almost opposite to Ff and therefore it is entirely demagnetizing in nature.
Ø Saturation doesn’t occur under short circuit conditions because (Ff-Fa) is almost equal to Fr and hence the resultant air gap flux is decreased.
TOPICS 2.4, 2.5
1) Draw the phasor diagram of cylindrical rotor alternator
2) Draw the circuit connections to draw O.C.C and S.C.C
3) Why O.C.C bends at higher values of field current but S.C.C does not bend?
2.6 Zero power factor characteristics
Ø Z.p.f characteristics is plotted between armature terminal voltage and its field current, when the speed and armature current of alternator is kept constant.
Ø The alternator is driven at rated and a purely inductive load is connected across the armature terminals. Then If is increased till full load armature current flows.
Ø Then the load is varied in steps such that always rated armature current flows by varying the field current.
Ø The z.p.f characteristics and phasor diagram under zero-power-factor over excited conditions
Terminal voltage VT=Er-IaXal and
Field m.m.f Ff =Fr+ Fa.
For field excitation equal to op, the no-load voltage is Pk (from o.c.c) when a purely inductive load is connected at the armature terminals, then the net excitation is of, which is less than the net excitation is of, which is less than the net excitation is of which is less than op by Fa (armature m.m.f) . The air gap voltage for field m.m.f of is Fc. When drop due to leakage reactance is subtracted from f we get terminal voltage which is equal to Fb. Since z.p.f characteristics is a plot between terminal voltage Vt and field m.m.f Ff it is not changed from no-load value of op. therefore point aliens on z.p.f. The triangle ABC is called potier triangle.
Where CB=IaXar
BA=Fa
Since the armature current at which point a is obtained can be known, the value of leakage reactance xal can be calculated from the potier triangle
Ø Draw DB parallel to f’o such that AD=f’o. Now draw DC parallel to air gap line.
Ø Therefore z.p.f characteristics are nothing but the o.c.c dropped by leakage reactance drop IaXal and shifted horizontal right by armature m.m.f Fa.
2.7 Concept of synchronous reactance and synchronous impedance
For a cylindrical rot synchronous machine as we studied earlier
Fr = Ff + Fa--------------> (1)
Er = Ef + Ear--------------> (2)
The m.m.f can be replaced by their corresponding e.m.f.
Since the corresponding e.m.f. lags their field m.m.f by 90(as they can expressed as
Ef = -jkFf -----> (3)
Er = -jkFr--------> (4) (where k=Ef/Ff)
Ear = -jkFa------> (5)
Here k is the slope of air-gap line.
But Fa = C Ia
Ear = -j Ck Ia --- à(6)
Substituting equation (6) in (7)
Er= Ef-j Ck Za
Ef =Er +j Ck Za -- à(7)
From the phasor
Er = Vt +Ia (ra+ jXal)
From (7) Ef = Er+j CkZa
Ef =Er-Vt+Ia ra+j Ia (xal+Ck) ----------- à(8)
Ck has dimensions of reactance due to Fa.
There fore Ear = -j K Fa
= -j k (C Ia)
= -j Xar Ia.
The total equivalent reactance Xal +Xar = Xs
Where Xs is called synchronous reactance.
The term ra = j Xs = Zs
Where Zs is called the synchronous impedance of cylindrical rotor synchronous machine.
From equation (8)
Ef = Vt +Iara +j Za Xs
= Vt + za (ra +j xs)
Ef = Vt + Ia Zs------- à(9)
Therefore the equivalent circuit for cylindrical rotor synchronous machine is drawn as shown below
Experimental determination of Xs and Zs
Ø O.C.C and SCC are required for the experimental determination of Zs.
In open circuit test Ia=0 which implies VT=Ef.
In short circuit test, entire e.m.f is wasted in circulating the short circuit current Is through the synchronous impedance
Zs = Ef \Isc = open circuit terminal voltage \short circuit current
The equivalent circuit under short circuit test and its phase or diagram is drawn below.
For a field current of OC amperes
Zs (ohms) = CD = AC (volts) \BC (amperes)
We can see that Zs curve is not a straight line. It decreases for high field current due to the onset of saturation in O. C. C.
From the formula Zs = √ra2+xs2
Xs = √zs2 –ra2
From the ammeter voltmeter method, d.c resistance r d.c can be calculated
rd.c = ½ voltmeter reading /ammeter reading (when armature winding is star connected)
And rd.c = 3/2 voltmeter reading / ammeter reading (when armature winding is delta connected)
ra = 1.3 rd.c
So by knowing Zs and ra, Xs can be calculated from the formula
Xs = √ (zs) 2 – (ra) 2
TOPICS 2.6, 2.7
1) What is potier triangle?
2) How leakage reactance can be determined using potier method?
3) How synchronous reactance can be determined experimentally?
4) Draw the equivalent circuit for cylindrical rotor synchronous machine shows synchronous reactance, leakage reactance
2.8 Load Characteristics
The load characteristics of an alternator is plot between armature terminal voltage VT and field current If for a constant armature current Ia.
The load characteristic of an alternator which is driven at constant speed is drawn below.
Figure 2.9
2.9 EXTERNAL LOAD CHARACTERISTICS
The external load characteristics of an alternator are a plot of armature terminal voltage Vt with armature current when field current, p.f and speed are kept constant.
TOPICS 2.8, 2.9
1) Draw the load characteristics of an alternator
2) Draw the circuit diagram to determine external load characteristics
LONG ANSWER QUESTIONS
1) Explain how2 harmonics are generated and how it can be suppressed?
Ans (refer section 2.1, 2.2)
2) Draw and explain the components of phasor diagram of cylindrical rotor alternator
Ans (refer section 2.4)
3) Write a short notes about the following
A) O.C.C
B) S.C.C
C) Z.p.f characteristics
Ans (refer section 2.5, 2.6)
4) Explain the concept of synchronous reactance and synchronous impedance and how they can be determined experimentally
Ans (refer section 2.7)
Multiple choice questions
1) The order of space harmonic is
a. (4k±1)
b. (6k±1)
c. (8k±1)
d. (12k±1) Ans: -(b)
2) Which of the following is not a space harmonic
a. 5th
b. 13th
c. 9th
d. 17th Ans: -(c)
3) The speed of 5th harmonic is
a. 5Ns
b. 25Ns
c. 1/5Ns
d. 1/25Ns Ans: -(c)
4) For a 3Φ,2-pole,12 slot alternator, the order slot harmonics are
a. 11,13
b. 5,7
c. 17,19
d. 23,25 Ans: -(a)
5) What adjustment is made in the field side of an alternator for the suppression of harmonics
a. Chamfering salient poles
b. Skewing armature slots
c. Distributed winding
d. Short pitch winding Ans: -(a)
6) _________harmonic can be eliminated by connecting the armature winding in star (or) delta.
a. Dominant
b. Higher order
c. Triple
d. Slot Ans: -(c)
7) Higher order harmonics are removed from generated e.m.f of alternator by using
a. Distributed winding
b. Chording winding
c. Fractional winding
d. Skewing armature slots. Ans: -(a)
8) The short pitch angle to eliminate both 5th and 7th harmonics is
a. 36°
b. 25°
c. 30°
d. 45° Ans: -(c)
9) The phase spread of distributed winding to eliminate 9th harmonic is
a. 20°
b. 40°
c. 60°
d. 90° Ans: -(b)
10) What is the coil span when the armature is short pitched to eliminate 5th harmonic
a. 36°
b. 72°
c. 216°
d. 144° Ans: -(d)
11) Which of the following about cylindrical rotor alternator is true
a. Vt = Er + Iara + jZaXal
b. Ef = Vt + Iara
c. Er = Vt + Iara +j ZaXal
d. Vt = Ef + Iara + jIaXal Ans: -(c)
12) Er____________ Fr by ___________ in a cylindrical rotor alternator
a. Lags, >90°
b. Leads, <90°>20A
c. <20a>30A
d. 40A Ans: -(c)
20) In the above question if the load is pure capacitive then the load current is
a. 30A
b. <30a>30A
d. 40A Ans: -(b)
21) While plotting z.p.f characteristics of alternator is not kept constant
a. If (field current)
b. Ia (armature current)
c. P.f (power factor)
d. N (speed) Ans: -(a)
22) From potier triangle we can find
a. Armature resistance
b. Armature leakage resistance
c. Synchronous reactance
d. Synchronous impedance Ans: -(b)
23) Z.p.f is plotted for
a. Ef Vs If
b. Vt Vs If
c. Ef Vs Ia
d. Vt VS Ia Ans: -(b)
24) Which of the following pertaining to cylindrical rotor alternator is true
a. Er = Vt+jIa Xar
b. Ef = Vt+Ia(ra+j Xs)
c. Xal + Xar = Xs
d. All the above Ans: -(d)
25) The z.p.f characteristics for the potier diagram can be obtained by loading the alternator with
a. Induction motor
b. Synchronous motor
c. Dc motor
d. Reluctance motor Ans: -(a)
26) OC test is conducted on an alternator at
a. Nominal flux
b. Reduced flux
c. Zero armature current
d. Nominal current Ans: -(a)
27) In a cylindrical motor alternator
a. Air-gap voltage lags the field by 90°
b. Armature reactance due to lagging p.f is purely magnetizing
c. Air-gap voltage leads terminal voltage
d. Ef lags Ff by an angle greater than 90° Ans:- (c)
28) Potier reactance of an alternator is same as that armature
a. Leakage reactance
b. Field winding leakage reactance
c. Synchronous reactance
d. Armature resistance Ans: -(b)
29) Load characteristics of an alternator is a plot between
a. Vt VS Ia
b. Vt VS If
c. Ef VS Ia
d. Ef VS If Ans:-(b)
30) Which of the following is not kept constant for plotting external load characteristics of alternator
a. Field current
b. Armature current
c. Power factor
d. Speed Ans: -(b)
31) The effect of armature reaction on in main field flux under U.p.f load is
a. Purely magnetization
b. Purely cross magnetization
c. Purely demagnetization
d. Partly magnetization + partly cross magnetization Ans:- (b)
32) The effect of armature reaction on main field flux under z.p.f lead for an alternator is
a. Partly magnetization + partly cross magnetization
b. Partly demagnetization + partly cross magnetization
c. Purely magnetization
d. Purely demagnetization Ans: -(c)
33) If the effect of armature reaction is purely demagnetization then the load connected is
a. U.p.f load
b. Purely inductive load
c. Purely capacitive load
d. 0.8 p.f lag Ans: -(b)
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