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Electrical Machines
Sheet-3 Synchronous
Synchronous Generator)
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A. The speed of rotation:
 = 
B. Equivalent ckt.
Fig.1 The per phase equivalent circuits
∅ =  − −  − 
 =  + 
∅ =  −  − 
+(  ) for lagging PF
 == ((∅ +− )

∅  )+(  ) for leading PF
 = ∅+() for unity P
C. Th
Thee ph
phas
asor
or di
diagr
agram
am:: Un
Uniity, lagging & leading PF
Fig.2 A simple circuits
D. Power Stage
E. POWER AND TORQUE IN SYNCHRONOUS GENERATORS
 =  … … … (1)
= 3
 = √3 …..
…..((3)
In phase quantity,
 = 3∅…….(4)
 = √3 
 = 3∅
ignored
d (sinc
(sincee Xs
Xs»
»R
RA
A ), then
then a very
very us
usefu
efull eq
equat
uation can be derived
If the armature
armature re
resista
sistance
nce RA is ignore
of the
the gene
nera
rato
torr. To
To der
deriive this
this equa
uattion, e amine the phasor
to ap
appr
prox
oxim
imat
atee the
the out
utpu
putt po
po er of
simplifie
fied
d phasor
phasor diag
diagram
ram of
of a genera
generator
tor with
with t e stator resistance
diagr
dia
gram
am in Fi
Figu
gure.
re. Figu
Figure
re sho
sho s a simpli
.
or
ignored. Notice that the vertical segment be can be expressed as either
Therefore,
= 
 an
  
substituting this expression into Equation 4 gives
∅…… ……………………………….(5)
 =  = 
Eq
produced
ed by a sync
synchrono
hronous
us gener
generato
atorr depe
depen
n s on the angle 
Equa
uati
tion
on 5 ssho
hows
ws th
that
at th
thee po
power produc
known aass the interna
internall ang
angle
le or torque
torque angle
angle of t e machine. Notice
between ∅ and  . The angle is known
also that the maximum power hat the generator can supply occurs when
degree , sin = 1, and
 = 90

 = 3∅
  ; where  =2 (
And,  = ∅ 
))
((
))   = 
 rad/sec

F. Measuring model paramet rs

  − )
 ==√(
egree. At
 = 90
G. Effect of load changes of Sy chronous Generator
H. Ef
Effe
fect
ct of fiel
field
d curr
curren
entt ch
chan
an es in a Synchronous Generator
1. Calculate the pitch factor for the under given windings: (a) 36 slots, 4 poles, coil span 1-8 (b) 72
stator slots, 6 poles, coil span 1 to 10 and (c) 96 slots, 6 poles, coil span 1 to 12.
Solution:
(a)
 =  = 9
Coil span 1-8 = 7

= 
 ×180

 = 
==
 40  = 0.9397
 = 12
(b)  =

Coil span 1-10 = 9
 =  ×180=45
 =  = 0.925
 =  = 16
Coil span 1-12 = 11
 =  ×180=56.25
 =  = 0.882
(c)

 = 
 

 =  − 


×180

 = 2
m = slots/pole/phase
 = 1 ×180


 = 
2. Calculate the distribution factor for a 36 slots, 4 pole,
pole, single layer 3 phase winding.
Solution:


 =  =9 =  =3
 ×

 =  =   = 0.96

=  ×180 =  ×180 = 20
3. Find the value of Kd for an alternator with 9 slots per pole for the following cases: (i) One winding
in all the slots (ii) One winding using only the first 2/3 of the slots/pole (iii) 3 equal windings placed
sequentially in 60 degree group.
Solution: Here,
(i)
 =  ×180 =  ×180 = 20 and values of m are 9, 6 and 3 respectively.
×
 = ×
  = 0.64
m= 9,  = 20,  =

 


(ii)
×

  = 0.83

m= 6,  = 20 ,  =

 
(iii)
× = 0.96
m= 3,  = 20,  =


4. An alternator has 18 slots/pole and the first coil lies in sslots
lots 1 to 16. Calculate the pitch factor for
rd
th
th
(i) fundamental, (ii) 3 harmonic, (iii) 5 harmonic (iv) 7 harmonic.
Solution:
=18
Here coil span is 16-1=15 slots, which falls short by 3 slots.
 
 = (18-15)x 

 =(i)  ×180
=  = 0.97 = 30
(ii)
 =  = 0.707
(iii)
 =   = 0.259
(iv)
 =  = 0.259
5. A 3 phase
phase 16 pole aalternator
lternator has a Y connected winding with
with 144 slots and 10 conductors per sslot.
lot.
The flux per pole is 0.03 Wb, sinusoidally distributed and the speed is 375 RPM. Find the frequency,
the phase and line EMF. Assume full pitched
p itched coil.
Solution:
 = ×
 = 
 =  ×180 = 20
 = 50 Hz.  = 
 =9 =  = 3
480 =240  = 1
 = 144×10
=480
=
3
2
×

 =   = 0.96

 =4.44∅ = 4.44×50×0.03×240×1×0.96 = 1534.46 V
534.46=
4 6 = 2658 
 = √3 × 1534
6. Find the no load phase and line voltage of a star connected
connected 3 phase, 6 pole alternator which runs at
1200 RPM having flux per pole of 0.1 Wb sinusoidally distributed. Its stator has 54 slots having
double layer winding. Each coil has 8 turns and the coil is chorded by 1 slot.
 = × = 60 Hz.
 = 

 = 20.
Since winding is chorded by one slot, it is short pitched by 1/9 or  =


×
 =  = 0.98;  =   = 0.96
Solution:
 =  = 9
m = 9/3 = 3,

 = × =144, =  = 72
 =4.44∅ = 4.44×60×0.1×72×0.98×0.96 = 1805 V
=
 √3 × 1805 = 3125 
7. Stator of a 3 phase, 16 pole alternator has 144 slots and there are 4 conductors per slot connected
in two layers and the conductors of each phase are connected in series. If the speed of the alternator is
375 rpm, calculate the emf induced per phase. Resultant flux in the air gap is
sinusoidally distributed. Assume the coil span as
electrical.
5×10 /
150
Solution:
 =  = 9  =  = 3 =144×4=576
×
 = 96
= ×
ℎ
ℎ


=
,

= 50 .



180



 = 180
180 −150 = 30
 = 9 = 20
×



 =   = 0.96
 =  = 0.966

 =4.44∅ = 4.44×50× 5 ×0.01×96×0.966
×0.96 = 988 V
For sinusoidal,
8. A 4-pole, 3-phase, 50-Hz, star-connected alternator has 60 slots, with 4 conductors per slot.
Coils are short-pitched by 3 slots. If the phase spread is 60º, find the line voltage induced for a flux
per pole of 0.943 Wb distributed sinusoidally in space. All the turns
turn s per phase are in series.
Solution:
 =  = 15 f = 50 Hz. m = 15/3 = 5  =  ×180=36  =  = 12
×



 =   = 0.957
 =  = 0.951
 = × = 80

80
 = 2 = 40
∅ = 0.943 

 =4.44∅ = 4.44×50×0.943×40×0.951×0.957 = 7621 V
 = √3 × 7621 = 13200 
ℎ
9.
A 4-pole, 50-Hz, star-connected alternator has 15 slots per pole and each slot has 10 conductors.
All
the conductors
of eachflux
phase
connected
in series'
the winding
factor
When are
running
on no-load
for a certain
perare
pole,
the terminal
e.m.f.
was 1825
volt.being
If the0.95.
windings
lapconnected as in a d.c. machine, what would be the e.m.f. between the brushes for the same speed and
the same flux/pole. Assume sinusoidal distribution of flux.
Solution:
n = 15, m = 15/3 = 5
E(phase) = 1825/1.7320 = 1053.695 V
No. of turn per phase = (20x10)/2 =
100
Kd = 0.95,
Kp = 1 (Assume)
∅10010.95
∅=49.96 .
When connected as dc generator, Eg = (∅/60)=(49.960.0016001500)/60=750  ;
No. of slot per phase = (15x4)/3 = 20,
E(ph) = 4.44x50x
Where, A=P, Z= 15x4x10 = 600, N = (120x50)/4 = 1500 RPM.
10. An alternator on open-circuit generates 360
360 V at
at 60 Hz when the field current is 3.6 A. Neglecting
saturation, determine the open-circuit e.m.f. when the frequency is 40 Hz and the field current is 2.4
Amp.
[BAPEX-16, DNCC-16, APSCL-16, AE]
Solution:
∝∅
 ∅
 = ∅

 


  

,
,∅∅ ∝  where  is the field current.


So,
 =  or  = .×
   .×
 = 160
160 
11. A 3-phase alternator has generated e.m.f. per phase of 230 V with 10 per cent third harmonic and 6
per cent fifth harmonic content. Calculate the r.m.s
r.m.s.. line voltage for (a) star connection (b) deltaconnection. Find also the circulating current in delta connection if the reactance per phase of the
machine at 50-Hz is 10 ohm.
[APSCL-16, SGFCL-17, AE]
Solution:
(a) Y connected, 230V, 10% 3rd harmonics, 6% 5th harmonics.
E1/phase = 230V
3rd harmonic cancel each other because they are
ar e co-phase.
E5/phase = 0.06x230 = 13.8V
E/phase =
(230 +13.8 = 230.42V
E/line = 230.42x1.7320 = 399V
(b) Delta connected,
E/phase = 230.42V = E/line
10 ohm at 50Hz.
10% at 3rd harmonic,
Circulating current = (230x0.1)/(3x10) = 0.77A.
12. A 1000 kVA, 3300V, 3 phase, star connected alternator delivers full load current at rated voltage
at 0.8 pf. Lagging. The resistance and synchronous reactance of the machine per phase are o.5 ohm
and 5 ohm respectively. Estimate the terminal voltage for the same excitation and same load current at
0.8 pf leading.
Solution:
Y connected, 1000 kVA, 3300V, 0.8 pf lagging, Ra=0.5 ohm & Xs=5 ohm
∅ =36.87 lagging

1000×10
 = √3 ×3300 =175
 =  + (∠−∅
∠−∅)) × (+
+))
∅ =36.87 leading,  =?
I = 175A,  =  =2582.74∠14.52
 =  + (∠−∅
∠−∅)) × (+
+))
 =2582.74∠14.52 V
 = 2957.11∠−2.034 V
(line) = 5121.67 V.
Case-1
 0.5+5))
 = .
 + (175∠−36.87 ) × (0.5+5
Case-2
0.5+5))
2582.74∠14.52 = +(175∠36.87) × (0.5+5
13. Find the synchronous impedance and reactance of an alternator in which a given
given field current
produces an armature current of 200A on short circuit and a generated emf of 50V on open circuit.
The armature resistance is 0.1ohm. To what induced voltage must the alternator be excited if it is to
deliver a load of 100A at a pf of 0.8 lagging, with a terminal voltage of 200V.
Solution:
Ra = 0.1 ohm,
=100,∅=0.8,∅=36.87  .
Vt = 200 V.
50 =0.25
 =  = 200
 = (0.25 −0.1) = 0.222929 ℎ
ℎ
 =  + (∠−∅
∠−∅)) × (+
+))
0.1+0.229))
= 200 + (100∠−36.87) × (0.1+0.229
= 222∠3
222∠3.18
.18 V
14. A 3 phase, 10kVA, 400V, 50Hz, Y connected alternator supplies the rated load at 0.8 pf lagging.
Armature resistance is 0.5 ohm and synchronous reactance is 10 ohms, find the power angle and
voltage regulation.
Solution:
Y connected, 10 kVA, 400V, Ra = 0.5 ohm, Xs = 10 ohm.
∅=0.8,∅=36.87  .

10×10
 = √3 × 400 =14.43
So, Power angle is 18.96
%   =
0.5+10))
 = √ + (14.43∠−36.87) × (0.5+10
= 341.85∠18.96
85∠18.96 V
./√

 ×100= /√ ×100=48.03
1kV,, 3 pha
phase
se,, Y connec
connected
ted alter
alternato
natorr whe
when
n supply
supplyiing 100A at zero pf
15. A given 3 MVA, 50 Hz., 11kV
12,3
,370
70V;
V; wh
when
en the
the load
load is remo
remove
ved,
d, the
the ter
ter inal voltage falls
lead
le
adin
ing
g has
has a line
line to line
line vo
volta
ltage of 12
regula
latio
tion
n of the
the alte
altern
rnat
ator
or when
when supp
supply
lying
ing full
full l ad at 0.8 lagging.
down to 11,000V. Predict the regu
Assume an effective resistance of 0.4 ohm per phase.
Solution: Y connected, 3MVA, 11kV, Ra = 0.4 ohm, Xs = ?
Case 1
Case 2
100A,
=  (leading).
∅
 ( 
))90= 
√ =7141.8 3 
 ( 
)) = 
√ =6350.85 
 =  + (∠−∅
∠−∅)) × (+))
ℎ
ℎ
71
7141
41.8.823
23== 63
6350
50.8.855 + (100∠9 ) × (0.4+
0.4+))
0.8= 36.87
∅ = 
× = 157
 = √××
.466 
 157.4
+))
 =  + (∠−∅
∠−∅)) × ( +
0.4+7.9))
 = 10
1000
00//√3 + (157.46∠ −36.87) × (0.4+7.9
Xs = 7.9 ohm.
=7211.43∠7.37 V
.. ×100=13.55
%  =   ×100= .
.
e OCC shown
in
16.
connec
nected
ted,, reactance
four-pole
four-pole of
ssync
ynchro
hronous
nousand
gene
generat
or has t esistance
of 0.015
Figur eA
. T480-V,
his gene60-Hz,
rator hadelta
s a s con
nchronous
0.1
ohm
anrator
armature
ohm, At full
full load
load,, the machin
machin suppl
supplie
iess 1200
1200 A at
at 0.8
0.8 P
PF
F la
lagg
ggin
ing.
g. U
Und
nder
er full
full-l
-l ad conditions, the
friction and windage losses are 40 kW,
kW, and
and the
the core
core loss
losses
es aare
re 30 kW.
kW. Ignore
Ignore any field circuit losses.
a. What
What is the speed
speed of rotation of this generator?
b. How much field c rre
rrent
nt mu
must
st be supplie
supplied
d to the
the ge
genera
nerator
tor to ma
make
ke t e terminal voltage
480 V at no load?
c. If the generator is ow connected to a load and the load draws 1200 A at 0.8 PF lagging,
how much field curren
rentt wil
willl be require
required
d to keep
keep the term
termina
inall vo
voltag
ltagee eq
equal to 480 V?
d. How much power is tile generator
generator now supplying?
supplying? How much powe is supplied to the
ge
gene
nerrato
torr by
by tth
he p
pri
rime mover? What is this machine's overall efficienc
effi cienc ?
e. If the ge
gene
nerrator'
or's l ad were suddenly
suddenly disconnecte
disconnected
d from the line, wh t would happen to
its terminal voltage?
f.
A at 0.8 PF
F
in
inaall
lly
y, How
su
supp
ppo
omuch
se t at
thecurrent
generator
is be
connected
loadVT
drawing
?
leading.
field
would
requiredto
to akeep
at 400 1200
[EGCB-14, AE]
Solution:
 =  = ×
 =1800 r/min.
(b) At no load,  = ∅ =  =480
(c) If the generator is supplying 1200A, then the armature current in the machine is,
 = 
√ =692.8.
(a)
 ==480+692.8∠−36.87
∅ +  +  (0.015+0.1
0.015+0.1))
=532∠5.3 V.
To keep the terminal voltage at 480V,  must be adjusted to 532V. From fig., the required
field current is 5.7A.
(d) The power that the generator is now supplying can be found from
 = √3  ∅.
= √3 ×480×1200×0.8
= 798 kw.
 =  +  ++ 
 
++  
 +  

=798+3×692.8 ×0.015×10 +30+40
889.6 kW.
 = =
. =89.75%
Stray losses were not
specified. So they are ignored.
ℎ
(e) If the load of the generator were suddenly dropped to zero, the terminal voltage would rise to
532V.
(f)
 = ∅ +  +  
(0.015+0.1
0.015+0.1))
=480+692.8∠36.87

=451∠7.1 V.
(g) To keep the terminal voltage at 480V,  must be adjusted to 451V. From fig., the required
field current is 4.1 A.
17. A 480-V, 50-Hz
50-Hz,, Y-connected, six-pole synchronous generator has a per-phase synchronous
synchronous
reactance of 1.0 D. Its full-load armature current is 60 A at 0.8 PF lagging. This generator has friction
and windage losses of 1.5 kW and core losses of 1.0 kW at 60 Hz at full load. Since the armature
resistance is being ignored, assume that the
losses are negligible. The field current has been
adjusted so that the terminal voltage is 480 V at no load.

a. What is the speed of rotation of this generator?
b. What is the terminal voltage of this generator if the following are true?
1. It is loaded with the rated current at 0.8 PF lagging.
2. It is loaded with the rated current at 1.0 PF.
3. It is loaded with the rated current at 0.8 PF leading.
c. What is the efficiency of this generator (ignoring the unknown electrical losses) when it is
operating at the rated current and 0.8 PF lagging?
d. How much shaft torque must be applied by the prime mover at full load? How large is the
induced counter torque?
e. What is the voltage regulation of this generator at 0.8 PF lagging? At 1.0 PF? At 0.8 PF
leading?
Solution:
 =  = ×
 =1000 r/min.
Alternatively, the speed expressed in radians per second is
  = 125.7 rad/s.
×
 =(1000 /min)× 




(a)
(b).
(1) for lagging PF,
 & ∅
Angle between
was not
mentioned. So, We can not use
the equation of
  ==(∅[ + +1.0×60×0.6
 )+( ))
  +    = ∅ +
277
When
+ [ 1.1.0 × 60 × 0.8)
+1.0×60×0.6)
∅
∅=0.8,∅=0.6
∅ =236.8 V
 ℎ 
   
,, = √3 × 236.8 = 410 
(2) for 1.0 PF,
= (∅ + )+()
277 = [ ∅ +1.0×60×0
+1.0×60×0)) + [ 1.1.0 × 60 × 1)
∅ =270.4 V
 = √3 × 270.4 = 468.4
(3)  = (∅ +   ) +(  )


+

∅ ==308.8
277
=√3 ×[ 3∅V0−1.0×60×0.6
−1.0×60×0.6)
8.8 = 535  ) [ 1.1.0 × 60 × 0.8)
(C). The output power of this generator
gen erator at 60A and 0.8 PF lagging is
 = 3 ∅.
= 3 ×236.8×60×0.8
= 34.1 kw.
 =  +   + 
  + 
ℎℎ 
 +  
=34.1+0+1+1.5
= 36.6 kW.
 = ..
 =93.2%
(d). The input torque to this generator is given by the equation,
 = 
. =271.3 .
 = .

(e). The voltage regulation of a generator is defined as
 =   ×100
×100=17.1%
1.1.

 ,
, = 

×100=2.6%
2. 
2.
,,  = 

3.3.
 
,,
 = 
 ×100=−10.3%
Stray losses were not
specified. So they are ignored.
18. Find the power angle When a 1500 KVA, 6.6KV, 3 phase, Y-connected alternator having a
resistance of 0.4 ohm and a reactance of 6 ohm per phase. Delivering a full load current at normal
rated voltage and 0.8 pf lagging. [EGCB-14, AE]
Soln:
V(Ph)=
.× = 3810.51 V ; I
√
a (Ph)=
× = 131.22 A
√×.×
Eg (Ph)= VPh+ Ia Zs
= 3810.51 °+(131.22
∠0

∠−36.87 )(0.4+j6) = 4366.1∠7.88
°
°
Load angle, = 7.88° Ans.
19. A 3 Phase 1
10
0 MVA, 400V, 50Hz Y connected alternator supplies the rated load 0.8 pf lagging. It
has synchronous reactance 0.635 ohm. Find the torque angle. [BGFCL-17,AE ]
Soln:
 = 213 V ; I = × = 14.43× 10 A
√×
√
tan(
tan( + )=  
VP=
a

sin = 0.6
cos = 0.8
××.
 
×.

.
tan(
tan( + )= ×
×.
.
 .
.×
××
tan(
tan( + )= 88.86
=88.86 − 36.86 = 52
°
°
°
°
20. A 10 MVA, 6.6KV, 50HZ,3 phase alternator
alternator has synchronous reactance of 0.635 ohm. It is
working full load at 0.8 lagging condition. Find its load angle. Neglect armature resistance.[ PGCB18,AE]
Soln:


Ia (Ph)=
√×.
×
× = 874.77 A ;
V(Ph)=
.√
×
= 3810.51 V
Eg (Ph)= VPh+ Ia Xs
∠0 +(874.77∠−36.87 )(j0.635) = 4167.56∠6.12
= 3810.51

°
Load angle, = 6.12° Ans.
°
°
21. A 3-Phase, 11-KVA, 10-MW, Y-Connected synchronous generator has synchronous im
impedance
pedance
of (0.8+j8.0) ohm per phase. If the excitation is such that the open circuit voltage is 14 KV, determine
(a) The maximum output of the generator (b)The current and (c) pf at the maximum output. [NPCL17, AE]
n
Sol :
(a) If we neglect Ra , the P max per phase = EV/Xs (Where V is the terminal voltage and E is the emf of
the machine.)
∴ Maximum Power of generator,(3-Phase) ,P = 3  
ℎ
ℎ
max
((

⁄√
√)×(
)×(

⁄√
√)) = 19.25 MW

√ = (/√)×(/√) = 1287
(b) I =
1287 



/√
(c) pf = =   =
√  (/√)×(/√) =0.786 ()
Pmax =
max
22. Can a generator rated 400V, 60 Hz be used in a 50 Hz system
system?? Yes or no? why?
[NWPGCL-14, AE]
Ans. Yes a 400 V, 60 Hz generator can be used in a 50 Hz system. Both voltage and apparent power
would have to be rated by factor of 50/60.
So its voltage rating reduced to (50/60)
× 400 = 333.33 V
23. At a location in Europe, it is nece
necessary
ssary to supply 1000 KW of 60 Hz power. The only power
source available operate at 50 Hz. It is described to generate the Power by means of a motor-generator
set consisting of a synchronous motor driving a synchronous generator. How many poles should each
of two machines have in order to convert 50 Hz power to 60 Hz power? [NWPGCL-17,AE]
Soln:
 = × = Synchronous speed of synchronous motor.


×


=
 =  = Synchronous speed of synchronous generator.
Ns1=
Ns2
∴N =N
120×50 = 120×60


 = 60
 ×
50 

 = ×= 
s1
s2
Therefore, a 10 synchronous motor must be coupled to a 12 pole synchronous generator to
accomplish, This frequency conversion.
24. A 750 KVA, 2300 V, Y connected , 3-phase alternator has open and short circuit test. Following
data are achieved from those test:
Short circuit test: If= 31.5 A, IL = rated
Open circuit test: If= 31.5 A, VL = 1050 V
Dc resistance across the pair terminals wave measured 0.42 ohm. Calculate the voltage regulation at
0.8 pf lagging. [PGCB-11,AE]
Soln:
× = 188.26 A
√×
 /  = /√ = 3.22 ohm
Z /Phase= 
 /  .
. = 0.21
DC resistance / Phase =

IF.L=
s
×
Ac effective resistance / Phase, Ra = 0.21 1.5 = 0.315 ohm
√3 −0.315 = 3.205 ohm
E /Phase = V + I Z = 1327.91 ∠0 + ( 188.28∠−36.87 )(0.315+j3.205) = 1793.98∠14.43
∴ % Voltage regulation =  ×100
=

..
.

 .
. ×100 = 35.09% Ans.
Xs =
o
ph
a
°
s
°
°
25. A generator is supplying a load of 1000 KW at 0.8 PF lagging.A second load of 800 KW at 0.707
PF lagging is to be connected in parallel with the 1st one. Generator has no load frequency is 61 Hz.
Determine the operating frequency before and after connecting the second load. Assume generator
slope Sp is 1MW/ Hz. [CPGCBL-18, AE]
Soln:
P= Sp (fNL
fsys = fNL
− f
− 
sys)
 = 60 Hz
= 61 . = 59.2 Hz
− −
Operating frequency before connecting second load, fsys(1) = 61
∴
Operating frequency after connecting second load, fsys(2)
Ans.
Ans.
26. Two synchronous generator are supplying a total load of 35MW at 0.8 pf
pf lagging. The no load
frequencies of generator-1 and generator-2 are 51.5 Hz and 51 Hz respectively . The slope of the
frequency is 1MW/ Hz. Determine the operating frequency of the system. [GTCL-18 AE]
Soln:
P1 = Sp1 (fNL1
P2 = Sp2 (fNL2
−f
−f
system)
system)
∴P +P = 35
∴35 = 1 MW/Hz (51.5 − f
1
2
system )
35 = 51.5 fsystem + 51
fsystem =49.5 Hz Ans.
⇒
−
−
+ 1 MW/Hz (51
fsystem
−f
system)
Practice Problem
27. A 3 phase, star connected alternator
alternator supplies a load of 10MW at 0.85 lagging and at 11 kV
terminal voltage. Its resistance is 0.1 ohm per phase and synchronous reactance 0.66 ohm per phase.
Calculate the line value of emf generated.
[11.486 kV]
28. The effective resistance of a 1200 kVA, 3.3kV, 50 Hz, 3 phase Y connected alternator is 0.25
ohm per phase. A field current of 35A produces a current of 200A on short circuit and 1.1kV (line to
line) on open circuit. Valculate the power angle and p.u change in magnitude of the terminal voltage
when the full load of 100kVA at 0.8 lagging pf is thrown off. Draw the corresponding phasor
diagram. [Exam 37.27,
Ans: 12 degree, 0.26]
29. A 2000 kVA, 11 kV, 3 phase Y connected alternator
alternator has a resistance of 0.3 ohm and reactance of
5 ohm per phase. It delivers full load current at 0.8 lagging pf at rated voltage. Compute the terminal
voltage for the same excitation and load current at 0.8 pf leading.
[Ans: 6975 V]
30. The following test results are obtained from a 3 phase, 6000kVA, 6600V, Y connected, 2 pole, 50
Hz turbo alternator:
With a field current of 125A, the open circuit voltage is 8000V at the rated speed; with the same field
current and rated speed, the short circuit current is 800A. At the rated full load, the resistance drop is
3%. Find the regulation of the alternator on full load and at a pf of 0.8 lagging.
[Ans: 62.2%]