Chapter 7 Additional Problems

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Chapter 7 Additional Problems
X7.1 A three-phase, 1200 HP, 2300 V, 60 Hz, RR synchronous machine has Rs  0.2  and
X s  5.6  . When operated with the mechanical load disconnected with rated voltage
and frequency, the field current is adjusted until line current has a minimum value. At
this point, I a  22.1 A , the measured input power is PT  17.5 kW , and the measured
field circuit values are V f  276 V and I f  53.2 A . (a) Determine the rotational losses
(core losses plus friction and windage) for this machine. (b) If the field current,
frequency, and impressed terminal voltage are unchanged, but a mechanical load
requiring 600 HP is attached, predict the power factor, line current, and efficiency.
(a) Using the no-load data,
PFW  PT  3I a2 Rs  17,500  3 22.1  0.2   17,793.05 W
2
(b) Since the no-load current was a minimum value, the machine was operating at
unity power factor. The excitation voltage corresponding to I f  53.2 A can be
determined.
E f  Van  I a  Rs  jX s  
2300
 22.1 0.2  j 5.6   1329.28  5.34 V
3
Since Rs X s , little error is introduced if Rs is neglected in calculation of the torque
 for the loaded condition. From energy balance,
Ps  PFW 
3Van E f
Xs
sin 
or


  Ps  PFW  X s 
1   600  746  17,793.05  5.6  
  sin 
 31.84
  sin
 3 2300 / 3 1329.28 
 3Van E f



1


Based on [7.36],
2300
0  1329.28  31.84
3
Ia 

 130.07   13.77 A
Rs  jX s
0.2  j 5.6
Van  E f
Hence,
I a  130.07 A
PFin  cos  Van  I a   cos 11.35  0.97 lagging
The losses are
Losses  PFW  3I a2 Ra  V f I f
Losses  17,793.05  3130.07   0.2   276 53.2  42,627.17 W
2
1

100  600  746 
100 Ps

 91.30%
Ps  losses 600  746  42,627.17
X7.2 If the synchronous motor of Problem X7.1 still operates the 600 HP load with terminal
voltage and frequency unchanged, but the field current is increased to 80 A, predict
(a) the input power factor and (b) the leading VARs supplied to the three-phase grid.
Assume that core losses (lumped with friction and windage losses) of Problem X7.1
are valid.
(a) For the new field current, the excitation voltage is
Ef 
80
1329.28  1998.92 V
53.2
Assuming Rs negligible,


  Ps  PFW  X s 
1   600  746  17,793.05  5.6  

sin
 19.10

 3 2300 / 3 1998.92  
 3Van E f



  sin 1 


2300
0  1998.92   19.10
3
Ia 

 153.45 42.86 A
Rs  jX s
0.2  j 5.6
Van  E f
PFin  cos  Van  I a   cos  42.86  0.733 leading
(b)
QT  3 VL I a sin   3  2300 153.45  sin  42.86 
QT  415.80 kVARs
X7.3 If the synchronous motor of Problem X7.1 now supplies rated power to a coupled
mechanical load with rated voltage and frequency applied to the stator and I f  80 A ,
calculate (a) line current and (b) efficiency. Assume that Rs can be neglected in torque
angle determination. Also, assume that core losses have changed negligibly from
Problem X7.1.
(a) The excitation voltage for I f  80 A is determined in Problem X7.2 to be
E f  1998.92 V .


  Ps  PFW  X s 
1 1200  746  17,793.05  5.6  
 39.97
  sin 


3 2300 / 3 1998.02 
 3Van E f



  sin 1 

2

2300
0  1998.02   39.97
3
Ia 

 229.46 11.75 A
Rs  jX s
0.2  j 5.6
Van  E f
or
I a  229.46 A
(b) From the values of V f and I f given in Problem X7.1,
Rf 
Vf
If

276
 5.19 
53.2
Losses  PFW  3I a2 Rs  I 2f R f
Losses  17,793.05  3  229.46   0.2   80   5.19   82,600.18 W
2

2
100 1200  746 
100 Ps

 91.55%
Ps  losses 1200  746  82,600.18
X7.4 The synchronous machine of Problem X7.1 is now operated as a 2300 V, 60 Hz
alternator supplying 800 kW to a 0.8 PF lagging load. Field current I f  53.2 A .
Determine the required input mechanical power to drive the rotor and coupled exciter
if the exciter has an efficiency of 91% when supplying the field voltage and current
given in Problem X7.1. Assume the lumped friction and windage losses and core
losses of Problem X7.1 are valid.
Since I f is unchanged from Problem X7.1, E f  1329.28 V is still valid.
Assume Van on the reference.
Ia 
PT
3 VL PF

800,000
 251.02 A
3  2300  0.8
Ps  PT  3I a2 Rs  PFW  V f I f / ex
Ps  800,000  3  251.02   0.2   17,793.05   276  53.2  / 0.91
2
Ps  871.73 kW
X7.5 Fig. X7.1 shows the phase a circuit of a three-phase synchronous generator that has a
per phase load consisting of a 5  pure inductive load. The machine is operating at
zero power factor lagging. Determine the ratio E f / Van .
3
By voltage division,
Van 
jX L
j5
Ef 
Ef
j XL  Xs 
j 5  j10
or
Ef
Van
3
X7.6 Fig. X7.2 shows the phase a circuit of a three-phase synchronous generator that has a
per phase load consisting of a 15  pure capacitive load. The machine is operating at
zero power factor leading. Determine the ratio E f / Van .
4
By voltage division,
jX C
 j15
Ef 
Ef
j  XC  X s 
 j15  j10
Van 
or
Ef
Van

1
3
X7.7 A 3600 rpm, 60 Hz, 13.8 kV synchronous generator has a synchronous reactance of
20  . The generator is operating at rated voltage and speed with the excitation voltage
E f  11.5 kV and the torque angle   15 . Calculate (a) stator current, (b) power
factor, and (c) total output power.
(a)
Ia 
E f  Van
jX s
13,800
0
3
 216.35  46.54 A
2090
11,50015 

I a  216.35 A
(b)
(c)
PF  cos  Van  I a   cos  46.54  0.688 lagging
PT  3VL I a PF  3 13,800  216.35 0.688  3557.8 kW
X7.8 For the synchronous generator of Problem X7.7, determine (a) the maximum power
that can be converted from mechanical to electrical form without loss of synchronism
if the field current is unchanged and (b) the value of current I a for this condition.
(a) Since   90 ,
3Pd max 
E f Van
Xs

11,500  13,800 /
20
3
  4.58 MW
(b)
Ia 
E f  Van
jX s
13,800
0
3
 699.5234.72 A
2090
11,50090 

I a  699.52 A
5
X7.9 A three-phase synchronous generator, rated at 240 V, 50 A, 60 Hz, has the stator
winding schematic of Fig. X7.3. All 12 phase group leads ( a1 , a1 , c2 , c2 ) are
available at the generator terminals. In addition to the given rating, list the balanced,
three-phase voltage-current ratings available from the generator by coil reconnection
without mix of phase coils – that is, use only phase a coils in phase a connections.
Three additional ratings are possible under the stated constraint.
1. Series connect phase coils and use wye configuration: 480 V, 25 A
2. Series connect phase coils and use delta configuration: 277 V, 43.3 A
3. Leave phase coils parallel-connected and use delta configuration: 139 V,
86.6 A
X7.10 The three-phase synchronous generator of Problem X7.9 is to be reconnected as a
single-phase alternator. What is the largest possible voltage that can be obtained?
Connect the phase coils in series with additive polarity to give
V1  Va1  Va 2  Vb1  Vb 2  Vc1  Vc 2
V1  2770  277  120  277120  5540 V
Or the largest voltage possible is V1  554 V .
6
X7.11 A 500 MVA, 24 kV, 60 Hz three-phase alternator is operating at rated voltage and
frequency with a terminal power factor of 0.8 lagging. The leakage reactance
X  0.15  and the synchronous reactance X s  0.8  . Stator coil resistance is
negligible. The excitation voltage E f  18 kV . Determine (a) the torque angle  ,
(b) the total output power, (c) the line current I a , and (d) the resultant voltage Er .
(a) The power converted must equal the output power.
3E f I a cos      3Van I a cos
or
Van cos 
 
 E f 
  cos1 
 24,000
 0.8  

3
  cos 1 
  cos 1  0.8   15.12
18,000




(b)
PT 
3E f Van
Xs
sin  

3 18,000  24,000 / 3
0.8
 sin 15.12  244 MW
(c)
Ia 
PT
3VL PF

244  106
 7337.4 A
3  24,000  0.8 
(d)
Er  Van  jX I a
Er 
24,000
0   0.1590  7337.4  36.87   14,5443.47 V
3
Er  14,544 V
X7.12 A 250 MVA, 24 kV, 60 Hz, three-phase alternator is operating at rated voltage, rated
frequency, and rated apparent power with a terminal power factor of 0.8 lagging.
Stator coil resistance is negligible. The excitation voltage E f  20 kV . Determine (a)
the torque angle  and (b) the value of synchronous reactance X s .
(a) Assume Van on the reference.
Ia 
ST
3VL
  cos 1  PF  
250  106
  cos 1  0.8  6014.2   36.87 A
3  24,000 
7
Power converted must equal power output.
3E f I a cos      3Van I a cos
or
Van cos 
 
 E f 
  cos1 
 24,000
 0.8  

3
  cos 1 
  cos 1  0.8   19.47
20,000




(b)
E f  Van  j I a X s
or
Xs 
E f  Van
j Ia
24,000
0
3

 1.38 
190 6014.2  36.87
20,00019.47 
X7.13 A 100 MVA, 13.8 kV, 60 Hz, salient-pole, synchronous generator with 8 poles has
X d  1.9  and X q  1.1  . The stator coil resistance is negligible. The generator is
supplying rated apparent power at a power factor of 0.866 lagging. Determine the
value of excitation voltage E f and torque angle  .
Ia 
ST
3 VL
  cos 1  PF  
100  106
 cos 1  0.866   4183.8  30 A
3 13,800 
By [3] of Example 7.10,
E f  Van  jX q I a 
13,800
0  1.190  4183.8  30 
3
E f  11,01521.21 V
Thus,
   E f  21.21
        30  21.21  51.21
I d  I a sin  4183.8sin  51.21  3261 A
I d  I d        A
8


E f  E f  j X d  X q Id
E f  11,01521.21   0.890  3261  68.79 
E f  13,623.821.21 V
E f  13,623.8 V
X7.14 For the salient-pole synchronous generator of Problem X7.13, calculate the developed
torque by two methods.
s 
2
2
   2 60   30 rad / s
p
8
By [7.53] with values from solution of Problem X7.13,
3Td 
3Van E f
s X d
sin  

3Van2 X d  X q
2 s X d X q
 sin 2
2
 13,800 
 13,800 
3
3
 13,623.8 
 1.9  1.1
3 
3 


3Td 
sin  21.21  
sin  2  21.21 
30 1.9 
2  30 1.9 1.1
3Td  918.8 kN  m
Alternatively for this machine with Rs  0 ,
3Td 
3Pd
s

PT
s
100 10   0.866  918.8 kN  m

6
30
X7.15 Determine the maximum power that the salient-pole synchronous generator of
Problem X7.13 can convert if the field circuit were open circuit.
In [7.53], set E f  0 ,   45 , and multiply by  s to give
3Pd max 

3Van2 X d  X q

2Xd Xq
2
3Pd max
 13,800 
3
 1.9  1.1
3 


 36.45 MW
2 1.9 1.1
9
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