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## fitzgerald

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Catégorie :Category: mViewer GX Creator App HP-Prime
Auteur Author: tassio...
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56

PROBLEM SOLUTIONS: Chapter 5

Problem 5.1
Basic equations are T ∝ ΦR Ff sin δRF . Since the ﬁeld current is constant, Ff
is constant, Note also that the resultant ﬂux is proptoortional to the terminal
voltage and inversely to the frequency ΦR ∝ Vt /f . Thus we can write
Vt sin δRF
T ∝
f
P = ωf T ∝ Vt sin δRF
part (a): Reduced to 31.1◦
part (b): Unchanged
part (c): Unchanged
part (d): Increased to 39.6◦
Problem 5.2
part (a): The windings are orthogonal and hence the mutual inductance is
zero.
part (b): Since the two windings are orthogonal, the phases are uncoupled
and hence the ﬂux linkage under balanced two-phase operation is unchanged by
currents in the other phase. Thus, the equivalent inductance is simply equal to
the phase self-inductance.
Problem 5.3
1
Lab = − (Laa − Lal ) = −2.25 mH
2

3
Ls = (Laa − Lal ) + Lal = 7.08 mH
2
Problem 5.4
part (a):

2 Vl−l,rms
Laf = √ = 79.4 mH
3ωIf
part (b): Voltage = (50/60) 15.4 kV = 12.8 kV.
Problem 5.5
part (a): The magnitude of the phase current is equal to

40 × 103
Ia = √ = 59.1 A
0.85 × 3 460
and its phase angle is − cos−1 0.85 = −31.8◦ . Thus
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57

Iˆa = 59.1e−j31.8

Then
460
Êaf = Va − jXs Iˆa = √ − j4.15 × 59.1e−j31.8 = 136  − 56.8◦ V

3
The ﬁeld current can be calculated from the magnitude of the generator
voltage

2Eaf
If = = 11.3 A
ωLaf
part (b):

Êaf = 266  − 38.1◦ V; If = 15.3 A

part (c):

Êaf = 395  − 27.8◦ V; If = 20.2 A

Problem 5.6
The solution is similar to that of Problem 5.5 with the exception that the
sychronous impedance jXs is replaced by the impedance Zf + jXs .
part (a):
Êaf = 106  − 66.6◦ V; If = 12.2 A
part (b):
Êaf = 261  − 43.7◦ V; If = 16.3 A
part (c):
Êaf = 416  − 31.2◦ V; If = 22.0 A
Problem 5.7
part (a):

2 Vl−l,rms
Laf = √ = 49.8 mH
3ωIf
part (b):

600 × 103
Iˆa = √ = 151 A
3 2300

Êaf = Va − jXs Iˆa = 1.77  − 41.3◦ V

2Eaf
If = = 160 A
ωLaf
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part (c): See plot below. Minimum current will when the motor is operating
at unity power factor. From the plot, this occurs at a ﬁeld current of 160 A.

Problem 5.8
part (a):
2
Vbase (26 × 103 )2
Zbase = = = 0.901 Ω
Pbase 750 × 106

Xs,pu Zbase
Ls = = 4.88 mH
ω
part (b):
Xal,pu Zbase
Lal = = 0.43 mH
ω
part (c):
2
Laa = (Ls − Lal ) + Lal = 3.40 mH
3
Problem 5.9
part (a):
AFNL
SCR = = 0.520
AFSC
part (b):
Zbase = (26 × 103 )2 /(800 × 106 ) = 0.845 Ω
1
Xs = = 2.19 pu = 1.85 Ω
SCR
part (c):
AFSC
Xs,u = = 1.92 pu = 1.62 Ω
AFNL, ag
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59

Problem 5.10
part (a):

AFNL
SCR = = 1.14
AFSC
part (b):
Zbase = 41602 /(5000 × 103 ) = 3.46 Ω
1
Xs = = 1.11 pu = 3.86 Ω
SCR
part (c):

AFSC
Xs,u = = 0.88 pu = 3.05 Ω
AFNL, ag
Problem 5.11
No numerical solution required.
Problem 5.12
part (a): The total power is equal to S = P /pf = 4200 kW/0.87 = 4828 kVA.
The armature current is thus
4828 × 103
Iˆa = √  (cos−1 0.87) = 670 29.5◦ A
3 4160
Deﬁning Zs = Ra + jXs = 0.038 + j4.81 Ω
4160
|Eaf | = |Va − Zs Ia | = | √ − Zs Ia | = 4349 V, line − to − neutral
3
Thus
 
4349
If = AFNL √ = 306 A
4160/ 3

part (b): If the machine speed remains constant and the ﬁeld current is not
reduced, the terminal voltage will increase to the value corresponding to 306 A
of ﬁeld current on the open-circuit saturation characteristic. Interpolating the
given data shows that this corresponds to a value of around 4850 V line-to-line.
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60

Problem 5.13

Problem 5.14
√ power, unity power factor, the armature current will be Ia =
At rated
5000 kW/( 3 4160 V) = 694 A. The power dissipated in the armature winding
will then equal Parm = 3 × 6942 × 0.011 = 15.9 kW.
The ﬁeld current can be found from
4160
|Eaf | = |Va − Zs Ia | = | √ − Zs Ia | = 3194 V, line-to-neutral
3
and thus
 
3194
If = AFNL √ = 319 A
4160/ 3

At 125◦ C, the ﬁeld-winding resistance will be
 
234.5 + 125
Rf = 0.279 = 0.324 Ω
234.5 + 75

and hence the ﬁeld-winding power dissipation will be Pfield = If2 Rf = 21.1 kW.
The total loss will then be

Ptot = Pcore + Parm + Pfriction/windage + Pfield = 120 kW

Hence the output power will equal 4880 kW and the eﬃciency will equal 4880/5000
= 0.976 = 97.6%.
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61

Problem 5.15
part (a):

part (b): AFNL = 736 A. AFSC = 710 A.
part (c): (i) SCR = 10.4, (ii) Xs = 0.964 per unit and (iii) Xs,u = 1.17 per
unit.
Problem 5.16
For Va = 1.0 per unit, Eaf,max = 2.4 per unit and Xs = 1.6 per unit

Eaf,max − Va
Qmax = = 0.875 per unit
Xs

Problem 5.17
part (a):
2
Vbase
Zbase = = 5.29 Ω
Pbase

1
Xs = = 0.595 per-unit = 3.15 Ω
SCR
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part (b): Using generator convention for current

part (c):
150
Eaf = = 0.357 per-unit
420
For Va = 1.0 per-unit,
Eaf − Va
Iˆa = = 1.08 90◦ per-unit = 1.36 90◦ kA
jXs
using Ibase = 1255 A.
part (d): It looks like an inductor.
part (e):

700
Eaf = = 1.67 per-unit
420
For Va = 1.0 per-unit,
Eaf − Va
Iˆa = = 1.12 − 90◦ per-unit = 1.41 − 90◦ kA
jXs
In this case, it looks like a capacitor.
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63

Problem 5.18

Problem 5.19
part (a)...

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