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PROBLEM SOLUTIONS: Chapter 5
Problem 5. Basic equations are T ∝ ΦRFf 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. T hus we can write
T ∝
Vt sin δRF 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. 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.
Lab = −
(Laa − Lal) = − 2. 25 mH
Ls =
(Laa − Lal) + Lal = 7. 08 mH
Problem 5. part (a):
Laf =
2 Vl−l,rms √ 3 ωIf
= 79.4 mH
part (b): Voltage = (50/60) 15.4 kV = 12.8 kV.
Problem 5. part (a): The magnitude of the phase current is equal to
Ia =
40 × 103
0. 85 ×
= 59.1 A
and its phase angle is − cos−^1 0 .85 = − 31. 8 ◦. T hus
Iˆa = 59. 1 e−j^31.^8 ◦
Then
Eˆaf = Va − jXs Iˆa =^460 √ 3
− j 4. 15 × 59. 1 e−j^31.^8
◦ = 136 − 56. 8 ◦^ V
The field current can be calculated from the magnitude of the generator voltage
If =
2 Eaf ωLaf
= 11.3 A
part (b):
Eˆaf = 266 − 38. 1 ◦^ V; If = 15.3 A
part (c):
Eˆaf = 395 − 27. 8 ◦^ V; If = 20.2 A
Problem 5. 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): Eˆaf = 106 − 66. 6 ◦^ V; If = 12.2 A part (b): Eˆaf = 261 − 43. 7 ◦^ V; If = 16.3 A part (c): Eˆaf = 416 − 31. 2 ◦^ V; If = 22.0 A
Problem 5. part (a):
Laf =
2 Vl−l,rms √ 3 ωIf
= 49.8 mH
part (b):
Iˆa =^600 ×^10
3 √ 3 2300
= 151 A
Eˆaf = Va − jXs Iˆa = 1. 77 − 41. 3 ◦^ V
If =
2 Eaf ωLaf
= 160 A
Problem 5. part (a):
SCR =
AFNL
AFSC
part (b): Zbase = 4160^2 /(5000 × 103 ) = 3.46 Ω
Xs =
SCR
= 1.11 pu = 3.86 Ω
part (c):
Xs,u =
AFSC
AFNL, ag
= 0.88 pu = 3.05 Ω
Problem 5. No numerical solution required.
Problem 5. part (a): The total power is equal to S = P /pf = 4200 kW/0.87 = 4828 kVA. The armature current is thus
Iˆa =^4828 ×^10
3 √ 3 4160
(cos−^1 0 .87) = 670 29. 5 ◦^ A
Defining Zs = Ra + jXs = 0.038 + j 4 .81 Ω
|Eaf | = |Va − ZsIa| = |
− ZsIa| = 4349 V, line − to − neutral
Thus
If = AFNL
= 306 A
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.
Problem 5.
Problem 5. At rated power, unity power factor, the armature current will be Ia = 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
|Eaf | = |Va − ZsIa| = |
− ZsIa| = 3194 V, line-to-neutral
and thus
If = AFNL
= 319 A
At 125◦C, the field-winding resistance will be
Rf = 0. 279
and hence the field-winding power dissipation will be Pfield = If^2 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/ = 0.976 = 97.6%.
part (b): Using generator convention for current
part (c):
Eaf =
= 0.357 per-unit
For Va = 1.0 per-unit,
Iˆa = Eaf^ −^ Va jXs
= 1. 08 90 ◦^ per-unit = 1. 36 90 ◦^ kA
using Ibase = 1255 A. part (d): It looks like an inductor. part (e):
Eaf =
= 1.67 per-unit
For Va = 1.0 per-unit,
Iˆa = Eaf^ −^ Va jXs
= 1. 12 − 90 ◦^ per-unit = 1. 41 − 90 ◦^ kA
In this case, it looks like a capacitor.
Problem 5.
Problem 5. part (a): It was underexcited, absorbing reactive power. part (b): It increased. part (c): The answers are the same.
Problem 5. part (a):
Xs =
= 0.268 per-unit
part (b): P = 0.875 and S = P/ 0 .9 = 0.972, both in per unit. The power- factor angle is − cos−^1 0 .9 = − 25. 8 ◦^ and thus Iˆa = 0. 875 − 25. 8 ◦.
Eˆaf = Va + jXs Iˆa = 1. 15 11. 6 ◦^ per-unit
The field current is If = AFNL| Eˆaf | = 958 A. The rotor angle is 11. 6 ◦^ and the reactive power is
Q =
S^2 − P 2 = 4.24 MVA
part (c): Now |Eaf | = 1.0 per unit.
δ = sin−^1
P Xs Va
|Eaf |
and thus Eˆaf = 1. 0 13. 6 ◦.
Iˆa =
Eˆaf − Va jXs
Q = Imag[Va Iˆ∗ a ] = − 0 .104 per-unit = − 1 .04 MVAR
and
Iˆa = Vte
jδt (^) − V∞ jX∞
= 0. 578 3. 93 ◦^ per-unit
Ibase = Pbase/(
3 Vbase) = 15.64 kA and thus Ia = 9.04 kA. (iii) The generator terminal current lags the terminal voltage by δt/2 and thus the power factor is
pf = cos−^1 δt/2 = 0.998 lagging
(iv)
| Eˆaf | = |V∞ + j(X∞ + Xs) Iˆa| = 1.50 per-unit = 36.0 kV,line-to-line
part (b): (i) Same phasor diagram (ii) Iˆa = 0. 928 6. 32 ◦^ per-unit. Ia = 14.5 kA. (iii) pf = 0.994 lagging (iv) Eaf = 2.06 per unit = 49.4 kV, line-to-line.
Problem 5. part (a):
part (b):
part (c):
Problem 5.
part (a): From the solution to Problem 5.15, Xs = 0.964 per unit. Thus, with V∞ = Eaf = 1.0 per unit
Problem 5.
Problem 5.
Problem 5. For Eaf = 0,
Pmax =
V (^) t^2 2
Xq
Xq
This maximum power occurs for δ = 45◦.
Id =
Vt cos δ Xd
= 0.786 per-unit
Iq =
Vt sin δ Xq
= 1.09 per-unit
and thus Ia =
I^2 d + Iq^2 = 1.34 per unit.
S = VtIa = 1.34 per-unit
Hence
Q =
S^2 − P 2 = 1.32 per-unit
Problem 5.
P =
V∞Eaf Xd
sin δ +
V ∞^2
Xq
Xd
sin 2δ
The generator will remain synchronized as long as Pmax > P. An iterative search with MATLAB can easily be used to find the minimum excitation that satisfies this condition for any particular loading. part (a): For P = 0.5, must have Eaf ≥ 0 .327 per unit. part (b): For P = 1.0, must have Eaf ≥ 0 .827 per unit.
Problem 5. part (a):
part (b): We know that P = 0.95 per unit and that
P =
V∞Vt Xbus
sin δt
and that
Iˆa = Vˆt − V∞ jXt
It is necessary to solve these two equations simultaneously for Vˆt = Vt δt so that both the required power is achieved as well as the specified power factor
part (c):
Problem 5.
f = n × poles 120
3000 × 6
= 150 Hz
Problem 5. part (a): Because the load is resistive, we know that
Ia =
P
3 Va
= 13.5 A
part (b): We know that Eaf = 208/
3 = 120 V. Solving
Eaf =
V (^) a^2 + (XsIa)^2
for Xs gives
Xs =
Eaf^2 − V (^) a^2 Ia
part (c): The easiest way to solve this is to use MATLAB to iterate to find the required load resistance. If this is done, the solution is Va = 108 V (line-to-neutral) = 187 V (line-to-line).
Problem 5.
I^ ˆa = Ea Ra + Rb + jωLa
ωKa Ra + Rb + jωLa
Thus
| Iˆa| = ωKa √ (Ra + Rb)^2 + (ωLa) 2
Ka
La
Ra +Rb ωLa
Clearly, Ia will remain constant with speed as long as the speed is sufficient to insure that ω >> (Ra + Rb)/La
