ee303 solutions
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Electrical engineering powerTRANSCRIPT
Electric Power Principles: Sources, Conversion, Distribution andUse
Solutions to Problems
James L. Kirtley Jr.c©2010 John Wiley & Sons
Introduction Herein are solutions to problems for each of the chapters of Electric Power Prin-
ciples: Sources, Conversion, Distribution and Use. I believe them to be correct, buterrors may have crept in. Use them with caution, and please check them before gradingstudent’s homework!
There are a number of Matlab scripts in an accompanying directory structure. There is asubdirectory for each chapter of the book for which there are scripts. (Chapters 2, 3, 5, 6and 8 through 15). (The scripts are named for the chapter and problem, so that ’p6 5.m’isa script that solves all or part of Problem 5 of Chapter 6. There are some auxiliary scriptsthat are required for some problems. They are located in the appropriate subdirectory andtheir identity is made clear in the main scripts that use them. The scripts are known to rununder Matlab Version 7.10.0.499 (R2010a).
Chapter 1
1. 240v × 50A = 12kW12kW × 3, 414BTU/kWh = 40, 968BTU/h
2. R = 3,414.5 = 6, 828BTU/kWh
3. Assume Coal energy content is 30,870 BTU/kg.If R=11,000 BTU/kWh, then coal consumption is:
m =11, 000BTU/kWh
30, 870BTU/kg≈ 0.3563kg/kWh
Then, if P = 1000MW = 106kW ,
M = 106kW × .3563 = 3.563 × 105kg/h
×365.25 × 24 = 3.12 × 109kg/yr = 3.12 × 106Tonnes/yr
4. If R = 30, 890BTU/kWh,
m =9, 500BTU/lWh
30, 890BTU/kg≈ .3075kg/kWh
×2.959kg CO2 /kg fuel = 0.9kg CO2/kWh
×600, 000 × 24 × 365.25 = 4.79 × 109kg CO2/yr = 4.79 × 106T CO2/yr
5. R = 3414.53 = 6, 441BTU/kWh
m =6, 441BTU/kWh
50, 780BTU/kg.127kg/kWh
= ×2.75 = .349kg CO2/kWh
×600, 000 × 24 × 365.25 = 1.83 × 109kg CO2/yr = 1.83 × 106Tonnes CO2/yr
6. The fraction of fuel converted to energy is:
f =1
2× .04 × 1
5× 1235 ≈ 2.128 × 10−5
Then energy released per kg of fuel is:
E = 2.128 × 10−5 × 9 × 1016 ≈ 1.915 × 1012J/kg
If R = 12, 000BTU/kWh, then thermal efficiency is η = 3,41412,000 ≈ .2845, and electrical
output per kilo of enriched fuel is:
Ee = 1.95 × 1012 × .2845 ≈ 5.548 × 1011J/kg
1, 000MW yr = 106kW yr = 3.6 × 1012 × 365.25 × 24 ≈ 3.16 × 1016J
So total fuel required is:3.16 × 1016J
5.548 × 1011J/kg≈ 56, 881kg
7. Power per unit area is:
P
A=
1
2ηρv3 =
1
2× 1.2 × .4 × 103 = 240W/m3
Then since power is P = PAπ4D
2, required diameter is:
D =
√
4
π
1.5 × 106
240≈ 89.2m
8. P = ρgvhη, so required flow is:
v =100 × 106
1000 × 20 × 9.812 × .8≈ 637m3/s
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 2
Chapter 2
1.
Rth = 4 + 8||8 = 4 + 4 = 8
Vth = 10 × 8||8 + 10 × 8
8 + 8= 40 + 5 = 45
2. The first (left-hand) circuit has the following impedance matrix:
Z =
[
3 22 3
]
The right-hand circuit has the following admittance matrix:
Y =
[
1R1
+ 1R2
− 1R2
− 1R2
1R3
+ 1R2
]
If we invert the impedance matrix for the first circuit:
Z−1 =1
5
[
3 −2−2 3
]
=
[
35 −2
5−2
535
]
This makes R2 = 2.5 and then
1
R1=
1
R3+
3
5− 2
5=
1
5
So R1 = R3 = 5.
3. This one is easily done by recognizing that the thevenin equivalent circuits for the sourcesand vertically aligned (totem pole style) resistors is as shown in Figure 1. The theveninequivalent voltage is derived from the voltage divider between the two resistors and theequivalent resistance is the same: 4||1 = 4
5 . Then the problem is reduced to what isshown in Figure 2. The output voltage is:
vo =
(
4
5− 1
5
)
× 18 × 1
1 + 85
=18
3= 6
4. The trick to this ’magic ladder’ problem is to see that the driving point impedance ofa section can be deduced to be 2R and the transfer relationship is defined by a simplevoltage divider to be 1
2 . This is true for each of the ’cells’ of the ladder network. Thus
the open-circuit output voltage V = V2 ×(
12
)5+ V1 ×
(
12
)7. Then, at that point, the
thevenin equivalent voltage is the open circuit voltage:
Vth =V1
128+V2
32
The thevenin equivalent resistance is
Rth = R||2R =2
3R
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 3
1818
1
4
18
4
1
45
45
45
5 181
Figure 1: Solution to Problem 3
−45
45
45
5 18118
1
vo+
Figure 2: Solution to Problem 3
5. While this one looks odd (the voltage source controls the voltage across the currentsource and the current source controls current through the voltage source, it is a prettygood approximation of the interface between solar and some wind generators and thepower system. In such situations, the system ’acts like’ a voltage source and the powerelectronics of the generators emulates a current source.
Real power is:
P =1
2V IRe
ejψ
=1
2cosψ
and Imaginary power is:
Q =1
2V IIm
ejψ
=1
2sinψ
Instantaneous power is, with the voltage phase being zero:
p = 2P cos2 ωt+Q sin 2ωt
The phasor diagrams corresponding with ψ = 0 and ψ = π2 are shown in Figure 3, and
instantaneous power is plotted for ψ = 0 in Figure 4 and for psi = π2 in Figure 5.
6. The voltage drops across the resistance and reactances are, respectively:
V R = 12010
10 + j20= 120
(
1
5− j
2
5
)
V X = 120j20
10 + j20= 120
(
4
5+ j
2
5
)
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 4
ψ = π/2
V
I
V
I
ψ = 0
Figure 3: Phasor Diagram for Problem 5
0 2 4 6 8 100
500
1000
1500
2000
2500Chapter 2, Problem 5, psi = 0
W
Phase om * t
Figure 4: Instantaneous real power for phase angle of zero
Current is:
I =VR10
= 12
(
1
5− j
2
5
)
Complex power is:
P + jQ = V I∗ = 120 × 12
(
1
5+ j
2
5
)
or P = 14405 = 288W and Q = 2880
5 = 576VARs
7. The resistance and reactance are in parallel, so:
IR =VsR
=120
10= 12
IX =VsjX
=120
j20= −j6
A phasor diagram that shows this is in Figure 6
Real and reactive power are:
P + jQ = V I∗ = 1440 + j720
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 5
0 2 4 6 8 10−1500
−1000
−500
0
500
1000
1500Chapter 2, Problem 5, psi = pi/2
W
Phase om * t
Figure 5: Instantaneous real power for phase angle of -90 degrees
S
I = 12
I = 12 − j 6
I = − j 6X
R
Figure 6: Solution to Problem 7
8. Maximum impedance magnitude will occur if the capacitive admittance balances (andthus cancels) the inductive admittance, so the condition for maximum voltage magnitudeis XC = −j10Ω, or C = 1
10×2×π×60 ≈ 265µF
The phasor diagram for the maximum voltage condition is shown in Figure 7
Impedance is:
Z = R||jωL|| 1
jωC=
11R
+ 1jωL
+ jωC
The magnitude of voltage is shown as a function of capacitance in Figure 8
9. This is the series analog of Problem 8. The capacitance to maximize voltage across theresistance is the one that balances (cancels) inductor impedance, and this is the same asin Problem 8, namely 265µF . The phasor diagram for voltages is, at resonance, shownin Figure 9.
Voltage across the resistance is given by a voltage divider:
VR = VSR
R+ jωL+ 1jωC
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 6
Ir = 10
Ic = j 10
I l = −j 10
Figure 7: Phasor Diagram for Maximum Voltage
0 100 200 300 400 500 60070
75
80
85
90
95
100Chapter 2, Problem 8
V, R
MS
C, microfarads
Figure 8: Voltage Magnitude
The magnitude of this is plotted in Figure 10
10. The two phasor diagrams are shown in Figure 11
Source voltage is:V = Vs + jXI
The locus of this voltage, with arbitrary phase angle of I is shown in Figure 12.
And the range of source voltage magnitudes is:
90 < |V | < 110
11. Inductive reactance is X = 2π × 60 × .02 ≈ 7.54Ω, so receiving end voltage is
Vr = VsR
R+ jX= Vs
R2 − jXR
R2 +X2≈ 76.6 − j57.7V
A phasor diagram of this case is shown in Figure 13.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 7
Vr
Vx
Vc
Figure 9: Voltage Phasors at Maximum Output Voltage
With the capacitor in place, the ratio of input to output voltages is:
Vr = VsR|| 1
jωC
R|| 1jωC
+ jωL= Vs
1
1 − ω2LC+jωL
R
To make the magnitude of output voltage equal to input voltage, it is necessary that:
(
1 − ω2LC)2
+
(
ωL
R
)2
= 1
Or noting X = ωL and Y = ωC
(XY )2 − 2XY +
(
X
R
)2
= 0
This is easily solved by:
Y =1
X±√
(
1
X
)2
− 1
R2
With X = 7.54Ω and R = 10Ω, this evaluates to Y = .0455S, so that C = .0455377 ≈ 120µF .
To construct the phasor diagram, start by assuming the output voltage is real (Vr = 120),Then the capacitance draws current Ic = .0455j×120 ≈ j×5.46A. Current through theinductance is Ix = 12+j5.46, and the voltage across the inductance is Vx = −41+j90.48.Source voltage is Vs = 78.8+ j90.48, which has magnitude of 120 V (all of this is RMS).The resulting phasor diagram is shown in Figure 14.
Maximum voltage at the outupt is clearly achieved when ω2LC = 1, when C = 351.8µF .Maximum output voltage is Vr = Vs
RωL
≈ 1.33 × 120 ≈ 159V. A plot of relative outputvs. input voltage is shown in Figure 15
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 8
0 100 200 300 400 500 6000
20
40
60
80
100
120Chapter 2, Problem 9
Res
isto
r V
olta
ge
Capacitance, Microfarads
Figure 10: Resistor Voltage Magnitude
Vx = 10
Vx = j10
V = 100 + j10
Vs = 100I=1
I=−j
V=110Vs = 100
Figure 11: Phasor Diagrams for Problem 10
12. The situation is shown in the phasor diagram of Figure 16. In complex terms, V =V s + jXI . In this situation, we know the magnitude of V s and the angle between Vand I. To find the magnitude of V , we invoke the law of cosines:
V 2s = V 2 + (XI)2 − 2V XI cos θ
Now, since θ = ψ + π2 ,
V 2s = V 2 + (XI)2 + 2V XI sin θ
This quadratic is solved by (for the most reasonable value of voltage:
V
Vs=
√
1 −(
XI
Vs
)2
+
(
XI
Vssinψ
)2
− XI
Vssinψ
This is plotted in Figure 17. To plot this against real power, all that needs to be notedis that P = V I cosψ. It should be noted that this system cannot make the specifiedamount of real power for some of the power factor cases. This is shown in Figure 18.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 9
Vs = 100
Locus of Input V
|I|=1
Figure 12: Locus of Current and Voltage Phasors
R
V = 120S
V = 76.6 −j 57.7
X V = 43.4 + j 57.7
Figure 13: Phasor Diagram: Uncompensated
RV = 120
xV = −41.2 +j 90.4
V = 78.8 + j 90.4S
Figure 14: Phasor Diagram: compensated to equal voltage
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 10
0 100 200 300 400 500 600 700 8000.7
0.8
0.9
1
1.1
1.2
1.3
1.4Chapter 2, Problem 11, vr vs. C
Vol
tage
mag
nitu
de r
atio
C, microfarads
Figure 15: Voltage transfer ratio vs. Capacitance
jXI
ψ
θ
Vs
V
I
Figure 16: Phasor Diagram: Terminal Voltage
0 200 400 600 800 10000
2000
4000
6000
8000
10000
12000Chapter 2, Problem 12
V, R
MS
A, RMS
Figure 17: Source Voltage vs. Current
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 11
0 2 4 6 8 10
x 106
0
2000
4000
6000
8000
10000
12000Chapter 2, Problem 12
V, R
MS
W
Figure 18: Source Voltage vs. Power
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 12
Chapter 3
1. Since Z0 =√
LC
and phase velocity is u = 1√LC
, L = uZ2s , or
L = .18355 × 108m/s × (30.3Ω)2 ≈ 0.165µH/m
In the steady state,
V = V+ + V− = 63.6kV
I =V+
Z0− V−Z0
= 325A
This solves for:
V+ = 36.7kV
V− = 26.9kV
I+ = 1212A
I− = 887A
At the instant of the switch opening, I+ + I− = 0, so I− = −1212A, and V− = 36.72kV.Total voltage is V = V+ + V− ≈ 73.4kV. When the excitation gets back to the sendingend, at time ∆T = 50×103m
1.83558m/s≈ 272.7µs, the forward going voltage is defined by
Vs = V+ + V−, or
V+ = Vs − V− = 63.6kV − 36.72kV ≈ 26.9kV
So current is:
I =V+
Z0− V−Z0
≈ −325A
This is shown in Figure 19
−325
∆ t∆2
t∆2t∆
t
t
Vr
Is
63.673.4
53.4
325t
Figure 19: Voltage Transients
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 13
2. Complex amplitude of voltage along the line is:
V = V +e−jkx + V −e
jkx
I =V +
Z0e−jkx − V −
Z0ejkx
If current is zero at x = 0 (line open) then V − = V +. At the sending end of the line,
x = −l, Vs = V +
(
ejkl + e−jkl)
= 2V + cos kl, and then receiving end (x = 0) voltage is:
Vl = 2V + =Vs
cos kl
At the source,
I =V +
Z0
(
ejkl − e−jkl)
= 2jV +
Z0sin kl = j
VsZ0
tan kl
In this case, wavelength is λ =1.84×108m/s
60s ≈ 3.1 × 106m, and l = 50km = 5× 104m, so
kl = 2π×5×104
3.1×106 ≈ 0.103. Then:
Vl ≈ 45.24kV (RMS)
Is ≈ 153A (RMS)
If the line is loaded with a unity power factor load with current IL, the relationshipbetween forward and reverse going components is:
V + − V − = Z0IL
orV − = V + − Z0IL
At the source end:
V + − V − = Z0IS
V + + V − = VS
Some algebra is required to find:
V + =VS + Z0ILe
−jkl
2 cos kl
V − =VS − Z0ILe
jkl
2 cos kl
Source current is:
IS =1
Z0
(
jVS tan kl +Z0ILcos kl
)
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 14
This would imply a limit on IL that is:
IL <
√
(Is cos kl)2 −(
VsZ0
sin kl
)2
This evaluates to IL < 285.1A. Power factor is:
cosψ =IL
cos kl
Is≈ 0.88
3. Inductance and capacitance are:
C =1
Z0c=
1
250 × 3 × 108≈ 1.33 × 10−11 = 13.3pF/m
L =Z0
c=
250
3 × 108≈ 8.333 × 10−7H/m
If the current is introduced in the middle of the line, we will have V+ = V− and I+ = −I−propagating away from the source, with I+ = −I− = 10, 000A and V+ = V− = 250 ×10, 000 2.5MV
At the shorted end, the current will double as voltage goes to zero. At the matched endthe voltage will appear as it is in the initial propagating wave, with no reflection. Theresult is shown in Figure 20.
1 ms
IL
VR 2.5 MV
20 kA20 sµ
Figure 20: Voltage Transients
4. Wavelength is λ = 3×108
60 = 5× 106m, so for a 300 kilometer line, kl = 2π × 3005000 ≈ .377.
Then open circuit sending end voltage is Vr = Vs
cos kl = 500.93 ≈ 537.8kV. Sending end
current is Is = Vs
Z0tan kl = 500kV
250Ω × tan 0.37 ≈ 792A
For source impedance of zero, voltage and current along the line are:
V (x) = Vs
ZL
Z0cos kx− j sin kx
j sin kl + ZL
Z0cos kl
I(x) =VsZ0
cos kx− j ZL
Z0sin kx
ZL
Z0cos kl − j sin kl
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 15
Evaluated for ZL
Z0= 1
0.8 , 1.0 and 11.2 ,and with source voltage of 500 kV, receiving end
voltage (at x = 0) is evaluated to have magnitude of 512.7 kV, 500 kV and 485.7 kV.Sending end current (at x = −l) is 1702 A, 2000 A and 2283 A, respectively.
Using the same formulae, with varying receiving end resistance, voltage is plotted inFigure 21.
0 1 2 3 4 5 6 7 8 9 10
x 108
5
5.05
5.1
5.15
5.2
5.25
5.3
5.35
5.4x 10
5 Chapter 3, Problem 4, Voltage vs. Loading
V, R
MS
Real Power, W
Figure 21: Receiving End Voltage
To estimate the effect of compensation, we assume a capacitance in parallel with thereceiving end, with a capacitive admittance of Yc = 2Q
V 2 . This is placed in parallel withthe receiving end resistance. The voltage at the receiving end is calculated in the normalway and is shown in Figure 22. Note there are three curves, corresponding to the threelevels of real load. Note also that the case of surge impedance loading (2,000 A) hasnominal voltage with zero compensation.
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
x 108
4.4
4.6
4.8
5
5.2
5.4
5.6x 10
5 Chapter 3, Problem 4, Compensation
V, R
MS
Compensation, VARs
Figure 22: Receiving End Voltage
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 16
Chapter 4
1.
ia ib icLoad A
√2 cosωt cos(ωt− 2π
3 ) cos(ωt+ 2π3 )
Load B√
2 cosωt cos(ωt− 2π3 ) cos(ωt+ 2π
3 )
Load C√
23 cos(ωt+ π
6 ) −√
23 cos(ωt+ π
6 ) 0
Load D√
2 cosωt cos(ωt− 2π3 ) cos(ωt+ 2π
3 )
Load E√
2 cosωt −√
23 cos(ωt+ π
6 ) −√
23 cos(ωt − π
6 )
Load F√
2 cosωt cos(ωt− 2π3 ) 0
2. Voltage magnitude is RI = 500 volts. The voltages across the three phase resistancesare just current times resistance. The voltage across the ground (neutral) resistor isthe resistance times the sum of the three phase currents, which is always either plus orminus the peak amplitude. The results are shown in Figure 23
g
ωt
ωt
ωt
ωt
500 v
500 v
500 v
500 v
π3
vb
va
vc
v
Figure 23: Resistor voltages
3.
vn = va1
2+ vb
23
23 + 2
+ vc
23
23 + 2
= va
(
1
2− 2
5
)
=1
10va
4. Neutral voltage is the average of the three sources, which will have amplitude of ±1003 V.
Voltage across the individual resistors will be the difference between phase voltage andneutral voltage, and this is shown in Figure 24.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 17
c
vn
va
i a
i b
i
Figure 24: Phase Currents
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 18
Chapter 5
1. Inductance is L = µ0N2Ag
, where g is total gap: g = 2 × .0005 = .001 m and A =
.02 × .025 = 5 × 10−4 m2. Then:
L = 4π × 10−7 × 1002 × 5 × 10−4/.001 ≈ 6.28mH
Since flux density in the gap is Bg = µ0NIg
, current required to make 1.8 T would be:
I =Bgg
µ0N=
1.8 × .001
4π × 10−7 × 100≈ 14.3A
To make an inductance of 10 mH, and noting that the gap on either side is half of thetotal gap:
gs =1
2g =
1
2
µ0N2A
L=
4π × 10−7 × 104 × 5 × 10−4
.01≈ .000314m
As a check: note that inductance is inversely proportional to gap dimension, so thatL1g1 = L2g2, or the required gap would be:
g2 =6.28mH
10mH× .0005m
2. Gap area is A = RθgL, where L is the axial length. Then maximum inductance is,noting that there are two gaps in series:
L = µ0N2RθgL
2g= 4π × 10−7 × 502 × .05 × π
6 × .1
2 × .0001≈ 10.3mH
If fringing can be ignored, the area for calculation of inductance falls linearly withrotational angle until the rotor pole is completely disengaged from the stator pole atθ = 30. The inductance vs. angle is shown in Figure 25.
o
10.3 mHy
30o 180150−30o o
Figure 25: Solution to Problem 2, part b
3. This problem has two gaps. The axial (variable) gap has reluctance:
Ra =x
µ0πR2
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 19
The radial clearance gap is, if the gap itself can be considered to be ’small’(for parts (a)and (c):
Rg1 =g
µ02πRW
If, on the other hand, the gap is not small, the reluctance is:
Rg2 =log Ro
Ri
µ02πW
The rest is documented in the atlab script p5 3.m. Inductances limited by the radialgap are:Part a) L = N2
Rg1≈ 15.79mH
Part b) L = N2
Rg2≈ 2.28mH
With nonzero axial gap, the inductances are L = N2
Ra+Rg1or L = N2
Ra+Rg2.
These are plotted in Figure 26.
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016Chapter 5, Problem 3
Indu
ctan
ce, H
Gap, m
Part cPart d
Figure 26: Solution to Problem 3, parts c and d
4. This problem involves a radius ratio large enough that a single path length cannot beassumed. Flux density is:
Bφ =µ0µrNI
2πr
Flux in the core is:
Φ = D
∫ Ro
Ri
Bφdr =µ0µrNI
2πlnR0
Ri
Inductance is then
L =NΦ
I=µ0µrN
2
2πlnRoRi
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 20
For the parameters and dimensions of this problem: Ri = .02m, Ro = .05m, D = .01m,µr = 200, N = 100, this evaluates to:
L = 3.665mH
Current required to saturate the core at radius r is:
I =1
N
2πrBsµ0µr
This evaluates to 6 A at r = .02 m and 15 A at r = .05 m.
5. This problem has three gaps, with reluctances:
RL =g
µ0D(x0 − x)
RR =g
µ0D(x0 + x)
RC =g
µ0Dx0
Straightforward circuit manipulation will give the flux in the center leg:
Φc = NI
[
1
RL + Rc||RR
RR
RC + RR− 1
RR + Rc||RL
RL
RC + RL
]
A bit of manipulation is required to put this into the form:
Φc = NI
[ RR −RL
Rc (RL + RR) + RRRL
]
Inserting the definitions (above) for the reluctances and manipulating,
Φc = NIµ0D
g
2
3x
Voltage induced in the central coil is
V = ωNcΦc = ωNc = NIµ0D
g
2
3x
Evaluated at x = .01 this is 105 V (peak). Plotted in Figure 27 is the absolute value(magnitude) of this. In a real application this voltage would be synchronously detectedso that the sign of x could be detected.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 21
x
105 V
.01.01
Figure 27: Solution to Problem 5
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 22
Chapter 6
1. Secondary IX = 24kVA240V = 100A
Primary IH = 24kVA24kV = 1A
Number of primary turns NH = 100 × 26 = 2, 600Loaded on the low voltage side, RX = 2402
24,000 = 2.4Ω
Referred to the high side, RH = N2 ×RX = 24kΩ
2. Note that this is an approximate analysis that is very close to being correct if the coreelements are large (low loss) and the series elements are very small (also low loss).
Referred to the high side,
Rc =8, 0002
100= 640kΩ
Xc =8, 0002
1000= 64kΩ
Referred to the low side,
Rc =2402
100≈ 576Ω
Xc =2402
1000≈ 57.6Ω
Rated current is 3 A on high side or 100 A on low side, so that if the series resistor ison the high side it would be: R = 1,200
32 ≈ 133Ω, or on the low side: r = 1,2001002 ≈ .12Ω
3. Voltage on the X side of each transformer is 128 kV. On the H side it is VH = 345√3≈
199kV, so that the turns ratio is N = 128199 ≈ 0.643.
The phasor diagram showing primary and secondary voltages, both line-line and line-neutral, is shown in Figure 28.
Magnitudes of the high side current will be IH = 1003×199 = 167.5A. Since the inverse
cosine of 0.8 is 36.9, and since the primary (H) side is at an angle of −30, the threeprimary currents will have angles of −6.9, −126.9 and 113.1. Currents in the lowside leads will have magnitude IX = 100
3×128 ≈ 451.2A, and their angles will be −36.9,−156.8 and 83.1.
4. Line-neutral voltage on the X side is 208√3
= 120, so the turns ratio is N = 600120 = 5.
The phasors for input and output voltage are shown in Figure 29.
If it is assumed that high side voltage in phase A has angle of zero, the secondary side
voltage is VAX = 120ejpi6 so that IAX = 12ej
π6 . then high side currents will be:
IAH =12
5ej
π6
IBH =12
5e−j
5π6
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 23
caH
VaX
V
V
V
bXV
cX
abX
bcXV
caX
V
VV
VV
V
aH
bH
cH
abHbcH
Figure 28: Solution to Chapter 6, Problem 3, Phasors
Real and reactive power on the high side are:
PAH =600√
3
12
5cos
π
6≈ 720W
PBH =600√
3
12
5cos−π
6≈ 720W
QAH = −600√3
12
5sin
π
6≈ −416VAR
QBH =600√
3
12
5sin
π
6≈ 416VAR
The phasor diagram for the high side voltage and current is shown in Figure 30.
5. Let’s assume that VA has an angle of zero. The three low-side voltages will be:
VA = 277
VB = 277e−j2π3
VC = 277ej2π3
Then the three currents on the load side are:
IA = 100(
ejπ6 + e−j
π6
)
=√
3 × 100 ≈ 173.2A
IB = −100ejπ6 = 100e−j
5π6
IC = −100e−jπ6 = 100ej
5π6
Script p6 5.m finishes the problem, and the detailed answers are:
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 24
cX
V
V
V
V
V
V
V
V
V
aH
bH
cH
abH
bcH
caH
aX
bX
Figure 29: Solution to Chapter 6, Problem 4, Part b
Problem 6_5
Secondary (LV)
VA = 277.128 + j 0 = 277.128 angle 0 deg
VB = -138.564 + j -240 = 277.128 angle -120 deg
VC = -138.564 + j 240 = 277.128 angle 120 deg
IA = 173.205 + j 0 = 173.205 angle 0 deg
IB = -86.6025 + j -50 = 100 angle -150 deg
IC = -86.6025 + j 50 = 100 angle 150 deg
Primary (HV)
VA = 6900 + j -3983.72 = 7967.43 angle -30 deg
VB = -6900 + j -3983.72 = 7967.43 angle -150 deg
VC = 9.09495e-13 + j 7967.43 = 7967.43 angle 90 deg
IA = 5.21739 + j -1.00409 = 5.31313 angle -10.8934 deg
IB = -5.21739 + j -1.00409 = 5.31313 angle -169.107 deg
IC = 0 + j 2.00817 = 2.00817 angle 90 deg
Secondary Complex Power
A = 48000 + j 0
B = 24000 + j 13856.4
C = 24000 + j -13856.4
Primary Complex Power
A = 40000 + j -13856.4
B = 40000 + j 13856.4
C = 16000 + j -1.82642e-12
Total Secondary = 96000 + j 0
Total Primary = 96000 + j 1.63635e-11
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 25
aH
V
V
aH
bH
I
IbH
Figure 30: Solution to Chapter 6, Problem 4, Part c
Figure 31 shows the voltage and current phasor diagrams for both primary and secondary.
See script p6 5.m, which uses two auxiliary functions dispc.m and dispp.m.
6. The turns ratio is N = 13,800√3×480
≈ 16.5988.
Using V0 as the magnitude of the primary voltage, the secondary voltages will be:
Va =V0√
3Ne−j
π6
Vb =V0√
3Ne−j
5π6
Vc =V0√
3Nej
π2
The currents have amplitude 100 A and so are:
Ia = 100e−jπ6
Ib = 100e−j5π6
Ic = 100ejπ2
Currents on the primary side of the transformers will then be:
IB = 1NIb =
100
Ne−j
5π6
IC = − 1NIa =
100
Ne−j
π6
These are shown in Figure 32
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 26
High Voltage (H)
100 V
100 A
VA
V
V
I
I
I
B
C
C
B
VV
V
I
II
c
ab
b a
c
1 kV
1A
Low Voltage (X)
Figure 31: Solution to Chapter 6, Problem 5
Currents in Primary
ab
c
100
I C
I B
6
Currents in Load
Figure 32: Solution to Problem 6, Problem 6, Part c
The primary currents and their phasor relationship to the primary voltage is shown inFigure 33. Primary power is:
P = 3 × 100 × 480√3≈ 83138W
Secondary power is:
P = 2 × 13, 800√3
× 100
16.5988cos
π
6≈ 83138W
Incidentally, the resistors must have value R = 277100 ≈ 2.77Ω, so when the ground is
lifted, all of the primary voltage appears across the two transformer legs, putting currentthrough two of the resistors: I = −j13,800
16.59881
2.77 ≈ −j150A Primary current is −j15016.5988 ≈
−9.0368A
7. This problem is done by Matlab script p6 7.m. There are three cases to be solved:
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 27
C
1A
1 kV
VA
V
VC
B
I
I B
Figure 33: Solution to Problem 6, Problem 6, Part d
(a) Star point of the resistors connected to the neutral of the supply, in which case thecurrents can be calculated independently and the problem is simple,
(b) Star point of the resistors is unconnected to the neutral of the supply. In this case,it is straightforward but tedious to convert the wye to a delta, calculate line-linevoltages, obtain current in the legs of the delta, add those to get terminal currents,transform them across the transformer and add transformer currents together to getterminal currents on the delta side, and
(c) Star point is grounded through a resistor. This is handled by calculating theimpedance matrix:
VaVbVc
Ra +Rg Rg RgRg Rb +Rg RgRg Rg Rc +Rg
iaibic
In principal, this matrix can be inverted to find the currents, since voltages areknown. To check, it is possible to set Rg = 0, in which case the third case shouldequal the first case. Or to set Rg to a very large number, in which case the thirdcase should approach the second case.
The answer for the problem as posed is:
Chapter 6, Problem 7
Part a: solidly grounded
Secondary
ia = 5.54256 + j 0 = 5.54256 angle 0 deg
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 28
ib = -4.6188 + j -8 = 9.2376 angle -120 deg
ic = -4.6188 + j 8 = 9.2376 angle 120 deg
Primary
iA = 2.03077 + j -1.59882 = 2.58462 angle -38.2132 deg
iB = -2.03077 + j -1.59882 = 2.58462 angle -141.787 deg
iC = 0 + j 3.19763 = 3.19763 angle 90 deg
Part b: ungrounded
Secondary
ia = 6.39526 + j 0 = 6.39526 angle 0 deg
ib = -3.19763 + j -8 = 8.61538 angle -111.787 deg
ic = -3.19763 + j 8 = 8.61538 angle 111.787 deg
Primary
iA = 1.91716 + j -1.59882 = 2.49634 angle -39.8264 deg
iB = -1.91716 + j -1.59882 = 2.49634 angle -140.174 deg
iC = -1.77504e-16 + j 3.19763 = 3.19763 angle 90 deg
Part c: Grounded through 1000 ohms
Secondary
ia = 6.38554 + j 1.75423e-15 = 6.38554 angle 1.57402e-14 deg
ib = -3.21384 + j -8 = 8.62141 angle -111.887 deg
ic = -3.21384 + j 8 = 8.62141 angle 111.887 deg
Primary
iA = 1.91846 + j -1.59882 = 2.49734 angle -39.8074 deg
iB = -1.91846 + j -1.59882 = 2.49734 angle -140.193 deg
iC = 3.55008e-16 + j 3.19763 = 3.19763 angle 90 deg
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 29
Chapter 7
1. Reactance X = ω (L−M) ℓ = 377 × 9 × 10−7 × 100 × 103 ≈ 34Ω.
Resistance R = 1.2 × 10−6 × 105 = 0.12Ω
Impedance Z = 0.12 + j34Ω
2. Base current is: IB = PB√3VB
= 100MVA√3138kV
= .418kA = 418A
Base impedance is: ZB =V 2
B
Pb= 1382
100 =≈ 190.4Ω
3. Per-Unit Impedance is: z = 0.12+j34190.4 = .0006 + j.1786
4. Reactance X = 0.4Ω/km × 50km = 20Ω
Line-neutral voltage V = 138√3
= 79.67kV
Current |I| = VX
= 79.6720 ≈ 3.984kA = 3984A
5. Line impedance is ZL = 50 × (j.35 + .02) = j17.5 + 1Ω
Current I = 79.671+j17.5 ≈ 0.259 − j4.538kA
Current magnitude |I| ≈ 4545A
6. Put this on 100 MVA base:
Generator: x = 100200 × .25 = .125
The transformer is already on this base: x = .05
Line impedance Z = j17.5 + 1 and base impedance ZB = 1382
100 = 190.44Ω, so per-unitline impedance is: zℓ = j.092 + .005
Total impedance is z = j (.125 + .05 + .092) + .005 = j.267 + .005
Fault current is iF = 1j.267+.005 ≈ .07 − j3.745
|iF | ≈ 3.745
Base currents are:
At generator: 100√3×13.8
≈ 4184A
On the line: 100√3×138
≈ 418.4A
So fault currents are:
At generator: 418.4 × 3.745 ≈ 1566.7AOn the line: 4184 × 3.745 ≈ 15667A
7. Put this on on 100 MVA base. The impedances are:Generator: xg = .125First Transformer: xt1 = .05Line: zℓ = j.092 + .005Second Transformer: xt2 = 100
20 × .07 = 0.35
Base currents are:Generator: 4184 ALine: 4182 AAt fault: 24,056 A
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 30
Impedance to the fault is: z = j (.125 + .05 + .092 + .35) + .005 = j.617 + .005|z| ≈ 0.617
Fault current is |iF | ≈ 1.621per-unit
In amperes:
IF = 6782A (generator)
= 678.2A (line)
= 38995A (at fault)
8. Put this one on a 100 MVA base. The impedances are: Generator: xg = 100500 × .25 = .05
Transformer: xt = 100500 × .05 = .01
50 km of line (see problem 6): zℓ = j.092 + .005
The problem can be represented as shown in the circuit diagram of Figure 34. Thegenerator and transformer are lumped together to form a reactance of 0.6 per-unit. Theupper line and right-hand part of the lower line are in series with an impedance of threetimes the left-hand side of the lower line. Total impedance from the source to the faultis: z = j.06 + zℓ||3zℓ ≈ j.129 + .00375. Currents through the two line segments aredetermined by a current divider:
i1 =1
4iF
i2 =3
4iF
−
j.06 .015 j.276
.005 j092
is i1
i2
if1+
Figure 34: Fault Situation
Then the per-unit currents are:
iF =1
j.129 + .00375≈ .225 − j7.743
i1 =1
4iF ≈ .05625 − j1.93575
i2 =3
4iF ≈ .16875 − j5.80725
To convert to ordinary variables, we need base currents:
IBH =100√3345
= 167A
IBG =100√324
= 2406A
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 31
Then the currents are:
fault (345 kV) 37.6 − j1293A
transformer (345 kV) 37.6 − j1293A
upper line (345 kV) 9.4 − j323A
lower left line (345 kV) 28.2 − j970A
generator (24 kV) 541 − j18630A
9. GMD =√.78 × .06 × .5 ≈ .140m
L = µ0
2π logR0R′ = µ0
2π log 100.140 ≈ 1.314 × 10−6H/m
10. GMD = 4√.78 × .06 × 13 ≈ .456m
11. GMD of the bundles is 0.140 m (see Problem 9)
(L−M)adjacent =µ0
2π× log
10
.14≈ 8.54 × 10−7H/m
(L−M)outside =µ0
2π× log
20
.14≈ 9.92 × 10−7H/m
(L−M)average =2
3(L−M)adjacent +
1
3(L−M)outside ≈ 9/times10−7H/m
Resistance of the aluminum conductors is: R = 12
1π×.032×3×107 ≈ 5.895 × 10−6Ω/m
Then, since 10 km is 104 m, Z = .05985 + j3.393Ω
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 32
Chapter 8
1. 3,000 RPM is 314.16 Radians/second, so Torque is:
T =1, 000W
314.16≈ 3.183N −m
Since T = 2πR2Lτ , and if L = 2R, and if τ = 4, 000Pa,
R = 3
√
3.183
4π × 4, 000≈ .03986m
Then D = L = 7.97 cm.
2. L = µ0N2Ag
and F e = i2
2∂L∂g
= − i2
2µ0N
2Ag2
sof e = 251.3N
3. Inductance is L = N2
Rg+Rx, where Rg = g
µ02πRW and Rx = xµ0πR2 . Then force is found
to be:
f e =µ0πR
2N2i2
(x+ gR2W )2
Since flux is Φ = NiRg+Rx
and Bx = ΦπR2 , current is:
i =Bxµ0N
(
x+gR
2w
)
A Matlab script p8 3.m calculates force. The naive calculation is shown in Figure 35.Logarithmic coordinates are used because the force goes so high.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
x 10−3
100
101
102
103
104
105
Chapter 8, Problem 3
For
ce, N
Displacement x, m
Figure 35: Solution to Problem 3: Naively derived force
Current to achieve flux big enough to approach saturation of the magnetic circuit isshown in Figure 36.
With this figure, you should ’smell a rat’, because the magnetic circuit is very highlysaturated with 10 A at small gaps. The force is limited to about what would be achievedwith 1.8 T, just over 100 Newtons. With that limit, the actual achievable force is shownin Figure 37.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 33
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
x 10−3
0
10
20
30
40
50
60
70
80Chapter 8, Problem 3
Cur
rent
, A
Displacement x, m
Figure 36: Solution to Problem 3: current to achieve saturation flux density
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
x 10−3
0
20
40
60
80
100
120Chapter 8, Problem 3
For
ce, N
Displacement x, m
Figure 37: Solution to Problem 3: more realistic force
4. If L = L0 + L2 cos 2θ, then
T =I2
2
∂L
∂θ= −I
2
2L2 sin 2θ
Then, if I = I0 cosωt,
T = −I20
2L2 cos2 ωt sin 2(ωt + δ)
What is interesting about this is the time average: using cos2 x = 12 + 1
2 cos 2ωt andsinx cos y = 1
2 sin(x+ y) + 12 sin(x− y), time average torque is found to be:
< T >=I20
4L2 sin 2δ
5. The inductance was estimated in Chapter 5, and is, for θ > 0,
L =µ0R(θ0 − θ)N2
2g
Torque is
T e =I2
2
∂L
∂θ= ±µ0RN
2I2
4g
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 34
or zero.
This is shown for the region of angle around zero in Figure 38. For the values given inthe problem statment, torque amplitude is:
Tm =µ0 × .02 × 10002
.002= 4π ≈ 12.57Nm
m
30o−30o
L max
T
Figure 38: Solution to Problem 4: Inductance and Torque
6. Surface current K = I0D
and force can be found using any of several methods (PrincipleOf Virtual Work, Maxwell Stress Tensor being the most convenient). It is:
f e =µ0
2K2WD =
µ0
2I20
W
D
Voltage is found using velocity of the block (projectile) u:
V =dΦ
dt= µ0KWu = µ0I0
W
Du
That velocity will be u = fe
Mt, so block position is x(t) = 1
2F e
Mt2.
Power converted into mechanical motion is:
Pm = f eu =µ0
2I20
W
Du
Power out of the source is
P e = V I0 = µ0I20
W
Du
Comparing the two,
η =Pm
P e=
1
2
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 35
7. Core area is A = .25m×.25m = .0625m2, and peak terminal voltage is Vp = 2400×√
2 ≈3394V , so that peak flux density in the core is:
Bp =3394
377 × .0625 × 96≈ 1.5T
Eyeballing the data given in Figures 8.19 and 8.20, we may estimate core loss to be about3.2 watts/kilogram and exciting power to be about 19.3 VA/kilogram. The volume ofactive material in the core is just about:
Vc = 1m× .75m× .25m− 2 × (.25m× .5m× .25m) = .125m3
If the core material density is 7,700 kg per cubic meter, this means the mass of active corematerial is 962.5 kilograms. Thus the core loss is Pc = 962.5×3.2 ≈ 3041 watts. Excitingpower is Pa = 962.5 × 19.3 ≈ 18576 VA. Exciting current is then about 7.7 Amperes.
8. Ampere’s Law in the gap region gives the relationship:
g∂Hy
∂x= Ks
or−jkgHy = Kzs
then
Hy = Re
j
kgKzse
j(ωt−kx)
Force on the lower plate will be vertically upward:
Tyy =1
2µ0H
2y
and will have the average value of:
< Tyy >=µ0
4
|Kzs|2(kg)2
9. From the prior problem, y- directed magnetic field in the gap is seen to be:
Hy = Re
j
kg
(
Kzs +Kzrejkx0
)
ej(ωt−kx)
Vertical force on the lower surface is:
Tyy =µ0
2|Hy|2
And this will have average value:
< Tyy >=µ0
4
1
(kg)2Re(
Kzs +Kzrejkx0
)(
Kzs +Kzre−jkx0
)
=µ0
4(kg)2
(
K2zs + k2
zr + 2kzskzr cos kx0
)
Shear stress is Txy = µ0HxHy. The y- directed field is found already. The x- directed
field at the lower surface is Hx = −Re
Kzrejkx0
. Shear stress is then found to be:
< Txy >= − µ0
2kgRe
j(
Kzs +Kzrejkx0
)
Kzre−jkx0
= − µ0
2kgKzsKzr sin kx0
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 36
10. DC resistance per unit length is Rℓ
= 1σA
= 1.01×.05×5.81×107 ≈ 3.44 × 10−5Ω/m =
34.4µΩ/m
At 60 Hz the skin depth is
δ =
√
2
377 × 4π × 10−7 × 5.81 × 107≈ 0.852cm
For really deep linear material, resistance and reactance are equal:
R
ℓ=X
ℓ=
1
σδw=
1
5.81 × 107 × .00852 × .01≈ 2.02 × 10−4 = 202µΩ/m
For material with some limited depth, use the expression for surface impedance:
Zs = jµ0ω
γcoth γh
In this case, where the wavenumber k can be taken to be zero, the propagation constantis:
γ =√
jωµ0σ =1 + j
δ
and the surface impedance is:
Zs =1 + j
σδcoth(1 + j)
h
δ
The script that calculates this as a function of frequency is p8 10.m. The results areshown in Figure 39. Not surprisingly, since this is actually a fairly deep slot (comparedwith the skin depth), the resistance and reactance are not far from the infinitely deepcase, with R = 201.914µΩ/m and X = 201.919µΩ/m.
100
101
102
103
10−5
10−4
10−3
Chapter 8, Problem 10
Ohm
s/m
eter
Frequency, Hz
ResistanceReactance
Figure 39: Solution to Problem 10
11. With saturating iron, the skin depth is: δ =√
2H0ωσB0
. With the data given in the problem,
δ =
√
20, 000
377 × 6 × 106 × 1≈ .00297m = 2.97mm
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 37
Surface impedance is
Zs =8
2π
1
σδ(2 + j) ≈ 4.76 × 10−5(2 + j)Ω
Then power per unit area is
1
2× 10, 0002 × 2 × 4.758 × 10−5 ≈ 4758W/m2
The solution to this is plotted for a range of current density from 10,000 to 100,000amperes per square meter in Figure 40.
1 2 3 4 5 6 7 8 9 10
x 104
0
2
4
6
8
10
12
14
16x 10
4 Chapter 8, Problem 11
W/m
2
A/m
Figure 40: Solution to Problem 11
See Matlab script p8 11.m.
12. Skin depth in Aluminum is:
δ =
√
2
377 × 3 × 107 × 4π × 107≈ .01886m
Then surface impedance is:1
σδ≈ 2.81 × 10−6Ω
With surface current dentisy of 1, 000A/m2, loss is about 140.5 watts per square meter.
Loss density in linear material is proportional to the square of current density, as isshown in Figure 41
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 38
104
105
102
103
104
105
Chapter 8, Problem 12
W/m
2
A/m
Figure 41: Solution to Problem 12
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 39
Chapter 9
1. Peak phase voltage is Vph,pk =√
23 × 26, 000 ≈ 21, 229V , and since this is Vph,pk =
ωMIfnl,
M =21, 299
377 × 1, 200≈ 46.9mH
Per-unit synchronous reactance is xd =Ifsi
Ifnl= 2.0.
Base impedance is ZB =V 2
Bℓ−ℓ
PB= 262
1,200 ≈ 0.5663Ω, so synchronous reactance is: Xd =2 × .5633 ≈ 1.127Ω and then
Ld =1.127
377≈ 2.99mH
2. Driven by current, torque is T e = −32MIaIf sin δi and this is:
T e = 1.5 × .056 × 1, 000 × 3, 1113 sin δi ≈ −2, 613, 492 sin δi
Driven by voltage, power is P e = −32VaEaf
Xdsin δ and torque is T e = p
ωP e.
Synchronous reactance is Xd = ω(La − Lab) = 377 × .0036 ≈ 1.3573Ω, to power is:
P e =1.5 × 21, 229 × 21112
1.3572sin δ ≈ −4.95 × 109 sin δ
Torque is then:
T e = −4.95 × 108
377sin δ ≈ −1, 313, 908 sin δ
The rest of this problem is implemented Matlab script p9 2.m, which generates thefollowing output:
Chapter 9, Problem 2: 60 Hz
Phase Voltage = 15011.2 RMS
Phase Current = 22205.7 A, RMS
Phase Reactance X = 1.35717 Ohms
Internal Voltage Eaf = 33668.5 RMS
Field Current I_f = 2255.38 A
Voltage Torque Angle = 63.5221 degrees
Current Torque Angle = 206.478 degrees
Check on power = 1e+09 and 1e+09
Torque = 2.65258e+06 N-m
A phasor diagram of this machine operation is shown in Figure 42
3. The solution to this problem is implemented in Matlab script p9 3.m. Phasor diagramsfor unity power factor operation are shown in Figure 43 and Figure 44.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 40
af
δ=63.5o
=206.5o
iδ
V =30,136 vx
Internal Flux
Current (motor sense)
Current (generator Sense)V=15011 v
E = 33668 v
Figure 42: Solution to Chapter 9, Problem 2
Chapter 9, Problem 3 f = 60
Part a:Ifnl = 49.9806
Part b:Ifsi = 102.009
Power Factor = 1
Power Factor Angle = 0 degrees
Angle delta = -53.7004 degrees
Current Angle = 53.7004 degrees
Terminal Voltage = 2424.87
Internal Voltage E1 = 4096.02
Internal Voltage Eaf = 5424.17
Current I_d = -110.787
Current I_q = 81.3799
Angle of Max Torque = -78.12 degrees
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 41
Breakdown Torque = 11902.6 N-m
d axis
δ
Eaf
V
j X I aq
I a
I d
I q
Figure 43: Solution to Chapter 9, Problem 3: Unity Power Factor
4. Peak phase voltage is Vph,pk =√
23 × 13, 800 = 11, 267.7V, Peak.
M =11, 267.7
377 × 100≈ 299mH
Base impedance is: ZB = 13.82
100 ≈ 1.9044Ω
Then, base inductance is: LB = 1.9044Ω377 ≈ 5.04mH
Thus:
Ld = 2 × 5.05mH = 10.1mH
Lq = 1 × 5.05mH = 5.05mH
To understand torque stability, note that:
T =veafxd
sin δ +v2
2
(
1
xq− 1
xd
)
sin 2δ
Then the stability point is defined by:
∂T
∂δ= −veaf
xdcos δ − v2
(
1
xq− 1
xd
)
cos 2δ
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 42
ψ
E af
j X Iq a
V
I a
I d
I q
E1
δ
Figure 44: Solution to Chapter 9, Problem 3: 0.8 Power Factor, Overexcited
At δ = 0 and v = 1, this yields eaf = −(
xd
xq− 1
)
= −1. The resulting vee curve is
shown in Figure 45.
5. For the specified operating condition, eaf =√
(1 + 2 × .6)2 + (2 × .8)2 = 2.72 ThusIf = 2720A.
Since, for a round rotor machine, p =veaf
xdsin δ, and for a round rotor machine the
stability limit is when sin δ = 1,
So, for a given power level, the stability limit is reached when sin δ = 1, and thenveaf = pxd.
The rest of this problem is worked in Matlab script p9 5.m. The Vee curves are shownin Figure 46.
6. First, we need to get current to make the motor produce exactly 1,000 kW. At unitypower factor, we can define a voltage ’inside’ the stator resistance: call it Vi. Power willbe P = 3ViI = 3Vi − 3RaI
2, then required current is:
I =V
2Ra−√
(V
2Ra)2 − P
3Ra
The rest of this problem is worked in Matlab script p9 6.m. Note that to produce theplot of efficiency vs. load, the core loss and friction and windage are added to mechanicalload. That efficiency vs. load is shown in Figure 47. Summary output is:
Chapter 9, Problem 6
Converted Power = 1.003e+06 W
Phase Current = 138.67 A
Output Power = 1e+06 W
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 43
300If
5920.8
|I |a
−100 100 200
Figure 45: Solution to Chapter 9, Problem 4: Zero Power Vee Curve
0 500 1000 1500 2000 2500 30000
0.5
1
1.5
2
2.5x 10
4 Chapter 9, Problem 5, Vee Curve
Arm
atur
e C
urre
nt, A
RM
S
Field Current, A DC
Figure 46: Solution to Chapter 9, Problem 5: Vee Curve
Torque Angle = -45.4144 degrees
Internal voltage E1 = 3434.61 V
Internal voltage Eaf = 4305.68
Field Current = 177.563 A
Armature Loss = 5768.79 W
Field Loss = 9458.6 W
Core Loss = 2000 W
F and W loss = 1000 W
Input Power = 1.01823e+06 W
Full Load Efficiency = 0.982099
7. Referring to that figure, note that xad = xd − xaℓ = 1.9per-unit.
Transient reactance is x′d = xaℓ + xad||xfℓ, or:
xadxfℓxad + xfℓ
= 0.3
Using xad = 1.9 in this,
xfℓ =0.3 × 1.9
1.9 − 0.3= 0.35625
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 44
0 2 4 6 8 10
x 105
0.94
0.945
0.95
0.955
0.96
0.965
0.97
0.975
0.98
0.985
0.99
Effi
cien
cy
Power Output (W)
Chapter 9, Problem 6
Figure 47: Solution to Chapter 9, Problem 6: Synchronous Motor Efficiency
and xf = 1.9 + 0.35625 = 2.25625.
Field resistance is:
rf =xf
ω0T ′do
=2.25625
377 × 5≈ 0.0012
In ordinary variables, the rotor elements, referred to the stator will be related by thebase impedance, which is:
ZB =242
500= 1.152Ω
then
Lad =ZBxadω0
=1.152 × 1.9
377= 5.81mH
Lfℓ =ZBxfℓω0
=1.152 × 0.35625
377= 1.09mH
To get these parameters on the field side, we need to find the field circuit base impedance.To start, note that VfBIfB = 3
2VBIB = PB . This means that the field circuit base
impedance Zfb = PB
I2fB
.
To find the field circuit base current, note that ifnlxad = 1, so that ifnl = 1xad
. Thismeans that base current for the field circuit is IfB = Ifnlxad = 500 × 19 = 950A. This
means ZfB = 500MVA.95KA2 = 554Ω.
Then field inductance and resistance are:
Lf =554Ω × 2.25635
377= 3.760H
Rf =3.760H
5s= 0.753Ω
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 45
8. See the Matlab script p9 8.m for the solution to this problem. Some iteration wasrequired to find the critical clearing time, which turns out to be about 252 mS, asopposed to the equal area criteria time of about 203 mS.A near-critical swing followedby a short setup time is shown in Figure 48.
0 1 2 3 4 5 60.5
1
1.5
2
2.5
3Transient Simulation: Clearing Time = 0.252
Tor
que
Ang
le, r
adia
ns
seconds
Figure 48: Solution to Chapter 9, Problem 8: Near-Critical Swing
Transient Stability Analysis
Initial Conditions:
Torque Angle delta = 0.830584
Direct Axis Flux psid = 0.674445
Quadrature Axis Flux psiq = -0.738325
Direct Axis Current I_d = 0.912004
Quadrature Axis Current I_q = 0.410181
Torque = 0.95
Required Internal Voltage E_af = 2.49845
Field Flux psif = 1.0122
Equal Area T_c = 0.202796
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 46
Chapter 10
1. part a : I1 = I3 , I2 = I
3 , I0 = I3
part b : I1 = I3a
2, I2 = I3a, I0 = I
3
part c : I1 = − j√3I, I2 = j√
3I, I0 = 0
2. positive sequence : Ia = I, Ib = a2I, Ic = aI
negative sequence : Ia = I, Ib = aI, Ic = a2I
zero sequence : Ia = I, Ib = I, Ic = I
Phasor diagrams are in Figure 49.
0
a
b
c
a a
b
b
c
c
Part a: I 1 2Part b: I Part c: I
Figure 49: Phasor Diagrams for Chapter 10, Problem 2
3. Ia = 27710 ≈ 27.7A, so I1 = I2 = I0 = 27.7
3 ≈ 9.23A
4. Vbc = −j480 so Ib = −j48 and Ic = j48Then, noting that −ja2 = ej
π6 and ja = ej
π6 and ej
π6 + e−j
π6 =
√3
then:
I1 = 483
√3 = 27.7
I2 = −483
√3 = −27.7
I + 0 = 0
5. Assume that we can set the time reference so that phase A voltage on the ’X’ side tohave a phase angle of zero. Then, on the ’X’ side, Ia = 277
10 ≈ 27.7AThen I1 = I2 = I0 = 27.7
3 ≈ 9.23A.The voltage ratio is N = 2400
480 = 5
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 47
So on the primary side, positive and negative sequence currents are rotated by 30 andthus are: I1 = 1.85e−j
π6 and I2 = 1.85ej
π6 Then, on the ’H’ side:
Ia = I1 + I2 = 3.2A
Ib = a2I1 + aI2 = −3.2A
Ic = 0
6. Assume tht we can set the time reference so that Phase A voltage on the ’X’ side tohave a phase angle of zero. Then, on the ’X’ side, Ia = 0, Ib = −j48A and Ic = j48A.The symmetrical component currents are:
I1 = 13
(
aIb + a2Ic)
= 27.7
I2 = 13
(
a2Ib + aIc)
= −27.7
I0 = 0
On the ’H’ side since the voltage ratio is 2400480 = 5, the symmetrical component currents
are:
I1 = 5.54e−jπ6
I2 = 5.54ejπ6
I0 = 0
Reconstructing phase currents:
Ia = I1 + I2 = −j5.54Ib = a2I1 + aI2 = −j5.54Ic = aI1 + a2I2 = j11.08
7. Since the neutral of the source is directly connected to the neutral of the resistors,currents are found directly:Ia = 27.7 = 23.1 + 4.6, Ib = 23.1e−j
2π3 , Ic = 23.1ej
2π3
The symmetrical component currents are simply:
I1 = 23.1 + 13 × 4.6 = 24.63
I2 = 13 × 4.6 = 1.53
I0 = 1.53
8. If the star point is grounded, its voltage is:
Vn = VaRb||Rc
Ra +Rb||Rc+ Vb
Ra||RcRb +Ra||Rc
+ VcRa||Rb
Rc +Ra||RbTaking advantage of the b-c symmetry:
Vn = Va
(
Rb||RcRa +Rb||Rc
− Ra||RcRb +Ra||Rc
)
= 277.1×(
6
16− 5.54
17.5
)
≈ 0.063×277.1 ≈ 17.4V
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 48
Then the three phase currents are:
Ia = 277.1−17.410 = 25.97 = 23.09 + 2.88A
Ib = 277.1e−j 2π3 −17.4
12 = 23.09e−j2π3 − 1.45A
Ic = 277.1ej 2π3 −17.4
12 = 23.09ej2π3 − 1.45A
Then the symmetrical components are:
I1 = 13
(
Ia + aIb + a2Ic)
= 23.09 + 13 (2.88 + 1.45) ≈ 24.53A
I2 = 13 (2.88 + 1.45) ≈ 1.44A
I0 = 0
9. The voltages can be written as: Va = 277 + 3, Vb = 277e−j2π3 and Vc = 277ej
2π3 then the
symmetrical component currents will be:Grounded:I1 = 27.7A, I2 = 0.1A, I0 = 0.1AUngrounded:I1 = 27.7A, I2 = 0.1A, I0 = 0
10. The transmission line has phase impedance:
Zph = jω
20 8 58 20 85 8 20
Matlab script that p10 10.m solves this problem. The solution proceeds as follows:First, get the symmetrical component impedance matrix by doing Zs = TZphT
−1. Thisis readily inverted to get the line admittance matrix. Note that in this situation, realpower is P = V1V2Y sin δ, where the admittance variable Y is the reactive admittance(this is a lossless situation) for positive sequence. By inverting that expression we find
phase angle δ. Then positive sequence current across the line is just Vd = V(
ejδ − 1)
,
and that is used with the full admittance matrix to find currents. The script is also usedto find real power to confirm that the angle is right. Here is the summary output:
Xs =
13.0000 - 0.0000i 1.0000 + 1.7321i -0.5000 + 0.8660i
1.0000 - 1.7321i 13.0000 - 0.0000i -0.5000 - 0.8660i
-0.5000 - 0.8660i -0.5000 + 0.8660i 34.0000
Ys =
0.0000 - 0.0790i -0.0107 + 0.0062i -0.0023 - 0.0013i
0.0107 + 0.0062i 0.0000 - 0.0790i 0.0023 - 0.0013i
0.0023 - 0.0013i -0.0023 - 0.0013i -0.0000 - 0.0296i
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 49
delt = 0.0958
Vdiff = -0.5276 +11.0027i
Ic =
0.8696 + 0.0417i
-0.0737 + 0.1146i
0.0136 + 0.0263i
S = 1.0000e+02 + 4.7948e+00i
>> abs(I2) = 0.1362
11. With a single phase fault, total reactance is x = x1 + x2 + x0 = 0.9, and then faultcurrent is if = 1/(j0.9) ≈ −j1.11 per-unit. Current in Phase A is ia = i1 + i2 + i0 = 3.33
Base current is IB = 100√3×138
≈ .4184kA = 418.4A.
Then phase A current is: Iaf = 3.33 × 418 ≈ 1361.3A
For the line-line fault, x = x1 + x2 = 0.5. Fault current is if = −j2. The threephase currents are ia = 0, |ib| = |ic| = 2
√3 ≈ 3.46 per-unit. Then the fault current is
|Ib| = |Ic| ≈ 1, 449A
12. Symmetrical component reactances are x1 = 0.55, x2 = 0.55 and x0 = 0.45.
Fault current for a line-ground fault is: i1 = i2 = i0 = 1j1.55 = −j.645.
At the fault, ia = i1 + i2 + i0 = −j1.935, ib = ic = 0. At the generator side of thetransformer,
ia = −j.645(
e−jπ6 + ej
π6
)
= −j1.117
ib = −j.645(
a2e−jπ6 + aej
π6
)
= j1.117
ic = −j.645(
ae−jπ6 + a2ej
π6
)
= 0
For the line-line fault at the fault: total reactance is x = x1 + x2 = 1.1, so that i1 =−i2 = 1
j1.1 ≈ −j.91. then, at the fault,
ia = 0
ib = −j.91(
a2 − a)
= −j.91 ×−j√
3 = −1.575
ic = −j.91(
a− a2)
= −j.91 × j√
3 = 1.575
On the generator side
ıa = −j.91(
e−jπ6 − ej
π6
)
= −.91
ıb = −j.91(
a2e−jπ6 − aej
π6
)
= −.91
ıc = −j.91(
ae−jπ6 − a2ej
π6
)
= 1.82
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 50
Base current at the fault is 418.4 A and at the generator is 4184 A, so the currentmagnitudes are:
Line Line-Neutral (A) Line-Line (A)Phase A 809.6 0Phase B 0 659.0Phase C 0 659.0GeneratorPhase A 4674 3807Phase B 4674 3807Phase C 0 7615
13. Do this one on the line base of 100 MVA, 345 kV on the line and 24 kV at the generator.On that base, generator reactance is xg = 100
600 × .25 ≈ .042 and transformer reactanceis xt = 100
600 × .07 ≈ .012. Positive and negative sequence reactances are then x1 = x2 =.25 + .042 + .012 = .314 Zero sequence reactance is x0 = .40 + .012 = .412.
For the line-neutral fault, z1 = z2 = j.314 and z0 = j.412. Total impedance to thefault is z = j1.04, so that fault current is if = 1
j1.04 ≈ −j.961. So at the fault: ia =i1 + i2 + i0 = −j2.88, ib = ic = 0.
On the generator side:
ia = −j.961(
e−jπ6 + ej
π6
)
=√
3 ×−j.961 = −j1.66
ib = −j.961(
a2e−jπ6 + aej
π6
)
= −√
3 ×−j.961 = j1.66
ic = −j.961(
ae−jπ6 + a2ej
π6
)
= 0
For the line-line fault, z = j.628 so that fault current is if ≈ j1.59. At the fault, thephase currents are:
ia = 0
ib = −j1.59(
a2 − a)
≈ −2.76
ic = −j1.59(
a− a2)
≈ 2.76
On the generator side, the currents are:
ia = −j1.59(
e−jπ6 − ej
π6
)
= −1.59
ib = −j1.59(
a2e−jπ6 − aej
π6
)
= −1.59
ic = −j1.59(
ae−jπ6 − a62ej
π6
)
= 3.18
Base currents are 2406 A at 24 kV and 167 A at 345 kV, so the currents, in amperesare:
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 51
Line Line-Neutral (A) Line-Line (A)Phase A 481 0Phase B 0 461Phase C 0 461GeneratorPhase A 3993 3826Phase B 3993 3826Phase C 0 7651
14. Since this is a rather routine calculation, we resort to using Matlab to work it. See Scriptp10 14.m. The results are:
Problem 10_14
Base Currents: Generator 4183.7 Line 167.348 Fault 1673.48
Per-Unit Currents, line-neutral in Phase a
Fault i_a = 0+j-2.34375
Line i_a = 0+j-1.35316 i_b = 0+j1.35316 i_c = 0+j-1.73472e-16
Generator i_a = 0+j-0.78125 i_b = 0+j1.5625 i_c = 0+j-0.78125
Currents in Amperes
Fault I_a = 3922.22 Ib = 0 Ic = 0
Line I_a = 226.449 I_b = 226.449 I_c = 2.90302e-14
Generator I_a = 3268.51 I_b = 6537.03 I_c = 3268.51
Problem 10_14
Per-Unit Currents, line-line in Phases b and c
Fault i_a = 0+j0 i_b = -1.41971+j0 i_c = 1.41971+j0
Line i_a = -0.819672+j0 i_b = -0.819672+j0 i_c = 1.63934+j0
Generator i_a = -1.41971+j0 i_b = -5.46011e-16+j0 i_c = 1.41971+j0
Currents in Amperes
Fault I_a = 0 Ib = 2375.86 Ic = 2375.86
Line I_a = 137.17 I_b = 137.17 I_c = 274.341
Generator I_a = 5939.65 I_b = 2.28435e-12 I_c = 5939.65
15. This problem is worked by Matlab script p10 15.m. The answers are:
Problem 10_15
Base Currents: Generator 4183.7 Line 167.348 Fault 1673.48
Per-Unit Currents, line-neutral in Phase a
Fault i_a = 0.0731707+j-2.34146
Line i_a = 0.0422451+j-1.35184 i_b = -0.0422451+j1.35184 i_c = 5.41572e-18+j-1.73303e-
Generator i_a = 0.0243902+j-0.780488 i_b = -0.0487805+j1.56098 i_c = 0.0243902+j-0.780488
Currents in Amperes
Fault I_a = 3920.3 Ib = 0 Ic = 0
Line I_a = 226.339 I_b = 226.339 I_c = 2.90161e-14
Generator I_a = 3266.92 I_b = 6533.84 I_c = 3266.92
Problem 10_15
Per-Unit Currents, line-line in Phases b and c
Fault i_a = 0+j0 i_b = -1.41819+j-0.046498 i_c = 1.41819+j0.046498
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 52
Line i_a = -0.818792+j-0.0268456 i_b = -0.818792+j-0.0268456 i_c = 1.63758+j0.0536913
Generator i_a = -1.41819+j-0.046498 i_b = -5.48173e-16+j0 i_c = 1.41819+j0.046498
Currents in Amperes
Fault I_a = 0 Ib = 2374.59 Ic = 2374.59
Line I_a = 137.097 I_b = 137.097 I_c = 274.193
Generator I_a = 5936.46 I_b = 2.29339e-12 I_c = 5936.46
16. This one is solved by Matlab script p10 16.m. The solution is in the output of thatscript is:
Problem 10_16
Base Currents: Generator 4183.7 Line 418.4
Per-Unit Currents, line-neutral in Phase a
Fault i_a = 0.000+j -2.727
Generator i_a = 0.000+j -1.575 i_b = 0.000+j 1.575 i_c = 0.000+j -0.000
Currents in Amperes\
Fault I_a = 1141.0 Ib = 0.0 Ic = 0.0
Generator I_a = 6587.6 I_b = 6587.6 I_c = 0.0
Per-Unit Currents, line-line in Phases b and c
Fault i_a = 0.000+j 0.000 i_b = -2.038+j 0.000 i_c = 2.038+j 0.000
Generator i_a = -1.176+j 0.000 i_b = -1.176+j 0.000 i_c = 2.353+j 0.000
Currents in Amperes
Fault I_a = 0.0 Ib = 852.5 Ic = 852.5\
Generator I_a = 4922.0 I_b = 4922.0 I_c = 9844.0
17. This problem is solved by Matlab script p10 17.m.
Chapter 10, Problem 17: Currents in Per-Unit
Line-Neutral Fault
Close
i_a = 1.667 i_b = 0.000 i_c = 0.000
Far
i_a = 1.111 i_b = 0.000 i_c = 0.000
Line-Line Fault
Close
i_a = 0.000 i_b = 1.575 i_c = 1.575
Far
i_a = 0.000 i_b = 1.083 i_c = 1.083
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 53
Chapter 11
1. Real power flow through the line is P = V1V2X
sin δ = 106
10 sin δ
So maximum power flow is 100 kW.
Since the sine of 30 is 1/2, real power flow is 50 kW.
Reactive power flow with equal voltage magnitudes isQ = V 2
X(1 − cos δ) and 1−cos30 ≈
.134, then reactive power flow is about 13.4 kVAR.
To get 75 kVAR to flow in the line, sin δ = .75 or δ ≈ 48.6
2. Real power flow in this three-phase line is P = 1382
40 sin δ ≈ 476.1 sin δ(MW)
So when δ = 10, P ≈ 82.7MWWhen δ = 30, P ≈ 238MWFor 100 MW, δ = sin−1 100
476.1 ≈ 12.1
With that angle, Q = 476.1 (1 − cos 12.1) ≈ 10.6MVAR
3. If sending and receiving end power are the same, real and reactive power at sending andreceiving ends are:
(P + jQ)S =V 2
R− jX
(
1 − ejδ)
(P + jQ)R =V 2
R− jX
(
e−jδ−1)
These are easily evaluated by Matlab script p11 3.m. Note the solution to the problem offinding the proper power angle for a given receiving end real power is nonlinear, but theMatlab routine fzero() can be used to solve that problem with an auxiliary function.The answers are:
Chapter 11, Problem 3
Angle = 10 degrees
Sending end P = 82.5715 MW Q = -1.02412 MVAR
Receiving end P = 81.1392 MW Q = -15.3469 MVAR
Chapter 11, Problem 3
Angle = 30 degrees
Sending end P = 242.008 MW Q = 39.5845 MVAR
Receiving end P = 229.378 MW Q = -86.7231 MVAR
Seeking 100 MW at receiving end
Chapter 11, Problem 3
Angle = 12.3843 degrees
Sending end P = 102.194 MW Q = 0.858907 MVAR
Receiving end P = 100 MW Q = -21.0783 MVAR
4. With a capacitance of 6.6µF , admittance is Yc = jωC = j2.49 × 10−3S, or QC =1382 × 2.49 × 10−3 ≈ 47.4MVAR
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 54
For a phase angle of 30,
P = V 2
Xsin δ = 238MW
Q = V 2
X(1 − cos δ) = 16.4MVAR
To get 100 MW in the line, δ = sin−1 100476.1 ≈ 12.125, for whichQ = V 2
X(1 − cos δ)−Qc ≈
10.6 − 47.4 = −36.8MVAR
5. The power circle called for in the problem is shown in Figure 50. It was generated byMatlab script p11 5.m.
−1000 −800 −600 −400 −200 0 200 400 600 800 1000
−800
−600
−400
−200
0
200
400
600
800
Chapter 11, Problem 5
Watts
Var
s
sending
receiving
Figure 50: Solution to Chapter 11, Problem 5: Power Circle
6. This problem is solved by Matlab script p11 6.m. The situation in which it is carrying10 kW is shown in Figure 51. the specific numbers are:
Chapter 11, Problem 6
Center of Power Circle = 1980.2 + j 19802
Radius of Power Circle = 19900.7
Seeking 10 kW at receiving end
Angle = 31.3025 degrees
Sending end P = 10576.5 MW Q = 1853.63 MVAR
Receiving end P = 10000 MW Q = -3911.28 MVAR
7. The phasor diagram without compensating capacitors is shown, to pretty good scale, inFigure 52
To find the required capacitance for receiving end voltage to be of the same magnitudeas sending end, see that:
VR = VSR||jXc
R||jXc + jXl
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 55
−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 3 3.5
x 104
−2
−1.5
−1
−0.5
0
0.5
1
1.5
x 104 Chapter 11, Problem 6
VA
RS
Watts
Figure 51: Solution to Chapter 11, Problem 6: Partial Power Circle
X
V = 8000s
|V |=7941 R
|V |=986
Figure 52: Solution to Chapter 11, Problem 6: Uncompensated Line
If we note Yc = 1Xc
and take into account the sign of complex numbers, the conditionfor VR = VS is:
Yc =1
Xl−√
(
1
Xl
)2
− 1
R2
This evaluates to Yc ≈ 7.843 × 10−4S or C = 2.1µF .
Voltage vs. capacitance is shown in Figure 53. This was calculated by Matlab scriptp11 7.m.
8. The Matlab scripts that evaluates this load flow program are p11 8.m, p11 8a.m andp11 8b.m. An auxiliary script, p11 8disp.m creates output for all variants of the script.The first part of the program is represented by the following:
Simple Minded Load Flow Problem
Line Impedances:
Z( 1) = 0.005 + j 0.1
Z( 2) = 0.01 + j 0.1
Z( 3) = 0.005 + j 0.15
Z( 4) = 0.001 + j 0.05
Z( 5) = 0.005 + j 0.1
Z( 6) = 0.005 + j 0.2
Z( 7) = 0.01 + j 0.3
Z( 8) = 0.005 + j 0.05
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 56
0 0.5 1 1.5 2 2.5 3 3.5 47920
7940
7960
7980
8000
8020
8040
8060Chapter 11, Problem 7
Res
isto
r V
olta
ge
Compensating Capacitor, µ F
Figure 53: Solution to Chapter 11, Problem 6: Receiving End Voltage vs. Compensating capaci-tance
Here are the voltage Magnitudes and angles
Voltage at Bus( 1) = 0.928 angle 8.313 (deg)
Voltage at Bus( 2) = 0.921 angle -0.817 (deg)
Voltage at Bus( 3) = 1.000 angle 5.155 (deg)
Voltage at Bus( 4) = 1.000 angle 0.000 (deg)
Voltage at Bus( 5) = 0.947 angle 2.018 (deg)
Voltage at Bus( 6) = 0.944 angle -0.298 (deg)
Complex Power at the buses
At Bus( 1) P = 2.000 Q = -0.000
At Bus( 2) P = -2.000 Q = -0.500
At Bus( 3) P = 1.000 Q = 0.800
At Bus( 4) P = 0.023 Q = 0.184
At Bus( 5) P = 0.000 Q = 0.000
At Bus( 6) P = -1.000 Q = 0.000
Line Current Magnitudes are
Line( 1) = 1.471
Line( 2) = 0.532
Line( 3) = 0.698
Line( 4) = 0.767
Line( 5) = 0.747
Line( 6) = 0.539
Line( 7) = 0.186
Line( 8) = 0.503
So if we insert Q = 0.5 into bus 5, the answer becomes:
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 57
Simple Minded Load Flow Problem
Line Impedances:
Z( 1) = 0.005 + j 0.1
Z( 2) = 0.01 + j 0.1
Z( 3) = 0.005 + j 0.15
Z( 4) = 0.001 + j 0.05
Z( 5) = 0.005 + j 0.1
Z( 6) = 0.005 + j 0.2
Z( 7) = 0.01 + j 0.3
Z( 8) = 0.005 + j 0.05
Here are the voltage Magnitudes and angles
Voltage at Bus( 1) = 0.961 angle 7.740 (deg)
Voltage at Bus( 2) = 0.952 angle -0.782 (deg)
Voltage at Bus( 3) = 1.000 angle 5.016 (deg)
Voltage at Bus( 4) = 1.000 angle 0.000 (deg)
Voltage at Bus( 5) = 0.981 angle 1.872 (deg)
Voltage at Bus( 6) = 0.971 angle -0.291 (deg)
Complex Power at the buses
At Bus( 1) P = 2.000 Q = -0.000
At Bus( 2) P = -2.000 Q = -0.500
At Bus( 3) P = 1.000 Q = 0.325
At Bus( 4) P = 0.020 Q = 0.094
At Bus( 5) P = 0.000 Q = 0.500
At Bus( 6) P = -1.000 Q = 0.000
Line Current Magnitudes are
Line( 1) = 1.422
Line( 2) = 0.534
Line( 3) = 0.676
Line( 4) = 0.764
Line( 5) = 0.573
Line( 6) = 0.478
Line( 7) = 0.096
Line( 8) = 0.425
And then, when Line 1 is removed, the distribution is:
Simple Minded Load Flow Problem
Line Impedances:
Z( 1) = 0.01 + j 0.1
Z( 2) = 0.005 + j 0.15
Z( 3) = 0.001 + j 0.05
Z( 4) = 0.005 + j 0.1
Z( 5) = 0.005 + j 0.2
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 58
Z( 6) = 0.01 + j 0.3
Z( 7) = 0.005 + j 0.05
Here are the voltage Magnitudes and angles
Voltage at Bus( 1) = 0.818 angle 28.186 (deg)
Voltage at Bus( 2) = 0.859 angle -4.363 (deg)
Voltage at Bus( 3) = 1.000 angle 6.107 (deg)
Voltage at Bus( 4) = 1.000 angle 0.000 (deg)
Voltage at Bus( 5) = 0.885 angle 3.694 (deg)
Voltage at Bus( 6) = 0.893 angle -1.058 (deg)
Complex Power at the buses
At Bus( 1) P = 2.000 Q = 0.000
At Bus( 2) P = -2.000 Q = -0.500
At Bus( 3) P = 1.000 Q = 1.693
At Bus( 4) P = 0.067 Q = 0.356
At Bus( 5) P = -0.000 Q = -0.000
At Bus( 6) P = -1.000 Q = -0.000
Line Current Magnitudes are
Line( 1) = 1.246
Line( 2) = 2.446
Line( 3) = 1.481
Line( 4) = 1.214
Line( 5) = 0.798
Line( 6) = 0.362
Line( 7) = 1.207
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 59
Chapter 12
1. Since diode current is (ignoring resistance) i = I0(
eqvkT − e
)
, solving for v we find:
v =kT
qlog
(
i
I0+ 1
)
At 299 K, kTq
= 1.38×10−23×2991.6×10−19 ≈ 25.2mV
For part a, log( iI0
+1) = log(5×1015) ≈ 36.15 and for part b, log( iI0
+1) = log(5×1016) ≈38.45
So the answers are: Part a) v = .0252 × 36.15 ≈ .911V and Part b) v = .0252 × 38.45 ≈.969V.
Matlab script p12 1.m was written to get voltage vs. current and the resulting plot isshown in Figure 54.
10−3
10−2
10−1
100
101
102
0.75
0.8
0.85
0.9
0.95
1
1.05
1.1
1.15Chapter 12, Problem 1
For
war
d V
olta
ge
Current, A
Figure 54: Solution to Chapter 12, Problem 1: Diode Voltage vs. Current
2. At 40 C, T = 313K and kTq
= 27mV. At 0 C, T = 273K and kTq
= 23.5mV, so Part a)v = .027 × 36.15 ≈ .976V, and Part b) v = .0235 × 36.15 ≈ .851V
Matlab script p12 2.m generates voltage vs. temperature as shown in Figure 55.
3. Part a): Vo = DVin = 12 × 48 = 24V
Part b): ∆I = (Vin − Vo)DTL
= 24 × .5×10−4
6×10−3 = 0.2A. This is sketched in Figure 56.
To get voltage ripple, see that the difference between input and output current (to thecapacitor) is a triangle wave. For the half period starting when the current reaches amaximum until it reaches a minimum,
dvcdt
=1
CIm
(
1 − 2
Tt
)
Where Im = Vin−Vo
LDT is the maximum value of ripple current: the difference between
inductor current and output current. Capacitor ripple voltage is:
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 60
−40 −20 0 20 40 60 80 1000.75
0.8
0.85
0.9
0.95
1
1.05
1.1
1.15
1.2
1.25Chapter 12, Problem 2
For
war
d V
olta
ge
Temperature, C
Figure 55: Solution to Chapter 12, Problem 2: Diode Voltage vs. Temperature
∆iL
tT
DT I
Figure 56: Solution to Chapter 12, Problem 3: Ripple Current
vR =1
CIm
(
t− t2
T
)
.
For the next half cycle the situation is just reversed, with a negative voltage excursion.
Matlab script p12 3.m does the evaluation and plots both ripple voltage and current.The maximum voltage ripple excursion is about 0.25 V.
4. Load voltage is VL = DVs and change of current from start to end of the ’on’ part ofthe cycle is:
∆I =DT
L(Vs − VL) = Vs
T
L
(
D −D2)
This is evaluated by the Matlab script p12 4.m and a plot, for this converter is shownin Figure 59
5. To find the limits to the ripple, solve the simple circuit problem:
vmax = vmine−DT
τ + Vs(
1 − e−DTτ
)
vmin = vmaxe− (1−D)T
τ
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 61
0 1 2 3 4 5 6
x 10−4
−0.4
−0.2
0
0.2
0.4Chapter 12, Problem 3: Ripple
Rip
ple
Vol
tage
0 1 2 3 4 5 6
x 10−4
−0.2
−0.1
0
0.1
0.2R
ippl
e C
urre
nt
Time, s
Figure 57: Solution to Chapter 12, Problem 3: Ripple Voltage and Current
where the time constant is just τ = LR
. This set can be solved:
vmax = Vs1 − e−
DTτ
1 − e−Tτ
Once the limits are found, voltage as a function of time is straightforward: During theON interval:
v = vmine− t
τ + Vs(
1 − e−t
tau
)
and during the OFF interval:
v = vmaxe− t
τ
Matlab script p12 5.m carries out these calculations and repeats the waveform for a fewcycles to make it more easily visible. The resulting output volgate is shown in Figure 60.
6. The output voltage is just the input voltage rectified, and the rectifier has the effect oftaking the absolute value of the input voltage. At the same time, it converts the current,so that, for this case,
vl = |Vs sinωt|is = IDCsign(sinωt)
Shown in Figure 61 are output voltage and input current for the full wave bridge rectifier.In Figure 62 are input and output power (neglecting diode forward drop). Plotted onthe same scale, they are the same.
According to the model of Problem 1, forward drop in each of the diodes is about0.911 volts, leading to about 18 watts total dissipation, distributed over the four diodesof the bridge.
See Matlab script p12 6.m for details.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 62
0 1 2 3 4 5 6
x 10−4
0
5
10
15
20
25Chapter 12, Problem 3: Capacitor Voltage
Cap
acito
r V
olta
ge
Time, s
Figure 58: Solution to Chapter 12, Problem 1: Capacitor Voltage
7. There are actually three important numbers related to the load voltage output from thethree-phase rectifier. They are the peak voltage, which is the peak of line-line voltage:Vpeak =
√
(2)×480 = 679V , the average load voltage, which is VL = 3π
√2×480 ≈ 648V ,
and the minimum voltage which is Vmin = 480√
2 cos π6 ≈ 588V . The actual waveformsare shown in Figure 63. Current in Phase A is positive whenever Phase A is most positiveand negative when Phase A is most negative. Since there is a large filter reactor on theDC side, current is constant.
Accouning for commutation reactance, the reactive voltage drop appears to be accountedfor by the fictitious resistor:RX = 3
πωLℓ
3π× 377× .003 ≈ 1.08Ω. With a load current of
10 A, VL = 648 − 10 × 1.08 ≈ 637.2V Load voltage as a function of current is shown inFigure 64.
8. Assuming the leakage inductance is negligible, average load voltage is:
< VL >=1
π
∫ π
0Vp sinωtdωt =
2
πVp =
2
π
√2 × 120 ≈ 108V
In the single phase rectifier, load voltage is zero during commutation and rate of changeof current in the leakage inductance is:
diLdt
=VpLℓ
sinωt
And since current in the leakage inductance starts at −IL, current at the end of thecommutation interval is
iL = −IL +VpωLℓ
(cosωt− 1))
The commutation interval tc is then determined by:
Vp2Lℓ
(cosωtc − 1) = 2IL
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 63
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2Chapter 12, Problem 4: Inductor Current Ripple
A
Duty Cycle
Figure 59: Solution to Chapter 12, Problem 4: Ripple current vs. duty cycle
0 0.5 1 1.5 2 2.5
x 10−3
0
50
100
150
200
250
300
350
Chapter 12, Problem 5: Buck Converter Voltage Output
V
t, s
Figure 60: Solution to Chapter 12, Problem 5: Buck Converter Output
Average voltage is, since output voltage during the commutation interval is zero:
< VL >=1
π
∫ π
ωtc
Vp sinωtdωt =1
π
∫ π
0Vp sinωtdωt− 1
π
∫ ωtc
0Vp sinωtdωt =
2
π− 1
πVp (cosωtc − 1)
And, sinceVp
ωLℓ(cosωtc − 1) = 2IL,
< VL >=2
πVp −
2
πωLℓIL
Here, if Lℓ = 5mH, RX = 2π× 377 × .005 ≈ 1.2Ω
Voltage drop is shown in Figure 65
9. Average output voltage in continuous conduction is:
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 64
0 0.01 0.02 0.03 0.04 0.05 0.060
50
100
150
200Chapter 12, Problem 6: Full Wave Bridge Output Voltage
Vol
ts
0 0.01 0.02 0.03 0.04 0.05 0.06
−10
−5
0
5
10S
ourc
e C
urre
nt, A
t, s
Figure 61: Solution to Chapter 12, Problem 6: Full Wave Rectifier Output Voltage and InputCurrent
Vout =VS
1 −D=
12
.5= 24
Current ripple is:
∆I =VSLDT =
12 × .5 × 2 × 10−5
240 × 10−6= .5A
Voltage ripple is found from output current:
∆V =I0 (1 −D)T
C=
1 × .5 × 2 × 10−5
10 × 10−6= 1V
10. Equivalent load resistance is found from power
R =V 2
P=
1202
12= 1.2kΩ
Since R = 2LD2T
,
D =
√
2L
RT=
√
2 × 72 × 10−6
1200 × 10−5≈ 0.11
11. Matlab scripts p12 11a.m and p12 11b.m generate the fourier series amplitudes of thewaveform. Construction of the PWM waveform is shown in Figure 66 and harmonicamplitudes are plotted in Figure 67.
12. The commutation effective resistance is RX = 3π× 1.5 ≈ 1.432Ω. Then the voltage drop
is VX = 1.432×5, 000 ≈ 7, 162V . Rectified open circuit voltage is 3πVp = 3
π×√
2×330 =445.7kV . DC voltage is 400kV = 445.7 cos α− 7.162kV , or firing angle is:
cosα =407.2
445.7≈ .913
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 65
0 0.01 0.02 0.03 0.04 0.05 0.060
200
400
600
800
1000
1200
1400
1600
1800Chapter 12, Problem 6: Input and Output Power
Time, s
Figure 62: Solution to Chapter 12, Problem 6: Full Wave Rectifier Power
Then α = 24. Overlap angle is u = cos−1(
cosα− 2XILVp
)
− α. Since 2XILVp
= 2×1.5×5330×
√2≈
.032,u = cos−1 (.913 − .032) − 24 ≈ 4.25
At the inverter end:
cosα =400 − 7.162
445.7≈ .881
α = 28.2
u = cos−1 (.881 − .032) − 28.2 ≈ 3.7
Finally, time harmonics: the period of conduction for pulses on the AC side is 120, forwhich the harmonic amplitudes can be readily calculated:
In = IDC ∗ 4
nπsinn
π
2sinn
π
3
This is an odd harmonic series that evaluates to, for IDC = 5, 000A,
Harmonic Number 1 Amplitude 5513.3
Harmonic Number 3 Amplitude -0.0
Harmonic Number 5 Amplitude -1102.7
Harmonic Number 7 Amplitude -787.6
Harmonic Number 9 Amplitude 0.0
Harmonic Number 11 Amplitude 501.2
Harmonic Number 13 Amplitude 424.1
Harmonic Number 15 Amplitude -0.0
Harmonic Number 17 Amplitude -324.3
Harmonic Number 19 Amplitude -290.2
Harmonic Number 21 Amplitude 0.0
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 66
−0.015 −0.01 −0.005 0 0.005 0.01 0.0150
200
400
600
Chapter 12, Problem 7: Three Phase Rectifier
Dc
Sid
e V
olta
ge, V
−0.015 −0.01 −0.005 0 0.005 0.01 0.015−10
−5
0
5
10P
hase
A c
urre
nt, A
Time, s
Figure 63: Solution to Chapter 12, Problem 7: Three Phase Bridge Voltage and Current
10Idc
Vdc
648637.2
Figure 64: Solution to Chapter 12, Problem 7: Voltage vs. Load Current
Harmonic Number 23 Amplitude 239.7
Harmonic Number 25 Amplitude 220.5
Discounting the signs of the harmonics, the first four nonzero harmonics are 5, 7, 11 and13, with amplitudes of 1103, 788, 501 and 424 A, respectively. In a twelve pulse system,the fifth and seventh harmonics cancel as do the 17th and 19th. Each of two invertershandles half the current, so the surviving harmonics are of half amplitude, but they addso we get back the factor of two. Then the harmonics are of order 11, 13, 23 and 35,with amplitudes of 501, 424, 240 and 221 A, respectively.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 67
L
5A
108102
V
Figure 65: Solution to Chapter 12, Problem 8: Voltage vs. Load Current
0 0.005 0.01 0.015 0.02 0.025 0.030
0.5
1
Generation of PWM Signal
Com
para
tor
Inpu
ts
0 0.005 0.01 0.015 0.02 0.025 0.03
0
0.5
1
PW
M O
utpu
t
Time
Figure 66: Solution to Chapter 12, Problem 11: Generate PWM Waveform
0 500 1000 1500 2000 25000
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Har
mon
ic A
mpl
iude
Harmonic Frequency
Chapter 12, Problem 11
Figure 67: Solution to Chapter 12, Problem 11: Harmonic Amplitudes
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 68
Chapter 13
1. A simplification to the equivalent circuit is shown in Figure 68.
160.4s
.4s200
+
−
j2.5
j10
j2 j2j2
+
−
Figure 68: Solution to Chapter 13, Problem 1: Simplified Equivalent Circuit
Deriving a thevenin equivalent on the voltage source, armature leakage and magnetizingbranch as shown in Figure 68, it is clear the resistance R2
sis looking at a source impedance
magnitude of 4Ω. Dissipation is maximized when R2s
= 4Ω, and this happens when s =0.1. For a four pole machine operating at 50 Hz, speed is N = 0.9×1, 500 = 1, 350RPM.At that speed, torque is:
T =1602 × 4
42 + 42× 2
100π≈ 21.73N-m
2. Matlab script p13 a.m works problems 2 through 9. The first computation is the torque-speed curve, shown in Figure 69.
0 200 400 600 800 1000 1200 1400 1600 18000
50
100
150
200
250Chapter 13, Problem 2
Tor
que,
N−
m
RPM
Figure 69: Solution to Chapter 13, Problem 2: Torque-Speed
3. Current is calculated as a byproduct of torque-speed and that is shown in Figure 70
4. Breakdown torque is, for the purposes of this problem, generated by using Matlab’smax() function. It and associated current and power factor are:
Breakdown torque = 246.344 N-m at 1603.07 RPM
Current at Breakdown = 86.4064 Power Factor = 0.711948
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 69
0 200 400 600 800 1000 1200 1400 1600 18000
20
40
60
80
100
120
140Chapter 13, Problem 3
RPM
Cur
rent
, A
Figure 70: Solution to Chapter 13, Problem 3: Current-Speed
5. Running light could be calculated by looking at the smallest value of slip, but in thescript the rotor branch is discounted and impedance of the magnetizing and armaturebrances was used:
Problem 5: Running light current = 5.45245
Real Power = 402.042 Reactive Power = 4515.22
6. Locked rotor conditions involve s = 1, for which the machine impedance can be calcu-lated and a voltage to achieve specified current is then easy to estimate. The resultsare:
Problem 6: Locked rotor voltage = 49.6024
Locked Rotor Torque = 1.8263
Input Real and Reactive Power = 216.345 + j 1089.83
7. The trick to estimating machine operation with fixed voltage and frequency is to findthe limiting values of slip at either end. A crude search was made to find those valuesof slip. The rest is straightforward and the results are shown in Figure 71
8. This problem asks for multiple torque-speed calculations, and the only thing to rememberabout this is to adjust the reactive elements for frequency, but the resistors stay constant.Note the lower breakdown torques for low frequencies, shown in Figure 72.
9. This problem uses some brute-force computation, but even using an interpreter likeMatlab, computation is not expensive. In this calculation, operational curves similar tothose shown in Figure 72 were estimated for frequencies separated by 1 Hz. Then a searchwas made to find two points that bracket the desired torque. Then linear interpolationwas used to approximate operation at the desired torque. Not shown is a curve thatwas drawn of torque vs. speed to estimate how well this interpolation worked. Resultsfor input and output power are shown in Figure 73 and efficiency and power factor areshown in 74.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 70
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
x 104
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1Problem 7: Efficiency and Power Factor
Per
−U
nit
Output Power, Watts
EfficiencyPower Factor
Figure 71: Solution to Chapter 13, Problem 7: Efficiency and Power Factor
10. Matlab script p13 b.m does the calculations for problems 10 and 11. The torque-speedcurve for the motor operating with fixed voltage and frequency is shown in Figure 75
Small correction is required for stray load loss. Once the end points are determined,getting efficiency and power factor is done over slip. This is shown in Figure 76.
11. Several torque/speed curves for different frequency and voltage levels are calculated bythe same script and shown in Figure 77
And then it is not difficult to generate an idea of operation by sweeping over frequencyand finding the correct power point along each curve. Resulting efficiency and powerfactor are shown in Figure 78
12. The winding plan is shown in Figure 79. Note this did not really need to be a ’consequentpole’ winding since groups with turns of 17, 9 and 8 turns, respectively, could have beenwound around each of four poles.
To find the winding factor, we use the weighted average of the individual coil pitchfactors:
kwn =
∑Nk=1Ns(k) sin(γ2nNc(k))
∑Nk=1Ns(k)
where N is the total number of coils (6), n is the harmonic number, Nc is the coil throwfor each coil and Ns is the number of turns in each coil.
This evaluates to kw1 = .9720.
Synchronous reactance is:
Xs = ωLsωµ03
2
4
π
N2aRLkw1
p2g
And this evaluates to about 85.5 Ohms.
Peak flux density is found from the voltage expression:
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 71
0 500 1000 1500 2000 2500 30000
50
100
150
200
250Problem 8: Volts/Hz Curves
Tor
que,
N−
m
RPM
Figure 72: Solution to Chapter 13, Problem 8: Volts/Hz Control
Vp = 2ωmRLNakw1Bp
where ωm is mechanical rotational speed so that ωmR is surface speed. ωm = ωp. Then:
Bp =Vp
2ωmRLNakw1≈ .748T
13. This problem and the next are about the same machine. Referred to the stator side, theinductances are:
Magnetizing Lm = 32LaA
N= 1.5 × 16.59
3 ≈ 8.295mH
Armature Leakage Laℓ = La − Lab − Lm = 5.6 + 2.8 − 8.295 ≈ 0.105mh
Rotor Leakage Lrℓ = LA−LAB
N2 = 50.4+25.29 ≈ 0.105mh
The impedances at 60 Hz are Xa = ωLaℓ ≈ .0396Ω, Xm = ωLm ≈ 3.127Ω and, as itturns out, Xr ≈ Xa.
In this problem, we ignore any winding losses in the doubly fed machine, so, as weexpect, rotor input power Pr = sPs, where Ps is stator output power. Total generatedpower is Pm = Ps + Pr, so that Ps = Pm
1−s .
Stator current can be computed to be Is = Ps+jQs
Vs. Then we can compute voltage across
the magnetizing branch: Vm = Vs + jXaIs.
The next step is to compute current through the magnetizing branch: Im = Vm
jXm.
Finally rotor current, referred to the stator is: Ir = Is + Im, and rotor voltage isVr = Vm + jXrIr. Rotor input power is Pr + jQr = 3sVrI
∗r
Matlab script p13 c.m computes two discrete points for Problem 13 and two curves forProblem 14. The results are:
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 72
0 500 1000 1500 2000 2500 30000
0.5
1
1.5
2
2.5x 10
4 Problem 9: Input and Output Power
Wat
ts
RPM
Figure 73: Solution to Chapter 13, Problem 9: Input and Output Power
Problem 13: Referred Reactances
Magnetizing Inductance = 0.008295
Stator Inductance = 0.0084
Rotor Inductance = 0.0084
Stator Leakage = 0.000105
Rotor Leakage = 0.0084
Impedances at Rated Frequency
Stator Leakage = 0.0395841
Magnetizing = 3.12714
Rotor Leakage = 0.0395841
Rotor Input at 30 % slip
Positive Slip: P_r = 360000 Q_r = 444392
Negative Slip: P_r = -360000, Q_r = 444392
14. The results of the previous problem are generalized in the script that follows to a pictureof power balance, in Figure 80 and of reactive power input to the rotor, Figure 81
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 73
0 500 1000 1500 2000 2500 30000.8
0.82
0.84
0.86
0.88
0.9
0.92
0.94
0.96Problem 9: Efficiency and Power Factor
Per
−U
nit
RPM
EfficiencyPower Factor
Figure 74: Solution to Chapter 13, Problem 9: Efficiency and Power Factor
0 200 400 600 800 1000 1200 1400 1600 18000
200
400
600
800
1000
1200Problem 10, Part A
Tor
que
RPM
Figure 75: Solution to Problem 10: Torque-Speed
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 74
1 2 3 4 5 6 7 8 9
x 104
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95Problem 10, Part B: Efficiency and Power Factor
Per
−U
nit
Output Power, Watts
EfficiencyPower Factor
Figure 76: Solution to Problem 10: Efficiency and Power Factor
0 500 1000 1500 2000 2500 3000 3500 40000
200
400
600
800
1000
1200Problem 11: Volts/Hz Curves
Tor
que,
N−
m
RPM
Figure 77: Solution to Problem 11: Volts/Hz Torque-Speed
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 75
500 1000 1500 2000 2500 3000 3500 40000.84
0.85
0.86
0.87
0.88
0.89
0.9
0.91
0.92
0.93
0.94Problem 11: Efficiency and Power Factor
Per
−U
nit
RPM
EfficiencyPower Factor
Figure 78: Solution to Problem 10: Variable Frequency Efficiency and Power Factor
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 76
Figure 79: Solution to Problem 12: Winding Plan
1200 1400 1600 1800 2000 2200 2400−2
0
2
4
6
8
10
12x 10
5 Problem 14: Doubly Fed Machine Power
Wat
ts
RPM
Stator OutputRotor InputTotal
Figure 80: Solution to Problem 14: Real Power Balance in the DFM
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 77
1200 1400 1600 1800 2000 2200 24000
1
2
3
4
5
6x 10
5 Problem 14: Doubly Fed Machine Power
VA
RS
into
Rot
or
RPM
Figure 81: Solution to Problem 14: Rotor Reactive Power in the DFM
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 78
Chapter 14
1. Since Va = GΩIf , Ω = 1101×1 = 110 radians/second. N = 60
2π × 110 ≈ 1050.4 RPM.
If torque is 10 N-m, armature current must be Ia = 10 A. Internal voltage is: Eb =GΩIf = 110 − 10 = 100 V, so that Ω = 100radians/second, or just about 955 RPM.
Power in is about Pmboxin = 110×11 = 1210 Watts while power out is Pout = 100×10 =1000 Watts. This implies efficiency of 82.6 %.
For the last part,
Pin = V If + VT
GIf
Pout = ΩT =
(
V
GIf−R
T
GIf
)
T
η =PoutPin
This is plotted in Figure 82
0 2 4 6 8 10 12 14 16 18 200
1000
2000
3000Chapter 14, Problem 1, Part d
Wat
ts
0 2 4 6 8 10 12 14 16 18 200
0.5
1
Effi
cien
cy
N−m
Figure 82: Solution to Problem 1: Power and Efficiency
2. Back voltage is Eb = 100 12 ×10 = 95V, so that G = 95
180 ≈ 0.528. Torque is T = GIIf =5.28n-m
3. Back voltage must be Eb = 50,000100 = 500V, so that resistance is R = 600−500
100 = 1Ω, andmotor constant is G = 500
100×200 = 0.026H
4. Output power is
Pout = GΩI2 =GΩV 2
(R+GΩ)2=.625 × 6002
.625 + .125= 400kW
Current is I = 600.625+.125 = 800A, so
Pin = 600 × 800 = 480kW
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 79
5. If the motor is producing 400 kW, its back voltage must be:
Eb = GΩI =400kW
800A= 500V
And, since 1,000 RPM is 104.7 Radians/second,
G =500
104.7 × 800≈ .00596Hy
To get speed vs. voltage we must make power converted by the motor equal to powerabsorbed by the load:
GV 2Ω
(R+GΩ)2= P0
(
Ω
Ω0
)3
Matlab script p14 5.m uses the Matlab function fzero() to solve the nonlinear equationand plot speed vs. voltage (see Figure 83). It also does a check calculation (figure notshown here).
100 200 300 400 500 600 700300
400
500
600
700
800
900
1000
1100Chapter 14, Problem 5
RP
M
Applied Voltage, V
Figure 83: Solution to Problem 5: Speed vs. Voltage
6. The back voltage is Eb = GΩIf = 550 − 80016 = 500V. So GIf = 500
104.7 ≈ 4.776Wb
In this problem the equality is:
GΩIf
(
V −GΩIfRa
)
= P0
(
Ω
Ω0
)3
This is solved by Matlab script p14 6.m and the results are shown in Figure 84.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 80
100 150 200 250 300 350 400 450 500 550 600100
200
300
400
500
600
700
800
900
1000
1100Chapter 14, Problem 6
RP
M
Applied Voltage, V
Figure 84: Solution to Problem 6: Speed vs. Voltage
7. The initial part of the piecewise linear voltage vs. field current curve has a slope of
V
If=
Ω
Ω0× 200 =
N
N0× 200
Then, the speed that will result in self-excitation will be:
N = N0 ×75
200=
1200
200× 75 = 450RPM
At 1,500 RPM, if the machine is on the third segment (If > 2A), V = 270.8 + 12.5If .Running light:V = 270.8 + 12.5If = 75If , or:
If =270.83
62.5≈ 4
1
3A
and V = 352V.
Loaded,
V = V0 +GΩIf −RaIa
If =V
Rf
Over a limited range of Ia, this becomes:
V =V0 −RaIa
1 − GΩRf
This evaluates toV = 325 − 2Ia Ia < 6.25A
To compound the generator, GsΩ = 2. Then, comparing with the shunt field, Ns =Nf × 2
12.5 = 80Turns.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 81
8. 800 RPM = 83.775 Radians/Second, so that
G =600
87.775 × 2≈ 3.581Hy
Connected with no series field winding,
Ia =V −GΩIf
RaT e = GIf Ia
In long shunt:
If =V
Rf
Ia =V −GsIaΩ −GIfΩ
Ra +Rs
T e = GsI2a +GfIf Ia
In short shunt, we have this set of linear expressions:
If (Rf +Rs) +RsIa = V
If (Rf −GΩ) − (GsΩ +Ra) Ia = 0
And, as before,T e = GIf Ia +GsI
2a
These calculations are carried out by Matlab script p14 8.m. Torques are shown inFigure 85 and currents in Figure 86.
500 550 600 650 700 750 8000
200
400
600
800
1000
1200
1400
1600
1800
2000
N−
m
RPM
Comparison of Torques
Long ShuntShort ShuntNo Shunt
Figure 85: Solution to Problem 8: Torque
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 82
500 550 600 650 700 750 8000
100
200
300
400
500
600
700
800
900
N−
m
RPM
Comparison of Currents
Long ShuntShort ShuntNo Shunt
Figure 86: Solution to Problem 8: Current
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 83
Chapter 15
1. The next several problems all use the following:
Hmhm +Hgg = 0
BmAm = BgAg
Bg =hmg
Br1 + Pu
So for this one, hm = 5mm.
2. hm = 403
3. Am = 50cm2
4. The previous problem just happened to hit the optimum:
hm = 10mm
Am = 50cm2
5. Br =√
4 × 50 × 106 ≈ 1.414T
6. Emax = 14,0002
4×1.05 ≈ 46.66Mg-Oe
7. V = KΩ, so Ω = 12.001 = 12, 000Rad/s ≈ 114, 591RPM.
8. 6,000 RPM is about 628 Radians/second, so K = 12628 ≈ .0191Wb.
9. No load speed is Ω = VK
= 12.02 = 6, 000Radians/second (about 5,730 RPM).
10. Since P = ΩK V−ΩKR
, we have a quadratic to find speed:
(ΩK)2 − ωKV + PmR = 0
There are actually two speeds at which the thing will make 12 watts: want the fasterone:
Ω =1
K
V
2+
√
(
V
2
)2
− PmR
≈ 473Rad/s
That is about 4519 RPM.
When making 10 W, ΩK = 6 +√
62 − 20 = 10V. Then Ω = 10.02 = 500Radians/s (4775
RPM), and current is I = 1A. Torque is T = /frac10500 = .02N-m.
Voltage is V = 10 + 2 = 12V, so efficiency is η = 1012 = 0.8333.
11. 3,000 RPM is 314.16 Radians/second, so K = 12314.16 ≈ .0382Wb. With 10 A, torque is
.382 N-m.
With a 12 volt supply the maximum converted power is Pmax = V 2
4R = 36W. At thatcondition, I = 6A, and Pin = 12 × 6 = 72W, so η = 0.5. Speed is 1,500 RPM.
12. 4,000 RPM = 66 2/3 Hz. 3×662/3 = 200, so the thing has six poles. Electrical frequencyis ω = 2π × 200 ≈ 1257Radians/s
Flux λ0 =√
2×1201257 ≈ 0.135Wb.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 84
13. T = 32pλ0I = 1.5 × 2 × 0.4 × 4 = 4.8N-m.
4,000 RPM is 418.9 Radians/second. So PB = 4.8 × 418.9 ≈ 2011W.
At 4,000 RPM, ω = 837.8, so internal voltage is Eb = .4 × 837.5 ≈ 335.1V (peak).Reactance is X = .05 × 837.8 ≈ 41.9Ω. 4 × 41.9 ≈ 167.6V. Assuming we are drivingthe thing for maximum torque per unit of current, internal power factor is unity andterminal voltage is V =
√335.12 + 167.62 ≈ 375V (peak).
Note xd = .05×4.4 = .5, which is less than one, so there will be a zero-torque speed. It is
when ω = 375.05×4 ≈ 1873Radians/s,or about 8944 RPM.
14. Here, we use the definitions given in the text:
Base Torque TB =3
2pλ0I0
d- axis reactance xd =LdI0λ0
q- axis reactance xq =LqI0λ0
per-unit torque te = (1 − (xq − xd) id) iq
Then use expressions 15.15 and 15.16 to find id and iq at the rating point (ia = 1). Atthe rating point:
ψd = 1 + xdid
ψq = xqiq
Voltages are:
Vd = ω0λ0 (raid − ψq)
Vq = ω0λ0 (rqiq + psid)
Voltage is V =√
V 2d + V 2
q and input power is Pin = 32 (VdId + VqIq).
Output power is Pout = ωpTBte.
All of this has been implemented in Matlab script p15 13.m and the results for the twocases are:
Chapter 15, Problem 13
Internal Flux = 0.01
xd = 7.5 xq = 22.5
id = -0.690637 iq = 0.723202
Developed Torque = 11.0906
Power Rating = 3484.21
Efficiency = 0.962699
Terminal Voltage = 159.827
Power Factor = 0.503211
>> p15_13
Chapter 15, Problem 13
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 85
Internal Flux = 0.1
xd = 0.75 xq = 2.25
id = -0.559816 iq = 0.828617
Developed Torque = 20.5798
Power Rating = 6465.32
Efficiency = 0.979546
Terminal Voltage = 186.376
Power Factor = 0.786977
15. The whole story is told by Matlab script p15 14.m, which uses the expressions cited forthe previous problem. The optimal locus for the axis currents is shown in Figure 87.The other elements of operation are:
Chapter 15, Problem 14
part b) Base Speed = 1642.93 RPM
part c) Maximum Torque = 37.0145 N-m
part d) Power Factor at Base = 0.687784
−0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 00
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8Chapter 15, Problem 14
q− a
xis
per−
unit
d− axis per−unit
Figure 87: Solution to Problem 14: Optimal Current Locus
The torque/speed and power/speed curves are shown in Figure 88.
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 86
0 1000 2000 3000 4000 5000 600010
20
30
40Chapter 15, Problem 14
N−
m
0 1000 2000 3000 4000 5000 60000
5000
10000
W
RPM
Figure 88: Solution to Problem 14: Torque and Power vs. speed curves
J.L. Kirtley Jr: Electric Power Principles: Sources, Conversion, Distribution and Use 87