EE303 Solutions

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


1818

1

4

18

4

1

45

45

45

5 181

Figure 1: Solution to Problem 3

−45

45

45

5 18118

1

vo+


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

)


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


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


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


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.


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


0 100 200 300 400 500 6000

20

40

60

80

100


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.


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


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


V, R

MS

A, RMS

Figure 17: Source Voltage vs. Current


0 2 4 6 8 10

x 106

0

2000

4000

6000

8000

10000


V, R

MS

W

Figure 18: Source Voltage vs. Power


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


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

)


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


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


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.


c

vn

va

i a

i b

i

Figure 24: Phase Currents


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


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


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.


x

105 V

.01.01



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


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:


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


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


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:


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


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


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


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


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


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


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.


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


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


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


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


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


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


Ohm

s/m

eter

Frequency, Hz

ResistanceReactance


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


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


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


104

105

102

103

104

105


W/m

2

A/m



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.


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


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


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:


Converted Power = 1.003e+06 W

Phase Current = 138.67 A

Output Power = 1e+06 W


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


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)


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


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


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


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


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


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


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:


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


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


Problem 10_15


Fault i_a = 0+j0 i_b = -1.41819+j-0.046498 i_c = 1.41819+j0.046498


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


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



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\


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


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:


Angle = 10 degrees

Sending end P = 82.5715 MW Q = -1.02412 MVAR

Receiving end P = 81.1392 MW Q = -15.3469 MVAR


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


Angle = 12.3843 degrees


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


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


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:


Center of Power Circle = 1980.2 + j 19802

Radius of Power Circle = 19900.7

Seeking 10 kW at receiving end

Angle = 31.3025 degrees


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


−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


0 0.5 1 1.5 2 2.5 3 3.5 47920

7940

7960

7980

8000

8020

8040


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)





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:



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









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( 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:


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


Z( 6) = 0.01 + j 0.3

Z( 7) = 0.005 + j 0.05









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( 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


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


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:


−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


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


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.


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


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:


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


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











−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



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.


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


Figure 67: Solution to Chapter 12, Problem 11: Harmonic Amplitudes


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


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


0 200 400 600 800 1000 1200 1400 1600 18000

20

40

60

80

100

120


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.


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:


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:


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


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


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


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


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


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


Figure 78: Solution to Problem 10: Variable Frequency Efficiency and Power Factor


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


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


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


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


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.


100 150 200 250 300 350 400 450 500 550 600100

200

300

400

500

600

700

800

900

1000


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.


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


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


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.


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:


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


Power Factor = 0.503211

>> p15_13



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


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:


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


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.


0 1000 2000 3000 4000 5000 600010

20

30


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


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