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PROBLEMS - chapter 1
PROBLEMS
This lecture note is based on the textbook # 1. Electric Machinery - A.E. Fitzgerald, Charles Kingsley, Jr., Stephen D. Umans- 6th edition- Mc Graw Hill series in Electrical Engineering. Power and Energy
1.1 A magnetic circuit with a single air gap is shown in Fig.1.1. The core dimensions are:
Cross-sectional area Ac =
1 . 8x 10 − 3 m 2 1 . 8x 10 − 3 m 2 size 12{1 "." 8x"10" rSup { size 8{ - 3} } m rSup { size 8{2} } } {}
Mean core length lc = 0.6 m
Gap length g = 2.3 x
10 − 3 10 − 3 size 12{"10" rSup { size 8{ - 3} } } {}
N = 83 turns
 Figure 1 |
Figure 1.1 Magnetic circuit.
Assume that the core is of infinite permeability (
μ → ∞ μ → ∞ size 12{μ rightarrow infinity } {} R C R C size 12{R rSub { size 8{C} } } {} R g R g size 12{R rSub { size 8{g} } } {} φ φ size 12{φ} {} λ λ size 12{λ} {}
1.2 Repeat Problem 1.1 for a finite core permeability of
μ = 2500 μ 0 μ = 2500 μ 0 size 12{μ="2500"μ rSub { size 8{0} } } {}
1.3 Consider the magnetic circuit of Fig.1.1 with the dimensions of Problem1.1. Assuming infinite core permeability, calculate (a) the number of turns required to achieve an inductance of 12 mH and (b) the inductor current which will result in a core flux density of 1.0 T.
1.4 Repeat Problem 1.3 for a core permeability of
μ = 1300 μ 0 μ = 1300 μ 0 size 12{μ="1300"μ rSub { size 8{0} } } {}
1.5 The magnetic circuit of Problem 1.1 has a nonlinear core material whose permeability as a function of
B m B m size 12{B rSub { size 8{m} } } {}
where
B m B m size 12{B rSub { size 8{m} } } {}
a. Using MATLAB, plot a dc magnetization curve for this material (
B m B m size 12{B rSub { size 8{m} } } {} H m H m size 12{H rSub { size 8{m} } } {} B ≤ B ≤ size 12{ <= B <= {}} {}
b. Find the current required to achieve a flux density of 2.2 T in the core.
c. Again, using MATLAB, plot the coil flux linkages as a function of coil current as the current is varied from 0 to the value found in part (b).
1.6 The magnetic circuit of Fig.1.2 consists of a core and a moveable plunger of width
l p l p size 12{l rSub { size 8{p} } } {}
You may neglect any fringing fields at the air gap and use approximations consistent with magnetic-circuit analysis.
a. Assuming that
μ → ∞ μ → ∞ size 12{μ rightarrow infinity } {} B g B g size 12{B rSub { size 8{g} } } {}
 Figure 2 |
Figure 1.2 Magnetic circuit for Problem 1.6.
plunger position is varied (
0 ≤ x ≤ 0 . 8X 0 0 ≤ x ≤ 0 . 8X 0 size 12{0 <= x <= 0 "." 8X rSub { size 8{0} } } {}
b. Repeat part (a) for a finite permeability
μ μ size 12{μ} {}
1.7 The magnetic circuit of Fig.1.2 and Problem 1.6 has the following dimensions"
a. Assuming a constant permeability of
μ = 2800 μ 0 μ = 2800 μ 0 size 12{μ="2800"μ rSub { size 8{0} } } {}
b. Repeat the calculation of part (a) for the case in which the core and plunger are composed of a nonlinear material whose permeability is given by
where
B m B m size 12{B rSub { size 8{m} } } {}
c. For the nonlinear material of part (b), use MATLAB to plot the air-gap flux density as a function of winding current for x = 0 and x = 0.5
X 0 X 0 size 12{X rSub { size 8{0} } } {}
1.8 An inductor of the form of Fig.1.1 has dimensions:
Cross-sectional area
A C = 3 . 6 cm 2 A C = 3 . 6 cm 2 size 12{A rSub { size 8{C} } =3 "." 6 ital "cm" rSup { size 8{2} } } {}
Mean core length
l C l C size 12{l rSub { size 8{C} } } {}
N - 75 turns
Assuming a core permeability of
μ = 2100 μ 0 μ = 2100 μ 0 size 12{μ="2100"μ rSub { size 8{0} } } {}
1.9 The magnetic circuit of Fig.1.3 consists of rings of magnetic material in a stack of height h. The rings have inner radius Ri and outer radius Ro. Assume that the iron is of infinite permeability (
μ → ∞ μ → ∞ size 12{μ rightarrow infinity } {}
h =2cm
g = 0.2 cm
calculate:
a. the mean core length
l C l C size 12{l rSub { size 8{C} } } {} A C A C size 12{A rSub { size 8{C} } } {}
b. the reluctance of the core
R C R C size 12{R rSub { size 8{C} } } {} R g R g size 12{R rSub { size 8{g} } } {}
For N = 65 turns, calculate:
c. the inductance L.
d. current i required to operate at an air-gap flux density of
B g B g size 12{B rSub { size 8{g} } } {}
e. the corresponding flux linkages
***SORRY, THIS MEDIA TYPE IS NOT SUPPORTED.***
of the coil.
 Figure 3 |
Figure 1.3 Magnetic circuit.
1.10 Repeat Problem 1.9 for a core permeability of
***SORRY, THIS MEDIA TYPE IS NOT SUPPORTED.***
.
1.11 Using MATLAB, plot the inductance of the inductor of Problem 1.9 as a function of relative core permeability as the core permeability varies for
***SORRY, THIS MEDIA TYPE IS NOT SUPPORTED.***
= 100 to
μ x μ x size 12{μ rSub { size 8{x} } } {}
1.12 The inductor of Fig. 1.4 has a core of uniform circular cross-section of area
A C A C size 12{A rSub { size 8{C} } } {}
1.13 The inductor of Fig.1.27 has the following dimensions:
g = 0.8 mm
N = 480 turns
Neglecting leakage and fringing and assuming
μ x μ x size 12{μ rSub { size 8{x} } } {}
 Figure 4 |
Figure 1.4 Inductor.
 Figure 5 |
Figure 1.5 Pot-core inductor .
1.14 The inductor of Problem 1.13 is to be operated from a 60-Hz voltage source. (a) Assuming negligible coil resistance, calculate the rms inductor voltage corresponding to a peak core flux density of 1.5 T. (b) Under this operating condition, calculate the rms current and the peak stored energy.
1.15 Consider the magnetic circuit of Fig. 1.5. This structure, known as a pot-core, is typically made in two halves. The N-turn coil is wound on a cylindrical bobbin and can be easily inserted over the central post of the core as the two halves are assembled. Because the air gap is internal to the core, provided the core is not driven excessively into saturation, relatively little magnetic flux will "leak" from the core, making this a particularly attractive configuration for a wide variety of applications, both for inductors such as that of Fig. 1.27 and transformers.
Assume the core permeability to be
μ = 2500 μ 0 μ = 2500 μ 0 size 12{μ="2500"μ rSub { size 8{0} } } {}
h = 0.75cm g = 0.5mm
a. Find the value of
R 3 R 3 size 12{R rSub { size 8{3} } } {}
b. Although the flux density in the radial sections of the core (the sections of thickness h) actually decreases with radius, assume that the flux density remains uniform. (i) Write an expression for the coil inductance and (ii) evaluate it for the given dimensions.
c. The core is to be operated at a peak flux density of 0.8 T at a frequency of 60 Hz. Find (i) the corresponding rms value of the voltage induced in the winding, (ii) the rms coil current, and (iii) the peak stored energy.
d. Repeat part (c) for a frequency of 50 Hz.
 Figure 6 |
Figure 1.6 Inductor.
1.16 A square voltage wave having a fundamental frequency of 60 Hz and equal positive and negative half cycles of amplitude E is applied to a 1000-turn winding surrounding a closed iron core of 1.25 x 10-3
m 2 m 2 size 12{m rSup { size 8{2} } } {}
a. Sketch the voltage, the winding flux linkage, and the core flux as a function of time.
b. Find the maximum permissible value of E if the maximum flux density is not to exceed 1.15 T.
1.17 An inductor is to be designed using a magnetic core of the form of that of Fig.1.6. The core is of uniform cross-sectional area
A C = 5 . 0 cm 2 A C = 5 . 0 cm 2 size 12{A rSub { size 8{C} } =5 "." 0 ital "cm" rSup { size 8{2} } } {} I C I C size 12{I rSub { size 8{C} } } {}
a. Calculate the air-gap length g and the number of turns N such that the inductance is 1.4 mH and so that the inductor can operate at peak currents of 6 A without saturating. Assume that saturation occurs when the peak flux density in the core exceeds 1.7 T and that, below saturation, the core has permeability
μ = 3200 μ 0 μ = 3200 μ 0 size 12{μ="3200"μ rSub { size 8{0} } } {}
b. For an inductor current of 6 A, use Eq.
W fld = ∫ V B 2 2μ dV W fld = ∫ V B 2 2μ dV size 12{W rSub { size 8{ ital "fld"} } = Int cSub { size 8{V} } { left [ { {B rSup { size 8{2} } } over {2μ} } right ]} ital "dV"} {} T fld = i 2 2 dL ( θ ) dθ T fld = i 2 2 dL ( θ ) dθ size 12{T rSub { size 8{ ital "fld"} } = { {i rSup { size 8{2} } } over {2} } { { ital "dL" \( θ \) } over {dθ} } } {}
1.18 Consider the inductor of Problem 1.17. Write a simple design program in the form of a MATLAB script to calculate the number of turns and air-gap length as a function of the desired inductance. The script should be written to request a value of inductance (in mH) from the user, with the output being the air-gap length in mm and the number of turns.
The inductor is to be operated with a sinusoidal current at 60 Hz, and it must be designed such that the peak core flux density will be equal to 1.7 T when the inductor current is equal to 4.5 A rms. Write your script to reject any designs for which the gap length is out of the range of 0.05 mm to 5.0 mm or for which the number of turns drops below 5.
Using your program find (a) the minimum and (b) the maximum inductances (to the nearest mH) which will satisfy the given constraints. For each of these values, find the required air-gap length and the number of turns as well as the rms voltage corresponding to the peak core flux.
 Figure 7 |
Figure 1.7 Toroidal winding.
1.19 A proposed energy storage mechanism consists of an N-turn coil wound around a large nonmagnetic (
μ = μ 0 μ = μ 0 size 12{μ=μ rSub { size 8{0} } } {} size 12{ <= {}} {}
For a coil with N = 1000 turns, r = 10 m, and a = 0.45 m:
a. Calculate the coil inductance L.
b. The coil is to be charged to a magnetic flux density of 1.75 T. Calculate the total stored magnetic energy in the toms when this flux density is achieved.
c. If the coil is to be charged at a uniform rate (i.e., di/dt = constant), calculate the terminal voltage required to achieve the required flux density in 30 sec. Assume the coil resistance to be negligible.
1.20 Figure 1.8 shows an inductor wound on a laminated iron core of rectangular cross section. Assume that the permeability of the iron is infinite. Neglect magnetic leakage and fringing in the two air gaps (total gap length = g). The N-turn winding is insulated copper wire whose resistivity is ρ
Ω Ω size 12{ %OMEGA } {} f w f w size 12{f rSub { size 8{w} } } {}
 Figure 8 |
Figure 1.8 Iron-core inductor.
a. Calculate the cross-sectional area and volume of the copper in the winding space.
b. Write an expression for the flux density B in the inductor in terms of the current density
J cu J cu size 12{J rSub { size 8{ ital "cu"} } } {}
c. Write an expression for the copper current density
J cu J cu size 12{J rSub { size 8{ ital "cu"} } } {}
d. Derive an expression for the electric power dissipation in the coil in terms of the current density
J cu J cu size 12{J rSub { size 8{ ital "cu"} } } {}
e. Derive an expression for the magnetic stored energy in the inductor in terms of the applied current density
J cu J cu size 12{J rSub { size 8{ ital "cu"} } } {}
f. From parts (d) and (e) derive an expression for the L/R time constant of the inductor. Note that this expression is independent of the number of turns in the coil and does not change as the inductance and coil resistance are changed by varying the number of turns.
1.21 The inductor of Fig. 1.8 has the following dimensions:
a = h = ω = 1.5cm b = 2cm g = 0.2cm
The winding factor (i.e., the fraction of the total winding area occupied by conductor) is
f w f w size 12{f rSub { size 8{w} } } {} 1 . 73 x 10 − 8 Ω . m 1 . 73 x 10 − 8 Ω . m size 12{1 "." "73"x"10" rSup { size 8{ - 8} } %OMEGA "." m} {}
where AWG is the wire size, expressed in terms of the American Wire Gauge, and
A wire A wire size 12{A rSub { size 8{ ital "wire"} } } {} m 2 m 2 size 12{m rSup { size 8{2} } } {}
1.22 The magnetic circuit of Fig.1.9 has two windings and two air gaps. The core can be assumed to be of infinite permeability. The core dimensions are indicated in the figure.
a. Assuming coil 1 to be carrying a current
I 1 I 1 size 12{I"" lSub { size 8{1} } } {}
b. Repeat part (a), assuming zero current in winding 1 and a current
I 2 I 2 size 12{I rSub { size 8{2} } } {}
c. Repeat part (a), assuming the current in winding 1 to be
I 1 I 1 size 12{I rSub { size 8{1} } } {} I 2 I 2 size 12{I rSub { size 8{2} } } {}
d. Find the self-inductances of windings 1 and 2 and the mutual inductance between the windings.
 Figure 9 |
Figure 1.9 Magnetic circuit.
 Figure 10 |
Figure 1.10 Symmetric magnetic circuit.
1.23 The symmetric magnetic circuit of Fig.1.10 has three windings. Windings A and B each have N turns and are wound on the two bottom legs of the core. The core dimensions are indicated in the figure.
a. Find the self-inductances of each of the windings.
b. Find the mutual inductances between the three pairs of windings.
c. Find the voltage induced in winding 1 by time-varying currents
i A i A size 12{i rSub { size 8{A} } } {} i B i B size 12{i rSub { size 8{B} } } {}
1.24 The reciprocating generator of Fig.1.11 has a movable plunger (position x) which is supported so that it can slide in and out of the magnetic yoke while maintaining a constant air gap of length g on each side adjacent to the yoke. Both the yoke and the plunger can be considered to be of infinite permeability. The motion of the plunger is constrained such that its position is limited to
{} 0 ≤ x ≤ ω 0 ≤ x ≤ ω size 12{0 <= x <= ω} {}
 Figure 11 |
Figure 1.11 Reciprocating generator.
There are two windings on this magnetic circuit. The first has
N 1 N 1 size 12{N rSub { size 8{1} } } {} I 0 I 0 size 12{I rSub { size 8{0} } } {} N 2 N 2 size 12{N rSub { size 8{2} } } {}
a. Neglecting any fringing effects, find the mutual inductance between windings 1 and 2 as a function of the plunger position x.
b. The plunger is driven by an external source so that its motion is given by
where
ε ≤ 1 ε ≤ 1 size 12{ε <= 1} {}
1.25 Figure 1.12 shows a configuration that can be used to measure the magnetic characteristics of electrical steel. The material to be tested is cut or punched into circular laminations which are then stacked (with interspersed insulation to avoid eddy-current formation). Two windings are wound over this stack of laminations: the first, with
N 1 N 1 size 12{N rSub { size 8{1} } } {} N 2 N 2 size 12{N rSub { size 8{2} } } {}
The accuracy of the results requires that the magnetic flux density be uniform within the laminations. This can be accomplished if the lamination width t =
R 0 − R 1 R 0 − R 1 size 12{R rSub { size 8{0} } - R rSub { size 8{1} } } {} i 1 = I 0 i 1 = I 0 size 12{i rSub { size 8{1} } =I rSub { size 8{0} } } {}
a. Find the relationship between the magnetic field intensity H in the laminations and current
i 1 i 1 size 12{i rSub { size 8{1} } } {}
b. Find the relationship between the voltage ve and the time rate of change of the flux density B in the laminations.
c. Find the relationship between the voltage
v 0 = G ∫ v 2 dt v 0 = G ∫ v 2 dt size 12{v rSub { size 8{0} } =G Int {v rSub { size 8{2} } } ital "dt"} {}
 Figure 12 |
Figure 1.12 Configuration for measurement of magnetic properties of electrical steel.
In this problem, we have shown that the magnetic field intensity H and the magnetic flux density B in the laminations are proportional to the current
i 1 i 1 size 12{i rSub { size 8{1} } } {} V 2 V 2 size 12{V rSub { size 8{2} } } {}
1.26 In order to test the properties of a sample of electrical steel, a set of laminations of the form of Fig.1.12 have been stamped out of a sheet of the electrical steel of thickness 3.0 mm. The radii of the laminations are
R i R i size 12{R rSub { size 8{i} } } {} R 0 R 0 size 12{R rSub { size 8{0} } } {}
a. The flux in the lamination stack will be excited from a variable-amplitude, 100-Hz voltage source whose peak amplitude is 30 V (peak-to-peak). Calculate the number of turns
N 1 N 1 size 12{N rSub { size 8{1} } } {}
b. With a secondary winding of
N 2 N 2 size 12{N rSub { size 8{2} } } {}
 Figure 13 |
Figure 1.13 Magnetic circuit.
1.27 The coils of the magnetic circuit shown in Fig. 1.13 are connected in series so that the mmf's