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

4.1 The rotor of a six-pole synchronous generator is rotating at a mechanical speed of 1200 r/min.

a. Express this mechanical speed in radians per second.

b. What is the frequency of the generated voltage in hertz and in radians per second?

c. What mechanical speed in revolutions per minute would be required to generate voltage at a frequency of 50 Hz?

4.2 The voltage generated in one phase of an unloaded three-phase synchronous generator is of the form v(t) = V0cosωtV0cosωt size 12{V rSub { size 8{0} } "cos"ωt} {}. Write expressions for the voltage in the remaining two phases.

4.3 A three-phase motor is used to drive a pump. It is observed (by the use of a stroboscope) that the motor speed decreases from 898 r/min when the pump is unloaded to 830 r/min as the pump is loaded.

a. Is this a synchronous or an induction motor?

b. Estimate the frequency of the applied armature voltage in hertz.

c. How many poles does this motor have?

4.4 A three-phase Y-connected ac machine is initially operating under balanced three-phase conditions when one of the phase windings becomes open-circuited. Because there is no neutral connection on the winding, this requires that the currents in the remaining two windings become equal and opposite. Under this condition, calculate the relative magnitudes of the resultant positive- and negative-traveling mmf waves.

4.5 What is the effect on the rotating mmf and flux waves of a three-phase winding produced by balanced-three-phase currents if two of the phase connections are interchanges?

4.6 In a balanced two-phase machine, the two windings are displaced 90 electrical degrees in space, and the currents in the two windings are phase-displaced 90 electrical degrees in time. For such a machine, carry out the process leading to an equation for the rotating mmf wave corresponding to Eq.4.39 (which is derived for a three-phase machine).

4.7 This problem investigates the advantages of short-pitching the stator coils of an ac machine. Figure 4.1a shows a single full-pitch coil in a two-pole machine. Figure 4.1b shows a fractional-pitch coil for which the coil sides are β radians apart, rather than π radians ( 18001800 size 12{"180" rSup { size 8{0} } } {}) as is the case for the full-pitch coil.

For an air-gap radial flux distribution of the form

B r = ∑ n odd B n cos nθ B r = ∑ n odd B n cos nθ size 12{B rSub { size 8{r} } = Sum cSub { size 8{"n odd"} } {B rSub { size 8{n} } } "cos"nθ} {}

where n = 1 corresponds to the fundamental space harmonic, n = 3 to the third space harmonic, and so on, the flux linkage of each coil is the integral of BrBr size 12{B rSub { size 8{r} } } {} over the surface spanned by that coil. Thus for the nth space harmonic, the ratio of the maximum fractional-pitch coil flux linkage to that of the full-pitch coil is

∫ − β / 2 β / 2 B n cos nθ dθ ∫ − π / 2 π / 2 B n cos nθ dθ = ∫ − β / 2 β / 2 cos nθ dθ ∫ − π / 2 π / 2 cos nθ dθ = ∣ sin ( nβ / 2 ) ∣ ∫ − β / 2 β / 2 B n cos nθ dθ ∫ − π / 2 π / 2 B n cos nθ dθ = ∫ − β / 2 β / 2 cos nθ dθ ∫ − π / 2 π / 2 cos nθ dθ = ∣ sin ( nβ / 2 ) ∣ size 12{ { { Int rSub { size 8{ - β/2} } rSup { size 8{β/2} } {B rSub { size 8{n} } "cos"nθ dθ} } over { Int rSub { size 8{ - π/2} } rSup { size 8{π/2} } {B rSub { size 8{n} } } "cos"nθ dθ} } = { { Int rSub { size 8{ - β/2} } rSup { size 8{β/2} } {"cos"nθ dθ} } over { Int rSub { size 8{ - π/2} } rSup { size 8{π/2} } {"cos"nθ dθ} } } = \lline "sin" \( nβ/2 \) \lline } {}

It is common, for example, to fractional-pitch the coils of an ac machine by 30 electrical degrees ( β=5π6=1500β=5π6=1500 size 12{β= { {5π} over {6} } ="150" rSup { size 8{0} } } {}). For n =1, 3, 5 calculate the fractional reduction in flux linkage due to short pitching.

Figure 4.1 (a) full-pitch coil and (b) fractional-pitch coil.

4.8 A six-pole, 60-Hz synchronous machine has a rotor winding with a total of 138 series turns and a winding factor krkr size 12{k rSub { size 8{r} } } {} = 0.935. The rotor length is 1.97m, the rotor radius is 58cm, and the air-gap length = 3.15 cm.

a. What is the rated operating speed in r/min?

b. Calculate the rotor-winding current required to achieve a peak fundamental air-gap flux density of 1.23 T.

c. Calculate the corresponding flux per pole.

4.9 Assume that a phase winding of the synchronous machine of Problem 4.8 consists of one full-pitch, 11-turn coil per pole pair, with the coils connected in series to form the phase winding. If the machine is operating at rated speed and under the operating conditions of Problem 4.8, calculate the rms generated voltage per phase.

4.10The synchronous machine of Problem 4.8 has a three-phase winding with 45 series turns per phase and a winding factor kwkw size 12{k rSub { size 8{w} } } {} = 0.928. For the flux condition and rated speed of Problem 4.8, calculate the rms-generated voltage per phase.

4.11The three-phase synchronous machine of Problem 4.8 is to be moved to an application which requires that its operating frequency be reduced from 60 to 50 Hz. This application requires that, for the operating condition considered in Problem 4.8, the rms generated voltage equal 13.0 kV line-to-line. As a result, the machine armature must be rewound with a different number of turns. Assuming a winding factor of kwkw size 12{k rSub { size 8{w} } } {} = 0.928, calculate the required number of series turns per phase.

4.12 Figure 4.2 shows a two-pole rotor revolving inside a smooth stator which carries a coil of 110 turns. The rotor produces a sinusoidal space distribution of flux at the stator surface; the peak value of the flux-density wave being 0.85 T when the current in the rotor is 15 A. The magnetic circuit is linear. The inside diameter of the stator is 11 cm, and its axial length is 0.17 m. The rotor is driven at a speed of 50 r/sec.

Figure 4.2 Elementary generator.

a. The rotor is excited by a current of 15 A. Taking zero time as the instant when the axis of the rotor is vertical, find the expression for the instantaneous voltage generated in the open-circuited stator coil.

b. The rotor is now excited by a 50-Hz sinusoidal alternating currrent whose peak value is 15 A. Consequently, the rotor current reverses every half revolution; it is timed to be at its maximum just as the axis of the rotor is vertical (i.e., just as it becomes aligned with that of the stator coil). Taking zero time as the instant when the axis of the rotor is vertical, find the expression for the instantaneous voltage generated in the open-circuited stator coil. This scheme is sometimes suggested as a dc generator without a commutator; the thought being that if alternative half cycles of the alternating voltage generated in part (a) are reversed by reversal of the polarity of the field (rotor) winding, then a pulsating direct voltage will be generated in the stator. Discuss whether or not this scheme will work.

4.13 A three-phase two-pole winding is excited by balanced three-phase 60-Hz currents as described by Eqs. 4.23 to 4.25. Although the winding distribution has been designed to minimize harmonics, there remains some third and fifth spatial harmonics. Thus the phase-a mmf can be written as

F a = i a ( A 1 cos θ a + A 3 cos 3θ a + A 5 cos 5θ a ) F a = i a ( A 1 cos θ a + A 3 cos 3θ a + A 5 cos 5θ a ) size 12{F rSub { size 8{a} } =i rSub { size 8{a} } \( A rSub { size 8{1} } "cos"θ rSub { size 8{a} } +A rSub { size 8{3} } "cos"3θ rSub { size 8{a} } +A rSub { size 8{5} } "cos"5θ rSub { size 8{a} } \) } {}

Similar expressions can be written for phases b (replace θaθa size 12{θ rSub { size 8{a} } } {} by θa−1200θa−1200 size 12{θ rSub { size 8{a} } - "120" rSup { size 8{0} } } {}) and c (replace θaθa size 12{θ rSub { size 8{a} } } {} by θa+1200θa+1200 size 12{θ rSub { size 8{a} } +"120" rSup { size 8{0} } } {}). Calculate the total three-phase mmf. What is the angular velocity and rotational direction of each component of the mmf?

4.14 The nameplate of a dc generator indicates that it will produce an output voltage of 24 V dc when operated at a speed of 1200 r/min. By what factor must the number of armature turns be changed such that, for the same field-flux per pole, the generator will produce an output voltage of 18 V dc at a speed of 1400 r/min?

4.15 The armature of a two-pole dc generator has a total of 320 series turns. When operated at a speed of 1800 r/min, the open-circuit generated voltage is 240 V. Calculate φPφP size 12{φ rSub { size 8{P} } } {}, the air-gap flux per pole.

4.16 The design of a four-pole, three-phase, 230-V, 60-Hz induction motor is to be based on a stator core of length 21 cm and inner diameter 9.52 cm. The stator winding distribution which has been selected has a winding factor kwkw size 12{k rSub { size 8{w} } } {} = 0.925. The armature is to be Y-connected, and thus the rated phase voltage will be 230/ 33 size 12{ sqrt {3} } {} V.

a. The designer must pick the number of armature turns so that the flux density in the machine is large enough to make efficient use of the magnetic material without being so large as to result in excessive saturation. To achieve this objective, the machine is to be designed with a peak fundamental air-gap flux density of 1.25 T. Calculate the required number of series turns per phase.

b. For an air-gap length of 0.3 mm, calculate the self-inductance of an armature phase based upon the result of part (a) and using the inductance formulas of Appendix B. Neglect the reluctance of the rotor and stator iron and the armature leakage inductance.

4.17 A two-pole, 60-Hz, three-phase, laboratory-size synchronous generator has a rotor radius of 5.71 cm, a rotor length of 18.0 cm, and an air-gap length of 0.25 mm. The rotor field winding consists of 264 turns with a winding factor of krkr size 12{k rSub { size 8{r} } } {} = 0.95. The Y-connected armature winding consists of 45 turns per phase with a winding factor kwkw size 12{k rSub { size 8{w} } } {} = 0.93.

a. Calculate the flux per pole and peak fundamental air-gap flux density which will result in an open-circuit, 60-Hz armature voltage of 120 V rms/phase (line-to-neutral).

b. Calculate the dc field current required to achieve the operating condition of part (a).

c. Calculate the peak value of the field-winding to armature-phase-winding mutual inductance.

4.18 Write a MATLAB script which calculates the required total series field- and armature-winding turns for a three-phase, Y-connected synchronous motor given the following information:

Rotor radius, R (meters)

Air-gap length, g (meters)

Electrical frequency, fefe size 12{f rSub { size 8{e} } } {}

Field-winding factor, kfkf size 12{k rSub { size 8{f} } } {}

Rotor length, l (meters)

Number of poles, poles

Peak fundamental air-gap flux density, BpeakBpeak size 12{B rSub { size 8{ ital "peak"} } } {}

Armature-winding factor, kwkw size 12{k rSub { size 8{w} } } {}

Rated rms open-circuit line-to-line terminal voltage, VratedVrated size 12{V rSub { size 8{ ital "rated"} } } {}

Field-current at rated-open-circuit terminal voltage, IfIf size 12{I rSub { size 8{f} } } {}

4.19 A four-pole, 60-Hz synchronous generator has a rotor length of 5.2 m, diameter of 1.24 m, and air-gap length of 5.9 cm. The rotor winding consists of a series connection of 63 turns per pole with a winding factor of krkr size 12{k rSub { size 8{r} } } {} = 0.91. The peak value of the fundamental air-gap flux density is limited to 1.1 T and the rotor winding current to 2700 A. Calculate the maximum torque (N.m) and power output (MW) which can be supplied by this machine.

4.20 Thermal considerations limit the field-current of the laboratory-size synchronous generator of Problem 4.17 to a maximum value of 2.4 A. If the peak fundamental air-gap flux density is limited to a maximum of 1.3 T, calculate the maximum torque (N.m) and power (kW) which can be produced by this generator.

4.21 Figure 4.3 shows in cross section a machine having a rotor winding f and two identical stator windings a and b whose axes are in quadrature. The self-inductance of each stator winding is LaaLaa size 12{L rSub { size 8{ ital "aa"} } } {} and of the rotor is LffLff size 12{L rSub { size 8{ ital "ff"} } } {}. The air gap is uniform. The mutual inductance between a stator winding depends on the angular position of the rotor and may be assumed to be of the form

M af = M cos θ 0 M af = M cos θ 0 size 12{M rSub { size 8{ ital "af"} } =M"cos"θ rSub { size 8{0} } } {} M bf = M sin θ 0 M bf = M sin θ 0 size 12{M rSub { size 8{ ital "bf"} } =M"sin"θ rSub { size 8{0} } } {}

where M is the maximum value of the mutual inductance. The resistance of each stator winding is RaRa size 12{R rSub { size 8{a} } } {}.

a. Derive a general expression for the torque T in terms of the angle θ0θ0 size 12{θ rSub { size 8{0} } } {}, the inductance parameters, and the instantaneous currents iaia size 12{i rSub { size 8{a} } } {}, IbIb size 12{I rSub { size 8{b} } } {}, and ifif size 12{i rSub { size 8{f} } } {}. Does this expression apply at standstill? When the rotor is revolving?

b. Suppose the rotor is stationary and constant direct currents IaIa size 12{I rSub { size 8{a} } } {} = I0I0 size 12{I rSub { size 8{0} } } {}, Ib=I0Ib=I0 size 12{I rSub { size 8{b} } =I rSub { size 8{0} } } {}, and If=2I0If=2I0 size 12{I rSub { size 8{f} } =2I rSub { size 8{0} } } {} are supplied to the windings in the directions indicated by the dots and crosses in Fig. 4.45. If the rotor is allowed to move, will it rotate continuously or will it tend to come to rest? If the latter, at what value of θ0θ0 size 12{θ rSub { size 8{0} } } {} ?

Figure 4.3 Elementary cylindrical-rotor, two-phase synchronous machine.

c. The rotor winding is now excited by a constant direct current If while the stator windings carry balanced two-phase currents

i a = 2 I a cos ωt i b = 2 I a sin ωt i a = 2 I a cos ωt i b = 2 I a sin ωt size 12{i rSub { size 8{a} } = sqrt {2} I rSub { size 8{a} } "cos"ωt" "i rSub { size 8{b} } = sqrt {2} I rSub { size 8{a} } "sin"ωt} {}

The rotor is revolving at synchronous speed so that its instantaneous angular position is given by θ0=ωt−δθ0=ωt−δ size 12{θ rSub { size 8{0} } =ωt - δ} {}, where δδ size 12{δ} {} is a phase angle describing the position of the rotor at t = 0. The machine is an elementary two-phase synchronous machine. Derive an expression for the torque under these conditions.

d. Under the conditions of part (c), derive an expression for the instantaneous terminal voltages of stator phases a and b.

4.22 Consider the two-phase synchronous machine of Problem 4.21. Derive an expression for the torque acting on the rotor if the rotor is rotating at constant angular velocity, such that θ0=ωt+δθ0=ωt+δ size 12{θ rSub { size 8{0} } =ωt+δ} {}, and the phase currents become unbalanced such that

i a = 2 I a cos ωt i b = 2 ( I a + I ' ) sin ωt i a = 2 I a cos ωt i b = 2 ( I a + I ' ) sin ωt size 12{i rSub { size 8{a} } = sqrt {2} I rSub { size 8{a} } "cos"ωt" "i rSub { size 8{b} } = sqrt {2} \( I rSub { size 8{a} } + { {I}} sup { ' } \) "sin"ωt} {}

What are the instantaneous and time-averaged torque under this condition?

4.23 Figure 4.4 shows in schematic cross section a salient-pole synchronous machine having two identical stator windings a and b on a laminated steel core. The salient-pole rotor is made of steel and carries a field winding f connected to slip rings.

Because of the nonuniform air gap, the self- and mutual inductances are functions of the angular position θ0θ0 size 12{θ rSub { size 8{0} } } {} of the rotor. Their variation with θ0 can be approximated as:

L aa = L 0 + L 2 cos 2θ 0 L aa = L 0 + L 2 cos 2θ 0 size 12{L rSub { size 8{ ital "aa"} } =L rSub { size 8{0} } +L rSub { size 8{2} } "cos"2θ rSub { size 8{0} } } {} L bb = L 0 − L 2 cos 2θ 0 L bb = L 0 − L 2 cos 2θ 0 size 12{L rSub { size 8{ ital "bb"} } =L rSub { size 8{0} } - L rSub { size 8{2} } "cos"2θ rSub { size 8{0} } } {} M ab = L 2 sin θ 0 M ab = L 2 sin θ 0 size 12{M rSub { size 8{ ital "ab"} } =L rSub { size 8{2} } "sin"θ rSub { size 8{0} } } {}

Figure 4.4 Schematic two-phase, salient-pole synchronous machine.

where L0L0 size 12{L rSub { size 8{0} } } {} and L2L2 size 12{L rSub { size 8{2} } } {} are positive constants. The mutual inductance between the rotor and the stator windings are functions of θ0θ0 size 12{θ rSub { size 8{0} } } {}

M af = M cos θ 0 M af = M cos θ 0 size 12{M rSub { size 8{ ital "af"} } =M"cos"θ rSub { size 8{0} } } {} M bf = M sin θ 0 M bf = M sin θ 0 size 12{M rSub { size 8{ ital "bf"} } =M"sin"θ rSub { size 8{0} } } {}

where M is also a positive constant. The self-inductance of the field winding, LffLff size 12{L rSub { size 8{ ital "ff"} } } {}, is constant, independent of θ0θ0 size 12{θ rSub { size 8{0} } } {}.

Consider the operating condition in which the field winding is excited by direct current IfIf size 12{I rSub { size 8{f} } } {} and the stator windings are connected to a balanced two-phase voltage source of frequency ωω size 12{ω} {}. With the rotor revolving at synchronous speed, its angular position will be given by θ0=ωtθ0=ωt size 12{θ rSub { size 8{0} } =ωt} {}.

Under this operating condition, the stator currents will be of the form

i a = 2 I a cos ( ωt + δ ) i b = 2 I a sin ( ωt + δ ) i a = 2 I a cos ( ωt + δ ) i b = 2 I a sin ( ωt + δ ) size 12{i rSub { size 8{a} } = sqrt {2} I rSub { size 8{a} } "cos" \( ωt+δ \) " "i rSub { size 8{b} } = sqrt {2} I rSub { size 8{a} } "sin" \( ωt+δ \) } {}

a. Derive an expression for the electromagnetic torque acting on the rotor.

b. Can the machine be operated as a motor and/or a generator? Explain.

c. Will the machine continue to run if the field current If is reduced to zero? Support you answer with an expression for the torque and an explanation as to why such operation is or is not possible.

4.24 A three-phase linear ac motor has an armature winding of wavelength 25 cm. A three-phase balanced set of currents at a frequency of 100 Hz is applied to the armature.

a. Calculate the linear velocity of the armature mmf wave.

b. For the case of a synchronous rotor, calculate the linear velocity of the rotor.

c. For the case of an induction motor operating at a slip of 0.045, calculate the linear velocity of the rotor.

4.25 The linear-motor armature of Problem 4.24 has a total active length of 7 wavelengths, with a total of 280 turns per phase with a winding factor kwkw size 12{k rSub { size 8{w} } } {} = 0.91. For an air-gap length of 0.93 cm, calculate the rms magnitude of the balanced three-phase currents which must be supplied to the armature to achieve a peak space-fundamental air-gap flux density of 1.45 T.

4.26 A two-phase linear permanent-magnet synchronous motor has an air-gap of length 1.0 mm, a wavelength of 12 cm, and a pole width of 4 cm. The rotor is 5 wavelengths in length. The permanent magnets on the rotor are arranged to produce an air-gap magnetic flux distribution that is uniform over the width of a pole but which varies sinusoidally in space in the direction of rotor travel. The peak density of this air-gap flux is 0.97 T.

a. Calculate the net flux per pole.

b. Each armature phase consists of 10 turns per pole, with all the poles connected in series. Assuming that the armature winding extends many wavelengths past either end of the rotor, calculate the peak flux linkages of the armature winding.

c. If the rotor is traveling at a speed of 6.3 m/sec, calculate the rms voltage induced in the armature winding.

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Last edited by vocw on Jul 8, 2009 4:39 am -0500.