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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

9.1 A 1-kW, 120-V, 60-Hz capacitor-start motor has the following parameters for the main and auxiliary windings (at starting):

Z main = 4 . 82 + j7 . 25 Ω size 12{Z rSub { size 8{ ital "main"} } =4 "." "82"+j7 "." "25" %OMEGA } {} main winding

Z aux = 7 . 95 + j9 . 21 Ω size 12{Z rSub { size 8{ ital "aux"} } =7 "." "95"+j9 "." "21" %OMEGA } {} auxiliary winding

a. Find the magnitude and the phase angles of the currents in the two windings when rated voltage is applied to the motor under starting conditions.

b. Find the value of starting capacitance that will place the main- and auxiliary-winding currents in time quadrature at starting.

c. Repeat part (a) when the capacitance of part (b) is inserted in series with the auxiliary winding.

9.2 Repeat Problem 9.1 if the motor is operated from a 120-V, 50-Hz source.

9.3 Given the applied electrical frequency and the corresponding impedances Z main size 12{Z rSub { size 8{ ital "main"} } } {} and Z aux size 12{Z rSub { size 8{ ital "aux"} } } {} of the main and auxiliary windings at starting, write a MATLAB script to calculate the value of the capacitance, which, when connected in series with the starting winding, will produce a starting winding current which will lead that of the main winding by 90 o size 12{"90" rSup { size 8{o} } } {} .

9.4 A 500-W, four-pole, 115-V, 60-Hz single-phase induction motor has the following parameters (resistances and reactances in Ω size 12{ %OMEGA } {} /phase):

R 1, main = 1 . 68 R 2, main = 2 . 96 X 1, main = 1 . 87 X m , main = 60 . 6 X 2, main = 1 . 72 size 12{R rSub { size 8{1, ital "main"} } =1 "." "68"" "R rSub { size 8{2, ital "main"} } =2 "." "96"" "X rSub { size 8{1, ital "main"} } =1 "." "87"" "X rSub { size 8{m, ital "main"} } ="60" "." 6" "X rSub { size 8{2, ital "main"} } =1 "." "72"} {}

Coreloss = 38 W Friction and windage = 11 . 8W size 12{"Coreloss"="38""W Friction and windage"="11" "." 8W} {}

Find the speed, stator current, torque, power output, and efficiency when the motor is operating at rated voltage and a slip of 4.2 percent.

9.5 Write a MATLAB script to produce plots of the speed and efficiency of the single-phase motor of Problem 9.4 as a function of output power over the range 0 P out size 12{<= P rSub { size 8{ ital "out"} }<= {}} {} 500 W.

9.6 At standstill the rms currents in the main and auxiliary windings of a four-pole, capacitor-start induction motor are I main size 12{I rSub { size 8{ ital "main"} } } {} = 20.7 A and I aux size 12{I rSub { size 8{ ital "aux"} } } {} = 11.1 A respectively. The auxiliary-winding current leads the main-winding current by 53 o size 12{"53" rSup { size 8{o} } } {} . The effective turns per pole (i.e., the number of turns corrected for the effects of winding distribution) are N main = 42 size 12{N rSub { size 8{ ital "main"} } ="42"} {} and N aux = 68 size 12{N rSub { size 8{ ital "aux"} } ="68"} {} . The windings are in space quadrature.

a. Determine the peak amplitudes of the forward and backward stator-mmf waves.

b. Suppose it were possible to adjust the magnitude and phase of the auxiliary-winding current. What magnitude and phase would produce a purely forward mmf wave?

9.7 The equivalent-circuit parameters of an 8-kW, 230-V, 60-Hz, four-pole, two-phase, squirrel-cage induction motor in ohms per phase are

R 1 = 0 . 253 X 1 = 1 . 14 X m = 32 . 7 R 2 = 0 . 446 X 2 = 1 . 30 size 12{R rSub { size 8{1} } =0 "." "253"" "X rSub { size 8{1} } =1 "." "14"" "X rSub { size 8{m} } ="32" "." 7" "R rSub { size 8{2} } =0 "." "446"" "X rSub { size 8{2} } =1 "." "30"} {}

This motor is operated from an unbalanced two-phase, 60-Hz source whose phase voltages are, respectively, 223 and 190 V, the smaller voltage leading the larger by 73 o size 12{"73" rSup { size 8{o} } } {} . For a slip of 0.045, find

a. the phase currents in each of the windings and

b. the internal mechanical power.

9.8 The induction motor of Problem 9.7 is supplied from an unbalanced two phase source by a four-wire feeder having an impedance Z = 0.32 + j1.5 Ω size 12{ %OMEGA } {} /phase. The source voltages can be expressed as

V ˆ α = 235 0 o { V ˆ α = 212 78 o size 12{ { hat {V}} rSub { size 8{α} } ="235"∠0 rSup { size 8{o} } " {" hat ital {V}} rSub { size 8{α} } ="212"∠"78" rSup { size 8{o} } } {}

For a slip of 5 percent, show that the induction-motor terminal voltages correspond more nearly to a balanced two-phase set than do those of the source.

9.9 The equivalent-circuit parameters in ohms per phase referred to the stator for a two-phase, 1.0 kW, 220-V, four-pole, 60-Hz, squirrel-cage induction motor are given below. The no-load rotational loss is 65 W.

R 1 = 0 . 78 R 2 = 4 . 2 X 1 = X 2 = 5 . 3 X m = 93 size 12{R rSub { size 8{1} } =0 "." "78"" "R rSub { size 8{2} } =4 "." 2" "X rSub { size 8{1} } =X rSub { size 8{2} } =5 "." 3" "X rSub { size 8{m} } ="93"} {}

a. The voltage applied to phase α size 12{α} {} is 220 0 o size 12{∠0 rSup { size 8{o} } } {} V and that applied to phase β size 12{β} {} is 220 65 o size 12{∠"65" rSup { size 8{o} } } {} V. Find the net air-gap torque at a slip s = 0.035.

b. What is the starting torque with the applied voltages of part (a)?

c. The applied voltages are readjusted so that V ˆ α = 220 65 o size 12{ { hat {V}} rSub { size 8{α} } ="220"∠"65" rSup { size 8{o} } } {} V and V ˆ β = 220 90 o size 12{ { hat {V}} rSub { size 8{β} } ="220"∠"90" rSup { size 8{o} } } {} V. Full load on the machine occurs at s = 0.048. At what slip does maximum internal torque occur? What is the value of the maximum torque?

d. While the motor is running as in part (c), phase β size 12{β} {} is open-circuited. What is the power output of the machine at a slip s = 0.04?

e. What voltage appears across the open phase- β size 12{β} {} terminals under the conditions of part (d)?

9.10 A 120-V, 60-Hz, capacitor-run, two-pole, single-phase induction motor has the following parameters:

L main = 47 . 2 mH R main = 0 . 38 Ω L aux = 102 mH R aux = 1 . 78 Ω size 12{L rSub { size 8{ ital "main"} } ="47" "." 2 ital "mH"" "R rSub { size 8{ ital "main"} } =0 "." "38" %OMEGA " "L rSub { size 8{ ital "aux"} } ="102" ital "mH"" "R rSub { size 8{ ital "aux"} } =1 "." "78" %OMEGA } {}

L r = 2 . 35 μH R r = 17 . Ω L main , r = 0 . 342 mH L aux , r = 0 . 530 mH size 12{L rSub { size 8{r} } =2 "." "35"μH" "R rSub { size 8{r} } ="17" "." 2μ %OMEGA " "L rSub { size 8{ ital "main",r} } =0 "." "342" ital "mH"" "L rSub { size 8{ ital "aux",r} } =0 "." "530" ital "mH"} {}

You may assume that the motor has 48 W of core loss and 23 W of rotational losses. The motor windings are connected with the polarity shown in Fig. 9.1 with a 40 μ size 12{μ} {} F run capacitor.

Figure 9.1 Permanent-split-capacitor induction-motor connections

a. Calculate the motor starting torque.

With the motor operating at a speed of 3490 r/min, calculate

b. the main and auxiliary-winding currents,

c. the total line current and the motor power factor,

d. the output power and

e. the electrical input power and the efficiency.

Note that this problem is most easily solved using MATLAB.

9.11 Consider the single-phase motor of Problem 9.10. Write a MATLAB script to search over the range of capacitor values from 25 μ size 12{μ} {} F to 75 μ size 12{μ} {} F to find the value which will maximize the motor efficiency at a motor speed of 3490 r/min. What is the corresponding maximum efficiency?

9.12 In order to raise the starting torque, the single-phase induction motor of Problem 9.10 is to be converted to a capacitor-start, capacitor-run motor. Write a MATLAB script to find the minimum value of starting capacitance required to raise the starting torque to 0.5 N. m.

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Source:  OpenStax, Electrical machines. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10767/1.1
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