barcode reader code in asp.net The no-load test of a single-phase induction in Software

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1823 The no-load test of a single-phase induction
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motor is made by running the motor without load at rated voltage and rated frequency Derive the equivalent circuit of a single-phase induction motor for the no-load test [Hint: The no-load slip is very small]
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1824 Derive the equivalent circuit of a single-phase
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induction motor for the locked-rotor test Neglect the magnetizing current
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1825 The design for a 1 -hp, two-pole, 115-V universal 8
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motor gives the effective resistances of the armature and series eld as 4 and 6 , respectively The output torque is 017 N-m when the motor is drawing rated current of 15 A (rms) at a power factor of 088 at rated speed Find: a The full-load ef ciency b The rated speed c The full-load copper losses d The combined windage, friction, and iron losses e The motor speed when the rms current is 05 A, neglecting phase differences and saturation
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1826 A 240-V, 60-Hz, two-pole universal motor
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operates at a speed of 12,000 rev/min on full load and draws a current of 65 A at 094 power factor lagging The series eld-winding impedance is 455 + j 32 ohms, and the armature circuit impedance is 615 + j 94 ohms Find a The back emf of the motor b The mechanical power developed by the motor c The power output if the rotational loss is 65 W d The ef ciency of the motor
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1827 A single-phase motor is drawing 20 A from a
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400-V, 50-Hz supply The power factor is 08 lagging
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18
Special-Purpose Electric Machines
What value of capacitor connected across the circuit will be necessary to raise the power factor to unity
the speed of the motor and the torque available at the maximum permissible load
1828 A three-phase induction motor is required to
operate from a single-phase source One possible connection is shown in Figure P1828 Will the motor work Explain why or why not
Speed, percent of full load 4 Resistor 2 1 0 280 240 200 160 120 80 0 4 3 2 1 0
v + ~
Field
(a) b c
40 80 120 160 Torque, percent of full load (b)
Figure P1831 Figure P1828
1832 Which single-phase motor would you choose for 1829 In performing a brake-load test upon a 1/4-hp
capacitor-start motor with its output adjusted to rated value, the following data were obtained: E = 115 volts; I = 38 amp; P = 310 W; rev/min = 1725 Calculate: a Ef ciency b Power factor c Torque in pound-inches the following applications a Inexpensive analog electric clock b Bathroom ventilator fan c Escalator which must start under all load conditions d Kitchen blender e Table model circular saw operating at about 3,500 rev/min f Hand-held circular saw operating at 15,000 rev/min g Water pump
Section 4: Motor Performance and Selection 1830 What type of motor would you select to perform
the following tasks Justify your selection a Vacuum cleaner b Refrigerator c Air conditioner compressor d Air conditioner fan e Variable-speed sewing machine f Clock g Electric drill h Tape drive i X-Y plotter
1833 The power required to drive a fan varies as the
cube of the speed If a motor driving a shaft-mounted fan is loaded to 100 percent of its horsepower rating on the top speed connection, what is the horsepower output in percent of rating: a At a speed reduction of 20 percent b At a speed reduction of 30 percent c At a speed reduction of 50 percent
1834 An industrial plant has a load of 800 kW at a
power factor of 08 lagging It is desired to purchase a synchronous motor of suf cient capacity to deliver a load of 200 kW and also serve to correct the overall plant power factor to 092 Assuming that the synchronous motor has an ef ciency of 91 percent, determine its KVA input rating and the power factor at which it will operate
1831 A 5-hp, 1,150-rev/min shunt motor has its speed
controlled by means of a tapped eld resistor as shown in Figure P1831 With the tap at position 3, determine
A P P E N D I X
Linear Algebra and Complex Numbers
A1 SOLVING SIMULTANEOUS LINEAR EQUATIONS, CRAMER S RULE AND MATRIX EQUATION
The solution of simultaneous equations, such as those that are often seen in circuit theory, may be obtained relatively easily by using Cramer s rule This method applies to 2 2 or larger systems of equations Cramer s rule requires the use of the concept of determinant The method of determinants is valuable because it is systematic, general, and useful in solving complicated problems A determinant is a scalar de ned on a square array of numbers, or matrix, such as det (A) = |A| = a11 a21 a12 a22
(A1)
In this case the matrix is a 2 2 array, with two rows and two columns and its determinant is de ned as det = a11 a22 a12 a21 (A2)
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