Math, asked by nannapanenigopi291, 7 months ago

A 6 pole lap wound dc motor takes 340 A when speed is 400 RPM.flux/pole 0.05 Wb & armature has 864 turns. Neglecting mechanical losses. Calculate the BHP of the motor​

Answers

Answered by anshusarkar5747
0

Answer:

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Answered by nihasrajgone2005
1

Answer:

Speed of a D.C. Motor

From the voltage equation of a motor (Art. 27.4), we get

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Speed Regulation

The term speed regulation refers to the change in speed of a motor with change in applied load torque, other conditions remaining constant. By change in speed here is meant the change which occurs under these conditions due to inherent properties of the motor itself and not those changes which are affected through manipulation of rheostats or other speed-controlling devices.

The speed regulation is defined as the change in speed when the load on the motor is reduced from rated value to zero, expressed as percent of the rated load speed.

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Torque and Speed of a D.C. Motor

It will be proved that though torque of a motor is admittedly a function of flux and armature current, yet it is independent of speed. In fact, it is the speed which depends on torque and not vice- versa. It has been proved earlier that

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It is seen from above that increase in flux would decrease the speed but increase the armature torque. It cannot be so because torque always tends to produce rotation. If torque increases, motor speed must increase rather than decrease. The apparent inconsistency between the above two equations can be reconciled in the following way :

Suppose that the flux of a motor is decreased by decreasing the field current. Then, following sequence of events take place :

1. Back e.m.f. Eb (= NF/K) drops instantly (the speed remains constant because of inertia of the heavy armature).

2. Due to decrease in Eb, Ia is increased because Ia = (V – Eb)/Ra. Moreover, a small reduction in flux produces a proportionately large increase in armature current.

3. Hence, the equation Ta µ FIa, a small decrease in f is more than counterbalanced by a large increase in Ia with the result that there is a net increase in Ta.

4. This increase in Ta produces an increase in motor speed.

It is seen from above that with the applied voltage V held constant, motor speed varies inversely as the flux. However, it is possible to increase flux and, at the same time, increase the speed provided Ia is held constant as is actually done in a d.c. servomotor.

Example 29.20. A 4-pole series motor has 944 wave-connected armature conductors. At a certain load, the flux per pole is 34.6 mWb and the total mechanical torque developed is 209 N-m. Calculate the line current taken by the motor and the speed at which it will run with an applied voltage of 500 V. Total motor resistance is 3 ohm.

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Now, speed may be found either by using the relation for Eb or Ta as given in Art.

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Example 29.21. A 250-V shunt motor runs at 1000 r.p.m. at no-load and takes 8A. The total armature and shunt field resistances are respectively 0.2 W and 250 W. Calculate the speed when loaded and taking 50 A. Assume the flux to be constant. (Elect. Engg. A.M.Ae. S.I. June 1991)

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Example 29.22. A d.c. series motor operates at 800 r.p.m. with a line current of 100 A from 230-V mains. Its armature circuit resistance is 0.15 W and its field resistance 0.1 W. Find the speed at which the motor runs at a line current of 25 A, assuming that the flux at this current is 45 per cent of the flux at 100 A. (Electrical Machinery – I, Banglore Univ. 1986)

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Example 29.23. A 230-V d.c. shunt motor has an armature resistance of 0.5 W and field resistance of 115 W. At no load, the speed is 1,200 r.p.m. and the armature current 2.5 A. On application of rated load, the speed drops to 1,120 r.p.m. Determine the line current and power input when the motor delivers rated load. (Elect. Technology, Kerala Univ. 1988)

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Example 29.24. A belt-driven 100-kW, shunt generator running at 300 r.p.m. on 220-V bus- bars continues to run as a motor when the belt breaks, then taking 10 kW. What will be its speed ? Given armature resistance = 0.025 W, field resistance = 60 W and contact drop under each brush = 1 V, Ignore armature reaction. (Elect. Machines (E-3) AMIE Sec.C Winter 1991)

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