JNTU
JNTU B.Tech II ELECTRICAL MACHINES-1 Previous Question Paper (October - 2020 - R18)
Answer any five questions. All questions carry equal marks.
Main
Explain the following terms as applied to a DC armature winding:
- Front Pitch
- Back Pitch
- Pole pitch
- Commutator Pitch
Answer
The terms related to DC armature windings are defined as follows:
- Front Pitch (Yf): The number of armature conductors spanned by the back connection of a coil, representing the distance between the second conductor of one coil and the first conductor of the next coil in the same layer.
- Back Pitch (Yb): The distance, measured in terms of armature conductors, between the two sides of the same coil at the back of the armature.
- Pole Pitch: The peripheral distance between the centers of two adjacent poles, equal to the total number of armature conductors divided by the number of poles.
- Commutator Pitch (Yc): The number of commutator segments spanned by the two ends of a coil connected to the commutator, which is 1 for simplex lap winding and equal to the number of coils divided by the number of pole pairs for wave winding.
A 50 KW, 500 V, 4 – pole generator has a 2 layer simplex lap winding in 36 slots with 10 conductors in each layer. If the brushes are given an actual lead of 10°, calculate:
- Demagnetizing AT / pole
- Cross – magnetizing AT / Pole
- Number of turns per pole on the compensating winding if the pole arc to pitch is 0.8 and brushes are placed on geometric neutral plane.
Answer
Given: P = 4, V = 500 V, Power = 50 kW, Slots = 36, Conductors/slot = 20 (2 layers of 10), Total conductors (Z) = 36 × 20 = 720. For simplex lap winding, A = P = 4.
Full load current (Ia) = 50,000 / 500 = 100 A. Current per conductor (I) = Ia / A = 100 / 4 = 25 A.
Total turns (T) = Z / 2 = 720 / 2 = 360. Turns per pole (Tp) = 360 / 4 = 90.
Brush lead (θ) = 10°. Demagnetizing AT/pole = (θ / 180) × Z × I / 2 = (10 / 180) × 720 × 25 / 2 = 500 AT/pole.
Cross-magnetizing AT/pole = (Z × I / 2P) × (1 - 2θ / 180) = (720 × 25 / 8) × (1 - 20 / 180) = 2250 × 0.888 = 1998 AT/pole.
For compensating winding, pole arc to pitch ratio = 0.8. The compensating winding covers the pole arc. Number of turns per pole = (Pole arc / Pole pitch) × (Z / 2P) = 0.8 × (720 / 8) = 0.8 × 90 = 72 turns per pole.
Sketch and explain the load characteristics of the following types of DC Generators:
- Series Generator
- Differentially compounded generator
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The external characteristic of a series generator from zero to 70 V at 350 A is a straight line. This generator is connected as a booster between a station bus – bar and a feeder of 0.35 Ω resistance. Find the voltage between the bus-bar and far end of the feeder at a current of 200 A.
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Derive an expression for the torque of a DC Motor.
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A 250V DC Shunt motor has an armature resistance of 0.5 Ω and a field resistance of 250 Ω. When driving a constant torque load at 600 rpm, the motor draws 21 A. What will be the new speed of the motor if an additional 250 Ω resistance is inserted in the field circuit?
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Derive the condition for maximum efficiency of a DC Motor.
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A 500 V series motor takes a current of 200A to develop 89.52 KW. The armature resistance is 0.1 Ω and series field resistance is 0.05 Ω. If the output is reduced to 44.76 KW, find the input current and efficiency.
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A 210 V shunt motor develops 18 KW when taking 21 KW. The field resistance is 60 Ω and armature resistance 0.05 Ω. What is the efficiency and power input when the output is 8 KW?
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What is Hopkinson’s test and why is it called regenerative test? Draw and explain the connection diagram to conduct this test.
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Distinguish between a core type and a shell type transformer.
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Draw and explain the equivalent circuit of a Transformer.
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Explain the significance and procedure of a Sumpner’s test and draw the necessary connection diagram for it and how it is different from OC and SC test.
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