JNTU
JNTU B.Tech II ELECTRICAL MACHINES-1 Previous Question Paper (August/September - 2022 - R18)
Answer any five questions. All questions carry equal marks.
Main
Explain the effect of armature reaction on the performance of DC generator.
Answer
Armature reaction is the effect of the magnetic field produced by the armature current on the main field flux. It causes two primary effects: demagnetization and cross-magnetization. Cross-magnetization distorts the main flux distribution, shifting the Magnetic Neutral Axis (MNA) in the direction of rotation, which leads to sparking at the brushes. Demagnetization reduces the total flux per pole, resulting in a drop in the generated EMF and a decrease in the terminal voltage of the generator. To mitigate these effects, compensating windings are used to neutralize armature flux, and interpoles are employed to improve commutation.
A dc shunt generator has the following open circuit magnetization curve running at 800 rpm. Field current (A): 0, 0.5, 1.0, 2.0, 3.0, 4.0, 5.0. EMF (V): 10, 50, 100, 175, 220, 245, 262. Find graphically the critical resistance of shunt field circuit. If the field resistance is changed to 75 ohms, what will be the critical speed for the machine to build up?
Answer
To find the critical resistance, draw a tangent to the open-circuit magnetization curve starting from the origin. The slope of this tangent represents the critical resistance (Rc). Using the data points (0, 10), (0.5, 50), (1.0, 100), (2.0, 175), (3.0, 220), (4.0, 245), and (5.0, 262), the tangent line passes through approximately (1.0, 100). Thus, Rc = 100 / 1.0 = 100 Ω.
For the second part, the critical speed (Nc) is related to the actual speed (Na) and the ratio of the field resistance (Rf) to the critical resistance (Rc) at the original speed. Given Na = 800 rpm, Rf = 75 Ω, and Rc = 100 Ω:
The critical speed for the machine to build up with a field resistance of 75 Ω is 600 rpm.
How demagnetizing and cross magnetizing ampere turns per pole are calculated in a DC machine?
Answer
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A d.c. shunt generator is supplying load connected to a bus-bar voltage of 220 V. It has an armature resistance of 0.025 Ω and field resistance of 110 Ω. Calculate the value of load current and load power when it generates an emf of 230 V. Neglect the effect of armature reaction. Draw circuit diagram.
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Derive the torque equation of DC Motor.
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A 6-pole DC motor has a wave connected armature with 87 slots, each slot containing 6 conductors. The flux per pole is 20 mWb and the armature has a resistance of 0.13 ohm when the motor is connected to 240V supply and the armature draws a current of 80A driving a load of 15KW. Calculate i) Speed ii) Armature Torque and iii) Shaft Torque.
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With the help of a neat sketch, explain the construction and working of 3-point starter?
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A 250V dc series motor has armature and series field resistance of 0.25 and 0.15 ohms respectively. (i) Calculate the current for developing a torque of 80 Nm at 1200 rpm. (ii) Calculate the percentage reduction in flux when the motor runs at 1800 rpm at half the current obtained in part (i).
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Briefly explain the procedure to conduct Swinburne’s test on DC machine and explain its significance?
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In a brake test on a DC shunt motor, the load on one side of the brake band was 35 kg and the other side 5kg. The motor was running at 1300 rpm; its input being 70 A at 420 V DC. The pulley diameter is 1 m. Determine the torque, output of the motor and the efficiency of the motor.
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Explain the Direct method of testing a dc machine.
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Hopkinson’s test on two similar DC shunt machines gave the following data: Line voltage 230 V, line current excluding both the field currents is 40 A, motor armature current 350 A, field currents 5 A and 4.2 A. Calculate the efficiency of each machine. Armature resistance of each machine is 0.02 Ω.
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Draw and explain the phasor diagram of a single-phase transformer with lagging p.f. load and leading p.f. load.
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A single-phase transformer is rated at 120 kVA, 5000/250V. The full – load copper losses are 2200W and iron losses are 1400 W. Find efficiency at i) full load 0.8 power factor leading ii) full – load 0.6 power factor lagging.
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With neat diagram, explain the various tests conducted on transformer to predetermine the efficiency of the transformer without directly loading the transformer?
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In a Sumpner’s test on two identical single-phase transformers rated 750 kVA, 11/0.5 kV, 50 Hz the wattmeter reading on h.v side is 8000 W and on the l.v side is 16000W. Find the efficiency of each transformer on half full load and 0.78 power factor. What will be its maximum efficiency?
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