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JNTU B.Tech III ELECTRICAL MACHINES-III Previous Question Paper (February - 2022 - R13)

B.Tech | III | III-I | R13 | February - 2022

Answer PaperPaper Code: 115AEMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Briefly explain the significance of chording and distribution factors? Also derive the expressions for chording factor and distribution factor.

7 marks

Answer

Chording and distribution factors are used to account for the reduction in induced EMF due to the physical arrangement of windings in an alternator. They are essential for improving the waveform of the generated voltage by suppressing harmonics.

Chording Factor (kp): A winding is chorded if the coil span is less than the pole pitch. If the coil span is (180° - α) electrical degrees, the chording factor is:

kp = cos(α/2)

This factor reduces the magnitude of the induced EMF by a factor of cos(α/2) but effectively eliminates or reduces specific harmonics.

Distribution Factor (kd): In a distributed winding, coils are placed in multiple slots per pole. The induced EMFs in these coils are not in phase. If there are m slots per pole per phase and the slot angle is β, the distribution factor is:

kd = sin(mβ/2) / (m sin(β/2))

This factor represents the ratio of the actual EMF to the EMF that would be induced if all coils were concentrated in a single slot.

b

Find the synchronous impedance and reactance in an alternator in which a given field current produces an armature current of 250 A on short circuit and generates an open circuit voltage of 1500 volts. The effective armature resistance is 0.5 Ω/ph. Hence calculate the terminal voltage, when a load of 250A, 6600 V at a power factor of 0.8 lagging is switched off.

8 marks

Answer

Given: Isc = 250 A, Voc = 1500 V, Ra = 0.5 Ω.

First, calculate the synchronous impedance (Zs):

Zs = Voc / Isc = 1500 / 250 = 6 Ω

Calculate the synchronous reactance (Xs):

Xs = √(Zs2 - Ra2) = √(62 - 0.52) = √(36 - 0.25) = 5.979 Ω

To find the terminal voltage after the load is switched off, we calculate the induced EMF (Ef) under the load condition (250 A, 6600 V, 0.8 pf lagging):

V = 6600 / √3 = 3810.5 V per phase. I = 250 ∠ -cos-1(0.8) = 250 ∠ -36.87° A.

Ef = V + I(Ra + jXs)
Ef = 3810.5 + (250 ∠ -36.87°)(0.5 + j5.979)
Ef = 3810.5 + (200 - j150)(0.5 + j5.979) = 3810.5 + (100 + j1195.8 - j75 + 896.85)
Ef = 4807.35 + j1120.8 = 4935.7 ∠ 13.13° V

When the load is switched off, the terminal voltage becomes equal to the generated EMF per phase. The line-to-line terminal voltage is:

VL = √3 × 4935.7 = 8550.5 V
215 marks
a

Explain the operation and effect of load power factor on the performance of alternator.

8 marks

Answer

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b

In a three phase Y-connected alternator, there are 2 coil sides per slot and 16 turns per coil. Armature has 288 slots on its periphery. When driven at 250 rpm, it produces 6600 V between the lines at 50 Hz. The pitch of the coil is 2 slots less than full pitch. Calculate the flux/pole.

7 marks

Answer

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315 marks
a

Describe the slip test and explain how the reactances are calculated using slip test.

8 marks

Answer

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b

Find the synchronous impedance and reactance in an alternator in which a given field current produces an armature current of 250 A on short circuit and generates an open circuit voltage of 1500 volts. The effective armature resistance is 0.5 Ω/ph. Hence calculate the terminal voltage, when a load of 250A, 6600 V at a power factor of 0.8 lagging is switched off.

7 marks

Answer

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4

A 1 MVA, 11 kV, 3-ϕ, star connected synchronous machine has following OCC test data:

If(A)50110140180
EOL(KV)712.513.75180

(Where EOL is the line voltage at no load)

The short circuit test yielded full load current at a field current of 60A. The ZPF yielded a full load current at terminal voltage for a field current of 150 A. The armature resistance is negligible. Calculate the voltage regulation at full load 0.866 pf lagging by Potier triangle method.

15 marks

Answer

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515 marks
a

Explain the effect of change in excitation on the parallel operation of two alternators.

8 marks

Answer

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b

A 5 MVA, 10 kV, 1500 RPM, 3-Phase, 50 Hz alternator is running in parallel with other machines. Its synchronous reactance is 20 %. Find synchronizing power and synchronizing torque per degree mechanical displacement, for no load and full load, 0.8 pf lagging.

7 marks

Answer

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615 marks
a

Explain the term synchronization, and hence explain, synchronizing power.

6 marks

Answer

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b

Two identical 3ph alternators work in parallel and supply a total load of 1500 kW at 11kV at a power factor of 0.867 lag. Each machine supplies half the total power. The synchronous reactance of each is 50 Ω per phase and the resistance is 4 Ω per phase. The field excitation of the first machine is so adjusted that its armature current is 50 A lagging. Determine the armature current of the second and the generated voltage of the machine.

9 marks

Answer

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715 marks
a

Derive an expression for power developed in a cylindrical rotor synchronous motor in terms of load angle and synchronous impedance?

8 marks

Answer

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b

A 3-ϕ, 400 V, 40 KVA, star connected synchronous motor is supplying 15 kW load with 0.8 pf lagging. The windage and friction losses are 1.5 kW and core losses are 1.0 kW. Calculate Armature current and Excitation voltage.

7 marks

Answer

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815 marks
a

Explain the working principle of single-phase induction motor with the help of double field revolving theory.

8 marks

Answer

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b

Explain the principle of operation of ac series motor.

7 marks

Answer

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This page contains the solved question paper with answers of ELECTRICAL MACHINES-III from JNTU for B.Tech EEE III III-I, February - 2022, R13.

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