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JNTU B.Tech III ELECTRICAL MACHINES-III Previous Question Paper (September - 2021 - R13)

B.Tech | III | I | R13 | September - 2021

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

Answer any five questions

All questions carry equal marks

Main

115 marks
a

Derive an expression for EMF induced of an alternator from fundamentals showing clearly the expressions for pitch and distribution factors.

8 marks

Answer

Consider a single conductor of length l moving with velocity v in a magnetic field of flux density B. The instantaneous EMF induced is e = Blv. For an alternator with Z conductors in series per phase, the average EMF is derived from the rate of change of flux linkages. If Φ is the flux per pole and f is the frequency, the average EMF per conductor is 4fΦ. For N turns in series, the total EMF is E = 4fΦN. To account for the actual winding arrangement, we introduce the pitch factor (kp) and distribution factor (kd).

The pitch factor (kp) accounts for the reduction in EMF due to short-pitching, where the coil span is less than the pole pitch. It is defined as:

kp = cos(α/2)

where α is the coil span deficiency angle. The distribution factor (kd) accounts for the fact that coils are distributed in multiple slots per pole per phase, leading to a phase difference between the EMFs of individual coils. It is defined as:

kd = sin(mβ/2) / (m sin(β/2))

where m is the number of slots per pole per phase and β is the slot angle. The final EMF equation is:

Eph = 4.44 f Φ Tph kp kd
b

The phase emf of a three phase, 50 Hz alternator consists of a fundamental, a 15 % third harmonic and a 5 % fifth harmonic. The amplitude of the fundamental voltage is 800 V. Calculate the rms line voltage when the alternator windings are in (i) star and (ii) delta.

7 marks

Answer

The phase voltage is given by v(t) = 800 sin(ωt) + 0.15(800) sin(3ωt) + 0.05(800) sin(5ωt). The RMS phase voltage is:

Eph(rms) = √[(800/√2)2 + (120/√2)2 + (40/√2)2] = √[320000 + 7200 + 800] = √328000 ≈ 572.71 V

(i) In star connection, the line voltage is √3 times the phase voltage. However, the third harmonic components in the three phases are in phase and cancel out in the line voltage. Thus, the line voltage is:

EL = √3 × √[(E1)2 + (E5)2] = √3 × √[(800/√2)2 + (40/√2)2] = √3 × 567.09 ≈ 981.9 V

(ii) In delta connection, the line voltage is equal to the phase voltage. The third harmonic current circulates within the delta, but the line voltage contains only the fundamental and fifth harmonic components:

EL = √[(800/√2)2 + (40/√2)2] ≈ 567.09 V
215 marks
a

What are the causes of harmonics in the emf waveforms of synchronous generators and what means are adopted to minimize them?

8 marks

Answer

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b

A 1000 kVA, 6 kV, 3-phase star connected alternator has a synchronous reactance of 20 Ω / phase with negligible resistance. It supplies full load current of 0.85 p.f. lagging and at rated terminal voltage. Compute the terminal voltage for the same excitation, When the generator supplies full load current at 0.8 p.f. leading.

7 marks

Answer

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315 marks
a

How the MMF method is different from EMF method in finding voltage regulation of an alternator? Explain the drawbacks of each method.

7 marks

Answer

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b

A 3-Phase star connected synchronous generator is rated at 1600 kVA, 13500 V. The armature resistance and synchronous reactance are 1.5 Ω and 30 Ω per phase. Calculate the percentage regulation for a load of 1280 kW at power factors of 0.8 lead and 0.8 lag.

8 marks

Answer

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4

The open and short circuit test readings for a three star connected 1000kVA, 2000V and 50 Hz alternator are:

VOC(Line)80015001760200023502600
ISC(A)-200250300--
IF(A)102025304050

The armature effective resistance is 0.2 Ω/phase. Estimate the full load voltage regulation using M.M.F. method for 0.8 pf lag.

15 marks

Answer

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515 marks
a

A synchronous generator is connected to infinite bus. Discuss with the help of phasor diagrams effect of changing the input at constant excitation.

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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6

Briefly explain the analysis of a short circuit current waveform of an alternator with the help of neat waveform.

15 marks

Answer

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715 marks
a

Develop an expression for the power developed in a synchronous motor.

8 marks

Answer

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b

A 400 V, 3 phase delta connected synchronous motor has negligible resistance and a synchronous reactance of 10 Ω per phase. At a certain load the motor works with an induced emf of 560 V and draws a power of 3.6 kW. Find the current and power factor?

7 marks

Answer

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815 marks
a

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

8 marks

Answer

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b

Explain the construction and working of universal 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 I, September - 2021, R13.

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