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JNTU B.Tech II ELECTRICAL MACHINES II Previous Question Paper (December - 2018 - R13)

B.Tech | II | II | R13 | December - 2018

Answer PaperPaper Code: 114ABMaximum Marks: 75Time: 3 Hours

Note: This question paper contains two parts A and B. Part A is compulsory which carries 25 marks. Answer all questions in Part A. Part B consists of 5 Units. Answer any one full question from each unit. Each question carries 10 marks and may have a, b, c as sub questions.

PART A

125 marks
a

Derive the condition for maximum efficiency of a transformer.

2 marks

Answer

For a transformer, the efficiency (η) is given by:

η = (Output Power) / (Input Power) = (Output Power) / (Output Power + Losses)

Let the output power be P_o and the total losses be P_L. The losses in a transformer consist of core losses (P_i) and copper losses (P_cu). The copper losses vary with the square of the load current (I_1^2 * R_1eq), where R_1eq is the equivalent resistance referred to the primary.

So, P_L = P_i + P_cu = P_i + I_1^2 * R_1eq.

The efficiency can be written as:

η = P_o / (P_o + P_i + I_1^2 * R_1eq)

For maximum efficiency, the denominator must be minimum for a constant output power. This occurs when the variable losses (copper losses) are equal to the constant losses (core losses).

d(Losses)/d(I_1) = 0

Differentiating the total losses with respect to current and setting it to zero:

d(P_i + I_1^2 * R_1eq)/d(I_1) = 0 + 2 * I_1 * R_1eq = 0

This implies that the condition for maximum efficiency is when the copper losses are equal to the core losses:

P_cu = P_i

This condition is independent of the power factor.

b

Draw and explain variation of efficiency with power factor.

3 marks

Answer

The efficiency of a transformer varies with the power factor of the load. The efficiency (η) is given by:

η = (Output Power) / (Input Power) = (Output Power) / (Output Power + Losses)

Let P_o be the output power, P_cu be the copper losses, and P_i be the core losses.

η = (V_2 * I_2 * cosϕ_2) / (V_2 * I_2 * cosϕ_2 + P_cu + P_i)

Where V_2, I_2, and cosϕ_2 are the secondary voltage, secondary current, and load power factor, respectively.

Variation of Efficiency with Power Factor:

1. At Unity Power Factor (cosϕ_2 = 1): For a given load current I_2, the output power (V_2 * I_2) is maximum. The copper losses (I_2^2 * R_2eq) and core losses (P_i) are constant. Thus, the efficiency is highest at unity power factor.

2. At Lagging Power Factor (cosϕ_2 < 1): For the same load current I_2, the output power (V_2 * I_2 * cosϕ_2) is less than that at unity power factor. Since the losses (P_cu + P_i) remain the same, the efficiency decreases as the power factor decreases (becomes more lagging).

3. At Leading Power Factor (cosϕ_2 < 1): Similar to lagging power factor, the output power is reduced, leading to lower efficiency compared to unity power factor. However, the behavior might differ slightly due to voltage regulation effects, but generally, efficiency is lower than at unity PF.

Diagram:

A typical graph plotting efficiency against power factor for a transformer at full load would show efficiency peaking at unity power factor and decreasing symmetrically as the power factor moves towards zero (lagging or leading). The curve would be concave downwards.

c

A 230/115 V, single phase transformer is supplying a load of 5 A, at power factor 0.866 lagging. The no load current is 0.2 A at power factor 0.208 lagging. Calculate the primary current and primary power factor.

2 marks

Answer

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d

Write down the conditions of parallel operation of a single phase transformer.

3 marks

Answer

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e

Explain procedure for conversion 2-winding transformer into auto transformer.

2 marks

Answer

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f

Write down the advantages and disadvantages of three phase transformer over single phase.

3 marks

Answer

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g

Define slip. And explain necessity of slip in an induction motor.

2 marks

Answer

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h

Define frequency of rotor current and its condition during starting and synchronous speed.

3 marks

Answer

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i

What information can be obtained from circle diagram of 3-φ Induction Motor?

2 marks

Answer

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j

What is the necessity of speed control of a three phase induction motor?

3 marks

Answer

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PART B

210 marks
a

Derive EMF equation of a single phase transformer.

5 marks

Answer

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b

The number of turns on the primary and secondary windings of a single phase transformer are 350 and 35 respectively. If the primary is connected to a 2.2 KV supply, determine the secondary voltage on no load.

5 marks

Answer

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OR
310 marks
a

Draw and explain phasor diagrams of a single phase transformer with R, L and C load.

5 marks

Answer

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b

A single phase transformer has 1000 turns on the primary and 200 turns on the secondary. The no load current is 3 A at a pf of 0.2 lagging. Calculate the primary current and power factor when the secondary current is 280 A at a pf of 0.80 lagging.

5 marks

Answer

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410 marks
a

Explain the procedure for conducting OC tests with neat diagrams.

5 marks

Answer

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b

A 25-KVA transformer has 500 turns on the primary and 50 turns on the secondary winding. The primary is connected to 3000V, 50Hz supply. Find the full load primary and secondary currents, the secondary emf and the maximum flux in the core. Neglect drops and no load primary current.

5 marks

Answer

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OR
5

Explain the parallel operation of a single phase transformer in equal and unequal voltage ratio.

10 marks

Answer

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610 marks
a

Explain the importance and operation of a scott connection in a 3-φ transformer.

5 marks

Answer

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b

A scott connected (2 to 3-phase) transformer links a 6,000 V, 2-phase system with a 440 V 3-Phase system. The frequency is 50 Hz, the gross core area is 300 Cm², while the maximum flux density is to be about 1.2 Wb/m². Find the number of turns on each winding and the point to be tapped for neutral wire on the 3-phase side. If the load is balanced on the one side of such a transformer, find whether it will also be balanced on the other side.

5 marks

Answer

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OR
7

Explain the operation of 3-φ to 2-φ and 3-φ to 3-φ conversion using scott connection.

10 marks

Answer

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810 marks
a

Derive the expression for the rotor EMF and rotor current of an induction motor.

5 marks

Answer

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b

A 3-phase, 50Hz, 6 pole induction motor has a full load speed of 960 rpm. Calculate:
i) Synchronous speed
ii) Slip speed
iii) Slip frequency.

5 marks

Answer

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OR
910 marks
a

Explain why the rotor of a poly phase induction motor can never attain synchronous speed.

5 marks

Answer

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b

Explain the concept of crawling of an induction motor.

5 marks

Answer

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1010 marks
a

Explain the pole changing method of speed control in Induction motor.

5 marks

Answer

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b

Explain the concept of No load test on a Induction motor.

5 marks

Answer

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OR
11

Explain the principle of operation of Induction generator with characteristics.

10 marks

Answer

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About this paper

This page contains the solved question paper with answers of ELECTRICAL MACHINES II from JNTU for B.Tech EEE II II, December - 2018, R13.

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