Reg.No.:

VIVEKANANDHA COLLEGE OF ENGINEERING FOR WOMEN(Autonomous)

Elayampalayam – 637 205, Tiruchengode, Namakkal Dt., Tamil Nadu.

Question Paper Code:

ANSWER KEY

PART – A

1. b -Increasing temperature and pressure

2. b – L/R

3. b – aircraft

4. b – Increase

5. d – f0/2

6. b – 2 KW

7. a- Selectivity

8. d- Both a and b

9. a – Active power

10. a - 1

11. a - an undamped

12. a - 0.25A

13. a - Stray loss

14. a - Shunt branch parameters

15. a- Main winding

16. d - R<2√L/C17. b– 55-65%

18. a- Less than meter resistance

19. c- damping

20. c- The change in same reading when the input is first increasing and then decreasing

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PART – B

21. d - 1.634

22. d – 0.866

23 c - 173.2 Nm

24.d - 0.073 H

25.a- 35-40%

26.b – 300 rad/sec

27.d- 0.5

28.c-62.9Hz

29.d-198

30.c-23 A and 5290 W

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Part-C

31 a) Structure of power system

It contains a generating plant, a transmission system, a sub transmission system and a distribution system. These subsystems are interconnected through transformers T1 , T2 and T3 . Let us consider some typical voltage levels to understand the functioning of the power system. The electric power is generated at a thermal plant with a typical voltage of 22 kV (voltage levels are usually specified line-to-line). This is

boosted up to levels like 400 kV through transformer T1 for power transmission. Transformer T2 steps this voltage down to 66 kV to supply power

31 b) Steam and nuclear power generation

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Steam generators are heat exchangers used to convert water into steam from heat

produced in a nuclear reactor core. They are used unpressurised between the primary and

secondary coolant loops.

32 a) EMF generation

A transformer is a device for stepping-up, or stepping-down, the voltage of an alternating electric signal. Without efficient transformers, the transmission and distribution of AC electric power over long distances would be impossible.. There are two circuits. Namely, the primary circuit, and the secondary circuit. There is no direct electrical connection between the two circuits, but each circuit contains a coil which links it inductively to the other circuit. In real transformers, the two coils are wound onto the same iron core. The purpose of the iron core is to channel the magnetic flux generated by the current flowing around the primary coil, so that as much of it as possible also links the secondary coil. The common magnetic flux linking the two coils is conventionally denoted in circuit diagrams by a number of parallel straight lines drawn between the coils.

32 b) RLC circuit

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Instantaneous Voltages for a Series RLC Circuit

33 a) i) Equivalent circuit for transformer

Equivalent Circuit of Transformer Referred to SecondaryIn similar way, approximate equivalent circuit of transformer referred to secondary can be drawn.

Where equivalent impedance of transformer referred to secondary, can be derived as

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33a) ii) ŋmax =(po/(p0 + 2Pi)

Pi =0.153 kW

K2=(Pi/Pc)

Pc=0.7=272 kW

Pin=Po+Pi+K2Pc

20+0.153+0.272= 20.425

Ŋ energy= €Wo/€Win= (240/246.9)=97.2%

33 b) i) Induction motor

Two main types of Induction Motor. An induction motor also known as an asynchronous motor or squirrel-cage motor is a type of an AC (alternating current) motor. In an induction motor, power is supplied to the rotor via an electromagnetic induction

33 b) ii) Tfl=(3/ѡs)x(v2 (R’2/Sfl)/(R’/sfl)2+(X’22)

Tmax=(3/ѡs(0.5 V2/X’2)

2/1=0.5(1+K2)X’22/(kX2’2)

K=0.268

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34a)

34b i) Speed control

34 b)ii)Armature reaction In a DC machine, the main field is produced by field coils. In both the generating and

motoring modes, the armature carries current and a magnetic field is established, which is called the armature flux. The effect of armature flux on the main field is called the armature reaction

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35a)

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35 b) Ea =KaФѡm

T=Ka ФIa =KIa

At no load

250-5x0.6=K x ((2πx1000)/60)

K=2.36

When a driving a load of 100 Nm

Ia = (T/K) = (100/2.36)=4.24A

ѡm=Ea/K = 95.15 rad/s

n=(60/2π) ѡm=909rpm

output =10kW at 1200 rpm

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Assuming linear magnetization Ea=K’aIf ѡm

T=K’aIf Ia

From the data of no load operation If=(250/150)=1.67 A

K’a=(K/If)=1.413 250-0.6 I’a=10x1000I’a=44.8 A Rf(total )=(250/1.257)=199Ω

Rf(external )=199-150=49 Ω

36a)

The permanent magnet moving coil instrument or PMMC type instrument uses two

permanent magnets in order to create stationary magnetic field. These types of instruments are

only used for measuring the dc quantities as if we apply ac current to these type of instruments

the direction of current will be reversed during negative half cycle and hence the direction of

torque will also be reversed which gives average value of torque zero. The pointer will not

deflect due to high frequency from its mean position showing zero reading. However it can

measure the direct current very accurately.

36b)

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An oscilloscope, previously called an oscillograph, and informally known as a scope, CRO (for cathode-ray oscilloscope), or DSO(for the more modern digital storage oscilloscope), is a type of electronic test instrument that allows observation of constantly varying signal voltages, usually as a two-dimensional plot of one or more signals as a function of time. Non-electrical signals (such as sound or vibration) can be converted to voltages and displayed.

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