Electrical Machine Interview Question And Answer

[1] Define the term synchronous speed [Dec-2003]
For synchronous machines there exists a fixed relationship between number of poles P, frequency (f) and the speed of the machine. The speed of the synchronous machine for the given number of poles and the rated frequency is called the synchronous speed mentioned as NS.

[2] Write an expression for synchronous speed (Dec-2004)
An expression for synchronous speed is NS = 120f/P.
Where
f = frequency
P = No of poles of the machine.

[3] What does speed voltage mean? (Dec-2007)
When the magnetic flux is constant as well as stationary and the coil rotates to cut the flux then EMF gets induced due to relative speed between flux and coil. This EMF is called speed EMF, rotational EMF or dynamically induced EMF.

[4] Mention the factors on which hysteresis loss depends (Dec-2008)
The hysteresis loss is directly proportional to the area under the hysteresis curve ie area of the hysteresis loop.
It is directly proportional to frequency ie number of cycles of magnetization per second.
It is directly proportional to volume of the material.

[5] Distinguish between statically induced and dynamically induced EMF (Dec - 2010, 2011, 2009)
How is EMF induced dynamically ( May-2010)
An induced EMF which is due to physical movement of coil, conductor with respect to flux or movement of magnet with respect to stationary coil, conductor is called dynamically induced EMF or motional induced EMF.
The change in flux lines with respect to coil can be achieved without physically moving the coil or the magnet. Such induced emf in a coil which is without physical movement of coil or a magnet is called statically induced EMF.

[6] Define torque (May-2010)
A turning or a twisting force about an axis is called as torque.

[8] What are the three types of basic rotating electric machines? (May-2011)
DC machines
Induction Machines
Synchronous Machines

[9] What are the causes of core loss? what are the components of core loss?
When a core is subjected to an alternating flux then it undergoes the cycles of magnetisation and demagnetisation. This produces hysteresis effect which causes hysteresis loss in the core.
Similarly core is under the influence of the changing flux and under such condition according to the Faraday's law of electromagnetic induction, EMF gets induced in the core. Such currents in the core which are due to induced emf in the core are called as eddy current loss. Thus eddy current and hysteresis are the two components of the core loss.

Transformer Interview Questions With Answer

 Hello Friends. How are you ?

  We shared a lot of interview questions on electrical machines, power system and power electronics. Today we are sharing interview questions on transformer. These transformer interview questions is collected from various sources. lets start transformer interview questions with answer.

Q.1. How is magnetic leakage reduced to a minimum in commerical transformers ?
Ans.By interleaving the primary and secondary windings.

Q.2. Mention the factors on which hysteresis loss depends ?
Ans.(i) Quality and amount of iron in the core
       (ii) Flux density and
       (iii) Frequency.
Must Read :

1.  Electrical Engineering Project Ideas
2. Construction Of DC Machine

Q.3. How can eddy current loss be minimised ?
Ans.By laminating the core.

Q.4. In practice, what determines the thickness of the laminae or stampings ?
Ans.Frequency.

Q.5. Does the transformer draw any current when its secondary is open ?
Ans.Yes, no-load primary current.

Q.6. Why ?
Ans.For supplying no-load iron and copper losses in primary.

Q.7. Is Cu loss affected by power factor ?
Ans.Yes, Cu loss varies inversely with power factor.

Q.8. Why ?
Ans.Cu loss depends on current in the primary and secondary windings. It is well-known that current required is higher when power factor is lower.

Q.9. What effects are produced by change in voltage ?
Ans.1. Iron loss.........varies approximately as V2.
       2. Cu loss.........it also varies as V2 but decreases with an increase in voltage if constant kVA output is assumed
       3. Efficiency.........for distribution transformers, efficiency at fractional loads decreases with increase in voltage while at full load or overload it increases with increase in voltage and viceversa.

       4. Regulation.........it varies as V2  but decreases with increase in voltage if constant kVA output is assumed.
     5. Heating.........for constant kVA output, iron temperatures increase whereas Cu temperatures decrease with increase in voltages and vice-versa
.
Q. 10.How does change in frequency affect the operation of a given transformer ?
Ans. 1. Iron loss .........increases with a decrease in frequency. A 60-Hz transformer will have nearly 11%      higher losses when worked on 50Hz instead of 60 Hz. However, when a 25-Hz transformer is worked on 60 Hz, iron losses are reduced by 25%.

2. Cu loss.........in distribution transformers, it is independent of frequecy.
3. Efficiency.........since Cu loss is unaffected by change in frequency, a given transformer efficiency is less at a lower frequency than at a higher one.
4. Regulation.........regulation at unity power factor is not affected because IR drop is independent of frequency. Since reactive drop is affected, regulation at low power factors decreases with a decrease in frequency and vice-versa.

For example, the regulation of a 25-Hz transformer when operated at 50-Hz and low power factor is much poorer.

5. Heating.........since total loss is greater at a lower frequency, the temperature is increased with decrease in frequency.

If you face any problem then comment below

Searches related to Transformer Interview Questions With Answer

transformer interview questions and answers pdf

cognos transformer interview questions and answers

electrical interview questions and answers on transformers

interview questions on transformer protection

interview questions on transformer stage in datastage

Electrical Engineering Project Ideas For Final Year

1. UPFC - Unified Power Factor Control.
2. Construction of Central Control Unit for Irrigation water pumps.
3. Cost effective method to control entire villager’s water pumps with user level authentication. Illiterate’s friendly system.
4. Centralized energy meter for display distance at 30meters - Avoiding tampering of electrical energy meter in houses
5. Embedded System Integrated Into a Wireless Sensor Network for Online DynamicTorque and Efficiency Monitoring in Induction Motors
6.  Design and Implementation of Intelligent Energy Distribution Management with Photovoltaic System
7. Sensor Network Based Oil well Health Monitoring and Intelligent Control
8. . Solar induction motor driver. 
9. Speed control of universal motor with automated ON load and OFF load sensing
10. . Solar tracking system with automated water pump controlling
11.  Touch screen based AC motor speed Monitoring and control system
12.  Touch screen based wireless AC motor speed Monitoring and control system
13.  Intelligent Transformer Isolation system using PC
14. Real time Atomization of Indian Agricultural system.
15. Automated Solar Street lighting system using LED’s
16. GSM based Energy meter with tampering alert
17. Event driven automation with Android backup control
18. Transformer over load alert through voice announcement.
19. Wireless Power Factor monitoring and controlling through PC
20. Protection of power transformer using microcontroller-based relay
21. Industrial parameters monitoring and crane controlling using Zigbee
22. Transformer load sharing with SMS alerting.
23. Multi zone temperature monitoring with voice announcement system.
24. Maximum demand controller & fault alerting for industries.
25. speed control of exhaust fan using RF remote.
26. Solar Tracker Robot using Microcontroller
27. Design and Implementation of intelligent Urban Irrigation System
28. Remote Controlled Screw Jack
29. Microcontroller based smart charge controller for standalone solar photovoltaic power systems.
30. GPRS based single phase fault monitoring and SMS alert 
31. Design and construction of parabolic solar reflectors
32. Servo Motor Control using mobile phone.
33. Wireless power theft monitoring with automatic circuit breaker system and indication at local substations
34. Energy Tapping Identifier through Wireless Data Acquisition System
35. Foot step power generation system.
36. Multi channel voltage scanner
37. Controlling of exhaust fan based on Smoke and Gas intensity in industries.
38. Four channel fault annunciation for industries.
39. Feeder Protection From Over Load.
40. Temp based speed control of exhaust fan using TRIAC.
41. Petrochemical Level Indicator and Controller for Automation of cotton purification process in spinning mills.
42. Autonomous energy meter with auto announcement system
43. Solar based self powered high efficient Line following robot with obstacle avoidance
44. Zigbee based power management system
45. Speech recognition based Wheel chair with elevated features
46. Water level indicator on LCD
47. DTMF 3-phase irrigation Control with Feedback
48. I-Button & keypad based DC motor door lock system

Searches related to electrical engineering projects

electrical engineering projects for college students

electrical engineering project ideas

electrical engineering projects for high school students

electrical engineering projects for beginners


electrical engineering projects for kids

electrical engineering science fair projects

computer engineering projects

electrical engineering projects list

Basic MCQ Interview Questions On Alternator

 Hello Reader. Today we collected basic mcq types questions of alternator.

Ques.[1] At lagging loads, armature reaction in an alternator is
(a) cross-magnetizing
(b) demagnetizing
(c) non-effective
(d) magnetizing.
Answer : D

Ques.[2] At leading p.f., the armature flux in an alternator....... the rotor flux.
(a) opposes
(b) aids
(c) distorts
(d) does not affect.
Answer : B

Ques.[3] The voltage regulation of an alternator having 0.75 leading p.f. load, no-load induced e.m.f.of 2400V and rated terminal voltage of 3000V is............... percent.
(a)20
(b)−20
(c) 150
(d)−26.7
Answer : B

Ques.[4] If, in a 3-phase alternator, a field current of 50A produces a full-load armature current of 200 A on short-circuit and 1730 V on open circuit, then its synchronous impedance is ....... ohm.
(a) 8.66
(b)4
(c)5
(d) 34.6
Answer : C

Ques.[5] The power factor of an alternator is determined by its
(a) speed
(b) load
(c) excitation
(d) prime mover.
Answer : B

Must Read: What is thermal Power Plant

Ques.[6] For proper parallel operation, a.c. polyphase alternators must have the same
(a) speed
(b) voltage rating
(c) kVA rating
(d) excitation.
Answer : B

Ques.[7] Of the following conditions, the one which does not have to be met by alternators working in parallel is
(a) terminal voltage of each machine must be the same
(b) the machines must have the same phase rotation
(c) the machines must operate at the same frequency
(d) the machines must have equal ratings.
Answer : D

Ques.[8] After wiring up two 3-φ alternators, you checked their frequency and voltage and found them to be equal. Before connecting them in parallel, you would
(a) check turbine speed
(b) check phase rotation
(c) lubricate everything
(d) check steam pressure.
Answer : B

Ques.[9] Zero power factor method of an alternator is used to find its
(a) efficiency
(b) voltage regulation
(c) armature resistance
(d) synchronous impedance.
Answer : B

Ques.[10] Some engineers prefer `lamps bright' synchronization to ‘lamps dark’ synchronization because(a) brightness of lamps can be judged easily
(b) it gives sharper and more accurate synchronization
(c) flicker is more pronounced
(d) it can be performed quickly.
Answer : B

Ques.[11] It is never advisable to connect a stationary alternator to live bus-bars because it
(a) is likely to run as synchronous motor
(b) will get short-circuited
(c) will decrease bus-bar voltage though momentarily
(d) will disturb generated e.m.fs. of other alternators connected in parallel
Answer : B

Ques.[12] Two identical alternators are running in parallel and carry equal loads. If excitation of one alternator is increased without changing its steam supply, then
(a) it will keep supplying almost the same load
(b) kVAR supplied by it would decrease
(c) its p.f. will increase
(d) kVA supplied by it would decrease.
Answer : A

Ques.[13] Keeping its excitation constant, if steam supply of an alternator running in parallel with another identical alternator is increased, then
(a) it would over-run the other alternator
(b) its rotor will fall back in phase with respect to the other machine
(c) it will supply greater portion of the load
(d) its power factor would be decreased.
Answer : C

Ques.[14] The load sharing between two steam-driven alternators operating in parallel may be adjusted by varying the
(a) field strengths of the alternators
(b) power factors of the alternators
(c) steam supply to their prime movers
(d) speed of the alternators.
Answer : C

Ques.[15] Squirrel-cage bars placed in the rotor pole faces of an alternator help reduce hunting
(a) above synchronous speed only
(b) below synchronous speed only
(c) above and below synchronous speeds both
(d) none of the above. (Elect. Machines, A.M.I.E. Sec. B, 1993)
Answer : C

Ques.[16] For a machine on infinite bus active power can be varied by
(a) changing field excitation
(b) changing of prime cover speed
(c) both (a) and (b) above
(d) none of the above .
Answer : B

Q.[17] The frequency of voltage generated by an alternator having 4-poles and rotating at 1800 r.p.m.is .......hertz.
(a)60
(b) 7200
(c) 120
(d) 450.
Answer : A

Q.[18] A 50-Hz alternator will run at the greatest possible speed if it is wound for ....... poles.
(a)8
(b)6
(c)4
(d)2.
Answer : D

Q.[19] The main disadvantage of using short-pitch winding in alternators is that it
(a) reduces harmonics in the generated voltage
(b) reduces the total voltage around the armature coils
(c) produces asymmetry in the three phase windings
(d) increases Cu of end connections.
Answer : B

Q.[20] Three-phase alternators are invariably Y-connected because
(a) magnetic losses are minimized
(b) less turns of wire are required
(c) smaller conductors can be used
(d) higher terminal voltage is obtained.
Answer : D

Q.[21] The winding of a 4-pole alternator having 36 slots and a coil span of 1 to 8 is short-pitched by ....... degrees.
(a) 140
(b)80
(c)20
(d) 40.
Answer : D

Q.[22] If an alternator winding has a fractional pitch of 5/6, the coil span is ....... degrees.
(a) 300
(b) 150
(c)30
(d) 60.
Answer : B

Q.[23] The harmonic which would be totally eliminated from the alternator e.m.f. using a fractional pitch of 4/5 is
(a) 3rd
(b)7th
(c) 5th
(d) 9th.
Answer : C

Q.[24] For eliminating 7th harmonic from the e.m.f. wave of an alternator, the fractional-pitch must be
(a) 2/3
(b)5/6
(c) 7/8
(d) 6/7.
Answer : D

Q.[25] If, in an alternator, chording angle for fundamental flux wave is α, its value for 5th harmonic is
(a)5α
(b)α/5
(c)25α
(d)α/25.
Answer : A

Q.[26] Regarding distribution factor of an armature winding of an alternator which statement is false?(a) it decreases as the distribution of coils (slots/pole) increases
(b) higher its value, higher the induced e.m.f.per phase
(c) it is not affected by the type of winding either lap, or wave
(d) it is not affected by the number of turns per coil.
Answer : B

Q.[27] When speed of an alternator is changed from 3600 r.p.m. to 1800 r.p.m., the generated e.m.f./phases will become
(a) one-half
(b) twice
(c) four times
(d) one-fourth.
Answer : A

Q.[28] The magnitude of the three voltage drops in an alternator due to armature resistance, leakage reactance and armature reaction is solely determined by
(a) load current, Ia
(b) p.f. of the load
(c) whether it is a lagging or leading p.f. load
(d) field construction of the alternator.
Answer : A

Q.[29]Armature reaction in an alternator primarily affects
(a) rotor speed
(b) terminal voltage per phase
(c) frequency of armature current
(d) generated voltage per phase.
Answer : D

Q.[30] Under no-load condition, power drawn by the prime mover of an alternator goes to
(a) produce induced e.m.f. in armature winding
(b) meet no-load losses
(c) produce power in the armature
(d) meet Cu losses both in armature and rotor windings.
Answer : B

Q.[31] As load p.f. of an alternator becomes more leading, the value of generated voltage required to give rated terminal voltage
(a) increases
(b) remains unchanged
(c) decreases
(d) varies with rotor speed.
Answer : C


Q.[32] With a load p.f. of unity, the effect of armature reaction on the main-field flux of an alternator is
(a) distortional
(b) magnetizing
(c) demagnetizing
(d) nominal
Answer : A
If you have any problem then comment below.

Searches related to Basic MCQ Interview Questions On Alternator

electrical machines objective questions and answers pdf

electrical machines objective questions pdf download

mcq on alternator with answers

power factor of alternator at no load

objective questions on synchronous machines pdf

open wire test is used for?

synchronous machines questions and answers

synchronous machines objective questions with answers

Searches related to alternator working

alternator working principle pdf

alternator construction

alternator working animation

alternator working video

alternator pdf

alternator generator

car alternator working

brushless alternator working principle

---

Construction of a DC Machine And working of dc motor

As stated earlier, whether a machine is d.c. generator or a motor the construction basically remains the same as shown in the Fig. 1.

Fig.1 A cross section of typical d.c. machine

It consists of the following parts :

1.1 Yoke

a) Functions :

1. It serves the purpose of outermost cover of the d.c. machine. So that the insulating materials get protected from harmful atmospheric elements like moisture, dust and various gases like SO2, acidic fumes etc.
2. It provides mechanical support to the poles.
3. It forms a part of the magnetic circuit. It provides a path of low reluctance for magnetic flux. The low reluctance path is important to avoid wastage of power to provide same flux. Large current and hence the power is necessary if the path has high reluctance, to produce the same flux.

b) Choice of Material : To provide low reluctance path, it must be made up of some magnetic material. It is prepared by using cast iron because it is cheapest. For large machines rolled steel, cast steel, silicon steel is used which provides high permeability i.e. low reluctance and gives good mechanical strength.

1.2 Poles

       Each pole is divided into two parts namely, I) Pole core and II) Pole shoe.

       This is shown in the Fig. 2.

Fig. 2 Pole Structure

a) Functions of pole core and pole shoe :

1. Pole core basically carries a field winding which is necessary to produce the flux.
2. It directs the flux produced through air gap to armature core, to the next pole.
3. Pole shoe enlarges the area of armature core to come across the flux, which is necessary to produce larger induced e.m.f. To achieve this, pole shoe has been given a particular shape. 

b) Choice of Material : It is made up of magnetic material like cast iron or cast steel. As it requires a definite shape and size, laminated construction is used. The laminations of required size and shape are stamped together to get a pole which is then bolted to the yoke.

1.3 Field Winding (F1-F2)

       The field winding is wound on the pole core with a definite direction.

a) Functions : To carry current due to which pole core, on which the field winding is placed behaves as an electromagnet, producing necessary flux.

       As it helps in producing the magnetic field i.e. exciting the pole as an electromagnet it is called Field winding or Exciting winding.

b) Choice of material : It has to carry current hence obviously made up of some conducting material. So aluminium or copper is the choice. But field coils are required to take any type of shape and bend about pole core and copper has good pliability i.e. it can bend easily. So copper is the proper choice.

Note : Field winding is divided into various coils called field coils. These are connected in series with each other and in such a direction around pole cores, such that alternate 'N' and 'S' poles are formed.

       By using right hand thumb rule for current carrying circular conductor, it can be easily determined that how a particular core is going to behave as 'N' or 'S' for a particular winding direction around it. The direction of winding and flux can be observed in the Fig 3. 

Fig. 3

1.4 Armature 

       It is further divided into two parts namely,

I) Armature core and II) Armature winding

I) Armature core : Armature core is cylindrical in shape mounted on the shaft. It consists of slots on its periphery and the air ducts to permit the air flow through armature which serves cooling purpose.

a) Functions :

1. Armature core provides house for armature winding i.e. armature conductors.
2. To provide a path of low reluctance to the magnetic flux produced by the field winding.

b) Choice of Material : As it has to provide a low reluctance path to the flux, it is made up of magnetic material like cast iron or cast steel.

       It is made up of laminated construction to keep eddy current loss as low as possible. A single circular lamination used for the construction of the armature core is shown in the Fig. 4.

Fig. 4 Single Circular lamination of Armature core

II) Armature winding : Armature winding is nothing but the interconnection of the armature conductors, placed in the slots provided on the armature core periphery. When the armature is rotated, in case of generator, magnetic flux gets cut by armature conductors and e.m.f. gets induced in them.

a) Functions :

1. Generation of e.m.f takes place in the armature winding in case of generators.
2. To carry the current supplied in case of d.c. motors.
3. To do the useful work in the external circuit. 

b) Choice of material : As armature winding carries entire current which depends on external load, it has to be made up of conducting material, which is copper.

       Armature winding is generally former wound. The conductors are placed in the armature slots which are lined with tough insulating material.

1.5 Commutator 

We have seen earlier that the basic nature of e.m.f. induced in the armature conductors is alternating. This needs rectification in case of d.c. generator, which is possible by a device called commutator.

a) Functions :

1. To facilitate the collection of current from the armature conductors.
2. To convert internally developed alternating e.m.f. to unidirectional (d.c.) e.m.f.
3. To produce unidirectional torque in case of motors.

b) Choice of material : As it collects current from armature, it is also made up of copper segments.

       It is cylindrical in shape and is made up of wedge shaped segments of the hard drawn, high conductivity copper. These segments are insulated from each other by thin layer of mica. Each commutator segment is connected to the armature conductor by means of copper lug or strip. This connection is shown in the Fig. 5.

Fig. 5 Commutator

1.6 Brushes and Brush Gear

       Brushes are stationary and resting on the surface of the commutator.

a) Function : To collect current from commutator and make it available to the stationary external circuit.

b) Choice of material : Brushes are normally made up of soft material like carbon.

       Brushes are rectangular in shape. They are housed in brush holders, which are usually of box type. The brushes are made to press on the commutator surface by means of a spring, whose tension can be adjusted with the help of lever. A flexible copper conductor called pig tail is used to connect the brush to the external circuit. To avoid wear and tear of commutator, the brushes are made up of soft material like carbon.

1.7 Bearings 

       Ball-bearings are usually used as they are more reliable. For heavy duty machines, roller bearings are prederred.

Searches related to Construction of a DC Machine

construction of dc machine pdf

working principle of dc machine

basic construction of dc machines

commutation of dc generator


parts of dc machine

direct current machine

general principles of dc machines

principle operation of dc generator

Gas Turbine Power Plant

Gas Turbine Power Plant

A generating station which employs gas turbine as the prime mover for the generation of electrical
energy is known as a gas turbine power plant

In a gas turbine power plant, air is used as the working fluid. The air is compressed by the compressor and is led to the combustion chamber where heat is added to air, thus raising its temperature. Heat is added to the compressed air either by burning fuel in the chamber or by the use of air heaters. The hot and high pressure air from the combustion chamber is then passed to the gas turbine where it expands and does the mechanical work. The gas turbine drives the alternator which converts mechanical energy into electrical energy.

It may be mentioned here that compressor, gas turbine and the alternator are mounted on the same shaft so that a part of mechanical power of the turbine can be utilised for the operation of the compressor. Gas turbine power plants are being used as standby plants for hydro-electric stations, as a starting plant for driving auxiliaries in power plants etc.

Advantages
(i) It is simple in design as compared to steam power station since no boilers and their auxiliaries are required.
(ii) It is much smaller in size as compared to steam power station of the same capacity. This is expected since gas turbine power plant does not require boiler, feed water arrangement etc.

(iii) The initial and operating costs are much lower than that of equivalent steam power station.
(iv) It requires comparatively less water as no condenser is used.
(v) The maintenance charges are quite small.
(vi) Gas turbines are much simpler in construction and operation than steam turbines.
(vii) It can be started quickly form cold conditions.
(viii) There are no standby losses. However, in a steam power station, these losses occur because boiler is kept in operation even when the steam turbine is supplying no load.

Disadvantages
(i) There is a problem for starting the unit. It is because before starting the turbine, the compressor has to be operated for which power is required from some external source. However, once the unit starts, the external power is not needed as the turbine itself supplies the necessary power to the compressor.
(ii) Since a greater part of power developed by the turbine is used in driving the compressor, the net output is low.
(iii) The overall efficiency of such plants is low (about 20%) because the exhaust gases from the turbine contain sufficient heat.
(iv) The temperature of combustion chamber is quite high (3000oF) so that its life is comparatively reduced.

Schematic Arrangement of Gas Turbine Power Plant

The schematic arrangement of a gas turbine power plant is shown in . The main components
of the plant are :
(i) Compressor (ii) Regenerator (iii)Combustion chamber (iv) Gas turbine (v) Alternator (vi) Starting motor

(i) Compressor.  The compressor used in the plant is generally of rotatory type. The air at atmospheric pressure is drawn by the compressor viathe filter which removes the dust from air. The rotatory blades of the compressor push the air between stationary blades to raise its pressure. Thus air at high pressure is available at the output of the compressor.

(ii) Regenerator. A regenerator is a device which recovers heat from the exhaust gases of the turbine. The exhaust is passed through the regenerator before wasting to atmosphere. A regenerator consists of a nest of tubes contained in a shell. The compressed air from the compressor passes through the tubes on its way to the combustion chamber. In this way, compressed air is heated by the hot exhaust gases.

(iii) Combustion chamber. The air at high pressure from the compressor is led to the combustion chamber viathe regenerator. In the combustion chamber, heat*is added to the air by burning oil. The oil is injected through the burner into the chamber at high pressure to ensure atomisation of oil and its thorough mixing with air. The result is that the chamber attains a very high temperature (about 3000oF). The combustion gases are suitably cooled to 1300o F to 1500o F and then delivered to the gas turbine.

(iv) Gas turbine. The products of combustion consisting of a mixture of gases at high temperature and pressure are passed to the gas turbine. These gases in passing over the turbine blades expand and thus do the mechanical work. The temperature of the exhaust gases from the turbine is about 900oF.

(v) Alternator. The gas turbine is coupled to the alternator. The alternator converts mechanical energy of the turbine into electrical energy. The output from the alternator is given to the bus-bars through transformer, circuit breakers and isolators.
(vi) Starting motor. Before starting the turbine, compressor has to be started. For this purpose, an electric motor is mounted on the same shaft as that of the turbine. The motor is energised by the batteries. Once the unit starts, a part of mechanical power of the turbine drives the compressor and there is no need of motor now.

Introduction To Nuclear Power Station

Nuclear Power Station

A generating station in which nuclear energy is converted into electrical energy is known as a nuclear power station.
In nuclear power station, heavy elements such as Uranium (U235) or Thorium (Th232) are subjected to nuclear fission.  in a special apparatus known as a reactor.The heat energy thus released is utilised in raising steam at high temperature and pressure. The steam runs the steam turbine which converts steam energy into mechanical energy. The turbine drives the alternator which converts mechanical energy into electrical energy.
The most important feature of a nuclear power station is that huge amount of electrical energy can be produced from a relatively small amount of nuclear fuel as compared to other conventional types of power stations. It has been found that complete fission of 1 kg of Uranium (U235) can produce as much energy as can be produced by the burning of 4,500 tons of high grade coal. Although the recovery of principal nuclear fuels (i.e., Uranium and Thorium) is difficult and expensive, yet the total energy content of the estimated world reserves of these fuels are considerably higher than those of conventional fuels, viz., coal, oil and gas.

At present, energy crisis is gripping us and, therefore, nuclear energy can be successfully employed for producing low cost electrical energy on a large scale to meet the growing commercial and industrial demands.

Advantages
(i) The amount of fuel required is quite small. Therefore, there is a considerable saving in the cost of fuel transportation.
(ii) A nuclear power plant requires less space as compared to any other type of the same size.
(iii) It has low running charges as a small amount of fuel is used for producing bulk electrical energy.
(iv) This type of plant is very economical for producing bulk electric power.
(v) It can be located near the load centres because it does not require large quantities of water and need not be near coal mines. Therefore, the cost of primary distribution is reduced.
(vi) There are large deposits of nuclear fuels available all over the world. Therefore, such plants can ensure continued supply of electrical energy for thousands of years.
(vii) It ensures reliability of operation.

Disadvantages
(i) The fuel used is expensive and is difficult to recover.
(ii) The capital cost on a nuclear plant is very high as compared to other types of plants.
(iii) The erection and commissioning of the plant requires greater technical know-how.
(iv) The fission by-products are generally radioactive and may cause a dangerous amount of radioactive pollution
(v) Maintenance charges are high due to lack of standardisation. Moreover, high salaries of specially trained personnel employed to handle the plant further raise the cost.
(vi) Nuclear power plants are not well suited for varying loads as the reactor does not respond to the load fluctuations efficiently.
(vii) The disposal of the by-products, which are radioactive, is a big problem. They have either to be disposed off in a deep trench or in a sea away from sea-shore.

Schematic Arrangement of Nuclear Power Station

The schematic arrangement of a nuclear power station is shown in Fig.. The whole arrangement can be divided into the following main stages :
(i) Nuclear reactor (ii) Heat exchanger (iii) Steam turbine (iv) Alternator.

(i) Nuclear reactor.  It is an apparatus in which nuclear fuel (U 235 ) is subjected to nuclear fission. It controls the chain reaction that starts once the fission is done. If the chain reaction is not controlled, the result will be an explosion due to the fast increase in the energy released.
A nuclear reactor is a cylindrical stout pressure vessel and houses fuel rods of Uranium, moderator and control rods . The fuel rods constitute the fission material and release huge amount of energy when bombarded with slow moving neutrons. The moderator consists of graphite rods which enclose the fuel rods. The moderator slows down the neutrons before they bombard the fuel rods. The control rods are of cadmium and are inserted into the reactor. Cadmium is strong neutron absorber and thus regulates the supply of neutrons for fission. When the control rods are pushed in deep enough, they absorb most of fission neutrons and hence few are available for chain reaction which, therefore, stops.

However, as they are being withdrawn, more and more of these fission neutrons cause fission and hence the intensityof chain reaction (or heat produced) is increased. Therefore, by pulling out the control rods, power of the nuclear reactor is increased, whereas by pushing them in, it is reduced. In actual practice, the lowering or raising of control rods is accomplished automatically according to the requirement of load. The heat produced in the reactor is removed by the coolant, generally a sodium metal. The coolant carries the heat to the heat exchanger

(ii) Heat exchanger. The coolant gives up heat to the heat exchanger which is utilised in raising the steam. After giving up heat, the coolant is again fed to the reactor

(iii) Steam turbine. The steam produced in the heat exchanger is led to the steam turbine through a valve. After doing a useful work in the turbine, the steam is exhausted to condenser. The condenser condenses the steam which is fed to the heat exchanger through feed water pump.

(iv) Alternator. The steam turbine drives the alternator which converts mechanical energy into electrical energy. The output from the alternator is delivered to the bus-bars through transformer, circuit breakers and isolators.

Selection of Site for Nuclear Power Station

The following points should be kept in view while selecting the site for a nuclear power station :

(i) Availability of water.  As sufficient water is required for cooling purposes, therefore, the plant site should be located where ample quantity of water is available, e.g., across a river or by sea-side.

(ii) Disposal of waste.  The waste produced by fission in a nuclear power station is generally radioactive which must be disposed off properly to avoid health hazards. The waste should either be buried in a deep trench or disposed off in sea quite away from the sea shore. Therefore, the site selected for such a plant should have adequate arrangement for the disposal of radioactive waste.

(iii) Distance from populated areas.  The site selected for a nuclear power station should be quite away from the populated areas as there is a danger of presence of radioactivity in the atmosphere near the plant. However, as a precautionary measure, a domeis used in the plant which does not allow the radioactivity to spread by wind or underground waterways.

(iv) Transportation facilities. The site selected for a nuclear power station should have adequate facilities in order to transport the heavy equipment during erection and to facilitate the movement of the workers employed in the plant. From the above mentioned factors it becomes apparent that ideal choice for a nuclear power station would be near sea or river and away from thickly populated areas.