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Triveni -Rtmnu-Engg-AEE-Sum17-Sol


Subject:  AEE

Semester: 1


Q. 1) Explain with heat block diagram, The operation of Nuclear power generation. Give any three differences between Non-conventional and conventionality.

Sol: – Diagram ( What is nuclear fission )

When an atom of radioactive material is bombarded by neutrons, its nucleus gets split. This process is called as nuclear fission. This energy is liberated during heat. This heat is neutralized to generate electrical energy.

Working of a nuclear Generating Station.

  • In nuclear power stations, nuclear fission takes place in nuclear reactors.
  • The heat is produced in reactor is used to heat fluid called coolants.
  • These coolants transfer the heat obtained from nuclear fission to the heat exchanges where feed water is heated.
  • Thus the feed water becomes superheated steam.
  • Then this superheated steam falls on steam turbine fan under pressure. Which gives mechanical energy.
  • And as turbine is coupled with alternate, the steam turbines drives the generator. In this way, the mechanical Energy converted into electrical energy.

Functions of various components of a nuclear power station.

1) Nuclear reactor

Heat is produced in reactor core due to nuclear fission chain reaction of radioactive material such as uranium Thorium, plutonium, etc.

  • The nuclei of uranium are bombarded by neutrons which split the nuclei and lot of energy and fast-moving neutrons are generated.
  • The energy liberated is used to heat the coolant.
  • Then that coolant is transferred to a heat exchanger

2) Coolants.

  • Coolants flow through and around the reactor core they transfer heat produced in chain reaction to the feed water through heat exchanger.
  • Coolants used should have low melting point high boiling and high heat transfer co-efficient.
  • Another function of coolant is to keep the fuel assemblies at safe temperature to avoid melting and destruction.
  • Co2, H2, air, water, heavy water, gas pressurized water can be used or coolant.
  • Commonly the liquid metal live sodium-potassium, cadmium is

3) Heat Exchanger

  • In heat exchanger, heat carried by the coolant is fed to the feed water which gets converted into superheated steam.
  • this steam is fed to the steam turbine converting heat energy into an live mechanical energy. The turbine drives the generator. Which converts mechanical energy into electrical energy.

Difference Between Conventional and non-conventional

Conventional Non- Conventional
1) The sources of energy which have been in use for a long time. (ancient times) 1) The resource which are yet in the process of development over the past few years.
2) e.g. coal, petroleum, natural gas and water power, peat Thermal, Hydro, Nuclear, Gas-based, Diesel bared Generation. 2) e.g. Solar, Hydroelectric wind, Tidal, own wave Geothermal, ocean thermal Biomass, Biofuel.
3) Generate electrical energy in bulk form. 3) It cannot generate etc. energy in bulk form.
4) They are exhaust-able except water. 4) They are generally pollution-free.
5) They cause pollution when used, as they exist smoke and ash. 5) They are generally pollution-free.
6) They are very expensive to maintain, store, and transmitted as they are carried over long distances through grids and lines. 6) hers expensive due to local used and easy to maintain.

Q. 1) B) What are different types of fuse. Explain in brief with heat sketch of HRC fuse.

Ans: Types of Fuse.

fuses are classified as ;

  1.  Semi-enclosed or re-wirable type.
  2.  Totally enclosed or cartridge type.
  3. Dropout fuse
  4. Expulsion fuse
  5. H.R.C. Fuse
  6. Striker fuse
  7. Switch fuse

HRC fuse ( High Rupturing Capacity Fuse )


HRC fuse ( High Rupturing Capacity Fuse)

  • Fig. Shows the essential part of typical HRC.
  • HRC fuse mainly consists of heat-resisting ceramic body both the ends of the ceramic  body consists  of metal end caps
  • A silver current carrying element is welded to these metal end caps.
  • The current-carrying element is completely surrounded by the filling powder.
  • The filling material may be plaster of parries, chalk, quartz, or marble dust.
  • Filling material acts as an arc quenching and cooling medium when fuse element blows off due to excessive heat generated under abnormal conditions.
  • Under abnormal conditions, the fuse element is at a temperature below its melting point. Therefore it carries the normal current without overheating.
  • When a fault occurs, the current increases, and heat produced is sufficient to melt fuse element. Fuse element6 melt before the fault current reaches its first peak value.
  • Vapourisation of metal silver elements chemically reacts with filling powder and results in the formation of high resistance substance and helps in quenching the are.

Advantages ;

  • Speed of operation is very high.
  • Maintenance cost is practically zero.
  • They are capable of clearing high as well as low fault current.
  • They do not deteriorate the age.
  • They provide reliable operation.
  • They are cheaper than other circuit interrupting devices of equal breaking capacity.


1) Heat produced by the arc may affect the associated switches.
2) they have to be replaced after each operator.


Q. 2] a) Draw and explain the single line diagram, for generation, transmission and distribution.

Ans: The increased demand of electricity needs more generation of electrical power. As the generation takes place at remote places, an efficient distribution is necessary.

  • The problems of AC transmission particularly in long-distance transmission has led to the development of DC transmission
  • The simple field block diagram of the extra-high voltage AC transmission system is shown in


Single line diagram. of basic EHV AC System

This system can be broadly divided into two parts.

a) Transmission system

b) Distribution system

  • Each part is again sub-divided into two parts.

a) 1) Primary Transmission

2) Secondary transmission


b) 1)P Primary Distribution

2) Secondary Distribution

*  a)  1) Primary Transmission.

1) In Fig. The central section CS generates power using a three-phase alternator at 6.6 or 11kv or 13.2 or 32kv.

2) This voltage is then stepped up by a suitable three-phase transformer to 132 kv as shown.

3) Such a high voltage of transmission required conductors of a smaller cross-sectional area which results in the reduction in the cost of copper or aluminum.

( Note : Reason behind step-up voltage from 11 to 132kv )

If v ↑    I↑     R↑

If R↑ means     As, R\alpha \frac{\rho l}{A}

1) Size of conductor ↓

2) Cost of conductor ↓

3) Power loss ↓

4) Efficiency ↑

4) However cost of insulation decreases, the transmission voltage is therefore determined by economic consideration Higher transmission voltage also reduces the line losses and improves efficiency.

5) The three-phase three wire overhead high voltage transmission line gets terminated in the step-down transformer in a substation known as receiving station.

6) Receiving station is usually situated outside the city in order to ensure safety.

7) Here the voltage is stepped down to 33 kv from 132 kv.

a) 2) Secondary Transmission

1) From the RS, the power is then transmitted at 33kv, by underground cable ( and occasionally by overhead lines ) to the various substation ‘SS’ located at various points in the city

2) These is known as secondary or low voltage transmission

3) At the substation (SS), this voltage is further reduced from 33kv to 3.3 kv using step-down transformer.

b) 1) Primary Distribution

The output of substation at 3.3 kv can be directly given to a customer whose demand exceeds 50 KVA using a special feeder. This is primary distribution.

b) 2) Secondary Distribution

1) Secondary distribution is alone at 440 V or 230 V

2) The reduction in the voltage level form 3.3 KV to 440 V or 230 V is done by step down transformer at the distribution substations.

3) The single-phase residential load is connected between any one line and the neutral whereas 3-phase, 440 V, motor load is connected across three-phase line directly.

4) The standard frequency in India is 50 Hz and in USA 60 Hz

Q. 2) B) Explain the necessity of Earthing.

Ans: fig.

Connecting metallic form of electrical machines, conduct pipes, etc. to the ground is known as earthing.

  • Any electrical equipment is mainly made up of minimum two parts.
  • Conducting material and insulating part.
  • Conducting part ( winding ) is an inner part which is covered by insulating material
  • This entire assembly is kept into metallic body.
  • If due to certain reasons, insulation between the conducting part and metallic body fails, then metallic body also carries certain current and becomes conducting.
  • In such situation suppose any person comes into contact with the instrument. Then the circuit gets completed through his body to earth as shown in fig. 1, if the system is not earthed, the person gets the shock.
  • IF the system is effectively grounded, then leakage current will flow through the earth connection to ground as shown in fig. 2
  • And thus the person gets protected against the electric shock.

Necessity of Earthing.

  • To avoid electric shock to the human
  • to avoid risk of fire due to earth leakage current though the unwanted path
  • To provide the discharge of short circuit current through earthing resistance.

Q. 3) a) Derive the EMF equation of DC Generator.

Ans: Let,

∅ = Flux per pole in wb

P = Number of poles

Z = Total no. of conductors an armature

N = Speed of rotation of armature in rpm

A = Number of parallel paths in armature

Eg = EMF generated / path in armature.

Now, emf generated by faraday low,

E=No of conductor\times \frac{d\phi }{dt}volts

∴ Total EMF generated = \frac{zd\phi }{dt}volts

As the armature is rotating in a magnetic field, conductor will cut the flux present in the air gap.

∴ Flux cut by one conductor in one rotation will be the total flux

= flux/pole × No. of poles

dΦ= Φ P webers ………………(1)

  • Speed in rpm =N rpm

speed in rps c=\frac{N}{60}rpm

Time required for one rotation will be


\therefore =\frac{60}{N}Second……………………(2)

\therefore\frac{d\phi }{dt}=\frac{\phi P}{\frac{60}{N}}=\frac{\phi PN}{60}

∴ EMF generated for one conductor = 1\times \frac{d\phi }{dt}=\frac{\phi PN}{60}volts

Now total Z conductors are present If armature winding has a parallel paths, then conductors per path =\frac{Z}{A}\frac{conductor}{path}

∴EMF Generated /path = Conductor/path × EMF/ Conductor

Eg=\frac{Z}{A}\times \frac{\phi PN }{60}

= \frac{\phi PNZ}{60A}

=\frac{\phi PN Z}{60A}

1) For Lap wdg. A = P

\therefore Eg=\frac{\phi NZ}{60}

2) For wave wdg A= 2

\therefore Eg=\frac{\phi PNZ}{120}


Q. 3) b An 8 pole armature has 96 slots with 8 cord per slots. It is driven at 600 rpm. The useful flux per pole is 10 mwb calculate the EMF in armature wdg when it is

a) lap connected     b) Wave connected

Ans : P= 8

Slots = 96

conductors/slots = 8

Z = slots × conductor / slots

= 96× 8

= 768

N= 600 rpm

Φ = 10 mwb = 10 ×  10-3wb

Eg = ? For Lap

Eg = ? For wave

1) For lap winding


Eg=\frac{\phi NZP}{60A} Eg=\frac{10\times 10^{-3}\times 600\times 768\times 8 }{60\times 8}

Eg = 76.8 volt

2) For wave wdg


Eg=\frac{\phi NZP}{60A} Eg=\frac{10\times 10^{-3}\times 600\times 768\times 8 }{60\times 2}

Eg = 307.2 volts


Q. 4 a) A 250V, DC shunt motor runs at 1000rpm at no load and takes 8A. The total armature and shunt field resistance are 0.2 Ωand 250 Ωres. Calculate the speed when loaded and taking 50 A. Assume flux to be constant.

Ans. = Given

Vt = 250V

At No load

N1 = 1000 rpm

IL1 = 8A

Ra = 0.2Ω

Rsh = 250Ω

At load

N2 = ?

IL2 = 50A

Φ = constant

Step 1)  I_{sh}=\frac{V_{t}}{R_{sh}}=\frac{250}{250}=1A

Step 2) Ia1 =IL1-Ish = 8-1 = 7A

Ia2 = IL2-Ish = 50-1 = 49A

Step 3) Eb1 = Vt-Ia1Ra =  250-7(0.2) = 248.6 volts.

Eb2 = Vt – Ia1Ra = 250-49(0.2) = 240.2 voles.

Step 4) N\propto \frac{Eb}{\phi }\therefore \phi Constant

\therefore \frac{N_{2}}{N_{1}}=\frac{E_{b2}}{E_{b1}} \therefore N_{2}=\frac{E_{b2}}{E_{b1}}\times N_{1} \therefore N_{2}=\frac{240.2}{248.6}\times 1000

∴ N2 = 966.2rpm


Q. 4 b) What is the need of a starter in DC Motor. What are the types of starter.

Ans. : Function of DC Motor starter.

  • At starting the DC Motor draws heavy current from the supply as some time is required to gain speed and hence the back emf build up.
  • Starters are required to limit this initial inrush current because this may cause overheating of armature and also may cause voltage drop in supply. Hence function of starter is to limit the starting current; in armature circuit during starting and to limit the accelerating time of motor.

*Necessity of DC Motor Starter.

For a DC shunt motor,

Vt = Eb + IaRa


But at the time of starting condition, Eb=0


Since practically the armature resistance of a motor is very small, generally less than 1Ω

Therefore Ia would be very large.

Example :- If Motor with Ra = 0.5 is connect directly to 230 v supply then

I_{a}=\frac{V_{t}}{Ra} =\frac{230}{0.5}

= 460A

  • Thus, this large current can damage the armature winding, brushes,  commutator or insulation, thus short-circuiting the armature winding.
  • To avoid this excessive starting current, starter is used for DC motor.
  • In starter, additional resistance is connected in series with armature at the time of starting and when motor attains speed, resistance is cut

* Types of Starter

  1. Two-point starter
  2. Three-point starter
  3. Four-point starter

Q. 5 a) What are different types of tariff. Explain simple rate tariff?

Ans: There are various types of tariff and are as follows.

  1. Flat rate tariff
  2. Block rate tariff
  3. Simple rate tariff
  4. Two part tariff
  5. Maximum Demand tariff
  6. Power factor tariff

* Simple Rate tariff

  • If the rate per unit of consumed electricity is fixed, then it is called a simple rate tariff
  • This tariff is also called a uniform rate tariff
  • The cost per unit remains always constant. And independent or increase or decrease in number of units consumed.
  • the energy consumption is measured by means of an energy meter installed at the consumer’s place.
  • This is the simplest type of tariff and easy for consumers to understand

* Disadvantages

  • Every consumer is charged equally. There is no discrimination.
  • This does not encourage the consumers to use more electricity
  • The cost of energy per unit is high.

Q. 5 b) Numerical on tariff.

Find the electric bill for the month of August, the consumer has to pay.

Ans : Step 1) Energy consumption for a day

E.C. = (4×100)+(4×500)+(4×500)+(5×4000)+(4×2500)

Daily E.C.= 34,400 whr. = 34.4 kwhr

Step 2) E.C. For month of August = 34.4×31 days

Total units consumed = 1066.4 kwhr.

Step 3) Block rate tariff.

0-100 unit = 100×3 = 300

101-200 unit = 100×5.5 = 550

201-300 unit = 100×75= 750

301and above = (1055.4-300)×10

= 766.4×10 = 7664

Electric bill for month of August =( 300+550+750+7664)

= 9,264 Rs. For Consumer


Q. 6 a) Explain the construction and working of sodium vapor lamps.

Ans: Fig.

  Sodium vapor lamp


  1.  It consists of an inner tube of V-shaped and made up of a special sodium vapor resisting glass.
  2. It consists of two coated tungsten electrodes which are connected across an autotransformer.
  3. The inner U-tube contains neon gas at a pressure about 10mm mercury and a small amount of sodium.
  4. The inner U-tube is enclosed within a double-walled vacuum flask type envelop and is well insulated in order to conserve the heat required to vaporize sodium.

Working :

  1. When supply is switched ‘ON’, high leakage reactance autotransformers produce and 50 to 480V volts across the electrodes.
  2. This voltage is sufficient to initiate the discharge through neon gas
  3. The heat of discharge through neon gas is sufficient to vaporize the sodium and the lamp stark operation. This requires few minutes.
  4. The light emitted is of yellow glow.
  5. Capacitor ‘C’ improves the power factor of the circuit
  6. Following are few points which are to be remembered in case of sodium vapor lamp i.e.
  • Efficiency of lamp is about 110 lumens /watt.
  • It cannot be restart immediately after stoppage.
  • Lamp should be operated only in a horizontal position to avoid the collection of sodium at one end

Application :

It can be used for street lighting, railway yard lighting, park lighting.


Q. 6 b) A hall of 25×40 m is to be illuminated using two fluorescent lamp fitting consisting if two 40 watt tubes. The luminous efficiency of lamp is 60 lumen, Assume a depreciation factor of 0.8 and co-efficient of utilization as 0.75. Calculate total no. of lamps required for illumination of 250lux on working plane.

Sol. : Given

A = 25×40 m2

P = 2×40 watt = 8Øwatt

d.f = 0.8

U.F = 0.75

N =?

E = 250

Formula :

lumen flux form each lamp

1) \eta =\frac{\phi }{pi}


2) GF=\frac{F}{UF\times DF}

F = E×A

3) N=\frac{GF}{\phi }

Setp 1) N=\frac{GF}{\phi }

\eta =\frac{\phi }{pi}

∴ Ø=η×pi

= 60×80

Ø=4800 lumens

Step 2) G.F=\frac{E\times A}{U.F\times D.F}

G.F=\frac{250\times 25\times 40}{0.75\times 0.8}


Step 3) N=\frac{GF}{\phi }

N=\frac{416.6\times 10^{3}}{4800 }

No. of pair of tubes = N= 87


Q. 7 a) Give the difference between squired cage rotor and slip ring rotor.

Squirrel Cage Rotor Wound or Slip Ring Rotor
1) Rotor is in the form of bars which are shorted at the ends with the help of end rings. 1) Rotor is in the form of 3 phase winding.
2) No-slip rings or brushes 2) Slip rings and brushes are used
3) Simple construction 3) Construction is complicated
4) External resistances cannot be connected 4) It is possible to connect the external resistance to the rotor
5) Small and moderate starting torque 5) High starting torque can be obtained
6) Starting torque cannot be adjusted. 6) Starting torque can be adjusted by adjusting the external les.
7) Frequent maintenance is not required due to absence of slip rings and brushes 7) Frequent maintenance is necessary due to the use slip rings and brushes.
8) Rotor resistance starter cannot be used 8) Rotor resistance starter can be used.
9) Less rotor copper loss 9) High rotor copper loss
10) Higher frequency 10) Low efficiency
11) Applications include lathes fans, water pumps, blowers, etc. 11)Application includes cranes elevators, compressors, lifts, etc.

Q.7 b) A three-phase induction motor is running at 900 rpm whose stator has 6 poles and is operated from 50Hz power supply. Calculate 1) Speed of RMF. 2) Relative speed 3) %slip  4)Frequency of rotor current

Sol. : Given

Nr = 900rpm

P= 6

F = 50 Hz

1) Ns =?     2) Ns-Nr =?

3) %S = ?   4) F’ =?

Step 1) Ns=\frac{120F}{P}

Ns=\frac{120\times 50}{6}

= 1000rpm

Ns = 1000rpm

Step 2) Relative speed =?

Relative speed = Ns – Nr


Relative speed = 100rpm

Step 3) %Slip=\frac{Ns-Nr}{Ns}\times 100[latex]</p> [latex] %Slip=\frac{1000-900}{1000}\times 100

%S = 0.1

Step 4) F'= ?

F'= SF

F'= 0.1×50Hz

F'= 5Hz


Q. 8 a) Why 1Φ IM. is never self-starting. What provision is made to make it self starting?

Ans. : 1) Con-structurally, this motor is, more or less, similar to a polyphase induction motor, except that a) its stator is provided with a single-phase winding and b) a centrifugal switch is used in some types of motors, in order to cut out a winding used only for starting purpose.

2) It has distributed stator winding and a squirrel cage rotor.

3) When fed from a single-phase supply, its stator winding produces a flux (or Field) which is only alternating, i.e. one which alternates along one space axis only. It is not asynchronously revolving (or rotating ) Flux, as in the case of a two or a three-phase stator winding fed from a 2 or 3 phase supply.

4) Now an alternating or pulsating flux acting an a stationary squirrel cage rotor cannot produce rotation ( only a revolving flux can ). That is why single-phase motor is not self-starting.

What provision is made to make it self starting

1) to make a single-phase I.M. self-starting, the magnetic field produced by the stator winding should be changed from alternating type to rotating type.

2) This is possible only if more than one single-phase winding are present in the stator core with space. Displacement between them

3) Also that the current flowing in the windings should have a phase displacement (time displacement) between them

4) If two single phase windings are wound in the starter core then ideally the windings should be spaced 90° physically apart and the phase displacement between the currents in the two windings should be as large as electrical to create maximum torque on the rotor.

Q. 8 b) Explain with neat labeled diagram, the working of capacitor start-capacitor run induction motor.

Ans. =Diagram

Cr = Running capacitor      Im= Main wdg current

Cs = Starting capacitor       Is = Starting wdg current

1) In this type of motor, the auxiliary winding and main wdg is connected with two capacitor Cs and Cr connected in parallel.

2) The auxiliary winding and capacitor remain connected in circuit all the times.

3) Since the capacitance which will give optimum running performance is not the same as the capacitor which will give best starting torque. Also, capacitor used for normal running (Cr) should be of continuous duty rating an another capacitor (Cs) used for starting should be short duty rating.

4) Centrifugal switch is used to take out Cs our of auxiliary winding circuit when motor has attained pre-determined speed.

5) During starting when the switch 'S' is closed, capacitor Cs and Cr in parallel so that the total capacitance in auxiliary winding circuit in sum of their values.

6) When the motor reaches about 75% of full speed, the centrifugal switch 'S' open and cuts the starting capacitor out of auxiliary winding circuit.

7) Only CR remains in the auxiliary wdg.

8) Such a motor operates a two-phase motor from single-phase supply, thereby, producing a constant torque.

9) As shown in fig. Current drawn by the main windings lags the supply voltage 'V' by some angle as it is inductive in nature whereas Is Leads V by certain angle as it is capacitive in nature.

Main wdg = Inductive in nature + Internal Resistance

L, r hence lags by some angles to V

Auxiliary winding = capacitive in nature + Internal Resistance Cr + Cs + r hence Leads by some angles to V


10) The two currents Is and Im are out of phase by near about 80°. Their resultant current I is small.

The main advantage of these motors is high starting torque that they can produce. The Ist can be as high as 300 to 400% of the full load torque.

Role of Capacitor.

1) The capacitor always remains in the circuit. Therefore it decides the performance of the motor at the time of starting as well as during the running mode.

2) The selection of the value of C is a comprise between two factors namely starting torque and better running performance.

3) Fig. Shows torque-speed characteristics of capacitor start capacitor run motor.

4) Note that starting torque is less than that of capacitor start motor.

5) Advantage of this motor is its high power factor which is due to the continuous presence of capacitor in the circuit.

6) Disadvantage is its higher cost.

Tst is very high

P.F improved less noisy

Efficiency high




Due to high Ist, capacitor start capa. run L.M used for following applications.

  1. Grinders
  2. Fan and air conditioners
  3. Compressors
  4. Refrigerators
  5. Conveyors

The end.

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