LinkedIn Insight Snehal-RTMNU-Engg-AEE-Win17-sol - Grad Plus

Snehal-RTMNU-Engg-AEE-Win17-sol

Q1.a. Explain the necessarily of equipment earthing. Explain pipe type earthing.

Ans:- For necessity of equipment earthing

Connecting metallic from 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 a inner part which is covered by insulting material
  • This entire assembly is kept into metallic body.
  • If due to certain reason, 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 get 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 unwanted path
  • To provide the discharge of short circuit current through earthing resistance.

Pipe type earthing

EE

1.First of all diag 5×5 ft(1.5×1.5) pit about 20-30 f (6-9m) in the ground.[Note that ,depth & width depends on the nature and structure of the ground.

2.Bury the pipe of dimension of the pipe is usually 40mm(1.5m)in diameter &2.75m(9 ft) in length for ordinary soil or greater for rocky soil. The moisture of the soil will determine the length of the pipe to be burried but usually it should be 4.75 m(15.5 ft)

3.Tight earth lead through nut bolts from two different places or earth pipe.

4.Use two earth lead through with each earth pipe& tight them.

5.To protect the joints from corrosion ,put grease around it.

6.Collect all the wires in a metallic pipe from the earth electrodes make sure the pipe is 1 ft (30cm) above the surface of the ground.

7.To maintain the moisture condition around the earth pipe , put a 1ft layer of powdered  charcoal(powdered wood coal) & lime mixture around the earth pipe.

8.Use  nut  bolts  to connect tightly wires to the bed plates of machine. Each machine should be earthed from two different places.

9.Earth continuity conductor which is connected to the body & metallic parts of all installation should be tightly connected to earth lead.

10.At last, test the overall earthing system through earth tester .If everything is going about the planning ,then fill the pit with soil. The maximum allowable resistance for earthing is 1Ω. If it is more than 1Ω than more the size of earth lead & earth continuity conductors .Keep the external ends of the pipe open & put the water time to time to maintain the moisture condition around the earth electrode which is important for the better earthing system.

Q1.b. Write comparison between overhead & underground system.

Condition Overhead cabeling system  Underground cabeling system
1.Public Safety This system is not preferable This system is preferred from the point of view of public safety.
2. Possibility of fault More Less or compared to OH system
3.Maintainance cost More Much lower because chances of fault are very less.
4.Useful life Half OH system Twice of OH system
5. Flexibility More flexible less flexible
6.Initial Investment  Less more
7.Selection of rout length Difficult Easy and simple
8. Time taken for repair Less time taken for repairing Too much time taken
9.Power factor less high
10. Taking branch line easy Difficult
11. Affection  affect the appearance of city It does not affect the appearance of the city.

Q.2a Draw a neat single line diagram of generation ,transmission and distribution through different voltage levels at each point.

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 development of DC transmission
  • The simple field block diagram of the extra high voltage AC transmission system is show in fig.

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

Similarly.

b) 1)P Primary Distribution

2) Secondary Distribution

*  a. 1) Primary Transmission.

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

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

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

( 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 ↑

5) 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.

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

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

8) 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 occasionaly 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 voltages is further reduced from 33kv to 3.3 kv using step down trasnformar.

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 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 where as 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.2b.What do  you mean by Fuse? Explain Rewireable and HRC fuse.

Ans:- FUSE:

1.Fuse is a simplest device, which breaks the circuit under abnormal condition i.e. it is only a current interrupting device under fault condition & is not able to make or break the circuit under normal condition.

2. It is used for overload or short circuit protection in medium voltage upto 33kv and low voltage upto 400v installations.

3.Fuses are widely used in power ,heating and lighting purpose.

REWIREABLE FUSE

1.This fuse is most commonly used in house wiring and small current circuits.

2.The fuse wire is fitted on protection “carrier” which is fitted in the porcelain base.

3. Whenever the fuse wire blows off due to overload or short circuit ,the fuse carrier can be pulled out, the new wire can be placed and service can be restored.

4.The Fuse wire may be of lead tinned ,copper or an alloy of tin lead.

5.The fuse wire may be replaced by a wire of correct size & correct specification otherwise it may prove dangerous with the possibility of equipment burning out.

Working

Since the fuse wire is exposed to the atmosphere ,it gets oxidised and deteriorated ,resulting in a reduction of the wire section with the passage of time. This causes the resistance causing operation of the fuse at lower currents.

2. HRC FUSE

HRC fuse ( High Rupturing Capacity Fuse )

Diagram

Image result for hrc fuse

Fig. Shows the essential part of typical HRC.

  • HRC fuse mainly consists of heat resisting ceramic body both the ends of 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 act as an arc quenching and cooling medium when fuse element blows off due to excessive  heat generated under abnormal condition.
  • 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.
  • Vaporisation 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.
  • Maintainance cost is practically zero.
  • They are capable of clearing high as well as low fault current.
  • They do not disteriorate the age.
  • They provide reliable operation.
  • They are cheaper than other circuit interrupting divices of equal breaking capacity.

Disadvantages.

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

Q3.a. State the function of following DC machine parts.

1.Commutator
2. Yoke
3.Armature winding.

Ans:-

The construction of a DC machine can be broadly divided into three parts.

  1. Stator (Stationary part)
  2. Armature(rotor) rotating part
  3. Commutator and Brushes

A. Stator:- Stator consists of following main pats

  1. Yoke
  2. Pole
  3. Pole shoes
  4. Field winding

1. Yoke- It is a cylindrical outer frame made of steel sheet or fabricated steel. It serves two purposes.

  1. It provides a mechanical support to the poles and acts as a protective cover for the whole machine.
  2. It provides a return path of low reluctance to the magnetic flux produced by the poles.

2. Poles – The poles are one of the most important component of DC machine. In very small machine, these poles may be permanent magnetic but usually they are electromagnets excited by a DC current.

3. Pole Shoes- The pole face shaped to fit the curvature of the rotor is known as shoe of the pole. To maintain the uniform width air gap between the poles and armature, these pole shoes are elongated on both sides as shown.

4. Field Winding- This is a simple concentrated coil which is wound over the poles. The basic purpose of this winding is to excite the magnetic poles whenever a DC current is passed through it. It is so wound that the equal number of north and south poles are formed. This winding is generally made of less cross section copper wire.

5. Commutator- 1. The various parts of commutator are shown in fig. They include commutator segment an equal no. of mica segments and an iron core consisting of two ends rings and a connecting shell on which the mica segments and bars are placed.

2. Commutator are device with many copper segments which  are equal to the no. of coils of armature windings is called as commutators.

3. Commutators is always mounted on the same shaft on which rotor core is present.

4. It helps in the transfer of electrical energy between the stationary electrical load and the rotating armature winding.

6. Brushes- The brushes whose function is to collect current from commutator and made up of carbon and are in the shape of rectangular block. The brushes are housed in brushes holders. They are made to rest on commutator by a spring whose tension can be adjusted.

B) Armature Construction – The winding in which output emf is generated is present on rotor hence it is called armature winding and rotor is also called as armature. It consists of following parts.

  1. Shaft 2. Armature winding 3. Armature core

c) Armature Core- The armature core is made up of thin circular lamination of sheet steel and is mounted on the shaft. It is cylindrical in shape and is slotted on the  outer periphery.

D) Armature Winding – The armature winding consists of a many number of coils. The conductor coils embedded in the rotor slots.

They are two winding

  1. Lap Winding
  2. Wave Winding

Q3 b. A 4 pole lap connected generator has 80 slots with 10 conductor per slot rotates at 1000 rpm which induces 400v . At what speed generator is rotate when it induces  200v.

Ans:-P=4
Lap winding  ∴A=P=4
slots=80
conductors per slot=10
∴ z=80×10=800
N1=1000 rpm
EV1=400 v

Step 1. The emf equation of generation is given by

\[ E_{g_{1}}=\frac{\phi N_{1}ZP}{60A} \]
\[   \phi =E_{g_{1}}\frac{60A}{N_{1}ZP}  \]

\[ \phi=\frac{400\times 60\times 4}{1000\times 800\times 4} \]

\[ \phi=0.03 wb \]

Step 2.  \[ E_{g_{2}} =\frac{\phi N_{2}ZP}{60A} \]
\[ N_{2}=\frac{E_{g_{2}}60 A}{\phi ZP} \]
\[=\frac{200\times 60\times 4}{0.03\times 800\times 4} \]
\[ N_{2}=500\; RPM \]

Q 4 a. Derive torque equation of a DC motor.

Ans:-

For a dc motor , Vt= Eb+IaRa

Multiplying both sides by Ia, we get,

VtIa=EbIa+Ia2Ra ———(1)

Where,

VtIa– Electrical input to the armature

Ia2Ra– Armature copper loss

Pm= ebIa  – Electrical Equivalent of mechanical power Pm developed in the armature.

If T is the torque in N-m, which is defined as the turning moment of a force about an axis, then the gross mechanical output power developed can be written as, Pm=Tw —– (2)

where, w- Angular velocity in radians/sec

N- Rotor speed in RPS

W= 2ΠN

w=\frac{2\pi N}{60} N is in rpm

Equation (1) and (2) can be equated as

T.w = EbIa

\[T=\frac{1}{\omega }\left ( \frac{\phi PN}{60}\times \frac{Z}{A} \right )I_{a} \]

\[ T=\frac{60}{2πN }\left ( \frac{\phi PN}{60}\times \frac{Z}{A} \right )I_{a}\]

\[T=\frac{1}{2π }\left ( \frac{\phi PZ}{A} \right )I_{a}\] N-m

Thus, the torque developed in armature is directly proportional to the product of flux and armature current.

T∝ Φ

T∝ I

Q 4 b. A 4 pole lap wound motor consumes 20 A at terminal voltage of 250 V .It has a field and armature resistance of 50 Ω and 0.05Ω respectively. Neglect brush drop . Determine 1. Armature current 2. Emf induced.

Ans:- p=4
IL=20 A
Vt=250 v
Ra=0.05Ω
Rsh=50Ω
1.Ia=?
2.Eg=?

Step 1.- For motor

IL=IA + Ish
Ia=IL-Ish
=20-Vt/Rsh
=20-250/50
=15 A

STEP 2-

Eb = Vt – IaRa
=250-15×0.0
=249.25v

Q5.a Explain the working and construction of mercury vapour lamp

Ans:- Circuit Diagram

Image result for low pressure mercury vapour lamp circuit diagram

Mercury vapour lamp
It is an electric discharge lamp in which light is produced by electrical conduction through mercury vapour. It emits a very bright greenish blue light.

Construction
1. As shown in figure , it consists of two tubes ,one inside the other.
2.The smaller inner tube is made of hard glass(or quartz) and is surrounded by the larger glass tube or bulb.
3.The space between the two tubes is completely evacuated to prevent heat loss.
4.The outer tube also absorbs harmful ultraviolet radiations.
5.The inner tube contains argon gas and a small quantity of mercury. It is called as arc tube.
6.It contains two main electrodes A and B and auxiliary electrode C. The  main electrode are and of tungsten and are coated with barium oxide.
7.The auxiliary electrode c is placed near the main electrode A and it is connected to electrode B through resistance  R.
8.The choke is connected in series to limit the current .
9.The capacitor C, is connected across the mains supply line to improve the power factor.

Operation
1.When switching ON the supply an initial discharge is established in argon gas between the main electrode A and auxiliary electrode C.
2. After few seconds , the discharge take place between the two main electrode and spreads throughout the inner tube.
3.The heat produced by this gaseous discharge causes mercury to get vapourized .
4.The potential across the electrodes A and B rises from about 20V to 150V . It takes about 4-5 minute for mercury arc to build up to full brightness.
5.The mercury vapour lamps must be operated in its vertical
6.After the lamp is switched OFF , before it can restart it must be cooled to reduce the vapour pressure.

Advantages
1.It life is of about 9000 working hours. Life is long.
2.It has high efficiency of about 50 lumens/watt.
3.Mercury lamps are available in various size upto 2 kw.

Disadvantage
1.Mercury vapour lamp cost is high as compared to incandescent lamp.

Application
It has widely used in shopping centers, industrial installation, railway yards ,street lighting, highway etc.

Q.5 b Define the following terms
1.Luminous Flux

a) It is defined as that radiant energy from hot body which produce the visual sensation upon the human eye.

b) It is denoted by F and expressed in Lumuns.

c) The whole of electrical power supplied to a lamp is not converted ‘Luminous Flux’ but some of the power is lost by heat conduction.

2) Luminous Intensity

 

nn

 

  • Consider a point sources of light ‘O’. let df be luminous flux crossing any section of a narrow cone of solid angle dw steradians. The apex of the cone so formed is at source
  • Solid angle It is subtended at a point in space by an area and is the angle enclosed in the volume formed by infinite number of lines lying on the surface of volume and meeting at the point. It is denoted by w and expressed in steradians.
  • Hence luminous intensity is defined as ratio of luminous flux df emitted by source to solid angle dw.
  • It is denoted by I and expressed in candela
  • If ‘df ⇒ luminous flux in lumens
  • dw  ⇒ Solid angle in steradian
I=\frac{df}{dw}

i.e. \frac{f}{w}

Lumens/Steradians    or candela.

3.Illuminance

  • When the light falls on a surface it is illuminated.
  • It is defined as the luminous flux no received by a surface per unit area.
  • It is denoted by E and represented by lumens per square meter or lux.

E=\frac{F}{A}
Lumens/m2   or Lux.

4.Candle power- Candle power of the source is defined as the number of lumens emitted by that source per unit solid angle in a given direction.
The term candle power is used interchangeably with luminous intensity .It is denoted by C.P.

C.P.=Lumens/ω    ω→ solid angle.

Q.6.a What is different types of tariff? Explain one part tariff?

Ans:- Different types of tariff and one part means simple rate tariff.

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 as simple rate tariff
  • This tariff is also called as uniform rate tariff
  • The cost per unit remain 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 consumers place.
  • This is the simplest type of tariff and easy for consumes to understand

* Disadvantages

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

Q6 b. A consumer has following load schedule for a day
Ans:-

Step1-

Energy consumption for a day =(6×200)+(3000×6)+(100×1)+(4000×3)+(2000×5)+(1000×3)
=443000 whr
Daily  EC   =44.300 k whr or 44.3 units

Step 2-
Daily bill by consumer pay= 44.3 units ×35 paise /kwhr
=  1550.5 paise
Bill= 15.5 Rs

Q .7 a. Explain speed-torque or torque slip characteristics of 3 phase induction motor.
Ans:- Torque Speed or Torque -slip characteristics of Induction motor.

1. The torque slip or torque speed characteristics of a induction motor is shown in the fig. The characteristics can be divided into three sections.

  1. Forward Motoring   2. Plugging 3. Regeneration
Mode Slip Range Slip Relation between Ns and Nr
1. Forward Motoring  s=0 to s=1 \[s= \frac{N_{s}-N_{r}}{N_{s}}\]

Let Nr= +ve = Ns, Nr<Ns

s=0, or s<1

2. Generating s=1 to s=0 \[s= \frac{N_{s}-N_{r}}{N_{s}}\]

Let Nr= 2Ns

s=-1

3. Plugging s=1 to s=2 \[s= \frac{N_{s}-N_{r}}{N_{s}}\]

Let Nr=-ve, Nr>Ns, s=2 or s>1

  1. Forward Motoring Region (s=0 to s=1)

1. The forward motoring region corresponds to the values of slip between 0 and 1

2. In the forward motoring region of the charcteristics shown in fig, the motor rotates in the same direction as that of rotating magnetic field.

3. The torque produced by the motor is zero at synchronous speed i.e. for s=0. This is the induced voltage in rotor is zero. when Nr=Ns.

4. The torque increases as the slip increases while the air gap flux remains constant.

5. Once the torque reaches its maximum value Tmax at a critical slip s=sm, the torque decreases, while increases in slip due to reduction in air gap flux.

6. In this forward motoring region(o≤s≤1), of the torque-speed characteristics of fig, the region ((o≤s≤Sm)is said to be a stable

Q7b. Explain the working and construction of shaded pole single phase Induction Motor.

Ans:-

 

1.Fig.shows the construction of a shaded pole induction motor.
2. Usually it has a squirrel cage rotor.
3.Every pole is divided into two parts
First part:- The laminated pole has a slot cut across the laminations approximately one-third distance from one edge .The short circuited copper coil is placed around this small part of the pole. This copper coil is known as shading coil and this part of the pole is called as shaded part.
4.When the stators winding carries current , the main poles produce a flux Φm.
5.This flux Φm with the shading band acts as a shorted secondary ,stator winding being its primary circulating currents induced in the band produce another flux Φs.
6.Means when an alternating current is passed through the exciting(or field ) winding surrounding the whole pole ,the exciting(or field) winding surrounding the whole pole, the axis of the whole shifts from the unshaded part a to shaded part b. This shifting of magnetic axis is equivalent to the actual physical movement of the pole. Means magnetic flux divides into two fluxes Φand Φs  and hence rotor starts rotating in the direction of thgis shift i.e. from unshaded part to the shaded part.

Characteristics

1.Such a motor develops low starting torque and it has a low power factor.

2.However it is cheaper in cost and commonly used in practice.

3.Torque speed characteristics of shaded pole induction motor is shown in fig.

Advantages

  1. Cheap
  2. Simple in construction

Disadvantages

1.Low starting torque
2.Low power factor

Application

Table fans , blowers , washing machine, Refrigerator

Q8 a. Write comparison between squirrel cage and slip ring induction motor.

Ans:-

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 an 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 toque 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.

Q8.b. A 3  phase 16 pole ,induction motor having synchronous speed of 400 rpm and rotor speed of 352 rpm. Calculate
1.Frequency
2. Rotor frequency
3.Stand still frequency
4.slip
5.slip speed

Solution- p=16

Ns=400rpm

Nr=352 rpm

To find-1. F=?
2. F’=?
3. F’at s=1
4.s
5.sns

Step 1-

Ns=120 F/p
F = P  Ns/120
= 16 ×400/120
F= 53.33 Hz

Step 2-

F’=sF
s =Ns-Nr/NS ×100
=400-300/400 ×100
s = 0.12

Step 3-

F’=sF
F’=0.12× 53.33
F’ =6.39 Hz

Step 4-

F’ at standstill 
F’ at standstill =s F  (s=1,F=50 Hz )
F’ at standstill =1× 50
F’ at standstill =50 Hz

Step 5-

slip speed = s ×Ns
=0.12 × 400
slip speed =48 rpm

F= 53.33 Hz
F’ =6.39 Hz
F’ at standstill =50 Hz
s = 0.12
slip speed =48 rpm

 

 

 

 

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