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Model Answer
Subject Code: 17507
Page 1 of 35
Important suggestions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more importance.
(Not applicable for subject English and communication skills)
4) While assessing figures, examiner may give credit for principle components indicated in a figure.
The figures drawn by candidate and model answer may vary. The examiner may give credit for any
equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model answer.
6) In case some questions credit may be given by judgment on part of examiner of relevant answer
based on candidate understands.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.
Q.1 a) Attempt any THREE of the following:
12 Marks
i) State two advantages and two disadvantages of: 1) Group drive and 2) Individual drive.
(Any Two Advantages Expected: 1/2 each)
Ans: Advantages of Group Drive:1. Initial Cost –
A cost of single motor of large capacity is less than cost of number of small
capacity motors for same H.P.
e.g. cost of 10HP motor is much less than that of 10 no.of motors of 1HP
2. Diversification of load –
All the machines and tools may not work at a time, so we can select main
motor of slightly small capacity (HP) than the total requirements of individual
machines.
3. over load capacity –
Group drive has higher over load capacity. E.g. 100% overload on individual
machine would cause only 8 to 10 % overload on main motor.
4. Space required – Less
5. Maintenance cost –
Maintenance cost of single motor of large capacity is less than maintenance
cost of number of small motors of total HP.
km
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Subject Code: 17507
Page 2 of 35
6. Efficiency and Power Factor –
If group drive is run at nearly equal to full load than Efficiency and Power
Factor of group drive will be higher
Disadvantages of Group Drive:-
(Any Two disadvantages Expected: 1/2 each)
1. Flexibility:Flexibility is lost due to common shaft for number of machines.
2. Safety: -
It is less safe.
3. Reliability:Its reliability is less at the time of breakdown and maintenance of single
large motor, Because, all the machines operations are required to be shut down at the
time of breakdown and maintenance of single large motor.
4. Mechanical power transmission losses:Considerable power loss takes place for transfer of mechanical energy from
shaft to machine.
5. Speed control:Speed control of individual machine is difficult, it requires special
arrangement.
6. Addition / Alteration:Possibility of addition or alteration in existing system is limited.
7. Efficiency and Power Factor: –
If group drive is run at reduced load then Efficiency and Power Factor of
group drive will be less.
Advantages of Individual Drive:-
(Any Two Advantages Expected: 1/2 each)
1. Flexibility:It has more flexibility that is machine can be placed in any desired position
and can be shifted whenever needed.
2. Safety: -
Working conditions are more safe.
km
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Subject Code: 17507
Page 3 of 35
3. Reliability:It has high reliability, because breakdown of single motor causes only one
machine operation required to be shut down and not all machines.
4. Speed control:Speed control is easily possible.
5. Addition / Alteration:Possibility of addition or alteration in existing system is easily possible.
6. Efficiency and Power Factor: –
If it is run at full load than Efficiency and Power Factor of group drive will
be high.
If there is no load it can be stopped thus no load losses can be eliminated.
Disadvantages of Individual Drive:-
(Any Two disadvantages Expected: 1/2 each)
1.Initial Cost –
Initial cost is high.
2. Diversification of load –
Diversification of load on individual machine is not possible.
3. Over load capacity –
4. Space required –
Over load capacity is less.
More
5. Maintenance cost –
Maintenance cost is more as number of drives are more.
6. Efficiency and Power Factor –
If it is run at reduced load then Efficiency and Power Factor of individual
drive will be less.
7. Mechanical power transmission losses:Considerable power loss takes place for transfer of mechanical energy from
shaft to machine
ii) State the causes of failure of heating element.
Ans: Following of the different causes of failure of heating element:
i) Formation of hot spot:
(Each Causes: 1 Mark, Total: 4 Mark)
Hot spot on heating element is the point which is at higher temperature than
remaining heating element portion. So there is possibility of breaking of heating element
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Model Answer
Subject Code: 17507
at hot spot.
ii) Due to oxidization:
At high temperature material gets oxidized which may cause failure of heating
element.
iii) Due to corrosion:
If heating element is directly exposed to chemical fumes then there is possibility of
rusting of heating element which causes failure of heating element.
iv) Mechanical consideration/Failure:
Measure heating element alloy contain iron which is brittle. Due to frequent heating
& cooling of heating element, it may break (fail) due to small mechanical injury also.
Describe with neat diagram construction and working of high pressure mercury vapour
discharge lamp.
(1 Mark)
Ans: Figure mercury vapour discharge lamp :iii)
MM n
+ mercury
LJ
<
fti
: :
\
'
*
r
^1
Ooke
[now
:. be
n
_ —_.
.
0
Outer lube with
fluorescent coating
1rom inside
M
.
R
A, B C are electrodes
OR
is resistance
Space is
evacuated
Construction:-
(1.5 Mark)
It consists of an inner bulb generally of silicon, to withstand high temperatures.
The bulb contains a small quantity of mercury and argon.
It is protected by outer glass, this may be cylindrical or elliptical.
The space between the two bulbs is filled with nitrogen at a pressure of half
atmosphere.
The discharge tube has three electrodes, namely two main electrodes A and B and
one starting electrode.
The starting electrodes are connected through a resistance of about 10-30 k ohm to
the main electrode, located at the far end.
The electrodes are of tungsten wire helices filled with electron emissive materials,
usually barium and strontium carbonates mixed with thorium.
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Model Answer
Page 5 of 35
OR Student may write
The construction & connection diagram is as shown in figure. As per this
construction there are following components.
Choke: The choke is acting as the ballast. At the time of supply voltage variation of
current flowing through the inner tube is maintained constant to keep uniform light
intensity. Sometimes choke can be designed for to get the higher voltages & to apply
the inner tube of mercury vapour lamp.
Starting resistance/limiting resistance: Whenever current flows through the starting
resistance there is a I2R loss which is converted into heat. If the temperature of this
heat goes near about 6000C then there will be heating effect & inert gases ionization
will be start.
Auxiliary electrode & Main electrode: It is made by high resistive element. The
ionization is taking place through the inert gases whenever current flows from
auxiliary electrode to main electrode.
Inner Tube: The various inert gases e.g. Argon, Nitrogen etc with mercury powder
are filled in the inner tube at low pressure or high pressure.
Outer Tube: The function of outer tube is to make the vacuum surrounding the inner
tube to avoid thermal dissipation or to maintain 6000C surrounding the inner tube.
Power factor improvement Capacitor: The function of power factor improvement
capacitor is to improve the power factor 0.5 to 0.95
Working:-
(1.5 Mark)
When supply is switched on an initial discharge lamp is established in the Argon gas
between main electrode A and aux. electrode C
The heat is produced due to the discharge through gas which causes warming up of
inner lamp
Thus mercury gets vaporized and increasing its pressure and thus the light output.
It takes about 5-7 min. for the mercury arc to buildup &give full light output.
After 3-4 min. mercury vapors is greenish blue light.
If the supply interrupted, the lamp must cool down and the vapour pressure be
reduced before it will start. It takes approximately 3-4 min.
The efficiency of this type of lamp is 30-40 lumens/W.
Mercury lamps are available in 125W; 250W & 400W rating for use 250V AC
Supply.
1-
^^
.
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Model Answer
Subject Code: 17507
iv) Describe the static capacitor method of power factor improvement.
(1 Marks for any one figure, Vector diagram 1 Mark, Formula 1 Mark &
Ans:
advantages & disadvantages (Any one) – 1 Mark: Total 4 Marks)
E± —
u
i
,_
3A
In
-
+
d
•>
**
*
s
v
AJY
^
<
^J P
J J^fW
I*M K *C
'
?!
*1
'
V?
Observation:
From above vector diagram & power triangle calculations, if capacitor is connected
across load then following observations are observed.
S.No.
1
Parameter
Magnetizing current ( I )
Effect
Reduces
2
5
4
Power factor
Total current
Lagging reactive power (KVAr)
Improves
Reduces
Reduces
Advantages of Static Capacitor: (Expected any one)
1. Initial cost is low.
2. Low operating cost.
3. Low maintenance cost.
4. Losses are very less (less than 0.5% )than that of rated value
5. Noise less operation as it is a static piece.
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Subject Code: 17507
Page 7 of 35
6. Less space is required. Therefore can be installed near load.
7. Greater reliability.
8. KVAr (leading) rating can be adjusted easily as per load condition.
Disadvantages of Static Capacitance: (Expected any one)
1. It has short life as compared to synchronous condenser.
2. Capacitors get easily damaged if the voltage exceeds than its rated value. Once the
capacitors are damaged its repair is uneconomical.
3. When capacitor is switched OFF then precaution is taken before making it ON. In
between OFF and ON time, time should be kept to discharge the capacitor, otherwise
capacitor may fail.
4. Switching current of capacitor is many times that of rated current; therefore cable size
should be double of the normal current carrying capacity, so its cost increases.
5. When there is no load or system is lightly loaded at that time capacitor bank must be
made OFF otherwise voltage across transformer increases.
Q.1b) Attempt any ONE of the following :
06 Marks
i) Describe any six factors governing selection of a motor for a particular application.
(Any Six Factors Expected Each Point: 1 Mark: Total :6 Marks )
Ans:
Following Factors governing / or are considered while selecting electric drive
(Motor) for particular application:
1. Nature of supply:
Whether supply available is
AC,
Pure DC
OR rectified DC.
2. Nature of Drive (Motor):
Whether motor is used to drive
Individual machine
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Model Answer
Page 8 of 35
OR group of machines.
3.
Nature of load:
Whether load required light or heavy starting torque
OR load having high inertia, require high starting torque for long duration.
OR Whether load torque increases with speed (T N)
OR decreases with speed (T 1/ N )
OR remains constant with speed (T = N)
OR increases with square of speed (T N2)
4.
Electric Characteristics of drive:
Starting,
running,
speed control
Braking characteristics of electric drive should be studied and it should be
matched with load requirements.
5.
Size and rating of motor:
Whether motor is short time
OR continuously
OR intermittently running
OR used for variable load cycle.
Whether overload capacity, pull out torque is sufficient.
6. Mechanical Considerations:
7.
Types of enclosure,
Types of bearing,
Transmission of mechanical power,
Noise
load equalization
Cost:
Capital,
running
And maintenance cost should be less.
ft
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Model Answer
Page 9 of 35
ii) Compare A.C. welding with D.C. welding on the basis of 1)Equipment, 2)Operating
efficiency, 3)Cost, 4)No-load voltage, 5)Heating and 6)Arc stability
( Each Point: 1 Mark: Total :6 Marks )
Ans:
Sr.No.
Points
AC Welding
DC Welding
Welding Transformer DC differential Compound
1
Supply equipment
Generator, or Rectifier
used
High, 85%
Low, 65%
2
Efficiency
Low
High
3
Cost:
72 to 100 volt
50 to 60 volt
4
No –Load Voltage
Not Uniform
Uniform
5
Heating
Use of series Reactor
D.C Differential
6
Arc Stability
OR Less arc stability component. Generator has
dropping characteristics.
OR More arc stability
Q.2
Attempt any FOUR of the following :
16 Marks
a) State the factors to be considered for selection of shape and size of the car of elevator.
Ans: The size and shape of elevator car depends are following two factors:
( Each factor : 2 Marks, Total : 4 Marks)
i) No. of passenger to be carried: While selecting the size of car it is a usual practice to
allow.
A Space of 2 Sq.fit/ person.
Average weight of passenger is assumed 68 kg/person.
Thus the maximum load capacity of elevator is considered 34 kg/sq.ft
There should be wide frontage and shallow depth
ii) Limitation in the building design:
Shape of elevator depends on space available in building.
b) State two advantages and two applications of dielectric heating.
Ans: Advantages of Dielectric Heating:( Any Two advantages Expected: 1 Mark Each)
1) This is only method for heating non-metallic material.(Di-electric)
2) Bad conductor of heat material can be heated by this method ( for e.g. porcelain)
3) As no flame, arc appears in the process, so material like plastic, wood cotton etc
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Model Answer
Page 10 of 35
heated safely.
4) As heat is produced inside material to be heated due to dielectric loss, so efficiency
of such type of heating is high.
5) Time required for heating is less as there is no heat transfer loss. Uniform heating
is possible.
6) Temperature control is easy by simply controlling voltage & frequency.
7) Low attention is required.
8) Clean and convenient method.
Applications of Dielectric Heating:-
(Any Two Application Expected: 1 Mark Each)
1) In food processing industry, dielectric heating is used for Baking of cakes &
biscuits in bakeries. Cooking of food without removing outer shell (eg-boiled egg)
and pasteurizing of milk.
2) For Rubber vulcanizing.
3) In Tobacco manufacturing industry for dehydration of tobacco.
4) In wood industry for manufacturing of ply wood.
5) In plastic Industry for making different types of containers.
6) In cotton industry for drying & heating cotton cloths for different processes.
7) In tailoring industry for producing threads.
8) For manufacturing process of raincoats & umbrellas.
9) In medical lines for sterilization of instruments & bandages.
10) For heating of bones & tissues of body required for certain treatment to reduce
pains & diseases.
11) For removal of moisture from oil.
12) For quick drying gum used for book binding purpose.
13) In foundry for heating of sand, core, which are used in molding processes.
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Model Answer
Subject Code: 17507
Draw a typical speed time curve for main traction line service. Show different time
periods on it.
Ans:
Typical speed time curve for main traction line service:
(4 Marks)
c)
f >*«* Kunr* ,nJ
p e i *ct
) jpr*d rurv*
' V( Krrilhy
Ruhn'r**!***
+
cl
P
6 *
fQOkS *"~*
*
^
PfT ocf
A
a
Ar
o
i
—
vma v
fircl
pfr'OO
”
J3
A
Uis±
ft I*
h'rmP CT )
3
OR
AfflO
p
Frtf Runr» r’j
H-
Pfji'o 4
A
$ peed
\j(tml >T)
*
8
Vftlfjt
>
••
<»
/
h :
i
—
Cl
<»
.Dt
T imeCt)
C
Compare electric locomotive over diesel locomotive on the basis of: (i) Centre of
d) gravity, (ii) Running / maintenance cost, (iii) Starting time and (iv) Regenerative
braking.
Ans:
( 1 Mark each point, Total 4 Points)
Sr.No.
Point
1
Centre of gravity
2
Running/Maintenance
cost
3
Starting Time
4
Regenerative Braking
Electric locomotive
Lower level
Less
Diesel locomotive
Higher level
More
Quick /less/start at any
time
Possible
Quick/It take more
time
Not possible
km
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Page 12 of 35
Describe the following terms with respect to traction mechanics: (i) Average speed and
(ii) Schedule speed.
Ans:
( Each Description: 2 Mark, Total:4 Marks)
e)
i) Average Speed: - It is defined as distance covered between two stops divided by actual
time of run is known as average speed. OR
Vav
3600D
T
Where T = is actual time of run in sec OR
Average Speed
Dis tan ce between stops or stations
Actual time of run
ii) Schedule Speed: - It is defined as distance covered between two stops divided by
schedule time is known as schedule speed. OR
Schedule Speed
Dis tan ce between stops or stations
( Actual time of run) (Stop time)
Schedule Speed
Dis tan ce between stops or stations
Schedule time
OR
Q.3
Attempt any TWO of the following :
16 Marks
What is electric braking? State the advantages of electric braking. Compare Rheostatic
a)
and Regenerative braking.
Ans:
Electric Braking:
( 2 Mark)
It is necessary to stop the vehicle when mechanical working is over or when required
within reasonable time. OR To reduce the speed of train electrical system is used for braking
e.g. Plugging, dynamic braking & Regenerative braking.
Advantages of electric Braking:
( Any Four advantages Expected: 1 Mark each, Total: 4 Mark)
Following are the advantages & disadvantages of electrical braking over
mechanical braking system.
1. It is most reliable braking system.
Because in mechanical braking heat is produced at break block & break shoes,
which may be source of failure of break.
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Page 13 of 35
In Electrical braking (dynamic) heat is produced at convenient place (external
rheostat) which is not harmful to braking system
2. Breaking actuation time is small as higher value of braking retardation is obtained.
3. Electrical braking is smooth & gradual.
Where as if mechanical breaks are not correctly adjusted then there are
chances of sudden braking which is discomfort able to passenger.
4. Life of braking system is more.
Because mechanical braking provides metal dust due to friction, No such dust
is formed in electrical braking.
5. There is less wear & tear of brake shoes, break block etc. so there is less
maintenance cost.
6. Higher speeds are possible even when train is going down the gradient, as breaking
system is reliable.
7. Trains having heavy loads can be stopped even when train going up the gradient.
8. Higher speeds of train is possible as braking system is reliable so pay load capacity
increases.
9. In case of electric regenerative braking we can utilize 60 to 80% of kinetic energy
to generate electricity which is not possible with mechanical braking.
Compare Rheostatic and Regenerative braking.
(Any Two point expected: 1 Mark each, Total: 2 Mark)
S.No
Rheostatic Braking
1
In this method at the time of
braking armature is
disconnected from supply &
connected across external
resistance at that time field
winding supply remain as it is.
2
Generated EMF is less than
supply voltage
3
Rheostatic braking is used in
any track position.
4
5
It is simple in operation
Kinetic Energy is wasted
Regenerative Braking
In this method at the time of braking
motors are made to work as a generator &
electrical energy is fed back to supply line.
Generated EMF is more than supply
voltage
Regenerative braking is possible when the
train going down the gradient or
Regenerative braking is not possible on
plane track
It is complicated in operation
80 % kinetic energy is utilized
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Model Answer
Subject Code: 17507
Describe the construction and working of direct arc heating furnace. Compare direct
b) arc furnace with indirect arc furnace on the basis of: (i) Temperature, (ii) Size, (iii)
Applications and (iv) Power requirement.
Ans: 1. Direct Arc Furnace:
Direct arc furnace explained on following points: (Any one Figure expected)
a) Bottom Conducting direct arc Furnace
b) 1-ph direct arc Furnace
—
—L* »a
\ ~-
- *V r\ tQd4
*
*
fin
r»
Ipod
Ptoc* lOC»
M
^
1
r •.
Cfofgg
rtyYttW' j
I
'
frUrrof *
ViflTO-
i*5
M-i& j
V
'
riu ' w
( TFVrfrof
Irxul ' O^ J
^
Construction of Arc Furnace:
Encfrina
O'tiof
OR
( Construction: 2 Mark)
Direct arc furnace consists of following main parts.
i) Heating Chamber:
Heating chamber is spherical in shape for providing minimum refractory material
and for easy rocking.
It has two openings, one is for pouring the charge and another is for taking out
molten metal. Both openings are closed during operation.
The wall of heating chamber is made from refractory material to withstand
heating chamber at high temperature and also to reduce heat loss.
ii) Furnace transformer:
iii) Series reactor:
iv) Automatic current regulator:
v) Circuit breaker:
vi) Rocking arrangement
vii) Electrode:
viii) Connecting lead:
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Page 15 of 35
Model Answer
Working of Arc Furnace:
( Working: 2 Mark)
When very high voltage is applied across any two electrodes separated by small air
gap then air between two electrodes gets ionized and ionized and as air is conducting,
so current starts flowing from one electrode to another electrode in the form of spark
(arc). OR The arc is produced between the electrode and charge and due to this heat
is developed.
This arc produces heat energy which is utilized for melting the charge.
Once arc is struck between two electrodes then low voltage is sufficient to maintain
the arc.
Compare direct arc furnace with indirect arc furnace on the basis of:
( 1 Mark each point: Total 4 Marks)
Sr.No.
Point
1
Temperature
2
Size
3
Applications
4
Power requirement
Direct Arc Furnace
High
More
1. Used for continuous and
large production of high
quality steel. 2. For Ferroalloy manufacturing
Less
Indirect Arc Furnace
Less
Less
It is suitable for melting
non-ferrous metals for
e.g. Copper, Brass,
bronze, gun metal etc.
More
(i) Suggest suitable electric drive for following application: 1) Paper mills, 2) Stone
c) crusher, 3) Textile mills and 4) Electric traction.
Ans:
( Each Drive: 1 Mark, Total: 4 Marks)
Sr.No
Application
1
Paper mills
2
Stone crusher
Suitable Electric Drive ( Each anyone expected)
DC Shunt motor, Synchronous motor, Scharge Motor
D.C series Motor, 1-Ph AC Series Motor, Slip-ring
induction motor, Double squirrel cage I.M
3
Group drive, Scharge Motor
Textile mills
4
Individual drive operated in parallel / D.C series
Electric traction
Motor, 1-Ph AC Series Motor, Multiple speed motor
OR
i) Paper Mills: - Group Drive
ii) Stone Crusher: - Individual drive
iii) Textile Mills: - Group Drive
iv) Electric Traction: - Individual drive
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Model Answer
Subject Code: 17507
c) (ii) Draw the curve and estimate suitable H.P. of motor having following duty cycle:
1) Rising load from 200 to 400 H. P.
- 4 minute
2) Uniform load of 300 H.P.
- 2 minute
3) Regenerative braking from 50 to zero H.P.
- 1 minute
4) Idle for
- 1 minute
Ans:
i) Load rising from 200 to 400 HP :- 4 min
ii) Uniform load of 300 HP
:- 2 min
iii) Regenerative braking from 50 to zero : 1 min
iv) idle for
: 1 min
11
c
400
i
i
i
i
'
\
1
1
t
H P 200
1
1
t
*
4
\
— -*
-
«
6
1
HP
Where,
1
/ T~t
3
H
2
1
or Equivalent fig----------------(1/2 Mark)
2
2
H1 H 2 H 2 t1 H 3 t2 1
2
3
H 4 t3
T
---------------------- (1 Mark)
T = t1 + t2 + t3 + t4
T= 4+2+1+1
T = 8 min.-------------------------------------------------------------------- (1 Mark)
HP
1
HP
HP
1 H 12 H1 H 2 H 2 2 t 1 H 3 2 t 2 1 H 4 2 t 3
3
3
8
3
200
2
200 400 4002 4 3002 2 1
8
3
(1/2Mark)
502 1
------------- (1/2 Mark)
I
1662500
24
HP 263 HP -----------------------------------------------Answer------------------ (1/2 Marks)
Nearest standard rating of motor is to be selected.
km
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Subject Code: 17507
Page 17 of 35
Q.4 a) Attempt any THREE of the following:
12 Marks
i) Define electric welding. State the factors deciding selection of electric welding system.
Ans: Meaning of electric Welding:
(1 Mark)
It is the process of joining two similar or dis-similar metals by application of
heat with or without application of pressure and addition of filler material.
Following Factors are considered while selecting of electric welding system:
( Any Three factors Expected:1 Mark each: Total 3 Marks)
1) Type of Material:Whether similar metal is to be welded or dis-similar metal is to be welded.
2) Property of Material:Whether ferrous or non-ferrous metal is to be welded.
3) Thickness of job:It is also depends on thickness of job to be welded.
e.g. for thick material- Arc welding is used. And for thin material –
Resistance welding is used.
4) Temperature required:Whether job required high or low temperature to weld the job.
e.g. For high Temperature - Arc welding is used. And for low Temperature
– Resistance welding is used.
5) Pressure required:If job is need of pressure at the time of welding in that case resistance welding is
used. And if pressure is not required Arc welding is used.
6) Type of Supply Available:Whether AC or DC or both supply are available.
7) Application:In case of mass production, resistance welding is used & for repair work Arc
welding is used.
:i= »i.
&Z
J
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Model Answer
Subject Code: 17507
ii) State and explain laws of illumination.
Ans: Laws of illumination: (Each Figure 1 Mark & Statement of each law 1 Mark, Total: 4 Mark)
a) Inverse Square Law :Radius
3x
Radius m 2 X
.
Radius
x
\
\
S ma c
A
sVt
3
/
X
I
/
-/
/
/
/
i
/
/
/
or Equivalent fig.
Illumination is inversely proportional to the Square of distance between source and
1
plain of the surface. E 2
r
b) Lambert’s cosine law :-
*
ii
r
D
or Equivalent fig.
The illumination of a surface is directly proportional to cosine of angle made by
the normal to the illuminated surface with the direction of the incident flux.
EB E A cos3 1 ,
EC E A cos 3 2 ,
E D E A cos 3 3 and so on.
Define the following terms related to illumination: 1) Utilization factors 2) Maintenance
factors 3) Depreciation factor and 4) Luminous efficiency
Ans:
(Each Defination: 1 Mark, Total: 4 Marks)
1) Utilization factor:iii)
It is defined as the ratio of total lumens reaching the working plane to the total
lumens given out by the lamp.
km
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Model Answer
Page 19 of 35
2) Maintenance factor:It is defined as the ratio of illumination under normal working conditions to the
illumination when everything is clean. OR
Ma int enacefactor
Illu min ation under normal working condition
Illu min ation under every thing is clean
3) Depreciation Factor:
It is defined as the ratio of initial illumination to the ultimate maintained
illumination on the working plane. OR
1
Depreciation factor
Ma int ennace Factor
4) Luminous Efficiency:It is ratio of energy radiated by the source in the form of light to the total
energy radiated by the source.
iv) State four requirements of Tariff.
Ans: Following are the requirements of Tariff:
(Any Four points expected: 1 Mark each: Total: 4 Marks)
i) It should be easy to understand to consumer.
ii) Easy to calculate.
iii) Tariff should be attractive; It should not be too high or too low. It should be
reasonable.
iv) Tariff should be economical as compare to other types of energy sources.
v) Tariff must be fair, so that different types of consumers are satisfied with rate of
electrical energy charges.
vi) Tariff should be framed into two parts i.e. fixed charges + running charges.
vii) While calculating tariff it should cover all expenses and reasonable profit,
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Model Answer
Subject Code: 17507
Q. 4 b) Attempt any ONE of the following :
06 Marks
a) Describe with neat sketch, construction and working of seam welding machine.
Ans: Sketch of Seam welding:
(1 Mark)
— -—
'i
;>
. 4, nFft
-
v 3 * ••’,
,
iv 1
d
m n’
.
>
.
4i
mj
. /ruin
iinniuu
— 4—
ffliwrf
RE
sE
? jfpy
^
- *
aln i:|{,
-
i
3Y
^
H|opfct E «TJ—
- Wifi < jjpt
'
—
4
''
r» >1 J
uiiJ
U-
l
* *
yff
—
\
v
*
|VtldiiM
|
I
2
TVaniWi
sr
J
wtld
>
SMl S !sl
W
-frnH
r
'Jin
OR
(loo to 4 C 5
sp s pvLtoitti - o . js hfrmtttL / min')
^
Construction:
(2.5 Mark)
Transformer used for seam welding is designed for low voltage and high current
secondary.
Transformer is oil cooled
There are two electrodes in this type beam or roller type electrodes are used.
Working:
(2.5 Mark)
Job is kept in between two electrodes under pressure. This pressure is kept constant
throughout.
In this type intermittent current is used, it means current is ON for definite time and
OFF for another time interval with the help of timer.
If current is continuously passes then heat produced may cause burning of job.
Heat is produced due to I2R losses where ‘R’ is the contact resistance.
This heat is utilized to obtain welding temperature (to become a plastic state)
When welding temperature is reached supply is cut down and external pressure is
applied simultaneously across the job to complete weld.
km
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Model Answer
Page 21 of 35
What are the different tariffs used by electricity supply authority? Describe any two of
them in brief.
Ans: Types of Tariff:(Any three types are expected: 1 Mark each; Total 3 Marks)
b)
i) Flat-demand Tariff
ii) Simple-demand Tariff or Uniform Tariff
iii) Flat-rate Tariff
iv) Step-rate Tariff
v) Block-rate Tariff
vi) Two-part Tariff:
vii) Maximum demand Tariff
viii) Three-part Tariff
ix) Power factor Tariff :- a) KVA maximum demand Tariff
b) Sliding Scale Tariff or Average P.F. Tariff
c) KW and KVAR Tariff
x) TOD (Time of Day) Tariff
Explanation:
(Any Two Explanation Expected: 3 Marks)
1) Flat Demand Tariff:
It is used where energy consumption is fixed per day i.e. where load is fixed and is
used for fixed hours.
e.g. Street lighting, Road Signal system and advertising board.
In this type no energy meter is connected, so meter reading, billing, accounting,
Stationary, bill distribution and collection etc. expenses are eliminated and also
save time for billing.
Calculation:
Energy consumed per day = Load (KW) X No. of working hrs. (H)
Energy consumed per Month =
Total energy consumed in one year 12 Month Load
Energy consumed in one year = Energy consumed per day 365 days
2) Simple or uniform demand tariff:-
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Page 22 of 35
In this type of tariff cost of energy charges is calculated on the basis of actual
energy consume energy meter is connected in consumer premises.
Calculation:
Energy Bill ( Current reading Pr evious reading )
Energy Bill (Total energy consumed ( KWH) Tariff ( Rs) / unit
3) Flat Rate Tariff:
In this type of tariff there are two energy meter in one premises.
One energy meter is for lighting circuit load and another meter is for power
circuit load.
Tariff rate for lighting and power load are different.
Tariff rate for lighting is higher than tariff for power load.
Disadvantages:
1. Two energy meters are required.
2. Meter reading & Billing cost increases
Calculation:
Energy Bill for lighting circuit ( Current reading Pr evious reading) Tariff (Rs) / unit for lighting circuit
Energy Bill for Power circuit ( Current reading Pr evious reading) Tariff (Rs) / unit for Power circuit
4) Step rate Tariff:
In this tariff there are steps for unit’s consumption and cost/unit is less for more
consumption of unit.
The main disadvantage of this tariff is that the consumer unnecessary wastes the
power to enter the next stage.
For example 1) Step- I- Rs.2/KWH :- If consumption not to exceed 50 unit
2) Step-II – Rs. 1.75/KWh:- If consumption not to exceed 200 unit.
3) Step-II- Rs. 1.50/KWh:- If consumption exceeds above 200 units.
5) Block Rate Tariff:
In case of block rate tariff there are blocks of units consumed and each block
tariff rate/unit (KWH) is different.
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Model Answer
Page 23 of 35
If generation is less than utilization than tariff rate/unit in each block goes on
increasing and vice versa.
S.No
E.g.
Block of units consumed
from…… To
1
2
3
4
5
0 to 100 unit
101 to 300 unit
301 to 500 unit
501 to 1000 unit
Above 1000 unit
If, Generation is
less than utilisationTariff (Rs)
3
5
7.15
8.30
8.55
If, Generation is more
than utilisation-Tariff
(Rs)
8.55
8.30
7.15
5
3
6) Two Part Tariff: In this type of tariff energy bill is split into two parts.
Only one energy meter is used to measure no. of units consumed it recovers a
fixed charge which depends on load (KW).
This type of tariff system is used for residential and commercial consumers.(up to
20 KW)
This type of tariff is not used for industrial consumers.
ENERGY BILL= FIXED CHARGE +RUNNING CHARGE
Advantages:
1. It recovers fixed charges which depends on load (KW), so it automatically
recovers capital investment of Supply Company
Disadvantages:
1. The consumer has to pay fix charges per month whether he has to consume or
not consume the electrical energy.
Application:
1. This type of tariff system is used for residential and commercial consumers.(up
to 20 KW)
2. This type of tariff is not used for industrial consumers.
7) Maximum Demand Tariff/KVA MD Tariff: It is similar to two part tariff except that maximum demand (KVA) is actually
measured by installing maximum demand (in KVA)
M.D. Meter is installed in the premises of consumer, in addition to energy meter.
km
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Model Answer
Page 24 of 35
Industrial consumer is trying to improve power factor to reduce maximum demand
charges.
This type of tariff is applicable to industrial consumer/H.T. consumer.
Maximum Demand Tariff =
M.D. ( KVA ) Rs ' X ' {Number of units ( KWH) Actual consumer} Rs ' Y"
8) Three part Tariff: Fixed charges per month depend on connected load.
Semi-fixed charges depend on KVA maximum demand.
Running charges depend on actual energy consume.
This type tariff is used for HT consumer.
ENERGY BILL= FIXED CHARGE+SEMI-FIXED CHARGE+RUNNING CHARGE
9) Power Factor Tariff: The tariff in which P.f. of industrial consumer is taken into consideration.
Power factor tariff is used for industrial consumer /H.T. consumer.
If the P.F. of consumer is less than P.F. declare by Supply Company (say below
0.92 Lag.) than penalty will be charged in energy bill.
If The P.F. of consumer is more than P.F. declare by Supply Company (say above
0.96 lag.) than discount will be given in energy bill.
There are three types :
a) KVA maximum demand Tariff:
It is similar to two part tariff except that maximum demand (KVA) is actually
measured by installing maximum demand (in KVA) meter.
M.D. Meter is installed in the premises of consumer, in addition to energy
meter.(Now a days there is only one meter which measures all parameters)
Industrial consumer is trying to improve power factor to reduce maximum demand
charges. Since KVA I 1/ pf
This type of tariff is applicable to industrial consumer/H.T. consumer.
b) Sliding Scale Tariff or Average P.F. Tariff:
km
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Model Answer
Page 25 of 35
In this case an average power factor say in the range of 0.92 lag to 0.96
lag is taken as a reference by electric supply company
If power factor of industry falls below (say 0.92 lag) this reference
power factor then penalty will be charged or suitable additional charges
will be added in the energy bill.
On the other hand, if the P.f. of consumer is above (Say 0.96 Lag) this
reference P.f. then there is discount as suitable amount reduced from the
energy bill.
So consumer is trying to improve P.f. to get discount in energy bill. So
overall P.f. of power system increases.
As usual consumer has to pay actual energy consumption charges
c) KW and KVAR Tariff:
In this type both active (KW) & reactive power (KVAr) supplied are
charged separately and actual energy consumption charges,
A consumer having low power factor draw more reactive power and
shall have to pay more charges and vice-versa.
So consumer is trying to improve power factor to reduce KVAr charges
in energy bill, so power factor of power system increases.
Energy Bill {Rs ' A ' (KW) Ch arg es} {Rs ' B' (KW) Ch arg es} {Rs ' C' (KWH) Ch arg es}
10) Time of Day (TOD) Tariff or OFF-load Tariff: TOD energy meter is installed in the HT consumer premises.
This meter is specially designed to measure energy consumption w.r.t. time.
This type of tariff is such that energy consumption charges/unit are less at during
OFF-load period
There is a higher tariff rate energy consumption charge during peak-load period.
This type of tariff is introduced to encourage industrial consumers to run their
maximum load during OFF-load period.
Due to this load factor of generating station increases. Such type of tariff is used
for HT Consumer.
e.g.
S.No
Block
Rate / KWH Rs
Remark
1
9.00 am to 12.00 noon
Rs. 6.00 per unit
Peak load period
2
12.00 noon to 6.00 pm
Rs. 5.00 per unit
OFF load Period
3
6.00 pm to 10.00 pm
Rs. 6.00 per unit
Peak load period
4
10.00 pm to 9.00 am
Rs. 5.00 per unit
OFF load period
Q.5
Attempt any FOUR of the following :
16 Marks
a) A 400 V, 50 Hz, 3-phase line delivers 200 KW at 0.7 p.f. lagging. It is desired to improve
km
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Subject Code: 17507
Page 26 of 35
the line power factors to unity by using shunt capacitors. Calculate value of capacitance
of each unit if they are connected in delta.
Ans: Volt : line volts V = 400V, f= 50 Hz P= 200kW cos1 =0.7
cos 2 =1
1 45.5729 0
Cos 1 0.7
tan 1 tan 42.5729 0
tan 1 = 1.020 ----------------------------------------------------------- (1/2 Mark)
tan 2 = 0 ------------------------------------------------------------- (1/2 Mark)
Q1 = P tan 1
= 200 x 1.020
= 204 KVAR
Q2= P tan 2
= 200 x 0
= 0 KVAR
QC = Q1- Q2
= P tan 1 - P tan 2
--------------------(1 Mark)
= 204-0
= 204 KVAR
---------------------------------------------------- (1/2 Mark)
Capacitor when connected in Delta:-
C per phase
QC
-------------------------------------------------------- (1 Mark)
3 V2
C per phase
204 10 3
3 2 50 400 2
C per phase
204 10 3
3 50.265 10 6
C per phase 1.3528 10 3 F ------------------------------------------------ (1/2 Mark)
"Precautions are taken for ill effects due to negative resistance characteristics of an
electric arc in electric arc welding." Justify the statement.
Ans: Effect of characteristics:
(2 Mark)
b)
Characteristics of Arc: Arc is conducting and it has negative temperature coefficient of resistance. i.e. its resistance decreases as temperature increases.
Effect of characteristics: This decreasing resistance will increase current further
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Model Answer
Subject Code: 17507
due to this arc blow increases i.e.arc does not remain steady. It goes on increasing
and increasing. Due to this job may burn.
Precautions:-
(2 Mark)
In case of D.C Welding Stability of Arc: To stabilized arc D.C differential
compound generator is used. It has dropping voltage characteristics, i.e, as load
increases voltage suddenly decreases. Due to this characteristics arc remains steady.
In case of A.C Welding Stability of Arc:
To stabilize the arc in case of metal arc welding series reactor is used.
c) Draw a labeled block diagram of A.C. electric locomotive.
Ans: Block diagram of A.C. electric locomotive
-
ftl ftr w
— sftr—
f ftnlot i
~
“
^m^mrnrn
ll
d f a o1 ucn^ aL- Ai- lnfnmn» N < ^
Y
rA ~.
H
Tr • T» o it k a oa f T
;
1
—
*
jiJxjpc
'
(4 Marks)
-J L
^
I
ILL
H
1 '
-
mo V rvt
k
E
7
-rk~
_
_
r
^=
W! L
i
:
*
—
-
—
- Of
—
COQJL
si
f
or
equivalent figure
d) Describe any four points that proves the suitability of D.C. series motor in traction system.
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Model Answer
Subject Code: 17507
Ans:
(Any Four Point Expected 1 Mark Each Point, Total: 4 Marks )
DC series Motor is suitable for traction purpose because of following points:
1)
DC series motor has High torque at low speeds, low torque at high speeds, this is the
basic requirement of traction unit.
2) DC Series motor robust in construction and capable to withstand against continuous
vibration.
3) DC series motor weight is 1.5 times less than 1-Ph AC series motor for same H.P.
4) DC Series motor has high starting torque.
5) DC Series motor has high rate of acceleration and retardation.
6) DC Series motor variable speed motor.
7) DC Series motor speed-torque characteristics are such that as torque increases speed
decreases. (Due to this characteristics motor is protected against overload)
8) DC Series motor maintenance cost is less.
9) When DC series motor are running in parallel the all motors share almost equal load.
10) Torque obtained by DC series motor is smooth and uniform, so it improves riding
quality.
e) Describe with neat sketch series-parallel control of traction motor
Ans: With neat sketches series control of traction motor:
(2 Mark)
-—
—
I
“
-
- —1— -1— — I
.
VV V' 1
rttp
r.
L
a
V V/V
—
'
wv
[
w
1
wv*
Sk
DCS Up p I y
5|
h.
F»1
J
SM/ > plu
ii
>
r* \
nj
O C
Su J' p / y
B
ttp J5
O c.
— J— -I— ^ — ^ ^ ^
i wJ
i*
-7
.Step
*
wv'
VW
n
'
'
- PT
J
C
4w _ <
o
I
r -'7
U
Up 6
^
B S
r
Wi
OC
+rt *1
^ ^ — J.
L
'VWV-
L
i
—.
'V V V'
Supply
*
f
L
—
—
—
^^
lvwv 1
-
c
» c
t> c
DC
Sv< p
ply
«
\A^
Sup ply
^^
— L^ ^
yyi
^ — - yv I
S»A.p pl
-
<yvv\
fy)
rw ]
^
or equivalent figure
——
nr*TT'>
r
JM
J>
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Model Answer
Subject Code: 17507
With neat sketches parallel control of traction motor:
* rvx-c
^
^
—.
jy> a>
T^\
. cJLf'prvf
DC
S
Jb*> AAA-snr^-!r* ^_^
— L^
.—
r
»
k\
_
-
,
Su^ p
L. A (
VVA-L-^AT
Supply
^^
JXWIJMVVJ-4/V
I
c
^
_
^rrm
3-
1
r-
— rrlrri.
_
_
S M P p|
H
~
— Cmnnr^
tn
p| v/
0<
t.f
^^
^^
n Wt Ml
1
& <-
>-1
— rr*rrr^_
c
t
^, I
jAJfyV^\-NfVVV- j-AVv
anrrm
Supply
n
DC
—
' ja
LCvwLA-vJ-« * v-f-
B5 C
<
rO
5
rm
=•1
y J—
N/\
- UP p lM
T 't
'W^-L-VWV-XJW^
(2 Mark)
^
Si
SL
0C
S M_p p
or equivalent figure
Q.6
Attempt any TWO of the following :
16 Marks
Model Answer
WINTER– 2014 Examinations
Page 30 of 35
MAHARASHTRA STATE BOARAD OF TECHNICAL EDUCATIOD
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Subject Code: 17507
-
1
i
9
T
-
'
(1 Mark)
*
/
r*
or Equivalent fig.
rs
Jl/i A
secondary short circuited.
metal remains in narrow ‘V’ notch from previous operation, which will help to keep
1) As furnace has narrow ‘V’ shape crucible at bottom. So small quantity of molten
Advantages Ajax Wyatt’ furnace: (Any Four Point Expected : 1 Mark each, Total: 4 Mark)
Where, R = Resistance of charge & I secondary current
utilized to melt the charge.
this current is responsible to produce heat in charge due to I2R losses. This heat is
As charge forms a close circuit (secondary) heavy current flows through charge
charge.
electromagnetic induction emf will be induced in secondary winding that is in the
winding i.e. charge through magnetic core. Hence according to faraday’s law of
winding which creates alternating flux in magnetic core this flux links to the secondary
When AC Supply is given to primary winding current flows through primary
charge which is to be melted is kept in crucible.
which is wound around one limb of magnetic core but secondary winding is actually
It is based on principle of transformer. In this type of primary winding is as usual
(3 Mark)
1
s
.it
B
i
;i
\
13
I
*
i
Up
. .
8
S
rf
Draw the neat sketch of Ajax Wyatt Furnace. Describe its operation. State its four
OR
II5 !I
-
.<
i
».
isa
i
advantages.
£
Sf
a)
:
Ans: Neat sketch of ‘Ajax Wyatt’ vertical core furnace:
UP
C.
Operation of Ajax Wyatt’ vertical core furnace:
' fa 3 £
ikf
HI
'
-i W \
i Vi
5-
£ !s
J 2
r
v
fS)
km
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Model Answer
Subject Code: 17507
So no extra care is required to start the furnace
2) Magnetic coupling between secondary & primary winding is better because both
windings are on central limb of magnetic core
So there is less leakage flux, Hence leakage reactance is less, so power factor is
better than horizontal crucible direct core type induction furnace.
3) Due to pinch effect in ordinary core type induction furnace there are chances of
temporary interruption in secondary circuit when current density exceeds above
500A/mm2
But in this type of induction furnace there are no chances of interruption in
secondary circuit even if current density exceeds 500A/mm2 because tendency
of weight of charge keep them in contact due to narrow ‘V’ shape.
So we can increase current density above 500A/mm2 to obtain more heat in
less time.
4) Vertical crucible is always better than horizontal crucible for pouring and taking out
the metal. Also space required is less.
5) As heat is produced directly in the charge there is no heat transfer loss. So efficiency
of furnace is more.
6) As heat is directly produced in the charge time required for melting metal is less. So
energy consumption is less.
7) As current is directly induced in the charge there is automatic stirring action taking
place in the charge due to electromagnetic forces developed in the charge due which,
Through mixing of molten metal is possible.
Uniform heating is possible
8) Temperature can be controlled easily.
The distance between two stations is 2 km. It is desired to have scheduled speed of 40
km/hr. with duration of stop of 20 sec. Assuming trapezoidal speed - time curve
b) calculate: (i) The maximum speed required when the acceleration is to be limited to 1.2
km/hr./sec. and braking retardation to be 3 km/hr./sec. and (ii) The distance covered
during acceleration and retardation.
Ans:
Given Data :D = 2 KM,
Schedule speed (Vsch) = 40KM / Hr,
Acceleration ( ) = 1.2 Km/Hr/sec;
Stop Time = 20 sec.
Retardation ( ) = 3 Km/Hr/sec.
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Model Answer
Subject Code: 17507
Trapezoidal speed time curve :*
ted
Fr » # Run"'*' *
RfJJ a I
Z Qpppcl
^
\f (t / r> / / yr )
Wr» <* *
-
<
»
Vsch
-n _ < 9
r
H * Pat < /
e
- -*->
‘]
p* <>
T im / Ct )
f
—
(
or Equivalent fig.-(1 Mark)
3600 D
----------------------------------------------------------- (1 Mark)
Schedule Time
Schedule Time
3600 D
Vsch
Schedule Time
3600 2
40
Schedule Time
7200
40
ScheduleTime 180 sec.----------------------------------------------------- (1/2 Mark)
SchudeleTime Actual Time of Run Stop time
Actual Time of Run ScheduleTime Stop time
Actual Time of Run 180 20
Actual Time of Run 160 sec.--------------------------------------------------- Maximum Speed =
V
But,
max
K
(1/2 Mark)
7
T T 2 4 K 3600D
--------------------------------------------- (1/2 Mark)
2K
.----------------------------------------------------------------2
(1/2 Mark)
km
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Subject Code: 17507
K
Page 33 of 35
1.2 3
21.2 3
K 0.5833 -------------------------------------------------------------- ( 1 /2Mark)
V
max
V
V
160 1602 4 0.5833 3600 2
2 0.5833
max
56.7345 KM/Hr --------------------Answer----------------- ( 1/2 Mark)
Distance covered during Acceleration ( D ) =
2
D
D
Vmax
-------------------------------------------------- (1 Mark)
7200
56.7345 2
7200 1.2
D 0.3725 Km ----------------------Answer------------- ( 1/2 Mark)
Distance covered during Retardation ( D ) =
2
V
D max ----------------------------------------------- (1 Mark)
7200
D
56.73452
7200 3
D 0.1490 Km----------------------------Answer--- ( 1 /2 Mark)
A consumer draws 500 KW power steadily at 0.8 p.f. lagging for 3650 hours per
annum. The tariff is Rs. 1300 per KVA of maximum demand plus Rs. 1.00 per kwh.
c)
The annual cost of phase advancing plant is Rs. 150 per KVAR. Find the annual saving
if the power factor of load is improved.
Ans: Given Data: - P= 500KW, cos = 0.8 lag, working Hours = 3650/annum
Tariff = Rs.1300 / KVA of max imum demand / annum Rs. 1.00 / Kwh
The annual cost of phase advancing plant = Rs.150 / KVAR
Saving in max. demand charges:
Rs. X.P. (
1
1
) Rs. X.P. (sec 1 sec 2 ) -------------------------(4Marks)
Cos1 Cos 2
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Model Answer
Subject Code: 17507
Rs. 1300 500 (
1
1
)
0.8 0.9933
Rs. 1300 500 (1.25 1.006)
Rs. 158128.80 ---------------------------------(3Marks)
Net Saving per year = saving –expenditure--------------------------------------(1Marks)
= 158128.80-150
= 157978.80 Rs.
OR
KW
Cos ---------------------------------------------------(1Mark)
500
KVA
0.8
KVA 625 --------------------------------------------------- (1 Mark)
KVA
No. of Units consume in One Year
M .D( KW ) No.of working hours in one year
1 500 3650
1825000 Kwh ---------------------------------------------------- ------ (1 Mark)
Annual Energy Bill := Rs.1300 / KVA of max imum demand / annum Rs. 1 .0 / Kwh
= Rs.1300 625 Rs.1 1825000
= Rs.8125000 Rs.1825000
= 2637500.00 Rs. -----------------------Equation No. 1------------- (1 Mark)
Saving will be maximum only at most Economical Power factor hence: Most
Economical Power factor is calculated.
2
Y
Most Economical P.f. = 1 ------------------------------------- (1Mark)
X
2
150
= 1
1300
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Page 35 of 35
= 0.99332 lagging --------------------------------------- (1/2Mark)
New Maximum Demand =
KVA
KW
------------------------------------------ (1Mark)
New Cos
New KVA
500
0.99332
New KVA 503.3620 --------------------------------------- (1/2Mark)
Annual Energy Bill if the Power factor load is improved:= Rs.1300 / KVA of max imum demand / annum Rs.1.0 / Kwh
= Rs.1300 503 .3620 Rs.1 1825000
= Rs.654370 .60 Rs.1825000
= 2479370 .60 Rs. -------------------- Equation No. ------2----- (1/2 Mark)
Annual Saving = Equation 1 – Equation 2
= 2637500 Rs. – 2479370.60 Rs.
= 158129.40 Rs. Saving if the power factor of load is improved. (1/2 mark)
------------------------------------------------------END-------------------------------------------------------
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Page 1 of 31
Model Answer
Important suggestions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more importance.
(Not applicable for subject English and communication skills)
4) While assessing figures, examiner may give credit for principle components indicated in a figure.
The figures drawn by candidate and model answer may vary. The examiner may give credit for any
equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model answer.
6) In case some questions credit may be given by judgment on part of examiner of relevant answer
based on candidate understands.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.
Q.1 a) Attempt any THREE of the following:
a) Define electric drive. List at least four advantages of electric drive.
Ans: Electric Drive:
12 Marks
( 2 Marks)
It is a machine which gives mechanical power. e.g. drives employing electric
motors are known as electric drives.
Following advantages of electric drive:
1. It is more economical.
2. It is more clean.
3. No air pollution.
4. It occupies less space.
5. It requires less maintenance.
6. Easy to start and control.
7. It can be remote controlled.
8. It is more flexible.
9. Its operating characteristics can be modified.
10. No standby losses.
11. High efficiency.
12. No fuel storage and transportation cost.
(Any Four point expected: 1/2 each)
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Model Answer
13. It is reliable source of drive.
14. Less maintenance cost.
15. It has long life.
State the principle of induction heating. Write at least four applications of induction
heating.
( 2 Mark)
Ans: Principle of Induction heating:
b)
The basic principle of induction heating is that, supply is given to primary winding of
furnace transformer & heat is produced in the secondary (charge) due to electromagnetic
action.
OR
Principle of Induction heating:
It is based on principle of transformer. In this type primary winding is as usual
which is wound around one limb of magnetic core but secondary winding is actually
charge which is to be melted is kept in crucible.
When AC Supply is given to primary winding current flows through primary
winding which creates alternating flux in magnetic core this flux links to the secondary
winding i.e. charge through magnetic core. Hence according to faraday’s law of
electromagnetic induction emf will be induced in secondary winding that is in the charge.
As charge forms a close circuit (secondary) heavy current flows through charge this
current is responsible to produce heat in charge due to I2R losses. This heat is utilized to
melt the charge.
Where, R = Resistance of charge & I secondary current
Following are applications of induction heating:
(Any Four point expected: 1/2 each)
1. Melting of steel and non ferrous metals at temperatures up to 1500 °C.
2. Heating for forging to temperatures up to 1250 °C.
3. Annealing and normalizing of metals after cold forming using temperatures in the range
of 750 – 950 °C.
4. Surface hardening of steel and cast iron work pieces at temperatures from 850 – 930 °C
(tempering 200-300 °C)
5. Soft and hard soldering at temperatures up to 1100 °C,
6. Moreover, special applications such as heating for sticking, sintering
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Model Answer
Subject Code: 17507
c) State the laws of illumination.
Ans: Laws of illumination: (Each Figure 1 Mark & Statement of each law 1 Mark, Total: 4 Mark)
a) Inverse Square Law :Radius s 3i
Radius = 2 x
Radius
=x
\
Source
''
X\ /
I
h
/
/
/
/
/
/
/
/
or Equivalent fig.
Illumination is inversely proportional to the Square of distance between source and
I
plain of the surface and directly proportional to light intensity. E 2
r
b) Lambert’s cosine law :L
fl
H
A
3
b
or Equivalent fig.
The illumination of a surface is directly proportional to cosine of angle made by the
normal to the illuminated surface with the direction of the incident flux.
EB E A cos3 1 ,
EC E A cos 3 2 ,
E D E A cos 3 3 and so on.
d) State any four causes of low power factor.
Ans: Following are the Causes of low power factor: - (Any Four causes expected: 1 Mark each)
1.
Magnitude of Magnetizing Current (I ):As magnetizing current increases, power factor reduces.
2.
Due to use of Induction Motor:Most of industrial drives, agriculture pumps, lift, irrigation pump set uses I.M.
which works at lagging power factor, and so power factor reduces.
3.
Due to use of Transformer: All transformers works at lagging power factor, so power factor of system reduces.
4.
Due to welding transformer: -
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Model Answer
Subject Code: 17507
Welding transformers are operated at low p.f. which reduces p.f. of the system.
5.
Due to inductance of transmission & distribution Line: In case of AC transmission & distribution lines, inductance is present which the
main cause of low power factor is.
6.
Series Reactor:Series reactor is used in substation to minimize fault current causes low power
factor.
7.
Industrial electrical heating furnaces:Induction and arc furnace used in steel manufacturing industry works at low p.f.
which reduces p.f. of the system.
8. Arc Lamp:Arc lamp & electric discharge lamps operates at low p.f.so p.f. of the system
reduces.
9. Equipments operated at light load:P.f. falls if equipments like alternator, transformer, I.M etc are not operated at
full load.
10. Improper repairs and maintenance:P.f. falls if proper maintenance or repairs of equipments are not done.
Q.1B)
Attempt any ONE :
06 Marks
a) i) State any six requirements of ideal braking system.
Ans: Following are different requirements of an ideal braking system:
(Any Six requirement are expected: 1/2 Mark each, Total 3 Mark)
1. The braking system should be simple easy to control & operate.
2. It should be most reliable.
3. Braking actuation time should be as small as possible.
4. Braking force should be very gradual and smooth in case of emergency braking also
to avoid discomforting to passenger & damage of goods.
5. Braking force applied to each axel should be proportional to axel load.
6. Reputed quick application of break should be possible without needing any
normalizing time in between two successive operations.
7. It should have less maintenance.
8. It should have long life.
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a) ii)
Model Answer
Page 5 of 31
State the advantages and disadvantages of electric braking over mechanical braking.
Ans: (Any three advantages expected:1/2 Mark each & Any three disadvantages expected: 1/2
Mark each, Total : 3 Mark)
Following are the advantages & disadvantages of electrical braking over mechanical
braking system.
Advantages: (Any three point expected)
1. It is most reliable braking system.
Because in mechanical braking heat is produced at break block & break shoes,
which may be source of failure of break.
In Electrical braking (dynamic) heat is produced at convenient place (external
rheostat) which is not harmful to braking system
2. Breaking actuation time is small as higher value of braking retardation is obtained.
3. Electrical braking is smooth & gradual.
Where as if mechanical breaks are not correctly adjusted then there are chances
of sudden braking which is discomfort able to passenger.
4. Life of braking system is more.
Because mechanical braking provides metal dust due to friction, No such dust is
formed in electrical braking.
5. There is less wear & tear of brake shoes, break block etc. so there is less maintenance
cost.
6. Higher speeds are possible even when train is going down the gradient, as breaking
system is reliable.
7. Trains having heavy loads can be stopped even when train going up the gradient.
8. Higher speeds of train is possible as braking system is reliable so pay load capacity
increases.
9. In case of electric regenerative braking we can utilize 60 to 80% of kinetic energy to
generate electricity which is not possible with mechanical braking.
Disadvantages: (Any three point expected)
1. In addition to electrical braking there must be arrangement of mechanical braking for
final stop.
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Page 6 of 31
2. Special arrangement of circuit is to be provided which makes electrical braking
system costly.
3. Operation in substation becomes complicated at the time of regenerative breaking
when generated energy is surplus.
4. Electrical braking need electric supply.
b) i) State the principle of resistance welding.
Ans: Working principle of resistance welding:
( 2 Mark)
In resistance welding, sufficiently heavy current at low voltage is passed directly
through two metals in contact to be welded.
Heat is produced due to I2R losses where ‘R’ is the contact resistance. This heat is
utilized to obtain welding temperature (to become a plastic state)
When welding temperature is reached supply is cut down and external pressure is applied
simultaneously across the job to complete weld.
According to joules law,
Heat produced H =I2 R t …………….. Watt-sec
From this equation it is clear that heat produced depends on
Square of current (I2)
Contact resistance (R)
Duration of current (t)
b) ii) State the various types of resistance welding.
Ans: Types of Resistance Welding:- (Any Two types expected)
1) Spot welding
2) Seam welding
3) Projection Welding
4) Butt Welding
5) Flash Butt welding
( 2 Mark)
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Model Answer
Subject Code: 17507
b) iii) Describe with neat sketch the operation of seam type resistance welding.
( Figure: 1 Mark & operation: 1 Mark)
Ans: Sketch of Seam welding:
l
mod ?
A fiF F
is
.'
J
4
OH
tot o Mi\ jtf u ,
>
Ulf
n rIMfvtf 41 M *
1
0(
|
Lti
J
—J \34 hr__
LJ.
l-H-U
Uft
iu nM
'
.^
^
rr Wnjpr
?A
S
^ffWS
Tvo'
u
4o be \iw \ del
^
\
|
ld | p|
)( ]
- OR
IVnflqM /fifi
r
1
VI
snifs spi
*
S? Weliinj
^" 1 r
p
('
W4
r
h
77 M
x
r
1
, uiac
r
ifW&
w s tcH
vfost
'
1 imn
h;
La,
~T—
s
/ IifciT
.
_
v
^ i/- Ws /
feoo ft 4
0
Tf
-
£4 rufa b
i
.’
-isbjflmeldt / mn )
1
i
l
l'
Operation:
Job is kept in between two electrodes under pressure. This pressure is kept constant
throughout.
In this type intermittent current is used, it means current is ON for definite time and
OFF for another time interval with the help of timer.
If current is continuously passes then heat produced may cause burning of job.
Heat is produced due to I2R losses where ‘R’ is the contact resistance.
This heat is utilized to obtain welding temperature (to become a plastic state)
When welding temperature is reached supply is cut down and external pressure is
applied simultaneously across the job to complete weld.
Q.2
Attempt any FOUR :
a) Describe in brief the size and shape of elevator car.
Ans:
16 Marks
( 4 Marks)
Size and shape of elevator car depends on following points:i) No. of passenger to be carried: While selecting the size of car it is a usual practice to
allow.
A Space of 2 Sq.fit/ person.
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Model Answer
Subject Code: 17507
Average weight of passenger is assumed 68 kg/person.
Thus the maximum load capacity of elevator is considered 34 kg/sq.ft
There should be wide frontage and shallow depth
ii) Limitation in the building design:
Shape of elevator depends on space available in building.
State the principle and nature of supply used for eddy current heating. State the
advantages and disadvantages of eddy current heating.
Ans: Principle of Eddy Current Heating:(1 Mark)
b)
Maartotic Field
*
*
nduocd Current In Part
J
t
b
r
lo
t
t
—r
'Dll1£
~"X
Currant In Coil
*
*
J
t
ID
PP' 1
kV ,
JO n> tfO KML
or Equivalent fig.
Principle:-
(1 Mark)
Heat produced eddy current loss B 2 f 2
1
F
The job which is to be heated is wound by coil as shown in figure.
Depth of penetration of heat
Supply of high voltage (10KV) & high frequency (10-40 KHz) is given to coil
which induces eddy current in job according to Faraday’s law of Electromagnetic
induction & these eddy currents are responsible to produce heat in job itself due to eddy
current loss.
In high frequency eddy current heating the phenomenon of skin effect plays an
important role.
Skin effect at high frequency is more pronounced (effective). Due to this surface of
job is more heated as compared to its depth.
km
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Model Answer
Nature of supply used for eddy current heating:
Page 9 of 31
(1 Mark)
High voltage (10KV)
High frequency (10-40 KHz)
Advantages eddy current heating:- (Any one point expected)
(1/2 Mark)
1) No heat transfer loss as heat is produced in job itself. So it has high efficiency.
2) As heat is produced in job itself so time required for heating is less. For e.g. in some
cases operating time taken for heating is of only one second.
3) By simply controlling frequency, we can control temperature accurately.
4) By simply controlling frequency, depth of penetration of heat can be controlled easily.
5) Very thin material surface can be heated easily.
6) Operation is simple & automatic.
7) For heating low attention is required.
8) Heating can be taken place in vacuum or other special atmospheric condition where
other methods are not possible.
9) It is clean and convenient method.
Disadvantages of eddy current heating:-
(1/2 Mark)
1) High initial cost because of high voltage high frequency supply equipment is required.
c) State any six requirements of an ideal traction system.
Ans: Any six requirements from the following: (First any Two Points: 1 Mark each & other
any Four point : 1/2 each, Total : 4 Marks)
Ideal Traction system should processes following requirement:- (Any Six Point expected)
1. It should have low capital, Running, maintenance cost.
2. Quick starting time.
3. It should have high rate of acceleration & retardation.
4. Highest speeds are possible.
5. Easy speed control method.
6. Braking system should be reliable.
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Model Answer
Subject Code: 17507
7. Absence of unbalance forces i.e. coefficient of adhesion should be more.
8. Centre of gravity should be lower.
9. Better riding quality (less vibration)
10. Traction system should be clean & long life.
11. It should be self contained.
12. No standby losses.
13. It should have high efficiency
14. Regenerative braking should be possible.
Draw a neat labelled block diagram AC electric locomotive. State the function of each
part.
Ans: Block diagram of A.C. electric locomotive
(2 Marks)
d)
# pTlr
»<
1 0> rC K
di
ji
g_i'_Q
6_J
2_v
a_Lo
10
C o r ^ 6r i
J
-
Bi-L— Ln r. omrrt
of
^
~
"
s
1
, .
.
,o
'
/. » Current Collector
"
•
H
. «.i 11
r
-
/
/
—
rA i s
_
±
4|
F
^14^
,
%
=
nrsTr>
TT
a
KJ
| Td
|
r
dh
\
_
—
\
r o n4 o v c o a t
_
:M l
-
P v_
\
^ To p C h o nj * *
1— 1
^
’
-
^
•
00
ftu
—
rw-
^
se
-
T
or equivalent fig.
Function of each part:
(2 Mark)
1) Overhead contact wire:
Supply of 1-ph, 25KV, 50Hz, AC is given to overhead conductor.
2) Current collecting device:
It collects current from overhead contact wire and passes it to tap changing
transformer through circuit breaker.
3) Circuit breaker (C.B):
It is connected in between current collecting and no-load tap changing transformer.
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Page 11 of 31
Model Answer
SF6 circuit breaker is used.
To disconnect locomotive equipments whenever there is fault.
It opens automatically when train passes from zone No.1 to Zone No.2 i.e. neutral
zone
4) On load tap changing transformer:
It changes the tap without disconnecting the load on transformer. Its purpose is to
vary the voltage for speed control of traction motor.
5) Traction Transformer:
It step down input voltage 25 KV to working voltage of traction motor
(1500V/3000V).
6) Rectifier:
It converts secondary voltage of transformer into DC supply.
7) Filter circuit (smoothing reactor):
It is used to obtain pure DC
8) Traction Motor:
It gives mechanical power to run the train DC series motor is used as traction motor.
9) Current Collector:
To collect current from overhead line.
"DC series motor is used for traction purpose". Justify your answer with any six
characteristics.
Ans: (First any Two Points: 1 Mark each & other any Four point : 1/2 each, Total : 4 Marks)
e)
DC series motor is used for traction purpose because of following characteristics:
1)
DC Series motor has high starting torque.
r
2)
DC series motor has High torque at low speeds, low torque at high speeds, this is the
basic requirement of traction unit.
4
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Model Answer
Subject Code: 17507
©
X*
.
Ta
3)
DC Series motor speed-torque characteristics are such that as torque increases speed
decreases. (Due to this characteristics motor is protected against overload)
w*
a-
v
rx
i
4)
5)
6)
o
DC Series motor variable speed motor.
DC Series motor has high rate of acceleration and retardation.
Torque obtained by DC series motor is smooth and uniform, so it improves riding
quality.
OR Student may write following additional Point
7)
8)
9)
When DC series motor are running in parallel the all motors share almost equal load.
DC Series motor maintenance cost is less.
DC Series motor robust in construction and capable to withstand against continuous
vibration.
10) DC series motor weight is 1.5 times less than 1-Ph AC series motor for same H.P.
Q.3
Attempt any TWO :
a) i) State the factors governing selection of electric drive.
Ans: Following are the factors governing selection of electric drive:
16 Marks
(Any Four factor expected: 1 Mark each, Total : 4 Mark)
1. Nature of supply:
Whether supply available is
AC,
Pure DC
OR rectified DC.
2. Nature of Drive (Motor):
Whether motor is used to drive
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Model Answer
Page 13 of 31
Individual machine
OR group of machines.
3.
Nature of load:
Whether load required light or heavy starting torque
OR load having high inertia, require high starting torque for long duration.
OR Whether load torque increases with speed (T N)
OR decreases with speed (T 1/ N )
OR remains constant with speed (T = N)
OR increases with square of speed (T N2)
4.
Electric Characteristics of drive:
Starting,
running,
speed control
Braking characteristics of electric drive should be studied and it should be matched
with load requirements.
5.
Size and rating of motor:
Whether motor is short time
OR continuously
OR intermittently running
OR used for variable load cycle.
Whether overload capacity, pull out torque is sufficient.
6. Mechanical Considerations:
7.
Types of enclosure,
Types of bearing,
Transmission of mechanical power,
Noise
load equalization
Cost:
Capital,
running
And maintenance cost should be less.
^ ^-
.
1
HI
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Model Answer
Subject Code: 17507
a) ii) Define load equalisation for electric motors. Explain how it is obtained for electric
motors.
Ans:
Define Load equalization:
(2 Mark)
There are many types of load which are fluctuating in nature e.g. wood
cutting m/c, Rolling mill. Etc. For such type of loads, load equalization is necessary to
draw the constant power from supply. Because,
When there is sudden load on motor, it will draw more current from supply at
start to meet additional power demand. Due to this heavy current there is large voltage
drop in supply system. This will affect electrical instrument, equipment, m/c, other
consumer etc. which are connected across same supply line.
Also to withstand heavy current, size of input cable increases so cost of cable
increases, Hence it is necessary to smooth out load fluctuations on motor.
The process of smoothing out load fluctuation is called load equalization.
How load equalization is done?
(2 Mark)
Load equalization is done by means of flywheel. It is mounted on motor shaft.
Flywheel stores kinetic energy when there is light or no load & it supplies kinetic energy
when there is sudden heavy load on motor. In this way load demand on supply remains
practically constant.
plu
LO W|
L>
Pl » +^3 Y ~
&
J & ad
b) A 40 kW, 3-phase, 400 V resistance oven uses nickel-chromium strip of 0.3 mm thickness.
The heating elements are star connected. If wire temperature is to be 1127°C and that of
charge is to be 727°C, estimate suitable width and length of the wire required.
Given: radiation efficiency = 0.6, specific resistance of Ni-Cr = 1.03x10 -6 ohm m,
emissivity = 0.9.
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Model Answer
Subject Code: 17507
Ans: Given Data:
T1 = 1127 0C = 1127 +273 = 1400 0K
T2 = 7270C = 727 +273 = 1000 0K
Radiation efficiency = 0.6, specific resistance of Ni-Cr = 1.03x10-6 ohm m, emissivity = 0.9.
H 5.72 104 k .e [ (
T1 4
T
) ( 2 ) 4 ] w / m2
1000
1000
----------------------------- (1 Mark)
1400 4 1000 4
H 5.72 104 0.6 0.9 [ (
) (
) ] w / m2
1000
1000
H 87771.3408 w / m 2
-------------------------------------------------------- (1 Mark)
Thickness : 0.3 mm 0.3 103 m
l
V2
wt P
-------------------------------------------------------- (1 Mark)
V
400
Voltgae across each resis tan ce
3
3
Voltgae across each resis tan ce 230.94 volt
Power
40 KW
3
13.3333 KW 13.3333 103 watt
l
V2
wt P
l
(230.94)2
w ( 0.3 10 3 ) (13.3333 103 ) 1.0310 6 )
l
1.59999 103
w
0.0137332
l
1165.0442 ................Equation........I
w
t 2H
l2
V2
---------------------------------------------------- (1 Mark)
t 2 ( 1.03 106 ) 87771.3408
l2
(230.94)2
t 0.1808089
l 2 53333.284
-------------- (1 Mark)
km
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Model Answer
Subject Code: 17507
t
3.39017 10 6
l2
t
l2
6
3.39017 10
0.3 103
l2
6
3.39017 10
l 2 88.491138
l 9.4069728 mtr
--------------------------------------------- (1 Mark)
----------------------------------------------- (1 Mark)
Putting in Equation : I
l
1165.0442
w
9. 4069728
w
1165.0442
w 0.00807434 mtr
----------------------------------------- (1 Mark)
Answer : Length
l 9.4069728 mtr
Widgth w 0.00807434 mtr
c) Describe the concept of load cycle with their graphical representation.
i) Continuous loading
ii) Short time loading
iii) Long time (intermittent) loading
iv) Continuous operation with short time loading.
(Each Graphical representation : 2 Mark, Total: 8 Mark)
Ans: i) Continuous loading:V\
P
—
/
/
T?
,
r r P e t a -K j i e
-f i s «-
/
/
^
z3
/
(
fl
,
i'
i
.
..
r
f1
T| m f
..
~
«
11
or equivalent figure
In this case motor is operated continuously non-stop for few days or month also
without exceeding the permissible temperature limit. e.g. water pumping motor, generating
power house auxiliary motors etc.
km
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Model Answer
Subject Code: 17507
ii) Short time loading:- Graphical representation
VA ?
A, fvrlp
Temperature
tfeotii-
f-
/
/
/
1
/
\
+
1
IfYI
—
*
rime
i
*i
or equivalent figure
In short time loading motor is operated for short time continuously without
exceeding the permissible temperature limit.e.g. 15min., 20min., 30min. etc than it is made
OFF This OFF load interval is sufficient to cool the motor temperature to its normal value.
iii) Long time ( Intermittent) Loading :- Graphical representation
f
t
p1
1
5
Temperature
*
\
Cvrlp
\ ? r v n 4P
/
'
s
or equivalent figure
In long time loading motor is operated for long time continuously without
exceeding the permissible temperature limit than it is made OFF for short time This OFF
load interval is not sufficient to cool the motor temperature to its normal value so
temperature of drive continuously increases.
iv) Continuous operations with short time loading:-
Graphical representation
.
1
v\ ? - iTemperature
/
/
Hf
'
s
":/
-$
/
cv c l,e
-t t p e o f ^ s
v
»/
/
/
,
~
+1 r r s t
|
<
or equivalent figure
In this case motor is operated continuously for short time and interval between
two load is not OFF- load but motor runs at no load for long time. So temperature of drive
continuously increases. So
Temperature rise is more than short-time loading.
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Model Answer
Subject Code: 17507
Q.4 A) Attempt any THREE :
12 Marks
a) Compare D.C. welding and a.c. welding on any six points.
Ans:
((Any Four Point Expected: 1 Mark each: Total :4 Marks)
S.No
Points
DC Welding
AC Welding
1
2
3
4
Supply equipment
used
Heating Effect
Temperature Obtain
Arc Blow
DC differential Compound
Generator, or Rectifier
Uniform
More
Pronounced (effective)
Welding Transformer
5
Stability of Arc
6
Type of Electrode
7
8
9
10
11
12
13
Voltage Required
Capital Cost
Running cost
Maintenance cost
Stand by losses
Efficiency
Application
D.C Differential component.
Generator has dropping
characteristics.
Non Coated Electrode is
used
50 to 60 volt D.C
High
High
High
High by 25%
Low, 65%
Resistance Arc Welding &
Metal Arc Welding
Not Uniform
Less
Not So Pronounced
(Not effective)
Use of series Reactor
Coated Electrode is
compulsory
72 to 100 volt A.C
Low
Low
Low
Low
High, 85%
Carbon Arc Welding
b) Describe the construction of high pressure mercury vapour lamp with neat sketch.
Ans: Figure mercury vapour discharge lamp :(2 Mark)
Argon
s' A
“HI
*
M
/+ mercuiy
1
1
Choke
h-
II r
‘V
m
lube
i_|
A
3
C
Outer tube with
tiuoroscent coating
H
"
^
Construction:-
Inner
i
irom inside
R
OR
A, B, C are electrodes
R is resistance
Space is
evacuated
(2 Mark)
It consists of an inner bulb generally of silicon, to withstand high temperatures.
km
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Model Answer
Page 19 of 31
The bulb contains a small quantity of mercury and argon.
It is protected by outer glass, this may be cylindrical or elliptical.
The space between the two bulbs is filled with nitrogen at a pressure of half
atmosphere.
The discharge tube has three electrodes, namely two main electrodes A and B and one
starting electrode.
The starting electrodes are connected through a resistance of about 10-30 k ohm to the
main electrode, located at the far end.
The electrodes are of tungsten wire helices filled with electron emissive materials,
usually barium and strontium carbonates mixed with thorium.
OR Student may write
The construction & connection diagram is as shown in figure. As per this
construction there are following components.
Choke: The choke is acting as the ballast. At the time of supply voltage variation of
current flowing through the inner tube is maintained constant to keep uniform light
intensity. Sometimes choke can be designed for to get the higher voltages & to apply
the inner tube of mercury vapour lamp.
Starting resistance/limiting resistance: Whenever current flows through the starting
resistance there is a I2R loss which is converted into heat. If the temperature of this heat
goes near about 600 0C then there will be heating effect & inert gases ionization will be
start.
Auxiliary electrode & Main electrode: It is made by high resistive element. The
ionization is taking place through the inert gases whenever current flows from auxiliary
electrode to main electrode.
Inner Tube: The various inert gases e.g. Argon, Nitrogen etc with mercury powder are
filled in the inner tube at low pressure or high pressure.
Outer Tube: The function of outer tube is to make the vacuum surrounding the inner
tube to avoid thermal dissipation or to maintain 6000C surrounding the inner tube.
Power factor improvement Capacitor: The function of power factor improvement
capacitor is to improve the power factor 0.5 to 0.95
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Page 20 of 31
Model Answer
c) Compare block-rate tariff and flat rate tariff. (any four points)
Ans:
(Each Point: 1 Mark: Total :4 Marks)
S. No
Block-rate Tariff
Flat rate Tariff
1
2
3
4
In case of block rate tariff there are
blocks of units consumed
Each block tariff rate and each block
tariff rate/unit (KWH) is different.
If generation is less than utilization than
tariff rate/unit in each block goes on
increasing and vice versa.
Used for Residential ,commercial
consumers
In case of flat rate tariff there are no
blocks of units consumed
There is flat tariff rate/unit (KWH).
Tariff rate/unit (KWH) not depends on
generation and utilization.
Where is constant for a fixed load for
fixed hours.
d) State any four advantages of good power factor for electric supply system.
Ans: Advantages of Good (High) power factor for electric supply system:
(Any Four Point Expected: 1 Mark each: Total :4 Marks)
1. P.F. increases current reduce so; cross section of conductor decreases hence its cost is
reduces.
2. P.F. increases current reduce so, cross section of conductor decreases hence weight
decreases. So design of supporting structure becomes lighter.
3. Copper losses Decreases, Hence transmission efficiency increases.
4. Voltage drop reduces, hence voltage regulation becomes better
5. Handling capacity (KW) of each equipment increases as p.f. increases.
6. Less capacity (KVA) rating of equipments are required so capital cost decreases.
7. Cost per unit (KWH) decreases.
OR
Following advantages are observed due to good power factor:
1. Generation:
a)
Low KVA rating of equipment (alternator) is required.
b) Handling capacity of equipment (alternator) increases.
c)
Cost per unit increases.
2. Transmission:
a)
Low KVA rating of equipment (Transformer) is required.
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b)
c)
d)
e)
f)
g)
h)
Model Answer
Page 21 of 31
Handling capacity of equipment (Transformer) increases.
Cost of conductor decreases.
Cost of supporting structure decreases.
Copper losses decreases
Transmission efficiency increases.
Voltage drop decreases.
Regulation becomes improved
Q. 4B)
Attempt any ONE of the following :
06 Marks
Describe electric arc welding in brief. How arc is formed in electric arc welding? State the
a)
characteristics of electric arc.
Ans: Electric Arc Welding:(2 Mark)
The processes in which two metal parts to be welded are brought to a molten state and
then allowed to solidify is called as arc welding.
Melting of metal is obtained due to heat developed by an arc struck between an electrode
and metal to be welded (job) OR
In this type of welding heat is developed due to arc produced in between electrode &
job.
High voltage is required to produce arc at starting .once arc is struck low voltage is
sufficient to maintain arc.
No mechanical pressure is required, so this type of welding is also known as nonpressure welding.
Temperature obtained by arc is very high (35000C to 6000 0C)
Arc welding power factor is poor. And power consumption is high.
At the time of welding external filler material is required.
Arc is formed in electric arc welding by any one of the following method:
a)
b)
(2 Mark)
By applying High Voltage
By separation of two current carrying electrodes suddenly
Following are the characteristics of electric arc: (Any Two Point expected)
1. To produce arc high voltage is required then to maintain the arc.
2. Arc is conducting.
3. Arc has negative temperature coefficient of resistance.
(2 Mark)
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Model Answer
Subject Code: 17507
b) A factory has a maximum demand of 300 kW with a load factor of 0.6. The following
tariffs are offered :
a) Two part tariff z 80/kW of M.D./year + 5 paise / kWh.
b) A flat rate of 12 paise/kWh.
Calculate tariff in both cases and write with reason. Which tariff will be cheaper?
Ans:
No. of Units consume in One Year
Load Factor M .D( KW ) 8760 -------------------------------- (1/2 Mark)
0.6 300 8760
15768000 Kwh --------------------------------------------------------(1 Mark)
Case-I: Energy Bill := Tariff given Rs. 800 of M .D. / year Rs. 5 paise / Kwh ------ (1/2 Mark)
= (300 80 ) ( 1576800 5 / 100)
= Rs. 24000 Rs.78840
= 102840 Rs. ------------------------------------------------------------ (1 Mark)
Case-II: Energy Bill := Tariff given flat rate of 12 Paise / Kwh --------------------------- (1 Mark)
= 1576800 12 / 100 )
= 189216 Rs. -------------------------------------------------------------- (1 Mark)
According to energy bill Case-I is economical ---------------------------------- (1 Mark)
For industrial consumer Case-I is economical
Q.5
Attempt any FOUR :
16 Marks
a) Compare metal arc welding and carbon arc welding. (any four points)
Ans:
(Any Four Point Expected: 1 Mark each: Total :4 Marks)
S.N
Point
Metal Arc Welding
1
Type of supply
used
Type of Electrode
Both AC/DC supply can
be used.
Coated electrode of same
metal to be welded is used
Welding Transformer
2
3
4
5
6
Supply
Equipment used
Arc Stability
Temperature
obtain
Capital Cost
Carbon Arc Welding
Only DC supply is used.
Carbon Electrode are used.
Less
D.C Differential component
Generator or Rectifier
D.C Differential component.
Generator has dropping
characteristics.
More
Less
More
Use of series Reactor
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Model Answer
Subject Code: 17507
7
8
9
10
11
12
Running
Maintenance
Stand by losses
Efficiency
Voltage required
Types
13
14
Application
Limitation
Less
Less
Less
More
72 to 100 volt A.C
Shielded & unshielded
welding
For Ferrous Metal
Can be used for vertical &
overhead welding
More
More
More
Less
50 to 60 volt D.C
Flux is used and flux is not used
for non ferrous metals
Not suitable for overhead welding
Describe through illustration following types of lighting scheme :
i) Direct lighting ii) Indirect lighting
Ans:
1) Direct lighting ( Figure: 2 Mark & Explanation: 2 Mark)
b)
or
/
ICO %
chled |ux
Ifghtinq source
li
“
A
X
'
7r
AA
A
K
_
1
7
\
lwi„|Tfnj
~
plnni°
-
l liJ P)
or equivalent fig.
In this method, the reflector is used on the lighting source. The 100% light is
reflected by this reflector on the working plane. So efficiency of direct lighting scheme
is very high and it is economical also. But limitation of direct lighting scheme is
that glare & shadows are more.
Application:
The direct lighting scheme is widely used in drawing room, workshop and flood
lighting etc.
2) Indirect lighting scheme:Tn 4 iVeci
LiqkhrLq Jacketme
'
.
.i
IOQ
7a
7r
-
fid££Kf&,
7
.
s/
OL> a (|
__L
lUX
1 i R
^
A
A
JL Ceili'nj
A
reilecfor
A
7R
or equivalent fig.
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Model Answer
Page 24 of 31
In this method the 100% light is reflected on ceiling and walls by the reflector
and these reflected light will be available on working plane. It is less efficient and
uneconomical scheme but glare and shadows are very less. i.e. Why surrounding may
be pleasant
Application:
Widely used in hotels, guest room, Show room etc.
c) Write the different systems of track electrification.
Ans: Following are the different track electrification system:
( Any Four System Expected: 1 Mark each)
1. Direct current track electrification:
600V, 750V DC for tramways
1500V, 3000V DC for Train (Urban service)
2. 1-Phase, low frequency AC Supply system:
1-Ph, 25KV, 16 2/3 Hz or 25 Hz (formerly it was 1-ph 15/16 KV)
3. 1-Ph, High frequency AC supply system:
1-Ph, 3.3 KV, 2500 Hz or 3000 Hz
4. 3-Ph, Low frequency AC supply system;
3-Ph, 3.3 KV, 16 2/3 Hz or 25 Hz
5. Composite system:
1-Ph AC (1-ph, 25KV) – DC Supply System ( 1500 / 3000V DC)
Kando System ( 1-Ph AC – 3-Ph AC)
d) Write any six desirable characteristics of traction motors.
Ans: Following are the desirable characteristics of traction motors:
(First any Two Points: 1 Mark each & other any Four point : 1/2 each, Total : 4 Marks)
A) Mechanical Properties or characteristics: ( Any Six Point Expected)
1) It should be simple in design
2) It should be robust in construction to withstand against continuous vibrations.
3) Weight of motor per HP should be minimum in order to increase pay load capacity.
4) It must be small in overall dimensions, especially in overall diameter.
5) It must have totally enclosed type enclosure to provide protection against entry of dirt,
^
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Model Answer
Subject Code: 17507
dust in drive.
6) When motors are running in parallel they should share almost equal load. (even where
there is unequal wear & tear of wheels)
B) Electrical Properties or characteristics:
7) It should have high starting torque.
8) It should possess high rate of acceleration & retardation.
9) It should be variable speed motor.
10) Its speed-torque characteristics should be such that high torque at low speed and less
toque at high speed.
11) Motor must be capable of taking excessive overload in case of emergency.
12) It should have simple speed control methods.
13) Electrical braking system should be reliable, easy to operate and control, especially
regenerative braking is possible.
14) Motor should draw low inrush (Starting current) current.
15) It should withstand for voltage fluctuation without affecting its performance.
C) General Properties or characteristics:
16) It should have less maintenance cost.
17) It should have high efficiency.
18) It should have long life.
19) It should have low initial cost.
e) Draw speed time curve. Show and list various time periods associated with it.
Ans: Typical speed time curve for main traction line service:
(2 Marks)
rurvf
**
V ( nlhy
I?Uhrpnj
tipred
i nf
**
Pygg
Runn' r*]
COOS * ini
period
PfTibd
e
r
|
P >
*
Arc/ n
trr,! ht )
4 \! (
¥
°
—
—
v-l *rl
r
4 #V r > » f T )
XfiP
OR
0
*
zr
/
| ptrioa
Fm RunV' r ]
Vy7
gpfpd
o
r
*
Min
pnifd
•
v<
-
—
LSX g
^
1\
1
Vm # >i
1
\
V.
,
.—
,
pf > / Oc/
.
.X
1
"/
M
<a
* Tirne / r ) ate
>
km
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Model Answer
There are five periods in the run of train as shown in speed time curve:
Page 26 of 31
(2 Marks)
i) Constant acceleration period ( o to A) :During this period starting resistance in motor circuit are gradually cut down. At
point ‘A’ all the starting resistance in motor circuit has been cut down.
ii) Acceleration on speed –Time curve ( A to B) For T2 sec. :Now train is continuous to accelerate & torque gradually falls until speed of train
exactly balance train resistance during this period.
iii) Free Running or constant speed tun period ( B to C) For T3 sec. :At the end of acceleration period train attend maximum speed. During this free
tuning period train runs at constant speed & constant power is taken from supply by
train.
iv) Coasting period (C to D) For T4 sec. :At the end of free running period the supply to traction motor is cut down & train
allow to run under its own movement. The speed of train goes on decreasing due to
resistance to motion of train. Rate of decreasing of speed during costing period is
known as costing acceleration (Pc)
v) Braking period (D to E) For T5 sec. :At the end of costing period brakes are applied to bring the train to rest (stop)
During this period speed of train rapidly decreases & reduces to zero. The rate of
decreases of speed during braking period is known as ‘Braking retardation’ ( )
Q.6
Attempt any TWO of the following :
16 Marks
a) i) State any four advantages and any two disadvantages of electric heating.
Ans: Advantages of Electric heating:
(First any Two Points: 1 Mark each & other any Two point : 1/2 each, Total : 3 Marks)
1. It can be put into service immediately.
2. No standby losses.
3. High efficiency.
4. More economical than other conventional types of heating system.
5. Easy to operate and control.
6. No air pollution.
km
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Model Answer
Page 27 of 31
7. System is clean, as there is no waste produced.
8. No fuel transportation cost.
9. No space is required for storage of fuel and waste.
10. Noiseless operation.
11. Uniform heating is possible, heating at particular point is also possible (spot
welding)
12. Dielectric material can be heated.
13. Electrical heating equipments are generally automatic, so it requires low attention
and supervision.
14. Protection against overheating can be provided by suitable switch gear.
Disadvantages of Electric heating: ( Any Two expected)
1. It depends on electricity, so not self-content.
2. This system is used only where electricity is available
3. Electrical system is costly
( 1 Mark)
a) ii) Describe any four causes of failure of heating elements.
Ans: Following of the different causes of failure of heating element:
i) Formation of hot spot:
(Each Causes: 1 Mark, Total: 4 Mark)
Hot spot on heating element is the point which is at higher temperature than
remaining heating element portion. So there is possibility of breaking of heating element at
hot spot.
ii) Due to oxidization:
At high temperature material gets oxidized which may cause failure of heating
element.
iii) Due to corrosion:
If heating element is directly exposed to chemical fumes then there is possibility of
rusting of heating element which causes failure of heating element.
iv) Mechanical consideration/Failure:
Measure heating element alloy contain iron which is brittle. Due to frequent heating
& cooling of heating element, it may break (fail) due to small mechanical injury also.
b) A train has schedule speed of 60 kmph between stops which are 6 km apart. Determine
the crest speed over the run assuming :
i) Duration of stops as 60 sec
ii) Acceleration as 2 kmphps
iii) Retardation as 3 kmphps, The speed time curve is trapezoidal
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Ans:
Page 28 of 31
Model Answer
Subject Code: 17507
Vsch = 60Km/hr
D= 6 M
Tstop 50 sec = 2 km/hr-sec
= 3 Km/hr.sec
Solution:
3600 D
Schedule Time (Tsch ) -----------------------------------------------------(1 Mark)
Vsch
Schedule Time (Tsch )
3600 D
Vsch
Schedule Time (Tsch )
3600 6
60
Schedule Time (Tsch )
21600
60
Schedule Time(Tsch ) 360 sec.--------------------------------------------- (1 Mark)
Schudele Time(Tsch ) Actual Time of Run (T ) Stop time (Tstop )
Actual Time of Run (T ) ScheduleTime (Tsch ) Stop time (Tstop )
Actual Time of Run (T ) 360 60
Actual Time of Run (T ) 300 sec. ---------------------------------------
(1 Mark)
Maximum Speed =
V
But,
T T 2 4K 3600D
2K
------------------------------------------------------- (1 Mark)
K
2 -------------------------------------------------------------------- (1 Mark)
max
K
23
22 3
K 0.4167 ------------------------------------------------------------------- (1 Mark)
Now,
V
V
V
max
T T 2 4K 3600D
2K
------------------------------------------------------ (1 Mark)
max
300 3002 4 0.4167 3600 6
2 0.4167
7
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Model Answer
Subject Code: 17507
300 232.372
2 0.4167
V
max
V
max
81.1471 Km/hr --------------------Answer------------------------ ( 1 Mark)
c) i) Derive the equation for most economical power factor.
Ans: Derivation for most economical power factor:
_
(4 Mark)
<§> c
f
1
-
fb 1
*
_
^ ^
Lj
or equivalent figure
1
Let,
P = Active power KW
S1, S2 = KVA Maximum demand before and after improving power factor
Q1, Q2 = Lagging reactive power before & after improving power factor
QC = Leading Reactive power drawn by Capacitor
Cos1 = Initial Power factor
Cos 2 = Improved Power factor
Rs X = Tariff charges towards M.D. (KVA) /year
Rs Y = Expenditure towards KVAr to be neutralized per year (Expenditure towards P.F.
improving apparatus)
1) Before improving Power factor:
Q1 P tan 1
P
Cos 1
S1
P
S1
Cos 1
KVA 1 (S1 ) P sec 1
2) After improving Power factor:
Q 2 P tan 2
Cos 2
P
S2
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Model Answer
Subject Code: 17507
S2
P
Cos 2
KVA 2 (S2 ) P sec 2
3) Saving in KVA charges:
= Rs X (S1 –S2)
= Rs X ( P sec 1 P sec 2 )
= Rs X .P ( sec 1 sec 2 )
4) Expenditure towards KVAr to be neutralized:
= Rs Y (Q1 –Q2)
= Rs Y ( P tan 1 P tan 2 )
= Rs Y .P ( tan 1 tan 2 )
5) Net Saving:
= Saving in KVA charges - Expenditure towards KVAr to be neutralized
= [Rs X .P ( sec 1 sec 2 )] - [ Rs Y ( P tan 1 P tan 2 )]
Saving will be maximum when differentiate above equation with respect to 2 and equate
to zero
ds
d
Rs X P (sec 1 sec 2 ) Rs Y P (tan 1 tan 2 )
d 2 d 2
0 X P sec 2 tan 2 0 Y P sec 2 2
0 Rs X P sec 2 . tan 2 0 Rs Y P sec 2 2
Rs X P sec 2 . tan 2 Rs Y P sec 2 2
Rs X tan 2 Rs Y sec 2
Rs X
sin 2
1
Rs Y
Cos 2
Cos 2
Rs X sin 2 Rs Y
sin 2 Rs
6)
Y
X
sin 2 2 Cos 2 2 1
Cos 2 2 1 Sin 22
Cos 2 1 (Y / x ) 2
Most economical Power factor Cos 2 1 (Y / x ) 2
Most economical power factor at which maximum saving will occurs
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Model Answer
Subject Code: 17507
c) ii) A factory takes 300 kW at 110 volts from a 3-phase supply and power factor of 0.7
lagging. A synchronous motor is installed which takes an additional 150 kW. What must
be the kVA rating of this motor to raise the power factor of the system to 0.85 lagging?
Ans:
STTO
^
/
<S)~m
oj
o -7
^
w
'<
o <U9 C f O
\©
s
1 „JU
Given Data:
PL = 300 KW
Power factor
'
-
IstKtfCV)
„
I
^rieuP ^
\
^5
nfui
L
or equivalent diagram
i
Cos 0.7 lag SinL 0.7 tan L 1
improved to 0.85 lag tan new 0.6197
Pm 150 KW
Reactive Power taken by load (QL) = PL tan L ------------------------------------ (1/2 Mark)
= 300 1
= 300 KVAR (lag ) -------------------------- (1/2 Mark)
Reactive Power taken after synchronous motor is connected (Qnew) =
= ( PL Pm ) tan new -------------------------- (1/2 Mark)
= (300 150) 06197
= 450 06197
= 278.8849 KVAR (lag ) ------------------ (1/2 Mark)
Reactive Power taken by synchronous motor to improve P.f =
= (L new )
= 300 278.8849
= 21.115 KVAR ( leading ) -------------- (1/2 Mark)
KVA Rating of Synchronous Motor Sm =
V( P )
m
m
2
--------------------------- (1/2 Mark)
=
/(150)
=
/22500 445.845
2
(21.115) 2
22945.845
= 151.4788 KVA --------------------- (1/2 Mark)
=
Power factor of Synchronous Motor Cosm =
Pm
150
S m 151.4788
= 0.9902 leading ------------------------- (1/2 Mark)
------------------------------------------------------END-------------------------------------------------------
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Page 1 of 43
Model Answer
Important suggestions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more importance.
(Not applicable for subject English and communication skills)
4) While assessing figures, examiner may give credit for principle components indicated in a figure.
The figures drawn by candidate and model answer may vary. The examiner may give credit for any
equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model answer.
6) In case some questions credit may be given by judgment on part of examiner of relevant answer
based on candidate understands.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.
Q.1 a) Attempt any THREE of the following:
a) Compare a group drive and an individual drive.
Ans:
12 Marks
(Any four point expected: 1 Mark each)
S.No.
1.
2.
3
4
5
6
7
8
9
10
Point
Initial Cost
Flexibility
Safety
Reliability
Space required
Overload Capacity
Maintenance cost
Speed control
Mechanical Power
transmission losses
Addition/Alternation
Group Drive
Individual Drive
Less
Less Flexibility
It is less Safe
It has less reliability
Less
Higher
Less
Difficult
More Losses
High
More Flexibility
It is more safe
It has high reliability
More
Less
More
Easily possible
Less losses
Easily not possible
Easily possible
OR
S.No.
Point
1
Definition
Group Drive
In a group drive single large
capacity electric drives is used
to run number of machine
through a long common shaft.
Individual Drive
In this type of drive each
machine has its own
separate electric drive
(motor). It may be directly
coupled or indirectly
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Model Answer
2
3
4
5
6
7
Cost
Total HP
Appearance
Safety
Flexibility
Performance
Less
Less
Not good
It is less Safe
Less Flexibility
Better if operated at full load
8
Any one application of Textile Industry (Similar
each
application will be consider)
coupled
High
More
Good
It is more safe
More Flexibility
Better if operated at full
load
Lathe Machine (Similar
application will be
consider)
b) State various applications of dielectric heating.
Applications of Dielectric Heating:(Any four application expected: 1 Mark each)
Ans:
1) In food processing industry, dielectric heating is used for Baking of cakes & biscuits in
bakeries.
2) Cooking of food without removing outer shell (e.g.-boiled egg) and pasteurizing of
milk.
3) For Rubber vulcanizing.
4) In Tobacco manufacturing industry for dehydration of tobacco.
5) In wood industry for manufacturing of ply wood.
6) In plastic Industry for making different containers.
7) In cotton industry for drying & heating cotton cloths for different processes.
8) In tailoring industry for producing threads.
9) For manufacturing process of raincoats & umbrellas.
10) In medical lines for sterilization of instruments & bandages.
11) For heating of bones & tissues of body required for certain treatment to reduces pains
& diseases.
12) For removal of moisture from oil.
13) For quick drying gum used for book binding purpose.
14) In foundry for heating of sand, core, which are used in molding processes.
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Page 3 of 43
OR
Dielectric heating is used in following purposes:
1. Plywood industry
2. Sand core baking
3. Plastic industry
4. Tobacco industry
5. Bakeries
6. Electronic sewing
7. Dehydration of food
8. Electro medical application
9. Book binding
c) Define the following terms: i) Luminous Intensity ii) Lumen iii) Candle power iv) Waste light factor.
( Each Definition: 1 Mark)
The luminous intensity in any particular direction is the luminous flux emitted by source
per unit solid angle is called the luminous intensity of the source. And its unit is Candela
Ans: i) Luminous intensity:-
OR I
(Where lu min ous flux , w Solid Angle)
w
ii) Lumen:
It is defined as the luminous flux emitted by a source of one candle power per unit solid
angle in all directions
OR
It is unit of luminous flux. One lumen is defined as luminous flux emitted per unit solid
angle from a point source of candle power.
iii) Candle power:
The candle power is the radiation capacity of the light source in the given direction.
The candle power is always given in lumens output per unit solid angle of the given light
source.
C .P
Lummens
, ( Where w Solid Angle)
w
km
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Model Answer
iv) Waste light factor:
When a surface is illuminated by several numbers of the sources of light, there is
certain amount of waste due to overlapping of light waves,
The waste of light is taken into account depending upon the type of area to be
illuminated.
The value of waste Light factor 1 to 1.5
d) Explain any three disadvantages of low power factor and state three methods to improve it.
Disadvantages of Low power Factor: -
Ans:
(Any three disadvantages are expected: 1 Mark each)
1) Cross section of conductor increases: C/s of conductor I 1/ ( pf )
As power factor reduces current increases, cross section of conductor increases. Hence
its cost increases.
2) Design of supporting structure: As power factor reduces, cross section of conductor increases, so its weight
increases. To handle this weight design of supporting structure becomes heavier, so its
cost increases.
3) Cross section of terminals increases: As power factor reduces, current increases, Hence cross section of switch gear,
bus bar, contacts, and terminals increases. So its cost increases.
4) Copper losses increases: As power factor reduces current increases. So copper losses increases. As an
effect efficiency reduces.
Copper losses I 2
1
( P. f ) 2
5) Voltage drop increases: As P.F. reduces current increases. Therefore voltage drop increases, so regulation
becomes poor.
Voltage drop I
1
P. f
km
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Model Answer
6) Handling Capacity of equipment reduces:
Handling capacity (KW) of each equipment such as Alternator, transformer
reduces as power factor reduces.
7) High KVA rating of equipment required:- KVA I 1/ pf ,
As power factor decreases KVA rating of all equipments increases, so that its cost
increases.
1
KVA rating I
P. f
8) Cost/unit increases: - From all above disadvantages it is seen that cost of generation,
transmission & distribution increases. Also its performance efficiency & regulation
reduces, So that cost/unit increases.
Following are the methods of improving power factor:
( Any one methods expected: 1 Mark )
1) By use of static capacitor (Condenser)
2) By use of over excited synchronous motor (Synchronous condenser)
3) By use of over excited Schrage motor
4) By use of phase advancer.
Q.1B)
Attempt any ONE :
06 Marks
a) What is electrical braking ? Explain regenerative braking for D.C. series motor.
Ans:
Meaning of Electrical braking :
( 2 Mark)
It is necessary to stop the vehicle when mechanical working is over or when
required within reasonable time. OR To reduce the speed of train electrical system is used
for braking e.g. Plugging, dynamic braking & Regenerative braking.
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Model Answer
Subject Code: 17507
Schematic diagram of regenerative braking of D.C. series:
( Figure: 2 Mark & explanation: 2 Mark)
Conductor
OH
Current
3
'
<& c v i c e
§
f
.
f
-
<>
r- r
r-
?V
CS)
*
Q
T
Undet
*
(
T
X
normaJ
OH
To I •f o • I
e *- c *> 4 o
To O'
To l i m i t
Cx» f f
rondoctot
nt
f
' >
'
T
1 <x.
condition
Cor> ncc 4 i o r» during
Fig. --- A
EP8
Fig. --- B
Explanation of regenerative braking:
During normal running motors are connected in parallel with field winding in series
w.r.t. armature as shown in figure A.
At the time of regenerating braking all the armature are connected in parallel without
series field winding and all series field winding are connected in series with external
resistance & are separately excited as shown in fig.B
At this time motor acts as a generator and excitation current is so adjusted that generated
voltage (Eg) is greater than supply voltage (V), so that power will be fed back to supply.
This process is continued up to the speed of train reaches up to 20 to 16 km/hr. after that it
is difficult to maintain generated voltage greater than supply voltage. So, electric
regenerative braking is stopped
For final stop mechanical braking is applied.
External Resistances are connected to limit the current.
b)
Ans:
Describe any two methods of current flow control in welding transformer.
Methods of current flow control in welding transformer:
1. Tapped Choke (Reactor) method
2. Moving coil method
3. Magnetic shunt method
.^1
^-
if HI
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Model Answer
Subject Code: 17507
4. Moving core method
5. Saturable reactor method
Explanation: (Any Two method Expected: Each figure: 1Mark & each Explanation: 1Mark )
1. Tapped chock ( Reactor) method:
/
SECONDARY
TAPPED REACTOR
ECiCTHOOE
HOlDfR
/
—•
TRANSFORMER
CORE
E
m
LINE
a
YVOFIt
pi
PRIMARY |l| TAPPED REACTOR
0
n
or equivalent figure
In this method tapping on reactor in the secondary circuit.
Use a tapped reactor, which does not gives a continuous current output demands for certain
important applications.
The limited number of tapes restricts the values of output current available.
However this system is relatively efficient and suitable general fabrication and repair work.
2. Moving coil method:
PRIMARY
LINE
SECONDARY
ADJUSTING SCREW
1
[li] MOVING
TRANSFORMER
CORE
ELECTRODE
HOLDER
COIL
WORK
or equivalent figure
In this method changing magnetic coupling between primary and secondary by physically
changing the position of coil.
Change the reactance of the transformer by changing the relative position s of the coils.
Moving one coil away from the other increases the amount of leakage flux flow between
them, thereby increasing the leakage reactance of the coils. This reduces the current output.
The change in positions of coils is brought about by a lead screw which facilities
continuous adjustment of current.
With such a design, the coils may at times loosened and vibrate when the transformer is in
,
STi
HUH
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Model Answer
Subject Code: 17507
use, causes noise.
The connections to coils may also create service problem because continues flexing.
3. Magnetic shunt method:
ADJUSTING SCREW
PRIMARY
LINE
SECONDARY
NT I
ELECTRODE
HOLDER
-
TRANSFORMER*CORE
WORK
or equivalent figure
In this method changing magnetic coupling between primary and secondary by putting a
movable magnetic shunt.
In this method uses a different principle of changing the reactance of the coils.
It employs path which diverts part of total lines of force linking with the secondary coils.
The movement of this magnetic shunt causes the leakage flux to vary and thereby adjusts
the output current.
This method also gives rise to vibration of movable parts with attendant noise if the parts
wear out and become loose in service.
Sometimes mechanical parts such as the lead screw can dirty and difficult to move.
These drawbacks can be taken care of through carefully assembly and efficient
workmanship in manufacture and through good maintenance at the user ends.
On large machines, movement of magnetic shunt can be conveniently motorized.
4. Moving core method:
ADJUSTING SCREW
REACTOR
TKANS -
FORMEf*
CORE
.
m
ELECTRODE
HOLDER
REACTOR
COIL
LX
RY
SECONDARY
*
UNE
PRIMARY
1
WORK
or equivalent figure
In this method moving the iron core in the reactance instead of in the main core.
The moving core changes the air gap which changes the reactance.
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Model Answer
Subject Code: 17507
The larger the air gap, the smaller the impedance and higher the output.
This method also gives rise to vibration of movable parts with attendant noise if the parts
wear out and become loose in service.
The inside of the typical core moving transformer show in above figure.
5. Saturable reactor method:
RHEOSTAT
^ RECTIFIER /
SATURATED
V\AA<
^
REACTOR
ELECTRODE
HOLDER
LINE
PRIMARY
HT
SECONDARY
TRANSFORMER CORE
WORK
or equivalent figure
In this method putting saturable reactor unit secondary circuit.
In this Method eliminates all the moving parts is with their service problems, but is more
expensive.
In this system, secondary reactor impedance is controlled by regulating the saturation level
of the core electrically.
The system uses a rectifier bridge and a rheostat to control the DC current in the control
coil.
When there is no DC current flowing through the control winding, the impedance is
maximum and the output is minimum. Reverse is the case when the maximum DC current
is flowing in the control winding.
With this method, remote control of welding current is possible, i.e. the reactor current
adjustment can be located near the welder by using an extension cord. The adjustment can
be by the welder's hand or foot.
Electrical circuit diagram of a typical heavy-duty welding transformer having maximum
continuous hand welding current of 300 Amps. at 60% duty cycle is shown in above
figure.
This machine uses moving core magnetic shunt method of current control.
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Model Answer
Subject Code: 17507
Q.2
Attempt any FOUR :
16 Marks
What is load equalization? Explain with neat diagram and graphs, the process of the load
a)
equalization.
Ans: Meaning of load equalization:
( Meaning : 2 Mark, Figure: 1 Mark & explanation: 1 Mark)
There are many types of load which are fluctuating in nature e.g. wood cutting m/c,
Rolling mill. Etc. For such type of loads, load equalization is necessary to draw the
constant power from supply. Because,
When there is sudden load on motor, it will draw more current from supply at start
to meet additional power demand. Due to this heavy current there is large voltage drop in
supply system. This will affect electrical instrument, equipment, m/c, other consumer etc.
which are connected across same supply line.
Also to withstand heavy current, size of input cable increases so cost of cable
increases, Hence it is necessary to smooth out load fluctuations on motor.
The process of smoothing out load fluctuation is called load equalization.
Diagram of Load Equalization:
plu
Speed without
flywheel
Un
s s
s^
co £?
*
1
Plopo Y
U
I oac(
11
Motor ;
: torque
;
Load
Speed with
flywheel
torque
Time ( r )
How load equalization is done?
Load equalization is done by means of flywheel. It is mounted on motor shaft.
Flywheel stores kinetic energy when there is light or no load & it supplies kinetic energy
when there is sudden heavy load on motor. In this way load demand on supply remains
practically constant.
not necessary to use large size of cable as it will not draw more current from supply.
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Model Answer
Subject Code: 17507
b) State the principle and application of eddy current heating.
Ans: Principle of Eddy Current Heating:( Figure: 1 Mark, Principle : 1 Mark & any four application : 1/2 Mark each, Total : 4
Mark )
Mannctic Field
EcLdg
CuxreoE
Heading
edcHj-
direction of
-
*
*
.
nduccd Currcni In Part
current
Current In Ceil
job
BtStHtS
frc
(j o K v
^ Hijjh Frequency
supply
,
to
+o
AOVLH
^
'
'
)
OR
t
Principle of operation for eddy current heating:
The job which is to be heated is wound by coil as shown in figure.
Supply of high voltage (10KV) & high frequency (10-40 KHz) is given to
coil which induces eddy current in job according to Faraday’s law of Electromagnetic
induction & these eddy currents are responsible to produce heat in job itself due to eddy
current loss.
In high frequency eddy current heating the phenomenon of skin effect
plays an important role.
Skin effect at high frequency is more pronouncedced (effective). Due to
this surface of job gets more heated as compared to its depth.
Application of Eddy Current Heating:- ( Any three application expected)
1) For Surface hardening of steel.
2) For analyzing of metals.
3) Sterilization of surgical instruments.
4) For electrolytic tin plating.
5) For soldering & welding.
6) For drying of paints & varnish.
7) Tempering of machine parts.
8) Forging of bolt heads & rivet heads etc
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c) Compare AC and DC system of traction.
Ans:
Sr.No
Points
( Any four point expected: 1 Mark each)
AC System of Traction
1
Supply given to O/H
conductor
1-ph, 25KV, AC 50Hz
2
Type of drive used
1-ph, AC series motor
3
Weight of traction motor
4
Weight of motor coach
5
Starting torque
6
7
Acceleration and
retardation
Overload capacity
8
Method of speed control
1.5 times more than D.C.
series motor for same HP
More Because of
transformer in motor
coach and high weight of
A.C. series motor.
Less starting torque than
D.C. series motor
Less than D.C. series
motor.
Less than D.C. series
motor
Not limited
9
Maintenance cost of
traction motor
Starting Efficiency
Chances of radio
interference
Ridding quality
Insulation cost
Cross section of
conductor
Design of supporting
straight
Distance between two
substation
No. of substation required
for same track distance
10
11
12
13
14
15
16
17
Page 12 of 43
Model Answer
Subject Code: 17507
DC System of Traction
DC 600/750V-Tromways
1500/3000V
Urban/suburban
DC series motor for
traction and DC
compound motor for
tramways.
1.5 times less then A.C.
series motor for same HP
less
High starting torque
High
High
More
Limited, except chopper
control method
Less
More
Yes
Less
No
Less better than D.C.
High
Less
Smooth (Better)
Low
More
light
Heavy
More
Less
Less
More
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18
19
20
21
Size (capacity) of traction
substation
Capital & maintenance
cost of substation
Cost track electrification
Applications
Page 13 of 43
Model Answer
Subject Code: 17507
More
Less
Less
More
Less
Main line services
More
Urban and suburban
services
d) Write any eight desirable characteristics of traction motors.
Ans: Traction motor should posses Following Characteristics:
( Any eight Characteristics expected: 1/2 Mark each)
A) Mechanical Properties or characteristics:
1) It should be simple in design
2) It should be robust in construction to withstand against continuous vibrations.
3) Weight of motor per HP should be minimum in order to increase pay load capacity.
4) It must be small in overall dimensions, especially in overall diameter.
5) It must have totally enclosed type enclosure to provide protection against entry of dirt,
dust, mud, water etc. in drive.
6) When motors are running in parallel they should share almost equal load. (even when
there is unequal wear & tear of driving wheels)
7) It should have high coefficient of adhesion.
8) It should have lower center of gravity.
B) Electrical Properties or characteristics:
9) It should have high starting torque.
10) It should possess high rate of acceleration & retardation.
11) It should be variable speed motor.
12) Its speed-torque characteristics should be such that it should produce high torque at low
speed and low toque at high speed.
13) Motor must be capable of taking excessive overload in case of emergency.
14) It should have simple speed control methods.
15) Electrical braking system should be reliable, easy to operate and control, especially
regenerative braking is possible.
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Model Answer
Subject Code: 17507
16) Motor should draw low inrush current (Starting current, and if supply is interrupted and
restore again.)
17) It should withstand for voltage fluctuation without affecting its performance.
18) It should have high power to weight ratio.
C) General Properties or characteristics:
19) It should have low initial cost.
20) It should have less maintenance cost.
21) It should have high efficiency
State the difference between actual speed and schedule speed of train. State the factors
affecting schedule speed of a train.
Ans:
( Any two point expected: 1 Mark each)
e)
S.No
Actual Speed
Schedule Speed
1
Distance / Time
Distance / Actual time of run + Stop time
2
Actual speed is the speed you are
traveling at any given moment at any
Schedule speed is a true speed which
includes stop time also
given point.
3
Railway or any time table is not based
on actual speed
Railway or any time table is based on
schedule speed
4
Actual Speed is more
Scheduled Speed is less
The following factors affect the schedule speed of a train:
( Any two point expected: 1 Mark each)
1. Acceleration:
By increasing acceleration we can reduce actual time of run, so schedule speed
increases.
2. Retardation:
By increasing retardation we can reduce actual time of run, so schedule speed
4
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Page 15 of 43
increases.
3. Both Acceleration and Retardation:
For a given run by increasing both acceleration and retardation we can reduce
actual time of run, so schedule speed increases.
4. Maximum speed:
By increasing maximum speed we can reduce actual time of run, so schedule
speed increases.
5. Stop time:
By reducing stop time we can reduce the schedule time so schedule speed increases.
6. Coasting period:
For a given run by reducing coasting period we can reduce actual time of
time. so schedule speed increases.
Q.3
a)
Attempt any TWO :
16 Marks
A motor has to perform the following duty cycle: 1) Load rising from 200 kW to 500 kW in
4 minutes. 2) Uniform load of 350 kW for 2 minutes.
3) Regenerative braking power returned to supply from 150 kW to zero in 2 minutes.
4) Remains ideal for 1 minute. Determine power rating of motor.
Ans:
i) Load rising from 200 to 500 Kw
:- 4 min
ii) Uniform load of 350 Kw
:- 2 min
iii) Regenerative braking from 150 Kw to zero :- 2 min
iv) idle for
:- 1 min
<30 o
o
•% %©
© O
AK\
^
l ft) W-
*
^
2 tv\
) v«
-
_
2 rv\ >0
or Equivalent fig-------(1 Mark)
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Page 16 of 43
Rating of Motor in KW =
1
KW
3
KW
2
1
2
2
2
2
KW1 KW2 KW2 t1 KW3 t2 1 KW4 t3 KW5 t4
3
----------- (1Mark)
T
Where,
T = t 1 + t 2 + t3 + t4
T= 4+2+2+1
T = 9 min. ------------------------------------------------------------- ------------ (1 Mark)
1
KW
KW
3
KW
1
2
2
2
2
2
KW1 KW2 KW2 t1 KW3 t2 1 KW4 t3 KW5 t 4
3
------ (1 Mark)
9
1 2002 200 500 5002 4 3502 2 1 (150) 2 2 0 2 1
3
3
------- (1 Mark)
9
KW
780000
9
KW 294.39 KW ------------------Answer----------------
(3 Mark)
So, Select nearest standard rating of motor available in the market.
b)
A 50 kW, three phase, 440 V resistance oven is to provide nickel-chrome strip 0.3 mm thick,
for the three-star connected heating elements. If the temperature of the wire is to be 1500°
C and that of the charge is to be 1000° C, calculate a suitable width of the strip. Take
emissivity as 0.91 and radiation efficiency as 0.6. The specific resistance of nichrome alloy is
1.016 X 106. What would be the temperature of the element, when charge is cold at 20° C ?
Ans: Given Data:
T1 = 1500 0C = 1500 +273 = 17730K
T2 = 1000 0C = 1000 +273 = 12730K
Radiation efficiency = 0.6, specific resistance of Ni-Cr = 1.016x10-6 ohm m, emissivity = 0.91.
(NOTE :_This problem is solved by taking value Specific resistance of Ni-Cr =
1.016 x 10 6 and also by taking value Specific resistance of Ni-Cr = 1.016 x 10 -6 :
Give marks to both answers)
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Model Answer
Subject Code: 17507
Solution By take Specific resistance of Ni-Cr = 1.016 x 10 6 :
T
T
H 5.72 104 k.e [ ( 1 ) 4 ( 2 ) 4 ] w / m 2
1000
1000
---------------------------------- (1/2 Mark)
H 5.72 104 0.6 0.91 [ (
1773 4 1273 4
) (
) ] w / m2
1000
1000
H 226602.97 w / m2
------------------------------------------------------------ (1 Mark)
Thickness : 0.3 mm 0.3 10 3 m
l
V2
wt P
----------------------------------------------------------- (1/2 Mark)
V
440
Voltgae across each resis tan ce
3
3
r -r
Voltgae across each resis tan ce 254.03 volt
------------------- (1 Mark)
Power
50 KW
3
16.6666 103 watt
-------------------------- (1 Mark)
2
l
V
wt P
l
(254.03)2
w ( 0.3 10 3 ) (16.66 103 ) 1.061106 )
l
(254.03) 2 ( 0.3 103 )
w (16.6666 103 ) 1.061106 )
l
1.1433 109 ................Equation........I
w
t 2H
l2
V2
-------------- (1 Mark)
------------------------------------------------------ (1/2 Mark)
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l2
l
l
Page 18 of 43
V2t
2 H
V2t
2 H
(254.0341)2 0.3 103
2 1.016 106 226602.97
l 6.4841 106 mtr
---------Equation II---------------------------- (1 Mark)
Putting in Equation : I
l
1.1433109
w
w
6.484110 6
1.1433 10 9
w 5671.3898 mtr
Answer : Length
l 6.4841 10
6
------------------------------------------------- (1 Mark)
mtr
Widgth w 5671.3898 mtr
PART - II: When charge is cold at 20° C?
Temperature of charge = T2 = 200C + 273 0C = 2930C
T
T
H 5.72 104 k .e [ ( 1 )4 ( 2 )4 ] w / m 2
1000
1000
H 5.72 104 0.6 0.91 [ (
T1 16400 K
T1 4
293 4
) (
) ] w / m2
1000
1000
--------------------------------------------------------- (1/2 Mark)
OR Student may solve this type
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Model Answer
Subject Code: 17507
Solution By take Specific resistance of Ni-Cr = 1.016 x 10 -6 :
T
T
H 5.72 104 k.e [ ( 1 ) 4 ( 2 ) 4 ] w / m 2
1000
1000
----------------------------- (1/2 Mark)
H 5.72 104 0.6 0.91 [ (
1773 4 1273 4
) (
) ] w / m2
1000
1000
H 226602.97 w / m2
-------------------------------------------------------- (1 Mark)
Thickness : 0.3 mm 0.3 10 3 m
l
V2
wt P
-------------------------------------------------------- (1/2 Mark)
V
440
Voltgae across each resis tan ce
3
3
r -r
Voltgae across each resis tan ce 254.03 volt
------------------- (1 Mark)
Power
50 KW
3
16.6666 103 watt
--------------------------- (1 Mark)
2
l
V
wt P
l
(254.03)2
w ( 0.3 10 3 ) (16.66 103 ) 1.06110 6 )
l
(254.03) 2 ( 0.3 103 )
w (16.6666 103 ) 1.06110- 6 )
l
1143.24 ................Equation........I
w
t 2H
l2
V2
----------------- (1 Mark)
------------------------------------------------------ (1/2 Mark)
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l2
l
l
Page 20 of 43
V2t
2 H
V2t
2 H
(254.0341) 2 0.3 10 3
2 1.016 10 6 226602.97
l 6.4841 106 mtr
---------Equation II---------------------------- (1 Mark)
Putting in Equation : I
l
1143.24
w
6.4841
w
1143.24
w 0.005671 mtr
Answer : Length
------------------------------------------------- (1 Mark)
l 6.4841 mtr
Widgth w 0.005671 mtr
PART - II: When charge is cold at 20° C?
Temperature of charge = T2 = 20 0C + 273 0C = 2930C
T
T
H 5.72 104 k .e [ ( 1 )4 ( 2 )4 ] w / m 2
1000
1000
H 5.72 104 0.6 0.91 [ (
T1 16400 K
T1 4
293 4
) (
) ] w / m2
1000
1000
--------------------------------------------------------- (1/2 Mark)
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Model Answer
Subject Code: 17507
c) What are the different safety and protective devices used in elevators? Also state functions
of each device.
Ans: Safety Devices used in elevators & its function:( Any four safety device & their function expected: 1 Mark each)
S.No
Safety Devices
1
Door safety switch
2
3
4
5
6
7
Over travel switch
Over speed control switch
Car safety switch
Car operating switch
Emergency STOP Switch
Fire Fighting Equipment
function
Door will not open when elevator is in operating
condition.
It avoids over travel
It controls the speed
It protects the car
It operates the car
In case of emergency this switch is operated
For Extinguishing Small Fire
Protective Devices used in elevators & its function:( Any four Protective device & their function expected: 1 Mark each)
S.No
Protective Devices
function
for a ON-OFF purpose
2
Main line service switch (main switch and
fuse)
CB and overload relay
3
Phase failure protective relay.(1-ph preventer)
4
Phase reversal protective relay
5
Over speed, slow down relay
1
for protection against over current
fault
It as good as single phase
preventer, It protects motor to run
on single phasing
It avoids motor to run in reverse
direction
It avoids to run motor in over speed
Q.4 A) Attempt any THREE :
a) Write classification of electric welding and its advantages.
Ans: Classification of electric welding :
i) Resistance Welding:-(Plastic / Non- Fusion / Pressure Welding )
1) Spot welding
2) Seam welding
3) Projection Welding
12 Marks
( 2 Mark)
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Model Answer
Subject Code: 17507
4) Butt Welding
5) Flash Butt welding
ii) Arc welding (Fusion/Non pressure welding):1) Carbon Arc Welding:
a) shielded welding b) unshielded welding
2) Metal Arc Welding:
a) shielded welding b) unshielded welding
Advantages of electric welding:
( Any Four advantages are expected: 1/2 Mark each)
1. It requires less time for joining (welding) two metals.
2. Two similar and dissimilar metals can be welded.
3. Control of current (temperature) and welding time can be controlled accurately and
automatically.
4. More perfect (sound) and uniform weld is obtained.
5. Properties of weld and joining material remain same.
6. It is more reliable.
7. Welding process is clean
8. Easy to operate and handle.
9. Welding equipments are portable.
10. No Standby losses. So high efficiency.
Explain the operating principle and working of a fluorescent lamp. Mention the function of
b) following components: i) Electrodes ii) Choke in) Capacitor in glow type starter. iv)
Capacitor connected across input terminals.
Ans:
cs
*
» r*er4
gases
ki \ u
oo
±
f
4^4 be.
Cnoke
C p,F
O
-J
^
sapless
SpatrtaT
*
no O
roscerd po^odear
^Cmfiycvi
^ y powd ^ )
i
C
H
P’ F
rgdi
_
^Laro- ni— —
fe
.
-
leqvientj
^
fm pxoxj^ n^en
c^ px^ ci Vo y -
-
'
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Model Answer
Subject Code: 17507
Operation:
( 2 Mark )
When switch is ON current flows through the choke.
At that time choke induces high voltage which is applied to two filaments
Due to this there will be ionization so that light will be emitted through the tube.
OR
Fluorescent lamps produce light by passing an electric current through a gas that emits
light when ionized by current. An auxiliary device known as a ballast supplies voltage to
the lamps’s electrodes, which have been coated with a mixture of alkaline earth oxides
to enhance electron emission.
Function of following components:
(Each function of components: 1/2 Mark)
i) Electrode:
It is made by high resistive element. to initiate an arc
ii) Choke: For providing high voltage at the time of starting and limit the current.
iii) Capacitor in glow type starter: To make and break the circuit to start the tube.
iv) Capacitor connected across input terminals: To improve the power factor, To
minimize the radio interference
c) Write short notes on the following: i) Two part tariff.
Ans: i) Two part tariff:
ii) Power factor tariff
( 2 Mark)
In this type of tariff energy bill is split into two parts.
ENERGY BILL= FIXED CHARGE which depends on load (KW)
+RUNNING CHARGE which depends on actual energy consume (KWH)
Fixed charge which depends on load (KW) which is declared by consumer on test
report.
There is no separate meter is installed to measure load.
Only one energy meter is used to measure number of units consumed.
This type of tariff system is used for residential and commercial consumers.(up to 20
KW)
This type of tariff is not used for industrial consumers. e.g.
Note :- Table not necessary
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Model Answer
Subject Code: 17507
t or .Residential consumers: e.s.
F i x e d/ D e m a n d C h a r g e
S f n g l e P h a s e : R s. 4 0 p e r
month
T h r e e P h a s e : R s. 1 3 0 p e r
month .
./ k W h )
Energy Charge (R s
Flat rate
e . s . Rs. 5 / Kwh for all units
consumed
OR
Block rate tariff
.
For Industrial / commercial consumers load up to 20 Kw
F i x e d/ D e m a n d C h a r g e
R s. 1 9 0 p e r m o n t h
e.s.
.
Energy Charge (R s / kWh)
Flat rate
e . s. Rs. 6 / Kwli for all units
consumed
OR
Block rate tariff
.
ii) Power Factor Tariff:-
( 2 Mark)
In addition to basic tariff (Maximum Demand Tariff / KVA Maximum Demand
Tariff / Load factor tariff) the tariff in which P.F. of industrial consumer is taken into
consideration. Is known as Power Factor Tariff.
If the P.F. of consumer is less than P.F. declare by Supply Company (say below 0.9
Lag.) than penalty will be charged in energy bill.
If The P.F. of consumer is more than P.F. declare by Supply Company (say above
0.95lag.) than discount will be given in energy bill.
As usual consumer has to pay actual energy consumption charges
Application :This type of tariff is applicable to industrial consumer/H.T/ commercial consumers
with contract demand above 80 kw/ 100Kva/107 hp consumer.
Note :- Table not necessary
Incentives and Penalties to Power factor tarrif :Power factor incentive:- e.g.
Power Factor
Percentage of incentive
0.95
0% of energy bill
Above 0.96
1% of energy bill
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Model Answer
Subject Code: 17507
Above 0.97
2% of energy bill
Above 0.98
3% of energy bill
Above 0.99
4% of energy bill
At unity P.F.
5% of energy bill
Note :- Table not necessary
Power factor penalty:- e.g.
Power factor lagging
Percentage of penalty
For 0.90 Power factor lagging
0% of energy bill
For 0.89 Power factor lagging
2% of energy bill
For 0.88 Power factor lagging
3% of energy bill
For 0.87 Power factor lagging
4% of energy bill
For 0.86 Power factor lagging
5% of energy bill
For 0.85 Power factor lagging
6% of energy bill
For 0.84 Power factor lagging
7% of energy bill
For 0.83 Power factor lagging
8% of energy bill
For 0.82 Power factor lagging
9% of energy bill
For 0.81 Power factor lagging
10% of energy bill
d) Derive an expression for the most economical value of power factor.
Ans: Derivation:
KVAR c Qc
p kW
4* 1
o.
y
H5 SJ
=
L
A
C
r
kVAR 2
Q2
1
*
- -
kVAR1
si
oi
B
or equivalent figure…
(1/2 Mark)
Let,
P = Active power KW
S1, S2 = KVA Maximum demand before and after improving power factor
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Subject Code: 17507
Page 26 of 43
Q1, Q2 = Lagging reactive power before & after improving power factor
QC = Leading Reactive power drawn by Capacitor
Cos1 = Initial Power factor
Cos 2 = Improved Power factor
Rs X = Tariff charges towards M.D. (KVA) /year
Rs Y = Expenditure towards KVAr to be neutralized per year (Expenditure towards P.F.
improving apparatus)
1) Before improving Power factor: ---------------------------------------------------- ( 1/2 Mark)
Q1 P tan 1
Cos 1
S1
P
S1
P
Cos 1
KVA1 (S1 ) P sec 1
2) After improving Power factor: ------------------------------------------------------- ( 1/2 Mark)
Q 2 P tan 2
Cos 2
S2
P
S2
P
Cos 2
KVA 2 (S 2 ) P sec 2
3) Saving in KVA charges: ----------------------------------------------------------------- ( 1/2 Mark)
= Rs X (S1 –S2)
= Rs X ( P sec 1 P sec 2 )
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Page 27 of 43
= Rs X .P ( sec 1 sec 2 )
4) Expenditure towards KVAr to be neutralized: -------------------------------------- ( 1/2 Mark)
= Rs Y (Q1 –Q2)
= Rs Y ( P tan 1 P tan 2 )
= Rs YxP ( tan 1 tan 2 )
5) Net Saving: ----------------------------------------------------------------------------- ( 1/2 Mark)
= Saving in KVA charges - Expenditure towards KVAr to be neutralized.
= [Rs X .P ( sec 1 sec 2 )] - [ Rs Y ( P tan 1 P tan 2 )]
Saving will be maximum when differentiate above equation with respect to 2 and equate
to zero.
ds
d
Rs X P (sec 1 sec 2 ) Rs Y P (tan 1 tan 2 )
d 2 d 2
0 X P sec 2 tan 2 0 Y P sec 2 2
0 Rs X P sec 2 . tan 2 0 Rs Y P sec 2 2
Rs X P sec 2 . tan 2 Rs Y P sec 2 2
Rs X tan 2 Rs Y sec 2
Rs X
sin 2
1
Rs Y
Cos 2
Cos 2
Rs X sin 2 Rs Y
sin 2 Rs
6)
Y
X
sin 2 2 Cos 2 2 1---------------------------- ( 1/2 Mark)
Cos 2 2 1 sin 2 2
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Model Answer
Subject Code: 17507
V
Most economical power factor = Cos 2 1 (Y / x) 2
--------------------------------- ( 1/2 Mark)
Most economical power factor at which maximum saving will occurs
Q. 4B) Attempt any ONE of the following :
06 Marks
a) With the help of neat sketch explain construction and working of spot welding machine.
Ans: 1) Spot Welding:
( Figure: 2 Marks, Construction: 2 Marks & Working: 2 Marks)
& P Ot
Pre
0|
^s y r e
'
(
Q
- electrode hollo©
Cu, 01 Cuallou
watee cooled! •
AC
Core
.
tip electrode
made fcromcd- Cue
oy
Gu aJJtxu
SU
ac
supply
spot
electrode holders
Electrode
arc
I
—
>
Pieces to
be welded
p tessate
>
OR
[AJalrlinn Irancfnrmor
Construction:
Spot welding means the joining of two metal sheets at suitable spaced interval.
It consists of:
Transformer used for spot welding is designed for low voltage and high current
secondary.
Transformer is oil cooled and portable
There are two electrodes one is fixed and other is movable
The electrodes are hollow and water cooled.
Electrodes are made from copper or copper alloys and tips of electrodes are made from
Cd-Cu or Cr-Cu.
Working:
As shown in fig. Job to be welded is placed one over the other between two electrodes
under pressure
Sufficiently heavy current at low voltage is passed directly through two metals in contact
to be welded.
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Model Answer
Page 29 of 43
Heat is produced due to I2R losses where ‘R’ is the contact resistance.
This heat is utilized to obtain welding temperature (to become a plastic state)
When welding temperature is reached supply is cut down and external pressure is applied
simultaneously across the job to complete weld.
Magnitude of current varies from 1000A to 10000A.and the voltage between electrodes is
usually less than 2V.
The period of flow of current and magnitude of current depends upon thickness of sheet
(job) to be welded.
b) The monthly reading of a Consumer's meter are as follow : Maximum demand = 50 kW
Energy consumed = 36,000 kWh, Reactive energy = 23,400 KVAR. If the tariff is Rs. 80 per
kW of maximum demand plus 8 paise per unit plus 0.5 paise per unit for each 1% of power
factor below 86%, calculate the monthly bill of the consumer.
Ans: Note:
( 4 Marks)
Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model
answer.
In case some questions credit may be given by judgment on part of examiner of relevant answer
based on candidate understands
Given Data:
Maximum demand = 50 kW
Energy consumed = 36,000 kWh
Reactive energy = 23,400 KVAR
Total Bill Annual Demand Ch arg es Annual energy ch arg es reactive energy / annum
Total Bill (50) ( Rs. 80) (36000) (0.08) (23400) (0.005)
Total Bill Rs.6997
OR
Total Bill Annual Demand Ch arg es Annual energy ch arg es reactive energy / annum
Total Bill (50) ( Rs. 80) (36000) (0.08)
Total Bill Rs.6880
Q.5
Attempt any FOUR :
16 Marks
a) Draw the following types of lamp fittings and lighting systems with the help of light
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Page 30 of 43
distribution graphs and its applications.
i) Direct lighting
h) Indirect lighting iii) Semi-direct lighting iv) Semi-indirect lighting.
Ans: 1. Direct lighting: ( Each Figure: 1/2 Mark & Each Application: 1/2 Mark, Total : 4 Mark)
Dracl
Direct : All the light goes downward or toward (90-10096)
Application:
The direct lighting scheme is widely used in drawing room, workshop and flood
lighting etc.
2. Indirect lighting:
Indirect: All the light goes upward or away (90 — 10096)
Application:
Which is useful for drawing offices and composing rooms. It is also used for decoration
purposes in cinema halls, hotels etc.
3. Semi direct lighting:
\ /
/ I \
Semi - direct : Most light goes downward (60 - 9096!)
Application:
It is mainly used for interior decoration.
m
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Model Answer
Subject Code: 17507
4. Semi indirect lighting:
\i/
7A
or equivalent figure.
Semi - indirect : Most of the light goes upward or away (60- 9096)
Application:
It is mainly used for interior decoration.
b) Describe carbon arc welding with neat sketch.
Ans: Figure carbon arc welding:
Ccurbon
( Figure : 2 Mark & Explanation: 2 Mark)
Weldmg .
Arc
,
l
welding
?ad
lead
cable
tapping
+ ve anode
•
carbon elect rode
5
- art
- return
-
Iead co bl e
erenHaJ
Compound genewadcJt.
DO dLi
^
or equiavlent figure
It is explain on following points
Principle of arc welding:
The process in which two metal parts to be welded are brought to a molten state and then
allowed to solidify is called as arc welding. Melting of metal is obtained due to heat developed
by an arc struck between carbon electrode and metal to be welded (Job) and the additional
metal is deposited in the weld from a filler rod.
Carbon arc welding is explain on following points :- (Any four points are expected)
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Model Answer
1. Type of supply used:
2. Type of Electrode:
3. Supply Equipment used:
Page 32 of 43
Only DC supply is used.
Carbon Electrode are used.
D.C Differential component Generator or
Rectifier
D.C Differential component. Generator has
dropping characteristics.
More
More
4. Arc Stability:
5. Temperature obtain:
6. Possibility of arc blow is
more.:
7. Capital Cost:
8. Running cost:
9. Maintenance cost:
10. Stand by losses:
11. Efficiency:
12. Voltage required:
13. Types:
14. Application:
15. Limitation:
More
More
More
More
Less
50 to 60 volt D.C
Flux is used and flux is not used
For welding non ferrous metals
Not suitable for overhead welding
c) What are the requirements of ideal traction system ? What are the different traction systems?
Ans: Ideal Traction system should processes following requirement:( Any Four requirement expected: 1/2 Mark each)
1. It should be Pollution free.
2. It should have low capital, Running and maintenance cost.
3. It should have quick starting time.
4. It should have high starting torque.
5. It should have high rate of acceleration & retardation.
6. Highest speeds are possible.
7. It should have easy speed control method.
8.
Its braking system should be reliable and causes less wear.
9. It should have better riding quality (less vibration)
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10. It should be free from unbalance forces i.e. coefficient of adhesion should be more.
11. It should have lower centre of gravity.
12. The locomotive should be self-contained and able to run on any route
13. There should be no standby losses.
14. It should have high efficiency
15. Regenerative braking should be possible.
16. The wear caused on the track should be minimum.
17. Equipment should be capable of overloads for short periods.
18. Capability of withstanding voltage fluctuations.
19. Parallel running usually more than one motor (2 or 4 motors) should be possible.
20. Traction system should be clean & long life.
21. There should be no interference to the communication lines running along the lines.
Following are the different types of traction system are used:
(2 Marks)
1.
2.
3.
4.
5.
Steam Engine Traction system
Diesel Engine (IC) Traction system
Diesel-Electric Traction system
Battery operated Traction system
Electric Traction System
OR
1. AC Traction System
2. DC Traction System
3. Composite traction system :a)1-Ph AC (1-ph, 25KV) – DC Supply System
b) Kando System (1-Ph AC – 3-Ph AC)
d) Draw a neat diagram of A.C. electric locomotive and explain function of each part in it.
Ans: Diagram of A.C. electric locomotive:
( Figure: 2Mark & Function : 2 Mark)
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Model Answer
Subject Code: 17507
Block Diagram of AC electrical locomotive
lph AC
25KV 50Hz
u supply
Pantograph
Contact wire
CB
Filter
T
Rectifier
Transformer
Motor Control
rrn
rrrr^
DC Series motor
Explanation:
1) Overhead contact wire:
Supply of 1-ph, 25KV, 50Hz, AC is given to overhead conductor.
2) Current collecting device:
It collects current from overhead contact wire and passes it to tap changing transformer
through circuit breaker.
3) Circuit breaker (C.B):
It is connected in between current collecting devices and tap changing transformer.
SF6 circuit breaker is used.
To disconnect locomotive equipments whenever there is fault.
It opens automatically when train passes neutral zone (from zone No.1 to Zone No.2)
4) On load tap changing transformer:
It changes the tap without disconnecting the load on transformer. Its purpose is to
vary the voltage for speed control of traction motor.
5) Traction Transformer:
It step down input voltage 25 KV to working voltage of traction motor
(1500V/3000V).
6) Rectifier:
It converts secondary voltage of transformer into DC supply.
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Model Answer
Subject Code: 17507
7) Filter circuit (smoothing reactor):
It is used to obtain pure DC supply.
8) Motor control unit: It controls operation of traction motor.
9) Traction Motor:
It gives mechanical power to run the train DC series motor is used as traction motor
e) With a suitable diagram explain series-parallel control of D.C. series motor.
Ans: Series parallel control of DC series motor:
1. For traction purpose, two motors are operated in following steps.
Series steps of traction motor:
( Steps for series control : 2 Mark)
Step 1 –
Two traction motors M1 and M2 are connected in series and started with all starting
resistances in series.
Step 2 to 7 –
The starting resistances are cut out one by one gradually and finally two motors are in
series without any resistance.
In series connection the supply voltage V is divided in two motors. (Both motors get half
or (V/2) volts). So speed is also half. (N/2)
L^ws^J — vw— L
.—
_<5>
Ml
— -e> j
±-WV— — wv— 1— VA— L
vac
step 2
^—
UAA
-
W\r-I
1
step 3
Ml
<
— w\ —
—-
i
—
Uw^— wv— — wv— L
-
^
S>J --wv—
vac
-— -cs^
Ml
AAA/
Vdc
^—
—
/
^
7
-®-|
M2
WV-l VA
1
— ^ — AAA — — Wv—
M2
—
M2
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Model Answer
Subject Code: 17507
step 4
^
VA
i
VW
Ml
SI
^
1 /W UAAA
U
M2
S2
M
-
Vdc
step 5
Ml
SI
^
i-MA/W
1
S2
—L
S2
AA
^ww-J^wv
WV
M2
-
Vdc
step 6
— I— VW J
—
-I Wv
VW
Ml
^
1
VW
I
VA
M2
Vdc
step 7
SI
Ml
A/W-MAAr
+
M2
WV-M/W-1— VW
Vdc
Voltage across each motor is Vdc / 2 and speed is N/ 2 RPM
Parallel steps of Control of D.C. series Motor:
S2
( Steps for parallel control : 2 Mark)
Step 1 –
After completion of series last step motors are now connected in parallel again with
series resistance otherwise motor will draw very high current and may damage itself.
Step 2 to 7 –
Both motors are now connected in complete parallel and starting resistances are cut out
one by one.
In parallel connection, voltage across M1 and M2 will be full i.e. V (voltage is always
same in parallel).
Voltage across each motor = V and speed of each motor = N
So, voltage is now increased from (V/2) to V.
Hence, speed also increases from (N/2) to N and motor runs with full speed.
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Model Answer
Subject Code: 17507
Parallel Steps
stepl
I
I
— WWI—
— L|
VW
—
-
^
>1
S1
M2
S2
J— wv-L
M
Vdc
step 2
J—
W\/
— I—
+
— ^—
Wv
Ml
<^>|
— I—
VW
J—
-— I—
\A/\
\/V\/
M2
— I—
M
Vdc
step 3
^
+
VW
Ml
S1
UAAA— vw—
-
^— vw— '
—
^
v w—
-
M2
S2
I
V W—
—
Vdc
step 4
J— VW— I—
-
^
VW— — VW—
+
Ml
^
—
— nrvri'-
S2
— vw— ' — vw— I— vw— L- ^- —
^
r
r
M2
—
1
Vdc
step 5
Ml
MA/VM/vV
ww—
I—
-
M
+
M2
v w J— v w—
—
Vdc
step 6
I— vw— '— vw— — vw—
S1
MI
-
+
— vw J ww-L1— vw— L'
^
ftAA
M2
~'—
—
yy>
Vdc
step7
— I—
+
S1
vw— — wv— vw— L^^-'—
Ml
^
vTv — vw I
^
w J— v
w—
Vd c
Voltage across each motor is Vdc and speed is N RPM
S2
M2
—
4
HW
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Model Answer
Subject Code: 17507
Q.6
Attempt any TWO of the following :
16 Marks
Describe the core type (Ajay Wyatt) induction furnace with a neat sketch and state its
a)
application and advantages.
Ans:
(Figure: 2 Mark, Explanation: 2 Mark, Application: 2 Mark & Advantages: 2 Marks)
Neat sketch of ‘Ajax Wyatt’ vertical core furnace:
Core
Verffcal
type Induction Furnace
j
ouflelfot moltan
metal •
(*
mag neiHc
Co re
¬
^
^—
ft
,
IIP OfWiin
^—
heat
i
nsulating material
-
Re Fra c +o r y
mall of furnace
pTimaiij ioindi n< j
,
'
^ central core
scrap
V - notch
Principle of Induction heating:
It is based on principle of transformer. In this type of Induction heating primary
winding is as usual which is wound around one limb of magnetic core but secondary
winding is actually charge which is to be melted is kept in crucible.
When AC Supply is given to primary winding current flows through primary winding
which creates alternating flux in magnetic core this flux links to the secondary winding i.e.
charge through magnetic core. Hence according to faraday’s law of electromagnetic
induction emf will be induced in secondary winding that is in the charge.
As charge forms a close circuit (secondary) heavy current flows through charge this
current is responsible to produce heat in charge due to I2R losses. This heat is utilized to
melt the charge.
Where, R = Resistance of charge & I secondary current.
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OR Student may write construction also instead of principle
Construction of ‘Ajax Wyatt’ vertical core furnace:
Vertical core type induction heating furnace is nothing but transformer. It consists of
Magnetic Core:
Primary winding
Secondary Winding:
Refractory Wall
Opening
Cooling arrangement
Tilting arrangement
Control panel
APFC
Application of ‘Ajax Wyatt’ vertical core furnace: ( Any Two expected)
1. It is used for melting metal having low resistivity.
2. It is used for heat treatment of silver, Copper, nickel etc.
3. Such type of furnace are used for continuous operations only and not used for intermittent
services.
Advantages:
( Any Two expected)
1) As furnace has narrow ‘V’ shape crucible at bottom. So small quantity of molten metal
remains in narrow ‘V’ notch from previous operation, which will help to keep secondary
short circuited.
So no extra care is required to start the furnace
2) Magnetic coupling between primary & secondary winding is better because both windings
are on central limb of magnetic core. So there will be less leakage flux, Hence leakage
reactance is less, so power factor is better than horizontal crucible direct core type
induction furnace.
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Due to pinch effect in ordinary core type induction furnace there are chances of
temporary interruption in secondary circuit when current density exceeds above
500A/cm2 OR 5Amp/mm2..
But in this type of induction furnace there are no chances of interruption in
secondary circuit even if current density exceeds 500A/cm2 OR 5Amp/mm2
Because tendency of weight of charge keep them in contact due to narrow ‘V’
shape.
So we can increase current density above 500A/cm2 OR 5Amp/mm2 to obtain
more heat in less time.
1) Vertical crucible is always better than horizontal crucible for pouring and taking out
the metal. Also space required is less.
2) As heat is produced directly in the charge there is no heat transfer loss. So efficiency
of furnace is more.
3) As heat is directly produced in the charge time required for melting metal is less. So
energy consumption is less.
4) As current is directly induced in the charge there is automatic stirring action taking
place in the charge due to electromagnetic forces developed in the charge due which,
Through mixing of molten metal is possible.
Uniform heating is possible
5) Accurate temperature control.
6) Ideal working condition in a cool atmosphere with no dirt, noise and fuel.
The speed-time curve of a train consists of :
i) Uniform acceleration of 5 km phps for 30 Sec. Free running for 10 min. iii) Uniform
b) retardation of 6 km phps to stop the train.
iv) A stop time of 5 min. Find the distance between the stations, the average and schedule
speed.
Ans: Given Data:
t1= 30 sec
t2= 10min = 600 sec
Tstop= 5 min = 300 sec
acceleration = 5 km phps
retardation = 6 km phps
Vrv/ Vj^' T
"
**
/V c c I
—
Dw
Fnre
-
YUr 'x ^ j
'
'
1
/
C> -
^
-V
^
~
De -f
* *"
£>
1
^
f 3
/
^ <“ -runnir 3
MW< Ca
^ cJ
(T)
----------------------------- (1 /2Mark)
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Page 41 of 43
Vmax
--------------------------------------------------------- (1 /2Mark)
t1
Vmax t1 30 5
Vmax 150 Km / hr ----------------------Answer-------------- ( 1/2 Mark)
Vmax
------------------------------------------------------ (1/2 Mark)
t3
t3
Vmax
150
6
t3 25 sec ----------------------------Answer------------- ( 1 /2 Mark)
Distance covered during Acceleration ( D ) =
2
D
D
Vmax
--------------------------------------------- (1 /2Mark)
7200
(150) 2
7200 5
D 0.625 sec ----------------------Answer------------- ( 1/2 Mark)
Distance covered during Retardation ( D ) =
2
D
Vmax
----------------------------------------------- (1/2 Mark)
7200
D
(150) 2
7200 6
D 0.5208 sec----------------------------Answer------- ( 1 /2 Mark)
D Free running
t 2 Vmax
--------------------------------------------- (1/2 Mark)
3600
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D Free running
Model Answer
Page 42 of 43
600 150
3600
D Free running 25 Km ----------------------------Answer-------- ( 1 /2 Mark)
Dis tan ce ' D ' D D D Free running
Dis tan ce ' D ' 0625 0.5208 25
Dis tan ce ' D ' 26.1458 Km ----------------------------Answer------- ( 1 /2 Mark)
Time ' T ' t1 t 2 t3 30 600 25
Time 'T ' 655 Sec --------------------------------------Answer-------- ( 1 /2 Mark)
Vav
3600 D
-------------------------------------------------------------------- (1/2 Mark)
Time
Vav
3600 26.1458
655
Vav 143.702 Km / hr ------------------------------------------------Answer--- ( 1 /2 Mark)
Vschv
3600 D
----------------------------------------------------------------- (1/2 Mark)
T Tstop
Vschv
3600 26.1458
655 300
Vschv 98.56 Km / hr ----------------------------------------------Answer------ ( 1 /2 Mark)
b) A 3-phase, 50 Hz, 400 V motor develops 100 HP, the power factor being 0.75 lagging and
efficiency 93%. A bank of capacitors is connected in delta across the supply terminals and
power factor raised to 0.95 lagging. Each of the capacitance unit is built of four similar 100
V capacitors. Determine the capacitance of each capacitor.
Ans:
Given Data
Volt : line volts V = 400V,
cos 1 =0.75
Cos 1 0.75
f= 50 Hz
cos 2 =0.95
P= 100 HP x 735.5/093 = 79.086 kW
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tan 1 = 0.88 ---------------------------------------------------------------------- (1 Mark)
tan 2 = 0.3286 ----------------------------------------------------------------- (1 Mark)
Q1 = P tan 1
= 79.086 x 0.88
= 69.595 KVAR
---------------------------------------------------- (1 Mark)
Q2= P tan 2
= 79.086 x 0.3286
= 25.9876 KVAR
---------------------------------------------- (1 Mark)
QC = Q1- Q2
= P tan 1 - P tan 2
------------------(1 Mark)
= 69.595 – 25.9876
= 436074 KVAR
---------------------------------------------- (1 Mark)
Capacitor when connected in Delta:-
C per phase
QC
-------------------------------------------------------- (1 Mark)
3 V2
C per phase
43.6074 103
3 2 50 4002
C per phase
436074 103
3 50.265 106
C per phase 2.891 104 F ------------------------------------------------- (1 Mark)
In ease delta connected phase 4 similar 100V capacitors are connected in series:
The capacitance of each capacitor = 2.891104 4
The capacitance of each capacitor = 11.564 104 ---------------------------- (1 Mark)
------------------------------------------------------END-------------------------------------------------------
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Model Answer
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Important suggestions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more importance.
(Not applicable for subject English and communication skills)
4) While assessing figures, examiner may give credit for principle components indicated in a figure.
The figures drawn by candidate and model answer may vary. The examiner may give credit for any
equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model answer.
6) In case some questions credit may be given by judgment on part of examiner of relevant answer
based on candidate understands.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.
Q.1 A) Attempt any three :
a) Define electric drive. State advantages and disadvantages of electric drive.
Ans: Drive:
(3x4=12)
( 1 Mark)
It is a machine which gives mechanical power. e.g. drives employing electric
motors are known as electric drives.
Following advantages of electric drive:
(Any Three advantages expected: 1/2 each)
1. It is more economical.
2. It is more clean.
3. No air pollution.
4. It occupies less space.
5. It requires less maintenance.
6. Easy to start and control.
7. It can be remote controlled.
8. It is more flexible.
9. Its operating characteristics can be modified.
10. No standby losses.
11. High efficiency.
12. No fuel storage and transportation cost.
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13. Less maintenance cost.
14. It has long life.
15. It is reliable source of drive.
Following disadvantages of electric drive:
(Any Three disadvantages expected: 1/2 each)
1. It is used only where electricity is available.
2. On failure of supply (electricity) it cannot be used.
3. It is not self-contain.
b) Explain in brief the causes of failure of heating elements.
Ans: Following of the different causes of failure of heating element:
( 4 Mark)
i) Formation of hot spot:
Hot spot on heating element is the point which is at higher temperature than remaining
heating element portion. So there is possibility of breaking of heating element at hot spot.
ii) Due to oxidization:
At high temperature material gets oxidized which may cause failure of heating element.
iii) Due to corrosion:
If heating element is directly exposed to chemical fumes then there is possibility of
rusting of heating element which causes failure of heating element.
iv) Mechanical Failure:
Measure heating element alloy contain iron which is brittle. Due to frequent heating &
cooling of heating element, it may break (fail) due to small mechanical injury also.
c) Define : i) Luminous flux ii) Luminous intensity iii) Space to height ratio iv) Utilization factor.
( Each definition : 1 Mark)
i) Luminous flux (F):The total energy radiated by a source of light in all directions in unit is called
Luminous flux. And its unit is Lumen
OR
Ans:
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Model Answer
Subject Code: 17507
Luminous flux is commonly called light output and is measured in lumens (lm).
ii) Luminous intensity:The luminous intensity in any particular direction is the luminous flux emitted by
source per unit solid angle is called the luminous intensity of the source. And its unit is
Candela
OR
I
(Where lu min ous flux , w Solid Angle)
w
iii) Space-Height ratio:
Space height ratio
Space between lamps
Height of lamps above working plane
iv) Utilization factor:It is defined as the ratio of total lumens reaching the working plane to the total
lumens given out by the lamp. Its value is always less than one.
d) State any four causes of low power factor.
Ans: Following are the Causes of low power factor: - ( Any four causes expected: 1 Mark each)
1. Magnitude of Magnetizing Current (I ):As magnetizing current increases, power factor reduces.
2. Due to use of Induction Motor:Most of industrial drives, agriculture pumps, lift, irrigation pump set uses I.M.
which works at lagging power factor, and so power factor reduces.
3. Due to use of Transformer: All transformers works at lagging power factor, so power factor of system reduces.
4. Due to welding transformer: Welding transformers are operated at low p.f. which reduces
p.f. of the system.
5. Due to inductance of transmission & distribution Line: In case of AC transmission & distribution lines, inductance is present
which the main cause of low power factor is.
6. Series Reactor:Series reactor is used in substation to minimize fault current causes low power factor.
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7. Industrial electrical heating furnaces:Induction and arc furnace used in steel manufacturing industry works at low p.f.
which reduces p.f. of the system.
8. Arc Lamp:Arc lamp & electric discharge lamps operates at low p.f.so p.f. of the system reduces.
9. Equipments operated at light load:P.f. falls if equipments like alternator, transformer, I.M. etc are not operated at full load.
10. Improper repairs and maintenance:P.f. falls if proper maintenance or repairs of equipments are not done.
Q.1B)
Attempt any ONE :
(1x6=6)
Define group drive and individual drive. State the advantages and disadvantages of each
a)
drive.
( Each Definition: 1 Mark, Advantages: 2 Mark & disadvantages : 2 Mark)
Ans:
1. Definition of Group drive:In a group drive single large capacity electric drives is used to run number of
machines through a long common shaft as shown in fig. is known as group drive.
2. Definition of Individual drive:In this type of drive each machine has its own separate electric drive ( motor ). It may be
directly or indirectly coupled.
Advantages of Group Drive:-
( Any two advantages expected: 1/2 Mark each)
1. Initial Cost –
A cost of single motor of large capacity is less then cost of number of small
capacity motors for same H.P.
2. Diversification of load –
All the machines and tools may not work at a time, so we can select main motor
of slightly small capacity (HP) than the total requirements of individual machines.
3. Over load capacity –
Group drive has higher over load capacity. E.g. 100% overload on individual
machine would cause only 8 to 10 % overload on main motor.
4. Space required –
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Less
5. Maintenance cost –
Maintenance cost of single motor of large capacity is less than maintenance cost of
number of small motors of total HP.
6. Efficiency and Power Factor –
If group drive is run at nearly equal to full load than Efficiency and Power Factor
of group drive will be higher
Disadvantages of Group Drive:-
( Any Two disadvantages expected: 1/2 Mark each)
1. Flexibility:Flexibility is lost due to common shaft for number of machines.
2. Safety:It is less safe.
3. Reliability:Its reliability is less at the time of breakdown and maintenance of single large
motor, Because, all the machines operations are required to be shut down at the time of
breakdown and maintenance of single large motor.
4. Mechanical power transmission losses:Considerable power loss takes place for transfer of mechanical energy from shaft to
machine.
5. Speed control:Speed control of individual machine is difficult, it requires special arrangement.
6. Addition / Alteration:Possibility of addition or alteration in existing system is limited.
7. Efficiency and Power Factor: –
If group drive is run at reduced load then Efficiency and Power Factor of
group drive will be less.
Advantages of Individual Drive:-
( Any Two advantages expected: 1/2 Mark each)
1. Flexibility:It has more flexibility that is machine can be placed in any desired position and can
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Model Answer
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be shifted whenever needed.
2. Safety:Working conditions are more safe.
3. Reliability:It has high reliability, because breakdown of single motor causes only one machine
operation required to be shut down and not all machines.
4. Mechanical power transmission losses:Less power loss takes place for transfer of mechanical energy from shaft to
machine.
5. Speed control:Speed control is easily possible.
6. Addition / Alteration:Possibility of addition or alteration in existing system is easily possible.
7. Efficiency and Power Factor: –
If it is run at full load than Efficiency and Power Factor of group drive will be
high. If there is no load it can be stopped thus no load losses can be eliminated.
Disadvantages of Individual Drive:-
( Any Two disadvantages expected: 1/2 Mark each)
1. Initial Cost –
Initial cost is high.
2. Diversification of load –
Diversification of load on individual machine is not possible.
3. Over load capacity –
Over load capacity is less.
4. Space required –
More
5. Maintenance cost –
Maintenance cost is more as number of drives are more.
6. Efficiency and Power Factor –
If it is run at reduced load then Efficiency and Power Factor of individual drive
will be less.
Model Answer
Summer– 2016 Examinations
Page 7 of 42
Ans:
( 2 Mark )
a) By applying High Voltage:- Figure:
Explanation:-
b) By separation of two current carrying electrodes suddenly
a) By applying High Voltage
How arc is formed :- for following method
Temperature obtained by arc is very high (35000C to 6000 0C)
and metal to be welded (job)
o
Q
L
u
0
.
c
-Q
a. OQ
L
i
-+0
S
X
<i V
,
-C
70
c
u
It*
rl
a
o
M
u
aj v
0
0
Q$ d
i
-0- +3
t
>
_x
—
ajLiLmia/
f
^rown.
2^0
•a.
c
_
<L
o
or equivalent figure
Operation:
When very high voltage is applied across any two electrodes separated by small air gap
then air between two electrodes gets ionized and ionized air is conducting, so current
starts flowing from one electrode to another electrode in the form of spark (arc).
This arc produces heat energy which is utilized for melting the charge.
High Voltage is required to produce arc and to maintain arc high voltage is not necessary.
Once arc is struck between two electrodes then low voltage is sufficient to maintain the
arc.
( 4 Mark )
Melting of metal is obtained due to heat developed by an arc struck between an electrode
required, so this type of welding is also known as non- pressure welding.
At the time of welding external filler material is required. No mechanical pressure is
allowed to solidify is called as arc welding or stick welding.
The processes in which two metal parts to be welded are brought to a molten state and then
Define electric arc welding:-
b) Define electric arc welding. How arc is formed? State the characteristics of electric arc.
Subject Code: 17507
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Model Answer
Subject Code: 17507
b) By Separation of two current carrying electrodes suddenly:Figure:
2. B y
Pn'ociple o f A r c Forrn (xtioo
Seperation of: ttioo Current Carrjinj
*
AC
electrode
Suddenly .
V
supply
3
1
I
hiah cuuer>1
carrying clechode ®
AC •*
•
electrode ©
ppUj \
LJ
QIC.
-
5
or equivalent figure
Operation: Another way to produce arc is to short circuit two current carrying electrodes as shown in
fig (a) and suddenly withdraw them, then there will be spark between two electrodes as
shown in figure (b)
This arc then produce heat energy which is utilized for melting the charge.
In this method high voltage is not necessary to produce the arc.
Characteristics of Arc:
1. Arc is conducting.
2. Arc has negative temperature coefficient of resistance.
Q.2
Attempt any FOUR :
(4x4=16 Mark)
a) Compare electric braking over mechanical braking.
Ans:
( Any Four point expected: 1 Mark each)
Sr.No.
Electric Braking
1
It is most reliable braking system.
2
Mechanical Braking
In mechanical braking heat is produced at
break block & break shoes, which may be
source of failure of break.
Breaking actuation time is small as Breaking actuation time is more as low
higher value of braking retardation value of braking retardation is obtained
is obtained
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3
Electrical braking is smooth &
gradual.
Where as if mechanical breaks are not
correctly adjusted then there are chances of
sudden braking which is dis -comfortable to
passenger.
Life of braking system is less.
There is more wear & tear of brake shoes,
break block etc. so there is more
maintenance cost.
4
5
Life of braking system is more.
There is less wear & tear of brake
shoes, break block etc. so there is
less maintenance cost.
6
Higher speeds are possible even
when train is going down the
gradient, as breaking system is
reliable.
Higher speeds are not possible when train
is going down the gradient, as breaking
system is less reliable.
7
Higher speeds of train is possible
as braking system is reliable so
pay load capacity increases.
Higher speeds of train is not possible as
braking system is not reliable so pay load
capacity decreases.
8
In addition to electrical braking
there must be arrangement of
mechanical braking for final stop.
No additional arrangement
is required
9
Special arrangement of circuit
extra complication makes
electrical braking system costly.
No special arrangement of circuit extra
complication required so system is less
costly.
b) Define electric heating. Classify the electric heating methods in detail.
Ans: Define electric heating:( Definition: 2 Mark & Classification: 2 Mark)
Electric heating is nothing but heat energy is obtain or created by the use of electrical
energy.
Electric heating are classified as below:
1) Power frequency electric heating:
i) Resistance heating:
a) Direct resistance heating
b) Indirect resistance heating
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ii) Arc Heating:
a) Direct arc heating (furnace)
b) Indirect arc heating
2) High frequency electric heating:
iii) Induction Heating:
a) Direct core type induction heating (furnace)
b) Vertical core type induction heating or Ajax Wyatt induction heating
c) Indirect core type induction heating
d) Core less induction heating
iv) Eddy Current heating
v) Dielectric heating
c) State any six requirements of an ideal traction system.
Ans: Ideal Traction system should processes following requirement:( Any First Two point : 1 Mark each & Reaming any four point : 1/2 Mark each)
1. It should be Pollution free.
2. It should have low capital, Running and maintenance cost.
3. It should have quick starting time.
4. It should have high starting torque.
5. It should have high rate of acceleration & retardation.
6. Highest speeds are possible.
7. It should have easy speed control method.
8.
Its braking system should be reliable and causes less wear.
9. It should have better riding quality (less vibration)
10. It should be free from unbalance forces i.e. coefficient of adhesion should be more.
11. It should have lower centre of gravity.
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12. The locomotive should be self-contained and able to run on any route
13. There should be no standby losses.
14. It should have high efficiency
15. Regenerative braking should be possible.
16. The wear caused on the track should be minimum.
17. Equipment should be capable of overloads for short periods.
18. Capability of withstanding voltage fluctuations.
19. Parallel running usually more than one motor (2 or 4 motors) should be possible.
20. Traction system should be clean & long life.
21. There should be no interference to the communication lines running along the lines.
d) Write the different systems of track electrification.
Ans: Following are the different track electrification system:
1. D.C. Supply system:1. Direct current track electrification:
600V, 750V DC for tramways
1500V, 3000V DC for Train (Urban and sub-urban services)
2. A.C. Supply system:2. 1-Ph, 25KV,standard frequency AC supply system:
1-Ph, 25 KV , 50 Hz
3. 1-Phase, low frequency AC Supply system:
1-Ph, 15/16 KV, 16.2/3 Hz or 25 Hz
4. 3-Ph, Low frequency AC supply system;
3-Ph, 3.3/3.7 KV, 16 2/3 Hz or 25 Hz
Composite system:5. 1-Ph AC (1-ph, 25KV) – DC Supply System
6. Kando System (1-Ph AC – 3-Ph AC)
( 4 Marks)
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Model Answer
Subject Code: 17507
e) Draw speed time curve. Show and list various time periods associated with it.
Ans: Diagram of speed time curve with various time periods :
t
. FREE
T
E
CURVE j y
RUNNING /
“a
/ 1 . RHEOSTATIC
/ 1 / ACCELERATION |
;
1
/
\( \
i
a) i)
K\
T\ COASTING
.
\
i
\\
t2
t3
TIME IN SECONDS
t4
,
BRAKING
RAK NG
' L l I i— \
*1
Q.3
/ RUNNING„
/
SPEED
cz
X
( 4 Marks)
tj
or equivalent figure
Attempt any TWO :
(2 x 8 =16 Marks)
Define: i) Continuous rating ii) Continuous maximum rating iii) Short time rating.
Ans: Definition of Following:i) Continuous Rating:-
( 1.5 Marks)
This is an output which a motor can deliver continuously without exceeding the
permissible temperature limit.
It can deliver 25% over load for two hours without rise in temperature.
ii) Continuous maximum Rating:-
( 1.5 Marks)
This is an output which a motor can deliver continuously without exceeding the
permissible temperature limit. It is similar to Continuous rating but not allowing overload.
iii) Short time Rating:-
( 1 Marks)
This is an output which a motor can deliver for a specific periods (short
duration) e.g. 15min., 20min., 30min. etc. without exceeding the permissible temperature limit.
a) ii) Classify electric elevators on the basis of : i) Service ii) Capacity iii) Speed iv) Power unit.
Ans:
Classify electric elevators on the basis of :
( 1 Mark for each classification)
i) According to Service :a) Passenger Elevators
b) Freight (goods) Elevators
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c) Combination of Elevators
ii) According to Speed of Elevator :a) Low speed Elevator
b) Medium speed Elevator
c) High speed Elevator
iii) According to capacity of Elevator :a) Light duty Elevator
b) Medium duty Elevator
c) Heavy duty Elevator
d) Extra Heavy duty Elevator
iv) According to power unit (elevator machine):a) Drum Elevator
b) Traction Elevator
b) i) Compare direct resistance heating and indirect resistance heating with suitable diagram.
Ans:
Sr.No.
1
Point
Working
Principle
2
3
Heat is transfer
loss.
Temperature
obtained:
4
Definition:
Direct resistance heating
Indirect resistance heating
When current is passed
through charge heat is
produced due to I2R losses
taking place in the charge.
Where, R is the resistance of
the charge and I be the current
passed through charge.
1. When current is passed
through heating element
then heat is produced due
to I2R losses taking place
in the heating element.
2. Heat is transferred towards
charge mainly be radiation
or sometimes by
conduction.
There is heat is transfer loss.
There is no heat is transfer
loss.
Temperature obtained is more
than in direct resistance
heating as heat is directly
produced in charge
As heat is produced in the
charge itself hence its name is
direct resistance heating.
Temperature obtained is less
than direct resistance heating
as heat is not directly
produced in charge
As heat is produced in the
heating element and then it is
transferred towards charge
which is to be heated. Hence
its name is indirect resistance
furnace/oven.
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It has high efficiency.
Efficiency:
Application:
1.This type of heating used
for industrial purpose
2.Salt bath heating: This
utilized for the purpose of
carbonizing, tempering,
quenching and hardening
of steel tools
3. Heating of water in boiler
Diagram
%
Ac / DCSU
"
^
%
pptu
su
~^ ~
L
'/ / / / / / / / / / /
' / ///////.
§
Heall'
I
A^
lecheries
^
Thermal
insu laiion
charg e
Heating element Electrodes are made up of
used:
carbon or graphite.
2
maj
4 Ther
i nsulaJi on
A
%A
AA
ng chembet
r 6 Sprinkled Hi n
tmmmy, IcaijkncjpouSdeT
8
PPS
penfng
loio vtg > h i h current
AC I O C
Overall efficiency is less
than direct resistance heating
due to heat transfer loss.
It is used for Room heater,
Electric Iron, Hair dryer, Hot
plate, Electric water heater,
Electric toaster and Electric
oven etc
-
Direct heaii n
'
7
Page 14 of 42
Model Answer
Subject Code: 17507
A
A
AA
A
2
;*
g
Cha rae
%
^4
'
H eal i og
Cl e rrven -t
•
2
4
Names of Material used for
manufacturing of heating
element i) Nichrome ii)
Constantan or Eureka iii)
Nickel-chromium iv) Ironchromium-Aluminimum
v) Silicon carbonate
b) ii) Explain the principle of dielectric heating.
Ans: Figure of dielectric heating:
D * elec-Kr i°c H e a t i n g^ .
0
me+oJ plcdte.
—
HV
C 50 to as fcv )
job of solid
dLielecVtic moiejriaJ
k Hf
C ) o Vo 3 MVt ^)
-
supp \
|
2
or equivalent figure
Principle of Dielectric heating:
For heating non-metallic material (dielectric material) for e.g. Glass, plastic, wood,
etc. dielectric heating is used.
Material to be heated is placed between two metallic plates as shown in figure (1)
across which a high voltage (20 to 25 KV) and high frequency (10 to 30 MHz) AC supply is
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Model Answer
Subject Code: 17507
given.
Material is heated due to dielectric loss taking place inside the job.
Operation :1. During (+) ve half cycle:
+
Hi
C 10 to 35 W)
k HP
. 10 to Soft Hz)
supply
pP et etj
electron rom
^
ouPat, orbit
shifct ( attract)
-fouiords uppw + Vt
plaie
•
or equivalent figure
Material to be heated is placed between two metallic plates, if upper plate is + Ve,
most of electrons from its outer orbit (of job) gets attracted towards + Ve plate.
2. During (-) ve half cycle:
HV
electron|rom
O to 35
HP
( \ 0 t o 30 MHt)
AC
oupttoibit
shi te
^
p CaM tools) towards
loiDet plate C-PVe)
•
iUpM
During - Ve half cycle field is reversed i.e. bottom plate becomes + Ve. At that
time most of electrons from its outer orbit gets attracted towards bottom electrode.
Effect:Due to inter atomic friction caused by repeated (due to frequency) deformation and
rotation of atomic structure, Dielectric loss takes place inside the job which produces heat.
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Model Answer
Subject Code: 17507
c) A electric motor has load variations as given below :
i) Torque 250 Nm for 20 min.
ii) Torque 150 Nm for 10 min.
iii) Torque 300 Nm for 10 min. iv) Torque 200 Nm for 20 min. If speed of the motor is 750
rpm find the power rating of the motor.
Ans:
i) 250 Nm
for 20 min.
Speed of motor : 750 rpm
ii) 150 Nm
for 10 min.
Rating of motor (KW) =?
iii)
300 Nm
for 10 min.
iv)
200 Nm
for 20 min.
Duty Cycle (T) = t1+ t2 + t3 + t4 ----------------------------------------------- (1 Marks)
= 20+10+10+20
= 60 Min. --------------------------------------------------- (1 Marks)
J
1
\
2
rating of motor (Torque)
2
2
2
T1 t1 T2 t2 T3 t3 T4 t4
---------------- (1 Marks)
T
rating of motor (Torque)
250 2 20 150 2 10 300 2 10 200 2 20
60
rating of motor (Torque)
3175000
60
rating of motor (Torque) 52916 .6666 Nm
rating of motor (Torque) 230 .0362 Nm ---------------------------------------- (1 Mark)
rating of motor ( watt )
rating of motor ( watt )
2 N T
60
----------------------------------------------- (1 Mark)
2 750 230 .0362
60
rating of motor ( watt ) 18057 .8439 Watt ------------------------------------------ (1 Mark)
rating of motor ( Kw)
18057 .8439
1000
rating of motor (kw) 18.057 kW i.e.
rating of motor ( kw) 18 kW
-------------------------------------------------------(2Mark)
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Model Answer
Subject Code: 17507
Q.4 A) Attempt any THREE :
(3 x 4 =12 Marks)
a) Define welding. State the requirements of good weld.
Ans: Define welding:
( Definition: 2 Marks & Requirement: 2 Marks)
It is the process of joining two similar or dis-similar metals by application of heat with or
without application of pressure and addition of filler material.
The good welding has following requirements:( Any Two requirement expected: 1 Marks each)
1) Welding joints must be strong and reliable
2) Joint (welding) is made by proper welding technique.
3) Surface of job should be uniformly welded.
4) Welding Should be free from following defects like:i) Cracks (cold crack or hot crack) ii)overlap iii) porous iv) blow holes
v) incomplete penetration vi)Excess penetration vii) incomplete fusion viii)Suck
buck ix)under flush x)burn through
5) Even counter & width of surface welding.
b) State the laws of illumination.
Ans: Laws of illumination:( Each laws explanation: 1 Mark & each Figure : 1 Mark )
1. Inverse Square Law:Intensity of illumination produced by a point source varies inversely as square of the
distance from source.
Distance from Source
4
2D
Area
1d \
'
4
D
'
9A
A
24
Intensity =
OR
E
Illuminance
(lux)
L
I
d
2
Where,
I = intensity
and
d = Distance
L/ 4
L/ 9
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Model Answer
Subject Code: 17507
2. Lamberts Cosine Law:
According to this law, Illumination at any point on a surface is proportional to the cosine
of the angle between the normal at that point and the direction of luminous flux
Cosine Law: E0
100%
= £ * cos(6)
0°
30°
A
30°
87%
100%
60°
87% \
^
2v
50%
\
4
>
Fig . 6.3 Lambert 's cosine law.
Fig. 6.5 Lambertian surface.
c) Compare block rate tariff and flat rate tariff (any four points)
Ans:
( Any Four Point expected: 1 Mark each)
S.No
1
2
3
Block rate tariff
Flat rate tariff
In case of block rate tariff there are
blocks of units consumed
Each block tariff rate/unit (KWH) is
different
Plus consumer has to pay fix charges.
In case of flat rate tariff there are no
blocks of units consumed
There is flat rate/unit for actual
energy consume
No fix charges have to be paid by
consumer.
If consumer has not consume the
energy he has not require to pay fix
charges
Number of units consume = Current
Reading – Previous Reading of
energy meter.
4
Even if consumer has not consume the
energy he has to pay fix charges
5
Rate Schedule:
Consumption
slab (kWh)
Fixed/Dema
nd Charges
0-100 units
101-300 units
301-500 units
501-1000
units
Single
Phase:
Rs…./Month
Three
Phase:
Rs…./Month
Above 1000
units (balance
units
Energy
Charges
(Rs.kWh)
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Page 19 of 42
6
The consumers which consume less
No such discrepancy in this type of
energy (KWH) shall have to pay fewer
Tariff
charges.,
The consumers which consume more
energy (KWH) shall have to pay more
charges.
7
Due to block rate tariff consumer will
This type of tariff is useful to
use minimum energy to reduce its
consumers whose power
energy bill. So this type of tariff is
consumption is less
helpful from energy conservation point
of view.
d) State any four advantages of good power factor for electric supply system.
Ans: Advantages of good power factor for electric supply system:
( Any Four Advantages are expected: 1 Mark each)
We know that,
P 3 VL I L Cos
For same power to be transmitted at same voltage over a same distance
1
1
I
Cos
P. f
From above equation it is seen that as power factor increases current decreases, due to
decreases in current, system has following advantages
( Any Four advantages expected)
1. Cross section of conductor reduces:
Cross section of conductor I
1
P. f
As P.F. increases current reduce so; cross section of conductor and its weight
reduces hence its cost reduces
2. Design of supporting Structure:
As weight of conductor reduces design of supporting structure (tower) becomes
lighter, so its cost reduces.
^
MW^
.
1
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3. Cross section of terminal (contacts) reduces:
As power factor increases, current reduces. hence cross section of switchgear bus
bar and contacts etc decreases.
4. Copper losses reduces:
As power factor increases current reduces. So copper losses reduces. As a effect
efficiency increase.
5. Voltage drop reduces:
As P.F. increases, current decreases. So voltage drop decreases, So regulation gets
improved (better)
6. Handling capacity (KW) of equipment increases:
As power factor increases, handling capacity of each equipment such as
Alternator, transformer increases
7. KVA rating of equipments reduces:
As P.F. increases, current decreases. So KVA rating of all equipments for egalternator, transformer etc decreases, so its capital cost reduces.
8. Cost per unit (KWH) reduces:
From all above advantages, it is seen that cost of generation, transmission &
distribution decreases, so cost/unit reduces.
Also performance i.e. efficiency & regulation gets improved at high power factor
Q. 4B) Attempt any ONE
06 Marks
a) Define resistance welding. State the types of resistance welding and explain any two in brief.
Ans: Definition of Resistance welding :( 1 Mark)
In resistance welding, sufficiently heavy current at low voltage is passed directly
through two metals in contact to be welded.
Heat is produced due to I2R losses where ‘R’ is the contact resistance. This heat is
utilized to obtain welding temperature (to become a plastic state)
4
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Model Answer
Subject Code: 17507
Types of Resistance Welding:
( Any Four Type expected: 2 Mark)
1. Spot Welding
2. Seam Welding
3. Projection Welding
4. Upset Butt welding
5. Flash Butt welding
Explanation of Types of Resistance Welding:-
( Any Two Type expected: 3 Mark)
1) Spot Welding:
Spot
Pr
Kleldli n
M re
.
'
electrode hoPou)
Cu, o v Cu. ai lou
ctfate?
cooled! •
Ac s « t
tipo electrode
fcrom cd Cu.
’nnadc
oy
Cfc- Gu CLUCUJ
job
^
-
spot
J
electrode holder
f \\e)d \ OA.
vne^.
p ressuxs
or equivalent figure
Explanation
Spot welding means the joining of two metal sheets at suitable spaced interval.
It consists of:
Transformer used for spot welding is designed for low voltage and high current
secondary.
Transformer is oil cooled and portable
There are two electrodes one is fixed and other is movable
The electrodes are hollow and water cooled.
Electrodes are made from copper or copper alloys and tips of electrodes are made from
Cd-Cu or Cr-Cu.
Working:
As shown in fig. Job to be welded is placed one over the other between two electrodes
4
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Model Answer
Subject Code: 17507
under pressure
Sufficiently heavy current at low voltage is passed directly through two metals in contact
to be welded.
Heat is produced due to I2R losses where ‘R’ is the contact resistance.
This heat is utilized to obtain welding temperature (to become a plastic state)
When welding temperature is reached supply is cut down and external pressure is applied
simultaneously across the job to complete weld.
Magnitude of current varies from 1000A to 10000A.and the voltage between electrodes is
usually less than 2V.
The period of flow of current and magnitude of current depends upon thickness of sheet
(job) to be welded.
2) Seam Welding:
decunn Weldi°r> c
1\ met
^
«
- eJ e c-p^de.
nrt> llet. ¥y p «
Acsoppf
<.
job
2
VMeld \ n
n ejts
\
or equivalent figure
Explanation:
Seam welding is nothing but series of continuous spot welding
It consists of:
Transformer used for seam welding is designed for low voltage and high current
secondary.
Transformer is oil cooled
There are two electrodes, in this type beam or roller type electrodes are used.
Working:
Job is kept in between two electrodes under pressure. This pressure is kept constant
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Model Answer
Subject Code: 17507
throughout.
In this type intermittent current is used, it means current is ON for definite time and OFF
for another time interval with the help of timer.
If current is continuously passes then heat produced may cause burning of job.
Heat is produced due to I2R losses where ‘R’ is the contact resistance.
This heat is utilized to obtain welding temperature (to become a plastic state)
When welding temperature is reached supply is cut down and external pressure is applied
simultaneously across the job to complete weld.
There are three types of seam welding:
1. Lap seam welding
2. Mesh seam welding
3. Metal finished seam welding
3) Projection Welding:
Rejection U elding
Ac suppl
7
ioppiog
XS
- job
eJec+>ode
CPIoJb)
pTOjecKon enre.
rrwxdC bc o-te
/
iPeJdbrv
V4elch' r\ cj
Xrne^'
^
^
or equivalent figure
Explanation:
It is modified form of spot welding, before welding projections are made to job on both
or one part to be welded by mechanical means. Hence it is called as a Projection Welding.
It consists of:
Transformer used for projection welding is designed for low voltage and high current
secondary.
Transformer is oil cooled
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Model Answer
Subject Code: 17507
There are two electrodes .In this type flat electrodes are used as shown in figure.
Therefore it is possible to join several welding points (spots) simultaneously
Working:
Job is kept in between two electrodes under pressure. This pressure is kept constant
throughout.
Sufficiently heavy current at low voltage is passed directly through two metals in contact
to be welded.
Heat is produced due to I2R losses where ‘R’ is the contact resistance.
This heat is utilized to obtain welding temperature (to become a plastic state)
When welding temperature is reached supply is cut down and external pressure is applied
simultaneously across the job to complete weld.
4. Upset Butt Welding:
Usl e J- d
IBo-tt
cvxmp
uoeidUog
guff
,
pres.su *-e is
)d
OLpp \ ted cohere V£> e
-tempe rcviure is ok> +oJ r> od.
-
1
1_5 _1SOLS
L5 5 -5 -
1
I
—
—
i L i9
Q
VTTTrtrrxTS irxTO
"
®
DO
elding.
x me^>
•
AC
soppU
^
or equivalent figure
Explanation:
Transformer used for welding is designed for low voltage and high current secondary.
Transformer is oil cooled
The job is clamped as shown in fig. two parts which are to be welded are brought
together
Sufficiently heavy current is passed through joints by welding transformer,
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Model Answer
Subject Code: 17507
which creates necessary heat at joints due to I2R
When welding temperature is reached supply is cut down.
And external pressure is applied simultaneously across the job to complete weld
5) Flash Butt welding: -
Waiding
flash Butt
Flash
cla/np
i
pressure-
—
tb
J°
SLSLXSUiS-saLSU^
nr-rsir
applied
tueld
t’s
.
cohere
-Kmp fs Chained
Anrv^
he.
£U
or equivlent figure
Explanation:
The job is clamped as shown in fig. two parts which are to be welded are brought
near to each other by keeping small air gap,
When welding transformer is made ON, due to heavy currents flash (arc) is
produced between joints.
This arc will produce heat which will create welding temperature.
When welding temperature is reached, supply is cut down and at the same time
mechanical pressure is applied for final weld.
Advantages:
1) Due to flash (arc) surface to be welded becomes clean and pure.
2) Weld obtained is better than butt welding.
3) It requires less power than butt welding.
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Model Answer
Subject Code: 17507
b) A factory has a maximum demand of 250 kW with a load factor of 0.6. The following tariffs
are offered.
a) Two part tariff Z 70/kW of M.D./Year + 4 paise/kWh. b) A flat rate of 10 paise/kWh.
Calculate tariff in both the cases and state with the reason which of the two will be cheaper.
Ans:
No. of Units consume in One Year
Load Factor M .D( KW ) 8760 -------------------------------- ( 1/2 Mark)
0.6 250 8760
1314000 Kwh --------------------------------------------------------(1 Mark)
Case-I: Energy Bill := Tariff given Rs. 70 of M .D. / year Rs. 4 paise / Kwh ------ (1/2 Mark)
= ( 250 70 ) ( 1314000 4 / 100 )
= Rs. 17500 Rs.52560
= 70060 Rs.
------------------------------------------------------------ (1 Mark)
Case-II: Energy Bill := Tariff given flat rate of 10 Paise / Kwh
= 1314000 10 / 100 )
= 131400 Rs.
--------------------------- (1 Mark)
-------------------------------------------------------------- (1 Mark)
According to energy bill Case-I is economical ---------------------------------- (1 Mark)
For industrial consumer Case-I is economical
Q.5
Attempt any FOUR :
(4 x4=16 Marks)
a) Explain in brief the construction and working of sodium vapour lamp.
Ans: Sodium Vapour Lamp diagram:
( Figure: 1 Mark, Construction: 1.5 Marks & Working : 1.5 Marks)
_
a
•
-
L
Ballast
s ^ry\
~
Uacmm
i fube
^
_ Capacitor
Ignitor
&
Lamp
krratoala
'
recic
Kerns dime)
OULUUAUU
N
%equivalent figure
.
^ fn ocle
inerr gas&s
SDcIfam
+
nrnrm
?f
Oi»W
-Hikt
1
1
(
HtWogen
mproue -
rneUl jowl
+ flaunt
etc
'
*
ment CjpqCrM .
OR
1 $ wovJ
Ac supply
or
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Model Answer
Construction
HPS lamps consist of an arc tube (inner) enclosed by an outer tube.
Vacuum is created between the inner & outer glass tube to prevent heat loss.
The arc tube is made from a special glass that can withstand to high temperatures
Arc tube is U Shape
The arc tube contains xenon / neon gas (starting gas), sodium and mercury and two
electrodes.
IT require a ballast to give high voltage at staring to produce the arc (The ballast provides
a high-voltage pulse (2,500 V) for one microsecond for lamp start.)
There is an igniter which sends a pulse to start the discharge.
To improve the power factor a capacitor is connected across the supply. (P.F. is low @
0.3 lag.)
HPS lamps do not have starting electrodes.
Working Principle:
When the lamp is turned on, a high voltage at staring is applied across two electrodes, to
initiate an arc which discharges and vaporizes xenon /neon gas (starting gas), sodium
and mercury.
The energized metal atoms emit light.
After 2 to 5 minutes lamp will glow 100 %.
For running the lamp low voltage of about 165 v is sufficient.
The color of light produce is yellowish.
b) Compare ac welding to dc welding (any four points).
Ans:
( Any Four point Expected: 1 Mark each)
S.No
1
2
3
4
Points
Supply equipment
used
AC Welding
Welding Transformer
DC Welding
DC differential Compound
Generator, or Rectifier
Heating Effect
Temperature Obtain
Possibility of Arc
Blow
Not Uniform
Less
No Possibility
Uniform
More
More Possibility
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Model Answer
Subject Code: 17507
5
Stability of Arc
Use of series Reactor
D.C Differential compound.
Generator has dropping
characteristics.
6
Type of Electrode
Non Coated Electrode is used
7
Voltage Required
Coated Electrode is
compulsory
72 to 100 volt
8
9
10
11
12
13
Capital Cost
Running cost
Maintenance cost
Stand by losses
Efficiency
Application
Low
Low
Low
Low
High, 85%
Resistance Welding , Metal
Arc Welding
High
High
High
High by 25%
Low, 65%
Carbon Arc Welding
50 to 60 volt
c) Draw a neat labelled diagram ofAC electric locomotive. State the function of each part.
Ans: labelled diagram of AC electric locomotive:
( Diagram: 2 Marks & Explanation: 2 Marks)
Block Diagram of AC electrical locomotive
lphAC
25KV 50Hz
u supply
Pantograph
Contact wire
C. B .
Filter |— Rectifier —
Transformer
I
Mater Control
( T T)
C)
DC Series motor
Explanation:
1) Overhead contact wire:
Supply of 1-ph, 25KV, 50Hz, AC is given to overhead conductor.
2) Current collecting device:
It collects current from overhead contact wire and passes it to tap changing transformer
through circuit breaker.
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Model Answer
Subject Code: 17507
3) Circuit breaker (C.B):
It is connected in between current collecting devices and tap changing transformer.
SF6 circuit breaker is used.
To disconnect locomotive equipments whenever there is fault.
It opens automatically when train passes neutral zone (from zone No.1 to Zone No.2)
4) On load tap changing transformer:
It changes the tap without disconnecting the load on transformer. Its purpose is to
vary the voltage for speed control of traction motor.
5) Traction Transformer:
It step down input voltage 25 KV to working voltage of traction motor
(1500V/3000V).
6) Rectifier:
It converts secondary voltage of transformer into DC supply.
7) Filter circuit (smoothing reactor):
It is used to obtain pure DC supply.
8) Motor control unit: It controls operation of traction motor.
9) Traction Motor:
It gives mechanical power to run the train DC series motor is used as traction motor.
d)
"DC series motor is used for traction purpose". Justify your answer with any four
characteristics.
Ans:
( 4 Marks)
Due to following characteristics and advantages, DC series motor is suitable for traction
purpose:
1) Characteristics:
We know that,
Tsh
/
/
/
/
cu
o
/
/
/
Armature current (la)
s
S
I
s.
Armature Current (la )
Characteristics of DC series motor
Torque (Ta)
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Model Answer
Page 30 of 42
Advantages:
1. DC Series motor robust in construction and capable to withstand against continuous
vibration.
2. DC series motor weight is 1.5 times less than 1-Ph AC series motor for same H.P.
3. DC Series motor has high starting torque.
4. DC Series motor has high rate of acceleration and retardation.
5. DC Series motor is variable speed motor. Due to these characteristics motor is protected
against overload.
6. DC Series motor speed-torque characteristics are such that as torque increases speed
decreases.
7. DC series motor has develops high torque at low speeds, low torque at high speed, this is
the basic requirement of traction unit.
8. Commutating property of series motor is good so we get sparkles commutation.
9. Torque is unaffected by variation in supply voltage.
10. DC Series motor maintenance cost is less.
11. When DC series motor are running in parallel the all motors share almost equal load.
12. Torque obtained by DC series motor is smooth and uniform, so it improves riding quality.
e) Write any six desirable characteristics of traction motors.
Ans:
( Any First Two point : 1 Mark each & Reaming any four point : 1/2 Mark each)
Traction motor should posses Following Characteristics :
A) Mechanical Properties or characteristics:
1) It should be simple in design
2) It should be robust in construction to withstand against continuous vibrations.
3) Weight of motor per HP should be minimum in order to increase pay load capacity.
km
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Model Answer
Page 31 of 42
4) It must be small in overall dimensions, especially in overall diameter.
5) It must have totally enclosed type enclosure to provide protection against entry of dirt,
dust, mud, water etc. in drive.
6) When motors are running in parallel they should share almost equal load. (even when
there is unequal wear & tear of driving wheels)
7) It should have high coefficient of adhesion.
8) It should have lower center of gravity.
C) Electrical Properties or characteristics:
9) It should have high starting torque.
10) It should possess high rate of acceleration & retardation.
11) It should be variable speed motor.
12) Its speed-torque characteristics should be such that it should produce high torque at low
speed and low toque at high speed.
13) Motor must be capable of taking excessive overload in case of emergency.
14) It should have simple speed control methods.
15) Electrical braking system should be reliable, easy to operate and control, especially
regenerative braking is possible.
16) Motor should draw low inrush current (Starting current, and if supply is interrupted and
restore again.)
17) It should withstand for voltage fluctuation without affecting its performance.
18) It should have high power to weight ratio.
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Model Answer
Subject Code: 17507
C) General Properties or characteristics:
19) It should have low initial cost.
20) It should have less maintenance cost.
21) It should have high efficiency.
22) It should have long life.
Q.6
Attempt any TWO of the following :
16 Marks
A resistance oven employing Nichrome wire is to be heated from 220V, 1-phase, supply and
is rated at 16 kW. If temperature of element is to be limited to 1170°C and average
a)
temperature of charge is 500°C, find diameter and length of wire. Radiating efficiency, K =
0.57 Emissivity e = 0.9, Specific resistance of Nichrome = 109x106 ohm cm..
Ans: Given Data:
T1 = 1170 0C = 1170+273 = 14430K
T2 = 5000C = 500 +273 = 773 0K
Radiation efficiency = 0.57, specific resistance of Ni-Cr = 1.016x10 -6 ohm m, emissivity = 0.9.
(NOTE :_This problem is solved by taking value Specific resistance of Ni-Cr =
1.016 x 10 6 and also by taking value Specific resistance of Ni-Cr = 1.016 x 10 -6 :
Give marks to both answers)
Solution By take Specific resistance of Ni-Cr = 1.016 x 10 6 :
T
T
H 5.72 104 k.e [ ( 1 ) 4 ( 2 ) 4 w / m 2
1000
1000
OR
T
T
H 5.72 k .e [ ( 1 ) 4 ( 2 ) 4 ] w / m 2
100
100
----------------------------- (1 Mark)
1443 4 773 4
H 5.72 0.57 0.9 [ (
) (
) ] w / m2
100
100
H 11 .4749 10 4 w / m 2
-------------------------------------------------------- (1 Mark)
V2
l
d2 4 P
---------- Equation No.1----------------------------- (1 Mark)
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Model Answer
Subject Code: 17507
l
(220) 2
d 2 4 16 1000 1.09 106
l
2.179660 10 6
d2
Equation No.2 ----------------------------- (1 Mark)
Heat Dissipated = Electrical Power I/p
dlH P
----------------------------- (1/2 Mark)
4
d l 11.6749 10 16000
d l 0.043623
---------------Equation No.3 ----------------------------- (1 Mark)
l
2 2.179660 10 6
d
l d 2 2.179660 106
----------------------------- (1/2 Mark)
By Simplify :
2.179 106 d 2 d 0.043623
0.043623
2.179 106
d 3 20022 .94
d 27.1545 mtr
d 3
--
--------------------------- (1/2 Mark)
Substitute Value of ‘d’ in Equation No.3 to calculate ‘l’ :
d l 0.043623
----------------------------- (1/2 Mark)
0.043623
27.154
l 1.60 103 mtr
l
------------------------------------------------ (1 Mark)
Answer: l 1.6 103 m
Diameter d 2 7.15 mtr
OR Student may solve this type
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Model Answer
Subject Code: 17507
Solution By take Specific resistance of Ni-Cr = 1.016 x 10 -6 :
T
T
H 5.72 104 k.e [ ( 1 ) 4 ( 2 )4 w / m 2
1000
1000
OR
H 5.72 k .e [ (
T1 4
T
) ( 2 )4 ] w / m2
100
100
----------------------------- (1 Mark)
1443 4 773 4
H 5.72 0.57 0.90 [ (
) (
) ] w / m2
100
100
H 11 .4749 10 4 w / m 2
-------------------------------------------------------- (1 Mark)
Thickness : 0.3 mm
V2
l
2
d
4P
0.3 10 3 m
---------- Equation No.1----------------------------- (1 Mark)
(220) 2
l
d 2 4 16 1000 1.09 10 6
l
2179660
d2
---------- Equation No.2 ----------------------------- (1 Mark)
Heat Dissipated = Electrical Power I/p
dlH P
----------------------------- (1/2 Mark)
4
d l 11.6749 10 16000
d l 0.043623
----------------------------- (1/2 Mark)
By Simplifying :
d 2 l 2 1.90297 10 3
d2
1.90297 10 3
l2
Substitute Value of d2 in Equation No.1 :
l
2179660
1.90297 10 3
l2
----------------------------- (1/2 Mark)
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Model Answer
Subject Code: 17507
l 3 2179660 1.90297 103
l 16.06 m
----------------------------- (1 Mark)
Substitute Value of ‘l’ in Equation No.2 to calculate ‘d’ :
l
2179660
d2
d 2 .7144 10
d 2 .714 mm
----------------------------- (1/2 Mark)
3
mtr
------------------------------------------------- (1 Mark)
Answer : Length
l 16 .06 mtr
Diameter d 2 .714 mm
A trapezoidal time curve of train consists of : i) Uniform acceleration of 6 kmphps for 25
seconds
ii) Free running for 10 minutes
iii) Uniform deceleration of 6 kmphps to stop
b)
the train iv) A stop time of 5 minutes.
Find the distance between the stations, average and scheduled speed.
Ans: Given Data:
cLciX c u > WL
Tr > c "Z-b \
^
1
V r\ c\%
<
_
t 3,
4
V
t1= 25 sec
t2= 10 min = 600 sec
acceleration = 6 km phps
\
Tstop= 5 min = 300 sec
retardation = 6 km phps
Vmax
--------------------------------------------------------- (1/2 Mark)
t1
V max t1 25 6
Vmax 150 Km / hr ----------------------Answer-------------- ( 1/2 Mark)
Vmax
------------------------------------------------------ (1/2 Mark)
t3
km
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t3
Page 36 of 42
Vmax
150
6
t3 25 sec ----------------------------Answer------------- ( 1 /2 Mark)
Distance covered during Acceleration ( D ) =
2
D
D
Vmax
7200
--------------------------------------------- (1/2 Mark)
(150) 2
7200 6
D 0.52083 km ----------------------Answer------------- ( 1/2 Mark)
Distance covered during Retardation ( D ) =
2
D
Vmax
----------------------------------------------- (1/2 Mark)
7200
D
(150) 2
7200 6
D 0.52083 km----------------------------Answer------- ( 1/2 Mark)
D Free running
D Free running
t 2 Vmax
--------------------------------------------- (1/2 Mark)
3600
600 150
3600
D Free running 25 Km
----------------------------Answer-------- ( 1/2 Mark)
Dis tan ce ' D ' D D D Free running
Dis tan ce ' D ' 0.52083 0.52083 25
Dis tan ce ' D ' 26.04168 Km ---------------------------Answer------- ( 1/2 Mark)
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Model Answer
Subject Code: 17507
OR Student may solve by using following formula also
V
T T 2 4 K 3600D
2K
But, K
2 --
max
Time ' T ' t1 t 2 t 3 25 600 25
Time ' T ' 650 Sec --------------------------------------Answer-------- ( 1 /2 Mark)
Vav
3600 D
-------------------------------------------------------------------- (1/2 Mark)
Time
Vav
3600 26.04168
650
Vav 144.2308 Km / hr ------------------------------------------------Answer--- ( 1 /2 Mark)
Vschv
3600 D
----------------------------------------------------------------- (1/2 Mark)
T Tstop
Vschv
3600 26.04168
650 300
Vschv 98.6842 Km / hr ---------------------------------------------Answer------ ( 1 /2 Mark)
c) i) Derive the equation of most economical power factor.
Ans: Derivation:
( 4 Mark)
KLVAJR. c Qc
P kw
ITX
7
?
-
S2^
T
c±
kVAR2
Q2
kVARi
SI
Let,
P = Active power KW
1
Q1
b5
km
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S1, S2 = KVA Maximum demand before and after improving power factor
Q1, Q2 = Lagging reactive power before & after improving power factor
QC = Leading Reactive power drawn by Capacitor
Cos1 = Initial Power factor
Cos 2 = Improved Power factor
Rs X = Tariff charges towards M.D. (KVA) /year
Rs Y = Expenditure towards KVAr to be neutralized per year (Expenditure towards P.F.
improving apparatus)
1) Before improving Power factor:
Q1 P tan 1
Cos 1
S1
P
S1
P
Cos 1
KVA1 (S1 ) P sec 1
2) After improving Power factor:
Q 2 P tan 2
Cos 2
S2
P
S2
P
Cos 2
KVA 2 (S2 ) P sec 2
3) Saving in KVA charges:
= Rs X (S1 –S2)
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Page 39 of 42
= Rs X ( P sec 1 P sec 2 )
= Rs X .P ( sec 1 sec 2 )
4) Expenditure towards KVAr to be neutralized:
= Rs Y (Q1 –Q2)
= Rs Y ( P tan 1 P tan 2 )
= Rs YxP ( tan 1 tan 2 )
5) Net Saving:
= Saving in KVA charges - Expenditure towards KVAr to be neutralized.
= [Rs X .P ( sec 1 sec 2 )] - [ Rs Y ( P tan 1 P tan 2 )]
Saving will be maximum when differentiate above equation with respect to 2 and equate
to zero.
ds
d
Rs X P (sec 1 sec 2 ) Rs Y P (tan 1 tan 2 )
d 2 d 2
0 X P sec 2 tan 2 0 Y P sec2 2
0 Rs X P sec 2 . tan 2 0 Rs Y P sec 2 2
Rs X P sec 2 . tan 2 Rs Y P sec 2 2
Rs X tan 2 Rs Y sec 2
Rs X
sin 2
1
Rs Y
Cos2
Cos2
Rs X sin 2 Rs Y
sin 2 Rs
6)
Y
X
sin 2 2 Cos 2 2 1
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Model Answer
Subject Code: 17507
Cos 2 2 1 sin 2 2
Most economical power factor = Cos 2 1 (Y / x ) 2
Most economical power factor at which maximum saving will occurs
c) ii) State the methods of power factor improvement. Explain any one of them.
Ans:
Methods of power factor improvement:-
( 2 Marks)
1) By use of static capacitor (Condenser)
2) By use of over excited synchronous motor (Synchronous condenser)
3) By use of over excited Schrage motor
4) By use of phase advancer.
Explanation:
1) The static capacitor method of power factor improvement.
Before connecting capacitor
Phasor diagram
.
4
vT
\o
V
v
CO
After connecting capacitor
.
i
i
i
+
**
c
\
<c
V
phasor diagram
r
c
ii>:
>Iw
>\s £5
gl
Xv
L
00
\
i
/
*v
12
Nil
\w
II
iiii )
Cos1 = Initial Power factor
( 2 Marks)
km
^ ^
=
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Model Answer
Subject Code: 17507
Cos 2 = Improved Power factor
Calculation from current vector diagram:
I C I1 I 2
IC [Iw tan 1 ] [Iw tan 2]
Now, I C
V
XC
V
IC
XC
XC
1
2 f c
1
2 f X C
C
Magnitude of new current:
I 2 ( Iw )2 (I2 )2
Calculation from power triangle:
Where ,
P = Active power KW
Q1, Q2 = Lagging reactive power before & after improving power factor
QC = Leading Reactive power drawn by Capacitor
S1, S2 = KVA Maximum demand before and after improving power factor
Cos1 = Initial Power factor
Cos 2 = Improved Power factor
KVAR c Qc
p kW
>2
(j
kVAHL
A-
"
>
*
S2
C
r
1
-1*-
Q2
WAR
,
Ql
3
QC = Q1 – Q2
QC [ P tan 1 ] [ P tan 2 ] KVAr rating of capacitor
4
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Model Answer
Subject Code: 17507
Observation:
From above vector diagram & power triangle calculations, if capacitor is connected
across load than following observations are observed.
S.No.
1
2
3
4
5
Parameter
Power factor
Magnetizing current ( I )
Improves
Reduces
Effect
Total current
Lagging reactive power (KVAr)
Apparent power (KVA)
Reduces
Reduces
Reduces
Connection diagram to connect capacitor to improve power factor (Delta connection)
3-0 Load
9-
C
or equivalent figure
KVAR
Farad
3 V 2
(Cph)
2 f
3-ph Star connected Capacitor Bank:
-
3 4» Load
C
yt
-T
C' k" : ,
or equivalent figure
(Cph )
KVAR
Farad
V2
2 f
------------------------------------------------------END-------------------------------------------------------
km
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Model Answer
Page 1 of 34
Important suggestions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more importance.
(Not applicable for subject English and communication skills)
4) While assessing figures, examiner may give credit for principle components indicated in a figure.
The figures drawn by candidate and model answer may vary. The examiner may give credit for any
equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model answer.
6) In case some questions credit may be given by judgment on part of examiner of relevant answer
based on candidate understands.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.
Q.1 A) Attempt any three :
(3x4=12)
a) State and explain any four factors governing the selection of electric drive.
Ans: Following Factors governing / or are considered while selecting electric drive (Motor) for
particular application:
( Any Four factor expected: 1 Mark each)
1. Nature of supply:
Whether supply available is
AC,
Pure DC
Or Rectified DC
2. Nature of Drive (Motor):
Whether motor is used to drive (run)
Individual machine
OR group of machines.
3.
Nature of load:
Whether load required light or heavy starting torque
OR load having high inertia, requirehigh starting torque for long duration.
OR Whether load torque increases with speed (T N)
OR decreases with speed (T 1/ N )
OR remains constant with speed (T = N)
OR increases with square of speed (T N2)
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Model Answer
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4. Electric Characteristics of drive:
Starting,
Running,
Speed control
and braking characteristics
of electric drive should be studied and it should be matched with load requirements(i.e.
machine).
5.
Size and rating of motor:
Whether motor is short time running
OR continuously running
OR intermittently running
OR used for variable load cycle.
Whether overload capacity, pull out torque is sufficient.
6.
Mechanical Considerations:
Types of enclosure,
Types of bearing,
Transmission of mechanical power,
Noise
and load equalization
7.
Cost:
Capital,
Running
and maintenance cost should be less.
List any 6 desired properties of heating element material. Write the names of any two
heating material.
Ans: Following desired properties of heating material :
b)
(Any Six Point are Expected : 1/2 Mark each)
1. High resistivity:
It should have high resistivity. So that is becomes compact in size and produces more
heat with small input current.
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Model Answer
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2. High melting point:
It should have high melting point to withstand at high temperature.
3. High Oxidizing temperature:
It should have high oxidizing temperature or it should not oxidize even at high
temperature.
4. High Resistance to corrosion:
It should have high resistance to corrosion to avoid rusting.
5. High Mechanical Strength:
It should have high mechanical strength to withstand from mechanical injury.
6. Ductile:
It should be ductile so that it can be manufactured into different size & shape.
7. Long Life:
It should have long life.
8. Less Costly:
It should be less costly and easily available.
9. Low temperature co-efficient of resistance:
For accurate temperature control, it should have low temperature co-efficient of
resistance.
10. It should not be brittle.
Names of Material used for manufacturing of heating element:
(Any Two are Expected : 1/2 Mark each)
i)
Nichrome
ii)
Constantan or Eureka
iii)
Nickel-chromium
iv)
Iron-chromium-Aluminimum
v)
Silicon carbonate
vi)
Tungsten vii) Platinum viii) Carbon
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Model Answer
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Page 4 of 34
Define the following terms related to illumination: 1) Luminous Intensity 2) Candle Power
3) MSCP 4) MHCP.
( Each definition : 1 Mark)
Ans:
1) Luminous intensity:The luminous intensity in any particular direction is the luminous flux emitted by source
per unit solid angle is called the luminous intensity of the source. And its unit is Candela
c)
OR I
(Where lu min ous flux , w Solid Angle)
w
2) Candle power:
The candle power is the radiation capacity of the light source in the given direction. The
candle power is always given in lumens output per unit solid angle of the given light source.
C .P
Lummens
, ( Where w Solid Angle)
w
3) MSCP (Mean Spherical Candle power):
It is the average of all candle powers in all direction in all planes.
OR
MSCP
Total Lu min ous lux in lumens
4
4) MHCP (Mean Horizontal candle Power (MHCP) :
MHCP is defined as the mean of the candle power of source in all directions in horizontal
plane.
d) Explain any four disadvantages of low power factor.
( Any Four disadvantages expected: 1 Mark each)
Ans: Disadvantages of Low power Factor: -
1) Cross section of conductor increases: As power factor reduces current increases, cross section of conductor increases. Hence its cost
increases.
2) Design of supporting structure: As power factor reduces, cross section of conductor increases, so its weight increases.
To handle this weight design of supporting structure becomes heavier, so its cost increases.
3) Cross section of terminals increases: -
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Model Answer
Subject Code: 17507
As power factor reduces, current increases, Hence cross section of switch gear, bus bar,
contacts, and terminals increases. So its cost increases.
4) Copper losses increases: As power factor reduces current increases. So copper losses increases. As an effect
efficiency reduces.
5) Voltage drop increases: As P.F. reduces current increases. Therefore voltage drop increases, so regulation becomes
poor.
6) Handling Capacity of equipment reduces:
Handling capacity (KW) of each equipment such as Alternator, transformer reduces as power
factor reduces
7) High KVA rating of equipment required:- ,
As power factor decreases KVA rating of all equipments increases, so that its cost increases.
8) Cost/unit increases: From all above disadvantages it is seen that cost of generation, transmission & distribution
increases. Also its performance efficiency & regulation reduces, So that cost/unit increases.
Q.1B)
Attempt any ONE :
(1x6=6)
Explain how Rheostatic Braking is achieved in case of :
a)
1) DC series motor 2) 3 phase Induction motor.
Ans: Rheostaic braking or dynamic braking of DC series Motor:
(Figure : 2 Mark & Explanation: 1 Mark)
5
.
S j.
— rrrnrrrvrn
Si
-
4
I- o
Ta
oc su PP
'J
•
^
fl
1
fi !
-
G
2
«
Under normal condition
_
Under Dynamic breaking condition
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Model Answer
Subject Code: 17507
Explanation:
In case of DC series motor field winding connections must be reversed when it acts as a
generator i.e. A1 is connected to S1 as shown in figure. And
Value of external resistance connected in armature circuit must be less the critical value
otherwise there will be no excitation.
Rheostaic braking of 3 phase Induction motor:-
,
Dgnomic
brakina
MotOt
30
slipping
(Figure : 2 Mark & Explanation: 1 Mark)
XP1
DC
Staton
Oct ion
*
.
^
opera!f on.
m leaJ toimtiw
t
^
pmOQ
SIi pm na Toia
HarmaL
SUppI
excitation
«
C 3 <p S U ppi
Dunn•g
Dynamjc br&k-incj
OR
«e
,
+r *
h
0<<
;
ESi
J
6
Explanation:
During generating supply 3-phase supply of stator is disconnected & excitation supply is
given to one of winding & external resistance is added in rotor circuit through slip-ring.
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Model Answer
Subject Code: 17507
For seam welding : 1) Draw its neat labelled sketch.
3) Write any two applications.
Ans: 1) Seam Welding its neat labelled sketch:
b)
SecLm Welcirnc
J
Acsoppl
2
Turie-t
2) Explain its working.
( 3 Mark)
^
,
( ~$y
. .
r
TBlIet Vype.
eJectWe
j job
i
lMeld i n
*
mejts
Working:
Job is kept in between two electrodes under pressure. This pressure is kept constant
throughout.
In this type intermittent current is used, it means current is ON for definite time and OFF for
another time interval with the help of timer.
If current is continuously passes then heat produced may cause burning of job.
Heat is produced due to I2R losses where ‘R’ is the contact resistance.
This heat is utilized to obtain welding temperature (to become a plastic state)
When welding temperature is reached supply is cut down and external pressure is applied
simultaneously across the job to complete weld.
Applications of Seam welding:-
1.
2.
3.
4.
5.
6.
7.
8.
(Any Two are Expected : 1/2 Mark each)
It gives leak-proof joints.
Hence used for welding of various types of containers,
Pressure tank,
Tank of transformer,
Gas line,
Air craft tank,
Condenser,
Evaporator and
Refrigerator etc.
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Model Answer
Subject Code: 17507
Q.2
Attempt any FOUR :
(4x4=16 Mark)
Compare Group Drive and Individual drive on the following parameters :
a)
1) Definition 2) Installation cost 3) Appearance 4) Safety and flexibility.
Ans:
( Each Point: 1 Mark)
S.No
1
Point
Definition
Group Drive
Individual drive
In a group drive single large
In this type of drive each
capacity electric drives is
machine has its own separate
used to run number of
electric drive(motor). It may
machine through a long
be directly coupledor
common shaft
indirectly coupled.
2
Installation cost
Less
High
3
Appearance
Good / Not good
Better/Good
4
Safety and flexibility
Less Safe
More Safe
b) Explain any four causes of failure of Heating Element.
Ans: Following of the different causes of failure of heating element:
( Any Four causes expected: 1 Mark each)
i) Formation of hot spot:
Hot spot on heating element is the point which is at higher temperature than
remaining heating element portion. So there is possibility of breaking of heating element
at hot spot.
ii) Due to oxidization:
At high temperature material gets oxidized which may cause failure of heating
element.
iii) Due to corrosion:
If heating element is directly exposed to chemical fumes then there is possibility of
rusting of heating element which causes failure of heating element.
iv) Mechanical Failure:
Measure heating element alloy contain iron which is brittle. Due to frequent heating
& cooling of heating element, it may break (fail) due to small mechanical injury also.
ft
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Model Answer
Compare 1-phase 25 kV AC and DC Track electrification on any eights factors (any four
point)
Ans: Compare 1-phase 25 kV AC and DC Track electrification :
(Any Four point expected: 1 Mark each)
c)
S.No
Points
1
Supply given to O/H
condition
2
Type of drive used
3
Weight of traction motor
4
Starting torque accln and
retardation
Accln and retardation
Overhead capacity
Method of speed control
5
6
7
AC System Traction
1-ph, 25KV, AC 25 Hz
1-ph, AC series motor
1.5 times more than d.c
series motor.
Less starting torque
More
More
Easy
Less better than d.c
High
Less
light
Less
Difficult
Smooth (Better)
Low
More
Heavy
More
Less
Less
More
More
Less
Less
More
19
20
Maintenance cost of
traction motor
Starting Efficiency
Regenerative braking
Ridding quality
Insulation cost
Cross section of conductor
Design of supporting
structure
Distance between two
substation
No.of substation required
Size (capacity) of traction
substation
Capital & maintenance
cost of substation
Cost track electrification
Electrolysis trouble
High
High
Limited, except chopper
method
Less
Less
No
21
Applications
Main line services
More
Yes, if ground is used as
return path
Urban and suburban area
8
9
10
11
12
13
14
15
16
17
18
Less than d.c series motor
Less than d.c series motor
Simple and smooth
DC System Traction
600/750V-Tromways
1500/300V urban/suburban
DC series motor for
tramways. DC compound
motor
1.5 times less then a.c series
motor
High starting torque
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Model Answer
Subject Code: 17507
d) Give any four justifying features which makes DC series motor suitable for traction work.
Ans: Due to following features, DC series motor is suitable for traction duty:
( Any Four Point expected: 1 Mark each)
1) Characteristics:
Ta
,
/
/
/
at
Tsh
/
/
/
'
S
/
o
—
,
/
/
/
/
/
/
/
'
~
a>
s>.
V
V
at
s.
(/
/
Armature current (la)
Armature Current ( la )
Torque ( Ta )
Characteristics of DC series motor
OR
Features of DC Series Motor :
1.
(Any Four Points Are Expected)
DC Series motor robust in construction and capable to withstand against continuous
vibration.
2.
DC series motor weight is 1.5 times less than 1-Ph AC series motor for same H.P.
3.
DC Series motor has high starting torque.
4.
DC Series motor has high rate of acceleration and retardation.
5.
DC Series motor is variable speed motor. Due to these characteristics motor is protected
against overload.
6.
DC Series motor speed-torque characteristics are such that as torque increases speed
decreases.
7.
DC series motor has develops high torque at low speeds, low torque at high speed, this is
the basic requirement of traction unit.
8.
Commutating property of series motor is good so we get sparkles commutation.
9.
Torque is unaffected by variation in supply voltage.
10. DC Series motor maintenance cost is less.
11. When DC series motor are running in parallel the all motors share almost equal load.
12. Torque obtained by DC series motor is smooth and uniform, so it improves riding quality.
IMJ
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Model Answer
Subject Code: 17507
e) Draw a neat labelled Block diagram of A. C. locomotive.
Ans: labelled diagram of A.C. locomotive:
( Diagram: 4 Marks )
Block Diagram of AC electrical locomotive
lph AC
25KV 50Hz
supply
¥
Paruoqraph
Contact wire
CB
Filter
Rectilier
Transformer
MoSOf Control
hrn
DC Series motoi
Q.3
Attempt any TWO :
a) i) Give any four ideal requirements of elevators.
Ans:
Ideal requirements of elevator:
(2 x 8 =16 Marks)
(Any Four Points are Expected : 1 Mark each)
1. There must be all safety features.
2. Compactable acceleration and retardation to avoid jerk.
3. It should have sufficient Speed (feet/min.) proportional to height of building.
4. There should adequate lighting and provision of fan.
5. There should better interior design of the car.
6. It should have minimum breaking period.
7. There should be wide-frontage for fast traffic.
8. It should have sufficient capacity to handle the weight (Average weight 68 Kg per person).
9. Sufficient space should be available for car (2 Sq,ft. per person).
10. There must be provision of back-up, when electric supply get’s failure like D.G. sets.
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Model Answer
Subject Code: 17507
a) ii) State the factors to be considered for selection of shape and size of elevator.
Ans: The size and shape of elevator car depends are following two factors:
(Any Four Factor are Expected : 1 Mark each)
No. of passenger to be carried: While selecting the size of car it is a usual practice to
allow:1.
A Space of 2 Sq.fit/ person.
2.
Average weight of passenger is assumed 68 kg/person.
3.
Thus the maximum load capacity of elevator is considered 34 kg/sq.ft
4.
There should be wide frontage and shallow depth
5.
Limitation in the building design
6.
Shape of elevator depends on space available in building.
7.
Type of building
b) i) Draw a neat labelled sketch to show construction of Ajax Wyatt furnace.
Ans: Neat sketch of ‘Ajax Wyatt’ vertical core furnace:
VerfTcal
Core
roeVal
•
magnetic
Core-
l
I%
l%
I n d u c t i o n Furnace .
opening
-t y p e
|p
j i
1
I
ou>lef
foe molten
( 3 Mark)
— In eat
—
(!
insulating, male rial
Re Fra cto r* y
tcall of furnace
pTimauj. lOindhng
N
s
central core
.scrap
•
Y- n o t c h
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Model Answer
b) ii) Explain its working.
Ans: Working of Induction heating:
Page 13 of 34
( 3 Mark)
It is based on principle of transformer. In this type of Induction heating primary winding
is as usual which is wound around one limb of magnetic core but secondary winding is
actually charge which is to be melted is kept in crucible.
When AC Supply is given to primary winding current flows through primary winding
which creates alternating flux in magnetic core this flux links to the secondary winding i.e.
charge through magnetic core. Hence according to faraday’s law of electromagnetic induction
emf will be induced in secondary winding that is in the charge.
As charge forms a close circuit (secondary) heavy current flows through charge this
current is responsible to produce heat in charge due to I2R losses. This heat is utilized to melt
the charge.
Where, R = Resistance of charge & I secondary current.
b) iii) Any 2 advantages and any 2 application of Ajax Wyatt furnace.
Ans: Advantages:
(Any Two Points Are Expected : 1/2 Mark each)
1) As furnace has narrow ‘V’ shape crucible at bottom. So small quantity of molten metal
remains in narrow ‘V’ notch from previous operation, which will help to keep secondary
short circuited. So no extra care is required to start the furnace
2) Magnetic coupling between primary & secondary winding is better because both windings
are on central limb of magnetic core. So there will be less leakage flux, Hence leakage
reactance is less, so power factor is better than horizontal crucible direct core type induction
furnace.
3) Due to pinch effect in ordinary core type induction furnace there are chances of temporary
interruption in secondary circuit when current density exceeds above 500A/cm2 OR
5Amp/mm2..
4) But in this type of induction furnace there are no chances of interruption in secondary
circuit even if current density exceeds 500A/cm2 OR 5Amp/mm2 Because tendency of
weight of charge keep them in contact due to narrow ‘V’ shape.
5) So we can increase current density above 500A/cm2 OR 5Amp/mm2 to obtain more heat in
less time.
6) Vertical crucible is always better than horizontal crucible for pouring and taking out the
4
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Model Answer
Subject Code: 17507
metal. Also space required is less.
7) As heat is produced directly in the charge there is no heat transfer loss. So efficiency of
furnace is more.
8) As heat is directly produced in the charge time required for melting metal is less. So energy
consumption is less.
9) As current is directly induced in the charge there is automatic stirring action taking place in
the charge due to electromagnetic forces developed in the charge due which,
Through mixing of molten metal is possible.
Uniform heating is possible
10) Accurate temperature control.
11) Ideal working condition in a cool atmosphere with no dirt , noise and fuel.
Application of ‘Ajax Wyatt’ vertical core furnace:
(Any Two Points Are Expected : 1/2 Mark each)
It is used for melting metal having low resistivity.
It is used for heat treatment of silver, Copper, nickel etc.
Such type of furnace are used for continuous operations only and not used for intermittent
services.
c) i) i) Describe the concept of load cycle with their graphical representation :
1) Continuous operation with Short Time Rating
2) Continuous operation with intermittent Rating.
Ans:
1). Continuous operations short time loading:- ( Graph : 1 Mark & Explanation: 1 Mark)
Graphical representation
ioadno
HP
-load
U cycle repecuts
Temperature
-
rise
HL
'
Time CO
4
HW
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Model Answer
Subject Code: 17507
Explanation :-In this case motor is operated continuously for short time and interval
between two load is not OFF- load but motor runs at no load for long time. So temperature of
drive continuously increases. So
Temperature rise is more than short-time loading.
2) Continuous operations intermittent loading : - ( Graph : 1 Mark & Explanation: 1 Mark)
Graphical representation
JoadmHP
- toad
£ cycle fepeais
<
/
/
t
til
/\
/
L-
- Temperature
rise
Time ct)
Explanation :-In this case motor is operated continuously for long time and interval
between two load is not OFF- load but motor runs at no load for short time. So temperature of
drive continuously increases.
i) A motor has
c )toii)ii) A motor has to perform the following duty cycle :
1) 100 HP for 10 minutes 2) No load for 5 minutes
3) 50 HP for 8 minutes
4) No load for 4 minutes
The duty cycle is repeated indefinitely. Draw the curve for the load cycle. Assuming that
the heating is proportional to the square of load. Determine the suitable size of continuously
rated motor.
1) 100 HP for 10 minutes
Speed of motor : 750 rpm
Ans:
2) No load for 5 minutes
Rating of motor (KW) =?
3) 50 HP for 8 minutes
4) No load for 4 minutes
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Model Answer
Subject Code: 17507
Graph:
loo Hf
looHP
Co HP
4o
HP
So
.
Hf
2.0 H f
2
4
&
g
\o
12.
_
/5 i c
Duty Cycle (T) = t1+ t2 + t3 + t4
2 o
—
* t2-
H
27
2
P
--
( 1 Mark)
----------------------------------------------- (1/2 Marks)
= 10+5+8+4
= 27 Min. --------------------------------------------------- (1/2 Marks)
Continuous rating of Motor:
2
2
2
2
T t1 T2 t 2 T3 t 3 T4 t 4
rating of motor 1
T
---------------- (1 Marks)
100 2 10 (0) 2 5 (50) 2 8 (0) 2 4
rating of motor
27
rating of motor
120000
27
rating of motor 66 .667 HP ---------------------------------------- (1 Mark)
Nearest Standard rating of motor should be selected
Q.4A)
Attempt any THREE :
(3 x 4 =12 Marks)
List any four welding equipment and Accessories used for protection and safety and describe
a)
each of them in brief.
Ans:
(Any four equipment and Accessories are expected : 1 Mark each)
1.
2.
Series reactor:
To stabilize the arc in case of metal arc welding series reactor is used.
Electrode holder:
Well insulated electrode holder is used to grip electrode. This holder is in operator’s
hand.
MU
..
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3.
4.
5.
6.
7
8
9
b)
Page 17 of 34
Model Answer
Subject Code: 17507
Earthing clamp:
It is essential to complete electric circuit and safety purpose.
Welding helmet and face shield :
Helmet is used for protection purpose and face shield protect face and eyes from rays of
arc which contains UV and infra-red rays.
Other:
Hand gloves, shoes, apparan, small hammer, file, small metal wire brush etc are essential
during welding
Safety equipments.
Welding shield (hood): This is the mask which is worn to protect the person welding
from the bright flash of the arc, and from sparks being thrown during welding and also
protect face and eyes from rays of arc which contains UV and infra-red rays.
Welding gloves : These are special, insulated leather gloves that reach about 6 inches (15.2
cm) above the wrists, and protect the hands and lower arms of the welder (the person
welding).
Welding leathers: This is an apron like leather jacket that covers the shoulders and chest
of the welder, used for overhead work where sparks might ignite the welder's clothing, or
cause burns.
Work boots : The person welding should wear at least a 6 inch (15.2 cm) lace-up type boot
to prevent sparks and hot slag from burning his feet. These boots should have insulating
soles made from a material which does not melt or burn easily.
Draw a neat labelled sketch to show the construction of sodium vapour lamp and explain its
working.
Sodium Vapour Lamp diagram:
( Figure: 2 Mark, & Working : 2 Marks)
i
Ballast
^\
Anode
Uacmm
rii
j
4ubc
L
Ans:
Qisttoj
. Capacitor
Ignitor
Lamp
' \\
H if
1
reac ince
tannicdime)
TttOJiOal
LtLLUUUU
c qUiode
siclfiim ( rnetal jawf
"
vrmrm
*
NiUoqen
P F iriproua menf copq- ^
a’kJ '
N
OR
equivalent figure
OR
; <$ 23 o \J
AC
supply
efc.
or
IMJ
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Model Answer
Subject Code: 17507
Electrode
Outer Tube
Alumina arc tube
V
Arc
Sodium-mercury amalga:
—
A.C. voltage
(Mixture )
dSMillr
h
Ballast
Working Principle:
When the lamp is turned on, a high voltage at staring is applied across two electrodes, to
initiate an arc which discharges and vaporizes xenon /neon gas (starting gas), sodium and
mercury.
The energized metal atoms emit light.
After 2 to 5 minutes lamp will glow 100 %.
For running the lamp low voltage of about 165V is sufficient.
The color of light produce is yellowish.
c) i) Define Tariff. ii) State any 4 desirable characteristics of tariff.
Ans:
Definition of Tariff:
( 2 Mark)
Tariff is the way of billing energy consumed by consumer. OR
The rate at which electrical energy is supplied to a consumer is known as tariff.
Following desirable characteristics of tariff:
(Any Four Point expected: 1/2 Mark)
1. It should be easy to understand to consumer.
2. Easy to calculate.
3. Tariff should be attractive i.e. It should not be too high or too low. It should be
reasonable.
4. Tariff should be economical as compare to other types of energy sources.
5. Tariff should be different for different types of consumers.
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Model Answer
Subject Code: 17507
6. Tariff must be fair, so that different types of consumers are satisfied with rate of
electrical energy charges.
7. Tariff should be framed into two parts i.e. fixed charges + running charges.
8. Tariff should be high during peak load period .
9. Tariff should be low during off load period.
10. For industrial consumer, in addition to basic tariff incentives and penalty related to
PF and LF should be considered.
d) Show the derivation for most economical power factor.
Ans: Derivation:
( 4 Mark)
ICVAR c Qc
P kW
T
*?2
kVA Rp
?v
•
x
'
>\
82
c
±
02
,
kVAR .
SI
1
Q1
R
Let,
P = Active power KW
S1, S2 = KVA Maximum demand before and after improving power factor
Q1, Q2 = Lagging reactive power before & after improving power factor
QC = Leading Reactive power drawn by Capacitor
Cos1 = Initial Power factor
Cos 2 = Improved Power factor
Rs X = Tariff charges towards M.D. (KVA) /year
Rs Y = Expenditure towards KVAr to be neutralized per year (Expenditure towards P.F.
improving apparatus)
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1) Before improving Power factor:
Q1 P tan 1
Cos 1
S1
P
S1
P
Cos 1
KVA1 (S1 ) P sec 1
2) After improving Power factor:
Q 2 P tan 2
Cos 2
S2
P
S2
P
Cos 2
KVA2 (S2 ) P sec 2
3) Saving in KVA charges:
= Rs X (S1 –S2)
= Rs X ( P sec 1 P sec 2 )
= Rs X .P ( sec 1 sec 2 )
4) Expenditure towards KVAr to be neutralized:
= Rs Y (Q1 –Q2)
= Rs Y ( P tan 1 P tan 2 )
= Rs YxP ( tan 1 tan 2 )
Page 20 of 34
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Model Answer
Subject Code: 17507
5) Net Saving:
= Saving in KVA charges - Expenditure towards KVAr to be neutralized.
= [Rs X .P ( sec 1 sec 2 )] - [ Rs Y ( P tan 1 P tan 2 )]
Saving will be maximum when differentiate above equation with respect to 2 and equate to
zero.
ds
d
Rs X P (sec 1 sec 2 ) Rs Y P (tan 1 tan 2 )
d 2 d 2
0 X P sec 2 tan 2 0 Y P sec2 2
0 Rs X P sec 2 . tan 2 0 Rs Y P sec 2 2
Rs X P sec 2 . tan 2 Rs Y P sec 2 2
Rs X tan 2 Rs Y sec 2
Rs X
sin 2
1
Rs Y
Cos 2
Cos 2
Rs X sin 2 Rs Y
sin 2 Rs
6)
Y
X
sin 2 2 Cos 2 2 1
Cos 2 2 1 sin 2 2
V
Most economical power factor = Cos 2 1 (Y / x ) 2
Most economical power factor at which maximum saving will occurs
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Model Answer
Subject Code: 17507
Attempt any ONE
06 Marks
a) For metal arc welding :
a)
1) Draw its neat labelled sketch. 2) Explain its operation. 3) Any 2 applications.
Ans: 1) Diagram of metal arc welding:
Q. 4 B)
_
f1stal Q rc JAIB Idm
-<ap select CLP-
( 2 Mark)
cooied elected*.
kC
'
MT '
*-
e
job
n
i
we) ding
uoeld
4
met
profccHvt
aj
^
Cnduct heaHoss)
OR
Welding machine AC dr DC
powif 30urcfl and control!
elect *
haldar
Electrode -.
~-
AI ;.
work
—
Work cable
Electro : wt>le
^
Operation of Metal Arc Welding :
( 3 Mark)-
The process in which two metal parts to be welded are brought to a molten state and then
allowed to solidify is called as arc welding. Melting of metal is obtained due to heat
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Subject Code: 17507
Page 23 of 34
developed by an arc struck between an electrode (Filler material) and metal to be welded
(job)
OR
Type of supply used:
Both AC/DC Supplies can be used but generally A.C. Supply is used because it has more
advantages.
Supply Equipment used: Welding Transformer designed for low voltage high current
secondary.
Arc Stability: Series Reactor is used for arc stability.
Temperature obtain: Less as A.C. supply is used.
Possibility of arc blow is less.
Capital Cost: Less since welding transformer is used as a supply equipment.
Running cost: Less
Maintenance cost :Less
Stand by losses: Less
Efficiency: More
Voltage required: 72 to 100 volt A.C
Types :Shielded & unshielded welding
( Any Two application expected: 1/2 Mark each)
Application: For welding Ferrous Metals, Can be used for vertical & overhead welding.
OR
Applications of Arc welding :
1. Extensively used in the construction of steel structures
2. In industrial fabrication
3. In manufacturing industry
4. In maintenance and repair industry
5. This method is used for welding ferrous metals (Iron , Steel , Stainless Steel ) as well as
for welding non ferrous metals (Aluminium , Nickel , copper alloys )
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b) Explain the following Tariff in brief: 1) Two Part Tariff
Ans:
1) Two Part Tariff :
Page 24 of 34
Model Answer
2) Time of Day Tariff.
( 3 Mark)
In this type of tariff energy bill is split into two parts.
ENERGY BILL= FIXED CHARGE which depends on load (KW)
+RUNNING (Variable) CHARGE which depends on actual energy
consume (KWH)
Fixed charge which depends on load (KW) which is declared by consumer on test
report.
There is no separate meter is installed to measure load.
Only one energy meter is used to measure number of units consumed.
This type of tariff system is used for residential and commercial consumers.(up to 20
KW)
This type of tariff is not used for industrial consumers.
2) Time of Day (TOD) Tariff or OFF-load Tariff:-
( 3 Mark)
In addition to basic tariff Consumer has to pay energy consumption charges according
to time for which energy is consumed.
TOD energy meter is installed in the consumer premises.
This meter is specially designed to measure energy consumption w.r.t. time.
This type of tariff is such that energy consumption charges/unit are less at during OFFload period
Energy consumption charges/unit are more during PEAK -load period
This type of tariff is introduced to encourage industrial consumers to run their maximum
load during OFF-load period.
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Model Answer
Subject Code: 17507
Q.5
Attempt any FOUR :
(4 x4=16 Marks)
Draw a neat labelled diagram of fluorescent Tube and give the function of the following :
a)
1) Choke 2) Starter present in it.
Ans: Diagram of fluorescent Tube :
( 2 Mark)
.
-
\
y\ a
OD
±
*
Uibe
I
-
-
^
-
pF
u*
ncr :2
pouidear
f mercsO «4
Cnoke
H
fc
F. Ivirosceni
radl
viency
^^
T
r e>4
3
\
-
c
poudfO
mpaMjmmT
COpQClVQT
I'
^
r
Function:
( Each Function: 1 Mark)
i) Choke: For providing high voltage at the time of starting and limit the current.
ii) Starter: To make and break the circuit to start the tube.
With the help of neat diagram, explain Tapped Reactor method for current control in
Welding Transformer.
Ans: Tapped Reactor method for current control in Welding Transformer:
( Diagram : 2 Mark & Working: 2 Mark)
b)
lap select
•
CLP-
coakd
elec+vde
AC
supply
r*
job
i
Iweldifvg
>
—
proKcKvtflai
OR
uoeldL
fwduce healloss)
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Model Answer
Subject Code: 17507
-fop select
coated
el echoic.
hC
£ UPpl
job
rIvoeldbfrv
^
dTC
*
uoeldL
4
me*-
pwkdlVe
CwdLuct heoMoss)
ai
^
OR
j
^ ,^
-y
*
ro
H'
^ j {i=
r
l
0
2
'
cue lc< ; cM
1—
=TH
rnKTann
'
"
., , ,
.
—
—
t
,
K
^ r^ eJ
Working : Reactor is used, to stabilize the arc. Arc has negative temperature co-efficient of resistance
i.e. its resistance decreases as temperature increases. So arc does not remain stable. To
stabilize the arc reactor is connected in series with arc furnace which control rise in current
hence arc.
Series reactor also serves as safety device by limiting current in circuit when there is short
circuited.
OR
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Model Answer
Subject Code: 17507
When reactor is connected in circuit then there is voltage drop across reactor. So voltage
across arc reduces. In this way the temperature control is obtained.
OR
Input power of arc furnace can be controlled by changing trappings of furnace transformer.
This is automatically done with the help of automatic current regulator.
c)
Compare urban service, suburban service and main line service on the following parameters :
1) Value of Acceleration 2) Value of Retardation 3) Maximum speed 4) Distance between stations.
Ans:
Compare urban service, suburban service and main line service :
Sr.No
Points
Urban line services
Suburban line
( Each Point : 1 Mark)
Main line services
services
1.
Acceleration ( )
High
High
low
2.
Retardation ( )
High
High
low
3.
Maximum Speed
120 km / hr
120 km / hr
160 km / hr
4.
Distance between
Low
Medium
High
two railway station
OR
Sr.No
Points
Urban line services
Suburban line
Main line services
services
1.
2.
Acceleration ( )
Retardation ( )
High = 1.5 to 4
High = 1.5 to 4
low = 0.6 to
km/hr-sec
km/hr-sec
0.8km/hr-sec
High = 3 to 4
High = 3 to 4
low = 1.5 km/hr-
km/hr-sec
km/hr-sec
sec
3.
Maximum Speed
120 km / hr
120 km / hr
160 km / hr
4.
Distance between
Low (1km)
Medium (2.5 to 3
High (above10km)
two railway station
km)
urn=
L-
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Model Answer
Subject Code: 17507
Draw the typical speed time curve for main traction line. Show the different time periods on
it.
Ans: Typical speed time curve for main traction line :
( 4 Mark)
d)
FREE
t
*
a
o
! RUNNING
SPEED
CURVE
RUNNING
COASTING
i
t
I
X
CL
i
.
X
o
MJ
RHEOSTATIC
ItJ
BRAKING
ACCELERATION
1
CL
I
I
1
*2TIME
U
IN SECONDS
OR
*n
ACCileration
Free Running|
*
k-
$
Retardation
r
a.
3
Vm
v.
a
ijj
LU
CL
L.O
—
-t5
tT
T
TIME IN SECONDS
km
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Model Answer
Subject Code: 17507
e) With the help of sketches, explain the various steps required for bridge transition system.
Ans: In bridge transition, series last step to parallel first step, is carried out by following steps :
( Each Step : 1 Mark)
Stepl: Bridge link is connected between two motors as shown in figure ( Series last step )
Ml
X
Br d8eLink
'
( M
Step2:Bridge link is so rotated that two motors are put in series without starting resistance. Which are
un-shorted at the same time.
Ml
Bridge Link
1 M2
0
h
Step 3:The portions of external resistance are connected in each motor circuit as shown in fig
Bridge Link
M
+-
M2
M
p'WV-pVWjAAAr-p
Step4 : In this last step bridge link is removed as shown in fig. This is nothing but parallel first step.
VI 2
.
pVWjAAArpVW-p
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Model Answer
Subject Code: 17507
Q.6
Attempt any TWO of the following :
16 Marks
A 20 kW single phase 220 V resistance oven employs a circular nichrome wire for its heating
element. If wire temperature is not to exceed 1170 0 C and temperature of charge to be 500 C ,
a) calculate the diameter and the length of the wire, Take K = 0.57, e = 0.95 and = 1.09 x 106
ohm meter.
Ans: Given Data:
T1 = 11700C = 1170+273 = 14430K
T2 = 5000C = 500 +273 = 7730K
Radiation efficiency = 0.57, specific resistance of Ni-Cr = 1.09x10 -6 ohm m, emissivity = 0.95.
T1 4
T
) ( 2 )4 w / m 2
1000
1000
OR
T
T
H 5.72 k .e [ ( 1 ) 4 ( 2 ) 4 ] w / m 2
100
100
----------------------------- (1 Mark)
H 5.72 104 k.e [ (
H 5.72 0.57 0.95 [ (
1443 4 773 4
) (
) ] w / m2
100
100
H 12 .3236 10 4 w / m 2
-------------------------------------------------------- (1 Mark)
Thickness : 0.3 mm
0.3 103 m
V2
l
d2 4 P
---------- Equation No.1----------------------------- (1 Mark)
l
(220) 2
d 2 4 20 1000 1.09 10 6
l
1743728.032
d2
---------- Equation No.2 ----------------------------- (1 Mark)
Heat Dissipated = Electrical Power I/p
dlH P
----------------------------- (1/2 Mark)
4
d l 12.3236 10 20000
d l 0.05165
----------------------------- (1/2 Mark)
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Model Answer
Subject Code: 17507
By Simplifying :
d 2 l 2 2.6686 10 3
d2
2.6686 10 3
l2
Substitute Value of d2 in Equation No.1 :
l
1743728.032
2.6686 10 3
l2
mm-------------------------- (1/2 Mark)
l 3 1743728 .032 2.6686 10 3
l 16 .69 m
--------------------------- (1 Mark)
Substitute Value of ‘l’ in Equation No.2 to calculate ‘d’ :
l
1743728 .032
d2
----------------------------- (1/2 Mark)
3
d 3 .66 10 mtr
d 3 .66 mm
------------------------------------------------- (1 Mark)
Answer : Length l 16 .69 mtr
Diameter d 3 .66 mm
The distance between two stations is 2 KM. It is desired to have scheduled speed of 40 km/hr
with duration of stop of 20 seconds. Assuming, trapezoidal speed time curve, Calculate :
b) 1) The maximum speed required when the acceleration is to be limited to 1.2 km/hr/sec and
braking retardation to 3 km/hr/sec.
2) The distance covered during acceleration and retardation.
Ans:
Given Data :D = 2 KM, Schedule speed (Vsch) = 40KM / Hr, Stop Time = 20 sec.
Acceleration ( ) = 1.2 Km/Hr/sec;
Retardation ( ) = 3 Km/Hr/sec.
Trapezoidal speed time curve :-
r
1
AfflO
pr* * * d
Fr* ? RunO' rjl
PfJ
A
ipeed
1
~T
/
/
1g
»\
y
\ f (t / n/ S )* )
o
' »4
1
W*’"*
!
•
-
|
-
<»
TirnrCT
rJ' > oq
‘
\
1
H
1
\
,
P^ , ~1
)
r
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Model Answer
Subject Code: 17507
Vsch
Page 32 of 34
3600 D
----------------------------------------------------------- (1 Mark)
Schedule Time
Schedule Time
3600 D
Vsch
Schedule Time
3600 2
40
Schedule Time
7200
40
ScheduleTime 180 sec.----------------------------------------------------- (1 Mark)
SchudeleTime Actual Time of Run Stop time
Actual Time of Run Schedule Time Stop time
Actual Time of Run 180 20
Actual Time of Run 160 sec.--------------------------------------------------- Maximum Speed =
V
But,
max
K
K
(1/2 Mark)
V
T T 2 4K 3600D
--------------------------------------------- (1/2 Mark)
2K
.----------------------------------------------------------------2
(1/2 Mark)
1.2 3
21.2 3
K 0.5833 -------------------------------------------------------------- ( 1 Mark)
V
max
V
160 1602 4 0.5833 3600 2
2 0.5833
max
56.7345 KM/Hr --------------------Answer----------------- ( 1/2 Mark)
Distance covered during Acceleration ( D ) =
km
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Subject Code: 17507
Page 33 of 34
2
D
D
Vmax
-------------------------------------------------- (1 Mark)
7200
56.7345 2
7200 1.2
D 0.3725 Km ----------------------Answer------------- ( 1/2 Mark)
Distance covered during Retardation ( D ) =
2
V
D max ----------------------------------------------- (1 Mark)
7200
D
56.73452
7200 3
D 0.1490 Km----------------------------Answer--- ( 1 /2 Mark)
c) A 300 HP, 3000 V, 50 c/s, 3 phase star connected induction motor has full load efficiency of
86% and pf of 0.707 lagging. If it is desired to improve the pf to 0.95 lagging by a bank of
three capacitors, find out the :
1) KVA Rating of the capacitor bank.
2) Capacitance of each unit when connected in i) Mesh ii) in star.
Ans:
Given Data
Volt : 3000 V,
f= 50 Hz
300 HP x 735.5 = 220650
P= 300 HP x 735.5 / 0.86 = 256569.7674 Watt = 256.57 kW
cos 1 =0.707
cos 2 =0.95
Cos 1 0.707
tan 1 = 1 -----------------------------
----------------------------------------- (1/2 Mark)
cos 2 =0.95
tan 2 = 0.328 -------------------- --------------------------------------------- (1/2 Mark)
Q1 = P tan 1
= 256.57 x 1
= 256.57 KVAR
---------------------------------------------------- (1/2 Mark)
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Subject Code: 17507
Page 34 of 34
Q2= P tan 2
= 256.57 x 0.328
= 84.15 KVAR
-
--------------------------------------------- (1/2 Mark)
i) KVA Rating of the capacitor Bank
QC = Q1- Q2
= P tan 1 - P tan 2
------------------(1 Mark)
= 256.57 – 84.15
= 172.42 KVAR
---------------------------------------------- (1 Mark)
Capacitor when connected in Mesh :C per phase
QC
3 V2
------------------------------------------------------- (1 Mark)
C per phase
172.42 10 3
3 2 50 (3000) 2
C per phase
172.42 10 3
8482300165
C per phase 2.032 10 8 F -
------------------------------------------------ (1 Mark)
Capacitor when connected in Star :C per phase
QC
V2
C per phase
172.42 10 3
2 50 (3000) 2
------------------------------------------------------- (1 Mark)
C per phase 6.098 10 8 F -
------------------------------------------------ (1 Mark)
------------------------------------------------------END-------------------------------------------------------
km
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Subject Code: 17507
Page 1 of 33
Important suggestions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more importance.
(Not applicable for subject English and communication skills)
4) While assessing figures, examiner may give credit for principle components indicated in a figure.
The figures drawn by candidate and model answer may vary. The examiner may give credit for any
equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model answer.
6) In case some questions credit may be given by judgment on part of examiner of relevant answer
based on candidate understands.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.
Q.1 A) Attempt any three :
(3x4=12)
a) State the factors governing selection of electric motors.
Following Factors governing / or are considered while selecting electric drive (Motor) for
Ans:
particular application:
( Any Four factor expected: 1 Mark each ,Total 4 Marks)
1. Nature of supply:
Whether supply available is
AC,
Pure DC
Or Rectified DC
2. Nature of Drive (Motor):
Whether motor is used to drive (run)
Individual machine
OR group of machines.
3.
Nature of load:
Whether load required light or heavy starting torque
OR load having high inertia, require high starting torque for long duration.
OR Whether load torque increases with speed (T N)
OR decreases with speed (T 1/ N )
OR remains constant with speed (T = N)
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Model Answer
Page 2 of 33
OR increases with square of speed (T N2)
4. Electric Characteristics of drive:
Starting,
Running,
Speed control
and braking characteristics
of electric drive should be studied and it should be matched with load requirements(i.e.
machine).
5.
Size and rating of motor:
Whether motor is short time running
OR continuously running
OR intermittently running
OR used for variable load cycle.
Whether overload capacity, pull out torque is sufficient.
6.
Mechanical Considerations:
Types of enclosure,
Types of bearing,
Transmission of mechanical power,
Noise
and load equalization
7.
Cost:
Capital,
Running
And maintenance cost should be less.
b) State four advantages of electric heating.
Ans: Advantages of Electric heating: ( Any Four Advantages expected : 1 Mark each, Total 4 Marks)
1. It can be put into service immediately.
2. No standby losses.
3. High efficiency.
4. More economical than other conventional types of heating system.
5. Easy to operate and control.
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Model Answer
Page 3 of 33
6. No air pollution.
7. System is clean, as there is no waste produced.
8. No fuel transportation cost.
9. No space is required for storage of fuel and waste.
10. Noiseless operation.
11. Uniform heating is possible; heating at particular point is also possible.
12. Dielectric material can be heated.
13. Electrical heating equipments are generally automatic, so it requires low attention and
supervision.
14. Protection against overheating can be provided by suitable switch gear.
c) Explain the factors to be considered while designing a lighting scheme.
Following factors to be considered while designing a lighting scheme
Ans:
( Any Four factors are expected : 1 Mark each, Total 4 Marks)
1.
Design of illumination scheme should be very simple.
2.
Area of the working plane.
3.
Find out application of working plane
4.
Decided lux level on working plane as per application.(Illumination level)
5.
Find out total lumens required on working plane.
6.
Decide the type and wattage of lamp which is to be used for that particular application
7.
Quality of light
8.
Assume waste light Factor
9.
Assume utilization Factor
10. Assume deprecation Factor
11. The illumination scheme is designed in such a way that there should be fewer glares.
12. The illumination scheme is designed in such a way that there should be minimum shadows.
13. The control of light intensity is possible.
14. Assume the illumination efficiency of those specific lamps which are to be used.
15. Provide safety and prevent accident.
4
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Model Answer
Subject Code: 17507
16. Take care to achieve desired energy saving.
17. Choose lamp with higher luminous efficiency, better color, and longer life
18. For any type of illumination scheme the maintenance & repairing should be less.
19. The cost of the designed illumination scheme should be low
d) d) Explain the principle of power factors improvement.
Principle of power factors improvement:We know that,
Ans:
( 4 Marks)
P 3 VL I L Cos
For same power to be transmitted
At same voltage
Over a same distance
I
1
1
Cos
P. f
From above equation it is seen that as power factor increases current decreases, due
to decreases in current, system has advantages.
..
I
I
-*k
v
i
OR
V
r
c
in:
A
«0
Illl
'
>Iw
Vo?'
\
X
12
II
m
OR
-> v
MU
..
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Model Answer
Subject Code: 17507
KVAR c Qc
, p
0
c
~
Ar
kVAR
2
-it: >
C
/
1Q 2
*
- -
kVAR 1
I
B
Qi
Q.1B)
Attempt any ONE :
a) i) (i) Define electrical braking.
Ans: Electrical braking:
(1x6=6)
( 1 Marks)
It is necessary to stop the vehicle when mechanical working is over or when required
within reasonable time by use of electrical energy.
a) ii) State its types electrical braking
Ans: Types of Electrical Braking: (Any two types expected---- 1 Mark)
Electrical braking system:
1. Plugging
2. Rheostatic (Dynamic) Braking
3. Regenerative Braking
4. Electromagnetic Braking
5. Eddy current braking
a) iii) Explain regenerative braking for D.C. series motor.
(Fig.—2 Marks, Explanation --- 2 Marks)
Ans:
Schematic diagram of regenerative braking of D.C. series :OH
conductor
o
O
Cuficol Li <riH»
devifli
:
OH
corxludot
“
f
To -Tj m 11
Current
.
ft
To I *'n> t
a
Ii" # o
C *f i
*
f
M
Mi
i“
»a
M
'
t
«> 5
3
1
t
Und^i.
normal condifion
Connection d u r i n g
Fig. --- A
E7PB
Fig. --- B
,
R
km
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Model Answer
Subject Code: 17507
Explanation of regenerative braking:
During normal running, motors are connected in parallel with field winding in series w.r.t.
armature as shown in figure A.
At the time of regenerating braking all the armature are connected in parallel without series
field winding and all series field winding are connected in series with external resistance &
are separately excited as shown in fig.B
At this time motor acts as a generator and excitation current is so adjusted that generated
voltage (Eg) is greater than supply voltage (V), so that power will be fed back to supply.
This process is continued up to the speed of train reaches up to 20 to 16 km/hr. after that it is
difficult to maintain generated voltage greater than supply voltage. So, electric regenerative
braking is stopped
For final stop mechanical braking is applied.
External Resistances are connected to limit the current.
b)
Explain with the diagram butt welding. State its applications.
Ans: (Diagram----- 2 Marks, Explanation------ 2 Marks, Applications------- 2 Marks, Total 6 Marks)
Diagram butt welding:-
_
3ujt.t
I/NJ e Lci
cVxmp
Bcii ^ coe\ dLx rv
-
^
.
is
preja su
*~> ere V£> eld
temp? toJu re. isob +omfti
-
T i l lSLSSL
'3
AC
su
to
OLpp \ ied
eldtrig.
X me
"
^
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Page 7 of 33
Explanation:
Transformer used for welding is designed for low voltage and high current secondary.
Transformer is oil cooled
The job is clamped as shown in fig. two parts which are to be welded are brought together
Sufficiently heavy current is passed through joints by welding transformer,
which creates necessary heat at joints due to I2R
When welding temperature is reached supply is cut down.
And external pressure is applied simultaneously across the job to complete weld
Application Butt Welding:
1) For welding rod, wire, pipe etc
2) For joining thick metal plates or bars at end
Q.2
Attempt any FOUR :
(4x4=16 Mark)
a) State four advantages of electrical braking over mechanical braking.
Ans:
( Any Four Advantages expected : 1 Mark each, Total 4 Marks)
Following are the advantages of electrical braking over mechanical braking system.
Advantages:
1. It is most reliable braking system.
2. Breaking actuation time is small as higher value of braking retardation is obtained.
3. Electrical braking is smooth & gradual.
4. Life of braking system is more.
5. There is less wear & tear of brake shoes, break block etc. so there is less maintenance cost.
6. Higher speeds are possible even when train is going down the gradient, as breaking system
is reliable.
7. Trains having heavy loads can be stopped even when train going up the gradient.
8. Higher speeds of train is possible as braking system is reliable so pay load capacity
increases.
9. In case of electric regenerative braking we can utilize 60 to 80% of kinetic energy to
generate electricity which is not possible with mechanical braking.
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Model Answer
Subject Code: 17507
b) State desirable properties of heating elements used in indirect resistance heating.
Ans:
( Any Four properties are expected : 1 Mark each, Total 4 Marks)
Following properties of good heating material in indirect resistance heating:1. High resistivity:
It should have high resistivity. So that it becomes compact in size and produces more heat
with small input current.
2. High melting point:
It should have high melting point to withstand at high temperature.
3. High Oxidizing temperature:
It should have high oxidizing temperature or it should not oxidize even at high
temperature.
4. High Resistance to corrosion:
It should have high resistance to corrosion to avoid rusting.
5. High Mechanical Strength:
It should have high mechanical strength to withstand from mechanical injury.
6. Ductile:
It should be ductile so that it can be manufactured into different size & shape.
7. Long Life:
It should have long life.
8. Less Costly:
It should be less costly and easily available.
9. Low temperature co-efficient of resistance:
For accurate temperature control, it should have low temperature co-efficient of
resistance.
10. It should not be brittle.
c) Compare A.C. and D.C. system of traction (any four points).
Ans: Compare A.C. and D.C. system of traction:
(Any Four point expected: 1 Mark each, Total 4 Marks)
S.No
Points
1
Supply given to O/H
condition
AC System Traction
1-ph, 25KV, AC 25 Hz
DC System Traction
600/750V-Tromways
1500/3000V urban/suburban
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Model Answer
Subject Code: 17507
2
Type of drive used
1-ph, AC series motor
3
Weight of traction motor
4
5
6
7
Starting torque
Accln and retardation
Overload capacity
Method of speed control
1.5 times more than d.c.
series motor.
Less starting torque
Less than d.c series motor
Less than d.c series motor
Simple and smooth
8
More
More
Easy
Less, better than d.c.
High
Less
light
Less
Difficult
Smooth (Better)
Low
More
Heavy
More
Less
Less
More
More
Less
Less
More
Less
More
20
Maintenance cost of
traction motor
Starting Efficiency
Regenerative braking
Ridding quality
Insulation cost
Cross section of conductor
Design of supporting
structure
Distance between two
substation
No. of substation required
for same track distance.
Size (capacity) of traction
substation
Capital & maintenance
cost of substation
Cost track electrification
for same track distance
Electrolysis trouble
DC series motor for
tramways. DC compound
motor
1.5 times less then a.c series
motor
High starting torque
High
High
Limited, except chopper
method
Less
No
21
Applications
Main line services
Yes, if ground is used as
return path
Urban and suburban area
9
10
11
12
13
14
15
16
17
18
19
d) Explain the suitability of 3-phase induction motor for traction service.
Ans: (Any Four points expected: 1 Mark each, Total 4 Marks)
Suitability of 3-phase induction motor for traction service because of following points:1.
It is robust in construction and capable to withstand against continuous vibration.
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2.
Simple in design & construction.
3.
Slip-ring induction has high starting torque when external resistance is added in rotor
circuit.
4.
Since the torque speed characteristic of the induction motor is markedly steeper, the
induction machine can take better advantage of maximum possible tractive effort.
5.
No restriction on speed of motor because of absence of commutators.
6.
Speed control methods are simple
7.
Power to weight ratio of induction motor is much higher than the DC motor.
8.
High efficiency.
9.
Require little maintenance. Apart from bearing, it has no parts subjected to wear. It is
not much effected by dust, vibration and heat
10.
Less maintenance.
11.
A high mean adhesion coefficient can be expected.
12.
The induction motor drives are about 20% energy efficient compared to DC drives.
13.
Three phase drives allow regeneration and unity power factor operation.
14.
The energy saving due to regeneration and improved power factor are sizable.
15.
It operates at high voltage ( 3.3 / 3.7 KV) consequently requiring less amount of current
16.
Automatic regeneration is the main advantage of I.M.
17.
Trouble free operation.
e) Draw single line diagram of 132 kV/25 kV traction substation.
Ans:
Single line diagram of 132 kV/25 kV traction substation:
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Model Answer
Subject Code: 17507
»32 kV
IMCOMINO
.
'
I INIt PROM
OFNIfMAT IMO
klAIION
TRACTION
U 3 ll AN
'/
&r
P
/
132 k V MAIN
*
UUSBAH
step
I \\
I *1
~
f
J
C
down transformer
C
^^
*
<NO
ISO
!A
cificuiT VHIMK C m
*i j
TO FeeONlf
POST ON Tower?
)PLCV
f
OHt A
~
SWITCH
*I%
US
f
1501 ATINO
*
Q.3
Attempt any TWO :
(2 x 8 =16 Marks)
a) Explain the requirements of elevator motor. State with reason best suitable motor for elevator.
Ans: Ideal requirements of elevator motor: (Any Six requirements are Expected: 6 Marks, suitable
motor for elevator – 2 Marks, Total 8 Marks.)
1. High starting torque along with high rate of acceleration and retardation.
2. Motors are compact in size especially smaller in diameter.
3. Speed of motor should not exceed 900 rpm
4. It should have sufficient overload capacity.
5. Low initial and maintenance cost.
6. Long life.
7. It should withstand for rapid fluctuation in supply voltage
Following motors are used because it covers all above requirements:1. DC Series Motor
2. Ac Series Motor (1-ph)
3. 3-Ph Slip ring I.M
4. Permanent magnet AC Motor.
5. Split Phase AC I.M.
6. Capacitor Start I.M.
b) Explain with neat diagram construction and working of Ajax Wyatt vertical core induction
furnace.
Ans:
(Diagram---3 Marks, Construction---- 2 Marks, Working----- 3Marks, Total 8 Marks)
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Model Answer
Subject Code: 17507
Neat sketch of ‘Ajax Wyatt’ vertical core furnace:
Ve r 4- fca. l
-t ype.
Core
j
J
ou> leF
fb -fc, mo l +on
rr.eVoJ
Induction
For nace.
i I P opening
_ j_
heol fosuiaJin
-^
AA
A
mocferiaX
e Fr a c +o r y
|\ o f
furnace
Q
p TI mcnr j.
^
3
ooi
nci f
moaneHc
cer> HoU core.
corescrcLp
•
•
-
V- n o t c h
Construction of ‘Ajax Wyatt’ vertical core furnace:
Vertical core type induction heating furnace is nothing but transformer. It consists of following
main parts: Magnetic Core:
Primary winding
Secondary Winding:
Refractory Wall
Opening
Cooling arrangement
Tilting arrangement
Control panel
APFC
Working:
It is based on principle of transformer. In this type of Induction heating primary winding is
as usual which is wound around one limb of magnetic core but secondary winding is actually
charge which is to be melted is kept in crucible.
When AC Supply is given to primary winding current flows through primary winding
which creates alternating flux in magnetic core this flux links to the secondary winding i.e.
charge through magnetic core. Hence according to faraday’s law of electromagnetic induction
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Model Answer
Subject Code: 17507
emf will be induced in secondary winding that is in the charge.
As charge forms a close circuit (secondary) heavy current flows through charge this current
is responsible to produce heat in charge due to I2R losses. This heat is utilized to melt the charge.
Where, R = Resistance of charge & I secondary current.
c) An electric motor has load variation as given below :
(i) Torque 140 Nm for 20 minutes (ii) 40 Nm for 10 minutes
(iii) 200 Nm for 10 minutes
(iv) 100 Nm for 20 minutes,
If the speed of the motor is 720 rpm, find the power rating of motor.
i) 140 Nm
for 20 min.
Speed of motor : 720 rpm
Ans:
ii) 40 Nm
iii)
200 Nm
iv) 100 Nm
for 10 min.
Rating of motor (KW) =?
for 10 min.
for 20 min.
Duty Cycle (T) = t1+ t2 + t3 + t4 ----------------------------------------------- (1 Marks)
= 20+10+10+20
= 60 Min. --------------------------------------------------- (1 Marks)
rating of motor (Torque)
1
J
rating of motor (Torque)
2
2
2
2
T1 t1 T2 t2 T3 t3 T4 t4
T
---------------- (1 Marks)
1402 20 40 2 10 2002 10 1002 20
60
rating of motor (Torque) V16800 Nm
rating of motor (Torque ) 129.61 Nm ------------------------------------------- (1 Mark)
rating of motor ( watt )
rating of motor ( watt )
2 N T
60
----------------------------------------------- (1 Mark)
2 720 129.61
60
rating of motor ( watt ) 9773.63 Watt ---------------------------------------------- (1 Mark)
rating of motor ( Kw)
9773.63
1000
rating of motor ( kw) 9.773 kW i.e.
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Model Answer
Subject Code: 17507
rating of motor (kw) 10 kW
-------------------------------------------------------(2Mark)
Q.4A)
Attempt any THREE :
(3 x 4 =12 Marks)
a) State four advantages of seam welding over spot welding.
Ans:
(Four advantages are expected 1 Mark each, Total 4 Mark)
Advantages of seam welding over spot welding:1.
2.
3.
4.
5.
b)
Due to seam welding we get continuous weld which is air-tight
Gas tight as well as liquid tight joints can be made.
The Overlap is less than spot or projection welding.
The production of single seam weld and parallel seams can be got simultaneously.
Efficient energy use.
Define the following : (i) Luminous intensity
(iv) Mean spherical candle power
(ii) Utilization factor
(iii) Depreciation factor
( Each definition : 1 Mark , Total 4 Mark)
1) Luminous intensity:The luminous intensity in any particular direction is the luminous flux emitted by source
per unit solid angle is called the luminous intensity of the source. And its unit is Candela
OR I
(Where lu min ous flux , w Solid Angle)
w
ii) Utilization factor:
It is defined as the ratio of total lumens reaching the working plane to the total
lumens given out by the lamp. Its value is always less than one.
(iii) Depreciation factor
Ans:
It is defined as the ratio of initial illumination to the ultimate maintained illumination on
the working plane. OR
1
Depreciation factor
Ma intenantece Factor
iv) MSCP (Mean Spherical Candle power):
It is the average of all candle powers in all direction in all planes.
OR
MSCP
Total Lu min ous lux in lumens
4
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Model Answer
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c) State and explain four types of tariff applicable to H.T. and industrial consumers.
Ans:
( Names of tariffs --- 1Mark, Explanation---3Marks, Total 4 Marks)
Following tariffs are applicable to industrial/ HT consumers:Types of Tariff:1. Maximum demand Tariff ( KVA maximum demand Tariff )
2. Power factor Tariff
3. TOD (Time of Day) Tariff
4. KW and KVAR Tariff
1. Maximum Demand Tariff/KVA Maximum Demand Tariff / Load factor tariff: It is similar to two part tariff except that maximum demand (KVA) is actually measured by
installing maximum demand meter (in KVA)
M.D. Meter (it is an electromagnetic or electronic trivector meter) is installed in the
premises of consumer, in addition to energy meter.
Maximum Demand Tariff / Load factor Tariff =
M .D. ( KVA) Rs ' X ' permonth { Number of units ( KWH ) Actual consumer} Rs ' Y "
2. Power Factor Tariff (Sliding Scale Tariff or Average P.F. Tariff):In addition to basic tariff the tariff in which P.F. of industrial consumer is taken into
consideration for billing.
If the P.F. of consumer is less than P.F. declare by Supply Company (say below 0.9 Lag.)
than penalty will be charged in energy bill.
If The P.F. of consumer is more than P.F. declare by Supply Company (say above 0.95lag.)
than discount will be given in energy bill.
As usual consumer has to pay actual energy consumption charges
3. Time of Day (TOD) Tariff or OFF-load Tariff:-
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Model Answer
Subject Code: 17507
In addition to basic tariff Consumer has to pay energy consumption charges according to
time for which energy is consumed.
TOD energy meter is installed in the consumer premises.
This meter is specially designed to measure energy consumption w.r.t. time.
This type of tariff is such that energy consumption charges/unit are less at during OFF-load
period
Energy consumption charges/unit are more during PEAK -load period
4.
KW and KVAR Tariff:
An electromagnetic or electronic trivector meter is installed in the consumer premises.
In this type both active (KW) & reactive power (KVAr) supplied are charged separately and
actual energy consumption charge.
Energy Bill {Rs ' A' ( KW ) Ch arg es} {Rs ' B ' ( KVAR ) Ch arg es} {Rs ' C ' ( KWH ) Ch arg es}
A 3-phase, 5 kW induction motor has a power factor of 0.75 lagging. Determine the size of
capacitor in kVAR require to improve the power factor to 0.90.
Ans:
Given Data
d)
cos 1 =0.75 lagging
P= 5kW
cos 2 =0.90 lagging
Cos 1 0.75
tan 1 = 0.881 ---------------------------------------------------------------------- (1 Mark)
tan 2 = 0.484
Q1 = P tan 1
= 5 x 0.881
= 4.405 KVAR
--------------------------------------------------------- (1 Mark)
Q2= P tan 2
= 5 x 0.484
= 2.42 KVAR
----------------------------------------------------------- (1 Mark)
QC = Q1- Q2
= P tan 1 - P tan 2
= 4.405 – 2.42
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= 1.988 KVAR
Page 17 of 33
Model Answer
-------------------------------------------------------- (1 Mark)
Attempt any ONE
06 Marks
a) Compare resistance welding and arc welding (any six points).
Ans:
(Any six points are expected, 1 Mark each point, Total 6 Marks.)
Q. 4 B)
Sr.No
Parameters
1
Type of welding
2
Principle of heat
developed
3
External filler material
required
External pressure
required
Type of supply used
4
5
Resistance Welding
Arc Welding
Plastic / Pressure / Nonfusion welding
Heat is developed due to I2R
losses where R is the contact
resistance
Not required during welding
Fusion / Non pressure
welding
Heat developed due to arc
produced in between
electrode and job
Required during welding
Required
Not required
Both AC, DC supply is used.
But generally Ac Supply is
used.
Metal arc welding – Both
AC, DC supply is used. But
generally Ac Supply is
used. and for
Carbon arc welding –only
DC supply are used
6
Voltage &current
Low voltage (2 to 20V AC)
Metal Arc welding Voltagerequired
and high current (40 to
70 to 100V AC and
400A, in some cases 5 to
Carbon arc welding
20KA ) supply is required
voltage50 to 60V DC,
Current- 50-600-800A
7
Energy consumption
Low (3 to 4 KWH/Kg of
High (5 to 10 KWH/Kg of
deposited material )
deposited material.)
8
Temperature obtained Temperature obtained is not Temperature obtained is
very high (up to 13500C)
very high (up to 35000C to
60000C)
9
Power factor
Low
Poor
10
Type of electrode
Non-consumable electrodes
Coated electrodes are used
are used.
for metal arc welding and
bare electrodes are used for
carbon arc welding.
(Electrodes may be
consumable or nonconsumable)
1.
Application
It is suitable for mass
It is suitable for heavy job,
production
maintenance and repair
work
b) Explain the factors affecting framing of tariffs.
1x6=6
Ans:
(Six factors are expected, 1 Mark each, Total 6 Marks)
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Model Answer
Page 18 of 33
Factors affecting framing of tariffs:1.
Total recovery of all taxes , duties and other charges
2.
Expenses on premium (installment) paid to insurance company.
3.
T&D losses.
4.
Electricity cannot be stored economically. It has to be consumed as soon as it is generated.
5.
Additional supply charges (ADC) to compensate the costly purchase energy (power) from
outside to reduce the load shading.
6.
Investment required for future expansion.
7.
Economics as compare to other types of energy sources.i.e.to encourage the consumers to use
electricity.
8.
Applying different tariff for different types of consumers.i.e. Proper return is secured from each
consumer.
9.
Applying tariff high during peak load period.
10.
Applying tariff low during off load period.
11.
For industrial consumer, in addition to basic tariff incentives and penalty related to P.F and L.F.
12.
The tariff should be simple cheap and capable of easy explanation to consumers.
Q.5
Attempt any FOUR :
(4 x4=16 Marks)
a) State the following : (i) Law of inverse squares (ii) Lambert's Cosine law
Ans:
(Each law—2Marks, Total-4 Marks )
Inverse Square Law:Intensity of illumination produced by a point source varies inversely as square of the distance
from source.
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Model Answer
Subject Code: 17507
1d
2d
1
Intensity
E
*
d
I
2
d
Where,
I = intensity
and
d = Distance
Distance from Source
3D
•4-
>
-
Area
D
4
9A
A
Illuminance
(lux)
L
L/ 4
L/ 9
(ii) Lambert's Cosine law:
According to this law, Illumination at any point on a surface is proportional to the cosine
of the angle between the normal at that point and the direction of luminous flux.
100%
Cosine Law: E0 = E * cos(0)
0°
30°
87%
100%
1
'
50%
.
.
Fig 6.3 Lambert 's cosine law
OR
Fig. 6.5 Lambertian surface.
.^1
^-
if HI
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Model Answer
Subject Code: 17507
b) Justify the use of saturable reactor to control the magnitude of welding current.
Ans:
(4 Marks)
A saturable reactor in electrical engineering is a special form of inductor where the magnetic core
can be deliberately saturated by a direct electric current in a control winding. Once saturated, the
inductance of the saturable reactor drops dramatically. This decreases inductive reactance and allows
increased flow of the AC current.
Saturable reactors often have multiple taps, allowing a small inductance to be used with a large load or
a larger inductance to be used with a smaller load. In this way, the required magnitude of the control
current can be also held roughly constant, no matter what the load.
c) Draw labelled speed-time curve for main line. Explain various parts of the curve.
Ans: Typical speed time curve for main traction line :
(Curve 2 Marks, Explanation 2 Marks, Total, 4 Mark)
. FREE
*
o
/
SPEED
RUNNING
CURVE
RUNNING
COASTING
X
a
Jt
.
a
LJ
UJ
a
%r <
i
RHEOSTATIC
ACCELERATION
t2
t3
TIME IN SECONDS
OR
BRAKING
urn=
L
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Model Answer
Subject Code: 17507
k-
ACCileratiatL
s
Free Running|
~ "
T
"
Retardation
r
CL
y
Vm
z
Q
LLi
UJ
CL
in
—
ta
tT
T
TIME IN SECONDS
Speed Time Curve Explanation:There are five periods in the run of train as shown in speed time curve.
i) Constant acceleration period (o to A):During this period starting resistance in motor circuit are gradually cut down. At point
‘A’ all the starting resistance in motor circuit has been cut down.
ii) Acceleration on speed –Time curve (A to B) For T2 sec.:Now train is continuous to accelerate & torque gradually falls until speed of train exactly
balance train resistance during this period.
iii) Free Running or constant period (B to C) For T3 sec.:At the end of acceleration period train attend maximum speed. During this free tuning
period train runs at constant speed & constant power is taken from supply by train.
iv) Coasting period (C to D) For T4 sec.:At the end of free running period the supply to traction motor is cut down & train allow to
run under its own movement. The speed of train goes on decreasing due to resistance to
motion of train. Rate of decreasing of speed during costing period is known as costing
retardation ( c)
v) Braking period (D to E) For T5 sec.:At the end of costing period brakes are applied to bring the train to rest (stop) during this
period speed of train rapidly decreases & reduces to zero. The rate of decreases of speed
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Model Answer
Subject Code: 17507
during braking period is known as ‘Braking retardation’ ( )
d) Explain which system is preferred for traction work in India.
Ans:
Single phase, 25 KV AC system is preferred for traction work in India
(2 Marks)
It is explained on following point:- (Any two Points are Expected)
1) Supply fed to traction sub-station:-
(2 Marks)
Traction sub-station receivers AC power from a 3-ph high voltage nearest transmission line.
For e.g. – 220/ 132 / 110 KV etc.
2) Equipment in sub-station:In traction sub-station input voltage is step down to utilization voltage (1-ph, 25KV,
AC) So main equipments in traction substation are
a. Step down transformer
b. Protective equipments,
c. Switchgear
d. Control panel
etc.
3) Number of Overhead conductor:Single conductor contact wire is used and return being through rail.
4) Normal Voltage fed to overhead conductor:
Voltage: - 1-ph, 25 KV A.C.
Frequency: - Standard ,50 Hz
5) Equipment in motor coach (locomotive):As working voltage of 1-ph AC series motor is 300/400 V AC. So supply voltage must be
step down in locomotive with the help of step down transformer. This is installed in motor coach.
6) Types of drives used:To obtain mechanical power to move the train 1-ph AC series motor is used. The
working voltage of motor is 1-ph AC, 300/400V.
7) Advantages:-
(Any two advantages are expected)
1. As system voltage is high (25KV) as compared to DC supply system (1500 /3000V) so current
drawn by overhead conductor is less because. (since I 1/V )
2. Due to low current cross section of overhead conductor reduces. So its weight reduces.
Sr.No.
Supply System
1
AC System
Weight of copper conductor
2.57 T / Km
2
DC System
5.265 T / Km
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Model Answer
Subject Code: 17507
3. As weight of overhead conductor reduces design of supporting structure becomes lighter.
4. Due to low current copper losses in transmission line reduces, so transmission efficiency
increases.
5. Due to low current voltage drop in transmission line decreases. Due to this distance between two
substation increases. So number of substation required is less than DC track electrification system
for same track distance. e.g.
S.No.
Voltage level
Distance between 2 substation
1
1-ph AC, 25KV
50 to 80 KM
2
3000V DC
12 to 30 KM
3
1500V DC
5 to 12 KM
4
750/600V DC
3 to 5 KM
6. Due to low current size (capacity) of AC substation is more than DC substation. So number of
substation required is less than DC track electrification system for same track distance.
Sr.No.
Supply System
Size of Substation
1
1-ph AC, 25KV
10 to 15 MW
2
3000V DC
2 to 6 MW
7. Due to all above advantages cost of track electrification less as compared to DC track
electrification system.
8. Since here 1-Ph AC series motor is used to obtain mechanical power and its Characteristics such
as high starting torque, variable speed are suitable for traction purpose.
9. Starting efficiency is high in case of AC supply system as voltage is reduced with the help of
transformer.
e) Explain with neat sketches series parallel control of traction motors.
Ans:
(Series steps---- 2Marks, Parallel steps------- 2Marks,Total 4 Marks)
Series parallel control of DC series motor
1. For traction purpose, two motors are operated in following steps.
Series steps of traction motor:
Step 1 –
Two traction motors M1 and M2 are connected in series and started with all starting
resistances in series.
Step 2 to 7 –
The starting resistances are cut out one by one gradually and finally two motors are in series
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Model Answer
Subject Code: 17507
without any resistance.
In series connection the supply voltage V is divided in two motors. (Both motors get half or
(V/2) volts). So speed is also half. (N/2)
Series Steps
stepl
M2
Ml
Vdc
step 2
S2
Vdc
step3
Ml
S2
M2
Vdc
step 4
si
Ml
S2
M2
S2
M2
Vdc
step5
si
Ml
Vdc
step6
Ml
step /
SI
Ml
-I-
Voltage across each motor is Vdc/ 2 and speed is N/ 2 RPM
M2
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Model Answer
Subject Code: 17507
Parallel steps of traction motor:
Step 1 –
After completion of series last step motors are now connected in parallel again with series
resistance otherwise motor will draw very high current and may damage itself.
Step 4 to 7 –
Both motors are now connected in complete parallel and starting resistances are cut out one
by one.
In parallel connection, voltage across M1 and M2 will be full i.e. V (voltage is always same
in parallel).
Voltage across each motor = V and speed of each motor = N
So, voltage is now increased from (V/2) to V.
Hence, speed also increases from (N/2) to N and motor runs with full speed.
Parallel Steps
Uv^ U/VV^WV ^—
-
-
-
^
^
MV U ^W\^ I^W
-
-
1
S2
1
M2
-
Vdc
I
step 2
—1
-1-
VVV
-
1
1
V V V—
VVV
1
51
^
IVI
- —1
m
i
•
Vdc
yyy
1
1
y y y—
*
yyy
1
S2
M2
*
IVI
m
1
-
1
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Model Answer
Subject Code: 17507
step3
I
Ml
S1
I
+
I
Vdc
I
S2
-
M2
*
-1
step4
—i
'VVv
1
'VW
1
'Wv
i
1
51
<^nnri
^
- 1
M
-i-
i|
*
*
1
,
1
A
.
A
1 iL.
-
Vdc
1
step5
—
1
'vW
-1-
1
—
'VW
,
'vvv
—~
1
(
-
ci
SI
'
r nr>
.
Ml
^
M
~M
CO
S
2
M2
-
_
I
1
Vdc
1
step6
SI
M1
+
S2
M2
Vdc
step7
SI
Ml
+
S2
M2
Vdc
Voltage across each motor is Vdc and speed is N RPM
Advantages:
1. This method has highest starting efficiency then rheostat method.
Starting efficiency of plain rheostat method = 50 %. By this method for two motor it is
66.66% & for 4 motors it is 72.72% and for 6 motors it is 75%
2. Different economical speeds are obtained:
For 2 Motor = 1 :2
For 4 Motor = 1:2:4
For 6 Motor = 1:2:3
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Model Answer
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3) For same power input torque of different magnitude is obtained.
Disadvantages:
1. If proper transition method is not used then
There is loss of torque when motors are disconnected from supply
There will be jerk when motors are reconnected in parallel
Q.6
Attempt any TWO of the following :
16 Marks
a) i) (i) Differentiate between core type and core less induction furnace.
Ans:
(Any Four Points are expected 1 Mark each, Total 4 Marks)
Sr.No. Core type induction furnace.
Core less induction furnace.
1. Less leakage flux
More leakage flux
2. Power factor is good
Power factor is poor
3. Works at normal frequency
High frequency supply is required
4. Weight & size is more
As there is no magnetic core weight & size
of furnace reduces.
5.
6.
7.
8.
9.
a) ii)
Design for high capacity
Crucible used is either Horizontal
or Vertical
Initial cost is less
Used to melt only conducting
metals
Time required for heating is more
as normal frequency is used.
Design for low capacity
Crucible of any shape is used
Initial cost is more as High frequency
supply is required
Both conducting and non-conducting
charge can be heated.
Due to high frequency, high voltage
supply, time required for heating is less.
Explain what is dielectric heating. State its four applications.
(Explanation---2Marks, Applications—2 Marks, Total 4Marks)
Ans:
Figure of dielectric heating:
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Model Answer
Subject Code: 17507
Dielecdr i°c
Heoting^ .
mefoJ plaJte,
L—
HV
C o teas w )
£
10 to 30 MHz
'
dlie\ ecVtJc noodenai
*
k ttf
job oP soil d
.
)
supply
or equivalent figure
Principle of Dielectric heating:
For heating non-metallic material (dielectric material) for e.g. Glass, plastic, wood, etc.
dielectric heating is used.
Material to be heated is placed between two metallic plates as shown in figure (1) across
which a high voltage (20 to 25 KV) and high frequency (10 to 30 MHz) AC supply is given.
Material is heated due to dielectric loss taking place inside the job.
Operation :1. During (+) ve half cycle:
M
C So to 35 Wi)
II HF
10 to 3o T^t Hz)
supply
+
pt etelj
electron rom
Outei orbit
^
shifct CatHocH)
-foulards UppW dVS
plaie
1
or equivalent figure
Material to be heated is placed between two metallic plates, if upper plate is + Ve, most
of electrons from its outer orbit (of job) gets attracted towards + Ve plate.
2. During (-) ve half cycle:
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Model Answer
Subject Code: 17507
Hv
Jrom
ouftt otbit shifts
electron
[Wt|
Cao t<>
tff
( \ 0 -W 30 MHt)
AC
suppt"
3
T
Colt tacts) towards
touted plate CtVe)
’o
During - Ve half cycle field is reversed i.e. bottom plate becomes + Ve. At that time
most of electrons from its outer orbit gets attracted towards bottom electrode.
Effect:Due to inter atomic friction caused by repeated (due to frequency) deformation and
rotation of atomic structure, Dielectric loss takes place inside the job which produces heat.
Applications of Dielectric Heating:-
(Any four application expected: 1/2 Mark each)
1) In food processing industry, dielectric heating is used for Baking of cakes & biscuits in
bakeries.
2) Cooking of food without removing outer shell (e.g.-boiled egg) and pasteurizing of milk.
3) For Rubber vulcanizing.
4) In Tobacco manufacturing industry for dehydration of tobacco.
5) In wood industry for manufacturing of ply wood.
6) In plastic Industry for making different containers.
7) In cotton industry for drying & heating cotton cloths for different processes.
8) In tailoring industry for producing threads.
9) For manufacturing process of raincoats & umbrellas.
10) In medical lines for sterilization of instruments & bandages.
11) For heating of bones & tissues of body required for certain treatment to reduce pains &
diseases.
12) For removal of moisture from oil.
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13) For quick drying gum used for book binding purpose.
14) In foundry for heating of sand, core, which are used in molding processes.
An electric train has a schedule speed of 25 km/hr between stations 800 m apart. The duration
of stop is 20 seconds, the maximum speed is 20% higher than average running speed and the
b)
braking retardation is 3 km/hr/sec. Determine rate of acceleration required to operate the train.
Ans:
Given:- Schedule speed of 25 km/hr,
Distance between stations 800 m
Stop time 20 Sec.
Maximum speed is 20% higher than average running speed, Braking retardation is 3
km/hr/sec
<J- lyiarkj
=
Scheduled speed
'
3600 x 0.8
= 115.20 sec
Actualtime cf run
= T = Txh - T: .p
= 115.20 - 20
I
= 95.20 sec .
----------------------------------------(1 Mark)
3600 D
A var age speed
T
3600 0.8
95.2
------------------------------------------------------- (1 Mark)
30.2521 km / hr.
Maximum Speed = 1.2 Average speed
=1.2 30.2521
=36.3025 km/hr. --------------------------------------------- (1 Mark)
1 1 7200 D Vmax
1 --------------------------------------------------- (1 Mark)
V 2 max Vac
7200 D
5760
----------------------------------------------(1 Mark)
1.2 1
2
(36 .3025)
1317 .87
1 1
0.8741
km
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1
1
0.8741
-------------------------------------------------------------------- (1 Mark)
1
0.8741
3
1
0.5407
1. 8495 kmphs ------------------------------------------------------------------ (1 Mark)
c) i) State any four advantages of high power factor.
Ans: Following Advantages of high power factor:
( Any Four Advantages are expected: 1 Mark each, Total 4 Marks)
1. Cross section of conductor reduces:
Cross section of conductor I
1
P. f
As P.F. increases current reduce so; cross section of conductor and its weight reduces
hence its cost reduces
2. Design of supporting Structure:
As weight of conductor reduces design of supporting structure (tower) becomes
lighter, so its cost reduces.
3. Cross section of terminal (contacts) reduces:
As power factor increases, current reduces. hence cross section of switchgear bus bar
and contacts etc decreases.
4. Copper losses reduces:
As power factor increases current reduces. So copper losses reduces. As a effect efficiency
increase.
5. Voltage drop reduces:
As P.F. increases, current decreases. So voltage drop decreases, So regulation gets
improved (better)
6. Handling capacity (KW) of equipment increases:
As power factor increases, handling capacity of each equipment such as Alternator,
km
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transformer increases
7. KVA rating of equipments reduces:
As P.F. increases, current decreases. So KVA rating of all equipments for eg- alternator,
transformer etc decreases, so its capital cost reduces.
8. Cost per unit (KWH) reduces:
From all above advantages, it is seen that cost of generation, transmission &
distribution decreases, so cost/unit reduces.
Also performance i.e. efficiency & regulation gets improved at high power factor
c) ii) A single phase 400 V, 50 Hz motor takes a supply current of 50 A at a P.F. of 0.6. The motor
P.F. has to he improved to 0.9 by connecting a capacitor
in parallel with it. Calculate the required capacity of capacitor in Farads
Ans: Given Data : V = 400 V , f = 50 Hz , I = 50 amp. At 0.6 P.F. To be improve 0.9
I I1 Cos1
I 50 0.6
I 30 Amp
---------------------------------------------------------(1 Mark)
I I tan 1
I 30 1.33
I 39.9 Amp
I 2 I tan 2
I 2 30 0.4843
I 2 14.529 Amp
I C I 1 I 2
I C 39.9 14.53
I C 25.37 Amp
____________________________ (1 Mark)
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IC
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V
XC
400
25.37
------------------------------------------------- (1 Mark)
X C 15.76 Ohm
XC
C
1
2 FX C
C
1
2 50 15.76 -------------------------------------------(1 Mark)
C 2.019 10 4 F
------------------------------------------------------END-------------------------------------------------------
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Important suggestions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more importance.
(Not applicable for subject English and communication skills)
4) While assessing figures, examiner may give credit for principle components indicated in a figure.
The figures drawn by candidate and model answer may vary. The examiner may give credit for any
equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model answer.
6) In case some questions credit may be given by judgment on part of examiner of relevant answer
based on candidate understands.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.
Q.1 A) Attempt any three :
12
State any four factors to be considered while selecting electric drives (motor) for a particular
a)
application.
Ans:
(Any Four Points From The Following Or Equivalent Points Are Expected 1 Mark
To Each Point, Total 4 Marks)
Following Factors governing / or are considered while selecting electric drive (Motor) for
particular application:
1. Nature of supply:
Whether supply available is
AC,
Pure DC
Or Rectified DC
2. Nature of Drive (Motor):
Whether motor is used to drive (run)
Individual machine
OR group of machines.
3.
Nature of load:
Whether load required light or heavy starting torque
OR load having high inertia, requirehigh starting torque for long duration.
OR Whether load torque increases with speed (T N)
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OR decreases with speed (T 1/ N )
OR remains constant with speed (T = N)
OR increases with square of speed (T N2)
4.
Electric Characteristics of drive:
Starting,
Running,
Speed control
and braking characteristics
of electric drive should be studied and it should be matched with load requirements(i.e. machine).
5.
Size and rating of motor:
Whether motor is short time running
OR continuously running
OR intermittently running
OR used for variable load cycle.
Whether overload capacity, pull out torque is sufficient.
6.
Mechanical Considerations:
Types of enclosure,
Types of bearing,
Transmission of mechanical power,
Noise
and load equalization
7.
Cost:
Capital,
Running
and maintenance cost should be less.
State any two applications of each for the following types of electrical heating :
b) (i) Direct resistance heating (ii) Indirect induction heating (iii) Direct.arc heating (iv) Dielectric
heating
Ans: (Any Two Application Are Expected Of Each Heating Type 1/2 Mark Each
Application, Total 4 Marks )
(i) Direct resistance heating:(Any Two Application Are Expected)
1. This type of heating used for industrial purpose
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Model Answer
2. Salt bath heating: This utilized for the purpose of carbonizing, tempering, quenching and
hardening of steel tools
3. Heating of water in boiler
(ii) Indirect induction heating:
(Any Two Application Are Expected)
1. For heating as well as melting
2. Production of carbon free ferrous alloys.
3. For vacuum melting.
4. For melting non-ferrous metals for e.g. copper, aluminum, nickel etc.
5. For duplexing steel products.
6. Heating of non-conducting material is also possible if crucible is made from conducting
material.
(iii) Direct.arc heating:-
(Any Two Application Are Expected)
1. Used for continuous and large production of high quality steel.
2. For Ferro-alloy manufacturing
(iv) Dielectric heating:-
(Any Two Application Are Expected)
1) In food processing industry, dielectric heating is used for Baking of cakes & biscuits in
bakeries.
2) Cooking of food without removing outer shell (e.g.-boiled egg) and pasteurizing of milk.
3) For Rubber vulcanizing.
4) In Tobacco manufacturing industry for dehydration of tobacco.
5) In wood industry for manufacturing of ply wood.
6) In plastic Industry for making different containers.
7) In cotton industry for drying & heating cotton cloths for different processes.
8) In tailoring industry for producing threads.
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9) For manufacturing process of raincoats & umbrellas.
10) In medical lines for sterilization of instruments & bandages.
11) For heating of bones & tissues of body required for certain treatment to reduces paints &
diseases.
12) For removal of moisture from oil.
13) For quick drying gum used for book binding purpose.
14) In foundry for heating of sand, core, which are used in molding processes.
Define following terms referred to illumination : (i) Space-height ratio (ii) Utilization factor (iii)
Maintenance factor (iv) Waste light factor
Ans: (Each Definition 1 Mark, Total 4 Mark)
c)
(i)Space-Height ratio:
Space height ratio
Space between lamps
Height of lamps above working plane
(ii) Utilization factor:It is defined as the ratio of total lumens reaching the working plane to the total
lumens given out by the lamp. Its value is always less than one.
(iii) Maintenance factor :It is defined as the ratio of illumination under normal working conditions to the illumination
when everything is clean.
OR
Ma int enacefactor
Illu min ation under normal working condition
Illu min ation under every thing is clean
(iv)Waste light factor:
When a surface is illuminated by several numbers of the sources of light, there is certain
amount of waste due to overlapping of light waves,
The waste of light is taken into account depending upon the type of area to be illuminated.
The value of waste Light factor 1 to 1.5
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d) State the four causes of low (poor) power factor.
Ans: (Any Four causes of low (poor) power factor From The Following Or Equivalent
Points Are Expected 1 Mark To Each Point, Total 4 Marks)
Following are the Causes of low power factor: 1.
Magnitude of Magnetizing Current (I ):As magnetizing current increases, power factor reduces.
2.
Due to use of Induction Motor:Most of industrial drives, agriculture pumps, lift, irrigation pump set uses I.M.
which works at lagging power factor, and so power factor reduces.
3.
Due to use of Transformer: All transformers works at lagging power factor, so power factor of system reduces.
4.
Due to welding transformer: Welding transformers are operated at low p.f. which reduces p.f. of the system.
5.
Due to inductance of transmission & distribution Line: In case of AC transmission & distribution lines, inductance is present which the main
cause of low power factor .
6.
Series Reactor:Series reactor is used in substation to minimize fault current Which causes low power factor.
7.
Industrial electrical heating furnaces:Induction and arc furnace used in steel manufacturing industry works at low p.f.
which reduces p.f. of the system.
8.
Arc Lamp:Arc lamp & electric discharge lamps operates at low p.f.so p.f. of the system reduces.
9.
Equipments operated at light load:P.f. falls if equipments like alternator, transformer, I.M.etc are not operated at full load.
10.
Improper repairs and maintenance:P.f. falls if proper maintenance or repairs of equipments are not done.
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Model Answer
Subject Code: 17507
Q.1B)
Attempt any ONE :
(1x6=6)
Draw the graph load vs. time and estimate suitable HP rating of electric drive (motor) having
a) following duty cycle : (i) Rising load from 200 to 400 HP : 4 minutes (ii) Uniform load of 300 HP : 2
minutes (iii) Regenerative braking from 50 to zero HP for : 1 minute (iv) Idle for :1 minute
Ans:
(When Final answer of Numerical is correct Give Full Marks & if final answer is
wrong give stepwise marks)
Graph:
—
I
(dOBtfp
300 HP
%6©
2=5«
— ioe>
p
Ann
'zmw,
J
rr,
50 t-p
Duty Cycle (T) = t1+ t2 + t3 + t4
m
--
( 2 Mark)
----------------------------------------------- (1/2 Marks)
= 4+2+1+1
= 8 Min. --------------------------------------------------- (1/2 Marks)
Continuous rating of Motor:
rating of motor
V
1
1
( HP12 HP2 2 HP1 HP2 )t1 (HP3 ) 2 t2 ( HP4 2 HP5 2 HP4 HP5 )T3 HP6 2 T4
3
3
T
-
----------------------------------- (1 Mark)
rating of motor
-
----
1
1
(2002 400 2 200 400) 4 (300) 2 2 (50 2 02 50 0)1 02 1
3
3
8
rating of motor 263.193 HP ----------------------------------------------------------- (2 Mark)
Nearest Standard rating of motor should be selected
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Model Answer
Subject Code: 17507
b) Compare between Resistance welding and Arc welding on any six points.
Ans: (Any Six Points From The Following Or Equivalent Points Are Expected 1 Mark To Each
Point, Total 6 Marks)
Sr.No
1
Parameters
Type of welding
Resistance Welding
Plastic/Pressure/Non-fusion
Arc Welding
Fusion/Non pressure welding
welding
2
3
Principle of heat
Heat is developed due to I2R
Heat developed due to arc
developed
losses where R is the contact
produced in between electrode
resistance
and job
Not required during welding
Required during welding
Required
Not required
Both AC,DC supply is used.
Metal arc welding – Both AC,DC
But generally Ac Supply is
supply is used. But generally Ac
used.
Supply is used.and for
External filler material
required
4
External pressure
required
5
Type of supply used
Carbon arc welding –only DC
supply are used
6
Voltage &current
required
Low voltage (2 to 20V AC)
and high current (40 to
400A, in some cases 5 to
20KA ) supply is required
Metal Arc welding Voltage70 to 100V AC and
Carbon arc welding voltage50 to 60V DC,
Current- 50-600-800A
7
Energy consumption
Low (3 to 4 KWH/Kg of
High (5 to 10 KWH/Kg of
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Temperature obtained
Page 8 of 35
Model Answer
Subject Code: 17507
deposited material )
deposited material.)
Temperature obtained is not
Temperature obtained is very
very high (up to 13500C)
high (up to 35000C to 60000C)
9
Power factor
Low
Poor
10
Type of electrode
Non-consumable electrodes
Coated electrodes are used for
are used.
metal arc welding and bare
electrodes are used for carbon arc
welding.(Electrodes may be
consumable or non-consumable)
1.
Application
It is suitable for mass
production
It is suitable for heavy job,
maintenance and repair work
Q.2
Attempt any FOUR :
(4x4=16 Mark)
a) State four advantages and four disadvantages of electrical braking over mechanical braking.
Ans:
(Any Four Points From The Following Or Equivalent Points Are Expected For
Advantages 1/2 Mark To Each Point, Total 2 Marks & For Disadvantages 1/2 Mark To Each
Point, Total 2 Marks, Total 4 Marks)
Following are the advantages & disadvantages of electrical braking over mechanical braking system.
Advantages: (Any Four Points From The Following Or Equivalent Points Are Expected)
1. It is most reliable braking system.
2. Breaking actuation time is small as higher value of braking retardation is obtained.
3. Electrical braking is smooth & gradual.
4. Life of electrical braking system is more.
5. There is less wear & tear of brake shoes, break block etc. so there is less maintenance cost.
6. Higher speeds are possible even when train is going down the gradient, as breaking system is
reliable.
7. Trains having heavy loads can be stopped even when train going up the gradient.
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Page 9 of 35
8. Higher speeds of train is possible as braking system is reliable so pay load capacity increases.
9. In case of electric regenerative braking we can utilize 60 to 80% of kinetic energy to generate
electricity which is not possible with mechanical braking.
Disadvantages: (Any Four Points From The Following Or Equivalent Points Are Expected)
1. In addition to electrical braking there must be arrangement of mechanical braking for final
stop.
2. Special arrangement of circuit and complication makes electrical braking system costly.
3. Operation in substation becomes complicated at the time of regenerative breaking when
generated energy is surplus.
4. Initial cost is more due to other control equipments & circuitry.
b) Derive an expression for design of heating element when heating element is circular wire.
Ans:
(Derivation up to equation 1 or part I--- 2Marks & For equation 2 or part II---2
Marks, Total 4 Marks)
Let,
P = electrical Input in watt (w)
V = Supply voltage in V
I = Currents in Amp
R= Resistance of heating element in
= Specific resistivity in /m
l = length of heating element in m
a = Cross section of heating element in m2
d = Diameter of heating element in m
Part-I:
P=VI
But , I
P
V
R
VV
R
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P
V2
----------------------------------------------- (A)
R
R
l
a
R
4 l
d2
But,
and
a
d2
4
Now Substitute value of ‘R’ in equation (A)
P
V2 d2
4l
Rearranged above equation,
l
V 2
d2 4 P
--------------------------------Equation I
Part-II:
When steady temperature in reached , electrical input is equal to heat output
Electrical input = Heat output
P = H (Surface area)
Surface area of circular heating element d l
P H ( d l)
Now substitute the value of “P”
V 2 d 2
H ( d l)
4l
Rearranging above equitation
d 4 H
l2
V2
Page 10 of 35
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Page 11 of 35
d 4H
l2
V2
--------------------------------Equation II
From equation I and II we can calculate length & diameter of heating element.
c) State eight requirements of an ideal traction system.
Ans: (Any Eight Points From The Following Or Equivalent Points Are Expected 1/2
Mark To Each Point, Total 4 Marks)
Ideal Traction system should processes following requirement:1. It should be Pollution free.
2. It should have low capital, Running and maintenance cost.
3. It should have quick starting time.
4. It should have high starting torque.
5. It should have high rate of acceleration & retardation.
6. Highest speeds are possible.
7. It should have easy speed control method.
8.
Its braking system should be reliable and causes less wear.
9. It should have better riding quality (less vibration)
10. It should be free from unbalance forces i.e. coefficient of adhesion should be more.
11. It should have lower center of gravity.
12. The locomotive should be self-contained and able to run on any route
13. There should be no standby losses.
14. It should have high efficiency
15. Regenerative braking should be possible.
16. The wear caused on the track should be minimum.
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Page 12 of 35
17. Equipment should be capable of overloads for short periods.
18. Capability of withstanding voltage fluctuations.
19. Parallel running usually more than one motor (2 or 4 motors) should be possible.
20. Traction system should be clean & long life.
21. There should be no interference to the communication lines running along the lines.
d) State various systems of track electrification.
Ans: (Any Four Systems Of Track Electrification From The Following Are Expected 1 Mark To
Each Systems Of Track Electrification, Total 4 Marks)
Following are the different track electrification system
D.C. Supply system:1. Direct current track electrification:
600V, 750V DC for tramways
1500V, 3000V DC for Train (Urban and sub-urban services)
A.C. Supply system:2. 1-Ph, 25KV,standard frequency AC supply system:
1-Ph, 25 KV, 50 Hz
3. 1-Phase, low frequency AC Supply system:
1-Ph, 15/16 KV, 16.2/3 Hz or 25 Hz
4. 3-Ph, Low frequency AC supply system;
3-Ph, 3.3/3.7 KV, 16 2/3 Hz or 25 Hz
Composite system:5. 1-Ph AC (1-ph, 25KV) – DC Supply System
6. Kando System (1-Ph AC – 3-Ph AC)
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Model Answer
Subject Code: 17507
Compare between urban line, sub-urban line and mainline services on following points :
(i) Distance between two Railway Station. (ii) Acceleration (iii) Retardation (iv) Maximum Speed
e)
(v) Specific energy consumption (vi) Free running period absent or present
(vii) Coasting period absent or present (viii) Shape of speed time curve
Ans:
(Eight Points From The Following Are Expected 1/2 Mark To Each Point, Total 4 Marks)
Sr.No
1.
Points
Urban line services
Suburban line services
Main line services
Distance
Low (1km)
Medium (2.5 to 3 km)
High (above10km)
Acceleration
High = 1.5 to 4
High = 1.5 to 4
low = 0.6 to 0.8km/hr-
( )
km/hr-sec
km/hr-sec
sec
Retardation
High = 3 to 4
High = 3 to 4 km/hr-
low = 1.5 km/hr-sec
( )
km/hr-sec
sec
Maximum
120 km / hr
120 km / hr
160 km / hr
Specific
High = 50 to 75
High = 50 to 75
Low = 18 to 31
energy
watt-hr/tone-km
watt-hr/tone-km
watt-hr/tone-km
Free running
Free running period
Free running period is
Free running period
period
is absent
absent
Present / long
Coasting
Coasting period is
Coasting period is
Coasting period is
period
Present/small
Present/small
Present/long.
Shape of
Quadrilateral
Quadrilateral
Trapezoidal
between two
railway
station
2.
3.
4.
Speed
5.
consumption
6.
7.
8.
speed-time
km
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Model Answer
Subject Code: 17507
Q.3 Attempt any TWO :
(2 x 8 =16 Marks)
a) i) State the factors to be considered for selection of shape and size of the car of the elevator.(
(Any Four Points From The Following Or Equivalent Points Are Expected 1 Mark To
Each Point, Total 4 Marks)
The size and shape of elevator car depends on following factors:
i) No. of passenger to be carried: While selecting the size of car it is a usual practice to allow.
A Space of 2 Sq.fit/ person.
Average weight of passenger is assumed 68 kg/person.
Ans:
Thus the maximum load capacity of elevator is considered 34 kg/sq.ft
There should be wide frontage and shallow depth
ii) Limitation in the building design:
Shape of elevator depends on space available in building.
iii) Type of building
iv) Application of elevator
a) ii)
Ans:
Draw graphical representation of load cycle :
(i) Continuous loading (ii) Short time loading (iii) Long time (intermittent) loading
(iv) Continuous operation with short time loading
( Each graphical representation of load cycle 1 Mark Each, Total 4 Marks)
(i) Continuous loading:JLoad » n HP
i
y'
i
-
4"
Temperature rise
/
Ct)
or eqivalent figure
km
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Model Answer
Subject Code: 17507
(ii) Short time loading:(J
load in HP
-load.
£ cycle repeals
- Temperdture nse
\
\/
‘Time Ct)
or eqivalent figure
(iii) Long time (intermittent) loading:ioaciin HP
4-oadL
8 cycle repeals
—
/
4-
•
Temperaturn rise
timeCt)
or eqivalent figure
(iv) Continuous operation with short time loading:-
jLoacbo
HP
—
XoacL
8 cycle repeals
X
f‘ ss
H-L
/
Temperature
,
*
xise
•
Time CO
or eqivalent figure
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b)
A 20 kW single phase 220 V resistance oven employs a circular nichrome wire for its heating
element. If wire temperature is not to exceed 1170 °C and temperature of charge is to be 500 °C.
Calculate diameter and the length of wire. Take k = 0.57, e = 0.95 and Resistivity = 1.09 x 10-6
ohm-mtz.
Ans:
(When Final answer of Numerical is correct Give Full Marks & if final answer is
wrong give stepwise marks)
Given Data:
T1 = 11700C = 1170+273 = 1443 0K
T2 = 500 0C = 500 +273 = 7730K
------------------------------------------- (1/2 Mark)
-----------------------------------------------(1/2 Mark)
Radiation efficiency = 0.57, specific resistance of Ni-Cr = 1.09x10 -6 ohm m, emissivity = 0.95.
T1 4
T
) ( 2 )4 w / m 2
1000
1000
OR
T
T
H 5.72 k .e [ ( 1 ) 4 ( 2 ) 4 ] w / m 2
100
100
----------------------------- (1 Mark)
H 5.72 104 k.e [ (
H 5.72 0.57 0.95 [ (
1443 4 773 4
) (
) ] w / m2
100
100
H 123236.0773w / m 2
-------------------------------------------------------- (1 Mark)
l
V2
d2 4 P
---------- Equation No.1----------------------------- (1 Mark)
(220) 2
l
d 2 4 20 1000 1.09 10 6
l
1743728.032
d2
l 1743728.032d 2
--- Equation No.2 ----------------------------- (1 Mark)
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Model Answer
Subject Code: 17507
d
4 H
2
l
V2
d
4 1.09 10 6 123236.0773
l2
(220) 2
d
1.1101 10 5
l2
d 1.1101 10 5 [1743728.032 d 2 ] 2
d 33754874.07 d 4
d
33754874.07
d4
1
33754874.07
d3
d 3 2.9625 10 8
Taking Cube root of both sides
d 3.0942 103 m
-------------- (1Mark)
d 3.0942mm
Substitute Value of ‘d’ in Equation No.2 to calculate ‘l’ :
l 1743728.032d 2
----------------------------- (1 Mark)
3 2
l 1743728.032 [3.0942 10 ]
l 16.69m
------------------------------------------------- (1 Mark)
Answer : Length l 16 .69 mtr
Diameter d 3 .0942 mm
c) i)
Ans:
Compare individual and group drive on any four points.
(Any Four Points From The Following Or Equivalent Points Are Expected 1 Mark To
Each Point, Total 4 Marks)
S.No.
Point
Individual Drive
Group Drive
1.
Definition
In this type of drive each
machine has its own separate
electric drive (motor). It may
be directly coupled or
indirectly coupled
In a group drive single large
capacity electric drives is used to
run number of machines through
a long common shaft.
2.
Initial Cost
High
Less
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Model Answer
Subject Code: 17507
3.
Flexibility
More Flexibility
Less Flexibility
4.
Safety
It is more safe
It is less Safe
5.
Reliability
It has high reliability
It has less reliability
6.
Space required
More
Less
7.
Overload Capacity
Less
Higher
8.
Maintenance cost
More
Less
9.
Speed control
Easily possible
Difficult
10. Mechanical Power
transmission losses
11. Addition/Alternation
Less losses
More Losses
Easily possible
Easily not possible
12. Total HP
13. Appearance
14. Any one application
of each
More
Good
Lathe Machine (Similar
application will be consider)
Less
Not good
Textile Industry (Similar
application will be consider)
c) ii) Why noise of motor is produced? How it can be reduced?
Ans: (Any Two Points From The Following Or Equivalent Points Are Expected, 1 Mark each
point Total 2 Marks)
Noise of motor is produced due to:1. Vibration
2. Bad foundation
3. Friction
4. Magnetic pulsation
(Any Two Points From The Following Or Equivalent Points Are Expected, 1 Mark each
point, Total 2 Marks)
Noise of motor can be reduced by –
1. Motor is mounted on cushion such as rubber pad instead of direct mounting on concrete foundation.
2. Motor can be mounted on spring so that it can absorb all the vibrating frequency.
3. By proper maintenance i.e. if bearings are worn out then replace it.
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Model Answer
Subject Code: 17507
4. Lubrication must be done regularly.
5. Electrical connection to motor should be given through PVC flexible pipe instead of metal flexible
pipe.
6. By connecting condenser of 0.1 to 0.25 microfarad across live and motor frame (earth) which will
reduce magnetic pulsation. Also it will reduce radio interference
Q.4A)
Attempt any THREE :
a) Give classification of electrical welding.
Ans:
(3 x 4 =12 Marks)
(Total 4 Marks)
i) Resistance Welding:1) Spot welding
2) Seam welding
3) Projection Welding
4) Butt Welding- i) Simple butt welding
ii) Flash butt welding
ii) Arc welding:1) Carbon Arc Welding: a) shielded welding b) unshielded welding
2) Metal Arc Welding: a) shielded welding b) unshielded welding
b) State the two laws of illumination.
(Inverse Square Law :- 2 Marks , Lamberts Cosine Law:- 2 Marks )
1) Inverse Square Law:Intensity of illumination produced by a point source varies inversely as square of the distance
from source.
d
Id
Ans:
*
:d
1
Intensity
E
I
d
2
Where,
I = intensity
and
d = Distance
4
HW
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Model Answer
Subject Code: 17507
2) Lamberts Cosine Law:
According to this law, Illumination at any point on a surface is proportional to the
cosine of the angle between the normal at that point and the direction of luminous flux
100%
Cosine Law: E0 = E * cos(0)
0°
30°
A
87%
100%
85°
S3k\
50%
Fig. 6.3 Lambert's cosine law.
Fig. 6.5 Lambertian surface.
OR
Following two tariffs are offered to consumers :
c) (i) Rs. 150 + 20 paise per unit. (ii) A flat rate of 40 paise per unit. State at what consumption which
tariff is economical.
Ans:
(When Final answer of Numerical is correct Give Full Marks & if final answer is
wrong give stepwise marks)
Given data: Two tariffs
(i) Rs. 150 + 20 paise per unit
(ii) A flat rate of 40 paise per unit
Let ‘X’ be energy consumption when both tariff give same energy bill:
Rs. 150
20
40
X
X
100
100
Rs. 150 0.2 X 0.4 X
Rs. 150 0.4 X 0.2 X
X
150
0.2
X 750 KWH ------------------------------
( 2 Marks)
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Model Answer
Page 21 of 35
State at what consumption which tariff is economical-------------------------------( 2 Marks)
When energy consumption is greater than 750 KwH the tariff one i.e. Rs. 150 + 20 paise/unit
is economical when energy consumption is less than 750 KwH than tariff Two i.e. flat rate 40
paise per unit is economical.
d) State disadvantages of low power factor.
Ans: (Any Four Points From The Following Or Equivalent Points Are Expected 1 Mark
To Each Point, Total 4 Marks)
Disadvantages of Low power Factor: 1) Cross section of conductor increases: As power factor reduces current increases, cross section of conductor increases. Hence its cost
increases.
2) Design of supporting structure: As power factor reduces, cross section of conductor increases, so its weight increases. To handle
this weight design of supporting structure becomes heavier, so its cost increases.
3) Cross section of terminals increases: As power factor reduces, current increases, Hence cross section of switch gear, bus bar,
contacts, and terminals increases. So its cost increases.
4) Copper losses increases: As power factor reduces current increases. So copper losses increases. As an effect efficiency
reduces.
5) Voltage drop increases: As P.F. reduces current increases. Therefore voltage drop increases, so regulation becomes poor.
6) Handling Capacity of equipment reduces:
Handling capacity (KW) of each equipment such as Alternator, transformer reduces as power
factor reduces. e.g.
6) High KVA rating of equipment required:As power factor decreases KVA rating of all equipment’s increases, so that its cost increases.
7) Cost/unit increases: - From all above disadvantages it is seen that cost of generation, transmission
& distribution increases. Also its performance efficiency & regulation reduces, So that cost/unit
increases.
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Model Answer
Subject Code: 17507
Solve any ONE of the following
06 Marks
Draw figure of (i) Seam welding and (ii) Flash Butt welding and write two applications of each
a)
type.
Ans: (figure of (i) Seam welding – 2 Marks, Application any Two 1/2 Mark each, figure
Q. 4 B)
of (ii) Flash Butt welding– 2 Marks, Application any Two 1/2 Mark each)
(i) Seam welding: (Figure- 2 Mark)
<
S> CCLm Welding.
Ti met.
Trbllet Vype.
Acsoppl
3
electrode
r
N ldhn
nvex>
or equivalent figure
Applications of Seam welding:- (Any Two Applications Are Expected ½ Mark Each,
Total 1 Mark)
It gives leak-proof joints.
1. Hence used for welding of various types of containers,
2. Pressure tank,
3. Tank of transformer,
4. Gas line,
5. Air craft tank,
6. Condenser,
7. Evaporator and
8. Refrigerator etc.
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Model Answer
Subject Code: 17507
(Figure- 2 Mark)
(ii) Flash Butt welding:-
f 1 CLS b £> mlt
-
W e\ d
Flaih
C \ cvnrif >
i
o
^
_
preisu >re.
j b
°
^
SULJLSLJL&-SIJ1-SLS-S
~c~
,
_rx~cmrc
'
~c
~ ~~
'
l
a
.
0 pf >1 ieci cohere
cueld
..
ooaldL r g
.
+* p » s otatoined
A
'
•
rr
^-
#vc
supply
Application Flash Butt welding: (Any Two Applications Are Expected ½ Mark Each,
Total 1 Mark )
1. For welding rod.
2. For weld shaft
3. Rail, ends
4. For welding chains
b) i) i) State the four requirements of Tariff.
Ans: Any Four Requirements From The Following Or Equivalent Points Are Expected
, Total 3 Marks)
Following are the requirements of Tariff :1. It should be easy to understand to consumer.
2. Easy to calculate.
3. Tariff should be attractive i.e. It should not be too high or too low. It should be
reasonable.
4. Tariff should be economical as compare to other types of energy sources.
5. Tariff should be different for different types of consumers.
6. Tariff must be fair, so that different types of consumers are satisfied with rate of
electrical energy charges.
7. Tariff should be framed into two parts i.e. fixed charges + running charges.
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Model Answer
Page 24 of 35
8. Tariff should be high during peak load period.
9. Tariff should be low during off load period.
10. For industrial consumer, in addition to basic tariff incentives and penalty related to
PF and LF should be considered.
b) ii) State two advantages of P.F. tariff and TOD tariff for the power system concern
Ans: (P.F. tariff advantages---- 1.5Marks & TOD tariff advantages---- 1.5Marks)
Advantages of P.F. tariff for the power system concern:-
(Any Two Advantages are Expected, 1.5 Marks )
Advantages:1) Industrial consumers were trying to run their industry above 0.95lagging power factor (At
High P.F.) To get discount in energy bill, so overall P.F. of power system increases.
2) As each industry run at high power factor then overall power factor of power system
increases. Due to this
3) Which will automatically beneficial from the economics of power system
Because at high power factor :Generation has following advantages:
a)
Low KVA rating of equipment (alternator) is required.
b)
Handling capacity of equipment (alternator) increases.
c)
Cost per unit decreases.
Transmission has following advantages:
a)
Low KVA rating of equipment (Transformer) is required.
b)
Handling capacity of equipment (Transformer) increases.
c)
Cost of conductor decreases.
d)
Cost of supporting structure decreases.
e)
Copper losses increases
f)
Transmission efficiency increases.
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g)
Voltage drop in transmission line decreases.
h)
Regulation gets improved
Page 25 of 35
Advantages of TOD tariff for the power system concern:-
(Any Two Advantages are Expected, 1.5 Marks)
Advantages:
1. Major industrial consumers are trying to run their maximum load during OFF load period, to get
rebate in their energy bill.
2. Major industrial consumers are trying to run their industry at reduced load during PEAK load
period to avoid additional charges charged in energy bill.
3. Due to above two reasons, it increases overall load factor as well as diversity factor of power
system.
4. As load factor and diversity factor of power system increases so overall cost per unit reduces.
5. Also due to this there will be maximum utilization of power plant & infrastructure.
So, TOD tariff helps to avoid the wastage of surplus energy generated during OFF load period.
In this way it helps to conserve energy.
Q.5
Attempt any FOUR :
(4 x4=16 Marks)
Compare sodium vapour lamp and metal halide lamp on following points : (i) Luminous efficiency,
a)
(ii) life of lamp, (iii) re-strike time, (iv) cost of installation.
Ans:
(Each Point 1 Mark, Total 4 Marks)
Sr.
Point
Sodium vapour lamp
Metal halide lamp
NO.
1. Luminous
Luminous efficiency Lm/w Luminous efficiency Lm/w 50-100
efficiency
80-100
2.
Life of lamp
Life more 12000-16000 hrs.
Life less than SV lamp 12000 hrs.
3.
Re-strike time
4.
Cost of
installation
Less Than MH lamp (2 To 5 More than SV lamp (5 To 10 min.)
min.)
Less Than MH lamp
More than SV lamp
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b) State any factors to be considered while selecting electrical welding system.
Ans: (Any Four Factors From The Following Or Equivalent Points Are Expected 1Mark To
Each Point , Total 4 Marks)
Following Factors are considered while selecting of electric welding system:1) Type of Material:Whether similar metal is to be welded or dis-similar metal is to be welded.
2) Property of Material:Whether ferrous or non-ferrous metal is to be welded.
3) Thickness of job:It is also depends on thickness of job to be welded.
e.g. for thick material- Arc welding is used. And for thin material – Resistance welding is used.
4) Temperature required:Whether job required high or low temperature to weld the job.
e.g. For high Temperature - Arc welding is used. And for low Temperature – Resistance welding
is used.
5) Pressure required:If job is need of pressure at the time of welding in that case resistance welding is used. And if
pressure is not required Arc welding is used.
6)Type of Supply Available:Whether AC or DC or both supply are available.
7) Application:In case of mass production, resistance welding is used & for repair work Arc welding is used.
iiii
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Model Answer
Subject Code: 17507
c) Sketch the various steps required for bridge transition system.
Ans:
(Each Step 1 Mark, Total 4 Marks)
In bridge transition, series last step to parallel first step, is carried out by following steps
Stepl: Bridge link is connected between two motors as shown in figure ( Series
last step )
Ml
Ml
X
BfldseLlrtk
U
Step 2: Bridge link is so rotated that two motors are put in series without starting resistance . Which are
un -shorted at the same time.
Mil
M
+
X
Bridge Link
1 M2
m
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Model Answer
Subject Code: 17507
Step 3:The portions of external resistance are connected in each motor circuit as shown in fig
pUw^WV^WV l
-
|
Mli
T'
X
1
fM
BTidgeUnk
2
^^
—^
|-VW-pW\rpVW-|
Step4: In this last step bridge link is removed as shown in fig. This is nothing but parallel first step.
ilM
Mil
w
00
1
2
pAAArjAAAq^AAq— 1
d) Compare AC and DC system of track electrification on any four points.
Ans:
(Any Four point expected: 1 Mark each, Total 4 Marks)
S.No
Points
1. Supply given to O/H
condition
AC System Traction
1-ph, 25KV, AC 50 Hz
DC System Traction
600/750V-Tromways
1500/3000V urban/suburban
2. Type of drive used
1-ph, AC series motor
3. Weight of traction motor
1.5 times more than d.c.
series motor.
Less starting torque than
d.c series motor
Less than d.c series motor
Less than d.c series motor
Simple and smooth
DC series motor for
tramways. DC compound
motor
1.5 times less than a.c series
motor
High starting torque
4. Starting torque
5. Accln and retardation
6. Overload capacity
7. Method of speed control
High
High
Limited, except chopper
method
ft
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Model Answer
Subject Code: 17507
8. Maintenance cost of
traction motor
9. Starting Efficiency
10. Ridding quality
More
Less
More
Less, better than d.c.
Less
Smooth (Better)
11. Insulation cost
High
Low
12. Cross section of conductor
Less
More
13. Design of supporting
structure
14. Distance between two
substation
15. No. of substation required
for same track distance.
16. Size (capacity) of traction
substation
17. Capital & maintenance
cost of substation
18. Cost track electrification
for same track distance
19. Applications
light
Heavy
More
Less
Less
More
More
Less
Less
More
Less
More
Main line services
Urban and suburban area
State any four desirable characteristics of ideal traction motor. State the names of different
traction motor used.
Ans: (Any Four Points From The Following Or Equivalent Points Are Expected ½ Mark
To Each Point, Name of any two motors 1 Mark Each, Total 4 Marks)
Desirable characteristics of ideal traction motors:-
(Any Four Points From The Following, Total 2 Marks)
A) Mechanical Properties or characteristics:
1) It should be robust in construction to withstand against continuous vibrations.
2) Weight of motor per HP should be minimum in order to increase pay load capacity.
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Model Answer
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3) It must be small in overall dimensions, especially in overall diameter.
4) It must have totally enclosed type enclosure to provide protection against entry of dirt, dust, mud,
water etc. in drive.
5) When motors are running in parallel they should share almost equal load. (even when there is
unequal wear & tear of driving wheels)
B) Electrical Properties or characteristics:
6) It should have high starting torque.
7) It should possess high rate of acceleration & retardation.
8) It should be variable speed motor.
9) Its speed-torque characteristics should be such that it should producehigh torque at low speed and
low toque at high speed.
10) Motor must be capable of taking excessive overload in case of emergency.
11) It should have simple speed control methods.
12) Electrical braking system should be reliable, easy to operate and control, especially regenerative
braking is possible.
13) Motor should draw low inrush current (Starting current,and if supply is interrupted and restore
again.)
14) It should withstand for voltage fluctuation without affecting its performance.
C) General Properties or characteristics:
19) It should have low initial cost.
20) It should have less maintenance cost.
21) It should have high efficiency.
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Page 31 of 35
22) It should have long life.
Some of the motors which find application are-
(Any Two Names of Motors From The Following, Total 2 Marks)
1. DC Series Motor
2. 1-Ph, AC series motor
3. 3-Ph, Slip-ring induction motor
4. Linear induction motor (LIM)
Q.6
Attempt any TWO of the following :
16 Marks
a) i) Give the definition of (1) Average Speed, (2) Schedule speed in a traction system.
Ans:
(Each Definition 2 Marks, Total 4 Marks)
1. Average Speed: - It is defined as distance covered between two stops divided by actual time of run is
known as average speed.
OR
Vav
3600 D
Km/hr
T
Where T = is actual time of run in sec OR
Average Speed
Dis tan ce between stops or stations
Actual time of run
2. Schedule Speed: - It is defined as distance covered between two stops divided by schedule time is
known as schedule speed. OR
Schedule Speed
Dis tan ce between stops or stations
Km/hr
( Actual time of run) ( Stop time)
OR
Schedule Speed
a) ii)
Dis tan ce between stops or stations
Schedule time
Draw figure of indirect arc furnace. State why indirect arc furnace is not built of large capacity.
(Fig. 2 Marks & why indirect arc furnace is not built of large capacity- 2 Mark,
Ans:
Total 4 Marks)
Figure of indirect arc furnace:-
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Model Answer
Subject Code: 17507
\ cf>
Furnace .
arc
Indirect
supply
from
furnace ;xrneJt^
A
r
>aJ I
electrode
chaioe
^u ^
H racto
J
T~ h e rm aU
"
(
pno-tot
ocfcin
/
^
I
arrangement
Why indirect arc furnace is not built of large capacity: Shape of heating chamber is more cylindrical to make rocking easily.
Due to this cylindrical construction of heating chamber there is limited to use only two
electrodes.
So this furnace is available only to use signal phase supply & power handling capacity of 1-ph
supply is less than 3-ph supply.
Hence indirect arc furnace is not built on large capacity
OR
Credit may be given by judgment on part of relevant answer based on candidate
understanding.
A train runs between two station is 2 km apart at average speed of 40 kmphr. Train accelerates at
2 kmphrpsec. and retards at 3 kmphrpsec. Assume trapezoidal speed time curve.
b)
Calculate: (i) Draw speed time curve and mark all. (ii) Maximum speed (iii) Distance travelled by
train before the breaks are applied.
Ans:
(When Final answer of Numerical is correct Give Full Marks & if final answer is
wrong give stepwise marks)
Given Data: Acceleration = 2 km phps , Retardation = 3 km phps , D = 2Km. , Vav = 40Kmph
Draw speed time curve and mark all :- (Speed Time Curve --- 3 Marks)
Ar
8
r
Free
i
Running
-
/
a
/
3
/
j\
Vm
\ datio
i \ n Peri
\ od
1
»
/
/
1
1
—— —
r
•
1
M
TIME IN SECONOS
B
%
£
T
\
|
—
#
=
\
l
/
Free Running
Retar
I
i
/ !1
Period
i\
7i
3C
S
a!
cn
.
Period'!
OR
—
-
' 4
1
Fig. 7.2
l2
4
|
t 3 ~*
T
Time in second
-
Trapezoidal Speed time Curve
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Page 33 of 35
To Calculate Maximum Speed :- (--- 4 Marks)
K
But,
K
2 -
------------------------------------------------------------- (1 Mark )
23
2 2 3
= 0.4167
Vav
40
3600 D
Time
-------------------------------------------------------------- (1 Mark )
3600 2
Time
Time = 180 sec.
V
V
max
max
V
T T 2 4 K 3600D
2K
-
------------------------------------------------------------- (2 Mark )
V
180 1802 4 0.4167 3600 2
2 0.4167
= 44.6061 kmph
To calculate Distance Travelled by train before breaks are applied:Distance covered during Retardation ( D ) =
2
D
D
Vmax
7200
(44.6061) 2
7200 3
D 0.0921 km
Distance Travelled by train before breaks are applied := D- D
= 2 – 0.0921
= 1.9079 Km.
( 1 Mark)
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Page 34 of 35
c) A three phase, 440 V, 50 Hz, 40 kW load has a P.F. 0.85 lagging. Calculate kVAR rating of
capacitor required to improve P.F. to 0.95 lagging.
What will valve of capacitor per phase, if (i) capacitors connected in Star? (ii) Capacitors
connected in Delta?
Ans:
(When Final answer of Numerical is correct Give Full Marks & if final answer is
wrong give stepwise marks)
Given Data
Volt : 440 V,
P= 40 Kw
cos 1 =0.85
f= 50 Hz
cos 2 =0.95
Cos 1 0.85
tan 1 = 0.6197 -----------------------------
----------------------------------------- (1/2 Mark)
cos 2 =0.95
tan 2 = 0.3286
-------------------- --------------------------------------------- (1/2 Mark)
Q1 = P tan 1
= 300 x 0.6197
= 24.788 KVAR
---------------------------------------------------- (1/2 Mark)
Q2= P tan 2
= 300 x 0.3286
= 13.144 KVAR
-
--------------------------------------------- (1/2 Mark)
i) KVAR Rating of the capacitor Bank
QC = Q1- Q2
= P tan 1 - P tan 2
------------------(1 Mark)
= 24.788 – 13.144
QC
= 11.644 KVAR
---------------------------------------------- (1 Mark)
Capacitor when connected in Star :-
C per phase
QC 103
or QC 2 FC V 2 2
V
------------------------------ (1 Mark)
km
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C per phase
Page 35 of 35
11.644 103
2 50 (440)2
C per phase 1.914 104 F -
------------------------------------------------ (1 Mark)
Capacitor when connected in delta :-
C per phase
QC 103
3 V 2
C per phase
11.644 103
3 2 50 (440) 2
C per phase 6.38 105 F -
------------------------------------------------------- (1 Mark)
------------------------------------------------ (1 Mark)
------------------------------------------------------END-------------------------------------------------------
km
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Model Answer
Page 1 of 42
Important suggestions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model
answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to
assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more importance.
(Not applicable for subject English and communication skills)
4) While assessing figures, examiner may give credit for principle components indicated in a figure.
The figures drawn by candidate and model answer may vary. The examiner may give credit for any
equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant
values may vary and there may be some difference in the candidate’s answers and model answer.
6) In case some questions credit may be given by judgment on part of examiner of relevant answer
based on candidate understands.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.
Q.1 A) Attempt any THREE :
3 x 4 = 12 Marks
a) Define electric drive. List at least four advantages of electric drive.
Ans: Electric Drive:
( 2 Marks)
It is a machine which gives mechanical power. e.g. drives employing electric motors are
known as electric drives.
Following advantages of electric drive:
( Any Four point Expected: 1/2 each point: 2 Marks)
1. It is more economical.
2. It is more clean.
3. No air pollution.
4. It occupies less space.
5. It requires less maintenance.
6. Easy to start and control.
7. It can be remote controlled.
8. It is more flexible.
9. Its operating characteristics can be modified.
10. No standby losses.
11. High efficiency.
12. No fuel storage and transportation cost.
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Model Answer
Subject Code: 17507
13. Less maintenance cost.
14. It has long life.
15. It is reliable source of drive.
b) State the causes of failure of heating element.
Ans: Following of the different causes of failure of heating element:
( Any Four causes expected: 1 Mark each)
i) Formation of hot spot:
Hot spot on heating element is the point which is at higher temperature than remaining
heating element portion. So there is possibility of breaking of heating element at hot spot.
ii) Due to oxidization:
At high temperature material gets oxidized which may cause failure of heating element.
iii) Due to corrosion:
If heating element is directly exposed to chemical fumes then there is possibility of
rusting of heating element which causes failure of heating element.
iv) Mechanical Failure:
Measure heating element alloy contain iron which is brittle. Due to frequent heating &
cooling of heating element, it may break (fail) due to small mechanical injury also.
c)
Suggest suitable electric drive for following application :(i) Paper mills (ii) Stone crusher (iii) Textile
mill and (iv) Electric traction
Ans:
( Each Suitable Any one Drive suggestion: 1 Mark each)
S.No
Application
Suitable electric drive
i)
Paper mills
Slip-Ring Induction Motor, Synchronous Motor
ii)
Stone crusher
A.C. Series Motor, Slip-Ring Induction Motor
iii)
Textile mill
Squirrel Cage Induction Motor
iv)
Electric traction
D.C. Series Motor, 1 Phase Slip-Ring Induction
Motor
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Model Answer
Subject Code: 17507
Draw the curve and estimate suitable H.P. of motor having following duty cycle :
(i) Rising load from 200 to 400 HP 4 minutes
d) (ii) Uniform load of 300 HP 2 minutes
(iii) Regenerative braking from 50 to zero H.P. — 1 minute
(iv) Idle for 1 minute
Ans:
(When Final answer of Numerical is correct Give Full Marks & if final answer is
wrong give stepwise marks)
i) Load rising from 200 to 400 HP
:- 4 min
ii) Uniform load of 300 HP
:- 2 min
iii) Regenerative braking from 50 to zero : 1 min
iv) idle for
: 1 min
111
1
/^
°°
4
S\
\
1P
H
X
ot
"
20
L
'
L
4
6
1
HP
3
H
2
1
——
_yrJ
~i
or Equivalent fig----------------(1/2 Mark)
2
2
H1 H 2 H 2 t1 H 3 t2 1
2
3
H 4 t3
T
Where,
---------------------- (1/2 Mark)
T = t1 + t2 + t3 + t4
T= 4+2+1+1
T = 8 min.-------------------------------------------------------------------- (1 Mark)
3
H
3
200
1
HP
1
HP
HP
2
1
2
2
2
H1 H 2 H 2 t 1 H 3 t 2 1 H 4 t 3
3
8
2
200 400 4002 4 3002 2 1
8
I
3
(1/2Mark)
50 2 1
------------- (1/2 Mark)
1662500
24
HP 263 HP -------------------------------------Answer------------------ (1 Marks)
Nearest standard rating of motor is to be selected.
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Model Answer
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Q.1B)
Attempt any ONE of the following :
6 Marks
Describe
any
six
factors
governing
selection
of
a
motor
for
a
particular
application.
a)
Ans: (Any Six Points From The Following Or Equivalent Points Are Expected 1 Mark To
Each Point, Total 6 Marks)
Following Factors governing / or are considered while selecting electric drive (Motor) for
particular application:
1. Nature of supply:
Whether supply available is
AC,
Pure DC
Or Rectified DC
2. Nature of Drive (Motor):
Whether motor is used to drive (run)
Individual machine
OR group of machines.
3.
Nature of load:
Whether load required light or heavy starting torque
OR load having high inertia, requirehigh starting torque for long duration.
OR Whether load torque increases with speed (T N)
OR decreases with speed (T 1/ N )
OR remains constant with speed (T = N)
OR increases with square of speed (T N2)
4.
Electric Characteristics of drive:
Starting,
Running,
Speed control
and braking characteristics
of electric drive should be studied and it should be matched with load requirements(i.e. machine).
5.
Size and rating of motor:
Whether motor is short time running
OR continuously running
OR intermittently running
OR used for variable load cycle.
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Page 5 of 42
Whether overload capacity, pull out torque is sufficient.
6.
Mechanical Considerations:
Types of enclosure,
Types of bearing,
Transmission of mechanical power,
Noise
and load equalization
7.
Cost:
Capital,
Running
Maintenance cost should be less.
b) State the factors to be considered for selection of shape and size of the car of elevator.
Ans: (Any four points are Expected 1.5 Marks to each Total 6 Marks)
The size and shape of elevator car depends on following factors:
i) No. of passenger to be carried: While selecting the size of car it is a usual practice to allow.
A Space of 2 Sq.fit/ person.
Average weight of passenger is assumed 68 kg/person.
Thus the maximum load capacity of elevator is considered 34 kg/sq.ft
There should be wide frontage and shallow depth
ii) Limitation in the building design:
Shape of elevator depends on space available in building.
iii) Type of building
iv) Application of elevator
Q.2
Attempt any FOUR :
4 x 4 = 16 Marks
a) Define load equalisation for electric motors. Explain how it is obtained for electric motors.
Ans: Define load equalization for electric Motor:
( Meaning : 2 Mark, Figure: 1 Mark & explanation: 1 Mark)
There are many types of load which are fluctuating in nature e.g. wood cutting m/c, Rolling
mill. Etc. For such type of loads, load equalization is necessary to draw the constant power from
m
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Model Answer
Subject Code: 17507
supply. Because,
When there is sudden load on motor, it will draw more current from supply at start to meet
additional power demand. Due to this heavy current there is large voltage drop in supply system.
This will affect electrical instrument, equipment, m/c, other consumer etc. which are connected
across same supply line.
Also to withstand heavy current, size of input cable increases so cost of cable increases,
Hence it is necessary to smooth out load fluctuations on motor.
The process of smoothing out load fluctuation is called load equalization.
Diagram of Load Equalization:
plu U) Vie e,|
.
L
*
-
i
Plo -h? Y
j /
ft
I sad
J
b)
Ans:
Define : (i) Continuous loading, (ii) Short time loading, (iii) Long time (intermittent) loading, (iv)
Continuous operation with short time loading.
( Each Definition 1 Mark , Total 4 Marks, Graphical Figure Not expected)
(i) Continuous loading:load in HP
Temperature rise
/
/
— *iime Ct)
or eqivalent figure
This is an output which a motor can deliver continuously without exceeding the permissible
temperature limit.
It can deliver 25% over load for two hours without rise in temperature.
m
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Model Answer
Subject Code: 17507
(ii) Short time loading:U
load m HP
n r
Icad
& cycle repeats
A
- Temperature
rise
• Time C1)
or eqivalent figure
In short time loading motor is operated for short time continuously without exceeding the
permissible temperature limit.e.g. 15min., 20min., 30min. etc than it is made OFF This OFF
load interval is sufficient to cool the motor temperature to its normal value.
(iii) Long time (intermittent) loading:JcadLin HP
-
toadL
£ cycle repeats
- Temperature rise
timeCt)
or eqivalent figure
Explanation :-In this case motor is operated continuously for long time and interval between two
load is not OFF- load but motor runs at no load for short time. So temperature of
drive continuously increases.
(iv) Continuous operation with short time loading:jLoaduo
HP
—
ToadL
& cycle pepecuts
Tempemkire
/
HL
Tise
/
Ti me CO
or eqivalent figure
Explanation :-In this case motor is operated continuously for short time and interval between two load
is not OFF- load but motor runs at no load for long time. So temperature of drive continuously
increases. So temperature rise is more than short-time loading.
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c) State the principle of induction heating. Write four applications of induction heating.
Ans: Principle of Induction heating:
( 2 Mark)
The basic principle of induction heating is that, supply is given to primary winding of furnace
transformer & heat is produced in the secondary (charge) due to electromagnetic action.
OR
Principle of Induction heating:
It is based on principle of transformer. In this type primary winding is as usual which is
wound around one limb of magnetic core but secondary winding is actually charge which is to be
melted is kept in crucible.
When AC Supply is given to primary winding current flows through primary winding
which creates alternating flux in magnetic core this flux links to the secondary winding i.e. charge
through magnetic core. Hence according to faraday’s law of electromagnetic induction emf will
be induced in secondary winding, that is in the charge. As charge forms a close circuit
(secondary) heavy current flows through charge this current is responsible to produce heat in
charge due to I2R losses. This heat is utilized to melt the charge.
Where, R = Resistance of charge & I secondary current
Following are applications of induction heating:
(Any Four point expected: 1/2 each, Total 2 Marks)
1. Melting of steel and non ferrous metals at temperatures up to 1500 °C.
2. Heating for forging to temperatures up to 1250 °C.
3. Annealing and normalizing of metals after cold forming using temperatures in the range of 750 –
950 °C.
4. Surface hardening of steel and cast iron work pieces at temperatures from 850 – 930 °C
(tempering 200-300 °C)
5. Soft and hard soldering at temperatures up to 1100 °C,
6. Moreover, special applications such as heating for sticking, sintering
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Model Answer
Subject Code: 17507
State the principle and nature of supply used for eddy current heating. State the advantages and
disadvantages of eddy current heating.
Ans: Figure of Eddy Current Heating:(1 Mark)
d)
-
etic Field
Induced Current In Part
vj
Current In Coil
J
tEtL tSt
)
H
lo
£
^
l
v
4.
,
—
•
<:
vy U i
jo
m
-t -
^o
i
Mt
or Equivalent fig.
Principle:-
(1 Mark)
Heat produced eddy current loss B2 f2
1
Depth of penetration of heat
F
The job which is to be heated is wound by coil as shown in figure.
Supply of high voltage (10KV) & high frequency (10-40 KHz) is given to coil which induces
eddy current in job according to Faraday’s law of Electromagnetic induction & these eddy currents
are responsible to produce heat in job itself due to eddy current loss.
In high frequency eddy current heating the phenomenon of skin effect plays an important role.
Skin effect at high frequency is more pronounced (effective). Due to this surface of job is
more heated as compared to its depth.
Nature of supply used for eddy current heating:
(1 Mark)
High voltage (10KV)
High frequency (10-40 KHz)
Advantages eddy current heating:- (Any one point expected)
(1/2 Mark)
1) No heat transfer loss as heat is produced in job itself. So it has high efficiency.
2) As heat is produced in job itself so time required for heating is less. For e.g. in some cases
operating time taken for heating is of only one second.
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Model Answer
Subject Code: 17507
3) By simply controlling frequency, we can control temperature accurately.
4) By simply controlling frequency, depth of penetration of heat can be controlled easily.
5) Very thin material surface can be heated easily.
6) Operation is simple & automatic.
7) For heating low attention is required.
8) Heating can be taken place in vacuum or other special atmospheric condition where other
methods are not possible.
9) It is clean and convenient method.
(1/2 Mark)
Disadvantages of eddy current heating:-
1) High initial cost because of high voltage high frequency supply equipment is required.
e) Compare Single phase 25 kV AC and 1500 V DC track electrification.
Ans:
(Any Four point expected: 1 Mark each, Total 4 Marks)
S.No
Points
Single phase 25 kV AC
track Electrification
1-ph, 25KV, AC 50 Hz
1500 V DC track
electrification
600/750V-Tromways
1500/3000V urban/suburban
DC series motor for tramways.
DC compound motor
1.5 times less than a.c series
motor
High starting torque
1.
Supply given to O/H
condition
2.
Type of drive used
1-ph, AC series motor
3.
Weight of traction motor
4.
Starting torque
5.
6.
7.
Accln and retardation
Overload capacity
Method of speed control
1.5 times more than d.c.
series motor.
Less starting torque than
d.c series motor
Less than d.c series motor
Less than d.c series motor
Simple and smooth
8.
Maintenance cost of traction
motor
Starting Efficiency
Ridding quality
Insulation cost
Cross section of conductor
Design of supporting
structure
9.
10.
11.
12.
13.
More
High
High
Limited, except chopper
method
Less
More
Less, better than d.c.
High
Less
light
Less
Smooth (Better)
Low
More
Heavy
km
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Model Answer
Subject Code: 17507
14. Distance between two
substation
15. No. of substation required
for same track distance.
16. Size (capacity) of traction
substation
17. Capital & maintenance cost
of substation
18. Cost track electrification for
same track distance
19. Applications
More
Less
Less
More
More
Less
Less
More
Less
More
Main line services
Urban and suburban area
f) State the various types of welding.
Ans:
(Total 4 Marks)
i) Resistance Welding:1) Spot welding
2) Seam welding
3) Projection Welding
4) Butt Welding- i) Simple butt welding
ii) Flash butt welding
ii) Arc welding:1) Carbon Arc Welding: a) shielded welding b) unshielded welding
2) Metal Arc Welding: a) shielded welding b) unshielded welding
Q.3 Attempt any TWO :
2 x 8 = 16 Marks
a) i) (i) State advantages and disadvantages of electric braking over mechanical braking.
(Any Four Points From The Following Or Equivalent Points Are Expected For
Advantages 1/2 Mark To Each Point, Total 2 Marks & For Disadvantages 1/2 Mark To
Each Point, Total 2 Marks, Total 4 Marks)
Following are the advantages & disadvantages of electrical braking over mechanical braking system.
Ans: Advantages: (Any Four Points From The Following Or Equivalent Points Are Expected)
1. It is most reliable braking system.
2. Breaking actuation time is small as higher value of braking retardation is obtained.
3. Electrical braking is smooth & gradual.
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4. Life of electrical braking system is more.
5. There is less wear & tear of brake shoes, break block etc. so there is less maintenance cost.
6. Higher speeds are possible even when train is going down the gradient, as breaking system is
reliable.
7. Trains having heavy loads can be stopped even when train going up the gradient.
8. Higher speeds of train is possible as braking system is reliable so pay load capacity increases.
9. In case of electric regenerative braking we can utilize 60 to 80% of kinetic energy to generate
electricity which is not possible with mechanical braking.
Disadvantages: (Any Four Points From The Following Or Equivalent Points Are Expected)
1. In addition to electrical braking there must be arrangement of mechanical braking for final
stop.
2. Special arrangement of circuit and complication makes electrical braking system costly.
3. Operation in substation becomes complicated at the time of regenerative breaking when
generated energy is surplus.
4. Initial cost is more due to other control equipments & circuitry.
a) ii) (ii) State any eight advantages of electric heating.
Ans: Advantages of Electric heating:
( Any Four Advantages expected : 1 Mark each, Total 4 Marks)
1. It can be put into service immediately.
2. No standby losses.
3. High efficiency.
4. More economical than other conventional types of heating system.
5. Easy to operate and control.
6. No air pollution.
7. System is clean, as there is no waste produced.
8. No fuel transportation cost.
9. No space is required for storage of fuel and waste.
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10. Noiseless operation.
11. Uniform heating is possible; heating at particular point is also possible.
12. Dielectric material can be heated.
13. Electrical heating equipments are generally automatic, so it requires low attention and
supervision.
14. Protection against overheating can be provided by suitable switch gear.
b)
Ans:
20 kW, 220 V resistance oven uses Nickel Chromium wire. If the temp. of charge is 727 °C and it
is to be heated to 1127 °C, find the suitable length and diameter of wire.
Assume : Emissivity = 0.9, Radiant efficiency = 0.6 & Sp. resistance = 1.03 x 10-6 SZ m.
(When Final answer of Numerical is correct Give Full Marks & if final answer is
wrong give stepwise marks)
Given Data:
T1 = 11270C = 1127+273 = 1400 0K
T2 = 727 0C = 727 +273 = 1000 0K
------------------------------------------- (1/2 Mark)
-----------------------------------------------(1/2 Mark)
Radiation efficiency = 0.6, specific resistance of Ni-Cr = 1.03x10-6 ohm m, emissivity = 0.9.
T1 4
T
) ( 2 )4 w / m 2
1000
1000
OR
T
T
H 5.72 k .e [ ( 1 ) 4 ( 2 ) 4 ] w / m 2
100
100
----------------------------- (1 Mark)
H 5.72 104 k.e [ (
H 5.72 0.6 0.9 [ (
1400 4 1000 4
) (
) ] w / m2
100
100
H 87771.3408w / m 2
-------------------------------------------------------- (1 Mark)
V2
l
d2 4 P
---------- Equation No.1----------------------------- (1 Mark)
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l
(220)2
d 2 4 20 1000 1.03 106
l
1845543.68
d2
l 1845543.68d 2
--- Equation No.2 ----------------------------- (1 Mark)
d
4 H
2
l
V2
d
4 1.03 10 6 87771.34
l2
(220) 2
d
7.471 6
l2
d 7.474 10 6 [1845543.678 d 2 ]2
d 2.5446 10 7 d 4
d
2.5446 10 7
4
d
1
2.5446 10 7
d3
d 3 39296.5 10 12
Taking Cube root of both sides
d 3.399 103 m
--d 3.399mm
----------- (1Mark)
Substitute Value of ‘d’ in Equation No.2 to calculate ‘l’ :
l 1845543.68d 2
----------------------------- (1 Mark)
3 2
l 1845543.68 [3.399 10 ]
l 21.209172 m
------------------------------------------------- (1 Mark)
Answer : Length l 21.209172 mtr
Diameter d 3 .399 mm
Model Answer
SUMMER– 2018 Examinations
Page 15 of 42
( 2 Mark )
c
4J
0
<
0
U
aL.
$1
q
0
X
Q
-. q-d
Q
u
C
$
rl
*1)
ty
u
o o o
a. > ^
4-
D
oL Xcn
^
.
ft
0)
"
+
rtnrfflinn
4
5
a
•a.*
-0
i
s
1
1»
0
>
*
UJXSJLUiU
3 \)
J
c
u
or equivalent figure
( 4 Marks )
Once arc is struck between two electrodes then low voltage is sufficient to maintain the arc.
High Voltage is required to produce arc and to maintain arc high voltage is not necessary.
This arc produces heat energy which is utilized for melting the charge.
one electrode to another electrode in the form of spark (arc).
between two electrodes gets ionized and ionized air is conducting, so current starts flowing from
When very high voltage is applied across any two electrodes separated by small air gap then air
Operation:
Explanation:- (Any one Method is expected 4 Marks)
a) By applying High Voltage:- Figure:
b) By separation of two current carrying electrodes suddenly
a) By applying High Voltage
How arc is formed :- for following method
allowed to solidify is called as arc welding or stick welding.
The processes in which two metal parts to be welded are brought to a molten state and then
Define electric arc welding:-
without application of pressure and addition of filler material.
It is the process of joining two similar or dis-similar metals by application of heat with or
c) What is electric welding ? Describe electric arc welding in brief. How arc is formed in electric arc
welding ?
Ans: Meaning of electric welding :
( 2 Marks)
Subject Code: 17507
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Model Answer
Subject Code: 17507
b) By Separation of two current carrying electrodes suddenly:Figure:
Principle
2.6
^
he
of
Seperation of
n,
supply
Arc
fcoo
Formation
Current
Oudol
1
V-hah cuuenl
1
Ac
1L
*- eledrode ©
r
;
electrode
.
I ranging electrode ®
3
Lr
-
inrj
^
^ —
Carr
O1
4
u
9
"*
o
1
or equivalent figure
r
Operation: Another way to produce arc is to short circuit two current carrying electrodes as shown in fig (a)
and suddenly withdraw them, then there will be spark between two electrodes as shown in figure
(b)
This arc then produce heat energy which is utilized for melting the charge.
In this method high voltage is not necessary to produce the arc.
Characteristics of Arc:
1. Arc is conducting.
2. Arc has negative temperature coefficient of resistance.
Q.4A)
Attempt any THREE :
3 x 4 = 12 Marks
a) Compare DC and AC welding on any four points.
Ans:
( Any Four point Expected: 1 Mark each, Total 4 Marks.)
S.No
1
Points
Supply equipment used
DC Welding
DC differential Compound
Generator, or Rectifier
AC Welding
Welding Transformer
2
3
4
Heating Effect
Uniform
Temperature Obtain
More
Possibility of Arc Blow More Possibility
Not Uniform
Less
No Possibility
5
Stability of Arc
Use of series Reactor
D.C Differential compound.
Generator has dropping
characteristics.
Voltage Required
Capital Cost
Running cost
Maintenance cost
Stand by losses
Efficiency
Application
7
8
9
10
11
12
13
High
High
High
High by 25%
Low, 65%
Carbon Arc Welding
50 to 60 volt
Non Coated Electrode is used
ot)
:
i;
3
t
d
0
r*
-m
-
rtmnnm
a
<
3
u
i
J
CL
|U
j2
0
Q-
c
T>
£
Z* K
o
If current is continuously passes then heat produced may cause burning of job.
Heat is produced due to I2R losses where ‘R’ is the contact resistance.
another time interval with the help of timer.
In this type intermittent current is used, it means current is ON for definite time and OFF for
£
Job is kept in between two electrodes under pressure. This pressure is kept constant throughout.
( 2 Mark)
or equivalent figure
( 2 Mark)
Low
Low
Low
Low
High, 85%
Resistance Welding , Metal Arc
Welding
72 to 100 volt
4J
Working:
Seam welding is nothing but series of continuous spot welding
Explanation:
w
Ans:
Page 17 of 42
Coated Electrode is compulsory
b) Describe with neat sketch operation of seam type resistance welding.
1) Seam Welding its neat labelled sketch:
Type of Electrode
6
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o
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Model Answer
Subject Code: 17507
This heat is utilized to obtain welding temperature (to become a plastic state)
When welding temperature is reached supply is cut down and external pressure is applied
simultaneously across the job to complete weld.
c) Describe the construction of high pressure mercury vapour lamp with neat sketch,
Ans: Figure mercury vapour discharge lamp :-
(2 Mark)
C
II
Atgoo
/•+ mercury
CM*
i
Inner
MM
Outer tube with
fluorescent coating
from inside
i
OR
A, B, Care electrodes
R is resistance
Spacers
evacuated
(2 Mark)
Construction:-
It consists of an inner bulb generally of silicon, to withstand high temperatures.
The bulb contains a small quantity of mercury and argon.
It is protected by outer glass, this may be cylindrical or elliptical.
The space between the two bulbs is filled with nitrogen at a pressure of half atmosphere.
The discharge tube has three electrodes, namely two main electrodes A and B and one starting
electrode.
The starting electrodes are connected through a resistance of about 10-30 k ohm to the main
electrode, located at the far end.
The electrodes are of tungsten wire helices filled with electron emissive materials, usually barium
and strontium carbonates mixed with thorium.
OR Student may write
The construction & connection diagram is as shown in figure. As per this
construction there are following components.
Choke: The choke is acting as the ballast. At the time of supply voltage variation of current
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flowing through the inner tube is maintained constant to keep uniform light intensity. Sometimes
choke can be designed for to get the higher voltages & to apply the inner tube of mercury vapour
lamp.
Starting resistance/limiting resistance: Whenever current flows through the starting resistance
there is a I2R loss which is converted into heat. If the temperature of this heat goes near about
6000C then there will be heating effect & inert gases ionization will be start.
Auxiliary electrode & Main electrode: It is made by high resistive element. The ionization is
taking place through the inert gases whenever current flows from auxiliary electrode to main
electrode.
Inner Tube: The various inert gases e.g. Argon, Nitrogen etc with mercury powder are filled in
the inner tube at low pressure or high pressure.
Outer Tube: The function of outer tube is to make the vacuum surrounding the inner tube to
avoid thermal dissipation or to maintain 6000C surrounding the inner tube.
Power factor improvement Capacitor: The function of power factor improvement capacitor is
to improve the power factor 0.5 to 0.95
d) Give the two laws of illumination.
Ans: (Inverse Square Law :- 2 Marks , Lamberts Cosine Law:- 2 Marks , Total 4 Marks.)
1) Inverse Square Law:Intensity of illumination produced by a point source varies inversely as square of the distance
from source.
1d
-d
Intensity
E
d5
I
d
2
Where,
I = intensity
and
d = Distance
m
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Model Answer
Subject Code: 17507
2) Lamberts Cosine Law:
According to this law, Illumination at any point on a surface is proportional to the
cosine of the angle between the normal at that point and the direction of luminous flux
Cosine Law: £e = E * cos(0)
0°
30° v
\
eo° \
85°
.
^
xv
100%
87%
10 D%
v
.
Fig 6.3 Lambert's cosine law
50%
.
.
Fig 6.5 Lambertian surface
Attempt any ONE of the following :
6 Marks
Describe through illustration the following types of lighting scheme : (i) Direct, (ii) Indirect, (iii)
a)
Semi-direct, (iv) Semi-indirect.
Ans: 1. Direct lighting:
(1.5 Marks)
Q. 4 B)
Direct: All the light goes downward or toward (90 -100 1)
^
Application:
The direct lighting scheme is widely used in drawing room, workshop and flood lighting etc.
2. Indirect lighting:
(1.5 Marks)
Indirect: All the light goes upward or away (90 -100%)
Application:
Which is useful for drawing offices and composing rooms. It is also used for decoration
purposes in cinema halls, hotels etc.
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Model Answer
Subject Code: 17507
(1.5 Marks)
3. Semi direct lighting:
\ /
/ I \
Semi - direct : Most light goes downward (60 - 9096)
Application:
It is mainly used for interior decoration.
(1.5 Marks)
4. Semi indirect lighting:
\l /
-
©
/ \
or equivalent figure.
Semi - indirect : Most of the light goes upward or away (60- 9096)
Application:
It is mainly used for interior decoration.
b) Describe with schematic diagram steps involved in series — parallel • control of traction motor.
Ans:
(Series steps--- 3 Marks, Parallel steps----- 3 Marks, Total 6 Marks)
Series parallel control of DC series motor
1. For traction purpose, two motors are operated in following steps.
Series steps of traction motor:
Step 1 –
Two traction motors M1 and M2 are connected in series and started with all starting resistances in
series.
Series Steps
stepl
si
+
Ml
M
S2
M2
M
Vdc
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Model Answer
Subject Code: 17507
The starting resistances are cut out one by one gradually from step 1 to step 7 and finally two
motors are in series without any resistance.
In series connection the supply voltage V is divided in two motors. (Both motors get half or (V/2)
volts). So speed is also half. (N/2)
step7
si
Ml
M2
S2
M
+
M
Vdc
Voltage across each motor is Vdc/ 2 and speed is N / 2 RPM
Parallel steps of traction motor:
Step 1 –
After completion of series last step motors are now connected in parallel again with series
resistance otherwise motor will draw very high current and may damage itself.
stepl
Parallel Steps
Ml
S2
M2
M
+
Vdc
Step 2 to 7 –
Both motors are now connected in complete parallel and starting resistances are cut out one by
one 2 To 7
In parallel connection, voltage across M1 and M2 will be full i.e. V (voltage is always same in
parallel).
Voltage across each motor = V and speed of each motor = N
So, voltage is now increased from (V/2) to V.
Hence, speed also increases from (N/2) to N and motor runs with full speed.
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step7
si
Ml
S2
M2
Vdc
Voltage across each motor is Vdc and speed is N RPM
Advantages:
1. This method has highest starting efficiency then rheostat method.
Starting efficiency of plain rheostat method = 50 %. By this method for two motor it is
66.66% & for 4 motors it is 72.72% and for 6 motors it is 75%
2. Different economical speeds are obtained:
For 2 Motor = 1 :2
For 4 Motor = 1:2:4
For 6 Motor = 1:2:3
3) For same power input torque of different magnitude is obtained.
Disadvantages:
1. If proper transition method is not used then
There is loss of torque when motors are disconnected from supply
There will be jerk when motors are reconnected in parallel
Q.5
Attempt any FOUR :
4 x 4 = 16 Marks
a) Write different systems of track electrification.
Ans: (Any Four Systems Of Track Electrification From The Following Are Expected 1 Mark
To Each Systems Of Track Electrification, Total 4 Marks)
Following are the different track electrification system
D.C. Supply system:1. Direct current track electrification:
600V, 750V DC for tramways
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1500V, 3000V DC for Train (Urban and sub-urban services)
A.C. Supply system:2. 1-Ph, 25KV,standard frequency AC supply system:
1-Ph, 25 KV, 50 Hz
3. 1-Phase, low frequency AC Supply system:
1-Ph, 15/16 KV, 16.2/3 Hz or 25 Hz
4. 3-Ph, Low frequency AC supply system;
3-Ph, 3.3/3.7 KV, 16 2/3 Hz or 25 Hz
Composite system:5. 1-Ph AC (1-ph, 25KV) – DC Supply System
6. Kando System (1-Ph AC – 3-Ph AC)
b) Write eight desirable characteristics of traction motor.
Ans: (Any Eight Points From The Following Or Equivalent Points Expected 1/2 Mark Each,
Total 4 Marks)
Desirable characteristics of ideal traction motors:1) It should be robust in construction to withstand against continuous vibrations.
2) Weight of motor per HP should be minimum in order to increase pay load capacity.
3) It must be small in overall dimensions, especially in overall diameter.
4) It must have totally enclosed type enclosure to provide protection against entry of dirt, dust, mud,
water etc. in drive.
5) When motors are running in parallel they should share almost equal load. (even when there is
unequal wear & tear of driving wheels)
6) It should have high starting torque.
7) It should possess high rate of acceleration & retardation.
8) It should be variable speed motor.
9) Its speed-torque characteristics should be such that it should produce high torque at low speed
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Model Answer
Subject Code: 17507
and low toque at high speed.
10) Motor must be capable of taking excessive overload in case of emergency.
11) It should have simple speed control methods.
12) Electrical braking system should be reliable, easy to operate and control, especially regenerative
braking is possible.
13) Motor should draw low inrush current (Starting current,and if supply is interrupted and restore
again.)
14) It should withstand for voltage fluctuation without affecting its performance.
15) It should have low initial cost.
16) It should have less maintenance cost.
17) It should have high efficiency.
18) It should have long life.
A train has schedule speed of 60 kmph between stops which are 6 km apart. Determine crest speed
c) over the run assuming : (i) Duration of stops as 60 sec. (ii) Acceleration as 2 kmphps (iii)
Retardation as 3 kmphps. The speed time curve is trapezoidal.
Ans: Given data:
Vsch 60 Km / hr
D 6 KM TStop 50sec 2km / hr sec 3km / hr.sec
Solution;
V
sch
3600 D
Schedule Time (Tsch )
------------------------------------------------------ (1 /2 Mark )
3600 D
Vsch
3600 6
Schedule Time (Tsch )
60
21600
Schedule Time (Tsch )
60
Schedule Time (Tsch )
Schedule Time (Tsch ) 360 sec
------------------------------------------------------- (1 /2 Mark )
Schedule Time (Tsch ) Actual Time of Run (T ) Stop time (Tstop )
Actual Time of Run (T ) Schedule Time (Tsch ) Stop time (Tstop )
Actual Time of Run (T ) 360 60
Actual Time of Run (T ) 300 sec --------
------------------------------------ ( 1/2 Mark)
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Model Answer
Subject Code: 17507
Maximum Speed
V
V
T T 2 4 K 3600D
2K
max
2 -
K
But,
K
------------- ------------------------------------------------ (1/2 Mark )
23
2 2 3
K = 0.4167
------------
----------------------------------------------- (1/2 Mark )
Now,
V
max
7
T T 2 4K 3600D
2K
-
----------------------------------------- (1/2 Mark )
i
max
300 (300) 2 4 0.4167 3600 6
2 0.4167
V
max
300 232.372
2 0.4167
V
max
81.1471 Km / hr
V
--
------------------------------------------------------------( 1 Mark)
d) Draw a neat labelled block diagram of AC electric locomotive. State the function of each part.
Ans: labelled diagram of AC electric locomotive:
( Diagram: 2 Marks & Function: 2 Marks)
Block Diagram of AC electrical locomotive
lph AC
25KV 50Hz
CJ
Contact
supply
Paruograph
wire
CB
Filter
Rectitiei
Trartsformer
Control
r
DC Series motor
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Function of AC electric locomotive Parts:
1) Overhead contact wire:
Supply of 1-ph, 25KV, 50Hz, AC is given to overhead conductor.
2) Current collecting device:
It collects current from overhead contact wire and passes it to tap changing transformer through
circuit breaker.
3) Circuit breaker (C.B):
It is connected in between current collecting devices and tap changing transformer.
SF6 circuit breaker is used.
To disconnect locomotive equipments whenever there is fault.
It opens automatically when train passes neutral zone (from zone No.1 to Zone No.2)
4) On load tap changing transformer:
It changes the tap without disconnecting the load on transformer. Its purpose is to vary the
voltage for speed control of traction motor.
5) Traction Transformer:
It step down input voltage 25 KV to working voltage of traction motor (1500V/3000V).
6) Rectifier:
It converts secondary voltage of transformer into DC supply.
7) Filter circuit (smoothing reactor):
It is used to obtain pure DC supply.
8) Motor control unit: It controls operation of traction motor.
9) Traction Motor:
It gives mechanical power to run the train i.e. DC series motor which is used as traction
motor.
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Model Answer
Subject Code: 17507
e) "DC series motor is used for traction purpose." Justify your answer with any six characteristics.
Ans:
( 4 Marks)
Due to following characteristics and advantages, DC series motor is suitable for traction purpose:
1) Characteristics:
We know that,
Ta
Tsh
a;
i
o
/
/
/
Armature current (la)
g
g
i
i
to
to
Armature Current ( la)
Torque (Ta)
Characteristics of DC series motor
Due To Following Reasons DC Series Motor Is Used For Traction Purpose:-
1. DC Series motor robust in construction and capable to withstand against continuous vibration.
2. DC series motor weight is 1.5 times less than 1-Ph AC series motor for same H.P.
3. DC Series motor has high starting torque.
4. DC Series motor has high rate of acceleration and retardation.
5. DC Series motor is variable speed motor. Due to these characteristics motor is protected against
overload.
6. DC Series motor speed-torque characteristics are such that as torque increases speed decreases.
7. DC series motor has develops high torque at low speeds, low torque at high speed, this is the
basic requirement of traction unit.
8. Commutating property of series motor is good so we get sparkles commutation.
9. Torque is unaffected by variation in supply voltage.
10. DC Series motor maintenance cost is less.
11. When DC series motor are running in parallel the all motors share almost equal load.
12. Torque obtained by DC series motor is smooth and uniform, so it improves riding quality.
^ ^-
.
1
HI
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Model Answer
Subject Code: 17507
e) Draw speed time curve. Show and list various time periods associated with it.
Ans: Typical speed time curve for main traction line :
( Total, 4 Mark)
. FREE
M
/ RUNNING
SPEED
CURVE
cr
RUNNING
COASTING
£
i
2
RHEOSTATIC
UJ
BRAKING
ACCELERATION
‘2
TIME
t3
ts
IN SECONDS
OR
ACCi I e ration
Free Runningj
r
Retardation
T
a
5
Vm
E
o
UJ
Ut
LL
1/1
_
t3
tj
ti
T
TIME IN SECONDS
—
-
Speed Time Curve list various time periods:There are five periods in the run of train as shown in speed time curve.
i) Constant acceleration period (o to A)
ii) Acceleration on speed –Time curve (A to B) For T2 sec.
iii) Free Running or constant period (B to C) For T3 sec.
iv) Coasting period (C to D) For T4 sec.
v) Braking period (D to E) For T5 sec.
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Q.6
Attempt any TWO :
8 x 2 = 16 Marks
A 400 V, 50 Hz, 3 Phase line delivers 200 kW at 0.7 p.f. lagging. It is desirable to improve the line
a) i) power factor to unity by using shunt capacitors. Calculate value of capacitance of each unit if they
are connected in delta.
Ans: Volt : line volts V = 400V, f= 50 Hz P= 200kW cos 1 =0.7
cos 2 =1
1 45.5729 0
Cos 1 0.7
tan 1 tan 42.5729 0
tan 1 = 1.020
----------------------------------------------------------- (1/2 Mark)
tan 2 = 0
------------------------------------------------------------- (1/2 Mark)
Q1 = P tan 1
= 200 x 1.020
= 204 KVAR
----------------------------------------------------------- (1/2 Mark)
Q2= P tan 2
= 200 x 0
= 0 KVAR
----------------------------------------------------------- (1/2 Mark)
QC = Q1- Q2
= P tan 1 - P tan 2
--------------------(1/2 Mark)
= 204-0
= 204 KVAR
---------------------------------------------------- (1/2 Mark)
Capacitor when connected in Delta:-
C per phase
QC
------------------------------------------------------- (1/2 Mark)
3 V2
C per phase
204 10 3
3 2 50 400 2
C per phase
204 10 3
3 50.265 10 6
C per phase 1.3528 10 3 F
------------------------------------------------ (1/2 Mark)
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a) ii)
Model Answer
Page 31 of 42
State four requirements of tariff.
(Any Four Requirements From The Following Or Equivalent Points Are Expected
, Total 4 Marks)
Following are the requirements of Tariff :1. It should be easy to understand to consumer.
2. Easy to calculate.
3. Tariff should be attractive i.e. It should not be too high or too low. It should be
reasonable.
4. Tariff should be economical as compare to other types of energy sources.
Ans:
5. Tariff should be different for different types of consumers.
6. Tariff must be fair, so that different types of consumers are satisfied with rate of
electrical energy charges.
7. Tariff should be framed into two parts i.e. fixed charges + running charges.
8. Tariff should be high during peak load period.
9. Tariff should be low during off load period.
10. For industrial consumer, in addition to basic tariff incentives and penalty related to
PF and LF should be considered.
b) (i) What are different tariffs used by electricity supply authority ? Describe any two in brief.
Ans: Types of Tariff:( Any Four Types expected: 1/2 each, Total 2 Marks)
1) Flat-demand Tariff
2) Simple-demand Tariff or Uniform Tariff
3) Flat-rate Tariff
4) Step-rate Tariff
5) Block-rate Tariff
6) Two-part Tariff
7) Maximum demand Tariff
8) Three-part Tariff
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9) Power factor Tariff :- a) KVA maximum demand Tariff
b) Sliding Scale Tariff or Average P.F. Tariff
c) KW and KVAR Tariff
10) TOD (Time of Day) Tariff
11) ABT:-This tariff system is called availability based tariff. As its name suggest it is a tariff
system which depends on the availability of power.
Explanation of Types of Tariff ( Any TWO Types explanation Expected: 1 Mark each,
Total 2 Marks)
1) Block Rate Tariff:
In case of block rate tariff there are blocks of units consumed and each block tariff rate/unit
(KWH) is different plus consumer has to pay fix charges e.g.
If generation is less than utilization than tariff rate/unit in each block goes on increasing and
vice versa. e.g.
2) Two Part Tariff:
In this type of tariff energy bill is split into two parts.
ENERGY BILL= FIXED CHARGE which depends on load (KW)
+RUNNING CHARGE which depends on actual energy consume (KWH)
Fixed charge which depends on load (KW) which is declared by consumer on test report.
There is no separate meter is installed to measure load.
Only one energy meter is used to measure number of units consumed.
This type of tariff system is used for residential and commercial consumers.(up to 20 KW)
This type of tariff is not used for industrial consumers.
Advantages:
1. It recovers fixed charges which depends on load (KW), so it automatically recovers capital
investment of Supply Company
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Disadvantages:
1. The consumer has to payfix charges per month whether he has to consume or not consume
the electrical energy.
Application:
1. This type of tariff system is used for residential and commercial consumers. (Up to 20 KW)
2. This type of tariff is not used for industrial consumers.
3) Maximum Demand Tariff/KVA Maximum Demand Tariff / Load factor tariff: This is basic tariff for all industrial / commercial consumers with contract demand above 80
KW/ 100KVA/107 HP
It is similar to two part tariff except that maximum demand (KVA) is actually measured by
installing maximum demand meter(in KVA)
M.D. Meter (it is an electromagnetic or electronic trivector meter) is installed in the premises
of consumer, in addition to energy meter.
Maximum Demand Tariff / Load factor Tariff =
M .D. ( KVA ) Rs ' X ' permonth { Number of units ( KWH ) Actual consumer } Rs ' Y "
Application: -This type of tariff is applicable to industrial consumer/H.T/ commercial consumers with
contract demand above 80 kw/ 100Kva/107 hp consumer.
Measurement of KVA M.D.:-
Actual Maximum Demand recorded in the month during 06am.To10pm. Is considered for
billing.
Incentives and Penalties to M.D. tarrif :-
Incentives :1) If consumer is used M.D. above 75 % to 85 % of saction contract demand than , consumer
will gate 0.75 % rebeat on the energy bill.
2) If consumer is used M.D. above 85 % to 100 % of saction contract demand than ,
consumer will gate 1 % rebeat on the energy bill.
Penalties :1) If consumer is used M.D. above 100 % of saction contract demand than , consumer has to pay
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more demand charges 150 % for use of extra M.D.
2) If consumer is used M.D. below 50 % of saction contract demand than , consumer has to pay
minimum demand charges 50 % of saction contract demand.
Advantages:1) Each industrial consumer is trying to use M.D. above 75 % to 100 % of sanction contract
demand to get discount in energy bill. So it will improve load factor of industry.
2) Industrial consumers were not utilizing their load simultaneously to avoid Penalties on exceed
of M.D. than saction contract demand. So it will improve diversity factor.
3) Industrial consumer is trying to improve power factor to reduce maximum demand charges.
Since KVA I 1/ pf
4) As each industry run at high load factor, diversity factor and power factor then overall load
factor, diversity factor and power factor of power system increases.
5) Which will automatically beneficial from the economics of power system and energy
conservation point of view.
4) Power Factor Tariff:In addition to basic tariff (Maximum Demand Tariff/KVA Maximum Demand Tariff /
Load factor tariff) the tariff in which P.F. of industrial consumer is taken into
consideration.Is known as Power Factor Tariff.
If the P.F. of consumer is less than P.F. declare by Supply Company (say below 0.9 Lag.) than
penalty will be charged in energy bill.
If The P.F. of consumer is more than P.F. declare by Supply Company (say above 0.95lag.)
than discount will be given in energy bill.
As usual consumer has to pay actual energy consumption charges
Application :This type of tariff is applicable to industrial consumer/H.T/ commercial consumers with
contract demand above 80 kw/ 100Kva/107 hp consumer.
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Model Answer
Incentives and Penalties to Power factor tarrif :Power factor incentive:- e.g.
Power Factor
Percentage of incentive
0.95
0% of energy bill
Above 0.96
1% of energy bill
Above 0.97
2% of energy bill
Above 0.98
3% of energy bill
Above 0.99
4% of energy bill
At unity P.F.
5% of energy bill
Power factor penalty:- e.g.
Power factor lagging
Percentage of penalty
For0.90Power factor lagging
0% of energy bill
For 0.89 Power factor lagging
2% of energy bill
For 0.88 Power factor lagging
3% of energy bill
For 0.87 Power factor lagging
4% of energy bill
For 0.86 Power factor lagging
5% of energy bill
For 0.85 Power factor lagging
6% of energy bill
For 0.84 Power factor lagging
7% of energy bill
For 0.83 Power factor lagging
8% of energy bill
For 0.82 Power factor lagging
9% of energy bill
For 0.81 Power factor lagging
10% of energy bill
Page 35 of 42
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There are three types of P.F. tariff ;a) KVA maximum demand Tariff: (All ready explain above)
b) Sliding Scale Tariff or Average P.F. Tariff:
If the P.F. of consumer is less than P.F. declare by Supply Company (say below
0.9 Lag.) than penalty will be charged in energy bill.
If The P.F. of consumer is more than P.F. declare by Supply Company (say above
0.95lag.) than discount will be given in energy bill.
As usual consumer has to pay actual energy consumption charges
c) KW and KVAR Tariff:
In this type both active (KW) & reactive power (KVAr) supplied are charged
separately and actual energy consumption charges
A consumer having low power factor draw more reactive power and shall have to
pay more charges and vice-versa.
So consumer is trying to improve power factor to reduce KVAr charges in energy
bill, so power factor of power system increases.
Energy Bill { Rs ' A' ( KW ) Ch arg es} { Rs ' B ' ( KVAR ) Ch arg es} { Rs ' C ' ( KWH ) Ch arg es}
5) Time of Day (TOD) Tariff or OFF-load Tariff: In addition to basic tariff (Maximum Demand Tariff / KVA Maximum Demand Tariff /
Load factor tariff also the tariff in which P.F. of industrial consumer is taken into consideration.)
Consumer has to pay energy consumption charges according to time for which energy is
consumed.
TOD energy meter is installed in the consumer premises.
This meter is specially designed to measure energy consumption w.r.t. time.
This type of tariff is such that energy consumption charges/unit are less at during OFF-load
period
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Energy consumption charges/unit are more during PEAK -load period
This type of tariff is introduced to encourage industrial consumers to run their maximum load
during OFF-load period.
e.g.
Sr.No
Block
Rate / KWH Rs
Remark
1
8.00 am to 12.00 noon
Rs. 6.00 per unit+0.80 Rs. Per unit
Peak load period
2
12.00 noon to 6.00 pm
Rs. 5.00 per unit+ 0 Rs. Per unit
Base load
3
6.00 pm to 10.00 pm
Rs. 6.00 per unit+ 1.10 Rs. Per unit
Peak load period
4
10.00 pm to 8.00 am
Rs. 5.00 per unit – 1.50 Rs. Per unit
OFF load period
Application :This type of tariff is applicable to industrial consumer/H.T/ commercial consumers with
contract demand above 80 kw/ 100Kva/107 hp consumer.
6) Three part Tariff: Fixed charges per month depend on connected load.
Semi-fixed charges depend on KVA maximum demand.
Running charges depend on actual energy consume.
b) (ii) State any four advantages of good power factor for electric supply.
Ans: Following Advantages of good power factor for electric supply:
( Any Four Advantages are expected: 1 Mark each, Total 4 Marks)
1. Cross section of conductor reduces:
Cross section of conductor I
1
P. f
As P.F. increases current reduce so; cross section of conductor and its weight reduces
hence its cost reduces
2. Design of supporting Structure:
As weight of conductor reduces design of supporting structure (tower) becomes lighter,
so its cost reduces.
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3. Cross section of terminal (contacts) reduces:
As power factor increases, current reduces. hence cross section of switchgear bus bar and
contacts etc decreases.
4. Copper losses reduces:
As power factor increases current reduces. So copper losses reduces. As a effect efficiency
increase.
5. Voltage drop reduces:
As P.F. increases, current decreases. So voltage drop decreases, So regulation gets
improved (better)
6. Handling capacity (KW) of equipment increases:
As power factor increases, handling capacity of each equipment such as Alternator,
transformer increases
7. KVA rating of equipments reduces:
As P.F. increases, current decreases. So KVA rating of all equipments for eg- alternator,
transformer etc decreases, so its capital cost reduces.
8. Cost per unit (KWH) reduces:
From all above advantages, it is seen that cost of generation, transmission & distribution
decreases, so cost/unit reduces.
Also performance i.e. efficiency & regulation gets improved at high power factor
c) (i) A Factory takes 300 kW at 110 V from a 3 phase supply and power factor of 0.7 lagging. A
synchronous motor is installed which takes an additional 150 kW. What must be the kVA rating of
this motor to raise the power factor of the system to 0.85 lagging ?
Ans:
Snr*)
3oo <u9 C °0
-
s r»« us
.
1
»
/
_
S ,L\
-
Q rn
ISDKtfCV)
Q MfU-f
®L
or equivalent diagram Given Data:
PL = 300 KW Cos 0.7 lag Sin 0.7 tan 1
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Model Answer
Power factor improved to 0.85 lag tan new 0.6197 Pm 150 KW
------------------------------------ (1/2 Mark)
Reactive Power taken by load (QL) = PL tan
= 300 x 1
= 300 KVAR (lag)
-------------------------- (1/2 Mark)
Reactive Power taken after synchronous motor is connected (Qnew) =
=(PL + Pm) tan new
-------------------------- (1/2
Mark)
= (300 +150) x 06197
= 450 x 06197
= 278.8849 KVAR (lag)
------------------
(1/2 Mark)
Reactive Power taken by synchronous motor to improve P.f =
= ( (L new )
= 300 - 278.8849
= 21.115 KVAR ( leading)
KVA Rating of Synchronous Motor Sm ( Pm m ) 2 -
---------------
( 1/2 Mark)
---------------------- ( 1/2 Mark)
4
Sm (150) 2 (21.115) 2
4
Sm 22945.845
Sm 151.4788 KVA ----------------------- ( 1/2 Mark)
Power Factor of Synchronous Motor Cosm
Pm
150
Sm 151.4788
Power Factor of Synchronous Motor Cosm 0.9902 leading --------------------- ( 1/2 Mark)
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Model Answer
Subject Code: 17507
c) (ii) Derive the equation of most economical power factor.
Ans: Derivation:
( 4 Mark)
KVAR. c Qc
P
kW
M
Q2
T
Q2
S 2 ^ r. j
_
kVARj
% V
kVAR
Sl
.
1‘
Q
u
Let,
P = Active power KW
S1, S2 = KVA Maximum demand before and after improving power factor
Q1, Q2 = Lagging reactive power before & after improving power factor
QC = Leading Reactive power drawn by Capacitor
Cos1 = Initial Power factor
Cos 2 = Improved Power factor
Rs X = Tariff charges towards M.D. (KVA) /year
Rs Y = Expenditure towards KVAr to be neutralized per year (Expenditure towards P.F.
improving apparatus)
1) Before improving Power factor:
Q1 P tan 1
Cos 1
S1
P
S1
P
Cos 1
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KVA1 (S1 ) P sec 1
2) After improving Power factor:
Q 2 P tan 2
Cos 2
S2
P
S2
P
Cos 2
KVA2 (S2 ) P sec 2
3) Saving in KVA charges:
= Rs X (S1 –S2)
= Rs X ( P sec 1 P sec 2 )
= Rs X .P ( sec 1 sec 2 )
4) Expenditure towards KVAr to be neutralized:
= Rs Y (Q1 –Q2)
= Rs Y ( P tan 1 P tan 2 )
= Rs YxP ( tan 1 tan 2 )
5) Net Saving:
= Saving in KVA charges - Expenditure towards KVAr to be neutralized.
= [Rs X .P ( sec 1 sec 2 )] - [ Rs Y ( P tan 1 P tan 2 )]
Saving will be maximum when differentiate above equation with respect to 2 and equate to
zero.
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Model Answer
ds
d
Rs X P (sec 1 sec 2 ) Rs Y P (tan 1 tan 2 )
d 2 d 2
0 X P sec 2 tan 2 0 Y P sec2 2
0 Rs X P sec 2 . tan 2 0 Rs Y P sec 2 2
Rs X P sec 2 . tan 2 Rs Y P sec 2 2
Rs X tan 2 Rs Y sec 2
Rs X
sin 2
1
Rs Y
Cos2
Cos2
Rs X sin 2 Rs Y
sin 2 Rs
6)
Y
X
sin 2 2 Cos 2 2 1
Cos 2 2 1 sin 2 2
V
Most economical power factor = Cos 2 1 (Y / x ) 2
Most economical power factor at which maximum saving will occurs
-------------------------------------------------- END---------------------------------------------------------------
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Page 1 of 32
Important suggestions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the
model answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may
try to assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more
importance. (Not applicable for subject English and communication skills)
4) While assessing figures, examiner may give credit for principle components indicated in a
figure. The figures drawn by candidate and model answer may vary. The examiner may
give credit for any equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed
constant values may vary and there may be some difference in the candidate’s answers
and model answer.
6) In case some questions credit may be given by judgment on part of examiner of relevant
answer based on candidate understands.
7) For programming language papers, credit may be given to any other program based on
equivalent concept.
Q.1 A)
Attempt any THREE :
12 Marks
a) Enlist any two functions of bearings. State and explain the types of bearings.
Ans: (Function of bearings 2 Marks, Types of bearing 1 Mark, Explain 1 Mark, Total 4 Marks)
Function of bearing:
(Any two points are expected)
(2 Marks)
1. It supports the rotating part of machine.
2. It helps to maintain moving member of a machine to a fixed physical location, relative
to stationary part.
3. It helps moving component to rotate with reduced friction.
4. It reduces noise
Types of Bearings:
1. Ball or Roller Bearing
2. Sleeve or bush Bearing (It is made from bronze)
(1 Mark)
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Explanation:-
Model Answer
Page 2 of 32
(Any two points are expected)
(1 Mark)
Advantages of ball or roller Bearing:
1. Low running and starting friction i.e. friction loss is less.
2. It produces less noise.
3. Maintenance cost is low.
4. Long Life.
5. It occupies less space.
Application of ball or roller Bearing:
1. It is used for I.M because friction loss is less, hence we can keep air gap between
stator and rotor minimum and kept constant.
2. It is used for shaft position other than horizontal to take axial thrust.
Disadvantages of sleeve bearing:
1. Large friction loss
2. It produces noise
3. Maintenance cost is more,
4. Lubricating oil is compulsory and it must be checked and replaced after six months
Application:1. Motors with sleeve bearings are always used with horizontal shaft.
b) State the advantages and applications of dielectric heating.
(Advantages 2 Marks, Applications 2 Marks, Total 4 Marks)
Ans:
Advantages of Dielectric Heating:- (Any two points are expected)
(2 Marks)
1) This is only method for heating non-metallic material. (Di-electric)
2) Bad conductor of heat material can be heat by this method (for e.g. porcelain)
3) As no flame or arc exists in the process, so material like plastic, wood cotton etc.
heated safely.
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4) As heat is produced inside material to be heated due to dielectric loss, so time
required for heating is less.
5) As there is no heat transfer loss so efficiency is high.
6) Uniform heating is possible.
7) Temperature control is easy by simply controlling voltage & frequency.
8) Clean and convenient method.
9) Operation is simple and automatic.
Applications of Dielectric Heating:- (Any two points are expected)
(2 Marks)
1) In wood industry for manufacturing of ply wood.
2) In plastic Industry for making different containers.
3) For manufacturing process of raincoats & umbrellas.
4) In medical lines for sterilization of instruments & bandages.
5) For quick drying gum used for book binding purpose.
6) In cotton industry for drying & heating cotton cloths for different processes.
7) For Rubber vulcanizing, tyre and tube manufacturing process
8) Cooking of food without removing outer shell (e.g.-boiled egg)
9) In milk industry for pasteurizing of milk.
10) In Tobacco manufacturing industry for dehydration of tobacco.
11) In food processing industry, dielectric heating is used for Baking of cakes & biscuits
in bakeries.
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Define : (i) Luminous intensity (ii) Illumination (iii) Space to height ratio (iv) Luminous
efficiency
( Each definition 1 mark , Total 4 Marks)
(1 Mark)
Ans: i) Luminous intensity:Lu min ousFlux
Luminous intensity = Illu min ation
Area
c)
OR I
w
(Where lu min ous flux , w Solid Angle)
OR
The luminous intensity in any particular direction is the luminous flux
emitted by source per unit solid angle is called the luminous intensity of the source.
ii) Illumination:(1 Mark)
When light falls on surface, it becomes visible, this phenomenon is called as
illumination OR The illumination is defined as the luminous flux falling on per unit area
of the given surface on the working plane
Lu min ousFlux
Illu min ation
lumens/ m2
Area
Unit-Lux
iii)Space-Height ratio:
(1 Mark)
It is the ratio of space between two lamps to height of lamp above working
plane.
Space height ratio
Space between two lamps
Height of the lamp above working plane
iv) Lamp η (lamp efficiency):-
(1 Mark)
It is defined as the ratio of the total luminous flux emitting from the source to
Its electrical power input in watts.
OR
total luminous flux emitting from the source
Lu min ousefficency
electrical power input
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d) Enlist any four disadvantages of low power factor.
Ans:
(Any four disadvantages 1 Mark each, Total 4 Marks)
We know that,
P /3 VL I L Cos
For same power to be transmitted at same voltage over a same distance
I
1
1
Cos
P. f
From above equation it is seen that as power factor decreases current
increases, due to increase in current, system has following disadvantages
Disadvantages of Low power Factor: 1) Cross section of conductor increases: C/s of conductor I 1/ ( pf )
As power factor reduces current increases, cross section of conductor increases. Hence its
cost increases.
2) Weight of conductor increases:As cross section of conductor increases its weight increases
3) Design of supporting structure: As weight of conductor increases design of supporting structure becomes heavier, so its cost
increases.
4) Cross section of terminals increases: As power factor reduces, current increases, Hence cross section of switch gear,
bus bar, contacts, and terminals increases. So its cost increases.
5) Copper losses increases: As power factor reduces current increases. So copper losses increases. As an effect efficiency
reduces.
Copper losses I 2
1
(P. f ) 2
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Page 6 of 32
6) Voltage drop increases: As P.F. reduces current increases. Therefore voltage drop increases, so regulation
becomes poor.
Voltage drop I
1
P. f
7) Handling Capacity of equipment reduces:
Handling capacity (KW) of each equipment such as Alternator, transformer reduces as
power factor reduces.
8) High KVA rating of equipment required:- KVA I 1/ pf ,
As power factor decreases KVA rating of all equipments increases, so that its cost
increases.
1
KVA rating I
P. f
9) Cost/unit increases: - From all above disadvantages it is seen that cost of generation,
transmission & distribution increases. Also its performance efficiency & regulation reduces,
So that cost/unit increases.
10) Energy bill (KWH) increases:So at low P.F. energy consumption increases so energy bill increases
KWH α I α 1/Cosφ
So at low P.F. KVA demand (M.D. charges) increases so energy bill increases
KVA α I α 1/Cosφ
Q.1B)
Attempt any ONE :
6 Marks
a) With the help of neat figure, describe the regenerative braking for D.C. shunt motor.
(5 Marks)
Ans: Regenerative braking: At the time of braking motors are made to work as a generator & generated electrical
energy is fed back to supply wire.
In this way instead of wasting kinetic energy during braking it is converted into
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electrical energy hence its name is regenerative braking.
Regenerative braking is only possible when train is going down the gradient exceeding
o.6 %.
Excitation current is so adjusted that generated voltage (Eg) is greater than supply
voltage (V), so that power will be fed back to supply.
It is possible to generate voltage greater then supply voltage only when D.C motor
field winding is separately excited and extra care must be taken to make it more stable.
Regenerative braking Of DC Shunt motor:
(1 Mark)
In this method, instead of being disconnected from the supply, it remains connected and
returns the braking energy to the line. Consider a shunt motor running as shown in Fig.
Suppose the load causes the speed to be increased above normal, the field current remaining
the same then the back e.m.f. becomes greater than the supply voltage ( Eb > V).
IL
+
+
L
5/1
I
1air
0
R
*
v
V
-
hr
(a) Running
(b) Braking
or equivalent figure
b) With the help of neat figure, explain ultrasonic welding. State its applications.
(2 Marks)
Ans: Figure ultrasonic welding:Pneumatic Press
—
Convertor or
Piezoelectric
Transducer
o
<D
s
Booster
QD
u
5
VI
ro
£
=>
Horn or Sonotrode
Plastic Materials
Fixture or Nest
Or Anvil
Ultrasonic Welding
or equivalent Figure
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Model Answer
Working Principle:-
Page 8 of 32
(2 Marks)
A high frequency (20 kHz to 40 kHz) ultrasonic vibration is used to join two plastic
pieces together. The high frequency vibration generates heat energy at the interface of the
two pieces and melts the material. The melted material fused with each other to form a
strong weld on cooling and solidification.
Application:-
(2 Marks)
1. It is most commonly used to weld thermoplastic materials and dissimilar materials.
2. Metal with thin section can also be welded.
Q.2
Attempt any FOUR :
16 Marks
a) What is group drive? State its four disadvantages.
Ans:
(What is group drive 2 Mark .disadvantages 1/2 Mark each, Total 4 Marks)
Group drive: -
(2 Marks)
In a group drive single large capacity electric drives is used to run number of
machines through a long common shaft is known as group drive.
Disadvantages of Group drive:-
(Any four disadvantages 1/2 Mark each, Total 2 Marks)
1. Flexibility:Flexibility is lost due to common shaft for number of machines.
2. Safety:It is less safe.
3. Reliability:Its reliability is less at the time of breakdown and maintenance of single large motor,
Because, all the machines operations are required to be shut down at the time of
breakdown and maintenance of single large motor.
4. Mechanical power transmission losses:Considerable power loss takes place for transfer of mechanical energy from shaft
to machine.
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Page 9 of 32
5. Speed control:Speed control of individual machine is difficult, it requires special
arrangement.
6. Addition / Alteration:Possibility of addition or alteration in existing system is limited.
7.Efficiency and Power Factor: –
If group drive is run at reduced load then Efficiency and Power Factor of
group drive will be less.
What are the requirements to heating elements materials? Enlist four names of heating
material.
( Requirements 2 Marks, Names of heating material 2 Marks, Total 4 Marks)
Ans:
b)
Following requirements of heating material:( Any Four requirements are expected ½ Mark each, Total 2 Marks)
1. High resistivity:
It should have high resistivity. So that is becomes compact in size and produces
more heat with small input current.
2. High melting point:
It should have high melting point to withstand at high temperature.
3. High Resistance to corrosion:
It should have high resistance to corrosion to avoid rusting.
4. Brittleness –
It should not be brittle.
5. High Oxidizing temperature:
It should have high oxidizing temperature or it should not oxidize even at high
temperature.
6. High Mechanical Strength:
It should have high mechanical strength to withstand from mechanical injury.
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Page 10 of 32
7. Ductile:
It should be ductile so that it can be manufactured into different size & shape.
8. Long Life:
It should have long life.
9. Less Costly:
It should be less costly and easily available.
10. Low temperature co-efficient of resistance:
For accurate temperature control, it should have low temperature co-efficient of
resistance.
Names of Material used for manufacturing of heating element:( Any Four Names of Material are expected ½ Mark each, Total 2 Marks)
i) Nichrome (Nickel-chromium) ii) Constantan or Eureka (Nickel-copper ) iii) Kanthal
(Iron-chromium-Aluminum) iv) Nickel-chromium-iron v) Carbon vi) Tungsten vii)
Platinum viii) Silicon carbonate ix) Iron-chromium-Aluminum
Refer the Fig. below, which is a speed time curve of a train and answer the questions :
(i) The curve is of which type of train service.
(ii) Give the names of following time periods : (I) 0 — t1 (2) t2 — t3 (3) t4 — t5
c)
t
Speed
/
' 1
^
/
t|
t;
Time
—
\
t.i
*
_
.
ti t
Ans: i) The curve is of which type of train service:- Main line services
ii) Names of following time periods :
(1) 0 — t1 :- Acceleration Period
(2) t2 — t3 :- Free running Period
(3) t4 — t5 :- Braking Period
(1 Mark)
(1 Mark)
(1 Mark)
(1 Mark)
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Page 11 of 32
d) State and explain in short any four desirable characteristics of traction motors.
(Any Four characteristics are expected 1 Mark each, Total 4 Marks)
Ans:
Traction motor should possess Following Characteristics:1) It should have high starting torque.
2) It should possess high rate of acceleration & retardation.
3) Its speed-torque characteristics should be such that it should produce high torque at
low speed and low toque at high speed.
4) It should be variable speed motor so motor is capable of taking excessive overload.
5) It should have simple speed control methods.
6) It should withstand for voltage fluctuation without affecting its performance.
7) It should have high power to weight ratio.
8) Weight of motor per HP should be minimum in order to increase pay load capacity.
9) It must be small in overall dimensions, especially in overall diameter.
10) It should have less maintenance cost.
11) It should have high efficiency.
12) It should have long life
13) It should be simple in design
14) It should be robust in construction to withstand against continuous vibrations.
15) It must have totally enclosed type enclosure to provide protection against entry of dirt,
dust, mud, water etc. in drive.
16) When motors are running in parallel they should share almost equal load. (even when
there is unequal wear & tear of driving wheels)
17) It should have high coefficient of adhesion.
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18) It should have lower center of gravity.
19) Electrical braking system should be reliable, easy to operate and control, especially
regenerative braking is possible.
20) Motor should draw low inrush current (Starting current, and if supply is interrupted
and restore again.)
21) It should have low initial cost.
e) Draw the diagram of AC electric locomotive and explain the function of each part
(Diagram 2 Marks, function of each part 2 Marks, Total 4 Marks)
Ans:
Block Diagram of AC electrical locomotive
lphAC
25KV 50Hz
y supply
c
Contact wire
Patvioejraph
C.B
Rectifier
Filter
Transformer
MoSOf Con!rol
.
JK
PS
\DC Series motor
OR Equivalent Figure
Explanation:
1) Overhead contact wire:
Supply of 1-ph, 25KV, 50Hz, AC is given to overhead conductor.
2) Current collecting device:
It collects current from overhead contact wire and passes it to tap changing
transformer through circuit breaker.
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3) Circuit breaker (C.B):
It is connected in between current collecting devices and tap changing transformer.
SF6 circuit breaker is used.
To disconnect locomotive equipment’s whenever there is fault.
It opens automatically when train passes neutral zone (from zone No.1 to Zone
No.2)
4) On load tap changing transformer /Tapping transformer:
It changes the tap without disconnecting the load on transformer. Its purpose is
to vary the voltage for speed control of traction motor.
Traction Transformer:
It step down input voltage 25 KV to working voltage of traction motor
(1500V/3000V).
5) Rectifier:
It converts secondary voltage of transformer into DC supply.
6) Filter circuit (smoothing reactor):
It is used to obtain pure DC supply.
7) Motor control unit:
It controls operation of traction motor.
8) Traction Motor:
It gives mechanical power to run the train DC series motor is used as traction motor.
Q.3
Attempt any TWO :
16 Marks
A certain motor has to perform the following duty cycle :
(i) 100 kW for 10 min (ii) 50 kW foe 8 min (iii) No load fear 10 min (iv) 150 kW for 5 min
a)
The above duty cycle is repeated continuously. Assuming heating is proportional to square
of the current, calculate the suitable size of a motor fitting the above requirement in HP.
KW t ( KW
2
Ans:
Continous rating of motor
Where,
1
1
T = t1 + t2 + t3 + t4
2
2
2
) t 2 ( KW3 ) t3 ( KW4 ) x t4
------ (1 Mark)
T
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T = 10 + 8 + 10 + 5
T = 33 min.---------------------------------------------------------- (2 Mark)
1002 10 (50) 2 8 ( 0) 2 10)
Continous rating of motor
(150) 2 x 5
33
100000 20000 112500
33
Continous rating of motor
kW 83.93 kW
(1 Mark)
--------------------------Answer---------------- (2 Marks)
W 83930 watt
Continous rating of motor in HP =
83930
735.5
Continous rating of motor in HP = 114.11 HP
-----------------Answer------------- (2 Marks)
Nearest standard rating of motor is to be selected.
b)
Explain with neat sketch the construction, working principle and two applications of
"Ajax Wyatt" vertical core furnace.
(1 Mark)
Ans: Neat Sketch of “Ajax Wyatt “ Vertical core furnace :
(Neat sketch 1 Mark, construction 2 Mark, working principle 4 Mark , Applications 1
Mark, Total 8 Marks)
type
Core
cal
Verf‘
j
I
i p
|
Induction
opening
1
N
ouiier
fbfe rnol +on
heauF
y.
y
rr»eVaJ
Q
mcxg neJHc
Furnace
^^
\
nsudoJ-iVi
. modericil
^
R e F r a cfor7
-oDal \
of fu mace,
.
^
pri mai
3
u) i nd i
cenVraJ core.
Cor«-
SCfCXp
V- n o t c h
or equivalent Figure
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Principle of Induction heating:
(4 Mark)
It is based on principle of transformer. In this type of Induction heating primary
winding is as usual which is wound around one limb of magnetic core but secondary
winding is actually charge which is to be melted is kept in crucible.
When AC Supply is given to primary winding current flows through primary
winding which creates alternating flux in magnetic core this flux links to the secondary
winding i.e. charge through magnetic core. Hence according to faraday’s law of
electromagnetic induction emf will be induced in secondary winding that is in the
charge.
As charge forms a close circuit (secondary) heavy current flows through charge this
current is responsible to produce heat in charge due to I2R losses. This heat is utilized to
melt the charge.
Where, R = Resistance of charge & I secondary current.
Construction of ‘Ajax Wyatt’ vertical core type furnace:
(2 Mark)
‘Ajax Wyatt’ type induction heating furnace is nothing but transformer. It consists of:a. Magnetic Core
b. Primary winding
c. Secondary Winding
d. Refractory Wall
e. Opening (There are two opening to the furnace.)
f. Cooling arrangement
g. Tilting arrangement
h. Control panel
i. APFC panel
Applications:- (Any two applications 1/2 mark each )
1.
For melting copper, copper alloys such as brass, bronze and zinc
2.
It is used for melting metal having low resistivity
3.
It is used for heat treatment of silver, Copper, nickel etc.
(1 Mark)
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c) With proper justification suggest suitable enclosures to be used for electric drives in
following locations of industries.
(i) Chemical plant (non-explosive) (ii) General industrial installation (non-explosive)
(iii) Mines or others hazardous locations (iv) General outdoor installation.
Ans: justification suggest suitable enclosures to be used for electric drives in following
locations of industries:
(i) Chemical plant (non-explosive):-
( 2 Marks)
Totally enclosed or Drip proof or moisture proof or equivalent
(ii) General industrial installation (non-explosive):-
( 2 Marks)
Totally enclosed type or pipe ventilated totally enclosed type or equivalent
(iii) Mines or others hazardous locations:-
( 2 Marks)
Fire (flame) proof type enclosure or equivalent or totally enclosed type.
(iv) General outdoor installation:-
( 2 Marks)
Totally enclosed type or pipe ventilated totally enclosed type or equivalent
Q.4A)
Attempt any THREE :
12 Marks
a) State any four requirements of good welding.
( Any four requirements are expected 1 Mark each, Total 4 Marks)
Ans:
The good welding has following requirements:1) Welding joints must be strong and reliable
2) Joint (welding) is made by proper welding technique.
3) Welds should have a reasonably smooth, uniform & consistent appearance.
4) Welding Should be free from any type of welding defects
5) To avoid oxidation welds should have an adequate shielding from the
atmosphere (Oxygen)
6) Welding should be done by only skilled welder
7) Correct welding technic should be used
8) The joint preparation of work pieces to weld should be done properly.
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9) The electrode of correct type and size should be used
10) According to the nature of job welding current should be selected.
11) In case of pressure welding, pressure should be correctly control in time.
12) In case of arc welding proper arc length should be maintained.
b) State and explain the law of inverse squares in illumination.
Inverse Square Law:(4 Marks)
Intensity of illumination produced by a point source varies inversely as square of the
distance from source.
Distance from Source
2D
Area
D
yA
A
Illuminance
(lux)
L
L/ 4
L/ 9
or equivalent figure
OR
Ans:
1d
:d
Intensity
1
y
or equivalent figure
E
I
d
1
E 2
d
2
Where,
I = intensity
and
d = Distance
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c) Compare two part tariff and three part tariff.
Ans: (Any four points from following or equivalent are expected 1 mark each , Total 4 Marks)
Sr.No
1
Two Part Tariff
Energy bill is divided into Two
Three part Tariff
Energy bill is divided into three parts
parts
2
3
Energy bill = Fixed charge +
Energy bill = Fixed charge + Semi fixed
Running charge
charge + Running charge
Fixed charge is directly
Fixed charge is directly proportional to
proportional to maximum demand
connected load.
in KW
4
No semi fixed charge
Semi fixed Charge directly proportional
to maximum demand in KVA
5
It is used for LT
It is used for HT consumer/industrial
consumer/residential
d) What are advantages of power factor improvement? (any four)
Ans: (Any four advantages from following or equivalent are expected 1 mark each , Total 4
Marks
We know that,
P 3 VL I L Cos
For same power to be transmitted at same voltage over a same distance
I
1
1
Cos
P. f
From above equation it is seen that as power factor increases current decreases, due
to decreases in current, system has following advantages
i
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1. Cross section of conductor reduces:
Cross section of conductor I
1
P. f
As P.F. increases current reduce so; cross section of conductor and its weight
reduces hence its cost reduces
2. Weight of conductor decreases:Weight of conductor I
1
P. f
As cross section of conductor reduces its weight reduces.
3. Design of supporting Structure:
As weight of conductor reduces design of supporting structure (tower)
becomes lighter, so its cost reduces.
4. Cross section of terminal (contacts) reduces:
As power factor increases, current reduces. Hence cross section of switchgear
bus bar and contacts etc decreases.
5. Copper losses reduces:
As power factor increases current reduces. So copper losses reduces. As an effect efficiency
increase.
Copper losses I 2
1
(P. f ) 2
6. Voltage drop reduces:
As P.F. increases, current decreases. So voltage drop decreases, So regulation
gets improved (better)
Voltage drop I
1
P. f
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7. Handling capacity (KW) of equipment increases:
As power factor increases, handling capacity of each equipment such as
Alternator, transformer increases.
8. KVA rating of equipment’s reduces:
As P.F. increases, current decreases. So KVA rating of all equipment’s for e.g.alternator, transformer etc. decreases, so its capital cost reduces.
KVA rating I
1
P. f
9. Cost per unit (KWH) reduces:
From all above advantages, it is seen that cost of generation, transmission &
distribution decreases, so cost/unit reduces.
Also performance i.e. efficiency & regulation gets improved at high power factor.
10. Energy bill (KWH consume) reduces:So at high P.F. energy consumption reduces so energy bill reduces.
KWH α I α 1/Cosφ
So at low P.F. KVA demand (M.D. charges) reduces so energy bill reduces
KVA α I α 1/Cosφ
Q. 4 B)
Attempt any ONE :
6 Marks
What is resistance welding ? Write its principle of operation. Applications (any two) and
a)
write its classification also.
Ans: (What is resistance welding 1 Mark, principle of operation 1 Mark, Applications (any two)
1 Mark classification 3 Marks, Total 6 Marks)
What is resistance welding:-
(1 Mark)
In this process welding is obtain by heating the metallic part to a plastic state &
then joining them together by applying external pressure is known as resistance
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welding. This type of welding is done without adding external filler material.
Principle of operation resistance welding:-
(1 Mark)
In resistance welding, sufficiently heavy current at low voltage is passed directly
through two metals in contact to be welded.
Heat is produced due to I2R losses where ‘R’ is the contact resistance. This heat is
utilized to obtain welding temperature (to become a plastic state)
When welding temperature is reached supply is cut down and external pressure is
applied simultaneously across the job to complete weld.
According to joules law,
Heat produced H =I2 R t …………….. Watt-sec
From this equation it is clear that heat produced depends on
Square of current (I2)
Contact resistance (R)
Duration of current (t)
Hence to obtain more heat in less time high current is necessary.
Applications:- (Any two applications are expected from following)
(1 Mark)
a) Applications of spot welding:1) Joining of automobile body section.
2) Joining sheet metal structure.
3) It is used for automatic welding process.
4) For spot welding to GI sheets, MS sheet, tinned, lead-coated sheets.
5) For spot welding to non-ferrous material such as brass, bronze, nickel, Cu,Al,etc.
6) In fabrication workshop for different applications.
OR
b) Applications of Seam welding:It gives leak-proof joints.
1. Hence used for welding of various types of containers,
2. Pressure tank,
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3. Tank of transformer,
4. Gas line,
5. Air craft tank,
6. Condenser,
7. Evaporator and
8. Refrigerator etc.
OR
c) Applications of Projection welding:1) For cross-wire welding.
2) Used for attachments of nut-bolts, stud, ring etc.
OR
d) Application Butt Welding:
1) For welding rod, wire, pipe etc
2) Butt welding is a resistance welding process for joining thick metal plates or bars
at end
OR
e) Application Flash Butt welding:
1) For welding rod.
2) For weld shaft
3) Rail, ends
4) For welding chains
Classification resistance welding:Resistance Welding:1) Spot welding
2) Seam welding
3) Projection Welding
4) Butt Welding
5) Flash Butt welding
(3 Marks)
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b) An industrial unit has maximum demand of 250 kW with a load factor of 0.6. The
following tariffs are offered: (i) Two part tariff? 70/kW of MD/year + 4 paisa/kWh
(ii) A flat rate tariff of 10 paisa/kWh. Which tariff is cheaper?
No. of Units consume in One Year
Ans:
Load Factor M .D( KW ) 8760
-------------------------------- ( 1/2 Mark)
0.6 250 8760
1314000 Kwh
--------------------------------------------------------(1 Mark)
Case-I: Energy Bill := Tariff given Rs. 70 of M .D. / year Rs. 4 paise / Kwh
= ( 250 70 ) ( 1314000 4 / 100)
------ (1/2 Mark)
= Rs. 17500 Rs.52560
= 70060 Rs.
------------------------------------------------------------ (1 Mark)
Case-II: Energy Bill := Tariff given flat rate of 10 Paise / Kwh
= 1314000 10 / 100)
= 131400 Rs.
--------------------------- (1 Mark)
-------------------------------------------------------------- (1 Mark)
Remark:-
---------------------------------- (1 Mark)
Two part tariff will be cheaper i.e. 70/kW of MD/year + 4 paisa/kWh because power
consumption is less
OR
According to energy bill Case-I is economical
For industrial consumer Case-I is economical
(5&
:
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Q.5
Attempt any FOUR :
(16 Marks)
With the help of neat sketch, explain construction and working principle of sodium
a)
vapour lamp.
Ans: (Neat sketch 1 Mark , construction 1.5 Marks, working principle 1.5 Marks , Total 4 Marks)
Sodium Vapour Lamp diagram:
(1 Mark)
Ballast
_rv
^rY\
~
L
Capacitor
Lamp
Ignitor
N
OR
*?* Au+o TransPo* mer
3£
*
RH41
do> ub\ S 'U>.cv,V\ ftc',
-H
-> ecVrocie.
—r JMj—
-
-
>
/
J
^
' -V \ iVsg
o< \ nnerA ube
* \J
>
OR Equiavelnt figure
Electrode
Outer Tube
Alumina arc tube
V
Arc
Sodium-mercury amalg
<S>
—
A.C. voltage
(Mixture)
n
ignitor
V
Ballast
U
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Construction:-
Page 25 of 32
(1.5 Mark)
HPS lamps consist of an arc tube (inner) enclosed by an outer tube.
Vacuum is created between the inner & outer glass tube to prevent heat loss.
The arc tube is made from a special glass that can withstand to high temperatures
Arc tube is U Shape
The arc tube contains xenon / neon gas (starting gas), sodium and mercury and two
electrodes.
It require a ballast to give high voltage at staring to produce the arc
There is an igniter which sends a pulse to start the discharge.
To improve the power factor a capacitor is connected across the supply. (P.F. is low @
0.3 lag.)
HPS lamps do not have starting electrodes.
Working Principle:
(1.5 Mark)
When the lamp is turned on, a high voltage at staring is applied across two electrodes,
to initiate an arc which discharges and vaporizes xenon /neon gas (starting gas),
sodium and mercury.
The energized metal atoms emit light.
After 2 to 5 minutes lamp will glow 100 %.
For running the lamp low voltage of about 165 v is sufficient.
The color of light produce is yellowish.
b) State any four safety features to be observed in welding work.
Ans: Following are safety features observed in welding work:( Any Four from following or equivalent Safety Features expected: 1 Mark each ; Total 4
Marks)
1. Select clothing to welder to provide maximum protection from spark & hot metals
2. Flame proof skull cap.
3. Hand held helmet with filter lens.
4. Safety shoes (Leather)
5. Fire resistant hand gloves.
6. Clean fire resistance clothing : i) Shirt full sleeves ii) No pockets iii) Collar buttoned
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iv) Long pants with no cuff v) Leather Apron
7. Dust Musk
8. Hearing Protection
9. PPE ( Personal Protective equipment)
10. First aid kit
11. Class ‘C’ fire extinguishers
12. Other equipment’s like small hammer, file, small metal wire brush etc. are essential
during welding.
13. Earthing clamp and proper earthing arrangement
c) How speed control of traction motor is done using shunt transition method.
Ans: speed control of traction motor is done using shunt transition method:
(Each step 1 Mark, Total 4 Marks)
In Shunt transition, speed control, is carried out by following step
Transition Steps
Stepl: Re insert starting resistances in motor circuit No.l.
SI
M1
S2
M2
M
+
Vdc
Step 2: Short circuit motor No.2 as shown in figure
Ml
S2
M2
S2
M2
Vdc
M
Step 3 - Open one end of short circuit motor & reinsert starting resistance in motor circuit No.2.
SI
Ml
M
+
Vdc
Step 4 -Connect open end of motor to supply terminal (positive). This is nothing but parallel first step.
+
Vdc
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d) Define schedule speed and state any four factors affecting schedule speed.
(Define 2 Marks, factors affecting schedule speed 2 Marks, Total 4 Marks)
Ans:
(2 Marks)
Schedule Speed: -
It is defined as distance covered between two stops divided by schedule time is known
as schedule speed. OR
Schedule Speed
Dis tan ce between stops or stations
( Actual time of run) ( Stop time)
Schedule Speed
Dis tan ce between stops or stations
Schedule time
Km/hr
The following factors affect the schedule speed:(Four factors affecting schedule speed are expected 1/2 Mark each , Total 2 Marks)
1. By increasing acceleration
2. By increasing retardation
3. By increasing both acceleration and retardation
4. By increasing maximum speed
5. By reducing stop time
6. By reducing coasting period
Why DC series motor is preferred for traction applications? Justify your answer with
characteristics.
Ans: (Characteristics 1 Mark, For justification points 3 Marks, Total 4 Marks)
Due to following characteristics and advantages, DC series motor is suitable for traction
duty:
1)
Characteristics:
(1 Mark)
e)
Ta
J sh
/
/
_
/
1
/
r
/
/
£
I
8
8
on
/
Armature current ( la )
Armature Current (la )
Characteristics of DC series motor
Torque (Ta )
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Advantages/ Justification points:(Any six points are expected ½ Mark each, Total 3 Marks)
1. DC Series motor has high starting torque.
2. DC Series motor has high rate of acceleration and retardation.
3. DC series motor has speed- Torque characteristics is such that it produces high torque
at low speeds, low torque at high speed.
4. DC Series motor is variable speed motor. Due to these characteristics motor is
protected against overload i.e. self –reliving property against over load
5. It has high power to weight ratio.
6. DC series motor consumes less power than AC motors for same HP
7. DC Series motor maintenance cost is less.
8. DC series motor weight is 1.5 times less than 1-Ph AC series motor for same H.P.
9. Torque is unaffected by variation in supply voltage.
10. Torque obtained by DC series motor is smooth and uniform, so it improves riding
quality.
11. DC Series motor robust in construction and capable to withstand against continuous
vibration.
12. When DC series motor are running in parallel the all motors share almost equal load.
Q.6
Attempt any TWO :
(16 Marks)
a) Enlist any four electrical equipment’s and their functions, which are used in arc furnaces.
(Any four electrical equipment’s are expected 2 Mark each, Total 8 Marks)
Ans:
Electrical equipment’s required for arc furnace:
1. Furnace Transformer:
It is ON- load tap changing transformer. Its secondary winding is designed for low
voltage and high current. Secondary winding has number of taps of different voltage
2. Series Reactor:
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Reactor is used, for two purposes:a. To stabilize the arc.
b. To limit current short circuit.
3. Circuit Breaker (CB):
To protect the furnace transformer against various types of fault.
4. Automatic current regulator (ACR):
It adjusts current automatically flowing through electrode for control of
temperature (arc).
5. Electrode:
Carbon, Graphite electrode are used.
6. Connecting rod:
It is used for making connection from secondary of transformer to electrode.
It carries very high current. So it consists of heavy copper rod or strip.
7. APFC: To improve the P.F. of furnace automatically to the most economical P.F.
A trapezoidal time curve of a train consists of (i) uniform acceleration of 6 kmphps for 25
sec. (ii) Free running for 10 min (iii) Uniform deceleration of 6 kmphps for stopping the
b)
train. (iv) Stop time of 5 min. Find the distance between the stations, average speed and
schedule speed.
Ans: Given Data:
TV^p C
cLcJv
\
C U.'*' WL
1
V
<
l-
t3
L
v
t1= 25 sec
t2= 10 min = 600 sec
acceleration = 6 km phps
\
vn t .
Tstop= 5 min = 300 sec
retardation = 6 km phps
Vmax
--------------------------------------------------------- (1/2 Mark)
t1
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Vmax t1 25 6
Vmax 150 Km / hr
Vmax
t3
t3
Vmax
---------------------Answer-------------- ( 1/2 Mark)
----------------------------------------------------- (1/2 Mark)
150
6
t3 25 sec ----------------------------Answer------------- ( 1 /2 Mark)
Distance covered during Acceleration ( D ) =
D
D
2
Vmax
7200
--------------------------------------------- (1/2 Mark)
(150) 2
7200 6
D 0.52083 km
----------------------Answer------------- ( 1/2 Mark)
Distance covered during Retardation ( D ) =
2
V
D max
7200
----------------------------------------------- (1/2 Mark)
(150) 2
D
7200 6
D 0.52083 km----------------------------Answer------- ( 1/2 Mark)
D Free running
t 2 Vmax
3600
D Free running
--------------------------------------------- (1/2 Mark)
600 150
3600
D Free running 25 Km
----------------------------Answer-------- ( 1/2 Mark)
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Dis tan ce ' D ' D D D Free running
Dis tan ce ' D' 0.52083 0.52083 25
Dis tan ce ' D' 26.04168 Km ---------------------------Answer------- ( 1/2 Mark)
OR
Student may calculate distance by using following formula also consider
V
max
V
T T 2 4 K 3600 D
2K
K
2 ---
Time ' T ' t1 t 2 t 3 25 600 25
Time ' T ' 650 Sec --------------------------------------Answer-------- ( 1 /2 Mark)
Vav
3600 D
-------------------------------------------------------------------- (1/2 Mark)
Time
Vav
3600 26.04168
650
Vav 144.2308 Km / hr ------------------------------------------------Answer--- ( 1 /2 Mark)
Vschv
3600 D
----------------------------------------------------------------- (1/2 Mark)
T Tstop
Vschv
3600 26.04168
650 300
Vschv 98.6842 Km / hr ---------------------------------------------Answer------ ( 1 /2 Mark)
c) An industrial unit consumes 250 kW at 110 V from a 3 phate supply and pf of 0.80 lagging.
A synchronous motor is installed which takes an additional 120 kW. What must be the
kVA rating of this motor to raise the pf of the system to 0.9 lagging ?
Given Data:
Ans:
PL = 250 KW, Cosφ = 0.8 tanφ1= 0.75
Power factorto be improved to 0.9 lag tanφ2 =0.4843
m
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CO 64 - 0 9 logging
P = 2 ?- O |<\A/
Co
0
S
lioVvj
4 >
- 80 lagging
S
s,
9
'
------------------------------ (1 Mark)
Reactive Power taken by load (Q1) = P1 tan 1
----------------------------- (1 Mark)
= 250 x 0.75
= 187.5 KVAR
---------------------- (1Mark)
Reactive Power taken after synchronous motor is connected (Q2) =
=(PL + Pm) tan 2
-------------------- (1 Mark)
= (250 +120) x 0.4843
= 179.191 KVAR
------------------
(1 Mark)
Reactive Power taken by synchronous motor to improve P.f =
= (Q1 – Q2 )
= 187.5 - 179.191
= 8.309 KVAR ( leading)
4
KVA Rating of Synchronous Motor Sm ( Pm m ) 2 -
------------ ( 1 Mark)
---------------------- ( 1 Mark)
4
Vschv (120) 2 (8.309) 2
Sm = 120.2873 KVA
---- ( 1 Mark)
-------------------------------------------------- END---------------------------------------------------------------
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Important suggestions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the
model answer scheme.
2) The model answer and the answer written by candidate may vary but the examiner may
try to assess the understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more
importance. (Not applicable for subject English and communication skills)
4) While assessing figures, examiner may give credit for principle components indicated in a
figure. The figures drawn by candidate and model answer may vary. The examiner may
give credit for any equivalent figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed
constant values may vary and there may be some difference in the candidate’s answers
and model answer.
6) In case some questions credit may be given by judgment on part of examiner of relevant
answer based on candidate understands.
7) For programming language papers, credit may be given to any other program based on
equivalent concept.
Q.1
Attempt any FIVE :
20 Marks
a) Define Electric drive. State the two advantages associated with electric drives.
Ans: Electric Drive:
( 2 Marks)
It is a machine which gives mechanical power. e.g. drives employing electric
motors are known as electric drives.
Following advantages of electric drive: ( Any Four point Expected: 1/2 each point: 2 Marks)
1. It is more economical.
2. It is more clean.
3. No air pollution.
4. It occupies less space.
5. It requires less maintenance.
6. Easy to start and control.
7. It can be remote controlled.
8. It is more flexible.
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9. Its operating characteristics can be modified.
10. No standby losses.
11. High efficiency.
12. No fuel storage and transportation cost.
13. Less maintenance cost.
14. It has long life.
15. It is reliable source of drive.
b) State its eight applications of dielectric heating.
Ans: Applications of Dielectric Heating:(Any eight points are expected: 1/2 Mark each point, Total 4 Marks)
1) In wood industry for manufacturing of ply wood.
2) In plastic Industry for making different containers.
3) For manufacturing process of raincoats & umbrellas.
4) In medical lines for sterilization of instruments & bandages.
5) For quick drying gum used for book binding purpose.
6) In cotton industry for drying & heating cotton cloths for different processes.
7) For Rubber vulcanizing, tyre and tube manufacturing process
8) Cooking of food without removing outer shell (e.g.-boiled egg)
9) In milk industry for pasteurizing of milk.
10) In Tobacco manufacturing industry for dehydration of tobacco.
11) In food processing industry, dielectric heating is used for Baking of cakes & biscuits
in bakeries.
c) Compare resistance welding and arc welding on any four points.
Ans: (Any Four Points From The Following Or Equivalent Points Are Expected 1
Mark To Each Point, Total 4 Marks)
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Sr.No
1
2
3
4
Parameters
Type of welding
Model Answer
Page 3 of 41
Resistance Welding
Arc Welding
Plastic/Pressure/Non-
Fusion/Non pressure
fusion welding
welding
Principle of heat
Heat is developed due to
Heat developed due to arc
developed
I2R losses where R is the
produced in between
contact resistance
electrode and job
External filler
Not required during
Required during welding
material required
welding
External pressure
Required
Not required
Both AC,DC supply is
Metal arc welding – Both
used. But generally Ac
AC,DC supply is used. But
Supply is used.
generally Ac Supply is
required
5
Type of supply used
used.and for
Carbon arc welding –only DC
supply are used
6
Voltage &current
required
7
8
Energy
Low voltage (2 to 20V
AC) and high current (40
to 400A, in some cases 5
to 20KA ) supply is
required
Low (3 to 4 KWH/Kg of
Metal Arc welding Voltage70 to 100V AC and
Carbon arc welding voltage50 to 60V DC,
Current- 50-600-800A
High (5 to 10 KWH/Kg of
consumption
deposited material )
deposited material.)
Temperature
Temperature obtained is
Temperature obtained is very
obtained
not very high (up to
high (up to 35000C to 60000C)
13500C)
9
Power factor
Low
Poor
10
Type of electrode
Non-consumable
Coated electrodes are used
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electrodes are used.
Page 4 of 41
for metal arc welding and
bare electrodes are used for
carbon arc
welding.(Electrodes may be
consumable or nonconsumable)
11
Application
It is suitable for mass
production
It is suitable for heavy job,
maintenance and repair work
d) State any four requirements of an ideal traction system.
Ans: (Any Four Points From The Following Or Equivalent Points Are Expected 1
Mark To Each Point, Total 4 Marks)
Ideal Traction system should processes following requirement:1. It should be Pollution free.
2. It should have low capital, Running and maintenance cost.
3. It should have quick starting time.
4. It should have high starting torque.
5. It should have high rate of acceleration & retardation.
6. Highest speeds are possible.
7. It should have easy speed control method.
8. Its braking system should be reliable and causes less wear.
9. It should have better riding quality (less vibration)
10. It should be free from unbalance forces i.e. coefficient of adhesion should be
more.
11. It should have lower center of gravity.
-
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12. The locomotive should be self-contained and able to run on any route
13. There should be no standby losses.
14. It should have high efficiency
15. Regenerative braking should be possible.
16. The wear caused on the track should be minimum.
17. Equipment should be capable of overloads for short periods.
18. Capability of withstanding voltage fluctuations.
19. Parallel running usually more than one motor (2 or 4 motors) should be
possible.
20. Traction system should be clean & long life.
21. There should be no interference to the communication lines running along the
lines.
e) Draw simplified speed time curve. Show and list various time periods associated with it.
Ans: Typical speed time curve for main traction line :
( Total, 4 Mark)
f
6
X
. FREE
SPEED
CURVE
r
g RUNNJNGXJ
j?
/
:r
o
?
y
Q
h
/ ^JF'FNING _
/
FV
*
»
t
|•
RHEOSTATIC
/»
/ 1 / ACCELERATION
Ar
!
*1
t2
COASTING
1
N.
;
I
i
J
J
t3
TIME IN SECONOS
OR
\
t
*
^
\ SBRAKING
/
\
t5
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ACCileration
Page 6 of 41
Free Running
|
F
8
Retardation
T
EL
Vm
2
C
LU
LU
LL
t3
t2
— —
T
TIME IN SECONDS
—
Speed Time Curve list various time periods:There are five periods in the run of train as shown in speed time curve.
i) Constant acceleration period (o to A)
ii) Acceleration on speed –Time curve (A to B) For T2 sec.
iii) Free Running or constant period (B to C) For T3 sec.
iv) Coasting period (C to D) For T4 sec.
v) Braking period (D to E) For T5 sec.
f) Describe the static capacitor method of power factor improvement.
Ans: The static capacitor method of power factor improvement.
Before connecting capacitor
Phasor diagram
vT
V
CO
i\\
+V
( 4 Marks)
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After connecting capacitor
’K
v
phasor diagram
Jc
r
c:
'
=P i l
X
L
4
Page 7 of 41
>Iw
- v
*
12
i
J
III I
/
II
oo
00)
Cos1 = Initial Power factor
Cos2 = Improved Power factor
Calculation from current vector diagram:
I C I1 I 2
IC [Iw tan 1 ] [Iw tan 2]
Now, I C
V
XC
1
2 f c
V
IC
XC
C
1
2 f X C
XC
Magnitude of new current:
I2 (Iw )2 (I2 )2
V
Calculation from power triangle:
Where ,
P = Active power KW
Q1, Q2 = Lagging reactive power before & after improving power factor
QC = Leading Reactive power drawn by Capacitor
S1, S2 = KVA Maximum demand before and after improving power factor
Cos1 = Initial Power factor
Cos2 = Improved Power factor
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Page 8 of 41
KVAK c Qc
P kW
n
r
sN
A
t
kVAR
.
L
?
1 Q2
c
kVAR,
B
JLQ
QC = Q1 – Q2
QC [ P tan 1 ] [ P tan 2 ] KVAr rating of capacitor
Observation:
From above vector diagram & power triangle calculations, if capacitor is connected
across load than following observations are observed.
S.No
.
1
2
3
4
5
Parameter
Effect
Power factor
Magnetizing current ( I )
Improves
Reduces
Total current
Reduces
Lagging reactive power (KVAr) Reduces
Apparent power (KVA)
Reduces
Connection diagram to connect capacitor to improve power factor (Delta connection)
>
3-< Load
Mk
c
or equivalent figure
(Cph)
KVAR
Farad
3 V 2
2 f
-?
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Model Answer
Page 9 of 41
3-ph Star connected Capacitor Bank:
=0
XV
'
or equivalent figure
KVAR
(Cph )
Farad
V2
2 f
OR Student may write this way
(1 Marks for any one figure, Vector diagram 1 Mark, Formula 1 Mark & advantages &
disadvantages (Any one) – 1 Mark: Total 4 Marks)
I
1f1
t•
E
5
t
'
,
L
Jj
J
°
6
—
Xi
*\
0
l
—
.
_
-
4
4. 1
t
^
f
f
1
)
?
IT
V
'WI
Observation:
From above vector diagram & power triangle calculations, if capacitor is connected
across load then following observations are observed.
Magnetizing current ( I )
1
Parameter
S.No
.
2
Effect
Reduces
Power factor
Improves
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5
4
Total current
Lagging reactive power (KVAr)
Page 10 of 41
Reduces
Reduces
Advantages of Static Capacitor: (Expected any one)
1. Initial cost is low.
2. Low operating cost.
3. Low maintenance cost.
4. Losses are very less (less than 0.5% )than that of rated value
5. Noise less operation as it is a static piece.
6. Less space is required. Therefore can be installed near load.
7. Greater reliability.
8. KVAr (leading) rating can be adjusted easily as per load condition.
Disadvantages of Static Capacitance: (Expected any one)
1. It has short life as compared to synchronous condenser.
2. Capacitors get easily damaged if the voltage exceeds than its rated value. Once the
capacitors are damaged its repair is uneconomical.
3. When capacitor is switched OFF then precaution is taken before making it ON. In
between OFF and ON time, time should be kept to discharge the capacitor, otherwise
capacitor may fail.
4. Switching current of capacitor is many times that of rated current; therefore cable size
should be double of the normal current carrying capacity, so its cost increases.
5. When there is no load or system is lightly loaded at that time capacitor bank must be
made OFF otherwise voltage across transformer increases.
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g) State any four factors governing selection of a motor for particular application.
Ans: (Any Four Points From The Following Or Equivalent Points Are Expected 1
Mark To Each Point, Total 4 Marks)
Following Factors governing / or are considered while selecting electric (Motor) for
particular application:
1. Nature of supply:
Whether supply available is
AC,
Pure DC
Or Rectified DC
2. Nature of Drive (Motor):
Whether motor is used to drive (run)
Individual machine
OR group of machines.
3.
Nature of load:
Whether load required light or heavy starting torque
OR load having high inertia, requirehigh starting torque for long duration.
OR Whether load torque increases with speed (T N)
OR decreases with speed (T 1/ N )
OR remains constant with speed (T = N)
OR increases with square of speed (T N2)
4.
Electric Characteristics of drive:
Starting,
Running,
Speed control
and braking characteristics
of electric drive should be studied and it should be matched with load requirements(i.e.
machine).
5.
Size and rating of motor:
Whether motor is short time running
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Page 12 of 41
OR continuously running
OR intermittently running
OR used for variable load cycle.
Whether overload capacity, pull out torque is sufficient.
6.
Mechanical Considerations:
Types of enclosure,
Types of bearing,
Transmission of mechanical power,
Noise
and load equalization
7.
Cost:
Capital,
Running
and maintenance cost should be less.
Q.2
Attempt any TWO :
a) i) (i) Give any four ideal requirements of elevators.
Ans: Ideal requirements of elevator:
16 Marks
(Any Four requirements are Expected: 1 Mark each, Total :4 Marks)
1. There must be all safety features.
2. Compactable acceleration and retardation to avoid jerk.
3. It should have sufficient Speed (feet/min.) proportional to height of building.
4. There should adequate lighting and provision of fan.
5. There should better interior design of the car.
6. It should have minimum breaking period.
7. There should be wide-frontage for fast traffic.
8. It should have sufficient capacity to handle the weight (Average weight 68 Kg per person).
9. Sufficient space should be available for car (2 Sq,ft. per person).
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10. There must be provision of back-up, when electric supply get’s failure like D.G. sets.
a) ii) (ii) State the factors to be considered for selection of shape and size of elevators.
(Any four points are Expected 1 Marks to each, Total 4 Marks)
The size and shape of elevator car depends on following factors:
i) No. of passenger to be carried: While selecting the size of car it is a usual practice to allow.
A Space of 2 Sq.ft/ person.
Average weight of passenger is assumed 68 kg/person.
Ans:
Thus the maximum load capacity of elevator is considered 34 kg/sq.ft
There should be wide frontage and shallow depth
ii) Limitation in the building design:
Shape of elevator depends on space available in building.
iii) Type of building
iv) Application of elevator
b) i) (i) Compare AC & DC system of track electrification on any four points.
Ans:
(Any Four point expected: 1 Mark each, Total 4 Marks)
S.No
1.
Points
Supply given to O/H
condition
AC track Electrification
DC track electrification
1-ph, 25KV, AC 50 Hz
600/750V-Tromways
1500/3000V urban/
suburban
2.
Type of drive used
1-ph, AC series motor
3.
Weight of traction motor
4.
Starting torque
5.
Accln and retardation
1.5 times more than d.c.
series motor.
Less starting torque
than d.c series motor
Less than d.c series
DC series motor for
tramways. DC compound
motor
1.5 times less than a.c series
motor
High starting torque
High
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6.
Overload capacity
7.
Method of speed control
8.
Maintenance cost of
traction motor
Starting Efficiency
Ridding quality
Insulation cost
Cross section of
conductor
Design of supporting
structure
Distance between two
substation
No. of substation
required for same track
distance.
Size (capacity) of traction
substation
Capital & maintenance
cost of substation
Cost track electrification
for same track distance
Applications
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
Model Answer
motor
Less than d.c series
motor
Simple and smooth
Page 14 of 41
High
More
Limited, except chopper
method
Less
More
Less, better than d.c.
High
Less
Less
Smooth (Better)
Low
More
light
Heavy
More
Less
Less
More
More
Less
Less
More
Less
More
Main line services
Urban and suburban area
A electric train has a schedule speed of 25 km/hr between stations 800 m apart. The
duration of stop is 20 seconds, the maximum speed is 20% higher than average running
b) ii)
speed and the braking retardation is 3 km/hr/sec. Determine rate of acceleration required
to operate the train.
Given:- Schedule speed of 25 km/hr,
Distance between stations 800 m
Stop time 20 Sec.
Ans:
Maximum speed is 20% higher than average running speed, Braking retardation is 3
km/hr/sec
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Scheduled speed
3600' x O.S
25
= 115.20 sec
=
Page 15 of 41
36 GGZ?
T =
^
Actualtime of run
A var age speed
3600 0.8
95.2
= T = Txh — T^
= 115.20 - 20
= 95.20 sec
----------------------------------------(1/2 Mark)
3600 D
T
------------------------------------------------------- (1 /2 Mark)
30.2521 km / hr.
Maximum Speed = 1.2 Average speed
=1.2 30.2521
=36.3025 km/hr. --------------------------------------------- (1/2 Mark)
1
1
1
1
7200 D Vmax
1 ---------------------------------------------------- (1 Mark)
V 2 max Vac
7200 D
5760
1.2 1
----------------------------------------------(1/2 Mark)
2
(36 .3025)
1317 .87
1
0.8741
0.8741
1
1
0.8741
3
1
0.5407
-------------------------------------------------------------------- (1/2 Mark)
1. 8495 kmphs ------------------------------------------------------------------ (1/2 Mark)
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Page 16 of 41
A 3ph, 440 V, 50 Hz, 40 kW load has a pl. 0.85 lagging. Calculate kVAR rating of capacitor
c) required to improve P.F. to 0.95 lagging. What will value of capacitor per phase, if (i)
capacitors connected in star? (ii) Capacitors connected in delta?
Given Data
Ans:
Volt : 440 V,
P= 40 Kw
f= 50 Hz
cos 1 =0.85
cos 2 =0.95
Cos 1 0.85
tan 1 = 0.6197 -----------------------------
----------------------------------------- (1/2 Mark)
cos 2 =0.95
tan 2 = 0.3286
-------------------- --------------------------------------------- (1/2 Mark)
Q1 = P tan 1
= 40 x 0.6197
= 24.788 KVAR
---------------------------------------------------- (1/2 Mark)
Q2= P tan 2
= 40 x 0.3286
= 13.144 KVAR
-
--------------------------------------------- (1/2 Mark)
i) KVAR Rating of the capacitor Bank
QC = Q1- Q2
= P tan 1 - P tan 2
------------------(1 Mark)
= 24.788 – 13.144
QC
= 11.644 KVAR
---------------------------------------------- (1 Mark)
Capacitor when connected in Star :-
C per phase
QC 103
or QC 2 FC V 2 2
V
C per phase
11.644 103
2 50 (440)2
C per phase 1.914 104 F Capacitor when connected in delta :-
------------------------------ (1 Mark)
------------------------------------------------ (1 Mark)
i
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C per phase
QC 103
3 V 2
C per phase
11.644 103
3 2 50 (440)2
------------------------------------------------------- (1 Mark)
C per phase 6.38 105 F Q.3
Page 17 of 41
------------------------------------------------ (1 Mark)
Attempt any FOUR :
16 Marks
a) State types of track electrification system.
(Any Four TYPES Of Track Electrification From The Following Are Expected 1
Mark To Each TYPE Of Track Electrification, Total 4 Marks)
Following are the different track electrification system
D.C. Supply system:1. Direct current track electrification:
600V, 750V DC for tramways
1500V, 3000V DC for Train (Urban and sub-urban services)
A.C. Supply system:2. 1-Ph, 25KV,standard frequency AC supply system:
Ans:
1-Ph, 25 KV, 50 Hz
3. 1-Phase, low frequency AC Supply system:
1-Ph, 15/16 KV, 16.2/3 Hz or 25 Hz
4. 3-Ph, Low frequency AC supply system;
3-Ph, 3.3/3.7 KV, 16 2/3 Hz or 25 Hz
Composite system:5. 1-Ph AC (1-ph, 25KV) – DC Supply System
6. Kando System (1-Ph AC – 3-Ph AC)
b)
State any one application of each of following :
i
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(i) Direct resistance heating (ii) Indirect induction heating (iii) Direct arc heating (iv)
Dielectric heating
Ans: (Any Two Application Are Expected Of Each Heating Type 1/2 Mark Each Application,
Total 4 Marks )
(i) Direct resistance heating:(Any Two Application Are Expected)
1. This type of heating used for industrial purpose
2. Salt bath heating: This utilized for the purpose of carbonizing, tempering,
quenching and hardening of steel tools
3. Heating of water in boiler
(ii) Indirect induction heating:
(Any Two Application Are Expected)
1. For heating as well as melting
2. Production of carbon free ferrous alloys.
3. For vacuum melting.
4. For melting non-ferrous metals for e.g. copper, aluminum, nickel etc.
5. For duplexing steel products.
6. Heating of non-conducting material is also possible if crucible is made from
conducting material.
(iii) Direct arc heating:-
(Any Two Application Are Expected)
1. Used for continuous and large production of high quality steel.
2. For Ferro-alloy manufacturing
(iv) Dielectric heating:-
(Any Two Application Are Expected)
1. In food processing industry, dielectric heating is used for Baking of cakes &
biscuits in bakeries.
2. Cooking of food without removing outer shell (e.g.-boiled egg) and pasteurizing
of milk.
i
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Page 19 of 41
3. For Rubber vulcanizing.
4. In Tobacco manufacturing industry for dehydration of tobacco.
5. In wood industry for manufacturing of ply wood.
6. In plastic Industry for making different containers.
7. In cotton industry for drying & heating cotton cloths for different processes.
8. In tailoring industry for producing threads.
9. For manufacturing process of raincoats & umbrellas.
10. In medical lines for sterilization of instruments & bandages.
11. For heating of bones & tissues of body required for certain treatment to reduces
paints & diseases.
12. For removal of moisture from oil.
13. For quick drying gum used for book binding purpose.
14. In foundry for heating of sand, core, which are used in molding processes.
c) State one applications of (i) Seam welding (ii) Butt welding (iii) Carbon arc welding (iv)
Metal arc welding
( Any one application expected: 1 Mark)
Ans: (i) Applications of Seam welding:It gives leak-proof joints.
1. Hence used for welding of various types of containers,
2. Pressure tank,
3. Tank of transformer,
4. Gas line,
5. Air craft tank,
6. Condenser,
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Page 20 of 41
7. Evaporator and
8. Refrigerator etc.
(ii) Application Butt Welding:
( Any one application expected: 1 Mark)
1) For welding rod, wire, pipe etc
2) Butt welding is a resistance welding process for joining thick metal plates or bars
at end
(iii) Application of Carbon arc welding:
(1 Mark)
1. For welding non ferrous metals
(iv) Application of Metal arc welding:
1.
(1 Mark)
For welding Ferrous Metals, Can be used for vertical & overhead welding.
d) Define : (i) Luminous Intensity (ii) Lumen (iii) Candle power (iv) MI-ICP
Ans: i) Luminous intensity:Lu min ousFlux
Luminous intensity = Illu min ation
Area
OR I
w
(1 Mark)
(Where lu min ous flux , w Solid Angle)
OR
The luminous intensity in any particular direction is the luminous flux
emitted by source per unit solid angle is called the luminous intensity of the source.
(ii) Lumen:
(1 Mark)
It is defined as the luminous flux emitted by a source of one candle power per unit
solid angle in all directions
OR
It is unit of luminous flux. One lumen is defined as luminous flux emitted per unit
solid angle from a point source of candle power.
iii) Candle power:
(1 Mark)
The candle power is the radiation capacity of the light source in the given
direction. The candle power is always given in lumens output per unit solid angle of the
given light source.
C.P
Lummens
, ( Where w Solid Angle)
w
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iv) MHCP (Mean Horizontal candle Power (MHCP) :
Page 21 of 41
(1 Mark)
MHCP is defined as the mean of the candle power of source in all directions in
horizontal plane.
e) State advantages of time off day tariff. (any four)
Ans: Advantages of time off day tariff: ( Any four point expected: 1 Mark each, Total 4 Mark)
1. Major industrial consumers are trying to run their maximum load during OFF load
period , to get rebate in their energy bill .
2. Major industrial consumers are trying to run their industry at reduced load during
PEAK load period to avoid additional charges charged in energy bill.
3. Due to above two reasons, it increases overall load factor as well as diversity factor of
power system.
4. As load factor and diversity factor of power system increases so overall cost per unit
reduces.
5. Also due to this there will be maximum utilization of power plant & infrastructure.
6. Due to TOD tariff major industrial consumers are trying to run their maximum load
during off load period
So, TOD tariff helps to avoid the wastage of surplus energy generated during OFF
load period.
In this way it helps to conserve energy.
f) Define : (i) Average speed (ii) Schedule speed Also, state the four factors affecting the
schedule speed.
(2 Marks)
Ans: (i) Average Speed: It is defined as distance covered between two stops divided by actual time of run is
known as average speed.
OR
Vav
3600 D
Km/hr
T
Where T = is actual time of run in sec OR
i
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Average Speed
Dis tan ce between stops or stations
Actual time of run
Page 22 of 41
(1 Marks)
(ii) Schedule Speed: -
It is defined as distance covered between two stops divided by schedule time is known
as schedule speed. OR
Schedule Speed
Dis tan ce between stops or stations
( Actual time of run) ( Stop time)
Schedule Speed
Dis tan ce between stops or stations
Schedule time
Km/hr
The following factors affect the schedule speed:(Two factors affecting schedule speed are expected 1/2 Mark each , Total 1 Marks)
1. By acceleration
2. By retardation
3. By both acceleration and retardation
4. By maximum speed
5. By stop time
6. By coasting period
Q.4
Attempt any FOUR :
16 Marks
a) Draw fig. of rheostatic braking in case of D.C. series motor and D.C. shunt motor.
(Figure : 2 Mark )
Ans: Rheostaic braking or dynamic braking of DC series Motor:
Si
—
Sa.
-xrrrrrrvnrn
Lrrmrrril
3b.
©
B1
DC SU pp
n2
Under normal condition
'^
Ja
t
Es
Under Dynamic breaking condition
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Page 23 of 41
Rheostaic braking or dynamic braking of DC shunt Motor:
(Figure : 2 Mark )
i
r<
5
M
u.
1
fa
ri
Under normal condition
Under Dynamic breaking condition
b) State four advantages of Ajax Wyatt vertical core induction furnace.
Advantages of Ajax Wyatt vertical core induction furnace:
(Any Four Points Are Expected : 1 Mark each, Total 4 Marks)
1) As furnace has narrow ‘V’ shape crucible at bottom. So small quantity of molten metal
remains in narrow ‘V’ notch from previous operation, which will help to keep
secondary short circuited. So no extra care is required to start the furnace
2) Magnetic coupling between primary & secondary winding is better because both
windings are on central limb of magnetic core. So there will be less leakage flux, Hence
leakage reactance is less, so power factor is better than horizontal crucible direct core
Ans:
type induction furnace.
3) Due to pinch effect in ordinary core type induction furnace there are chances of
temporary interruption in secondary circuit when current density exceeds above
500A/cm2 OR 5Amp/mm2..
4) But in this type of induction furnace there are no chances of interruption in secondary
circuit even if current density exceeds 500A/cm2 OR 5Amp/mm2 Because tendency of
weight of charge keep them in contact due to narrow ‘V’ shape.
5) So we can increase current density above 500A/cm2 OR 5Amp/mm2 to obtain more
heat in less time.
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Page 24 of 41
6) Vertical crucible is always better than horizontal crucible for pouring and taking out
the metal. Also space required is less.
7) As heat is produced directly in the charge there is no heat transfer loss. So efficiency of
furnace is more.
8) As heat is directly produced in the charge time required for melting metal is less. So
energy consumption is less.
9) As current is directly induced in the charge there is automatic stirring action taking
place in the charge due to electromagnetic forces developed in the charge due which,
Through mixing of molten metal is possible.
Uniform heating is possible
10) Accurate temperature control.
11) Ideal working condition in a cool atmosphere with no dirt , noise and fuel.
c) Compare between resistance welding and arc welding on any four points.
Ans: (Any Four Points From The Following Or Equivalent Points Are Expected 1
Mark To Each Point, Total 4 Marks)
Sr.No
1
2
3
4
Parameters
Type of welding
Resistance Welding
Arc Welding
Plastic/Pressure/Non-
Fusion/Non pressure
fusion welding
welding
Principle of heat
Heat is developed due to
Heat developed due to arc
developed
I2R losses where R is the
produced in between
contact resistance
electrode and job
External filler
Not required during
Required during welding
material required
welding
External pressure
Required
Not required
Both AC,DC supply is
Metal arc welding – Both
required
5
Type of supply used
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Page 25 of 41
used. But generally Ac
AC,DC supply is used. But
Supply is used.
generally Ac Supply is used.
And for
Carbon arc welding –only DC
supply are used
6
Voltage &current
required
7
8
Energy
Low voltage (2 to 20V
AC) and high current (40
to 400A, in some cases 5
to 20KA ) supply is
required
Low (3 to 4 KWH/Kg of
Metal Arc welding Voltage70 to 100V AC and
Carbon arc welding voltage50 to 60V DC,
Current- 50-600-800A
High (5 to 10 KWH/Kg of
consumption
deposited material )
deposited material.)
Temperature
Temperature obtained is
Temperature obtained is very
obtained
not very high (up to
high (up to 35000C to 60000C)
13500C)
9
Power factor
Low
Poor
10
Type of electrode
Non-consumable
Coated electrodes are used
electrodes are used.
for metal arc welding and
bare electrodes are used for
carbon arc
welding.(Electrodes may be
consumable or nonconsumable)
11
Application
It is suitable for mass
production
It is suitable for heavy job,
maintenance and repair work
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Page 26 of 41
d) State direct and indirect lighting scheme with one application.
( Figure Not compulsory)
Ans: 1. Direct lighting:
—
-
“
]
rfyedoy1
———
1
/
1
'
o o 6A
i i
dried |UK f i
~1
'.
4
|
i
i r
?T
\
*
p-\
’
—
A
scarce
|iiblirq
,
.
A
A
Ifunmi'nn
A
A
i
plnn<»
_1UIP)
4
Explanation:
(1 Marks)
As is clear from the name, in this system almost 90 to 95 % light falls directly on the
object or the surface. The light is made to fall upon the surface with the help of deep
reflectors.
Application of Direct lighting scheme:
(1 Marks)
1. This type of lighting scheme is most used in industries and commercial lighting.
Although this scheme is most efficient but it is liable to cause glare and shadows.
2. Indirect lighting: ( Figure Not compulsory)
•—
(TO
.
V
33i4 ttfld
*'
"
Leilecled
.
-
Liqhlibq
p
( MX
\
*
^
V
/
*
*\
_£khervie
k
-*
:
<
*
1
1
7
1
\
*
,
tI
i
*
rejecter
*l i* ti
*\
Ceili’na
/\
f\
tanl )
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Page 27 of 41
Explanation:
(1 Marks)
In this system, the light does not fall directly on the surface but more than 90 % of
light is directed upwards by using diffusing reflectors. Here the ceiling acts as a source of
light and this light is uniformly distributed over the surface and glare is reduced to
minimum.
Application of Indirect lighting scheme:
(Anyone expected: 1 Marks)
1. It provides shadow less illumination which is useful for drawing offices and
composing rooms.
2. It is also used for decoration purposes in cinema halls, hotels etc.
D.C. series motor is used for traction purpose. Justify your answer with any four
characteristics.
Ans: (Characteristics 1 Mark, For justification points 3 Marks, Total 4 Marks)
Due to following characteristics and advantages, DC series motor is suitable for traction
duty:
1)
Characteristics:
(1 Mark)
e)
/
/
/
3
F
/
/
|
K
/
Armature current ( la )
t/1
Armature Current (la )
Torque (Ta)
Characteristics of DC series motor
Advantages/ Justification points:(Any six points are expected 1/2 Mark each, Total 3 Marks)
1. DC Series motor has high starting torque.
2. DC Series motor has high rate of acceleration and retardation.
3. DC series motor has speed- Torque characteristics is such that it produces high torque
at low speeds, low torque at high speed.
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4. DC Series motor is variable speed motor. Due to these characteristics motor is
protected against overload i.e. self –reliving property against over load
5. It has high power to weight ratio.
6. DC series motor consumes less power than AC motors for same HP
7. DC Series motor maintenance cost is less.
8. DC series motor weight is 1.5 times less than 1-Ph AC series motor for same H.P.
9. Torque is unaffected by variation in supply voltage.
10. Torque obtained by DC series motor is smooth and uniform, so it improves riding
quality.
11. DC Series motor robust in construction and capable to withstand against continuous
vibration.
12. When DC series motor are running in parallel the all motors share almost equal load.
f) Sketch the various steps required for bridge transition system.
Ans:
(Each Step 1 Mark, Total 4 Marks)
In bridge transition, series last step to parallel first step, is carried out by following steps
Stepl: Bridge link is connected between two motors as shown in figure ( Series
—
UA VWMAVLI
- /
I
last step )
•
M2
@
^—
—
^V\ArrV\AnAVV-|
i
Step2:Bridge link is so rotated that two motors are put in series without starting resistance. Which are
un -shorted at the same time.
WWW
r-
Mil
•
-
N.
Bridge Link
MI
-
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Step 3:The portions of external resistance are connected in each motor circuit as shown in fig
lpUw UA
Mjl
1
WV 1
^,
-
1
ndgeun
I^W
.
.
-
WV- ^VW- — 3
^
I
J
Step4 : In this last step bridge link is removed as shown in fig. This is nothing but parallel first step.
Mil
(M)
(
^
J
M2
MJ
Q.5
Attempt any FOUR :
a) State the meaning of load equalization. How is it done ?
Ans: Meaning of load equalization:
16 Marks
( Meaning : 2 Mark, Figure: 1 Mark & explanation: 1 Mark)
There are many types of load which are fluctuating in nature e.g. wood cutting
m/c, Rolling mill. Etc. For such type of loads, load equalization is necessary to draw
the constant power from supply. Because,
When there is sudden load on motor, it will draw more current from supply at
start to meet additional power demand. Due to this heavy current there is large
voltage drop in supply system. This will affect electrical instrument, equipment, m/c,
other consumer etc. which are connected across same supply line.
Also to withstand heavy current, size of input cable increases so cost of cable
increases, Hence it is necessary to smooth out load fluctuations on motor.
The process of smoothing out load fluctuation is called load equalization.
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How load equalization is done?
Load equalization is done by means of flywheel. It is mounted on motor shaft.
Flywheel stores kinetic energy when there is light or no load & it supplies kinetic
energy when there is sudden heavy load on motor. In this way load demand on supply
remains practically constant.
b) Draw figure of indirect arc furnace. State its two advantages and two disadvantages.
(Figure: 2 Mark, advanatges:1 Mark & disadvantages: 1 Mark, Total 4 Marks)
Ans:
Figure of indirect arc furnace:
(2 Marks)
Indirect arc Furnace .
\ cp
supply from furnace
I
l1
arc
elec+rocLe
charge
Vt ra c + oruj
^
Th e
UJCU I
(
rmofot
rnrjaJ
—
X o , ( j J oJ
^
i
oo
rocf i
arrangement
Advantages: (Any Two advantages expected: 1/2 each: total 1 Mark)
1. No carbon particles of electrodes are mixed with molten metal. So we will get more
pure casting.
2. Rocking arrangement is compulsory in this type of furnace, due to this
Uniform heating is possible.
Through mixing of molten metal is possible
Life of refractory wall increases.
Disadvantages: (Any Two disadvantages expected: 1/2 each: total 1 Mark)
1. This type of furnace is not build of large capacity, because there is limitation to use
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only two electrodes because of its shape
2. Initial cost is more because rocking arrangement is compulsory required. (As there is
no automatic stirring action taking place in charge just like bottom conducting direct
arc furnace.
3. Due to indirect heating temperature obtained is less.)
4. Melting time required is more than direct arc furnace
5. Capacity of furnace is less.
c) Give classification of electrical welding.
Ans:
(Total 4 Marks)
i) Resistance Welding:1) Spot welding
2) Seam welding
3) Projection Welding
4) Butt Welding- i) Simple butt welding
ii) Flash butt welding
ii) Arc welding:1) Carbon Arc Welding: a) shielded welding b) unshielded welding
2) Metal Arc Welding: a) shielded welding b) unshielded welding
d) State any four factors to be considered while selecting electrical welding system.
Ans: (Any Four Factors From The Following Or Equivalent Points Are Expected 1Mark To Each
Point , Total 4 Marks)
Following Factors are considered while selecting of electric welding system:1) Type of Material:Whether similar metal is to be welded or dis-similar metal is to be welded.
2) Property of Material:Whether ferrous or non-ferrous metal is to be welded.
3) Thickness of job:-
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It is also depends on thickness of job to be welded.
e.g. for thick material- Arc welding is used. And for thin material – Resistance welding
is used.
4) Temperature required:Whether job required high or low temperature to weld the job.
e.g. For high Temperature - Arc welding is used. And for low Temperature –
Resistance welding is used.
5) Pressure required:If job is need of pressure at the time of welding in that case resistance welding is used.
And if
pressure is not required Arc welding is used.
6)Type of Supply Available:Whether AC or DC or both supply are available.
7) Application:In case of mass production, resistance welding is used & for repair work Arc
welding is used.
e) Explain with neat diagram metal halide lamp.
Diagram metal halide lamp:
Ans:
( Diagram: 2 Mark, Construction: 1 Mark & operation: 1 Mark, Total 4 Mark)
Circuit for Ballast-Ignitor-Capacitor -Lamp
BALLAST
Com
240V
<
200V
|GN
PL
i
Yellow
<
Kea
"I
( Phasae )
Black
( Neutral )
IGNITOR
or equivalent figure
Construction is similar to mercury lamp.
MH lamps consist of an arc tube (inner) enclosed by an outer tube.
Vacuum is created between the inner & outer glass tube to prevent heat loss.
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The inner arc tube contains the electrodes and various metal halides, along with
mercury and inert gases that make up the mix.
MH lamp has three electrodes – two for maintaining the arc and a third internal
starting electrode
OR Pulse-start MH lamps do not have a starting electrode. An igniter in the pulse
start system delivers a high voltage pulse (typically 3 to 5 kilovolts) directly across
the lamp’s operating electrodes to start the lamp
IT require a ballast to give high voltage at staring to produce the arc
The capacitor is used to improve the power factor.
Operation:
When the lamp is turned on, a high voltage at starting is applied across two electrodes,
to initiate an arc which discharges and vaporizes argon gas (starting gas), mercury
vapor and chemical components called “metal halides”
The energized metal atoms emit light.
f) State any four causes of failure of resistance heating element.
Ans: Following of the different causes of failure of resistance heating element:
( Any Four causes expected: 1 Mark each, Total 4 Marks)
i) Formation of hot spot:
Hot spot on heating element is the point which is at higher temperature than
remaining heating element portion. So there is possibility of breaking of heating
element at hot spot.
ii) Due to oxidization:
At high temperature material gets oxidized which may cause failure of heating
element.
iii) Due to corrosion:
If heating element is directly exposed to chemical fumes then there is
possibility of rusting of heating element which causes failure of heating element.
iv) Mechanical Failure:
Measure heating element alloy contain iron which is brittle. Due to frequent
heating & cooling of heating element, it may break (fail) due to small mechanical
injury also.
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Q.6
Attempt any FOUR :
a) Derive expression of most economical P.F.
Ans: Derivation:
16 Marks
( 4 Mark)
KVAR c Qc
o
kW
P
A
,
Q
%
s
T
'
T
kVAR.,
C
"
1 Q2
.
kVAR.
si
B
IT
Let,
P = Active power KW
S1, S2 = KVA Maximum demand before and after improving power factor
Q1, Q2 = Lagging reactive power before & after improving power factor
QC = Leading Reactive power drawn by Capacitor
Cos1 = Initial Power factor
Cos2 = Improved Power factor
Rs X = Tariff charges towards M.D. (KVA) /year
Rs Y = Expenditure towards KVAr to be neutralized per year (Expenditure towards
P.F. improving apparatus)
1) Before improving Power factor:
Q1 P tan 1
Cos 1
S1
P
S1
P
Cos 1
KVA1 (S1 ) P sec 1
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2) After improving Power factor:
Q 2 P tan 2
Cos 2
S2
P
S2
P
Cos 2
KVA2 (S2 ) P sec 2
3) Saving in KVA charges:
= Rs X (S1 –S2)
= Rs X ( P sec 1 P sec 2 )
= Rs X .P ( sec 1 sec 2 )
4) Expenditure towards KVAr to be neutralized:
= Rs Y (Q1 –Q2)
= Rs Y ( P tan 1 P tan 2 )
= Rs YxP ( tan 1 tan 2 )
5) Net Saving:
= Saving in KVA charges - Expenditure towards KVAr to be neutralized.
= [Rs X .P ( sec 1 sec 2 )] - [ Rs Y ( P tan 1 P tan 2 )]
Saving will be maximum when differentiate above equation with respect to 2 and
equate to zero.
ds
d
Rs X P (sec 1 sec 2 ) Rs Y P (tan 1 tan 2 )
d 2 d 2
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0 X P sec 2 tan 2 0 Y P sec 2 2
0 Rs X P sec 2 . tan 2 0 Rs Y P sec 2 2
Rs X P sec 2 . tan 2 Rs Y P sec 2 2
Rs X tan 2 Rs Y sec 2
Rs X
sin 2
1
Rs Y
Cos 2
Cos 2
Y
Rs X sin 2 Rs Y sin 2 Rs —
X
sin 2 2 Cos 2 2 1
6)
Cos 2 2 1 sin 2 2
Most economical power factor = Cos 2 1 (Y / x) 2
Most economical power factor at which maximum saving will occurs
b) Draw neat sketches of series parallel control of traction motors.
Ans: Series steps of traction motor:
Step 1 –
Two traction motors M1 and M2 are connected in series and started with all starting
resistances in series.
Series Steps
stepl
1
1
— ^V W— — \A/V—
1
i s1
. —
— v w-1^
|
V d c
1
1
1 _s_2__ —
V /—
^
*•
The starting resistances are cut out one by one gradually from step 1 to step 7 and
finally two motors are in series without any resistance.
In series connection the supply voltage V is divided in two motors. (Both motors get
half or (V/2) volts). So speed is also half. (N/2)
m
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step7
si
Ml
M2
S2
M
+
M
Vdc
Voltage across each motor is Vdc/2 and speed is N/2 RPM
Parallel steps of traction motor:
Step 1 –
After completion of series last step motors are now connected in parallel again with
series resistance otherwise motor will draw very high current and may damage
itself.
Parallel Steps
stepl
si
Ml
S2
M2
^
M
+
Vdc
Step 2 to 7 –
Both motors are now connected in complete parallel and starting resistances are cut
out one by one 2 To 7
In parallel connection, voltage across M1 and M2 will be full i.e. V (voltage is always
same in parallel).
Voltage across each motor = V and speed of each motor = N
So, voltage is now increased from (V/2) to V.
Hence, speed also increases from (N/2) to N and motor runs with full speed.
step7
si
Ml
S2
M2
^
M
+
Vdc
Voltage across each motor is Vdc and speed is N RPM
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c) State types of mechanical power transmission system.
Ans: Following types of mechanical power transmission system:-
Page 38 of 41
( 4 Marks)
i) Direct Transmission Drive: There are two types of direct transmission drive,
1) Fixed (rigid) Coupling.
2) Flexible Coupling.
ii) Indirect Transmission drive:1. Belt Drive:- The leather belt is most economical drive .It is used up to 300HP
and slip is 4%
2. Rope Drive: - This type of drive is used when power to be transmitted is
beyond the scope of belt drive. Slip is very small
3. Chain Drive: - It is more expensive than belt and rope drive. But it is more
efficient and can be used for greater speed ratio. It is used where no slip is
required.
4. Gear Drive: - With the help of this drive speed reduction is obtained. (Speed
increase is also obtained.)
5. Vertical Drive: - In this type of drive motor is arranged vertically.
d) State the different methods of temperature control of resistance furnace/oven.
( 4 Marks)
Ans: Following Methods of temperature control of resistance furnace/oven.:
A) By varying voltage across heating element:
1. With the help of autotransformer.
2. With the help of tap-changing transformer.
3. By use of series impedance or reactance.
4. Bucking and boosting secondary voltage.
5. By use of separate M.G. Set ( for large capacity resistance furnace/oven)
6. By series and parallel combination of resistances
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B) By varying the value of resistance of heating element:
1. By varying number of heating element.
2. By varying value of resistance by different configuration in circuit
C) By use of thermostat:
e) A factory has a maximum demand of 250 kW with a load factor of 0.6. The following tariffs
are offered: (i) 2 part tariff 70/kW of M.D./year + 4 paise/kWh (ii) A flat rate of 10
paise/kWh. Which tariff is economical ?
No. of Units consume in One Year
Ans:
Load Factor M .D( KW ) 8760
-------------------------------- ( 1/2 Mark)
0.6 250 8760
1314000 Kwh
-----------------------------------------------------(1/2 Mark)
Case-I: Energy Bill := Tariff given Rs. 70 of M .D. / year Rs. 4 paise / Kwh
= ( 250 70 ) ( 1314000 4 / 100)
------ (1/2 Mark)
= Rs. 17500 Rs.52560
= 70060 Rs.
------------------------------------------------------------ (1/2 Mark)
Case-II: Energy Bill := Tariff given flat rate of 10 Paise / Kwh
= 1314000 10 / 100)
= 131400 Rs.
--------------------------- (1/2 Mark)
-------------------------------------------------------------- (1/2 Mark)
Remark:-
---------------------------------- (1 Mark)
Two part tariff will be cheaper i.e. 70/kW of MD/year + 4 paisa/kWh because power
consumption is less
OR
According to energy bill Case-I is economical
For industrial consumer Case-I is economical
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f) State the two functions and types of enclosures provided to machine.
( Function: 2 Marks & Types: 2 Marks, Total 4 Marks)
Ans:
Functions of enclosures as follows:-( Any Two point expected: 1 Mark each, Total : 2 Mark)
1. It protects the operator against the contact with live and moving parts.
2. It provides the protection to internal parts of motor against mechanical injury.
3. It provides the protection against entry of moisture, dirt and dust particles
inside the motor.
4. It gives mechanical support.
5. It fold the machine
Types of enclosures and their function: ( Any Two types expected: 2 Mark each, Total : 4 Mark)
i) Open type enclosure:It is used where motor is installed in clean atmosphere and in closed room.
ii) Screen Protected enclosure:- (Guarded enclosure:)
Here screen is provided for rotating parts for better protection. It is also used
where motor is installed in clean atmosphere and in closed room.
iii) Drip (moisture) proof enclosure:- (Weather-protected type 1 enclosure, Weatherprotected type 2 enclosure, Waterproof enclosure,)
This type of enclosure is used in very damp atmospheric condition such as
water pumping station motor on ship sub-merssible motors, etc.
iv) Flame (Fire) proof enclosure:- (Splash-proof enclosure, Dust-ignition-proof enclosure)
It is used where motors are installed in explosive atmosphere like chemical
plants, mines etc.
v) Totally enclosed type enclosure:It is used where there is dusty atmosphere such as saw mill, stone crushing
plant, coal handling plant, cement manufacturing plant, cotton industry etc.
As it is totally enclosed it requires special cooling arrangement.
vi) Pipe ventilated totally enclosed type enclosure(Totally enclosed fan-cooled enclosure,
Totally enclosed pipe-ventilated enclosure, Totally enclosed water-cooled enclosure,
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Totally enclosed water–air-cooled enclosure, Totally enclosed air–air-cooled enclosure)
It is used where there is dusty atmosphere such as saw mill, stone crushing
plant, coal handling plant, cement manufacturing plant, cotton industry etc.
As it is totally enclosed it requires pipe ventilation, clean and cold air is
circulated through pipe forcefully for cooling of motors and hot air is taken out through
pipe.
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