MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION

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WINTER – 2023 EXAMINATION
Model Answer – Only for the Use of RAC Assessors

Subject Name: Utilization of Electrical Energy Subject Code: 22626

Important Instructions to examiners: XXXXX
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 principal components indicated in the 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 of some questions credit may be given by judgement on part of examiner of relevant answer based on
candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on equivalent concept.
8) As per the policy decision of Maharashtra State Government, teaching in English/Marathi and Bilingual (English +
Marathi) medium is introduced at first year of AICTE diploma Programme from academic year 2021-2022. Hence if
the students write answers in Marathi or bilingual language (English +Marathi), the Examiner shall consider the
same and assess the answer based on matching of concepts with model answer.

Q. No. Sub Answer Marking

Q. N. Scheme
1 Attempt any FIVE of the following: 10 Marks
a) State Lambert’s Cosine Law. 2 Marks
Ans:- According to this law the illumination at any point on the surface is proportional to the
cosine of angle between the normal at that point and direction of luminous flux.
Cos θ=AC/AB
E=I cos θ/d2
b) Define the following terms: 1 Mark
I. Utilization factor each
II. Illumination
Ans:- Utilization factor:-
Definition:-It is the ratio of total lumens falls on working plane to the total lumens
given out by the lamp
Note:-Its value always less than one
Illumination:-
Definition:-It is defined as the luminous flux falling on the surface per unit area
It is denoted by E
It is expressed in lumen/m2
c) Classify electric heating on basis of power frequency heating. 1 Mark
Ans:- Power frequency heating classified into two types each
1)Resistance Heating
a)Direct resistance heating
b)Indirect resistance heating

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2)Arc heating
a)Direct arc heating
b)Indirect arc heating
D) State any two advantages and disadvantages of Group Drive. ½ Mark
Ans Advantages of Group Drive: for any
1) Initial cost of single motor of large capacity is less than cost of number of small two
capacity motor for same HP. advantage
2) Overload capacity of group drive is higher than individual drive. and
3) It require less space disadvant
4) It gives better efficiency and power factor. age
Disadvantages of Group Drive:
1) It gives less flexibility due to common shaft for number of machine.
2) It is less safe.
3) Its reliability is less
4) It require special arrangement to control speed of machine
e) List any four characteristics of tariff. ½ Mark
Ans:- characteristic of tariff: each
1. Attraction of consumers
2. Simplicity in understanding the tariff
3. Reasonable and fair enough
4. Proper returns to company
5. Reasonable profit to company
f) List various voltages levels used for Electrical Traction. ½ Mark
Ans:- 1.5 KV DC, 3 kV DC, 15 kV AC, 25 kV AC, Other Levels, including 600 V DC and each
750 V DC for specific light rail and tram systems.
g) Write any four causes of low power factor. ½ Mark
Ans:- 1) Motors operating at low power factor each
2) Use of Arc lamp/Electric discharge lamp.
3) Improper maintenance of machines
4) Warn-out bearings.
5) Industrial heating furnaces
6) Repairs of motor
2 Attempt any THREE of the following: 12 Marks
a) Describe with a neat labeled diagram working of high pressure mercury lamp. 2 Marks
Ans:- High pressure mercury vapour lamp: Principle of working for high pressure mercury for
vapour lamp: "When mercury discharge under low pressure gives mainly ultraviolet diagram,
radiations, If pressure is increased to one or two atmospheres, its proportion of 2 Marks
radiations in the visible spectrum is increased and we get light having bluish tinge". for
explanatio
n

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Construction and Working of high pressure mercury vapor lamp:

1) It consists of two tubes. Inner tube contains neon or argon gas under low
pressure, two main electrodes in the form of oxide coated coils and a starting or
auxiliary electrode near the cap end side of the main electrode.
2) Inner tube is made of hard glass and in an evacuated outer tube, which
maintains high operating temperature of the inner arc tube.
3) When supply is given, an electric field is set up between the starting electrode
and the adjacent main electrode. This causes discharge first to take place
between them through limiting carbon resistance.
4) This discharge through argon gas evaporates all mercury and for the lamp to
reach full brilliance requires 4 to 8 minutes depending upon the design. Once
the arc tube is filled with mercury vapour a low resistance path is created for
current to flow between the main electrodes. The arc then shifts to main
electrodes. Path of the starting electrode automatically becomes inactive.
5) Electrodes are kept in electron emitting condition by the bombardment of heavy
mercury ions.
6) When the lamp goes out while In services, It will not restart till mercury vapour
pressure has fallen low enough to allow for restarting of discharge between
main and auxiliary electrodes.
b) Compare core type of furnace and coreless type of furnace (Induction) on the 1 Mark
following points. for each
i) Weight and size
ii) Frequency
iii) Leakage flux
iv) Crucible shape
Ans:- Parameter Core type furnace Coreless type furnace
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Weight and size More Less
Frequency Normal frequency required High frequency required
Leakage flux Less More
Crucible shape Horizontal or vertical Any shape
c) Describe with neat sketch regenerative braking applied for d. c. shunt motor. 2 Marks
Ans:- for
diagram,
2 Marks
for
explanatio
n

Or any other equivalent diagram

A machine operating as motor may go into regenerative braking mode if its speed
becomes sufficiently high so as to make back emf greater than the supply voltage i.e.,
Eb > V. Obviously under this condition the direction of Ia will reverse imposing torque
which is opposite to the direction of rotation. The situation is explained in figures. The
normal motor operation is shown in figure where armature motoring current Ia is drawn
from the supply and as usual Eb < V. Since Eb = kφ n1. The question is how speed on its
own become large enough to make Eb < V causing regenerative braking. Such a
situation may occur in practice when the mechanical load itself becomes active.
Imagine the d. c. motor is coupled to the wheel of locomotive which is moving along a
plain track without any gradient as shown in figure. Machine is running as a motor at a
speed of n1 rpm. However, when the track has a downward gradient, component of
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gravitational force along the track also appears which will try to accelerate the motor
and may increase its speed to n2 such that Eb = kφ n2 > V. In such a scenario, direction
of Ia reverses, feeding power back to supply. Regenerative braking here will not stop
the motor but will help to arrest rise of dangerously high speed.
Or any equivalent answer
d) State the types of Track Electrification used in India. Explain any one type. 02 marks
Ans:- The three main types of electric traction systems that exist are as follows: for state
1. Direct Current (DC) electrification system types and
2. Alternating Current (AC) electrification system 2 Marks
3. Composite system. explanatio
1. DC Electrification System n any one
The choice of selecting DC electrification system encompasses many advantages, such
as space and weight considerations, rapid acceleration and braking of DC electric
motors, less cost compared to AC systems, less energy consumption and so on.
In this type of system, three-phase power received from the power grids is de-escalated
to low voltage and converted into DC by the rectifiers and power-electronic converters.

This type of DC supply is supplied to the vehicle through two different ways:
3rd and 4 the rail system operate at low voltages (600-1200V)
Overhead rail systems use high voltages (1500-3000V)
The supply systems of DC electrification include;
300-500V supply for the special systems like battery systems.
600-1200V for urban railways like tramways and light metro trains.
1500-3000V for suburban and mainline services like light metros and heavy metro
trains.
Due to high starting torque and moderate speed control, the DC series motors are
extensively employed in the DC traction systems. They provide high torque at low
speeds and low torque at high speeds.
2. AC Electrification System
An AC traction system has become very popular nowadays, and it is more often used in
most of the traction systems due to several advantages, such as quick availability and
generation of AC that can be easily stepped up or down, easy controlling of AC motors,
less number of substations requirement, and the presence of light overhead catenaries
that transfer low currents at high voltages, and so on.
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The supply systems of AC electrification include single, three phase, and composite
systems. The Single phase systems consist of 11 to 15 KV supply at 16.7Hz, and 25Hz
to facilitate variable speed to AC commutation motors. It uses step down transformer
and frequency converters to convert from the high voltages and fixed industrial
frequency.
The Single phase 25KV at 50Hz is the most commonly used configuration for AC
electrification. It is used for heavy haul systems and main line services since it doesn’t
require frequency conversion. This is one of the widely used types of composite
systems wherein the supply is converted to DC to drive DC traction motors.

Three phase system uses three phase induction motor to drive the locomotive, and it is
rated at 3.3.KV, 16.7Hz. The high-voltage distribution system at 50 Hz supply is
converted to this electric motor rating by transformers and frequency converters. This
system employs two overhead lines, and the track rail forms another phase, but this
raises many problems at crossings and junctions.
3. Composite System
Composite System (or multi-system) trains are used to provide continuous journeys
along routes that are electrified using more than one system. One way to accomplish
this is by changing locomotives at the switching stations. These stations have overhead
wires that can be switched from one voltage to another. Another way is to use multi-
system locomotives that can operate under several different voltages and current types.
In Europe, it is common to use four-system locomotives. (1.5 kV DC, 3 kV DC, 15 kV
16⅔ Hz AC, 25 kV 50 Hz AC).
3 Attempt any THREE of the following: 12 Marks
a) Estimate the number & wattage of lamps which would be required to illuminate a (4 marks)
workshop 80 m by 20m, spaced 60 × 15 m by means of lamps mounted 6m above
the working plane. The average illumination required is about 100 lux, coefficient
of utilization is 0.4, luminous efficiency is 16 lumens per watt. Assume a space
Ans:- height ration of unity & candle power depreciation of 20 %.
Area to be illuminated
A = 80 m × 20 m = 1600 m2
E = 100 Lux (Given)
Total lumens on working plane or lumen utilized
Total lumens = A × E = 1600 × 100 = 160000
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M . F. = 1 – 0.2 = 0.8
Gross Lumen = (A × E) / (U. F. × M. F.)
= (1600 ×100) / (0.4 × 0.8)
Gross Lumen = 500000
Total Wattage = Gross Lumen / Luminous Efficiency
= 500000 / 16
Total Wattage = 31250 Wattage
If we consider 200 Watt per lamp then
Nos of lamps = 31250/200
= 156.25
Nos of lamps = 157 Lamps
b) Estimate with neat lebelled diagram construction & working of carbon arc Diagram 2
welding. marks
Ans:- Constructi
on &
Working
2 marks
(4 Marks)

Construction & Working

 The main purpose behind the Carbon Arc Welding process is to form a strong
bond between the distinct metals. Here, the carbon electrode is used for
producing an electric arc between the electrode and the metals being bonded.
 Electrode: The diameter of the electrode used in this process is around 3 to 22
mm.
 Power source: In the CAW process, direct current welding machines are used
as a power source. These machines can be either rotating or rectifier types.
 Electrode holder: . As the temperature involved in this process is very much
high, we can’t use the traditional electrode holder during this process.
 An electric arc is generated between the electrode and the parent metal.
 The heat generated due to the electric arc melts the base metal.
 After the solidification of the molten metal, the required weld is produced in the
given region.
 You can vary the size of the electrode used in the process depending on the
generated current.
c) Recommend relevant motor for the following application. 1 mark for
Ans:- 1. Refrigeration & Air Conditioners each (4
2. Electric Clock marks)
3. Vaccum Cleaner
4. Washing Machine
1. Refrigeration & Air Conditioners - DC Compound Motor
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2. Electric Clock - A. C. Synchronous Motor
3. Vacuum Cleaner – Universal Motor
4. Washing Machine - Single Phase Induction Motor
d) Derive the equation for most economical power factor. (04
Ans:-  When a consumer improves the power factor, there will be reduction in kVA Marks)
demand. So there will be annual saving over the maximum demand charges.
 Most economical power factor is defined as the value of which the p.f should be
improved so as to have maximum net annual saving.
 In Figure, cos Φ1 is original p.f. and cos Φ2 is p.f. after adding p.f. correction
equipment

2-Marks

figure

------ (1)

------(2)

Subtracting Eq. (1) from Eq. (2), we get 2-Marks

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Most economical power factor is given by

............... (3)
Here cos Φ2 in Eq. (3) depends on the relative costs of supply and p.f. correction
equipment but is independent of initial p.f. (cos Φ1).
4 Attempt ant THREE of the following: 12 marks
a) Explain with neat labeled diagram construction & working of direct resistance Diagram 2
heating. Marks and
Ans:- Explanati
on 2
Marks
(04
Marks)

Construction –
 It consists of thermally insulated chamber.
 It also consists a two electrode which is made up by graphite material, both are
placed vertically
 The insulated chamber filled up by charge to be heated

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 The electrode are immersed in the charge
 At the upper side high resistive powder provided to control excessive current
Working –
 When supply is ON then the current passes from one electrode to another
electrode through charge & high resistive powder.
 Due to this there is I2R power loss in the charge itself & therefore charge get
heated up.
 The upper layer of high resistive powder is required to limit the value of flow of
current otherwise heavy flow of current in the circuit.

b) State the factors to be consider for selection of motor. 1 mark for

Ans:- 1) Nature of Electric Supply:-AC, Pure DC or Rectified DC each (4
2) Nature of Load:- marks)
a) Load having high inertia required high starting torque for long duration.
b) Whether load torque increases with speed
c) Whether load torque decreases with speed
d) Whether load torque constant with speed
3) Electrical Characteristics of Motor:- Speed, Running, Speed control
4) Mechanical Consideration:-Type of enclosures, Type of bearing,
Transmission of mechanical power, Noise.
5) Size & Rating of Motor
6) Cost:- Capital, Running, Maintenance cost should be low.
c) Compare electric locomotive over non-electric locomotive for the following points. 1 mark for
1) Starting Torque each (4
2) Regenerative Breaking marks)
3) Starting Time
4) Maintenance
Sr. Non - electric
Parameter Electric Locomotive
Ans:- No. Locomotive
1. Starting Torque High Low
Regenerative
2. Possible Not Possible
Breaking
3. Starting Time Start at any time Take more time
4. Maintenance Less More
d) A single phase 400 V, 50 Hz motor takes a supply current of 50 A at power factor (04
of 0.8 lag. The motor p.f. has been improved to unity by connecting a condenser in Marks)
Ans:- parallel. Calculate the capacity of a condenser required.
Cos = I /I1
I I1 Cos = 50 0.8 = 40 A
I 1=I tan = 40 0.74 =29.6A (1-Mark)
Similarly I 2=I tan = 40 0=0
IC = I 1- I 2= 29.6 – 0 = 29.6 A

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IC = V / XC
XC = V / IC = 400 / 29.6 = 13.51 Ω (1-Mark)
C = 1 / (2 FXC)
C = 1 / (2 )
-4
C = 2.35 10 Farad (2-Marks)
e) Draw the block diagram of 25 KV, single phase, 50 Hz AC locomotive used for Diagram 2
traction system. State the function of each part. Marks,
Ans:- Function
2 marks
(4 Marks)

The various components of of 25 KV, single phase, 50 Hz AC locomotive used

for traction system.
1. Over Head conductor - Power at 25 kV is taken via a pantograph from the
overhead contact wire and fed to the step-down transformer in the locomotive
2. Current collecting device – It collect current from overhead wire & passes to
tap changing transformer through C. B.
3. Circuit Breakers – It provide protection under fault condition in locomotive.
4. On-load tap-changers – It is used to vary the voltage for speed control of
traction motor.
5. Transformer – It step down input voltage 25 KV into working voltage of
traction motor.
6. Rectifier – It convert secondary voltage of transformer into DC supply
7. Filter- It is used to obtain pure DC supply.
8. Traction motors – It give mechanical power to run the train.
5 Attempt any Two of the following

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a) Describe with neat sketch Ajax Watt furnace 03 Marks
Ans:- for
Diagram,
03 Marks
for
explanatio
n

Or any equivalent diagram

Construction:-
 A shell type transformer is used.
 The primary winding arranged on central limb and secondary formed by charge
itself.
 The furnace assembly is vertical so that vertical core is used.
 V notch is formed at bottom part where temperature is maximum.
 Refractory lining is provided from all sides to reduce heat loss.

Working:-
 Ac supply is applied to primary causing heavy current in the material which is
housed in V-channel due to convection
 Produced heat utilized for melting the charge.
 The circulation of molten metal in V channel due to convection.
 The high temperature metal goes up and low temperature metal goes down.
 Thus, the stirring action is automatically obtained.
 This stirring action causes homogeneous heating.

Application:-

 It is used for melting metal having low reistivity

 It is used for heat treatment of silver, copper
 Such type of furnace used for continuous operation only not used for
intermittent services

b) Write the function of bearing and its types. O3 Marks

Ans:- Function of Bearing:- for
 It support the rotating part of machine function,
 It helps to maintain moving member of machine to a fixed physical location, 03 Marks
relative to stationary part.
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 It helps to moving component to rotate with reduced friction. for types
 It reduces noise.
Types of bearing:-
 Bal bearing or roller bearing
 Bush bearing or sleeve bearing
 Ball Bearing:-
Explaination
 A ball bearing is a type of rolling element bearing that uses balls to maintain the
separation between races.
 The purpose of a ball bearing is to reduce rotational friction and support radial
and axial loads.
 It achieves this by using at least two races to contain the balls and transmit the
loads through the balls.
 Application:
 It is used in induction motor
 It is used for shaft position
 Bush bearing:-
Explaination:-
 A bushing, also known as a bush, is an independent plain bearing that is inserted
into a housing to provide a bearing surface for rotary application.
 This is most common form of plain bearing.
 Application
 Motor with sleeve bearing are always used with sleeve bearing are always used
with horizontal shaft.
c) A train has scheduled speed of 80kmph between are 8km apart. Determine the
Ans:- crest or maximum speed over the run. Assuming
i) Duration of stop 50 second
ii) Acceleration 2 kmphps
iii) Retardation 3 kmphps
The speed time curve is trapezoidal.
Vsch=80kmph D=8km Tstop=50sec
ά=2km/h-sec and β= 3km/h-sec
Vsch=3600×D/Tsch
Tsch=3600×8/80
Tsch=360Sec (1-Mark)
Tsch= Actual time +Stop time
Actual time=Tsch-Tstop
=360-50
Actual time =310sec (1-Mark)
Maximum speed
Vmax=T-√T2-4k3600D/2k (T-Actual Time) (1-Mark)
But, K= ά+ β/2(ά× β)
K=2+3/2(2×3)
K=0.4167 (1-Mark)
Vmax= 310-√3102-4×0.4167×3600×8 ∕ 2×0.4167
Vmax=310-219.308 ∕ 2×0.4167
Vmax=108.82km/hr (2-Marks)
6 Attempt any Two of the following
a) A 40 kw, 3-phase, 400V resistance oven uses Nickel chromium strip of 0.3mm
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thickness. The heating element is star connected. The wire temperature is to be
11270C and that of the charge is to be 7270, Estimate the width and length of the
wire required.
Given : radiation efficiency =0.6, specific resistance of Ni-Cr=1.03×10-6 ohm-m,
emissivity=0.9.
Given Data
Ans:- Radiation efficiency=0.6
Specific Resistance of Ni-Cr=1.03×10-6 Ohm-m
Emissivity=0.9
T1=11270C=1127+273=14000K
T2=7270C=727+273=10000K
Radiation Efficiency=0.6, specific resistance of Ni-Cr= 1.03×10-6 Ohm-m,
emissivity=0.9
H=5.72×104 k.e [(T1/1000)4-(T2/1000)4]w/m2
H=5.72×104 ×0.6 × 0.9[(1400/1000)4-(1000/1000)4]w/m2
(1-Mark)
H=87771.3408 w/m2
Thickness=0.3mm
Therefore,
0.3×10-3
Therefore,
l/wt=v2/pρ (t=Thickness and w=Width)
Therefore
Voltage across each resistance=V/√3=400/√3
Voltage across each resistance =230.94 Volt
Power= 40KW/3=13.333KW=13.3333×103 Watt (1-Mark)
Therefore
l/wt=V2/pρ
l/w(0.3×10-3)=(230.94)2/(13.3333×103)×(1.03×10-6)
l/w=1165.044
Therefore,
t/l2=2ρH/v2
t/l2 = 2×(1.03×10-6)×87771.3408/(230.94)2
t/l2= 3.39017×10-6
(t=0.3×10-3)
l2=t/3.39017×10-6
l=9.4069728mtr (2-Marks)
l/w=1165.0442
w=l/1165.0442
w=0.00807434mtr (2-Marks)
b) Describe the need of load equalisation in motors.
Ans:- Load equalization:- 06 Marks
 There are many types of load which are fluctuating in nature for
For Example:-Wood cutting m/c, Rolling mills etc descriptio
 For such type of loads, load equalization is necessary to draw the constant
n
power from supply.
 After sudden load on motor 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.
 To withstand heavy current, size of input cable increases.
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Hence it is necessary to smooth out load fluctuation on motor.
The process of smoothing out fluctuations in load is known as load equalization.
 In industrial drives, load varies between very wide limits in a very short interval
of times e.g. steel rolling mill in which the load is very large during the rolling
operation and very little when charges pass by.
 A typical example of Rolling Mill Load Cycle
 The average load on motor is quiet less.
 Even though we have to select motor rating as per maximum load.
 The maximum load is there for short interval of time.
 After that motor is on no load.
 So a flywheel is connected to motor shaft.
 The motor rotates the flywheel along with load.
 During no load period the flywheel accelerates and stores energy.
 During loading period, the flywheel gives out its kinetic energy to load.
 Thus there is fewer loads on the motor now. The motor of less rating that
maximum load can be selected because flywheel also shares the load.
 Speed fluctuations are less when flywheel is connected
c) Explain With neat sketch, the construction and working of Faiveley type
Ans:- pantograph 02 marks
Constructional Diagram:- for
diagram,
04 Marks
for
explanati
on

In the case of 25 kV a.c traction system the contact wire of OHE is of light section.
This would mean low contact pressure of 6.5 to 9 kg. Pantographs for a.c traction
system are, therefore, light weight single aided Faiveley pattern shown in Fig. It
consists of a
 sub frame or base,
 articulated system,
 pneumatic control system including throttle valve,
 two raising springs and
 four insulators.
The base is made up of welded sections and houses two ball bearings on which carried
the articulated system. Rubber stops are provided to limit the folding of the articulated
system at the lower part.
Articulated system consists of:
(a) horizontal spindle turning into two ball bearings which are the part of the base;
(b) lower arm integral with horizontal spindle and supporting at the upper end bearing

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on which are fitted two tubes of the upper frame;
(c) thrust rod articulated at lower end on fixed point of the base frame and on upper end
on bearing housed in the ‘yoke’ casting at the upper end of lower arm. Thrust rod
ensures the positioning of the upper frame as lower arm rotates;
(d) transversal tube rotating in the sleeves fitted on the tube ends of upper frame. Pan
positioning rod is connected to the transversal tube through shank which is welded to it.
At the end of transversal tubes are fitted spring boxes which form resilient support to
the bow;
(e) Positioning rod, articulated on a pin fitted on the thrust rod end on one side and on
the crank pin fitted to the shank as shown in Fig. on the other side, keeps the contact
plane horizontal during vertical movements of the bow.
Two up springs are hooked on one side to balancing member attached to the base and
on the other side to the arm of horizontal spindle.
Tension in these springs balances the mass of articulated system and also ensures
pantograph raising and producing con tact force.
Piston in the control cylinder is held back by the holding down springs.
Admission of air under pressure to the cylinder operates the piston rod which in turn
operates the lever arm.
This produces translator motion of the insulating rod and slotted rod.
Control cylinder is earthed and fitted separately from the base on the vehicle roof.
Compressed air enters the control. Cylinder through throttle valve.
The admission and escape of air are adjusted so that raising and lowering can be done
without shock either on the contact wire Bow as shown in Fig. 6.5(b) and (c) consists of
frame on which are bolted two collecting strips.
Intermediate parts or on the base. Join the frame with the horns.
The bow is fitted on resilient spring box through rods with fork. Whole of the
pantograph is fitted on the roof of the vehicle by means of four insulators.
Operation:-
When compressed air is admitted in the control cylinder, piston compresses the holding
down springs and slotted rod gets translatory motion which permits horizontal spindle
to rotate under the action of upsprings.
The pantograph then rises until collector touches the OHE.
The articulated system then stops and piston complete its stroke.
From this point onward the air motor plays no further part and piston remains stationary
during normal operation.
The pin of the horizontal spindle is permitted to move freely in the slot of slotted rod
and pantograph is operated purely by up springs.
This design permits free movement of articulated system throughout the full distance of
its rise and fall.
Opening of the control cylinder to atmosphere causes piston to return under the force of
holding down springs.

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 MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
Slotted rod presses on the pin of the horizontal spindle thereby lowering the articulated
system.

Page 17 of 17