Unit-5 Industrial Applications 5.

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IndustRiaL applications

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Part-A
Short Questions with Solutions
Q1. What is electronic timer?
Ans: Model Paper-III, Q1(i)

Electronic timer is a device used to automatically record the start and end operation of any element with high precision
and accuracy.
Q2. Compare electronic timers and mechanical timers.
Ans:
Electronic timers are used to control small time intervals and high controlling flexibility. Whereas, mechanical timers are
used to control large time intervals i.e., from 1 second to many hours.
Q3. Give the significance of
(i) Repeatability
(ii) Actuating speed
(iii) Isolation in direct load control switching elements.
Ans:
(i) Repeatability
The rate at which load of control switching elements responds to the changes in time i.e., load is switched on and off is
the repeatability time in load control switching elements.
In direct switching, repeatability is high and in indirect switching, repeatability is low.
(ii) Actuating Speed
Actuating speed is the rate at which the load is activated after providing a control signal as an input to circuit.
(iii) Isolation
The breakdown voltage at which the load and input can be separated is called isolation. In direct switching, isolation
breakdown factor is low and in indirect switching, isolation in relays is high.
Q4. What are digital timers?
Ans: Model Paper-I, Q1(i)

Digital timer or digital timing element is an electronic timing device used to trigger load control switching device and
activate load. This triggering is possible by switching on and off the circuit at a RC rate using oscillators and pulse generators.
A frequency divider is used as a counting circuit for division of frequency.
Q5. What is a linear time base generator? Why the time base generators are called sweep circuits?
Ans:
Linear Time Base Generator
A linear time base generator is an electronic circuit that provides an output waveform, a part of which exhibits linear
variation of voltage or current with respect to time.

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5.2 Industrial electronics [JNTU-Hyderabad]
There are two types of time base generators,
1. Voltage time base generators - Output voltage of waveform linearly varies with time.
2. Current time base generators - Output current of waveform linearly varies with time.
Time base generators play a significant role in CRO’s functioning to display the variation of an arbitrary waveform with
respect to time. A linearly varying voltage with time applied to one set of deflecting plates of the oscilloscope sweeps the electron
beam horizontally across the screen. Because of this, it is called sweep voltage and time base generators are called sweep circuits.
Q6. Differentiate voltage and current sweep circuits.
Ans:
Voltage Time Base Generator Current Time Base Generator
1. A voltage time base generator is one that produces 1. A current time base generator is one that produces output
an output voltage waveform, a portion of which exhi- current waveform, a portion of which exhibits a linear
bits a linear variation with respect to time. variation with respect to time.
2. All linear-voltage-sweep circuits have a common 2. Operation of current sweeps will closely parallel to
basic method of operation such as exponential the various voltage sweeps with, however, the
charging of capacitor followed by a change in deflection coil inductance replacing the capacitor as
the circuit to discharge the capacitor at a proper the basic timing element.
point of the cycle.
3. A basic voltage sweep circuit is shown in figure 3. A basic current sweep circuit is shown in figure below.
below.
Vs Rs
‘S’

RL
C RD
I R C RD
L

1 2
Figure: Basic Voltage Sweep Circuit Figure: Basic Current Sweep Circuit
4. In CROs, TV and radar displays, in precise time 4. Linearly varying currents are required for magnetic defl-
measurements and in time modulation. ection application. Such as, the electric deflection of a
high energy electron beam requires an excessively large
sweep amplitude. In such cases, magnetic deflection,
which is produced by a linearly increasing current in the
deflection coil is used.
Table
Q7. What is welding? Classify the types of welding.
Ans: Model Paper-II, Q1(i)

Welding
Welding is nothing but joining of the metals placed together by heating the surfaces to the point of melting and pressing
or hammering them together.
Types of Welding
Generally welding is classified into the following groups.
1. Resistance welding
(i) Butt welding
(ii) Seam welding
(iii) Spot welding
(iv) Projection welding.

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Unit-5 Industrial Applications 5.3
2. Arc welding Disadvantages
(i) Carbon arc welding 1. Initial cost of equipment is high.
(ii) Metal arc welding 2. Skilled persons are needed for the maintenance of
equipment and controls.
(iii) Atomic hydrogen arc welding
3. In some materials, special surface preparation is
(iv) Helium or Argon arc welding. required.
Q8. Explain the principle involved in resistance 4. Bigger job thicknesses cannot be welded as it requires
welding. larger amount of current.
Ans: Q11. List essential characteristics of a good heating
element.
In resistance welding, the heat required for fusion is
obtained by the electrical resistance to the passage of high Ans:
current at the joint. At the time of heating, sufficient pressure is The essential characteristics of a good heating element
applied at the joint at right time. The heat generated in resistance are as follows,
welding is given by, (i) High specific resistance
2
H = I Rt (ii) High melting point
Where, H = Heat generated in Joules (iii) Low temperature co-efficient of resistance.
I = Electric current in Amperes (iv) High mechanical strength
(v) Non-corrosive nature
R = Resistance in Ohms
(vi) Less oxidation
t = Time for which the electric current
passing though the joint. (vii) Economic in nature.
Q12. What are the advantages and disadvantages of
Q9. Specify reason why A.C is more suitable for high frequency heating?
resistance welding. List the applications of A.C
welding sets. Ans:
Advantages
Ans: Model Paper-III, Q1(j)
1. It is quick, clean and convenient method.
A.C provides desired combination of current and voltage, 2. Amount of heat wasted is less.
when compared to D.C. Hence it is more suitable for resistance
welding. 3. Control of temperature is easy.
4. The heat can be made to penetrate into metal surface to
Applications any desired depth.
1. Mostly used in industrial welding operations. 5. Unskilled labour can also operate the equipment.
2. Popular application of transformer is production welding 6. The amount of heat produced can be accurately
on heavy gauge steel. controlled with help of suitable timing devices.
7. With this type of heating, it is possible to heat different
3. Transformers are mostly used for flux shielded metal objects of different shapes and sizes with the same coil.
arc welding.
Disadvantages
Q10. List the advantages and disadvantages of re- 1. Generation of heat is costly.
sistance welding.
2. Efficiency of equipment is quite low.
Ans: 3. Initial cost of equipment is quite high.
Advantages Q13. State the principle of induction heating.
1. Fast rate of production Ans: Model Paper-I, Q1(j)

Induction heating is based on the principle of

2. Less skilled workers can do the job
transformers. When an A.C current is passed through primary
3. Both similar and dissimilar metals can be welded heating coil, an electric current is induced in the charge and
4. High reliability and reproducibility are obtained the value of the induced current is dependent on,
1. The magnitude of primary current.
5. Employs semi automatic equipments and hence
maintenance cost is less 2. The ratio of number of turns in the primary and
secondary circuit.
6. No rod is required for welding. 3. Coefficient of magnetic coupling.

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Q14. Write down the advantages of induction heating.
Ans:
Advantages of Induction Heating
(i) Installation and operating costs are low.
(ii) It is suitable for periodic operations.
(iii) This method is more precise for surface hardening of metals like steel objects, Billets rods etc., and also for non-ferrous
metals.
(iv) Flexible temperature control.
(v) In very less time it reach to the melting point.
(vi) There is no noise, dust, dirt and smoke in its operation.
(vii) It has very high accuracy.
(viii) For production of high grade alloys, induction heating is most suitable.
(ix) Technological advancement made the induction heating process as a useful tool in the industry of heat treatment plants.
Q15. Mention the properties of dielectric materials.
Ans: Model Paper-II, Q1(j)

The properties of dielectric materials which are heated by dielectric heating are as follows,
(i) For most of the dielectric materials, the power factor and dielectric constant do not change broadly with frequency.
(ii) For a given material, power factor and dielectric constant increases along with the content of moisture in it.

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Unit-5 Industrial Applications 5.5

Part-b
eSSAY Questions with Solutions
5.1 Industrial Applications - I

5.1.1 Industrial Timers – Classification, Types, Electronic Timers – Classification, RC and

Digital Timers, Time Base Generators

Q16. Classify the industrial timers and briefly explain them.

Ans: Model Paper-I, Q10

According to the technique used, the industrial timers are classified as

1. Thermal timer
2. Electromechanical timer
3. Electrochemical timer
4. Mechanical timer
5. Pneumatic timer
6. Electronic timer.
1. Thermal Timer
The property of linear thermal expansion of metals are used in thermal timer. Usually two metal strips are placed
near to the heating element. When a specified temperature is reached, the strips make contact with each other and
a switching action takes place. The most common example of thermal timer is electrical iron used in houses with
temperature control.
2. Electromechanical Timer
These type of timers uses a synchronous motor as timing element. When the motor rotates, it initializes the count and
investments for each cycle. Hence, when the counter reaches the specified value the desired action on the controlled
device is performed.
3. Electrochemical Timer
This type of timer uses electrolytic timing cells which are specially designed electrochemical devices and mercury
coulometers. In this timers, electrolytic cell is the timing element that undergoes plating and deplating operations.
4. Mechanical Timer
This type of timers does not require energy source, the timing is done absolute by some mechanical events. The best
example of mechanical timers is mechanical clocks.
5. Pneumatic Timer
This type of timers uses the pressure of air flow through a valve or a nozzle along with a pressure gauge as timing
device. As the pressure builds or falls above/below the predetermined level timing operation is performed.
6. Electronic Timer
Electronic timers uses RC circuits as timing devices. There are large variety of electronic timers available such as
digital timers and IC timers. The rise time of the RC circuit is used as timing intervals and hence by changing RC,
one can control the length of timing.
Q17. What are electronic timers? List the advantages of electronic timers.
Ans:
Electronic Timer
Electronic timer is a device used to automatically record the start and end operation of any element with high precision
and accuracy. It is used in many applications to control small time intervals and high controlling flexibility.

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5.6 Industrial electronics [JNTU-Hyderabad]
A basic electronic timer contains a time varying quantity and an electronic device that measures the instants at which time
varying quantity attains a fixed (reference) level.

In electronic timers, solid state electronic devices which has timing circuits and electromechanical switches are significantly
used.

For a complete electronic timer, SCR are used as switches and BJT, FET and UJT are used in the circuitry.

Advantages

The advantages of solid state electronic timers are as follows,

1. Filament heating time is not required.

2. A little amount of energy is consumed.

3. Highly reliable.

4. Due to electrical isolation from input to output, safety hazards are decreased and potential breakdown is removed.

5. Small size and robust.

6. Time period can be easily and accurately adjustable.

7. The affect of ambient temperature is neglected by providing temperature compensation.

8. Because of the usage of zener diode, no effect is observed due to changes in simply voltage.

Q18. Give the classification of electronic timers.

Ans:

Classification of Electronic Timers

1. According to the type of timing elements, electronic timers are classified into,

(i) RC timers

The capacitor in RC element charges and discharges through resistor and generates delay in the circuit.

(ii) Digital timers

The circuit is switched on and off at a RC rate using oscillators or pulse generators. The pulses generated are used to
trigger the load.

The functioning of above timers is same but differs by timing accuracy, repeat accuracy and operation satisfaction

2. According to type of triggering elements used, they are classified into,

(i) BJT triggered

(ii) FET triggered

(iii) UJT triggered

(iv) Op-amp triggered.

The triggering elements BJT, FET, UJT and Op-amp responds to variations created by timing circuit and use them to
trigger the load

3. According to the load control switching element, they are classified as,

(i) Direct switching type of load device

(ii) Indirect switching type through relay control.

These two timers differ by their timing repeatability, speed and isolation between input and load.

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Unit-5 Industrial Applications 5.7
The basic block diagram of an electronic timer with its elements is as shown in figure below.

Trigger element Load control Load

Timing (UJT, FET,BJT,Op switching element
element device
Amp SCR, TRIAC

R.C network Counter or

frequency
+ divider
–

Electronic
clock
mechanism

RC time
base

Figure: Block Diagram of Electronic Timer

Q19. Write a short note on RC timing element?

Ans:

RC timing elements operates on the principle of charging and discharging of capacitor.

The two basic RC circuits used for supplying timing voltages are as shown in figure (1).
S S

R
+ + vR
V C R
– C vc

(a) Charging of capacitor (b) Discharging of capacitor

Figure (1): RC Timing Circuits

In figure 1(a), resistor R and capacitor C are in series with switch S. When the switch is closed, the input voltage applied
to the circuit appears across the resistor R and at time t = 0, voltage at capacitor C is zero. As the time increases, the capacitor
starts charging and exponentially raises voltage across C. The voltage across capacitor ‘C’ is given by,

vC = V[1 – e–t/RC]

In figure 1(b), the switch ‘S’ is connected in between R and C. This circuit corresponds to the discharging of capacitor C.
When capacitor C is fully charged i.e., upto voltage V, it retains in same voltage till the switch is closed. At time t = 0, the switch
S is closed and the capacitor starts discharging with time through resistor R. Therefore, the voltage across resistor R is given by,

vR = V e–t/RC

Figure (2) illustrates the time constant curves for charging and discharging of capacitor C.

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100

75

Percentage of full voltage

Charge
50

Discharge
25

0 1 2 3 4 5
Time, t.t
Figure (2): Time Constant Curves
The charging and discharging times of capacitor are constrained to one time constant. The charging curves and discharging
curves linearly increases and decreases with time ‘t’. If time is above one time constant, then the linearity of curve decreases i.e.,
non-linear curve is observed.
The design of solid state timers extends the timing constants i.e., from linearity to normal non-linear curve. In solid state
timers, control switching devices are used to turn on and off the voltage source for shorter time periods. Thus, delay is produced
in charging and discharging actions and time period is extended as shown in figure (3).
100% Allowable trip point variation

Vc trip
point
Resulting timing
variation

0
Time, t
Figure (3): Extended RC Curve using Solid State Timers
Based on circuit values, the high and low levels of output is shown in figure (4).
Vin

Time, t
Vo

High

Low
Time, t

Figure (4): High and Low Levels of Output

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Unit-5 Industrial Applications 5.9
Q20. Discuss in detail on time base generator.
Ans: Model Paper-III, Q10

Time Base Generator

For answer refer Unit-V, Q5.
Two types of voltage time base generators are described below.
Schmitt Trigger
The circuit diagram of schmitt trigger using BJT is as shown in figure (1).
+VCC

RC RC
1 2
R1
+

+ Q1 Q2 output vo
input vi

Re R2

Figure (1): Circuit of Schmitt Trigger

Schmitt trigger is a voltage time base generator. For an input of 50 Hz sine wave, schmitt trigger can produce a 50 Hz
square waveform.

Schmitt trigger basically consists of two npn transistors Q1 and Q2 coupled through emitter. The feedback to the circuit is
obtained from the common resistance RE. Since it is a bistable multivibrator, it has got two stable states because of the provision
of positive feedback in the circuit and due to the loop gain greater than unity.

The operation of the circuit can be explained in two stages i.e., when Q1 is OFF, Q2 is ON and Q1 is ON and Q2 is OFF.

1. Q1 OFF, Q2 ON

At t = 0, when the Input Vs is applied to the circuit, Q2 is ON and is conducting whereas Q1 is OFF and hence, not
conducting i.e., no current flows through it. The current flowing through RE is just I E2 (emitter current of Q2),

\ I C1 = 0, the output voltage V0 = VCC – I C2 R2

For t > 0, the value of Vs starts increasing, such that the transistor Q1 is driven into the active region. This is possible only
when the input voltage reaches a value greater than UTP i.e., Upper Threshold Point.
2. Q1 ON, Q2 OFF

If Q1 is ON and Q2 is already conducting, the collector current of Q1 i.e., I C1 , starts flowing due to which there is a drop
in voltage across RC1 . This drop reduces the base voltage VB2 of the transistor Q2. The reduction in VB2 in turn reduces
the value VE. This process is cumulative and at one point the transistor Q2 is turned OFF. Then the output voltage V0 = VCC
and this remains constant even if the input is further increased.

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Since Q2 is OFF and Q1 is conducting, the current I C1 causes a drop across RC1 and hence, the output voltage reduces
to a value VCC – I C RC . As the input voltage VS starts decreasing and becomes equal to the base voltage of Q2 i.e., VB2 , Q2 is
1 1
again driven into active region and gets turned ON. The value of Vi at which the transistor Q2 is turned ON is called the Lower
Threshold Point (LTP). The output voltage is constant and remains at its lower value until the input reaches the upper threshold
point. Hence, at input voltage UTP, the output is maximum i.e., VCC and is maintained until the input reaches LTP.
The variation of output with respect to input is as shown in figure (2).

0
vi

Time, t
Output of Q1

VCC
Time, t

VCC
Output of Q2

Time, t

Figure (2): The Variation of Output with respect to Input

As the schmitt trigger circuit converts a sine wave to a square wave. Therefore, it is also known as sine to square
converter or squaring circuit.
The pulse repetition rate (PRR) and voltage accuracy of the output waveform depends on input frequency. The high
frequency input produces an output with greater precision.
Astable Multivibrator
Astable multivibrator is also a voltage time base generator, which can divide the frequencies other than fundamental
frequency 50 Hz. Since, the astable multivibrator produces an output waveform without any external source, it is also called as
free-running multivibrator, square wave generator and relaxation oscillator.
Figure (3) shows the circuit arrangement of astable multivibrator.

Figure (3): Astable Multivibrator

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Unit-5 Industrial Applications 5.11
The coupling elements used in this circuit are capacitors Q23. What are the disadvantages of electric welding?
i.e., both are ac coupling. It does not require any triggering Ans:
signal to change its state. 1. After welding the equipment it is necessary to inspect
The output waveform produced at the collectors of Q1 the welded area.
and Q2 completely depends on RC time constants. 2. Metals like mercury cannot be welded.
5.1.2electric welding 3. Metals having low strength or low melting point cannot
classification, types and methods be used in welding process.
of resistance and Arc welding, 4. Before welding a metal work-piece, it is necessary
electronic d.c motor control to check its physical and mechanical properties like
temperature, density, thermal conductivity, tensile
Q21. Write short notes on electric welding. strength, ductility etc., availability and price.
Ans: 5. A skilled labour is required for welding.
Electric Welding 6. A huge amount of electricity is needed.
Electric welding is a process by means of which two 7. The structure of the welded joint is not same as that of
metal pieces are joined together by the heat produced due to parent metal.
the flow of electric current.
8. It produces harmful radiation, (light) fumes and spatter.
Due to the reliability of welded joints in comparison to
9. Incomplete penetration, internal air pockets, such defects
rivetted or bolted joints, electric welding has been adopted in cannot be detected.
many engineering fields.
Q24. Explain the principle of resistive welding. List
There are two methods by which electric welding can the applications of resistance welding.
be carried out. These are,
Ans:
(i) Resistance Welding
Resistance Welding
For answer refer Unit-V, Q24. Resistance welding is that process in which a sufficiently
(ii) Arc Welding strong electric current is sent through the two metals in contact
to be welded which melts the metal by the resistance they offer
For answer refer Unit-V, Q25. to the flow of electric current.
Q22. What are the advantages of electric welding In resistance welding a heavy current (above 100 A) at
processes? a low voltage is passed through the work-piece and the heat
Ans: developed by the resistance to the flow of current is utilized.
1. Joining of the metals by welding is an easy method at The heat developed at the contact area between the pieces
low cost. to be welded reduces the metal to a plastic state, the pieces are
then pressed together to complete the weld. In the process, two
2. It is a great advantage for the welders that any type of
electrodes of low resistance are used and metals to be welded
metals can be used for welding.
are pressed between the electrodes. The electrical voltage
3. Welding is widely employed to manufacture or repair required ranges from 4-12 V depending upon the composition,
all products made of metal. area, thickness etc. of metal pieces to be welded. Alternating
4. Required flexible design can be achieved. current is found to be most suitable for resistance welding as
it can provide any desired combination of current and voltage
5. By the use of welding technique different metals can be by means of a transformer.
welded and it is possible to get a good quality products.
The heat developed is given by I 2 Rt
6. In the manufacture of mechanization and automation Where, I = Current flow (A)
welding is very advantageous.
R = Resistance (ohms)
7. To make joint it requires no holes. t = Time for which the current flows.
8. It is very economical, because the cost of labour and The resistance in the above equation is made of,
material required is reduced.
(i) Resistance of current path in the work.
9. Complete permanent joints can be provided with the (ii) Resistance between the contact surfaces or parts
welding process. being welded.
10. Rivetting process produces noise. But in welding process (iii) Resistance between the electrodes and surface of
noise is not produced. parts being welded.

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In order to develop higher temperature between the interfaces of the work to be welded rather than at surface of work in
contact with the electrodes, it is necessary to keep the resistance between the electrodes and surface of the body being welded
to a minimum.
In resistance welding, the magnitude of current is controlled by varying primary voltage of a transformer. The time for which
current flow is very important. Usually automatic arrangements are devised which switches off the supply after a predetermined
time.
Applications
Resistance welding is used for,
1. Joining sheets, bars, rods and tubes.
2. Making tubes and metal furniture.
3. Making cutting tools.
4. Welding aircraft and automobile parts.
5. Making fuel tanks of cars, tractors etc.
6. Making wire fabric, grids, grills, containers, etc.
Q25. Write short notes on arc welding.
Ans:

Arc welding is the process in which the pieces to be welded are brought to the proper welding temperature at a point of
contact by the heat liberated at the arc terminals and in the arc stream so that metals are completely fused into each other, form-
ing a single solid homogenous mass after it solidifies.

In this process an electric arc is produced by bringing conductors (electrode and metal piece) connected to a suitable
source of electric current, momentarily in contact and then separating by a small distance. The current continues to flow across
the small gap and gives intense heat. The heat developed is utilized to melt the part of work piece and filler metal and thus form
joint. So, arc welded joint is a union of metal parts made by localized heating without any pressure. That is why, this type of
welding is known as non-pressure welding.

Figure
The temperature is order of 3600°C at which mechanical pressure is not required for jointing. A.C and D.C both can be
used in arc welding. The arc voltage varies between 70-100 V on A.C whereas it is 50-60 V on D.C.
Electric arc welding is widely used for joining of metal parts, the repair of fractured casting and fillings by the deposition
of new metal on worn out parts. Arc welding is further classified as,
1. Carbon arc welding
2. Metal arc welding
3. Hydrogen arc welding
4. Inert gas metal arc welding .

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Unit-5 Industrial Applications 5.13
Q26. What is the importance of welding?

Ans:

Importance of Welding

Nowadays, metals are mostly used for manufacturing of products. Almost all the products are made of metal and the
process of welding is very commonly employed in manufacture and in repair of metal products. Welding is the only technique of
joining the two or more pieces of metal. Previously joints were made using rivets or bolts in the construction of bridges, ships,
etc. But these require holes in the metal to be used to make a joint which reduces the cross-sectional area (upto 10 percent) of
the metal piece. Gusset plates are also needed for the joint which increases the cost as well as the weight of the material. But in
welding both the cost as well as the material required is reduced. Thus welding has eliminated the above drawback and made the
work very easy and economical.

The introduction of automotive welders, laser welders, electron beam welders makes the welding highly accurate. These
things provide a stable inherently strong and permanent joint. Due to this reason welding skills are in high demand and welding
engineering has become more popular. Such improvements in welding makes an exciting time for the unemployed to join
the welding industry. Welding is very advantageous for industries and factories like shipbuilding, automobile manufacturing,
construction and oil industries. It has also branched out into areas like electronics, aerospace, medical device manufacture,
instrumentation and photonics. Today’s welding processes involve plastic, glass, fiberglass and ceramic in addition to metal.
Q27. Differentiate between resistance welding and arc welding?
Ans: Model Paper-I, Q11

Resistance Welding Arc Welding

1. A.C supply is used for resistance welding. 1. Both A.C as well as D.C supply is used in arc
welding.
2. No additional material is required to get the 2. Filler material is used to perfect strength in the
metal pieces joined. metal.
3. Resistance welding requires low voltage 3. An arc requires more
with high currents. voltage to strike than to be maintained.
4. In this type of welding external pressure 4. Here no external pressure is required, thus the
is applied. material can be welded very easily.
5. The heat produced is mainly due to the contact 5. The heat produced is due to the arc formed in
resistance. between the electrode and the metal piece
to be welded.
6. The temperature is less when compared to the 6. The arc produced will have a very high temperature
arc welding temperature. and may cause damage.
7. The power factor of resistance welding 7. The power factor of arc welding is very poor.
will be 0.25 or 0.3 lagging.
8. It is advantageous for mass production. 8. It is advantageous for repair work.
9. It is not used for repair work. 9. It is not used for mass production.
10. The electrodes used in resistance welding are 10. The electrodes used in arc welding are flux coated
roller, flat or bar type electrodes. metal electrodes.

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Q28. With a neat circuit, explain the process of A.C resistance welding.
Ans:
A.C Resistance Welding
The basic circuit arrangement of A.C resistance welding is shown in below figure.
Sequence
Timer Solenoid valve

Welding
Electrodes

SCR (or)
1-f A.C ~ Circuit Ignition
supply Breaker Circuit
Welding
spot Workpieces

Power Welding Transformer

Transformer
Welding Machine
Figure: Circuit Diagram of A.C Resistance Welding
In this type of welding, A.C power is supplied to the welding machine by means of a timing device (sequence timer)
through the power transformer, circuit breaker and a control circuit. The transformer inside the welding machine steps-down
the supply voltage to about 10 to 15 V and supplies a large current of several thousand amperes to the welding spot of the metal
pieces through electrodes. The electrode tips are cooled properly and should be kept clean.
Solenoid valve applies the required air pressure on the electrodes in order to bring them together and squeeze the workpieces
properly. These workpieces are then properly heated by high welding current and gets welded. The workpieces are held under
pressure between the electrodes until the weld hardens. The current should to be applied intermittently in order to weld the
workpieces. A sequence timer controls the timings of welding process and this welding cycle is divided into squeeze time, weld
time, hold time and OFF time. The total timing sequence gives the heat control timing of welding process. Earlier in the process
of welding, current could controlled by these sequence weld timers. But now-a-days, for greater accuracy the electronic circuits
are used for controlling the welding current.

5.2 industrial Applications-II

5.2.1 high frequency heating, principle – merits, applications, HIgh

frequency source for induction heating
Q29. Give a simple circuit for the speed control of a D.C separately excited motor.

Ans: Model Paper-II, Q10

A separately excited D.C motor fed from 1-φ semi-converter is shown below.
Ia

Ra
I DF
T1 T2
Supply La
Df

D1 D2 Separately
excited Field
D.C. Motor

IF

Figure (1): Speed Control of Separately Excited D.C Motor Fed from 1-φ Semi-converter

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Unit-5 Industrial Applications 5.15
Control over the speed of a separately excited D.C motor can be achieved by controlling the armature voltage (Ea) of the
motor. When a separately excited D.C motor is fed from 1-φ semi-converter, its armature voltage can be controlled by varying
the firing angle (α) of the thyristors. Depending upon the circuit parameters and conditions of operation, the motor armature
current (Ia) can be continuous or discontinuous. Ia is discontinuous due to large values of α and low values of torque (T). The
effects due to discontinuous armature current is,

(i) The ratio of r.m.s value to average value of armature current increases

(ii) Performance of the motor becomes worse

(iii) Poor regulation of speed.

Hence, it is desirable to operate motor always in continuous current mode.

Let us consider that the armature current (Ia) of the D.C separately excited motor is continuous throughout its operation.

The supply to the converter is sinusoidal. During positive peak, the thyristor T1 is triggered at a firing angle α. Now, thy-
ristor T1 and diode D2 conducts between the period α to π. The current flows from T1 and D2 and the voltage at the terminals of
motor (Ea) is equal to the supply voltage. During the negative peak, between the period π to (π + α), the voltage tends to reverse
and appears in the negative side of the waveform. But, due to the presence of freewheeling diode (DF) in the circuit, this reverse
voltage is restricted to appear in negative side. Now, the current which was previously flowing through T1 and D2, freewheels
through DF which commutates T1 and Ea becomes zero.

At a firing angle (π + α) the thyristor T2 is triggered and the current flows through T2 and D1. The process repeats for dif-
ferent firing angles.

The average armature voltage of D.C separately excited motor is given by,

Em (1 + cos α)
Ea = π
The voltage and current waveforms for continuous armature current of motor is shown in figure (1).
v
V
Supply voltage

0
π ωt
α (π + α) 2π 3π

Firing pulses

0 ωt

Ea

ωt

Ia

DF DF DF
T1 D2 T2 D1 T1 D2
ωt

Figure (2): Voltage and Current Waveforms of Continuous Armature Current

The speed-torque characteristics of a D.C separately excited motor is shown in figure (3).

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5.16 Industrial electronics [JNTU-Hyderabad]
Speed (rpm)

3500

3000

2500 α=0
α = 30
2000

1500 α = 60

1000
α = 90
500

0 Torque (N-m)
1 2 3 45 6 7 8

Figure (3) : Speed-Torque Characteristics

It can be observed from speed-torque characteristics that with increase in value of α the speed of the motor decreases.
Hence, by varying the firing angle α, speed of the D.C separately excited motor can be controlled.
The steady-state speed of a D.C separately excited motor fed from 1-φ semi-converter is given by,

(Ea – Ia Ra)
N= Ka φ
(1 + cos α) Em Ra T
= (Ka φ) π –
(Ka φ) 2
T
Where, armature current, Ia = K φ
a

Speed of motor at no-load,

(1 + cos α) Em
N(no-load) = (Ka φ) π [Where, T = 0]

Also, from speed-torque characteristics shown, it is observed that, continuous armature current is necessary to obtain good
speed regulation.
Q30. Explain the principle of induction heating and list the factors on which it is dependent.
Ans:
Induction Heating

Induction heating is based on the principle of A.C transformers. There is a primary winding through which an A.C current
is passed which is magnetically coupled to the charge to be heated. When an A.C current is passed through primary heating coil,
an electric current is induced in the charge and the value of the induced current is dependent on,
1. The magnitude of primary current.
2. The ratio of number of turns in the primary and secondary circuit.
3. Coefficient of magnetic coupling.

The heat developed depends upon the power drawn by the charge and therefore to develop heat sufficient to melt the
charge, the resistance should be low which is possible only with metals and the voltage must be higher which is obtained by
employing higher flux since the higher the flux linked, the higher is the voltage induced. Thus magnetic materials are found to
be suitable for this type of heating because of their higher permeability.

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Unit-5 Industrial Applications 5.17
The following diagram shows the electrical representation of induction heating in which the magnitude of current induced
is equal to 3Ip.
 NS I 1 IP 
 = P ⇒ = ⇒ I S = 3I P 
 N P I S 3 I S 

Figure
In case the charge to be heated is nonmagnetic, the heat generated is due to eddy current losses. Whereas, if it is a magnetic
material there will be hysteresis losses. In addition, eddy current loss is proportional to frequency and hysteresis loss is proportional
to square of frequency and these laws holds good up to a limited temperature (curie point) since the magnetic materials lose their
magnetic properties above curie temperature.
The high frequency required for induction heating is obtained from motor generator set, spark gap converter and vacuum
tube oscillator.
The various factors on which the induction heating depends are,
1. Magnitude of primary current Ip, if the primary current is high the flux is high and Is is high, also the heat developed is
high.
2. Frequency, since hysteresis and eddy current loss depends on frequency.
3. Reciprocal of distance between primary coil and charge, if the distance is less the magnetic coupling is more and thus
heat developed is also more.
4. Permeability of the charge (metal) and resistivity of the charge.
Magnetic materials generally have high permeability and resistivity than nonmagnetic materials. Thus, the induction
heating is more adaptable and economical for treating magnetic materials.
Q31. What are the factors to be considered for inductor design in induction heating?
Ans:
The following points are to be considered for inductor design.
1. Since magnetic flux density due to flow of current in a conductor is maximum near the conductor, hence closer the coil
to the part to be heated, greater will be the heating effect.
2. Eddy current in the parts to be heated flow in such a way so as to produce magnetic flux just to oppose the change in
magnetic flux of inductor coil. Contour of the coil has to be such as to produce the desired heating.
3. Due to skin effect, high frequency current in inductor coil usually flows through the outer portion. Thus, inductor coil is
made of copper tubing through which cooling water flows to remove the heat generated in it.
4. High concentration of heat at inaccessible zones can be achieved by single turn coils carrying heavy current. Electronic
heaters can load up to 300 A. Therefore for higher output, air core type transformers are used which are less efficient.
5. Lead wires from power supply to inductor coil should be kept as short and together as possible to avoid reducing power
factor. Even the turns of inductor coil should be kept close together and air-gap between the coil and work-piece should
be kept minimum to prevent low power factor.
6. In case of brazing, it is essential that joint should be first brought to correct temperature before brazing alloy melts and is
drawn into the joint.

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Q32. Mention the applications of induction heating.
Ans:
Applications of Induction Heating
The following are the various uses of induction heating.
1. Surface Hardening
The materials used for making parts such as spindle, saw blades, gears, axles etc., should be hard and tough to withstand
the wear which is possible with induction heating. It is possible to concentrate the heating effect to desirable portion.
2. Deep Hardening
With the help of induction heating, hardening of metal to any depth is possible and hence this type of heating is used for
deep hardening of articles such as screw driver, tools, drills, etc.
3. Tempering
In some mechanical processes, the work pieces become more hard than required and may need tempering to loose their
hardness. For tempering, accurate control of heat is required which is possible only with induction heating.
4. Melting
Induction heating at high frequency is preferred for extraction of metal from ore where the process is to be carried out in
some protective atmosphere or vacuum.
5. Soldering
For soldering it is essential that required amount of heat is to be developed at the soldering point whereas the remaining
portion of the solder may remain cold which can be achieved economically and efficiently by induction heating. With the
help of induction heating, it is possible to melt various metals in suitable furnaces.

Q33. Explain the working of HF power source for induction heating.

Ans:
The circuit diagram of high frequency power source for induction heating is, as shown in figure below.

Figure
The major components of the above circuit are step-up transformer, bridge rectifier, a filter circuit contain LC filter and a
colpitt’s oscillator. Step-up transformer is used in the circuit to step-up the A.C Voltage, coming from a single phase 230 V A.C
source. Bridge rectifier present in the circuit converts the A.C voltage into a high D.C voltage and output it to the filter circuit.
Here, ripple components present in D.C voltage are minimized and a pure D.C voltage is produced. This ripple free voltage,
i.e., pure D.C voltage, is given as an input to colpitt’s oscillator and a high frequency power is obtained. Because of this high
frequency high voltage supply, large amount of heat is generated, which in-turn decreases the overall conversion efficiency of
the system. To minimize the amount of heat generated, induction coils are designed like allow pipes and water is passed through
them to abosorb the excess heat.

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Unit-5 Industrial Applications 5.19

5.2.2 dielectric heating – principles, material properties, Electrodes and

their coupling to rf generator, thermal losses and applications

Q34. State the principle of dielectric heating.

Ans: Model Paper-III, Q11

Dielectric Heating
When nonmetallic materials i.e., insulators such as wood, plastic, china glass, ceramics etc., are subjected to high alternating
voltage, their temperature will increase after some time. This increase in temperature is due to the conversion of dielectric loss
to heat. The material to be heated is placed as slab between the metallic plates or electrodes connected to high frequency A.C
supply.
The electrodes used are selected and placed in such a manner that they fullfill the desired requirements and give the
optimum results.
Dielectric loss is dependent upon the frequency and high voltage therefore for obtaining adequate heating effect, high
voltage at about 20 kV and frequency of about 10-30 MHz are usually employed. High frequency is obtained from valve oscillator.

Figure
The current drawn by the capacitor when connected to an A.C supply voltage does not lead the supply voltage by exactly
90o since it is not possible to get a pure capacitor and there is always some resistance due to which heat is always produced in
the dielectric material placed in between the two plates of capacitor. The electric energy dissipated in the form of heat energy in
dielectric material is known as dielectric loss.
Q35. What are the merits and demerits of dielectric heating?
Ans:
Merits of Dielectric Heating
1. As the heat required is generated within the dielectric medium itself, this method of heating is useful in heating materials
with poor thermal conductivities.
2. In this dielectric heating, heat is produced throughout the material. Hence, uniform heating is obtained.
3. In this method, the materials are heated at a faster rate and with increase in frequency, the heating becomes more faster.
4. The materials heated by this method are combustible and they cannot be heated by flame.
5. This is the only method which is available for heating bad conductors of heat.
6. Dielectric heating can be stopped as and when required.
7. In this method, flame is not visible. Hence, inflammable articles such as plastics and wooden products can be heated safely.
Demerits of Dielectric Heating
1. This method requires high frequencies due to which cost of the equipment is quite high.
2. Due to requirement of high frequency, it is difficult to tune inductance in order to resonate with the charge capacitance.
3. At higher frequencies, it is difficult to obtain uniform distribution of voltage.
4. This method causes interference in communication apparatus such as radio station services at high frequencies. This
method should be employed only when other methods are impractical or slow.

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5.20 Industrial electronics [JNTU-Hyderabad]
Q36. Explain the electrodes used in dielectric heating and methods of coupling to the RF generator.
Ans:
Electrodes used in Dielectric Heating
For answer refer Unit-III, Q34.

In dielectric heating, the usage of electrodes depends on type and shape of material to be heated. For regular shape
materials, flat plate arrangement is employed to heat the material and for irregular materials we use special shaped electrodes
future depending on shape of that particular material. Generally, two plane parallel plates of adequate size with A.C voltage
applied across electrodes produce almost uniform electric field and high electrostatic field intensity in the region of material can
be achieved by applying high frequency-high magnitude voltage across the metallic electrodes. However, if the width (or)
diameter of dielectric sample is of an appreciable fraction of the applied voltage wavelength the voltage gradient becomes non-
uniform and the exterior region of the material will not get heated properly, further seeking compensation to correct field
distribution.
The voltage gradient, in pth layer is given as,

V n
dx
Vp = dTp / dx
x=1

Where,
VT = Total applied voltage in volts
VP = Voltage Gradient in pth layer in v/cm

Îp = Relative dielectric of pth layer

dx = Thickness of pth layer in cm

Îx = Relative dielectric constant of xth layer

Methods of Coupling to RF Generator
For answer refer Unit-III, Q37.

Q37. Explain any two methods of coupling of electrodes to the RF generator in dielectric heating.
Ans:
The commonly used methods for coupling the electrode to the tank circuit of the oscillator are,
(i) Electrode directly connected across the tank coil
(ii) Load across part of the tank coil
(iii) Auxiliary inductor in series with load
(iv) Load connected through the transmission line
(i) Electrode Directly Connected Across the Tank Coil

Figure (1) shows the circuit arrangement for the electrode connected directly across the tank coil.
Oscillator
tank

Dielectric with
electrodes

Figure (1)

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Unit-5 Industrial Applications 5.21
In this method, the load with the electrodes representing a capacitor becomes a part of the tank circuit capacitor and
appears in shunt with it. Hence, the load voltage equals the tank circuit voltage.

(ii) Load Across Part of the Tank Coil

Figure (2) shows the circuit arrangement for load across a part of the tank coil.
Oscillator
tank

Dielectric with
electrodes

Figure (2)

In this method, the load voltage is reduced to a desired value below the tank circuit voltage.

(iii) Auxiliary Inductor in Series with Load

Figure (3) shows the circuit arrangement for auxiliary inductor in series with load.

Auxiliary
Oscillator Inductor
tank
Dielectric with
electrodes

Figure (3)

In this method, the load itself represents a capacitor. This capacitor along with series inductor forms a series resonance
circuit. If the value of inductor is increased, the circuit becomes more and more resonant and the impedance value decreases
resulting in high voltage across the load i.e., above the oscillator voltage.

(iv) Load Connected through Transmission Lines

In this method, the load is connected away from R.F generator and the energy from the R.F generator is sent to the load
over the transmission line.

Impedance matching can be obtained with the help of reactive T or p sections and also with short and open circuited stubs,
if the required stub length is small.

Q38. What are the factors which decide the frequency and voltage of dielectric heating? Derive the expression
for the heat produced in the dielectric material.

Ans:

Heat produced in dielectric heating is given by,

P = 2πf CV 2 cos φ

= 2πf CV 2 δ
For any material, C and δ are constant. Thus, heating produced is proportional to V2 and f. But, there are various factors
which decide the frequency and voltage of the dielectric heating.

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Factors Affecting Frequency

v Impedance of work piece, which changes with change in frequency. Since for max power transfer source impedance must
be equal to load impedance.

v Mass of the work piece, as the optimum frequency decreases with increase in mass.

v Formation of standing waves, due to which the material will not be uniformly heated so, a suitable frequency must be
selected.

v Difficulty of constructing a tuning circuit.

Factors Deciding Voltage

v Insulation problem at high voltages.

v Personal safety, at high voltages breakdown may occur due to ionized air if air gap is left, hence voltage of 600 V to 3 kV are
more common.

Derivation for Heat Produced in Dielectric Heating

In dielectric heating, the material to be heated is placed as slab between the metallic plates or electrodes connected to a
high frequency A.C supply as shown in figure (1).

Figure (1): Parallel Plate Capacitor

Let,
f – Supply frequency in Hz
A – Surface area
t – Thickness of dielectric
C – Capacitance of the condenser.

The vector diagram is shown in figure (2),

Figure (2): Vector Diagram

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Unit-5 Industrial Applications 5.23
We have,

V V
IC = I = =
XC 1
2πfC

= 2πfCV
Power drawn from the supply,
P = VI cos φ
= V(2πfCV) cos φ
= 2πfCV2 cos φ
From the vector diagram, we have,
φ = 90º – δ
cosφ = cos(90°– δ)
= sin δ
= tan δ
=δ
Therefore, power drawn from supply, P = 2πfCV 2 δ watts.
Where, δ is very small and expressed in radians.
Here, C is the capacitance of the condenser given by,
∈o ∈r A
C=
t

Where,
∈o = Absolute permittivity = 8.854 × 10–12 F/m
∈r = Relative permittivity of the dielectric.
For any material, C and δ are constant. Thus, heating produced is proportional to (V2f ).
This is the reason for using high voltage at high frequency for dielectric heating.

Q39. Discuss the various sources of thermal losses in dielectric heating.

Ans:

During dielectric heating, the total power delivered to a dielectric material is,

Total power (Pt) = Ps + Pcs + Pcd + Pcv + Pr watts ... (1)

Where,

Ps – Specific heat power (or) thermal power

Pcs – Power required to transform the physical state of material.

Pcd – Conduction loss

Pcv – Convection loss

Pr – Radiation loss

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Various sources of thermal losses mentioned in equation (1) can be discussed as follows,

(i) Thermal Power or Specific Heat Power (PS)

For a given dielectric material, Ps is the power needed to increase its temperature to the desired final temperature

Specific heat power is,

Ps = 17.6 mc DT watts

Where, m – Proportion of material heated (expressed in pounds per minute)

C – Specified heat (in B.T.U pounds/degree Fahrenheit)
DT – Rise of temperature (in degree Fahrenheit)
(ii) Conduction Loss (Pcd)
The heat flow rate from the dielectric material to the electrodes is called conduction loss. This is proportional to the tem-
perature and hence this may not be stable during the whole interval of heating.
(iii) Convection loss (Pcv)
The formula for convection loss per sq inch is,
Pcv/sq.inch = 4.66 × 10–1 (DT)1.33 watts
Where,
DT – Rise of material temperature above room temperature in degree Fahrenheit.
This Pcv is considered for large areas only and hence this is minor for dielectric heating.
(iv) Radiation loss (Pr)
The radiation loss per square inch is,
RSJ 4V
SK T ON KJ T ON WW
4
Pr/sq.inch = 37! SSKK 2 OO – KK 1 OO WW watts
S 1000 W
TL P L 1000 P X
Where, T2 – Load temperature in degree Fahrenheit

T1 – Ambient temperature in degree Fahrenheit

d – Relative emissivity factor.

Q40. Mention the applications of dielectric heating.
Ans: Model Paper-II, Q11
Applications
The cost of the equipment required for dielectric heating is so high that it is employed where other methods are impracticable
and too slow. Some of the applications of this type of heating are,
1. Synthetics
The raw materials called plastic performs used for synthetics are required to be heated uniformly before putting them
in hot moulds so that the whole mass becomes fluid at a time, otherwise if the raw material is put directly in the moulds
usually heated by steam the outer surface of the perform will become hot and start curing while inner surface does not
reach the fluid temperature thereby resulting in unequal hardening of plastic. Moreover the plastic raw material once cured
cannot be softened again. Thus it is necessary to cure the plastic one at a time which is possible with dielectric heating as
the heat produced is uniform.
2. Diathermy
Dielectric heating is employed for heating tissues and bones of body required for the treatment of certain types of pains
and diseases.
3. Sterilization
Dielectric heating is quite suitable for sterilization of bandages, absorbent cotton, instruments etc.
4. Textile Industry
In textile industry, dielectric heating is employed for drying purposes.

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Unit-5 Industrial Applications 5.25
5. Baking of Foundary Cores
Dielectric heating is most suitable for baking foundary cores where thermo setting binders are employed as they set
instantaneously when brought to polymerizing temperature. The dielectric heating not only evaporates water rapidly from
the core but also raises the temperature of core to polymerization point.
6. Wood Processing Industries
Dielectric heating has been employed in wood industry for gluing of wooden sheets or boards because if animal glues are
used it requires long curing time as well as the parts to be joined are to be kept under mechanical pressure after application
of glue for nearly 24 hours whereas it is not necessary for dielectric heating. Thus dielectric heating is preferred for
obtaining curved wooden sections such as radio cabinets, furniture etc. The curves obtained by this process are stable and
are of long life.
7. Food Processing
The dielectric heating for food processing is one of the most modern method and set forth such processes which are outside
realm of cooking.
The dielectric heating is used for :
(a) Heating of general processing such as coffee roasting, chocolate industry etc.
(b) Cooking of sea foods such as oysters without removing the outer shell.
(c) Dehydration of fruits, milk, cream, vegetables (peas) and eggs.
(d) Defrosting of frozen foods such as meat and vegetables.
(e) For control of bacterial growth and production of germicidal reactions the food products are heated and to prevent
the product losing the flavour, they are dielectric heated.
Q41. The power required for dielectric heating of a slab of resin 150 sq cm in area and 2 cm thick is 200
watts at a frequency of 30 MHz. The material has relative permittivity of 5 and a pF 0.05. Determine the
voltage necessary and current flowing through the material. If the voltage is limited to 600 V. What will
be the value of the frequency to obtain the same heating?
Ans:
Given that,
Area of the slab, A = 150 sq.cm = 150 × 10–4m2
Thickness, t = 2 cm = 0.02 m
Power absorbed, P = 200 watts
Supply frequency, f = 30 MHz = 30 × 106 Hz
Relative permittivity, ∈r = 5
Power factor, cosφ = 0.05
Voltage required, V = ?
Current flowing through the material, I = ?
If the voltage obtained is limited to 600 V
Frequency, f2 = ?
To obtain the same heating.
The capacitance of the parallel plate condenser that the material forms is given by,
∈o ∈r A
C=
t
Where,
∈o = Absolute permittivity = 8.854 × 10–12 F/m
∈r = Relative permittivity
A = Area of the slab
t = Thickness.
Capacitance,
∈o ∈r A
C =
t

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5.26 Industrial electronics [JNTU-Hyderabad]

8.854 × 10 −12 × 5 × 150 × 10 −4

=
0.02
= 33.2 × 10–12 F = 33.2 pF

1
XC =
2πfC

1
=
2π(30 × 10 ) × 33 .2 × 10 −12
6

1
= = 160
6.25 × 10 −3

We have,

Power absorbed,

P = VI cos φ

 V 
= V  cos φ = V 22πfC cos φ
X 
 C
∴ Voltage required,

P
V2 =
2πfC cos φ

200
=
2π(30 × 10 ) (33.2 × 10 −12 ) × 0.05
6

= 639176.4783

Voltage required,

V = 631976.4783

∴ V = 799 .49 V

Current flowing through the material,

V 799.49
I = IC = = 160 = 4.99 ~ 5A
XC

∴ I = 5A

The voltage is limited to V2 = 600 V (given)

Frequency, f2 = ?

When less voltage is applied, we require more frequency.

We have,

Heat produced ∝ V 2 f

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Unit-5 Industrial Applications 5.27

V22 f 2 = V12 f1

⇒ (600)2 f2 = (799.49)2 × 30 × 106

(799.49) 2 (30 × 10 6 )
⇒ f2 =
(600) 2

= 53.27 × 106

∴ f 2 = 53.27 MHz

5.2.3 ultrasonics – generation and applications

Q42. Define ultrasonic waves. List its properties. Explain the various methods for generating ultrasonic
waves.

Ans:
Ultrasonic waves are defined as the sound waves whose frequency range is greater than the audible limit (i.e., 20 Hz to
20 kHz) of humans hearing. Because of higher operating frequencies, the energy of ultrasonic waves is very high, which inturn
suitable for long distance communication.

Properties of Ultrasonic Waves

1. The speed of propagation of ultrasonic waves increases with increase in frequency.

2. The wavelength of waves is very small and the waves exhibit negligible diffraction effects.

3. They can travel over long distances as a highly directional beam and without appreciable loss of energy.

4. They are highly energetic ultrasonic waves may have intensities upto 10 kW/m2. Generally, 1 to 2 kW/m2 intensities are
used.

5. They produce cavitation effect in liquids.

The various methods for generating ultrasonic waves are,

1. Mechanical generator or Galton whistle method.

2. Mechanical vibrator method.

3. Siren method.

4. Electromagnetic transducer method.

5. Magnetostriction method.

6. Piezoelectric method etc.

Among these the last two are the most important and commonly used methods.

Magnetostriction Method

The principle of magnetostriction effect is used in this method for production of ultrasonic waves. According to this
effect, a bar of ferromagnetic material like iron or nickel changes in its length when it is placed in a strong magnetic field applied
parallel to its length. A nickel rod placed in a rapidly varying magnetic field alternately expands and contracts with twice the
frequency of the applied magnetic field. This change in length of the ferromagnetic material in independant of the polarity of
applied magnetic field. The longitudinal expansion and contraction in ferromagnetic rod produces ultrasonic sound waves in the
medium surrounding the nickel rod. The range of frequencies generated depend on the mode of vibration of the ferromagnetic
material and may vary from the few hundred to 300 kHz.

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Piezoelectric Method
The principle of piezoelectric effect is used in this method for production of ultrasonic waves. When mechanical pressures
are applied to the opposite faces of certain crystal slices cut suitably, then equal and opposite electric charges are developed on
other faces resulting a difference of potential. The magnetic of potential difference so developed is proportional to the applied
pressure. This phenomena is called piezoelectric effect. The converse effect in also possible that is, if a potential difference is
applied to the opposite faces of a crystal. then it change in dimension i.e., a mechanical contraction or expansion in the other
faces would take place according to the direction of potential difference.

Crystals like Quartz Teurmaline Rocks salt will exhibit piezoelectric effect. When two opposite faces of a quartz crystal are
subjected to alternating voltage the other pairs of opposite faces can express stress and strain. The quartz crystal will continuously
expand and contract due to this elastic vibrations setup in the crystal. This expansion and construction produces ultrasonic sound
waves in the medium surrounding the quartz crystal. With quartz crystal ultrasonic waves, frequencies upto 540 kHz can be
produced.
Q43. List out the applications of ultrasonic waves.
Ans:
1. Ultrasonic waves are used to detect irregularities or discontinuities in metallic plates arised due to pressing or rolling.
2. They are used to identify obstructions or imperfections in concrete structure.
3. Detection of flaws in metals.
4. Detection of submarines, iceberg and other objects in ocean.
5. Depth of sea can be determined by the ultrasonic waves of high frequency. The time interval between the reflected wave
and incident wave on the sea is recorded. Finally the depth of the sea can be calculated by knowing the wave velocity.
6. In metallurgy they are used to the prevent the formation of cores and to release the trapped gases the ultrasonic waves are
used.
7. Abnormal growth in the brain, tumors which cannot be detected by X-rays can be detected by ultrasonic waves.

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