MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2013 Certified)
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SUMMER – 2023 EXAMINATION
Model Answer – Only for the Use of RAC Assessors
Subject Name: Thermal Engineering.
Subject Code:
22337
Important Instructions 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 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
Q.
N.
1.
Answer
Marking
Scheme
Attempt any Five of the following : (5x2=10)
10 Marks
2 Marks
a)
Define and Give one example of each i) Energy ii) Work
Ans
i) Energy : Energy is capacity to do the work and it is stored in the substance by means of
mechanical, electrical, chemical or internal energy.
½ Marks
e.g. Dry Battery stores energy by means of chemical energy flywheel stores mechanical for Each
Defination
energy.
&½
ii) Work : Work is said to be done when a force moves through a distance. OR Work done Marks for
each
is the product of the force and the distance it moves in the direction of force.
Exapmple
Work done = Force × Distance
e.g. piston moves under the action of force in the cylinder.
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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SUMMER – 2023 EXAMINATION
Model Answer
Subject Name: Thermal Engineering
Q. Sub
No. Q. N.
1.
b)
Subject Code:
Answer
22337
Marking
XXXXX
Scheme
Reprsent Isentropic process on P-V and T-S Diagram
2 Marks
Ans
1 Marks for
P-V
Diagram
and 1
Marks for
T-S Diagram
c)
A Sample of 10 kg of wet steam contains 0.5 kg of water , which is in suspension . Find its
Dryness fraction
Ans
1 Marks for
formula
1 Mark for
Answer
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1.
d)
What is the compounding of steam turbine ?.
2 Marks
Ans
Compounding of steam turbines is the method in which energy from the steam is
extracted in a number of stages rather than a single stage in a turbine.
2 Marks
A compounded steam turbine has multiple stages i.e. it has more than one set of nozzles
and rotors, in series, keyed to the shaft or fixed to the casing, so that either the steam
pressure or the jet velocity is absorbed by the turbine in number of stages.
e)
Write continuity equation of nozzle.
Ans
Flow energy Equation
2 Marks
For Nozzle
1 mark
1 mark
f)
Ans
Draw neat sketch of induced draught cooling tower
2 Marks
1 Mark for
drawing
and 1 Mark
for
Labelling
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1.
g)
Define and write unit of following i) Absolute Temperature ii) Heat
Ans
i) Absolute Temperature : Absolute temperature, also called thermodynamic
temperature, is the temperature of an object on a scale where 0 is taken as absolute zero.
Units are Kelvin and Rankine.
2 Marks
1 Mark for
ii) Heat : Heat is the form of energy that changes the temperature of any substance. Heat definition
involves the transfer of energy from an object or an energy source to another medium or and 1 Mark
for Unit
an object. Unit is Joules
12 Marks
Attempt any THREE of the following : (3X4=12)
2.
a)
Ans
Explain the first law of thermodynamic with an example
4 Marks
Definition : When a system undergoes a thermodynamic cycle, then the net heat supplied
to the system from the surroundings, is equal to net work done by the system on its
surroundings.
∮ dW = ∮ dQ
where, ∮ represents the sum for a complete cycle.
•
Heat and work are mutually convertible but the total energy remains constant as 2 Marks for
Explanation
per law of conservation of energy.
•
No machine can produce energy without corresponding expenditure of energy.
•
It is impossible to construct a perpetual motion machine of first kind.
e.g. Heat Engine. In a heat engine, the thermal energy is converted into mechanical 2 Marks for
Example
energy, and the process also is vice versa. Heat engines are mostly categorized as open
systems. The basic working principle of a heat engine is that it makes use of the different
relationships between heat, pressure and volume of a working fluid which is usually a gas.
Sometimes phase changes might also occur involving a gas to liquid and back to gas.
b)
Write statement of following laws and write equation of it. i) Ideal Gas law ii) Boyle’s
4 Marks
law iii) Charle’s Law
Ans
i) Ideal Gas Law
An ideal gas is defined as a gas having no forces of intermolecular attraction. The gases
which follow the gas laws at all ranges of pressures and temperatures are considered as
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‘ideal gases’.
PV = RT
where, P → Pressure of gas, V → Volume of gas, R → Constant ,T → Temperature of gas
ii) Boyle’s law
It states that volume of a given mass of a perfect gas varies inversely as the absolute
pressure when temperature is constant.
1 Marks for
each
Mathematically, If P is the absolute pressure of the gas and V is the volume occupied by
definition
the gas
and 1 Mark
for
equation
∴ PV = constant
If a gas changes its volume from V1 to V2 and pressure changes from P1 to P2 at constant
temperature P1V1 = P2 V2 = constant
iii) Charle’s law
Statement: If the pressure of the gas is maintained constant, during a process, the volume
of gas varies directly with the absolute temperature.
i. Mathematically
Pressure (P) = constant
V ∝ T (Volume ∝ Temperature)
∴ V/T = constant
If a gas changes its volume from V1 to V2 and absolute temperature from T1 to T2
without change in pressure, then
V1/T1 = V2/T2
2.
c)
4 Marks
Represent the following processes on P-V , T-S , H-S diagram
i) In a constant volume process A-B , Initial condition of steam is superheated and final
condition is wet
ii) In a constant volume process B-C , Initial condition of steam is wet and final condition
is liquid
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Ans
i) In a constant volume process A-B , Initial condition of steam is superheated and final
condition is wet
2 Marks
ii) In a constant volume process B-C , Initial condition of steam is wet and final condition
is liquid
2 Marks
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2.
d)
Ans
4 Marks
Explain the working of Babcock and Wilcox Boiler with neat sketch
2 Marks for
diagram
•
•
•
•
•
Babcock and Wilcox boiler: Babcock and Wilcox is a water-tube boiler is an
example of horizontal inclined tube boiler it also a High Pressure Boiler.
Working: The fire door the fuel is supplied to grate where it is burnt.
The hot gases are forced to move upwards between the tubes by baffle plates
provided.
2 marks for
The water from the drum flows through the inclined tubes via down take header working
and goes back into the shell in the form of water and steam via uptake header.
The steam gets collected in the steam space of the drum. The steam then enters
through the anti priming pipe and flows in the super heater tubes where it is
further heated and is finally taken out through the main stop valve and supplied to
the Steam turbine or Steam engine when needed.
12 Marks
Attempt Any THREE of the following. (3X4=12)
3.
Define Mach Number. How it affects the performance of steam nozzle.?
a)
•
It is the ratio of velocity of fluid to the sonic velocity of compressible fluid
M=V/a
Where,
2 Marks
M= Mach number
V=Velocity of fluid
a=Sonic velocity
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Significance of Mach number in performance of steam nozzle:
Depending on value of Mach number type of flow in nozzle is decided
3.
b)
•
M<1 subsonic nozzle, flow is called subsonic
•
M>1 supersonic nozzle, flow is called supersonic
•
M=1 subsonic and supersonic nozzle, flow is called sonic
2Marks
Explain Construction and working of Steam Turbine
2 Marks for
Sketch
Pressure velocity variation in Impulse turbine
Working of Impulse Turbine:
•
The impulse turbine consists of one set of Nozzle followed by one set of moving
blades as shown in above figure.
•
In Impulse turbine power is developed by impulsive force of high velocity steam jet
on moving blade
•
Steam from boiler enters in nozzle ring of impulse turbine where, high velocity jet
is obtained by expansion of steam in nozzle ring.
2 Marks for
explanation
High velocity steam jet then passed through moving blade ring with no pressure drop but
gradual reduction in velocity.
• Moving blades changes direction of steam jet thus the momentum of jet which
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rotates the shaft.
3.
c)
2 Marks
2 Marks
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3.
d)
2 Marks
2 Marks
4.
Attempt Any THREE of the following. (3X4=12)
a)
12 Marks
Define Vacuum. How it is necessary to operate condensers?
Vacuum: The term "vacuum" is used to describe the zone of pressure below
atmospheric pressure. Vacuum is a negative gauge pressure, usually referenced to the 2 Marks
existing standard barometric pressure where the equipment will operate.
Necessity of vacuum to operate condensers:
Primary Function Of condenser:
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•
•
The primary function of condenser to increase mechanical work done developed
by turbine in thermal power plant.
2 Marks
Due to vacuum in condenser, it can maintain low back pressure on exhaust side of
steam turbine, due to which steam is expanded to greater extent, which results in
increase in available energy for converting into mechanical work done.
4.
b)
4 Marks
(Schematic
representation
with details of
heat and work
interaction)
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4.
c)
2 Marks
2Marks
4.
d)
Explain Construction and working of Shell and Tube Type heat exchanger. Write its
different industrial applications.
•
Shell and tube heat exchanger is the most widely used heat exchanger and is among
the most effective means of heat exchange.
•
Shell and tube heat exchanger is a device where two working fluids exchange heat by
thermal contact using tubes housed within a cylindrical shell.
•
The fluid temperature inside the shell and tube are different and this temperature
difference is the driving force for temperature exchange.
•
Used for wide temperature and pressure ranges, Shell and tube heat exchangers are 2 Marks
compact in design, easy to construct and maintain and provide excellent heat
exchange.
Main Components of Shell And tube Type Heat Exchanger:
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1. Shell
2. Tube Bundle
3. Front and Rear Headers of Shell
4. Baffles
2 Marks
Industrial Applications of Shell and Tube type Heat Exchanger:
• Power Generation.
• HVAC.
• Marine Applications.
Refrigeration, Pharmaceuticals, Metals and Mining
4.
e)
Explain the Throttling Process with neat Sketch.
2 Marks
2 Marks
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5.
Attempt any TWO of the followings: (2X6=12)
a)
12 Marks
Describe the construction and working of i) Impulse turbine ii) Reaction turbine
Solution:
Impulse Turbine: It is commonly used in thermal power plants. In Impulse
turbine Nozzle is used to convert high pressure steam in to high velocity steam.
The construction of an Impulse turbine involve the following components:
1. Nozzle: The nozzle is a convergent-divergent type through which high-pressure steam
flows. It converts the pressure energy of the steam into kinetic energy.
2. Moving Blades: The high-velocity steam from the nozzle impacts the moving blades.
These blades are mounted on a wheel and they are cup shaped. The steam imparts a force
on the blades, causing them to rotate.
3. Stationary Blades: The stationary blades, also known as nozzle guide vanes, are
arranged in a circular casing around the moving blades.
4. Casing: The casing encloses the turbine and provides support to the blades.
3 Marks
Working Principle of Impulse Turbine:
1. High-pressure steam is admitted into the nozzle, where it undergoes a pressure drop
and accelerates to a high velocity.
2. The high-velocity steam jet strikes the moving blades of the turbine, causing them to
rotate.
3. As the steam impinges on the moving blades, its kinetic energy gets converted into
mechanical work. The steam's velocity decreases, and its pressure remains relatively
constant.
4. The rotation of the moving blades drives a shaft connected to a generator, converting
the mechanical work into electrical energy.
Reaction Turbine: The reaction turbine is another type of steam turbine used in thermal
power plants. In this pressure drop of steam takes place in blades.
The construction and working of a reaction turbine involve the following components:
1. Fixed Blades (Guide Vanes): The fixed blades, also known as guide vanes or nozzles, are
fixed in the turbine casing and serve to control the flow of steam. They are designed to
gradually expand the steam and convert its pressure energy into kinetic energy as it passes
3 Marks
through them.
2. Moving Blades (Rotor Blades): The moving blades, also called rotor blades, are attached
to a rotor or wheel. They are designed to extract the kinetic energy of the steam and
convert it into mechanical work.
3. Casing: The casing encloses the turbine and provides support to the blades.
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The working of a reaction turbine involves the following steps:
1. High-pressure steam enters the turbine through fixed blades or guide vanes. These
vanes gradually expand the steam and convert its pressure energy into kinetic energy.
2. The steam then passes through the moving blades or rotor blades, where it imparts a
force on the blades and causes them to rotate.
3. As the steam passes through the moving blades, both its kinetic energy and pressure
energy are converted into mechanical work. The steam's pressure and velocity decrease
gradually.
4. The rotation of the moving blades drives a shaft connected to a generator, converting
the mechanical work into electrical energy.
The initial condition of steam is 100% dry at 10 bar pressure. It expands to 1.00 bar by
PV= C.
find
i) find saturation temperature
ii) Quality of steam
5.
b)
Solution: Use steam table
Properties of steam at 10 bar pressure using Steam table
Temperature (T1) = 179.9 0C
Specific volume (V1) = 0.3118 m3/kg (superheated steam)
Now use equation P.V = C
Therefore
P1V1 = P2V2
10 * 0.3118 = 1 * V2
V2 = 3.118 m3/kg
2 Marks
Using steam table saturation temperature corresponding to 1 bar
Therefore Saturation Temperature T5 = 100 0C
2 Marks
To determine quality of steam at 1 bar
Specific volume of saturated liquid (Vf) = 0.001043 m3/kg
Specific volume of saturated vapour (Vg) = 1.6728 m3/kg
Now to find quality of Steam
X = (V2 – Vf ) / (Vg - Vf)
X = (3.118 – 0.001043) / (1.6728 – 0.001043)
X = 1.864
2 Marks
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5.
c)
An outer wall of office consists of 20 cm layer of brick. It is followed by 4 cm layer of
gypsum plaster and 6 cm of rockwool insulation. Estimate quantity of heat transfer
through wall.
Take thermal conductivity of brick = 0.7 W/(m.k.)
thermal conductivity of gypsum plaster = 0.5 W/(m.k.)
thermal conductivity of insulation = 0.065 W/(m.k.)
Solution: To estimate the quantity of heat transfer through the wall, we can use the concept of
thermal resistance.
The formula for thermal resistance is:
R=L/k
Let's calculate the thermal resistance for each layer of the wall:
Brick layer:
L1 = 20 cm = 0.2 m
k1 = 0.7 W/(m·K)
R1 = 0.2 / 0.7 = 0.2857 m²·K/W
2 Marks
Gypsum plaster layer:
L2 = 4 cm = 0.04 m
k2 = 0.5 W/(m·K)
R2 = 0.04 / 0.5 = 0.08 m²·K/W
Rockwool insulation layer:
L3 = 6 cm = 0.06 m
k3 = 0.065 W/(m·K)
R3 = 0.06 / 0.065 = 0.9231 m²·K/W
The total thermal resistance is the sum of the individual thermal resistances:
Total thermal resistance = R1 + R2 + R3 = 0.2857 + 0.08 + 0.9231 = 1.2888 m²·K/W
2 marks
Once we have the thermal resistance of the wall,
we can calculate the heat transfer (Q) through the wall using the formula:
Q = (ΔT) / Total Thermal resistance
Note: Since T1 and T2 temperatures are not given in question , students can assume any value
of T1 and T2. Accordingly marks can be awarded.
Let’s assume outside temperature of office 30 0C and inside Temperature of office 20 0C. (student
can assume suitable inside and outside temperature)
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T1 = 30+273 = 303 K and T2 = 20 + 273 = 293 K
Therefore Q = ( 303 – 293 ) / 1.2888
2 Marks
Q = 10 / 1.2888
Q = 7.76 W
12 Marks
Attempt any TWO of the followings: (2X6=12)
6.
a)
Write modes of heat transfer. Explain each with suitable example and neat sketch.
Solution: The three modes of heat transfer are conduction, convection, and radiation.
Conduction: Conduction is the transfer of heat through a solid or between objects in
direct physical contact. In this mode, heat is transferred by the collision of particles within
the material. The particles with higher energy (temperature) transfer their energy to
neighboring particles with lower energy.
2 marks
Example: Consider a metal rod being heated at one end. The heat conducted through the
rod will gradually increase the temperature of the other end. This can be observed when
one end of a metal rod is placed in a flame. The heat conducts through the rod, making
the other end hot as well.
Convection:
Convection is the transfer of heat through the movement of fluids (liquids or gases).
It involves the transfer of heat through the actual movement of the fluid, carrying thermal
energy from one location to another.
2 marks
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Example: A pot of water being heated on a stove demonstrates convection. As the water
at the bottom of the pot absorbs heat, it becomes less dense and rises to the top. This
creates a circular flow called convection current. The heated water near the bottom
transfers heat to the cooler water near the top, leading to the overall heating of the entire
pot.
Radiation: Radiation is the transfer of heat through electromagnetic waves. Unlike
conduction and convection, radiation does not require a medium for heat transfer. It can
occur in a vacuum or through transparent media.
2 Marks
Example: The sun heating the Earth through space is an example of radiation.
The sun emits electromagnetic waves, including infrared radiation, which travel through
the vacuum of space and reach the Earth. These waves are absorbed by the Earth's
surface, converting radiation into heat energy..
Describe construction and working of forced circulation cooling tower.
6.
b)
Solution: In a thermal power plant, a forced circulation cooling tower is an essential
component for efficient heat dissipation. It helps in cooling the circulating water that
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absorbs heat from the condenser, ensuring the thermal efficiency and proper functioning
of the power plant. The construction and working of a forced circulation cooling tower in a
thermal power plant involve the following components:
2 Marks
A forced circulation cooling tower in a thermal power plant consists of a Tower structure,
Cooling water basin, Cooling water pumps, Spray nozzles or Distribution system, Fill media
and Air circulation system with Fans or Blowers.
2 Marks
Working:
1.
2.
3.
4.
5.
6.
7.
8.
Cooling water is extracted from the condenser and circulated by pumps.
The water is distributed over the fill media using spray nozzles or a distribution
system.
As the water flows over the fill media, it comes into contact with the air drawn in by
fans or blowers.
Heat is transferred from the water to the air through evaporation, cooling the
water.
Warm, moisture-laden air is discharged from the tower by fans or blowers.
Cooled water collects in the basin and is recirculated to the condenser or heat
exchangers.
Control mechanisms regulate water flow, fan speed, and other parameters for
optimal cooling efficiency and temperature control.
The forced circulation cooling tower efficiently cools the circulating water,
maintaining the thermal efficiency and proper functioning of the thermal power
plant.
2 Marks
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6.
c)
Classify parallel flow and counter flow heat exchangers with respect to parameters
i) Definition, ii) Flow of fluids, iii) Capacity, iv) Maintenance, v) Applications, vi) Cost
Solution:
Sr.
No.
1
Parameters
Definition
2
Flow of Fluids
Parallel Flow Heat Exchangers
In a parallel flow heat
exchanger, both the hot and
cold fluids flow in the same
direction.
The hot and cold fluids flow in
the same direction, with the
hot fluid entering at one end
and the cold fluid entering at
the other end.
Parallel flow heat exchangers
generally have lower heat
transfer efficiency compared
to counter flow heat
exchangers.
3
Capacity
4
Maintenance
Parallel flow heat exchangers
are relatively easier to clean
and maintain compared to
counter flow heat exchangers.
5
Applications
They are used in some HVAC
systems, air conditioning
units, and low-temperature
heat transfer processes.
6
Cost
Parallel flow heat
exchangers are generally
less expensive to
manufacture and install
compared to counter flow
heat exchangers due to
their simpler design.
Counter Flow Heat
Exchangers
In a counter flow heat
exchanger, the hot and cold
fluids flow in opposite
directions.
The hot and cold fluids flow
in opposite directions, with
the hot fluid entering at one
end and the cold fluid
entering at the other end.
Counter flow heat
exchangers generally offer
higher heat transfer
efficiency compared to
parallel flow heat
exchangers.
Counter flow heat
exchangers can be more
challenging to clean and
maintain compared to
parallel flow heat
exchangers.
They are commonly found in
power plants, chemical
processes, refrigeration
systems, and hightemperature heat transfer
applications.
Counter flow heat
exchangers tend to be more
expensive to manufacture
and install compared to
parallel flow heat
exchangers. Their design
complexity.
1 Mark for
each point
----------------------------------------XXXX END XXXX---------------------------------------
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2022 SUMMER
21222
3 Hours / 70 Marks
22337
Seat No.
15 minutes extra for each hour
Instructions –
(1) All Questions are Compulsory.
(2) Answer each next main question on a new page.
(3) Illustrate your answers with neat sketches wherever
necessary.
(4) Figures to the right indicate full marks.
(5) Assume suitable data, if necessary.
(6) Use of Non-programmable Electronic Pocket
Calculator is permissible.
(7) Use of Steam tables, logarithmic, Mollier's chart is
permitted.
(8) Mobile Phone, Pager and any other Electronic
Communication devices are not permissible in
Examination Hall.
Marks
1. Attempt any FIVE of the following
10
a) List the factors making the process irreversible.
b) List any two assumptions for ideal gas.
c) State the function of
i) Superheater
ii) Blow off cock
d) Recite the significance of Mach number.
e) Define critical pressure in nozzle.
f)
State the functions of condenser.
g) Define thermal conductivity and State its unit.
P.T.O.
22337
[2]
Marks
2. Attempt any THREE of the following
12
a) Write steady flow energy equation and apply it to turbine
and condenser.
b) Differentiate between adiabatic and isothermal process.
(Four points)
c) Draw P-V and T-S diagram of Rankine cycle and list the
processes involved in it.
d) Determine if the steam is wet or superheated and calculate
the dryness fraction or the superheated steam temperature
for P = 8 bar and V = 0.28 m3/kg.
3. Attempt any THREE of the following
12
a) Describe with neat sketch nozzle control governing.
b) Classify turbines in details.
c) One Kg of air initially at 1 bar and 156°C is compressed
isothermally till the volume is reduced to 0.28 m3. Determine
the work done and change in internal energy.
d) Recite the steps involved in energy conservation of boilers.
4. Attempt any THREE of the following
a) The partial absolute pressure in the condenser is 11.56 KPa
when the barometer reads 1 bar. The condenser temperature is
40°C. Calculate partial pressure of air and vacuum efficiency.
b) 0.44 Kg of gas having a volume 0.28 m3 and a pressure of
1.4 bar is compressed to a pressure of 14 bar according to
pv1.3 = C. Find the change of internal energy.
Cp = 1.041 KJ/Kg-k
Cv = 0.743 KJ/Kg-k
c) A quantity of gas occupying 0.14 m3 at a pressure of 1400 KPa
and 300°C is expanded isentropically to 280 KPa calculate
i) Final temperature and
ii) Work transfer
12
22337
[3]
Marks
d) Define natural and forced convection and give two examples of
each.
e) Classify condensers in details.
5. Attempt any TWO of the following
12
a) Explain with neat sketch bleeding of steam. State its
advantages.
b) Explain with neat sketch automotive heat exchanger.
c) Draw a schematic diagram of Mollier chart and list its
features.
6. Attempt any TWO of the following
a) A refrigerator is loaded with fresh food and door is closed.
After some period, machine consumes 1.25 KWh of
electrical energy and internal energy of food items decreases
by 4500 KJ. Calculate the magnitude and direction of heat
transfer for the steam.
b) State the purpose of cooling tower and describe with neat
sketch natural draught cooling tower.
c) Sheets of brass and steel each 10 mm thick are placed in
contact. The outer surface of brass is kept at 100°C and outer
surface of steel is kept at 0°C. Estimate the temperature of
common interface if thermal conductivities of brass and
steel are in the ratio of 2:1.
12
WINTER 2022
12223
3 Hours / 70 Marks
Instructions –
22337
Seat No.
(1) All Questions are Compulsory.
(2) Answer each next main Question on a new page.
(3) Illustrate your answer with neat sketches wherever
necessary.
(4) Figures to the right indicate full marks.
(5) Assume suitable data, if necessary.
(6) Use of Non-programmable Electronic Pocket
Calculator is permissible.
(7) Mobile Phone, Pager and any other Electronic
Communication devices are not permissible in
Examination Hall.
(8) Use of Steam tables, logarithmic, Mollier’s chart is
permitted.
Marks
1. Attempt any FIVE of the following:
10
a) Define Gray body.
b) Represent Isochoric process on P-V and T-S chart.
c) State the function of :–
(i)
Economiser
(ii)
Fusible plug.
d) List four applications of nozzle.
e) What is the necessity of compounding of steam turbine?
f)
State Dalton’s Law of partial pressure.
g) State Fourier’s Law of heat conduction.
P.T.O.
22337
[2]
Marks
2. Attempt any THREE of the following:
12
a) State extensive property and intensive property with two
examples of each.
b) 2 kg of gas at 50°C is heated at constant volume until the
pressure is doubled.
Determine:
(i) Final temperature.
(ii) Change in internal energy.
Take Cv = 0.718 kJ/kgK.
c) State the main features of Indian Boiler Regulation. (IBR)
d) Explain the working of Cochran boiler with neat sketch.
3. Attempt any THREE of the following:
12
a) Define and state significance of Mach number.
b) Explain the working of impulse steam turbine with neat sketch.
Also show pressure and velocity variation for the same.
c) A gas at 7 bar and 400 k occupies a volume of 0.2m3.
The gas expands according to the law PV1.5 = C upto
pressure of 1.5 bar. Determine work transfer.
d) Wet steam at 10 bar pressure having total volume of 0.125 m3
and enthalpy content is 1800 kJ. Calculate mass and dryness
fraction of steam.
4. Attempt any THREE of the following:
a) Differentiate between natural draught and forced draught
cooling tower.
b) A quantity
and 300°C
Calculate.
(i) Final
(ii) Work
of gas occupying 0.14 m3 at a pressure of 1400KPa
is expanded isentropically to 280 KPa.
temperature and
transfer.
c) A balloon is considered to be sphere of 10m diameter. The
balloon is filled with hydrogen at 25°C and at atmospheric
pressure. The temperature of the surrounding air is 20°C and
at atmospheric pressure. Determine the load which can be lifted
by the balloon.
12
22337
d) State
(i)
(ii)
(iii)
(iv)
[3]
Marks
:–
Fourier’s Law.
Thermal conductivity.
Newton’s Law of cooling.
Radiation.
e) Draw a neat sketch of surface condenser and label it.
5. Attempt any TWO of the following:
12
a) Explain with neat sketch Regenerative feed heating. State its
advantages.
b) A refrigerator wall is constructed with two metallic plates
2mm. thick with 5cm. of glass wool insulation between them.
Find the heat transfer per m2 of area if inner and outer surface
temperature are –10°C and 40°C. Assume thermal conductivity
of metallic plates and glass wool are 50 W/mk and
0.1 W/mk respectively.
c) Draw and explain temperature-Entropy diagram for formation
of steam and show the following on it :–
(i) Saturated liquid line
(ii) Wet region
(iii) Critical point
(iv) Dryness fraction lines.
6. Attempt any TWO of the following:
12
a) (i)Explain the application of second law of thermodynamics
to heat engine.
(ii) Prove that, (C.O.P.) Heat pump = 1 + (C.O.P.) Refrigeration
b) (i) State the sources of air leakage in
(ii)Vacuum gauge on condenser reads
barometer reads 759 mm. of ‘Hg’.
standard barometer of 760 mm. of
condenser.
700 mm. of ‘Hg’ when
Correct the vacuum to
‘Hg.’
c) Suggest the type of heat exchangers for following applications:–
(i) Dairy plant. (Milk chilling plant)
(ii) Condenser of refrigeration system (Household system).
Justify your answers.
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
WINTER – 2022 EXAMINATION
Model Answer
Subject Name: Thermal Engineering
Subject Code:
22337
Important Instructions 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 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 in first year (first and second semesters) 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
Q. N.
1.
Answer
Marking
Scheme
Attempt any FIVE of the following:
10
a)
Define Gray Body.
Sol.
Gray body :- A grey body is defined as a body with constant emissivity over all wavelengths
and temperatures. It absorbs a definite percentage of incident energy irrespective of their
wavelengths.
2 Marks
for Def.
Represent Isochoric Process on P-V and T-S chart.
b)
Sol.
1 Marks
for each
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
WINTER – 2022 EXAMINATION
Model Answer
Subject Name: Thermal Engineering
Q.
No.
Sub
Q. N.
1
c)
Subject Code:
22337
Answer
Marking
Scheme
State Function of :i) Economizer
ii) Fusible plug
Sol.
(i) Economizer :- Function of economizers in steam power plants is to capture the waste 1 Marks
heat from boiler flue gases and transfer it to the boiler feed water. This raises the
temperature of the boiler feed water, lowering the needed energy input, in turn increase in
boiler efficiency.
(ii) Fusible plug-The function of the fusible plug is to put-off the fire in the furnace of the
1 Marks
boiler when the water level falls below an unsafe level and thus avoids the explosion which
may take place due to overhearing of the tubes and the shell.
d)
List Four Applications of nozzle
1) In flow measurement to measure discharge
2) Steam and gas turbine
3) Jet engines
4) Rocket motors
5) In flow measurement
6) In water sprinklers
7) In injectors for removing air from condensers.
e)
Sol.
½ Marks
for each
(any four
)
What is the necessity of compounding of steam turbine?






The compounding of steam turbine means the methods to reduce the speed of rotor
shaft.
To increase the thermal efficiency in power plants, high pressure and high temp.
steam is used.
If the entire pressure drop (from boiler pressure to condenser pressure)is carried
out one stage only.
Then the velocity of steam entering into the turbine will be extremely high.
This will make the rotor to run at a very high speed, which is not useful from
practical point of view.
Hence it becomes necessary to reduce the rotor speed of turbine by gearing or no.
of stages.
Any 2
points
01 mark
each
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MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
WINTER – 2022 EXAMINATION
Model Answer
Subject Name: Thermal Engineering
Q.
No
.
Su
b
Q.
N.
1.
f)
Subject Code:
22337
Answer
Marking
Scheme
State Dalton's law of partial pressure.
It states that’ “The pressure exerted by mixture of air and steam is equal to sum of partial
pressures, which each constitute would exert, if it occupies the same volume”.
1 Marks
In condenser total pressure is the sum of partial pressure of steam and air.
Mathematically,
Pc= Pa + Ps
1 Marks
Where;
Pc = Pressure in condenser containing mixture of air and steam
Pa = Partial pressure of air
Ps = Partial pressure of steam
g)
Sol
.
2.
State Fourier’s law of heat conduction
The law state that for homogeneous material the rate of heat transfer in steady state in any
direction is directly proportional to temperature gradient in that direction.
Q/A α dt/dx
Q/A = -k dt/dx
Where, Q/A is rate of heat transfer
dt/dx is temperature gradient
k conductivity of medium
Sol
.
1 Marks
12
Attempt any THREE of the following:
a)
1 Marks
State Extensive property and Intensive property with two example of each .
(i) Extensive Property:
It is defined as the property which depends upon the mass of the system.
Or
Extensive properties are those whose values are dependent of the mass possessed by the
system, such as volume, enthalpy, and entropy.
Ex. Total volume, Area, Enthalpy, Entropy etc.
(ii)Intensive Property:
It is defined as the property which is does not depend upon the mass of the system.
Or
2 Marks
for Each
Def.
2 marks
for
Exampl
e of
each
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__________________________________________________________________________________________________
Intensive properties are those whose values are independent of the mass possessed by the
system.
Ex. Pressure, Temperature, Density, Specific volume, specific Enthalpy, etc.
b)
2.
2 kg of gas at 50oc is heated at constant volume until the pressure is doubled. Determine
i)
Final Temp. ii) Change in internal energy. Take Cv = 0.718 KJ/Kgk.
An
2 Marks
for Final
Temp.
2 Marks
for
Change
in
Internal
Energy
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__________________________________________________________________________________________________
2.
c)
State the Main Features of Indian Boiler Regulation (IBR)
Sol
.
1. A boiler cannot be put to use unless it has been registered with the Chief Inspector of
Boilers.
2. The maximum working pressure of the boiler has to be determined by Boiler Inspector
who will issue certificate for this. Owner cannot exceed this pressure limit in any case.
3. In case of accident, it should be reported by owner within 24 hours with full details.
4. The rules, regulations and bye-laws governing the upkeep and maintenance of boilers,
procedure of registration, inspection and certification of maximum pressure, safety
conditions etc. are subject to a revision by a Central Board under control of Govt. of India.
Any
Four
Feature
s
1 Mark
for each
5. The boiler house plan, chimney design (Max height 30.48 m from floor) should be
approved by boiler inspector.
6. Owner should apply for registration in prescribed format, inspector should fix date of
inspection within 30 days, conduct inspection/examination of boiler, Issue the certificate of
registration not exceeding 12 months period.
7. Following inspections are carried out by Boiler Inspector at various stages/ levels /needInspection for registration, Hydraulic test, steam test, annual inspection, Inspection under
steam, Internal inspection, Accident inspection, Casual inspection
8. Violation of law is liable to prosecution and punishment with fine.
d)
Sol
.
Explain the working of Cochran boiler with neat sketch
The Cochran boiler is vertical, multi-tube boiler generally used for small capacity steam
generation. Cochran boilers are made in different sizes of evaporative capacities ranging
starting from 150 to 3000 kg/hr. and working pressure up to 15 bar.
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__________________________________________________________________________________________________
2 Marks
for fig.
Working:





First, the coal is fed to grate via a fire hole for a burn.
The ash formed in burning is collected in ash-pit below the grate and it removes
manually.
The hot gases from the grate pass through the combustion chamber to horizontal fire
tubes and transfer the heat to water by convection.
Exhaust gases out from fire tubes pass through smokebox and exhaust to the
atmosphere via a chimney.
There is a door in the smokebox for cleaning the fire tunes and smokebox. The
Cochran boiler has a working pressure of 6.5 bar and a steam capacity of 3500 kg/hr.
Attempt Any THREE of the following.
3
a
Sol
2 Marks
for
working
12
Define and state significance of Mach number.
•
It is the ratio of velocity of fluid to the sonic velocity of compressible fluid
M=V/a
Where,
Define
02
Marks
M= Mach number
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__________________________________________________________________________________________________
V=Velocity of fluid
a=Sonic velocity
Significance of Mach number:
If
b
02
•
M<1 subsonic nozzle
•
M>1 supersonic nozzle
•
M=1 subsonic and supersonic nozzle
Working of Impulse Turbine:
02
Pressure velocity variation in Impulse turbine
02
Working of Impulse Turbine:

The impulse turbine consists of one set of Nozzle followed by one set of moving blades
as shown in above figure.

In Impulse turbine power is developed by impulsive force of high velocity steam jet on
moving blade
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__________________________________________________________________________________________________

Steam from boiler enters in nozzle ring of impulse turbine where, high velocity jet is
obtained by expansion of steam in nozzle ring.

Moving blades changes direction of steam jet thus the momentum of jet which rotates
the shaft.

High velocity steam jet then passed through moving blade ring with no pressure drop
but gradual reduction in velocity.
c
02
02
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__________________________________________________________________________________________________
d
02
02
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__________________________________________________________________________________________________
4
Attempt Any THREE of the following.
a
12
Differentiate between natural draught and forced draught cooling tower.
Natural Draught cooling
tower
Forced draught cooling
tower
1. Circulation of air is provided by
pressure difference of air inside
cooling tower
1.for circulation of air forced draught
fan provided.
2.cooling capacity is less
2.cooling capacity is more
3.Operating cost is less
3. Operating cost is more
4. Maintenance Cost is less
4.Maintenance cost is more
5.Space Requirement is more
5.Space requirement is less
6. It is generally hyperbolic in shape
6.It is rectangular in shape.
Any
four
point
01 mark
each
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4
b
02
02
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c
01
01
02
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4
d
i)Fourier’s law:
“ Heat Transfer Rate per unit area is proportional to normal temperature gradient.”
01
ii)Thermal Conductivity:
Thermal conductivity of material is define as, “the amount of energy conduct through a body of
unit area and unit thickness in unit time when the difference in temperature between the face
causing heat flow is unit temperature difference.” K=Thermal conductivity.
01
State
i) Newton’s Law of cooling:
“The rate of cooling of a body is directly proportional to the difference in
temperature of the body (T) and surrounding (To), provided difference in temperature should
not be exceed by 30 0c.”
01
ii) Radiation :
“It is process of heat transfer between two bodies without any carrying medium
through different kind of electro-magnetic wave.”
01
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e
Draw Neat sketch of surface condenser and label it.
03
marks
for
sketch
and 01
mark
for
labellin
g
Attempt ant TWO of the following:
5
a
12
Regenerative feed heating
The process of draining steam from turbine at certain points during its expansion and using this steam for
heating feed water supplied to boiler is known as regenerative feed heating.
02
Marks
02
Marks
For Fig.
Fig. Regenerative feed heating
Advantages:
1. It increases the thermal efficiency of plant.
2. The temperature stresses in the boiler are reduced due to decreased range of working temperature.
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02
Marks
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__________________________________________________________________________________________________
b
Given data:
02
Marks
Figure
LA= 2 mm = 0.002 m
LB= 5 cm =0.05 m
LC=2 mm =0.002 m
KA = KC = 50 W/m0k
KB = 0.1 W/m0k
T1= -10 0C = -10 +273 = 263 0k
T4= 40 0C = 40 + 273 = 313 0k
01
Mark
for data
03
Marks
For cal.
c
Temperature and Entropy diagram:
04
Mark
for dia.
02
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A T-S diagram is the type of diagram most frequently used to analyze energy transfer system Marks
cycles. This is because the work done by or on the system and the heat added to or removed from For
the system can be visualized on the T-S diagram. By the definition of entropy, the heat expl.
transferred to or from a system equals the area under the T-S curve of the process. Figure is the
T-S diagram for pure water. A T-S diagram can be constructed for any pure substance. In the
liquid-vapor region in figure, water and steam exits together.
12
Attempt any TWO of the following:
6
ai
Heat Engine:
Heat Source
QA
E
1.5
W= QA-QR
QR
Heat Sink
Fig. Heat Engine
In heat engine, heat is extract from the high thermal reservoir or heat source same part of heat is
converted into work and remaining heat rejected to thermal reservoir or heat sink.
The performance of heat engine is measured in terms of efficiency.
So,
1.5
The efficiency of heat engine is always less than 1. It means that heat engine is not 100%
efficient.
-----------------------------------------------------------------------------------------a ii
Prove that, (C.O.P.) Heat pump = 1 + (C.O.P.) refrigeration.
(C.O.P.) Heat pump =
03
(C.O.P.) Refrigeration =
(C.O.P.) Heat pump + 1 =
+1
=
=
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=
bi
(C.O.P.) Heat pump + 1 = (C.O.P.)Refrigeration
T
The main sources of air leakage in condenser are given below:
1)
There is leakage of air from atmosphere at the joint of the parts which are internally under
a pressure less than atmospheric pressure.
2)
Air is also accompanied with steam from the boiler into which it enters dissolved in feed
water.
3)
03
In jet condensers, a little quantity of air accompanies the injection water.
b ii
03
Ci
Type of Heat Exchanger for following applications:
i.
C ii
Dairy Plant (Milk Chilling Plant): Plate Type Heat Exchanger
Because,
1. It is made up of aluminum alloy which provides higher rate of heat transfer.
2. Due to larger surface area, it has more heat transfer as compare to other heat exchanger
which is useful for dairy plant.
3. It is lighter in weight.
ii.
Condenser of Refrigeration System (Household system): Counter Flow tube type heat
Exchanger
Because,
1. High performance due to large surface area.
2. Compact and light in weight.
3. In tubes generally turbulent flow is develop which reduces scale deposition.
Less installation and maintenance cost.
03
03
END
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__________________________________________________________________________________________________
Page No: ____/ N
2019 WINTER
11920
3 Hours / 70 Marks
Instructions –
22337
Seat No.
(1) All Questions are Compulsory.
(2) Answer each next main Question on a new page.
(3) Illustrate your answers with neat sketches wherever
necessary.
(4) Figures to the right indicate full marks.
(5) Mobile Phone, Pager and any other Electronic
Communication devices are not permissible in
Examination Hall.
Marks
1. Attempt any FIVE of the following:
10
a) Define (i)
Intensive property
(ii)
Extensive property. Give one example of each.
b) Represent Isochoric Process on P-V and T-S chart.
c) A sample of 35 Kg of dry steam contains 0.7 Kg of water
is in suspension, find its dryness fraction.
d) Suggest the different methods to control the speed of rotation
of steam turbine constant at all varying loads.
e) Explain the functions of steam nozzle.
f)
Write the elements of forced draught cooling tower.
g) Define (i)
Thermal conductivity
(ii)
Thermal resistance
P.T.O.
22337
[2]
Marks
2. Attempt any THREE of the following:
12
a) Explain the concept of flow work associated with flow processes.
b) Two leg of gas contained in cylinder at a pressure of 7 bar
and temperature 27°C expands four times its original volume
at constant pressure. Calculate (i)
Work done by gas
(ii)
Heat added
c) In a constant pressure vapour process, the initial condition
of steam is wet and final condition is superheated. Represent
the process on P–V, T–S, and H–S chart.
d) Explain the working of Lamont boiler with neat sketch.
3. Attempt any THREE of the following:
a) Write the criteria for selection of nozzle for given situation.
b) Explain the need of compounding. Suggest the method of
compounding for reaction steam turbine with justification.
c) A nitrogen gas is expanded from 8 bar to 1 bar at 47°C
according to law PV = C. Plot the process on P–V and T–S
diagram and state the formulas to be used to find out workdone,
Amount of heat supplied and change in entropy.
d) Determine the amount of heat required to convert 2 Kg of
water at 25°C into steam at 5 bar and having 90% dry.
12
22337
[3]
4. Attempt any THREE of the following:
Marks
12
a) Explain Dalton’s law of partial pressure. How it is applicable
to condenser?
b) A system is composed of a gas contained in a cylinder
fitted with a piston. The gas expands from the state 1 for
which Internal energy U1 = 75 KJ to state 2 for which
U2 = –25 KJ. During the expansion the gas does 60 KJ of
work on the surrounding. Determine the heat transferred to
or from the system during the process.
c) 3 m3 of gas of 30°C and 6 bar pressure is expanded
isothermally to 1 bar. Find work done, change in internal
energy and heat transferred during the process.
d) Explain construction and working of shell and tube type
heat exchanger. A ice plant producing 2000 Kg ice per day
required the condenser. Suggest the type of condenser with
justification.
5. Attempt any TWO of the following:
12
a) (i)Suggest the methods to improve the performance of
steam turbine. Explain any one in brief.
(ii) Identity the different losses occoured in steam turbine.
b) An exterior wall of house consists 10.6 cm layer of common
brick. It is followed by 3.8 cm layer of gypsum plaster and
5.83 cm of rockwool insulation. Estimate the amount of heat
transferred through structure it.
Thermal conductivity of brick = 0.7 W/mk
Thermal conductivity of Plaster = 0.48 W/mk
Thermal conductivity of Insulation = 0.065 W/mk
c) The initial condition of steam is 15% wet at a pressure of 7 bar.
It expands to 1.2 bar by PV1-3 = C. Find
(i) Quality of steam at the end of expansion
(ii) Work done.
P.T.O.
22337
[4]
Marks
6. Attempt any TWO of the following:
a) A mass of 0.8 Kg of air at 1 bar and 25°C is contained in a
gas tight frictionless piston cylinder device. The air is now
compressed to a final pressure of 5 bar. During this process
the heat is transferred from air such that the temperature inside
the cylinder remains constant. Calculate the heat transferred and
work done during process and direction of each in the process.
b) For steam power plant having capacity 600 MW capacity a
cooling tower is required to set up with condenser. Suggest
the type of condenser and cooling tower with justification.
c) Suggest the type of heat exchangers for following applications (i)
Dairy plant (Milk Chilling Plant)
(ii)
Condenser of refrigeration system (House hold system)
Justify your answers.
12
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)
__________________________________________________________________________________________________
WINTER – 19 EXAMINATION
Subject Name: Thermal Engineering
Model Answer
Subject Code:
22337
Important Instructions 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 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.
Q.1.
Attempt any FIVE of the following:
10
Marks
a)
Define(i) Intensive property
(ii) Extensive property. Give one example of each.
(i)
Sol.Intensive Property:
01 mark
It is defined as the property which is does not depend upon the mass of the system.
Or
Intensive properties are those whose values are independent of the mass possessed by the
system.
Ex. Pressure, Temperature, Density, Specific volume, specific Enthalpy, etc.
(ii) Extensive Property:
It is defined as the property which depends upon the mass of the system.
01 mark
Or
Extensive properties are those whose values are dependent of the mass possessed by the
system, such as volume, enthalpy, and entropy.
Extensive properties are denoted by uppercase letters, such as volume (V), enthalpy (H) and
entropy (S).
Per unit mass of extensive properties are called specific properties and denoted by lowercase
letters. For example, specific volume v = V/m, specific enthalpy h = H/m and specific entropy
s = S/m
Ex. Total volume, Area, Enthalpy, Entropy etc.
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b)
Sol.
Represent Isochoric Process on P-V and T-S chart.
01 mark
each
c)
Figure: P-V and T-S representation of Isochoric process
A sample of 35 Kg of dry steam contains 0.7 Kg of water is in suspension,
find its dryness fraction.
Sol.
Mass of dry steam=35 kg
Mass water suspension=0.7 kg
Weight of wet steam=35+0.7=35.7 kg
So,
Dryness fraction X=Actual mass of dry steam/ weight of wet steam
01 mark
Formula
01 mark
= 35 / (35+0.7)
=0.098039
d)
Sol.
Suggest the different methods to control the speed of rotation of steam turbine
constant at all varying loads.
Following are the different methods to control the speed of rotation of steam turbine
constant at all varying loads;
a) Throttle governing
½ mark
each
b) Nozzle control governing
c) By pass governing
d) Combine throttle and nozzle control governing
e) Combine throttle and by pass governing
e)
Sol.
f)
Sol.
Explain the functions of steam nozzle.
The steam nozzle is a passage of varying cross section by means of which the thermal energy
of steam is converted into kinetic energy. When steam f lows through a nozzles expansion
process take place.(Only function is expected and not in detail working)
Write the elements of forced draught cooling tower.
Following are the elements of forced draught cooling tower;
a) Forced draught fan
b) Eliminator
c) Spray header
d) Spray nozzle
2 marks
½ mark
each
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e) Circulating pump
g)
Define-
Sol.
(i) Thermal conductivity
(ii) Thermal resistance
Thermal conductivity of material is define as ,”the amount of energy conduct through a body
of unit area and unit thickness in unit time when the difference in temperature between the
face causing heat flow is unit temperature difference.”
K=Thermal conductivity.
01 mark
01 mark
Thermal resistance is a property of a heat and measured by a temperature difference of a
substance resist heat flow.
Q.2.
a)
Sol.
Attempt any THREE of the following:
Explain the concept of flow work associated with flow processes.
A control volume may involve one or more forms of work at the same time. ... Work is
needed to push the fluid into or out of the boundaries of a control volume if mass flow is
involved. This work is called the flow work (flow energy). Flow work is necessary for
maintaining a continuous flow through a control volume.
12
Marks
02 marks
02 marks
b)
Two leg of gas
27°C expands
(i)
(ii)
contained in cylinder at a pressure of 7 bar and temperature
four times its original volume at constant pressure. CalculateWork done by gas
Heat added
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Sol.
Note:
1) Printing mistake: instead of mass(Kg), leg is printed
2) Values of CP, R are not mentioned.
(If student assume a data and solve the numerical with correct procedure then
give appropriate marks)
01 mark
01 mark
01 mark
01mark
c)
In a constant pressure vapour process, the initial condition of steam is wet
and final condition is superheated. Represent the process on P-V, T-S, and HS chart.
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Sol.
03 marks
For
charts
01 mark
for labels
Figure: P-V, T-S, and H-S chart
d)
Explain the working of Lamont boiler with neat sketch.
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Sol.
02 marks
Figure: Lamont boiler
1. This is a modern high pressure, water tube boiler working on a forced circulation.
2. The circulation is maintained by a centrifugal pump, driven by a steam turbine, using
steam from the boiler.
3. Feed water is supplied to economiser from hot well which is passed to separating and
storing drum.
4. Water from separating and storing drum, flows by gravity to circulating pump.
5. Circulating pump circulates the water to set of tubes known as convective evaporator
and then radiant evaporator.
02 marks
6. By the time, water leaves the radiant evaporator, it converts into steam.
7. This steam is passed through storage and separator drum.
8. From separator and storage drum steam is fed to super heater to superheat.
9. The superheated steam is passed to main stream to supply for required application.
Lamont boilers generates 45 to 50 tones steam per hour at 130 bar with 5000 C.
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Q.3.
a)
Sol.
Attempt any THREE of the following:
Write the criteria for selection of nozzle for given situation.
Following are the situation for selection criteria of nozzle.
Situation first:
It is used when the back pressure is equal or more than the critical pressure ratio. It is also used
for non – compressible fluids.
Convergent nozzle: Cross sectional area is decreases continuously from entrance to exit.
02 marks
Situation second:
When back pressure is less than critical pressure divergent nozzle is used.
Divergent nozzle: Cross sectional area is increases continuously from entrance to exit.
Situation third:
When back pressure is less than critical pressure convergent divergent nozzle is used.
Convergent and Divergent nozzle: Cross sectional area of nozzle first continuously decreases
and then increases from entrance to exit.
b)
Sol.
c)
12
Marks
02 marks
Explain the need of compounding. Suggest the method of compounding for reaction
steam turbine with justification.
Need of compounding:
✔ The compounding of steam turbine means the methods to reduce the speed of rotor 02 marks
shaft.
✔ To increase the thermal efficiency in power plants, high pressure and high temp. steam
is used.
✔ If the entire pressure drop (from boiler pressure to condenser pressure)is carried out
one stage only.
✔ Then the velocity of steam entering into the turbine will be extremely high.
✔ This will make the rotor to run at a very high speed, which is not useful from practical
point of view.
02 marks
✔ Hence it becomes necessary to reduce the rotor speed of turbine by gearing or no. of
stages.
Following are the methods of compounding:
i.
Velocity compounding
ii.
Pressure compounding
iii. Pressure-Velocity compounding
A nitrogen gas is expanded from 8 bar to 1 bar at 47°C according to law PV
= C. Plot the process on P- V and T-S diagram
and state the formulas to
be used to find out work done, Amount of heat supplied and change in
entropy.
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Sol.
01 marks
for
figure
01 marks
each
formula
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d)
Sol.
Determine the amount of heat supplied to 2kg of water at 25°C to convert
it into steam at 5 bar and 0.9 dry.
Note: Value of Cp of water is not given assuming it standard value.
01 marks
01 marks
02 marks
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Q.4.
a)
Sol.
Attempt any THREE of the following:
12 Marks
Explain Dalton's law of partial pressure. How it is applicable to condenser?
It states that’ “The pressure exerted by mixture of air and steam is equal to sum of partial
pressures, which each constitute would exert, if it occupies the same volume”.
02 marks
02 marks
Figure: Dalton’s law of partial pressure
In condenser total pressure is the sum of partial pressure of steam and air.
Mathematically,
P c= P a + P s
Where;
Pc = Pressure in condenser containing mixture of air and steam
Pa = Partial pressure of air
Ps = Partial pressure of steam
b)
A system is composed of a gas contained in a cylinder fitted with a piston.
The gas expands from the state 1 for which internal energy U 1 = 75 KJ to
state 2 for which U2 = -25 KJ. During the expansion the gas does 60 KJ of
work on the surrounding. Determine the heat transferred to or from the
system during the process.
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Sol.
02 marks
02 marks
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c)
3 m3 of gas of 30°C and 6 bar pressure is expanded isothermally to 1 bar. Find
work done, change in internal energy and heat transferred during the process.
Sol.
01 mark
01 mark
01 mark
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01 mark
d)
Explain construction and working of shell and tube type heat exchanger. A ice plant
producing 2000 Kg ice per day required the condenser. Suggest the type of
condenser with justification.
Sol.
02 marks
Shell and tube heat exchanger consist of a bundle of round tubes placed inside the cylindrical
shell.Tube axis parallel to that of shell. One fluid inside the tubes while the other over the
tubes.
The main components of this type of heat exchanger are:
i.
Shell
ii.
Tube bundle
iii. Front and rear headers of shell
iv.
baffles
The baffles provide the support to tubes and also deflect the fluid flow approximately normal
to tubes. This increase the turbulence of shell side fluid and improves heat transfer. The
various types of baffles are existing and their type, spacing, shape, will depend on the flow
rate, shell side pressure drop, required tube support, flow vibrations etc.
The fluid combination may be :
1 Liquid to liquid
2 Liquid to gas
3 Gas to gas
01 mark
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A ice plant producing
used.
Justification:
2000 Kg ice per day required the evaporative condenser is
01 mark
The evaporative condenser is essentially a combination of a water-cooled condenser and an
air-cooled condenser, utilizing the principle of heat rejection by the evaporation of water into
an air stream traveling across the condensing coil.
Q.5.
a)
Attempt any TWO of the following:
(i)
Suggest the methods to improve the performance of steam turbine.
Explain anyone in brief.
(ii)
Sol.
Identity
12 Marks
the different losses occurred in steam turbine.
i) Methods to improve turbine efficiency
1) Reheating of Steam
2) Regenerative feed heating
01 mark
3) Binary Vapour Plant
Regenerative feed heating System
The process of draining steam from turbine at certain points during it’s expansion and using this steam
for heating feed water supplied to boiler is known as regenerative feed heating. It increases the thermal
efficiency of plant, The temperature stresses in the boiler are reduced due to decreased range of
working temperature.
01 mark
01 mark
ii) Losses occurred in steam turbine
Residual velocity loss- The steam leaves the turbine with a certain absolute velocity which results in
loss of KE. This loss is about 10 to 12% .It can be reduced by multistaging.
Losses in regulating valves-Due to throttling action in valve , steam pressure drop occurs. Hence
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steam pressure at entry to turbine is less than the boiler pressure.
Losses due to friction in nozzle-Friction occurs both in nozzle and turbine blades. In nozzle, nozzle
efficiency is considered, whereas in turbines, blade velocity coefficient is taken into account. This loss
is about 10%
Loss due to leakage-The leakage occurs between the shaft, bearings and stationary diaphragms
carrying the nozzles in case of impulse turbines. In reaction turbine the leakage occurs at blade tips.
This is about 1-2%.
03 marks
(Any 3
Point)
Loss due to mechanical friction-This occurs in bearings and may be reduced by lubrication
Loss due to wetness of steam-In multistage turbine, condensation occurs at last stage ,so in dragging
water particles with steam, some KE of stem is lost
Radiation loss-As turbines are heavily insulated to reduce the heat loss to surroundings by radiation
and so these losses are negligible
b)
Sol.
An exterior wall of house consists 10.6 cm layer of common brick. It is
followed by 3.8 cm layer of gypsum plaster and 5.83 cm of rock wool
insulation. Estimate the amount of heat transferred through structure it.
Thermal conductivity of brick = 0.7 W/mK
Thermal
conductivity
of Plaster = 0.48 W/mK
Thermal
conductivity
of Insulation
= 0.065 W/mK
Note:
1.
Temperature gradient not mentioned.
(If student assume a data and solve the numerical with correct procedure then
gives appropriate marks)
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03marks
03marks
c)
The initial condition of steam is 15% wet at a pressure of 7 bar It expands
1.2 bar by PV1.3 = C. Find
(i)
Quality of steam at the end of expansion
(ii)
Work done.
to
Sol.
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03 marks
03 marks
Q.6.
a)
Attempt any TWO of the following:
A mass of 0.8 Kg of air at 1 bar and 25°C is contained in a gas tight
frictionless piston cylinder device. The air is now compressed to a final
pressure of 5 bar. During this process the heat is transferred from air such
that the temperature inside the cylinder remains constant. Calculate the heat
transferred and work done during process and direction of each in the
process.
12 Marks
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Sol.
01 mark
02 marks
02 marks
01 mark
b)
For steam power plant having capacity 600 MW capacity a cooling tower is
required to set up with condenser. Suggest the type of condenser and cooling
tower with justification.
Sol.
For Steam power plant having Capacity 600 MW the requirement of condenser and cooling tower is as
follow.
3 marks
1) Condenser:- Given Capacity is medium to low capacity for this we can use Jet Condenser
-Which cooling water and steam are mixed to each other ,
-Mainly it requires less quantity of cooling water.
-It is simple in construction and less costly.
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-Maintenance cost Is also less.
2) Cooling Tower :- For this Capacity we can use Force draught cooling tower
- Less space is required
-Cooling rate and efficiency of tower is high
3 marks
-Temperature of water coming out from tower can be controlled.
c)
Suggest the type of heat exchangers
for following applications
-
(i) Dairy plant (Milk Chilling Plant)
(ii) Condenser of refrigeration
Sol.
system. (House hold system) Justify your answers.
Types of Heat Exchanger Used for
1) Dairy Plant (Milk Chilling Plant)- Plate Type Heat Exchanger
1 mark
Because , It is made up of aluminum alloy which provides higher rate of heat transfer.
Due to larger surface area, It has more heat transfer as compare to other heat exchanger which
is useful for dairy plant.
2 marks
It is lighter in weight.
2) Condenser of Refrigeration System:- Counter Flow tube type heat Exchanger
Because , High performance due to large surface area
1 mark
2 marks
Compact and light in weight
In tubes generally turbulent flow is develop which reduces scale deposition.
Less installation and maintenance cost.
Page 19 of 19
2019 SUMMER
21819
3 Hours / 70 Marks
Instructions –
22337
Seat No.
(1) All Questions are Compulsory.
(2) Illustrate your answers with neat sketches wherever
necessary.
(3) Figures to the right indicate full marks.
(4) Assume suitable data, if necessary.
(5) Use of Steam tables, logarithmic, Mollier’s chart
is permitted.
(6) Mobile Phone, Pager and any other Electronic
Communication devices are not permissible in
Examination Hall.
Marks
1. Attempt any FIVE of the following:
10
a) Differentiate between Heat and Work.
b) State clausius statement of second law of thermodynamics.
c) Define dryness fraction and degree of superheat.
d) Define mach number and critical pressure.
e) Explain bleeding of steam.
f)
State Dalton’s law of partial pressure.
g) Define Fourier’s law.
P.T.O.
22337
[2]
Marks
2. Attempt any THREE of the following:
12
a) State extensive property and Intensive property with two examples
each.
b) Define isentropic process and plot it on P-V and T-S diagram.
c) Define:
(i)
Sensible heat
(ii)
Latent heat
d) Differentiate water tube boiler and fire tube boilers (any four)
3. Attempt any THREE of the following:
12
a) State the term governing of turbine and explain nozzle control
governing.
b) Explain principle of working of Impulse steam turbine with
neat sketch.
c) A gas occupying 0.26 m3 at 300°C and 0.4 MPa pressure expands
till volume becomes 0.441 m3 and pressure 0.26 MPa. Calculate
the change in internal energy per kg of gas.
CP = 1 kJ/kg K, CV = 0.71 kJ/kg K.
d) Determine the amount of heat supplied to 2kg of water at
25°C to convert it into steam at 5 bar and 0.9 dry.
4. Attempt any THREE of the following:
a) Differentiate between natural draught and forced draught cooling
tower.
b) A gas has a volume of 0.14 m3, pressure 1.6 bar and a
temperature 110°C. If the gas is compressed at constant pressure
until its volume becomes 0.112m3 Determine:
(i)
Work done in compression of gas
(ii)
Heat given out by gas
c) A certain gas has CP = 1.968 kJ/kg K CV = 1.507 kJ/kgK. Find
the molecular weight and the gas constant. A constant volume
chamber of 0.3m3 capacity contain 2 kg of this gas at 5°C.
Heat is transferred to the gas until the temperature is 100°C.
Find the work done and change in internal energy.
12
22337
[3]
Marks
d) Define:
(i)
Transmissivity
(ii)
Black body
(iii) Grey body
(iv) Reflectivity
e) Draw a neat sketch of surface condenser and label it.
5. Attempt any TWO of the following:
12
a) List out any six losses in steam turbine.
b) A steel pipe of inner and outer diameter 6 cm and 8 cm
respectively has inside temperature 140°C and outside temperature
50°C. The thermal conductivity of steel is 24 W/mk. Calculate
the rate of heat transfer through the pipe if length of pipe is
1.5 m.
c) List any six methods of energy conservation in boilers.
6. Attempt any TWO of the following:
a) Explain the necessity of compounding in steam turbine and
draw a neat sketch of pressure velocity compounding.
b) (i)Explain the application of second law of thermodynamics
to refrigerator.
(ii)
State any three functions of steam condenser.
c) Derive characteristic gas equation using Boyle’s and Charle’s
law.
12
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SUMMER – 19 EXAMINATION
Subject Name: Thermal Engineering
Model Answer
Subject Code:
22337
Important Instructions 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 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.
Q.1.
a)
Sol.
Attempt any FIVE of the following:
Differentiate between Heat and Work.
Differentiate between heat and work.
Ans:
Sr.No.
Parameter
Heat
1.
b)
Sol.
10 Marks
Definition
Work
Form of energy that is The amount of energy
transferred between system transferred by a force acting
and surrounding or two though a distance
systems due to temperature
difference
2.
Function
Heat is a function of the
state
Heat is a function of the Path
3.
Energy
Interaction
Due to Temperature
Difference
Other than Temperature
difference
Any 2 points
1 marks for
each Point of
difference
State Clausius statement of second law of thermodynamics.
Clausius statement: It state that “It is impossible for a self-acting machine working in
a cyclic Process without any external force, to transfer heat from a body at a lower
temperature to a body at a higher temperature. Thus external mechanical work
expenditure is necessary to transfer heat from a body at a low temperature to a body at
high temperature.
1 mark
1 mark
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c)
Sol.
Define d r y ne s s fraction and degree of superheat.
Dryness fraction:
It is defined as a fraction of dry steam that is present in a liquid vapour is called dryness
fraction.
Or
Dryness fraction is the ratio of the Mass of actual dry steam to the Mass of wet steam.
Any one
definition
1 mark
X = M s/ Ms+Mw
Where X – Dryness fraction
Ms – mass of vapour (dry steam) contain in steam
Mw = mass of water in suspension in steam
d)
Sol.
Degree of Superheat:
1 mark
The difference between the temperature of superheated steam and saturated steam (Tsup –
Tsat) is known as degree of superheat.
Define Mach number and critical pressure.
1. Mach Number: In fluid dynamics, the Mach number (M or Ma) is a dimensionless
quantity representing the ratio of flow velocity past a boundary to the local speed of sound.
1 mark
M = c/a
M is the Mach number, c is the local flow velocity, a is the speed of sound in the medium
1 mark
2. Critical Pressure: The Pressure for which the maximum discharge through nozzle
occurs is called as critical pressure . It is denoted as Pc
e)
Sol.
Explain bleeding of steam.
It is process of draining steam from turbine at certain point during its expansion and using
these steams for heating the feed water supplied to boiler is known as bleed and the
process is known as bleeding of steam.
1 mark
1 mark
Figure: Bleeding of steam
f)
Sol.
State Dalton's law of partial pressure.
This law states that “The total pressure exerted by a mixture of air and water vapour on the
walls of container is the sum of partial pressure exerted by air separated and that exerted
by vapour separately at common temperature of the condenser”.
P = Pa + Ps
Where
Pa= partial pressure exhausted by air
Ps = partial pressure exhausted by vapour
P = total pressure of mixture at temperature.
01 mark
01 mark
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g)
Sol.
Define Fourier's law.
The law state that for homogeneous material the rate of heat transfer in steady state in any
direction is directly proportional to temperature gradient in that direction.
01 mark
Q/A α dt/dx
01 mark
Q/A = -k dt/dx
Where, Q/A is rate of heat transfer
dt/dx is temperature gradient
k conductivity of medium
Q.2.
a)
Sol.
Attempt any THREE of the following:
State extensive property and Intensive property with two examples each.
Extensive property:- An extensive property of a system is one whose value depend upon
the mass of the system.
e.g. volume, energy, enthalpy, entropy, internal energy.
12 Marks
1 mark
1 mark
Intensive property:- An intensive property of a system is one whose value does not
depend upon the mass of the system.
1 mark
e.g. Density, Temperature , Pressure
b)
Sol.
1 mark
Define isentropic process and plot it on, P- V and T-S diagram.
Isentropic Process: The process in which working substance neither receives nor rejects
heat to its surrounding during expansion or compression is called as Isentropic process, it
is also known as adiabatic process.
Adiabatic process reversible when it is frictionless and the process is irreversible when it
involves friction .
Process is denoted by equation PVɣ =C
2 marks
2 marks
(1 Mark for
each Dig.)
PV Diagram
TS Diagram
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c)
Sol.
Define:
(i) Sensible heat
(ii) Latent heat
i) Sensible Heat:
The heat in which change in temperature of a substance can be observed but phase remains
unchanged that heat is known as Sensible heat.
1 mark
This heat can be sensed by ordinary thermometer, It is given by the equation
Sensible heat = m Cp (T2-T 1)
m is mass
Cp is Specific heat at constant pressure
1 mark
T1 is Initial Temperature
T2 is Final Temperature
1 mark
ii) Latent Heat:
It is define as amount of heat required for the change of phase of 1 kg of water at saturated
temperature to dry saturated steam at constant pressure .
It is denoted by L , Its value can be directly obtained from steam table
1 mark
Heat at which solid changes phase to liquid is known as Latent heat of fusion
Heat at which Liquid Changes Phase to vapour is known as Latent heat of vaporization
d)
Sol.
Q.3.
Differentiate water tube boiler and fire tube boilers (any four)
Sr.
Parameter
Water tube boiler
Fire tube boiler
No.
1.
Medium in tube
Water is circulated in tube Hot gases are circulated
and hot gases passed over through the tube and water
the tube
flows over tube.
2.
Steam Formation
Steam formation rate is
Steam formation rate is low
Rate
high
3.
Steam Pressure
It can generate steam at
Generate steam at lower
higher pressure more than pressure up to 25 bar
25 bar
4.
Operating cost
Operating cost high
Operating cost low
5.
Overall efficiency Overall efficiency high
Overall efficiency low
6.
Cleaning and
Cleaning and Inspection is Cleaning and Inspection is
Inspection
easy
difficult
7.
Application
High power Generation
Low to medium power
generation
8.
Example
Babcock & Wilcox boiler, Cochran Boiler, Lancashire
Loeffler Boiler
boiler
Attempt any THREE of the following:
1 mark for
each point
(Any 4 Points)
12 Marks
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a)
Sol.
State the term governing of turbine and explain nozzle control governing.
Governing of turbine:
The function of governor is to regulate the supply of steam to the turbine so that the
speed of rotation shall remain constant at all loads.
01 mark
Nozzle control governing:
02 marks for
Figure
01 mark
explanation






b)
Figure: Nozzle control governing
The arrangement of nozzle control governing is shown in figure.
The poppet type valve uncover as many steam passages as necessary to meet
the load, each passage serving a group of nozzle.
The control governor has the advantage of using steam at full boiler pressure.
The nozzles are divided into three groups N1, N2 and N3 and the control valves
V1, V2 and V3 controls the amount of steam supply to each nozzle group
respectively.
The number of nozzle group may vary from three to five or more. Various
arrangements of group nozzles and valves can be employed. Two
arrangements are shown in figure (b) & (c).
Under full load condition all the regulating valve are opened. When the load
on the turbine is reduced the supply of steam to a group nozzle is shut off.
Explain principle of working of Impulse steam turbine with neat sketch.
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Sol.
02 marks for
figure
Figure: Impulse Turbine
02 marks for
explanation
Construction:
Impulse turbine is simpler, less expensive and does not need to be pressure proof. It
can operate with any pressure stream but is considerably less efficient.
Impulse turbine consist of one fixed set of nozzle mounted on a stationary diaphragm that
orient the steam flow into high speed jets, which is followed by one set of moving blade
ring as shown in Fig. for a single stage impulse turbine.
c)
A gas occupying 0.26 m3 at 300°C and 0.4 MPa pressure expands till volume
becomes 0.441 m3 and pressure 0.26 MPa. Calculate the change in internal
energy per kg of gas.
Cp = 1 kJ/kg K, C y = 0.71 kJ/kg K.
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Sol.
02 marks
02 marks
d)
Determine the amount of heat supplied to 2kg of water at 25°C to
convert it into steam at 5 bar and 0.9 dry.
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Sol.
01 mark
01 mark
01 mark
01 mark
Q.4.
a)
Sol.
Attempt any THREE of the following:
Differentiate between natural draught and forced draught cooling tower.
Sr. No.
Natural draught
Forced draught
1
The air flows naturally without fan
Fan is located at the top of the
through tower and provides required
tower and enters the side of the
cooling
tower.
2
The air circulation through the tower
The air circulation through the
depends on wind velocity.
tower depends on fan speed.
3
The cooling Rate and efficiency of tower The cooling Rate and efficiency
is less.
of tower is high.
4
It requires large space for same capacity. It requires less space for same
capacity.
5
No power requires due to absence of fan. Fan requires more power as it
handles hot air.
6
The temp. of water coming out from
The temp. of water coming out
12 Marks
01 mark for
each
differentiation
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b)
tower cannot be controlled.
from tower can be controlled.
3
A gas has a volume of 0.14 rn , pressure 1.6 bar and a temperature
1100C. If the gas is compressed at constant pressure until its volume becomes
0.112 m3 Determine:
i. Work done in compression of gas
ii. heat given out by gas
Sol.
01 mark
01 mark
01 mark
01 mark
c)
A certain
gas has
Cp = 1.968 kJ/kg K Cy = 1.507 kJ/kgK. Find the
molecular weight and the gas constant. A constant volume chamber of
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0.3m3 capacity contain 2 kg of this gas at 5°C. Heat is transferred to
the gas until the temperature is 100°C. Fmd the work done and change
in internal energy.
Sol.
01 mark each
answer
d)
Sol.
Define:
i. Transmissivity
ii. Black body
iii. Grey body
iv. Reflectivity
Transmissivity:
It is the the fraction of energy which is transmitted through the body.
01 mark each
definition
Or
The ratio of amount of energy transmitted to the amount of energy incident on a body.
Black body: A black body is an object that absorbs all the radiant energy reaching its
surface from all the direction with all the wavelengths. Gray body:
Grey Body: A gray body is defined as a body whose absorptivity of a surface does not
vary with variation in temperature and wavelength of the incident radiation. It absorbs a
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definite percentage of incident energy irrespective of wavelength. Its absorptivity lies
between 0 to 1.
e)
Reflectivity:
It is defined as the ratio of amount of energy reflected to the amount of energy incident on
a body.
Draw a neat sketch of surface condenser and label it.
Sol.
02 marks for
Sketch
02 marks for
label.
Q.5.
a)
Sol.
b)
Sol.
Figure: Surface Condenser
Attempt any TWO of the following:
List out any six losses in steam turbine.
Losses in steam turbine
1. Residual velocity loss
2. Loss due to friction
3. Leakage loss
4. Loss due mechanical friction
5. Radiation loss
6. Loss due to moisture
7.carry over losses
A s t e e l pipe of inner and outer diameter 6 cm and 8 cm respectively
has inside temperature 140°C and outside' temperature 50°C. The thermal
conductivity o f steel is 24 W/mk. Calculate the rate of heat transfer through
the pipe if length of pipe is 1.5 m.
Given Data: Length of pipe = L=1.5 m , Inner diameter ,d1=6 cm , Outer diameter
,d2= 8 cm
Inside temp T1 =140º C , Outside temp T2 =40º C , Ksteel = 24 W/Mk
12 Marks
02 marks
02 marks
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02 marks
c)
Sol.
Q.6.
a)
Q =78616.35 Watts or 78.62 KW
List any SIX methods of energy conservation in boilers.
List any six methods of energy conservation in boilers
Following methods can conserve energy in boilers
1) Reduction radiation and convention losses
2) Waste heat recovery for heating to the feed water.
3) Continues monitoring of flue gases losses and other losses
4) Using standard efficient fuel firing equipments, burners, mechanical stockers.
5) Scheduling boiler operation to avoid fluctuation in boiler load
6) Installation of variable speed drives.
7) Optimise boiler stem pressure and temperature
8) Periodic energy audit.
Periodic preventive maintenance of all components.
Attempt any TWO of the following:
Explain the necessity of compounding in steam turbine and draw a neat
sketch of pressure velocity compounding.
Sol. Necessity of compounding in steam turbine: Compounding of steam turbines is
necessary 1) To reduce speed of rotor blades to practical limits.
2) To reduce centrifugal force and hence to prevent failure of blades.
3) To reduce velocity of steam leaving blades.
If entire pressure drop from boiler pressure to condenser pressure is carried out in a single
stage of nozzle then the velocity of steam entering the turbine blades will be very
high. The turbine speed has to be also very high as it is directly proportional to steam
velocity. Such high rpm of turbine rotor are not useful for practical purposes & there is a
danger of structural failure of blades due to excessive centrifugal stresses. Hence
compounding is carried out.
Neat sketch of pressure velocity compounding.
01 mark each
12 Marks
03 marks
03 marks for
figure
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b)
Sol.
Figure: Pressure and Velocity compounding
Explain the application of second law of thermodynamics to refrigerator.
State any three functions of steam condenser.
According to Clausius‘s statement “ heat cannot flow itself from cold body to a hot body
without help of external agency”
01 mark
diagram
Figure: Refrigerator
REFERIGERATOR is shown in the figure with cold body (T2) and hot body
(atmosphere) (T1). A refrigerator is a device which maintains the temperature of cold body
below the surrounding temp. The amount of heat taken from cold body which is to be
cooled is Q2. For doing this work , external energy is required to the refrigerator.
So heat rejected to atmosphere Q1= Q2 + W.
As per the statement of Statement of Second Law of Thermodynamics, it is
observed that refrigerator operates between the two different temperatures in a cyclic
manner. It also extracts the heat from cold body only (storage space) and does the
equivalent amount of work as shown.
03 marks
explanation
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In a full cycle of a refrigerator, three things happen:
1. Heat is absorbed from cold body, the heat can be called Q2.
2. Some of the energy from that input heat is used to perform work (W).
3. The rest of the heat is rejected to hot body (Q1).
An performance of the refrigerator can be calculated as: Efficiency = Q1 / work W
So it is cleared that the external energy is required to absorb heat from cold body and to
reject it to hot body.
c)
Sol.
Function of condenser:
02 marks
1) To maintain a very low back pressure so as to obtain the maximum possible
energy from steam and thus secure a high efficiency.
2) To condense the steam and reuse it to supply as pure feed water to the hot well
from where it is pumped back to the boiler.
3) To remove of air and non-condensable gases
Derive characteristic gas equation using Boyle'S and Charle's law.
Characteristic gas equation using Boyle’s & Charle’s law:
Let us consider a unit mass of an ideal gas to change its state in following two processes as 01 mark for
shown in fig.
each step
Here, process 1-2’ is at constant pressure
Process 2’-2 is at constant temperature
Now, applying Charle’s law for process 1-2’
We get
Now, applying boyle’s law for process 2’-2 ,
P2’ V2 =P2 V2
( T=C )
P1 V2’=P2 V2
( As P2’= P1)
01 mark
01 mark
Substituting eq (II) IN eq (I), We get
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i.e
consider m kg of gas , multiply eq (IIII) by m
Therefore PV = m R T------......................................CHARACTERISTIC
EQUATION
Page 15 of 15
2018 WINTER
11819
3 Hours / 70 Marks
Instructions –
22337
Seat No.
(1) All Questions are Compulsory.
(2) Answer each next main Question on a new page.
(3) Illustrate your answers with neat sketches wherever
necessary.
(4) Assume suitable data, if necessary.
(5) Use of Non-programmable Electronic Pocket
Calculator is permissible.
(6) Mobile Phone, Pager and any other Electronic
Communication devices are not permissible in
Examination Hall.
(7) Use of Steam tables, logarithmic, Mollier’s chart
is permitted.
Marks
1. Attempt any FIVE of the following:
10
a) Define gray body.
b) State the function of :
(i)
Fusible plug and
(ii)
Economiser
c) Define:
(i)
Boiler efficiency
(ii)
Latent heat
d) State Dalton’s law of partial pressure.
P.T.O.
22337
[2]
Marks
e) Explain choked flow condition in nozzle.
f)
What is universal gas constant?
g) Define:
(i)
Flow work
(ii)
Entropy
2. Attempt any THREE of the following:
12
a) Represent the following processes on P-V and T-S diagram.
(i)
Isentropic process
(ii)
Isobaric process
b) Define irreversible process. State the factors making process
irreversible.
c) In a steam power cycle, the steam supply is at 15 bar and
dry and saturated. The condenser pressure is 0.4 bar. Determine
dryness fraction and enthalpy of steam.
d) How steam turbines are classified?
3. Attempt any THREE of the following:
a) Using the mollier chart, find the heat drop and final condition
of steam when the steam from an initial pressure of 30 bar
and temperature 350°C is expanded adiabatically to a pressure
of 1 bar.
b) State steady flow energy equation and apply it to condenser
with block diagram.
c) 3 m3 of gas at 30°C and 5 bar pressure is expanded isothermally
to 1 bar with low PV = C. Find work done, change in internal
energy and heat transferred.
d) State the main features of Indian boiler regulations. (IBR)
12
22337
[3]
Marks
4. Attempt any THREE of the following:
12
a) A cylinder contains 0.12 m3 of air at 1 bar and 90°C. It is
compressed to 0.03 m3, the final pressure being 6 bar. Find the
index of compression and increase in internal energy.
b) Explain different losses in steam turbine.
c) State:
(i)
Fourier’s law
(ii)
Newton’s law of cooling
(iii) Radiation and
(iv) Thermal conductivity
d) State the advantages of regenerative feed heating.
e) A balloon of spherical shape, 10 m in diameter is filled with
H2 at 20°C and atm. pressure. The surrounding air is at 15°C
and barometer reads 75 mm of Hg. Determine the load lifting
capacity of the balloon.
5. Attempt any TWO of the following:
a) (i)
(ii)
12
Define throttling and state the purposes of it.
Steam at a 6.87 bar, 205°C, enters in an insulated nozzle
with velocity of 50 m/s. It leaves at a pressure of 1.37 bar
and a velocity of 500 m/s. Determine the final enthalpy.
b) Explain with neat sketch, construction and working of impulse
turbine.
c) Explain with neat sketch. Construction and working of plate
type heat exchanger. State its applications.
6. Attempt any TWO of the following:
a) Explain with neat sketch induced draught cooling tower.
b) Explain with neat sketch. Construction and working of Loeffler
boiler.
c) (i)
(ii)
State the sources of air leakage in condenser.
Steam enters a condenser at 36°C and with barometer
reading 760 mm. If the vacuum of 695 mm is produced,
find the vacuum efficiency.
12
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WINTER– 18 EXAMINATION
Subject Name: Thermal Engineering
Model Answer
Subject Code:
22337
Important Instructions 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 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.
Q.
No.
Sub
Q. N.
1
a
Answer
Marking
Scheme
Attempt any Five
2
(i) Gray body :- A grey body is defined as a body with constant emissivity over all
wavelengths and temperatures. It absorbs a definite percentage of incident
energy irrespective of their wavelengths.
Write functions of
(i) Fusible plug-The function of the fusible plug is to put-off the fire in the furnace of the
boiler when the water level falls below an unsafe level and thus avoids the
explosion which may take place due to overhearing of the tubes and the shell.
(ii) Economizer :- Function of economizers in steam power plants is to capture the
waste heat from boiler flue gases and transfer it to the boiler feed water. This
raises the temperature of the boiler feed water, lowering the needed energy
input, in turn increase in boiler efficiency.
b
Define(ii) Boiler efficiency :-It is the ratio of heat actually used in producing the steam to the
heat liberated in the furnace. It is also known as thermal efficiency of boiler.
(iii) Latent heat:-It is energy absorbed or released by a substance during a change in its
physical state (phase) that occurs without changing its temperature .e.g. latent
heat of fusion and latent heat of vaporization
.
c
1
1
1
1

d 
Dalton’s law of partial pressure:- Dalton's law of partial pressures states that in a
2
mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial
pressures of the individual gases.

or
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
In a mixture of perfect gases, total pressure exerted by the mixture is the sum of
partial pressures, which each gas would exert if it separately occupied the whole volume
and was at the same temperature as the mixture.
Ptotal = p1+p2+p3 +-------pn
where p1, p2, ..., pn represent the partial pressures of each component.
e
f
g
Choked flow condition in nozzle :-Choked flow is a fluid dynamic condition associated with
the Venturi effect. When a flowing fluid at a given pressure and temperature passes through
a constriction (such as the throat of a convergent-divergent nozzle or a valve in a pipe) into
a lower pressure environment the fluid velocity increases.
Choked flow is a limiting condition where the mass flow will not increase with a further
decrease in the downstream pressure environment while upstream pressure is fixed.
Universal gas constant :- Universal gas constant or molar constant (denoted by Ru) of a gas
is the product of the gas constant and the molecular mass of the gas. Ru is same for all
gases. It is 8.314 KJ/Kg-mol K
Ru =M X R
M = Molecular mass of the gas expressed in kg-mole; R= Gas constant
In general, M1, M2,M3 are the Molecular masses of different gases and R1,R2,R3 are their
gas constants respectively, then
M1R1=M2R2=M3R3 = - - - -= Ru
(i) Flow work- This is the work necessary to advance the fluid against the existing
pressure, . It is the work required to cause the flow of fluid in any passage.
Flow work= PV
where P= pressure of fluid, V= volume of fluid.
2
2
1
(ii) Entropy- Entropy is the extensive property of the system (depends on the mass of 1
the system) and its unit of measurement is J/K (Joule per degree Kelvin). Entropy
is heat or energy change per degree Kelvin temperature. Entropy is denoted by
‘S’, while specific entropy is denoted by ‘s’ in all mathematical calculations.
Entropy is defined as the property used to measure the quantity of energy or
irreversibility of a process.
Attempt any THREE
2
a
I ) Isentropic process
2
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(i) Isobaric process
2
b
A process is said to be irreversible if it cannot reach back to its original state without using
external work. The spontaneous process will not be a reversible or quasi-static process. It is
an irreversible process. There are many factors that make a process irreversible. Four of the
most common causes of irreversibility are friction, unrestrained expansion of a fluid, heat
transfer through a finite temperature difference, and mixing of two different substances.
2
Irreversibility is classified according their causes:
i) External: The irreversibility caused by external physical factors like friction, resistance,
viscosity, surface tension, finite temperature difference, etc.
2
The energy lost due to friction can never be regained. Hence, the direction of the process
cannot be reversed without supplying external work.
ii) Internal: Irreversibility caused by properties of the working fluid in a process like throttling
or free expansion.
When a gas expands, it uses its internal energy to do so. It cannot contract on its own and
reverse the process.
iii) Chemical: Irreversibility caused by internal chemical properties like structure, bonds, etc.
When a chemical reaction occurs in association with absorption or liberation of heat, it
cannot reverse spontaneously. Involvement of dissipative effect during the process will also
be a cause of irreversibility of a process. There are following various types of dissipative
effect during a process 1.
2.
3.
4.
5.
Mechanical friction
Magnetic hysteresis
Electrical resistance
Viscosity or fluid viscosity
Inelasticity
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Steam supply 15 bar (dry saturated)
Condenser pressure 0.4 bar Find dryness fraction and enthalpy of steam.
From steam tables,
At 15 bar,
hf1=844.89 KJ/Kg;hfg1=1947.3; hg1=2792.2KJ/Kg; x1=1(Given); sf1=2.315; sfg1= 4.129
KJ/KgK
At 0.4 bar,
hf2=317.58;hfg2=2319.2; hg2=2636.8KJ/Kg;; sf2=1.0259; sfg2= 6.645 KJ/KgK
Let x2 =Final dryness fraction
Considering steam power cycle as isentropic,
Initial Entropy= Final Entropy
Sf1+x1 sfg1= sf2 + x2 sfg2
2.315+ 1 X 4.129 =1.0259 + x2 X 6.645
x2 = 0.815 (Dryness fraction)
Final Enthalpy = hf2+ x2 X hfg2=2207.72 KJ/Kg
c
2
2
Steam turbines may be classified in following ways: ( Any four )
d
1. According to working principle or Action of steam over blade :
(a) Impulse Turbine
1 for
each
(b) Reaction Turbine
(c) Impulse Reaction Turbine
2. According to the stage of expansion of steam:
(a) Single stage turbine
(b) Multistage turbine
3. According to the position of shaft :
(a) Horizontal turbine
(b) Vertical turbine
4. According to pressure of steam supplied:
(a) High Pressure turbine
(b) Low Pressure turbine
5. According to direction of steam flow:
(a) Axial flow turbine
(b) Radial flow turbine
(c) Tangential flow turbine
6. According to exhaust steam pressure
(a) Condensing type steam turbine

Non-condensing type steam turbine
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3
a
From Mollier Chart
2
H2 = 2510 at point B and H1 = 3130 at point A
Heat drop = H1 - H2
= 3130 - 2510 = 620 kJ/kg
2
Final condition of steam
From Mollier chart at point B dryness fraction is 0.90
b
steady flow equation can be expressed as:
Internal Energy at 1 + Potential Energy at 1 + Kinetic Energy at 1 + Flow work at 1 + Heat
supplied = Internal Energy at 2 + Potential Energy at 2 + Kinetic Energy at 2 + Flow work at
2 + Work done
Hence the steady flow energy equation is,
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1
Where,
h1& h2 = Enthalpy at inlet and outlet in J /kg
C1& C2 = velocity at inlet and out of fluid---- m/s
Z1 and Z2 = height of inlet & outlet above datum
1
Q = heat supplied per ---------------Joule
W = work done by 1 kg of fluid----Joule
PV = Flow work---------N-m or Joule
Application :
Steam Condenser :- It is a device to condensed the exhaust steam
Heat- is lost hence q is – ve
Applying SFEE
2
Fig: Steam condenser
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c
Given :
V1 = 3 m3
T1 = 30 OC , P2 = 1 bar
1
For isothermal process
PV = C
P1V1 = P2V2
5 x 105 x 3 = 1 x 105 x V2
2
V2 = 15 X 105
Work done :
dW = P1V1 loge (V2/V1)
1
= 5 x 105 x 3 loge ( 15/3)
dW = 2.41 x 106 kJ/kg
For Isothermal process dW = dQ and dU = 0
d
Features of Indian Boiler regulation : ( Any Four points 1 mark each)
1. A boiler cannot be put to use unless it has been registered with the Chief Inspector of
Boilers.
2. The maximum working pressure of the boiler has to be determined by Boiler Inspector
who will issue certificate for this. Owner cannot exceed this pressure limit in any case.
3. In case of accident, it should be reported by owner within 24 hours with full details.
4. The rules, regulations and bye-laws governing the upkeep and maintenance of boilers,
procedure of registration, inspection and certification of maximum pressure, safety
conditions etc. are subject to a revision by a Central Board under control of Govt. of India.
5. The boiler house plan, chimney design (Max height 30.48 m from floor) should be
approved by boiler inspector.
6. Owner should apply for registration in prescribed format, inspector should fix date of
inspection within 30 days, conduct inspection/examination of boiler, Issue the certificate of
registration not exceeding 12 months period.
7. Following inspections are carried out by Boiler Inspector at various stages/ levels /need>Inspection for registration, Hydraulic test, steam test, annual inspection, Inspection under
steam, Internal inspection, Accident inspection, Casual inspection
8. Violation of law is liable to prosecution and punishment with fine.
01 for
each
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4
a
Given :
Initial Volume of air =V1 : 0.12 m3
Initial pressure of air = P1: 1 bar
Initial temperature of air = T1 : 90 + 273 = 360 K
Final Volume of air = V2 : 0.03m3
Final Pressure of air = P2 : 6 bar
P1V1 = mRT1
1 x 105 x 0.12 = m x 289 x 360
m = 1.15 kg
Assume compression to be polytrophic
P1/ P2 = (V2/V1)n
2
1/6 = ( 0.03/0.12)n
n = 1.29
We know that
P1V1/T1 = P2V2/T2
(1 x 0.12) / 360 = (6 x 0.03) / T2
T2 = 540 K
Increase in internal energy
2
= m Cv ( T2 - T1)
= 1.15 x 0.72 x (540 – 360)
= 149.04 kJ
b
Energy losses in steam turbines [Any four points with explanation 01 mark each]
(i) Residual velocity loss- The steam leaves the turbine with a certain absolute velocity which
results in loss of KE. This loss is about 10 to 12% .It can be reduced by multistaging.
01 for
(ii) Losses in regulating valves-Due to throttling action in valve , steam pressure drop occurs. each
Hence steam pressure at entry to turbine is less than the boiler pressure.
(iii) Losses due to friction in nozzle-Friction occurs both in nozzle and turbine blades. In
nozzle, nozzle efficiency is considered, whereas in turbines, blade velocity coefficient is
taken into account. This loss is about 10%
(iv) Loss due to leakage-The leakage occurs between the shaft, bearings and stationary
diaphragms carrying the nozzles in case of impulse turbines. In reaction turbine the leakage
occurs at blade tips. This is about 1-2%.
(v) Loss due to mechanical friction-This occurs in bearings and may be reduced by
lubrication
(vi) Loss due to wetness of steam-In multistage turbine, condensation occurs at last stage ,so
in dragging water particles with steam, some KE of stem is lost
(vii) Radiation loss-As turbines are heavily insulated to reduce the heat loss to surroundings by
radiation and so these losses are negligible
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c
State : ( 1 Mark for each definition )
i) Fourier’s law: “Fourier's law states that the rate of heat flow, dQ/dt, through a
homogeneous solid is directly proportional to the area, A, of the section at right angles to the 01 for
each
direction of heat flow, and to the temperature difference along the path of heat flow, dT/dx.
ii) Newtons Law of cooling : “The rate of cooling of a body is directly proportional to the
difference in temperature of the body (T) and surrounding (T o), provided difference in
temperature should not be exceed by 30 OC
iii) Radiation – It is process of heat transfer between two bodies without any carrying
medium through different kind of electro-magnetic wave.
iv) Thermal Conductivity : It is defined as amount of energy flow through a body of unit
area and unit thickness in unit time when the difference in temperature between the faces
carrying the heat flow is 1 0C . Thermal conductivity depends on molecular structure, specific
gravity etc .
d
Advantages of feed heating: (1 mark each)
01 for
each
1.The thermal efficiency of boiler increases as heat input decreases.
2. Capacity of Condenser changes
3. Reduce fuel consumption.
4. Thermal stress in the boiler reduces as temperature difference is decrease due to hot
feed water is supplied
5. Overall efficiency of the plant increase.
e
Volume of balloon = π r3
V=
π (5)3 = 523 .6 m3
MR = 8.3143
R=
= 4.15715 kJ/kg K
Pressure of hydrogen in the balloon = Atmospheric pressure
= 101.325 kN / m2
2
Applying gas equation , PV = mRT
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Mass of air hydrogen in balloon =
= 101.325 x 523.6 / 4.15715 x (20 +273)
= 43.56 kg
The volume of air displaced by the balloon = Volume of the balloon
m=
= (101.325 x 523.6) / 0.287 x ( 20 +273)
= 630 .91 kg
Total load lifted by the balloon = 630.91 - 43.56
2
= 587.35 kg
5
a(i)
Throttling:
When fluid or gas flow through the restricted passage like a plate with partially opened valve
1
or suddenly reduce the diameter of the pipe pressure drop occur. The kinetic energy at the
inlet and outlet is very small and there is no change in potential energy and there is no work
done and there is enough time to appreciate heat transfer. It can show that there is an abrupt
change in pressure (high pressure converted into low pressure) between the inlet and outlet at
constant enthalpy. It is called throttling process.
Purpose of throttling:
2
1. For determining the condition of steam
2. used in refrigeration plant
3. Liquefaction of gas
4. in many cryogenic application
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ii
3
b
2
Figure: Impulse Turbine
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Construction:
Impulse turbine consist of one fixed set of nozzle mounted on a stationary diaphragm that
2
orient the steam flow into high speed jets, which is followed by one set of moving blade
ring as shown in Fig. for a single stage impulse turbine.
Working:
In impulse turbine power is developed by the impulsive force of high velocity jet or jets
2
which contain significant kinetic energy which is converted in to shaft rotation by the
bucket-like shaped rotor blades, as the steam jet changes direction.
A pressure drop occurs across only the stationary blades, with a net increase in steam
velocity across the stage. As the steam flows through the nozzle its pressure falls from
inlet pressure to the exit pressure.
The high velocity steam jets are obtained by complete expansion of steam in the
stationary nozzles fitted in diaphragm then this velocity steam passes through moving
blade with no drop in pressure but a gradual reduction in velocity.
In pure impulse steam turbine the high velocity jet from nozzle strikes on the blades
mounted on the wheel attached to the shaft.
Theses blades change the direction of steam and hence momentum of the jet of steam which
rotate the shaft.
c
2
Figure: Plate type heat exchanger
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It consists of closely spaced parallel plates fins held in between. The plate separate the two
fluids which flow through passages alternately, formed between plates.it also has fin attached 2
over primary heat transfer surface so as to increase heat transfer areas. This improves the
effectiveness of heat exchanger.
The counter flow or parallel flow arrangement can also be possible. The fin may be plain fin
are attached to plate by brazing or soldering. They are more suitable for gas to gas
application.
Applications:
a. Milk chilling plants
2
b. Radiator in automobile
c. Air conditioning
d. Food industries
6
a
Induced draught cooling towers:
3
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Figure: Induced draught cooling tower
Construction:
In this, fan is located at the top of the tower and air enters the side of the tower. The hot
water from the condenser is sprayed in the tower from top. Drift eliminator are attached
3
below the fan to remove the water in the air. Louvers are attached both the side of the
tower for air.
Working:
Depending on the air inlet and flow pattern, induced draft towers are of two types, crossflow and counter flow towers.
Figure shows that schematic diagram of a induced draught cooling tower. In this system,
a fan is installed at the top of the cooling tower.
The hot water from the condenser is supplied at the top of the cooling tower which is sprayed
through the nozzles. Fan sucks the air from louvers and cools the water. The water in the air
is eliminated by drift eliminator.
b
2
Figure: Loeffler Boiler
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Construction :
1. Loeffler boiler consists of evaporator drum, which may be placed at any convenient
point outside the furnace setting.
2. The evaporator drum which is used provided with set of nozzle through which steam
2
enters in evaporator drum.
3. Nozzles are made of special design to avoid priming and noise.
4. The feed water pumps feed the water to economizer, which is placed in the path of
flue gases.
5. This is water tube boiler using a forced circulation.
6. In this boiler water is heated mainly by means of superheated steam.
7. The steam will act as heat carrying and heat absorbing medium.
8. Thus, boiler uses the circulation of steam instead of water and difficulty of deposition
of salt and sediment in boiler tubes is completely eliminated.
Working:
The economizer extracts sensible heat from flue gases and hot water
1. at temp. close to saturation temp. is passed to evaporator drum.
2. From super heater big portion of steam (about 3/4) is trapped off for external use and
remainder portion (about 1/4) is passed to evaporator drum.
2
3. The steam from evaporator drum is passed to super heater through circulating pump.
4. The air preheater maybe placed in path of flue gases to supply the hot air in
combustion chamber.
Loeffler boiler has steam-generating capacity of 100 tons/hour at 140 bar pressure.
C (i)
 The main sources of air leakage found in condenser are given below:
1)
There is leakage of air from atmosphere at the joint of the parts which are internally
under a pressure less than atmospheric pressure.
2)
Air is also accompanied with steam from the boiler into which it enters dissolved in
feed water.
3)
3
In jet condensers, a little quantity of air accompanies the injection water.
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(ii)
3
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