LAB MANUAL
THERMODYNAMIC
EXPERIMENT # 1
Objective:-
To Find the Co-efficient of Linear Expansion of the
Copper Material due to Steam
Apparatus:-
Steam Generator, Copper rod, DMM,
Thermocouple, Dial gauge, Beaker, Rubber tubing, Meter rod.
Theory:-
Definition:-
“The co-efficient of linear
expansion or thermal expansion of a substance is the fraction of its original length
by which the rod of the substance expands per degree rise in temperature.”
Mathematically:-
α= ∆L/(L1∆T)
Most materials expand somewhat when heated
through a temperature range that does not produce a change in phase. The added heat
increases the average amplitude of the vibrations of the atoms in the material which
increases the average separation between the atoms.
In
an isotropic material, the expansion occurs equally in all dimensions. If an object
of length “L” is heated through a small temperature exchange “∆T”, then the change
in length“∆L” is directly proportional to the original length “L” and to change
in temperature.
Hence,
∆L ∝ (L)( ∆T)
∆L= α (L)( ∆T)
This
proportionality constant “α” is known as co-efficient of linear expansion.
In this experiment α, is measured for
copper and copper is isotropic i.e. same expansion in all directions, so that “α”
needs only to be measured along one dimension. Also within the limits of the experiment,
α does not vary with temperature.
Procedure:-
·
Record the room temperature
and measure “L” length of copper rod at room temperature and mount the rod in expansion
base.
·
Place the
thermocouple sensor in immediate contact with rod or mid length and also
connect the thermocouple with DMM, where it is used to measure “∆T”.
·
Fill the steam generator
two-thirds with water.
·
Use tubing to attach
your steam generator to the end of copper rod. Attach it to the end farthest from
the dial gauge.
·
Lead the tubing from
expansion apparatus into a beaker well below the level of apparatus which is used
to catch the draining water.
·
Turn the outer
casing of dial gauge to a length “zero” point on the scale with the long
Indicator needle will move in a counter-clock wise direction.
·
Turn on the steam
generator. As steam begins to flow, watch the dial gauge and DMM. Also record the
expansion of rod length (∆L) as indicated by the displacement of indicator on dial
gauge.
·
(Each increment on
dial gauge is equivalent to 0.01mmof rod expansion)
·
Find out the “α”.
·
Disconnect the
apparatus, empty the steam generator and beaker leave everything in neat shape.
Calculation:-
Material
|
Initial
Temperature
T1
(C)
|
Final
Temperature
T2
(C)
|
Original
Length
(cm)
|
Change
in
Length
(cm)
|
Change
in
Temperature
∆T
(C)
|
Coefficient
of
Linear
Expansion
α
|
Copper
|
Conclusion
EXPERIMENT
# 2
Objective:-
To Find the Co-efficient of Linear Expansion of the Aluminium
Material due to Steam
Apparatus:-
Steam Generator, Aluminium rod, DMM,
Thermocouple, Dial gauge, Beaker, Rubber tubing, Meter rod.
Theory:-
Definition:-
“The co-efficient of linear
expansion or thermal expansion of a substance is the fraction of its original length
by which the rod of the substance expands per degree rise in temperature.”
Mathematically:-
α= ∆L/(L1∆T)
Most materials expand somewhat when heated
through a temperature range that does not produce a change in phase. The added heat
increases the average amplitude of the vibrations of the atoms in the material which
increases the average separation between the atoms.
In an isotropic material, the expansion
occurs equally in all dimensions. If an object of length “L” is heated through a
small temperature exchange “∆T”, then the change in length“∆L” is directly
proportional to the original length “L” and to change in temperature.
Hence,
∆L ∝ (L)( ∆T)
∆L= α (L)( ∆T)
This
proportionality constant “α” is known as co-efficient of linear expansion.
In this experiment α, is measured for
copper and copper is isotropic i.e. same expansion in all directions, so that “α”
needs only to be measured along one dimension. Also within the limits of the experiment,
α does not vary with temperature.
Procedure:-
·
Record the room temperature
and measure “L” length of copper rod at room temperature and mount the rod in expansion
base.
·
Place the
thermocouple sensor in immediate contact with rod or mid length and also
connect the thermocouple with DMM, where it is used to measure “∆T”.
·
Fill the steam generator
two-thirds with water.
·
Use tubing to attach
your steam generator to the end of copper rod. Attach it to the end farthest from
the dial gauge.
·
Lead the tubing from
expansion apparatus into a beaker well below the level of apparatus which is used
to catch the draining water.
·
Turn the outer
casing of dial gauge to a length “zero” point on the scale with the long
Indicator needle will move in a counter-clock wise direction.
·
Turn on the steam
generator. As steam begins to flow, watch the dial gauge and DMM. Also record the
expansion of rod length (∆L) as indicated by the displacement of indicator on dial
gauge.
·
(Each increment on
dial gauge is equivalent to 0.01mmof rod expansion)
·
Find out the “α”.
·
Disconnect the
apparatus, empty the steam generator and beaker leave everything in neat shape.
Calculation:-
Material
|
Initial
Temperature
T1
(C)
|
Final
Temperature
T2
(C)
|
Original
Length
(cm)
|
Change
in
Length
(cm)
|
Change
in
Temperature
∆T
(C)
|
Coefficient
of
Linear
Expansion
α
|
Aluminium
|
Conclusion:-
EXPERIMENT
# 3
Objective:-
To Find the Co-efficient of Linear Expansion of the Steel
Material due to Steam
Apparatus:-
Steam Generator, Steel rod, DMM,
Thermocouple, Dial gauge, Beaker, Rubber tubing, Meter rod.
Theory:-
Definition:-
“The co-efficient of linear
expansion or thermal expansion of a substance is the fraction of its original length
by which the rod of the substance expands per degree rise in temperature.”
Mathematically:-
α= ∆L/(L1∆T)
Most materials expand somewhat when heated
through a temperature range that does not produce a change in phase. The added heat
increases the average amplitude of the vibrations of the atoms in the material which
increases the average separation between the atoms.
In an isotropic material, the expansion
occurs equally in all dimensions. If an object of length “L” is heated through a
small temperature exchange “∆T”, then the change in length“∆L” is directly
proportional to the original length “L” and to change in temperature.
Hence,
∆L ∝ (L)( ∆T)
∆L= α (L)( ∆T)
This
proportionality constant “α” is known as co-efficient of linear expansion.
In this experiment α, is measured for
copper and copper is isotropic i.e. same expansion in all directions, so that “α”
needs only to be measured along one dimension. Also within the limits of the experiment,
α does not vary with temperature.
Procedure:-
·
Record the room temperature
and measure “L” length of copper rod at room temperature and mount the rod in expansion
base.
·
Place the
thermocouple sensor in immediate contact with rod or mid length and also
connect the thermocouple with DMM, where it is used to measure “∆T”.
·
Fill the steam generator
two-thirds with water.
·
Use tubing to attach
your steam generator to the end of copper rod. Attach it to the end farthest from
the dial gauge.
·
Lead the tubing from
expansion apparatus into a beaker well below the level of apparatus which is used
to catch the draining water.
·
Turn the outer
casing of dial gauge to a length “zero” point on the scale with the long
Indicator needle will move in a counter-clock wise direction.
·
Turn on the steam
generator. As steam begins to flow, watch the dial gauge and DMM. Also record the
expansion of rod length (∆L) as indicated by the displacement of indicator on dial
gauge.
·
(Each increment on
dial gauge is equivalent to 0.01mmof rod expansion)
·
Find out the “α”.
·
Disconnect the
apparatus, empty the steam generator and beaker leave everything in neat shape.
Calculation:-
Material
|
Initial
Temperature
T1
(C)
|
Final
Temperature
T2
(C)
|
Original
Length
(cm)
|
Change
in
Length
(cm)
|
Change
in
Temperature
∆T
(C)
|
Coefficient
of
Linear
Expansion
α
|
Steel
|
Conclusion:-
EXPERIMENT
# 4
Objective:-
To Study a Steam
Power Plant.
Theory:-
First of all we
understand the flow
chart
diagram
of the steam power
plant, after
that we define its
major parts.
Now in next few pages we define
major
working parts
of the steam power
plant which
are:
·
Steam Turbine,
·
Boiler,
·
Condenser,
· Generator.
Steam Turbine:-
A steam turbine
is a device which
extracts
thermal energy
from pressurized
steam and uses it
to do
mechanical
work on a rotating
output shaft.
Because the
turbine generates rotary motion, it is particularly suited to
be used to drive an electrical generator.
Steam turbine
working to
change the
heat
energy
contained in
the
steam
in to rotary motion. Steam with high
pressure and temperature were
directed
to push
turbine
blades mounted
on the shaft,
so the shaft rotates. Due to
perform
work
on the turbine, the pressure
and
temperature of steam coming
into the turbine down to saturated vapor. This steam then
flows to
the
condenser, while the rotary power is
used
to turn
a generator. Today almost all
of the steam turbine
is a
type of condensing turbine.The
steam turbine works on ideal Rankine cycle, which is defined below:
T-s diagram of Ideal Rankine Cycle:-
There are four
processes
in the Rankine cycle. These
states are
identified by
numbers
(in brown) in the
above T-s diagram.
· Process1:-
The working
fluid
is pumped
from
low to
high pressure.
As the fluid
is a liquid
at this stage, the pump requires
little input
energy.
· Process2:-
The high pressure liquid enters
a boiler where it
is heated
at constant pressure
by an external heat source
to become a dry
saturated
vapor. The input
energy
required can be easily
calculated
graphically, using an enthalpy-entropy chart (akah-schart or Mollier
diagram),or numerically, using steam tables.
· Process3:-
The dry saturated
vapor
expands
through
a turbine, generating power.
This
decreases the temperature and
pressure of the
vapor,
and
some
condensation may occur. The
output
in this process can
be easily calculated
using
the chart or tables noted
above.
· Process4:-
The
wet vapor then enters a condenser where
it is condensed at a constant pressure to become a saturated
liquid. The
efficiency of the Rank in the cycle is limited by the high heat of vaporization
of the working fluid. Also, unless the pressure and temperature reach supercritical levels in the steam boiler, the temperature range
the cycle can operate over is quite small steam turbine entry temperatures are
typically around 565°C and steam condenser temperatures are around 30°C.
Boiler:-
Boiler has
the
function to
convert
water
in to steam. The
process
of change
of water to
vapor done by
heating
the
water
in the pipes
with heat
from
burning fuel. Combustion processes
carried out
continuously in the combustion
chamber with fuel
and air flow from the
outside.
The resulting
steam
is superheat steam which
have high
temperature and high pressure.
Steam
production quantities
dependent on
the surface area of heat transfer, flow
rate,
and
the
heat of combustion is
given.
Boiler
construction
consisting of
water-filled pipes called a water tube
boiler.
Condenser:-
Condensers are devices
to convert steam into
water.
The
changes done
by the steam flowing to
a room containing
tubes. Steam flows outside
tubes,
while
the
cooling water flowing inside
the tubes.
This is called
surface condenser. Usually for
coolant
use
sea water.
Heat transfer rate depends
on the flow of cooling
water, sanitation
tubes
and
the
temperature difference
between the steam and
cooling
water.
The
process
of change into
water
vapor occurs
at saturated
pressure
and temperature,
in this
case
the
condenser is
under vacuum.
Because the cooling water temperature
equal to
the
outside temperature, the
maximum
temperature
condensate water near
the
outside
air
temperature. If the rate of heat
transfer
interrupted it
will affect the pressure and temperature.
Generator:-
The main purpose of
the activities at
a plant
is electricity.
Electrical energy generated
from the generator Function generator
converts
mechanical energy into electrical energy in
the
form
of around with
the
principle of
magnetic
induction. Generator
consists
of stator and
rotor. Stator
consists of
the casing which
contains coils and a rotor magnetic
field
station
consists of
a core containing
a coil.
EXPERIMENT
# 5
Objective:-
To Find the Pressure
by U Tube Manometer
U Tube Manometer:-
Technically a manometer is any device used to measure
pressure. However, the word manometer is commonly used to mean a pressure
sensor which detects pressure change by means of liquid in a tube. The U-tube
manometer is somewhat self-descriptive. In its basic form it consists of a
clear glass or plastic tube shaped into the form of a 'U'. The tube is
partially filled with a liquid, such as water, alcohol, or mercury (although for
safety reasons mercury is no longer commonly used). The lower the density of
the liquid is the higher the sensitivity of the manometer.
Theory:-
The
unknown pressure is applied in the one arm of the tube and the mercury in the
tube or manometeric liquid filled in the tube moves in the tube or rises to the
constant region and then the movement is stopped. The height of the liquid is
measured and noted. The pressure is calculated by using the formula:
P = pgh
Where,
P
is the required pressure
ρ
= density (kg/m3, lb/ft3)
g
= acceleration of gravity (9.81 m/s2, 32.174 ft/s2)
h =
liquid height (m, ft)
The
pressure (p) to be measured is to be compared with the height (h) of a liquid
column. If the pressure exerted on the two surfaces of the so-called confined
liquid is not the same, there is a deflection and consequently a difference in
height. The confined liquid continues to rise until the effect of the force of
the pressure differentials and the weight of the liquid columns are identical.
In accordance with the laws of physics, the effect of the liquid column on the
pressure in the liquid is, in essence, only dependent on height (h) of the
liquid column and on density (rm) of the liquid. Further influences are
relatively low and known. For highly precise measurements, correction
calculations can be made. Recalibration is not necessary.
Procedure:-
·
First check the
zero error of the instrument.
·
If zero error is
present then manage it by reconnecting the joints of tube until U tube 1 and U
tube 2 reaches at the same level.
·
Then take a
plinger and connect it to the tube.
·
Check the reading
of pressure gauge at zero millibar.
·
By increasing the
pressure by the plinger take the pressure reading from zero to 10 millibar on
the pressure gauge.
·
Note the values
in the table
·
Take the readings
from U tube 1 and U tube 2 and write them in the table.
·
Find the value of
h by using formula
Δh=U1-U2
·
Calculate
pressure P using formula
P =
pgh
·
Now take 5
consecutive readings by changing the pressure on pressure gauge
This measurement is called gauge pressure, and the
relationship for a positive pressure is expressed by:
Absolute pressure
= Atmospheric pressure + Positive gauge pressure
Calculation:-
For Positive Pressure
Sample #
|
U
Tube 1 (mm)
|
U
Tube 2 (mm)
|
Δh
U1-U2
|
P = pgh
|
Pressure
(millibar)
|
1
|
|||||
2
|
|||||
3
|
|||||
4
|
|||||
5
|
Advantages of U-tube
Manometer:-
² Simple in construction
² Low cost
² Very accurate and sensitive
² It can be used to measure other process
variables.
Disadvantages of U-tube
Manometer:-
² Fragile in construction.
² Very sensitive to temperature
changes.
² Error can happen while measuring the h.
Characteristics of liquid
used in U-tube Manometer:-
While choosing the manometer fluid for a particular
application we need to remember following things.
² Manometer fluid should not wet the wall
² Manometer fluid should not absorb gas
² Manometer fluid should not react chemically
² Manometer fluid should have low vapor pressure
² Fluid should move freely
Mercury is one of the most commonly
used manometer fluid
Conclusion:-
EXPERIMENT
# 6
Objective:-
To Find the Pressure
by U Tube Manometer above and below
U Tube Manometer:-
Technically a manometer is any device used to measure
pressure. However, the word manometer is commonly used to mean a pressure
sensor which detects pressure change by means of liquid in a tube. The U-tube
manometer is somewhat self-descriptive. In its basic form it consists of a
clear glass or plastic tube shaped into the form of a 'U'. The tube is
partially filled with a liquid, such as water, alcohol, or mercury (although
for safety reasons mercury is no longer commonly used). The lower the density
of the liquid is the higher the sensitivity of the manometer.
Theory:-
The
unknown pressure is applied in the one arm of the tube and the mercury in the
tube or manometeric liquid filled in the tube moves in the tube or rises to the
constant region and then the movement is stopped. The height of the liquid is
measured and noted. The pressure is calculated by using the formula:
P = pgh
Where,
P is the required pressure
ρ = density (kg/m3, lb/ft3)
g = acceleration of gravity (9.81
m/s2, 32.174 ft/s2)
h = liquid height (m, ft)
The
pressure (p) to be measured is to be compared with the height (h) of a liquid
column. If the pressure exerted on the two surfaces of the so-called confined
liquid is not the same, there is a deflection and consequently a difference in
height. The confined liquid continues to rise until the effect of the force of
the pressure differentials and the weight of the liquid columns are identical.
In accordance with the laws of physics, the effect of the liquid column on the
pressure in the liquid is, in essence, only dependent on height (h) of the
liquid column and on density (rm) of the liquid. Further influences are
relatively low and known. For highly precise measurements, correction
calculations can be made. Recalibration is not necessary.
Procedure:-
·
First check the
zero error of the instrument.
·
If zero error is
present then manage it by reconnecting the joints of tube until U tube 1 and U
tube 2 reaches at the same level.
·
Then take a
plinger and connect it to the tube.
·
Check the reading
of pressure gauge at zero millibar.
·
By increasing the
pressure by the plinger take the pressure reading from zero to 10 millibar on
the pressure gauge.
·
Note the values
in the table
·
Take the readings
from U tube 1 and U tube 2 and write them in the table.
·
Now reset the
tube setting to initial stage.
·
Now take 5
consecutive readings by changing the pressure on pressure gauge
·
By using the
plinger apply negative pressure (vacuum) till the dial gauge shows -10.
·
Note the readings
from U tube 1 and U tube 2.
·
Find the value of
h by using formula
Δh=U1-U2
·
Calculate
pressure P using formula
P =
pgh
·
Now take 5
consecutive readings by changing the pressure on pressure gauge
This measurement is called gauge pressure, and the
relationship for a positive pressure is expressed by:
Absolute
pressure = Atmospheric pressure + Positive gauge pressure
|
For a negative pressure (vacuum) measurement the
column heights reverse and the relationship is expressed by:
Absolute
pressure = Atmospheric pressure + Negative gauge pressure
|
Calculation:-
For Positive Pressure
Sample #
|
U
Tube 1 (mm)
|
U
Tube 2 (mm)
|
Δh
U1-U2
|
P = pgh
|
Pressure
(millibar)
|
1
|
|||||
2
|
|||||
3
|
|||||
4
|
|||||
5
|
For Negative Pressure
Sample #
|
U
Tube 1 (mm)
|
U
Tube 2 (mm)
|
Δh
U1-U2
|
P = pgh
|
Pressure
(millibar)
|
1
|
|||||
2
|
|||||
3
|
|||||
4
|
Advantages of U-tube
Manometer:-
² Simple in construction
² Low cost
² Very accurate and sensitive
² It can be used to measure other process
variables.
Disadvantages of U-tube
Manometer:-
² Fragile in construction.
² Very sensitive to temperature
changes.
² Error can happen while measuring the h.
Characteristics of liquid
used in U-tube Manometer:-
While choosing the manometer fluid for a particular
application we need to remember following things.
² Manometer fluid should not wet the wall
² Manometer fluid should not absorb gas
² Manometer fluid should not react chemically
² Manometer fluid should have low vapor pressure
² Fluid should move freely
Mercury is one of the most commonly
used manometer fluid.
Conclusion:-
0 Comments