Lab Manual for Heat Transfer Of MDU ROHTAK
L T P Sessional : 50 Marks
- - 3 Practical : 50 Marks
Total : 100 Marks
Duration of Exam : 3Hrs.
List of Experiments:
1. To determine the thermal conductivity of a metallic rod.
2. To determine the thermal conductivity of an insulating power.
3. To determine the thermal conductivity of a solid by the guarded hot plate method.
4. To find the effectiveness of a pin fin in a rectangular duct natural convective condition and plot temperature distribution along its length.
5. To find the effectiveness of a pin fin in a rectangular duct under forced convective and plot temperature distribution along its length.
6. To determine the surface heat transfer coefficient for a heated vertical tube under natural convection and plot the variation of local heat transfer coefficient along the length of the tube. Also compare the results with those of the correlation.
7. To determine average heat transfer coefficient for a externally heated horizontal pipe under forced convection & plot Reynolds and Nusselt numbers along the length of pipe. Also compare the results with those of the correlations.
8. To measure the emmisivity of the gray body (plate) at different temperature and plot the variation of emmisivity with surface temperature.
9. To find overall heat transfer coefficient and effectiveness of a heat exchange under parallel and counter flow conditions. Also plot the temperature distribution in both the cases along the length of heat of heat exchanger.
10. To verify the Stefen-Boltzmann constant for thermal radiation.
11. To demonstrate the super thermal conducting heat pipe and compare its working with that of the best conductor i.e. copper pipe. Also plot temperature variation along the length with time or three pipes.
12. To study the two phases heat transfer unit.
13. To determine the water side overall heat transfer coefficient on a cross-flow heat exchanger.
14. Design of Heat exchanger using CAD and verification using thermal analysis package eg. I-Deas etc.
Note:
1. At least ten experiments are to be performed in the semester.
2. At least seven experiments should be performed from the above list. Remaining three experiments may either be performed from the above list or designed & set by the concerned institute as per the scope of the syllabus.
Experiment No:1
Aim: To determine the thermal conductivity of a metallic rod.
Apparatus Used: Voltmeter, Ammeter, Stop watch, Copper – constantan thermocouple, Power supply, Heating element, Digital temperature indicator and Voltage regulator.
Theory: Thermal conductivity is an important thermo - physical property of conducting materials, by virtue of which the material conducts the heat energy through it. From Fourier’s law of conduction the thermal conductivity is defined as
K=(Q/A)*(dx/dT)= q(dx/dT) in W/m.K
Where,
Q = heat transfer rate, watts,
q = heat flux, W /m2
A = area normal to heat transfer, m2, and
dT / dx = temperature gradient in the direction of heat flow=(T2-T1)/(x2-x1).
The thermal conductivity for a given material depends on its state and it varies with direction, structure, humidity, pressure and temperature change. Thermal energy can be transported in solids by two means.
1. Lattice vibration,
2. Transport of free electrons.
In good conducting materials, a large number of free electrons move about their lattice structure of metal. These electrons move from higher temperature region to lower temperature region, thus transport heat energy. Further, the increased temperature increases vibration energy of atoms in the lattice structure. Thus in hotter portion of the solid, the atoms which have larger vibration energy, transfer a part of its energy to the neighboring low energy molecules and so on throughout the whole length of the body.
Fig: Thermal Conductivity of Rod
Procedure:
1. Switch on the power supply and adjust voltage and current so as to allow some 500 watts input to heater coil.
2. Wait for steady state. Steady state can be observed by the temperature reading at one or all points on the surface of the metal rod. Steady state is reached when these temperatures stop changing with time.
3. Under steady state conditions note the temperature of each point on the surface of the rod as well as temperature of surrounding. Repeat above procedure for several power input.
Observations & Calculations:
Specifications: -
• Room temperature (T) = --------------------°C
• Diameter of the metal rod (d) = ------------ m
• Length of the metal rod (L) = ---------------m
• Digital voltmeter (0-220) Volt
• Digital ammeter (0-2)Amp
• Digital temperature indicator (0-300°C)
• Thermocouple (copper constantan)
• Heating filament (500 watts)
• Variac (voltage regulator) (0-2A, 0-230 V, AC supply )
Observation Table:
The following readings are noted as shown in the table after reaching steady state condition.
Note down x1,x2,x3,x4,x5,x6 i.e the distance between the base plate and corresponding temperature. i.e for x1 is the distance between the base plate and T1 temperature thermocouple. All temperatures are in °C.
Sr. no
|
V (Volt)
|
I (Amp)
|
Q =
(V x I)
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T1
|
T2
|
T3
|
T4
|
T5
|
T6
|
dT/ dx = (T2 – T1) / (x2 – x1)
|
K (W/mk)
|
1.
| |||||||||||
2.
| |||||||||||
3.
| |||||||||||
4.
|
Conclusion: Hence the thermal conductivity of a metallic rod is ________.
Experiment No:2
Aim: To determine the thermal conductivity of an insulating powder.
Apparatus Used: There are two spherical shells and the insulating material whose conductivity is to be calculated is packed between these two shells. A heating coil is provided in the inner shell. The power is supplied from outside through auto-transformer for heating purpose. A few copper-constantan thermocouples are provided along the radius of the inner and outer spheres, one is fixed on the outer-surface of inner sphere and one is fixed on the outer spheres, one is fixed on the outer-surface of inner sphere and one is fixed on the outer surface of the outer sphere. Sometimes 4 – thermocouples are provided on the inner sphere surface and four are provided on the outer sphere surface for finding the average temperatures on inner and outer sphere surfaces.
Theory: Thermal conductivity is one of the important properties of the materials and its knowledge is required for analyzing heat conduction problems. Physical meaning of thermal conductivity is how quickly heat passes through a given material. Thus the determination of this property is of considerable engineering significance. There are various methods of determination of thermal conductivity suitable for finding out thermal conductivity of materials in the powdered form.
Considering the transfer of heat by heat conduction through the wall of a hollow sphere formed by the insulating powdered layer packed between two thin copper spheres.
For,
ri = inner radius in meters.
r0 = outer radius in meters.
Ti = average temperature of the inner surface in °C.
T0 = average temperature of the outer surface in °C.
Where,
Ti = (T1+T2+T3+T4)/4
T0 = (T5+T6+T7+T8+T9+T10)/6
Note: - That T1 to T10 denote the temperature of thermocouples (1) to (10). Applying the Fourier law of heat conduction for a thin spherical layer of radius r and thickness dr with temperature difference dT the heat transfer rate.
q = -K(4πr2 )(dT/dr) its units are Kcal / hr ……………..(1)
Where,
K = thermal conductivity.
Separating the variables (q/4π K)(dr/r2)= dT …....(2)
Integrating eq.(2) in the limits of ri, r0 and Ti, and T0
(q/4π K)[(1/ ri)-(1/ r0)]=(Ti - T0)
q=4π K ri r0(Ti - T0)/( r0 – ri) …....(3)
K=q( r0 – ri)/ 4π ri r0(Ti - T0) …....(4)
Fig: Thermal Conductivity of an Insulating Powder
Procedure:
1. Increases slowly the input to heater by the dimmer stat starting from zero volt position.
2. Adjust input equal to 40watt max. by voltmeter and ammeter / Wattage (W= V x I)
3. See that this input remains constant throughout the experiment.
4. Wait till a satisfactory steady state condition is reached. This can be checked by reading temperatures of thermocouples (1) to (6) and note changes in their readings with time.
5. Note down the readings in the observations table as given.
Observations & Calculations:
Inner sphere
Thermocouple no.
|
1
|
2
|
3
|
4
|
Mean temperature Ti
Ti = 1 + 2 + 3 + 4
4
| |||
Temperature
°C
| ||||||||
Outer sphere
Thermocouple no.
|
4
|
5
|
6
|
Mean temperature To
To = 4 + 5 + 6
3
| ||
Temperature
°C
| ||||||
Conclusion: Hence the thermal conductivity of an insulating power is ________.
Experiment No:3
Aim: To determine the thermal conductivity of a solid by the guarded hot plate method.
Apparatus Used: Apparatus consists of the hot plate, the cold plate, the heater assembly, thermocouples and the specimens in position.
Theory:
For the measurement of thermal conductivity (K) what is required is to have a one dimensional
heat flow through the flat specimen, an arrangement for maintaining its faces at constant temperature and some metering method to measure the heat flow through a known area. To eliminate the distortion caused by edge losses in unidirectional heat flow, the central plate is surrounded by a guard ring which is separately heated. Temperatures are measured by calibrated thermocouples either attached to the plates or to the specimens at the hot and cold faces. Two specimens are used to unsure that all the heat comes out through the specimen only. Knowing the heat input to the central plate heater, the temperature difference across the specimen. Its thickness and the metering are. One can calculate K of the specimen by the following formula:
Where,
K = (Q / A) [L / (Th – Tc)] Kcal / hr – m - °C
Th = temperature of hot plate i.e. (T1 + T2) /2 in °C Tc = temperature of cold plate i.e. (T5 + T6) /2 in °C
K = thermal conductivity of sample Kcal / hr – m - °C
q = Heat flow rate in the specimen. Kcal / hr
A = metering area of the specimen. m2
L = thickness of specimen in m.
If the specimen thickness are different and the respective hot and cold temperatures are different then
K = (q / A) [ {t L1/ (Th1 – Tc1)} + {t L2/ (Th2 – Tc2)} ]
Where suffix 1 is the upper specimen and 2 lower specimens.
Outer heater having high temperature heat flow from high to low, so if outer heater temperature is high heat then inner heat will not flow in this direction thus ensures unidirectional heat flow. The heater plate is surrounded by a heating ring for stabilizing. The temperature of the primary heat and prevent heat loss radially around its edges. The primary and guard heater are made up of mica sheets in which is wound closely spaced Nichrome wire and packed with upper and lower mica sheets. These heaters together from a flat which together with upper and lower copper plates and rings form the heater plate assembly.
Two thermocouples are used to measure the hot face temperature at the upper and lower central heater assembly copper plates. Two more thermocouples are used to check balance in both the heater input.
Specimen is held between the heater and cooling unit on each side of the apparatus. Thermocouple no 5 and 6 measures the temperature of the upper cooling plate and lower cooling plate respectively.
Fig: Thermal Conductivity of Gaurded Hot Plate
Procedure:
1. The specimens are placed on either side of the heating plate assembly. Uniformly touching the cooling plates. Then the outer container is fitted with loose fill insulation such as glass wool supplied in small cloth packets.
2. The cooling circuit is started then calculated input is given to central and guard heaters through separate single phase supply lines with a dimmer state in each line desired temperature.
3. The guard heater input is adjusted in such a way that there is no radial heat flow which is checked from thermocouple readings and is adjusted accordingly. The input to the central heater (current and voltage or watts) and the thermocouple readings are recorded every 10 minutes till a reasonably steady state condition is reached. The readings are recorded in the observation table.
4. The final steady state values are taken for calculations.
Observations & Calculations: (Values given below may vary a/c to equipment in lab)
Central heater input (Inner heater) = V1 x T1 = W1
Guard heater input (outer heater) = V2 x T2 = W2
Specimen used = composite material or hard board
Specimen thickness = 72.5mm
Diameter of specimen = 120mm
Metering area
A = π / 4 ( 10 + X)2 = Cm2
Where X = 0.125cm
K = (Q / A) [L / (Th – Tc)] Kcal / hr – m - °C
Temperature measurement
Sr. no
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Power input (watts)
|
Metal rod surface temperature °C
|
Average temperature
°C
| |||||
T1
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T2
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T3
|
T4
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T5
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T6
| |||
1
2
3
4
5
| ||||||||
Conclusion: Hence the thermal conductivity of a solid is ________.
Experiment No:4
Aim: To find the effectiveness of a pin fin in a rectangular duct natural convective condition and plot temperature distribution along its length.
Apparatus Used: A pin fin is screwed to a cylindrical heating pipe making the test section. The test section is kept in a rectangular duct, which is open of both ends for natural air flow. One acrylic sheet is also fitted in the duct so as to see the duct and test section. When electric current input and the test- section, heat transfer takes place from cylinder to pin fin, the temperature of fin can be measured by digital temperature indicator with the help of copper – constant thermocouples. Five thermocouples are embedded on pin fin at intervals of 25mm. One thermocouple is used to measure surrounding temperature of the duct.
Fig: Pin Fin Natural Convection
Theory:
Heat transfer due to free convection is described by Newton
Q = h A (TS-T∞) = h A ∆T
The rate of heat transferred to the surrounding fluid is proportional to the object's exposed area A, and the difference between the object temperature TS and the fluid free-stream temperature T∞.
An extended surface (also known as a combined conduction-convection system or a fin) is a solid within which heat transfer by conduction is assumed to be one dimensional, while heat is also transferred by convection (and/or radiation) from the surface in a direction transverse to that of conduction.
Fin effectiveness: Ratio of the fin heat transfer rate (qf) to the heat transfer rate that would exist without the fin.
Єf = (qf) / [hAb (Tb-T∞)]
Where, qf = (Tb-T∞) / Rt,f
Where, Rt,f = 1 / (hAfηf)
Fin efficiency (ηf): The ratio of the actual heat transfer rate from the fin to the maximum rate at which a fin could dissipate energy.
ηf = qf / qmax = qf /[hAf(Tb-T∞)]
Tb is temperature at base of fin
T∞ is fluid free-stream temperature
Ab is area of cross section at base of the fin
A fin of brass (iron or copper) of circular cross-section (square or triangular) or length L is fitted in rectangular duct. One end of the fin which is projected outside duct is provided with a heater for heating the fin. Five thermocouples are provided on the surface of the fin at equal distances. The duct is provided with a blower to control the air flow with the help of valve.
Procedure:
1. Switch on the power supply to control panel.
2. Increase power input to heating cylinder by means of dimmer state to desired heat input.
3. Wait till steady state condition reaches.
4. Take the readings of parameters required for calculations.
5. Repeat above for various heat input, say 50 watts, 100 watts and 150 watts.
Observations & Calculations:
Steady state fin temperature for electric input of ------------W
SI. no.
|
Distance from base of fin x L mm
|
Temperature along length of fin
|
TS
|
Average temperature
| ||||
T8
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T9
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T10
|
T11
|
T12
| ||||
1
2
3
4
5
6
| ||||||||
Fin length = 150mm
Diameter = 5 mm
Room temperature = T∞ – TS °C
Metal brass K = ?
h = given
P = peripheri= πD
A = Area= πD2/4
Distance base of fin x = 150mm
Temperature at x.T = °C
From graph of X averages I = Fin
base temperature TS = °C
Temperature grain [ dt / dx]x = 0
q = KAm (Ts - T∞) tan h m.L
m = √h.P./KA
Fin efficiency (η) = (tan h m.L) / m.L
Draw the graph between temperature and length of fin.
Conclusion: Hence the effectiveness of a pin fin in a rectangular duct under natural convection is ________.
Experiment No:5
Aim: To find the effectiveness of a pin fin in a rectangular duct under forced convective and plot temperature distribution along its length.
Apparatus Used: Pin fin apparatus for forced convection.
Theory:
Procedure:
Difference between natural and forced convection procedures:
Comparisons of Calculations for forced and natural convections:
Comparisons of Observations for forced and natural convection's:
Conclusion: Hence the effectiveness of a pin fin in a rectangular duct under forced convection is ______.
Experiment No:6
Aim: To determine the surface heat transfer coefficient for a heated vertical tube under natural convection and plot the variation of local heat transfer coefficient along the length of the tube. Also compare the results with those of the correlation.
Apparatus Used: Natural convection apparatus - Consists of a tube fitted vertically in a rectangular duct which is open at the top and the bottom as shown in figure below:
Fig: Tube Natural Convection
An electric heater is provided in the vertical tube which heats the tube surface. Heat is lost from the tube to the surrounding air by natural convection. Air around the tube gets heated up and becomes less dense, causing it to rise. This in turn gives the continuous flow of air upward in the duct. The temperature at the various locations on the surface of the vertical tube is measured using thermocouples.
Theory:
The rate of heat transfer Q/t at which a hot object transfers heat to a surrounding fluids by convection is approximately proportional to the area A of the object in contact with it and to the temperature difference ∆T between them.
Q/t = hA∆T
Where, h is the convection heat transfer co efficient, depends on the shape and orientation of the object . It is used as a factor for calculating heat transfer between a fluid and a solid, between fluids separated by solids and between solids. Its unit is Wm-2 K-1.
Procedure:
1. Adjusting the voltage and corresponding current given as the input power.
2. Power on button- This button is on when all the above initial adjustment is done.
3. Temperature indicator in setup should come to steady. Note down the T1 to T6 temperatures on thermocouples in degree Celsius.
Observations & Calculations:
Plot the variation of local heat transfer coefficient along the length of the tube.
Conclusion: Hence the surface heat transfer coefficient for a heated vertical tube under natural convection is ________.
Experiment No:7
Aim: To determine average heat transfer coefficient for a externally heated horizontal pipe under forced convection & plot Reynolds and Nusselt numbers along the length of pipe. Also compare the results with those of the correlations.
Fig: Tube Forced Convection
Theory:
SPECIFICATIONS:
1. Pipe diameter = (D0) = 33mm;
Pipe diameter = (Di) =28mm.
2. Length of test section (L) = 400mm.
3. Blower -1/2 HP motor
4. Orifice dia. = (d) = 14mm. connected to water manometer.
5. Dimmerstate 0 to 2Amp. , 260 Volts, A .C.
6. Temperature Indicator – Range 0 to 3000 C.
7. Voltmeter 0 – 100/200 V, Ammeter 0-2 A.
8 Heater –400 watt.
Procedure:
1. Average surface heat transfer coefficient for a pipe losing heat by forced convection to air flowing through it can be obtained forced convection to air flowing through it can be obtained for different air flow and heat flow rates.
2. Reynolds’s s number and Nusselt number for each experimental condition can be calculated. Plot on these values on log –log graph.
3. Keep the dimmerstate at zero position before switching ON the power supply.
4. Start the blower unit.
5. Increase the voltmeter gradually.
6. Do not stop the blower in between the testing period.
7. Do not disturb thermocouples while testing.
8. Operate selector switch of the Temperature Indicator gently.
9. Do not exceed 200 watts.
10. Start the blower and adjust the flow by means of gate valve to some desired difference in manometer level.
11. Start the heating of the test section with the help of dimmerstat and adjust desired heat input with the help of voltmeter and Ammeter.
12. Take reading of all the six thermocouples at an interval of 10 minutes until the steady state is reached.
13. Note down the heater input.
Observations & Calculations:
1. Outer diameter of the pipe (D0) = 33 mm.
2. Inner diameter of the test pipe (Di) =28 mm.
3. Length of the test (L) =400 mm.
4. Diameter of the orifice (d) =14 mm.
Conclusion: Hence the average heat transfer coefficient for a externally heated horizontal pipe under forced convection is _______.
Experiment No:8
Aim: To measure the emmisivity of the gray body (plate) at different temperature and plot the variation of emmisivity with surface temperature.
Apparatus Used: Emmisivity measurement apparatus: The experimental set up consists of two circular aluminum plates identical in size provide with heater coils at the bottom as shown in figure.
Fig: Emmisivity Apparatus
It is kept in an enclosure so as to provide undisturbed natural convection surroundings. The heat input to the heaters is varied by two regulators and is measured by an ammeter and voltmeter. Each plate is having three thermocouples; hence an average temperature is taken. One thermocouple is kept in the enclosure to read the chamber temperature. One plate is blackened by a layer of enamel black paint to form the idealized black surface whereas the other plate is the test plate. The temperatures of the plates are measured by using thermocouples.
Theory:
Emissivity of a surface is defined as ratio of the radiation emitted by the surface to the radiation emitted by the black body at the same temperature. If a sample is replaced by a black body of temperature of same area at same temperature, under thermal equilibrium, the emissivity of the body is equal to the absorptivity.
The same amount of power input is given to both test plate and black plate. After achieving steady state temperature for black plate, it continuously emits radiations and this radiation is completely absorbed by the test plate. But its emit radiation is slightly less than the black body because emissivity depends on nature of the material.
Procedure:
1. Adjusting the voltage and corresponding current given as input power.
2. Tuning Switch- Used to turn either Black plate (BP) or Test plate (TP) and thereby can change the corresponding voltage and current for both the plates.
Note: Power should be given for both the plates must be same.
3. Power on button- Using this button we can switch on the power when all the initial adjustments were done.
4. Temperature indicator should be reaching steady.
5. Note down the T1 to T7 temperature on thermocouples in degree Celsius.
Observations & Calculations:
Conclusion: Hence the emmisivity of the gray body (plate) at different temperature is ________.
Experiment No:9
Aim: To find overall heat transfer coefficient and effectiveness of a heat exchange under parallel and counter flow conditions. Also plot the temperature distribution in both the cases along the length of heat of heat exchanger.
Fig: Parallel and Counter Flow Heat Exchager
Theory: Heat exchangers are devices used to transfer heat energy from one fluid to another. Typical heat exchangers experienced by us in our daily lives include condensers and evaporators used in air conditioning units and refrigerators. Boilers and condensers in thermal power plants are examples of large industrial heat exchangers. There are heat exchangers in our automobiles in the form of radiators and oil coolers. Heat exchangers are also abundant in chemical and process industries. One of the fluids is usually passed through pipes or tubes, and the other fluid stream is passed round or across these. At any point in the equipment, the local temperature differences and the heat transfer coefficients control the rate of heat exchange. The fluids can flow in the same direction through the equipment, this is called parallel flow; they can flow in opposite directions, called counter flow; they can flow at right angles to each other, called cross flow.
In a parallel-flow heat exchanger fluids flow in the same direction. If the specific heat capacity of fluids are constant, it can be shown that:
Figure below shows a fluid flowing through a pipe and exchanges heat with another fluid through an annulus surrounding the pipe.
Fig: Parallel Flow
In a counter-flow heat exchanger fluids flow in the opposite direction. If the specific heat capacity of fluids are constant, it can be shown that:
Figure below shows a fluid flowing through a pipe and exchanges heat with another fluid through an annulus surrounding the pipe.
Fig: Counter Flow
Procedure:
1. Adjusting the voltage and corresponding current given as input power.
2. Power on button- Using this button we can switch on the power when all the initial adjustments were done.
4. Temperature indicator should be reaching steady.
5. Note down the temperature readings on thermocouples in degree Celsius.
Observations & Calculations:
Sl No
|
T1
|
T1’
|
T2
|
T2’
|
U = q / (A∆Tm)
Where q = heat transferred per unit time = m cp dT
Heat loss = heat gain
m1cp1(T1’-T2’)=m2cp2(T1-T2)
m= rate of mass flow of the fluid in kg/sec,
cp= specific heat in J/kg oC,
dT= T1-T2
A=πDL
∆Tm= log mean temperature difference=(∆T1-∆T2)/(ln[∆T1/∆T2])
∆T1=T1-T1’
∆T2=T2-T2’
Efficiency of heat exchanger, η=(1-e-α)/(1-{[m1cp1]/[m2cp2]} e-α)
Effectiveness for parallel flow & counter flow heat exchanger: Quite often, heat exchanger analysts are faced with the situation that only the inlet temperatures are known and the heat transfer characteristics (UA value) are known, but the outlet temperatures have to be calculated. Clearly, LMTH method will not be applicable here. In this regard, an alternative method known as the ε-NTU method is used.
Definition of effectiveness: The effectiveness, ε, is the ratio of the energy recovered in a HX to that recoverable in an ideal HX.
ε=Qactual/Qmax
Where Cmin=min. sp. Heat
Cmax=max. sp. heat
Plot for the temperature distribution in Parallel flow case along the length of heat of heat exchanger is as shown below:
Plot for the temperature distribution in Counter flow case along the length of heat of heat exchanger is as shown below:
Conclusion: Hence the overall heat transfer coefficient of a heat exchanger under parallel flow condition is _______, the overall heat transfer coefficient of a heat exchanger under counter flow condition is _______ and effectiveness of a heat exchanger under parallel flow condition is _______, effectiveness of a heat exchange under counter flow conditions is _______.
Experiment No:10
Aim: To verify the Stefen-Boltzmann constant for thermal radiation.
Apparatus Used: Heater, temperature-indicators, box containing metallic hemisphere with provision for water-flow through its annulus, a suitable black body which can be connected at the bottom of this metallic hemisphere.
Fig: Stefen-Boltzmann Experiment
Theory: A black body is an ideal physical body which absorbs all types of electromagnetic- radiation incident on it. Because of its 100% absorptivity, it is also the best emitter of thermal radiation. According to Stefan’s Boltzmann law (formulated by the Austrian physicists, Stefan and Boltzmann), energy radiated by a body per unit area per unit time is given by,
R=єσT4
where R =energy radiated per area per time,
Є =emissivity of the material of the body,
σ =Stefan’s constant=5.67x10-8 w/m2/K4,
T is the temperature in Kelvin scale.
For a black body, emissivity Є=1, and hence,
R=σT4
In the given experimental set up, the net heat transferred to the disc per second is,
(∆Q/∆t)=mcp(dT/dt)= σA(Th4- Td4)
Where,
Where m = mass of the disc in kg,
Cp=specific heat of the material of the disc,
A=area of the disc,
dT/dt=slope of the temperature –time
Td= steady state temperature of disc in Kelvin,
Td= steady state temperature of disc in Kelvin,
Th=hot temperature in Kelvin.
Form the above expression,
σ= [mcp(dT/dt)] / [A(Th4- Td4)]
Procedure:
1. Remove the disc from the bottom of the hemisphere and , switch on the heater and allow the water to flow through it.
2. Allow the hemisphere to reach the steady state and note down the temperature T1, T2, T3 .
3. Fit the disc (black body) at the bottom of the hemisphere and note down its rise in temperature, T4 with respect to time till steady state is reached.
4. A graph is plotted with temperature of disc along Y-axis and time along X-axis.
5. Find out the slope dT/dt from the graph.
Conclusion: Hence the Stefen-Boltzmann constant is _________.
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