te lab - ii manual.doc

7/30/2019 TE Lab - II Manual.doc

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1. THERMAL CONDUCTIVITY APPARATUS-GAURDED HOT

PLATE METHOD

AIM:

To find the thermal conductivity of the specimen by two slab guarded hot plate

method.

DESCRIPTION OF APPARTUS:

The apparatus consists of a guarded hot plate and cold plate. A

specimen whose thermal conductivity is to be measured is sand witched

between the hot and cold plate. Both hot plate and guard heaters are heated

by electrical heaters. A small trough is attached to the cold plate to hold

coolant water circulation. A similar arrangement is made on the other side

of the heater as shown in the figure. Thermocouples are attached to measure

temperature in between the hot plate and specimen plate, also cold plate and

the specimen plate.

A multi point digital temperature indicator with selector switch is

provided to note the temperatures at different locations. An electronic

regulator is provided to control the input energy to the main heater and guard

heater. An ammeter and voltmeter are provided to note and vary the input

energy to the heater.

The whole assembly is kept in an enclosure with heat insulating

material filled all around to minimize the heat loss.

SPECIFICATION:

Thickness of specimen = 2.5mm

Diameter of specimen (d) = 20cm


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MODEL CALCULATIONS:

FORMULA USED:

Since the guard heater enables the heat flow in uni direction

q = KA dT/dxWhere A = surface area of the test plate considered for heat flow = m2

dx = thickness of the specimen plate = m

dt = average temperature gradient across the specimen = c

q = Q/2 since the heat flow is from both sides of the heater = watts

Tavg1 = T1 + T2 / 2 ; Tavg2 = T3 +T4 / 2

Q = V.I. WattsQ = K1 A. dT / dx (for lower side)

Q = K1. d2/4 (Tavg1 T5)/dx

Where dx = 2.5mm = 0.0025m

Diameter of specimen


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d = 20cm = 0.2m

Q = K2 d2/4 . (Tavg2 T6)/dx ( for upper side)

KAvg = (K1 + K2 )/ 2

PROCEDURE:

1. Connect the power supply to the unit. Turn the regulator knob clockwise to

power the main heater to any desired value.

2. Adjust the guard heaters regulator so that the main heater temperature is

less than or equal to the guard heater temperature.

3. Allow water through the cold plate at steady rate. Note the temperatures atdifferent locations when the unit reaches steady state. The steady state is

defined, as the temperature gradient across the plate remains same atdifferent time intervals.

4. For different power inputs is in ascending order only the experiment may by

repeated and readings are tabulated as below.

RESULT:

The thermal conductivity of the specimen is found to be -------------

W/mK.


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2. HEAT TRANSFER FROM FINS

AIM:

To determine the temperature distribution of a PIN-FIN for forced convection and

FIN efficiency.

DESCRIPTION OF APPARATUS :

Consider a PIN-FIN having the shape of rod whose base is attached to a wall at a

surface temperature Ts, the fin is cooled along the axis by a fluid at temperature TAMB.

The fin has a uniform cross sectional area Ao is made of material having a uniform

thermal conductivity K and the average heat transfer co-efficient between the surface tothe fluid. We shall assume that transverse temperature gradients are so small so that the

temperature at any cross section of the fin is uniform.

The apparatus consists of a Pin-fin placed inside an open duct, (one side open) the

other end of the duct is connected to the suction side of a blower, the delivery side of a

blower is taken up through a gate valve and an orifice meter to the atmosphere. The

airflow rate can be varied by the gate valve and can be measured on the U tube

manometer connected to the orifice meter. A heater is connected to one end of the pin-

fin and seven thermocouples are connected by equal distance all along the length of the

pin and the eigth thermocouple is left in the duct.

The panel of the apparatus consists of voltmeter, ammeter and digital temperature

indicator. Regulator is to control the power input to the heater. U tube manometer with

connecting hoses.

SPECIFICATIONS:

Duct width b = 150 mm

Duct height w = 100 mmOrifice dia. do = 20 mm

Orifice co-efficient cd = 0.6

Fin length L = 14.5cm

Fin diameter df = 12mm


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(Characteristic length)

PROCEDURE:

1. Connect the three pin plug to a 230V, 50Hz, 15A power and switch on the unit.

2. Keep the thermocouple selector switch in first position.

3. Turn the regulator knob to clockwise and set the power to the heater to any

desired value by looking at the voltmeter and ammeter.

4. Allow the unit to stabilize.

5. Switch ON the blower.

6. Set the airflow rate to any desired value looking at the difference in U tube

manometer limb levels.7. Note down the temperatures indicated by temperature indicator.

8. Repeat the experiment by

a. Varying the airflow rate and keeping the power input to the

heater constant.

b. Varying the power input to the heater and keeping the air flow

rate constant.

9. Tabulate the readings and calculate for different conditions.

10. After all the experiment is over, put off the blower switch, turn the energy

regulator knob anti clockwise, put off the main switch and disconnect the power

supply.


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TABULATION:

Sl.No. Manometer reading

Fin surface temp.

Amb. temp.

h1 h2 T1 T2 T3 T4 T5 T6


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CALCULATIONS :

Volume of air flowing through the duct

Vo = cd a1a2 2gha / a12 a22

Where cd = co-efficient of orifice = 0.6

g = gravitational constant = 9.81 m/sec2

ha = heat of air = (lw /la)h

a1 = area of the pipe.

a2 = area of the orifice.

h = manometer differential head.Velocity of air in the duct = Vo / (W X B)

Where W = width of the duct.

D = breadth of the duct.

REYNOLDS NUMBER OF AIRFLOW:

Reynolds number Re = (L X Va X Pa) / a

Where Va = Velocity of air in the duct.

Pa = density of air in the duct.

a = Viscosity of air at to C.

L = length of fin.

PRANDTL NUMBER OF AIRFLOW

Prandtl number = (cpa X a ) / ka

Where cpa = specific heat of air.

a = viscosity of air


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ka = thermal conductivity of air.

HEAT TRANSFER CO-EFFICIENT CALCULATIONS

NUSSELT NUMBER (Nu)

For 40 < NRe < 4000

Nnu = 0.683 (NRe) 0.466 (NPr) 0.333

For 1 < NRe < 4

Nnu = 0.989 (NRe)0.33 (NPr)0.333

For 4 < NRe < 40

Nnu = 0.911 (NRe)0.385 (NPr)0.333

For 4000 < NRe < 40000Nnu = 0.193 (NRe)0.618 (NPr)0.333

For NRe > 40000

Nnu = 0.0266 (NRe)0.805 (NPr)0.333

Heat transfer co-efficient h = Nnu X (Ka / L)

Ka = thermal conductivity of air

L = length of fin.

Efficiency of the pin-fin = actual heat transferred by the fin(heat which would have

been transferred if entire fin where at the base temperature)

= Tan Hyperbolic ML/ML

Where, h = heat transfer co-efficient

L = length of the fin

M = hp/(kb X A)

P = perimeter of the fin

( X dia of the fin)

A = cross sectional area of the fin.

kb = thermal conductivity of brass rod.


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Temperature distribution = Tx = [cosh M (L-X) /cosh ML (To - Ta)] + Ta

X = distance between thermocouple and heater.

EVALUATION OF THE HEAT TRANSFER CO-EFFICIENT (h)

Natural convection (blower off)

Nuav = (hd)/k = 1.1 (Gr Pr)1/6 for 1/10 < Gr Pr < 104

Nuav = 0.53 (Gr Pr)1/4 for 104 < Gr Pr < 109

Nuav = 0.13 (Gr Pr)1/3 for 109 < Gr Pr < 1012

Where Nuav = average Nusselt number

= (hD) / k

D = Dia. of fin

K = thermal conductivity of air.

Gr = Grashof number = g T D3 / r2

= 1/ (Tav + 273)

T= (Tav Tamb)

Pr = Prandtl Number = ( Cp) / K

PIN-FIN

V A T1 T2 T3 T4 T5 T6 T7 T8 h1cm h2cm

135 0.6 67 61 59 56 49 47 46 29 75 15.5

Mean Temp = 51.75o C


10/45

Vol. of airflow thro duct = Q = Cd a1 a2 2gh / a12 - a22

a1 = /4 (0.04)2 = 1.256 X 10-3

a2 = /4 (0.02)2 = 3.14 X 10-4

h =w / a ( 0.155 0.075)

= 68.96m

Q = 8.704 X 10-6 / 1.216 X 10-3 = 7.158 X 10-3 m3/sec

Velocity of air flow thro duct = Q/A

A = Length X Breadth of the duct

= 0.15 X 0.1 = 0.015 m2

Velocity = 0.477 m/sec

Re =D / = D = Length of the Fin = 0.145

= 0.145 X0.477/ 17.95 X 10-6 = 3853

Using the correlation

For 40 > Re


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= 5.54 w/m2 o c

M = hp / k b A = 5.54 X 0.0376 / 110.7 X /4(0.012)2

= 4.078 kg/m

Fin efficiency = Tan G ML/ML =0.89 = 89o

Temp. distribution = [cosh M (L-X) /cosh ML (To - Ta)] + Ta

T2 = [cosh M (0.022) /1.17997 (67 - 29)] 0.85088(38)

= 61.33o c

T3 = 0.8588(32) + 29 = 56.22oc

T4 = 0.8588(30) + 29 = 54.5oc

T5 = 0.8588(27) + 29 = 51.9ocT6 = 0.8588(20) + 29 = 46oc

T7 = 0.8588(18) + 29 = 44oc

RESULT :

The efficiency of the fin is found to be ----------------------

Temperature at x = 20mm, T20 = -------------




3. TEST ON EMISSIVITY APPARATUS


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keep the toggle switch in position 2 and operate regulator 2 and feed power to the test

surface.

c) Allow the unit to stabilize. Ascertain the power inputs to the black and test surfaces are

at set values. i.e. equal.

d) Turn the thermocouple selector switch clockwise step by step and note down the

temperatures indicated by the temperature indicator from channel 1 to 7.

e) Tabulate the readings and calculate.

f) After the experiment is over turn both the energy regulators 1 & 2.

g) For various power inputs repeat the experiment.

TABULATION :

Sl.No.Black bodytemperature

AverageTemp.Tb

Polished bodytemperature

AverageTemp.Tp

Chamber

Temp. T4

Emmissivity

T5 T6 T7 T1 T2 T3


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CALCULATIONS:

Temperature of the black body in absolute unit T ba = T b + 273

Temperature of the polished body in absolute unit T pa = T p + 273

Temperature of the chamber in absolute unit T ca = T 7 + 273

Emissivity p = b X T4 ba - T4 ca / T4 pa - T4 ca

Where b, emissivity of black body which is equal to 1.Emmissivity apparatus :

V A T1 T2 T3 T4 T5 T6 T7

100 0.4 89 92 90 40 79 80 81

Avg. temp. of polished plate = 363.3o k = (89 + 92 + 90 / 3) + 273

Avg. temp. of Black plate = (79 + 80 + 81 / 3) + 273 = 353 o k.

Chamber temp. = 40 + 273 = 313

o

kPower Input Q = pA (Tp4 - Ta4) = bA (Tb4 - Ta4)

Since the power input is same for both heaters and area of radiating surface (A) is also

same, knowing the b =1. The emmissivity of polished surface

p = b (Tb4 - Ta4) / (Tp4 - Ta4)

= 3534 - 3134 / 3634 - 3134


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= 5.9 X 109 / 7.76 X 109

= 0.76

RESULT :

Emissivity of the specimen is found to be ---------------


16/45

4. HEAT TRANSFER THROUGH COMPOSITE WALLS

Aim:

To determine the rate of heat transfer through different layers of composite wall

Description of Apparatus:

When heat conduction takes place through two or more solid materials of

different thermal conductivities, the temperature drop across each material depends on

the resistance offered to heat conduction and the thermal conductivity of each material.

The experimental set-up consists of test specimen made of different materials

aligned together on both sides of the heater unit. The first test disc is next to a controlled

heater. The temperatures at the interface between the heater and the disc is measured by

a thermocouple, similarly temperatures at the interface between discs are measured.

Similar arrangement is made to measure temperatures on the other side of the heater. The

whole set-up is kept in a convection free environment. The temperature is measured

using thermocouples (Iron-Cons) with multi point digital temperature indicator. A

channel frame with a screw rod arrangement is provided for proper alignment of the

plates.

The apparatus uses a known insulating material, of large area of heat transfer to

enable unidirectional heat flow. The apparatus is used mainly to study the resistance

offered by different slab materials and to establish the heat flow is similar to that of

current flow in an electrical circuit.

The steady state heat flow Q = t/R

Where t = is the overall temperature drop and

R is the overall resistance to heat conduction.

Since the resistance are in series

R = R1 + R2

Where R1, R2 are resistance of each of the discs.


17/45

TABULATION:

Sl.No

.

Voltmeter

reading

Ammeter

readin

g

T1 T2 T3 T4 T5 T6 T7 T


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SPECIFICATION:

1. Thermal conductivity

Of sheet asbestos = 0.116 W/MK

Thickness = 6mm

2. Thermal conductivity of wood = 0.052W/MK

Thickness = 10mm

3. Dia. Of plates = 300mm

4. The temperatures are measured from bottom to top plate T1,T2,.T8.


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PROCEDURE

1. Turn the screw rod handle clockwise to tighten the plates.

2. Switch on the unit and turn the regulator clockwise to provide any desired heat input.

3. Note the ammeter and voltmeter readings.

4. Wait till steady state temperature is reached.

5. (The steady state condition is defined as the temperature gradient across the plates doesnot change with time.)

6. When steady state is reached note temperatures and find the temperature gradient across

each slab.

7. Since heat flow is from the bottom to top of the heater the heat input is taken as Q/2 and

the average temperature gradient between top and bottom slabs from the heater to be

taken for calculations. Different readings are tabulated as follows.

CALCULATION:

Now the resistance ( R ) offered by individual plates for heat flow.

R1 = L1/AK1 R2 = L2 / AK2 R3 = L3/AK3

Where A = Area of the plate

K = Thermal Conductivity

L = Thickness of the plate.

Knowing the thermal conductivities

Q = (T4 T1)/R =(T2 T1)/R1=(T3 T2)/R2=(T4 T3)/R3

COMPOSITE WALLS

V A T1 T2 T3 T4 T5 T6 T7 T8 Time for 1 Rev.182 0.5 76 75 72 71 66 67 50 51 E.M

heater ms 71.5 ashess 66.5 wood 50.5Area of the plate / 4 (0.3)2 = 0.07m2


20/45

Resistance of Asbestos (R1) = L1 /A1K1 = 0.005/0.07 X 69 X 10-3 =1.03

Resistance of Wood (R2) = L2/A2K2 = 0.008/0.07 X 52 X 10-3 = 2.19

Heat flow Q1 = Temp. across Asbestos / R1 = 5/1.03 =4.85 Watts

Q2 = Temp. across Wood / R2 = 16/2.19= 7.3 Watts

As per electrical anology Q1 = Q2 = Q3

Total Resistance R3 = 1.03 + 2.19 = 3.22

Q3 =(Temp. across Asbestos + Wood) / R3 = 21/3.22 = 6.521

As we have find the inside heat transfer co-efficient for heat flow from heater to

MS plate, we consider only the second and third layer.

RESULT:

The rate of heat transfer through different materials are found to be

a. MS section = ------------- Wb. Wood section = ------------- Wc. Asbestos section = --------------W


21/45

5. FREE CONVECTION

Date:AIM:

To find the heat transfer coefficient under natural convection environment.

DESCRIPTION OF APPARATUS:

Convection is a mode of heat transfer where by a moving fluid transfers heat from

a surface. When the fluid movement is caused by density differences in the fluid due to

temperature variations, it is called FREE orNATURAL CONVECTION.

This apparatus provides students with a sound introduction to the features of freeconvection heat transfer from a heated vertical rod. A vertical duct is fitted with a heatedvertical placed cylinder. Around this cylinder air gets heated and becomes less dense,

causing it to rise. This in turn gives rise to a continuous flow of air upwards in the duct.The instrumentation provided gives the heat input and the temperature at different pointson the heated cylinder.SPECIFICATION:

Length of cylinder = 50 cm

PROCEDURE:1. Switch on the unit and adjust the regulator to provide suitable power input.

2. Allow some time for the unit to reach steady state condition.

3. Note the temperature of inlet air, outlet air and temperatures along the heater rod.

4. Note ammeter and voltmeter readings.

5. For different power inputs the experiments may be repeated.

The readings are tabulated as below: -

FORMULA USED:The power input to heater = V X A = hAt

Where A = Area of heat transfer = dl

D = Dia. Of heater rod = 40mm

L = Length of heater rod = 500mm

t= Avg. temp. Of heater rod Avg. temp. of air.


22/45

H = Overall heat transfer co-efficient.

THEORETICAL METHODUsing free convection correlations for vertical cylinders.

Nu = hl / K = 0.53(GrPr)1/4 for GrPr < 105

Nu = hl / K = 0.56(GrPr)1/4 for 105 < GrPr < 108

Nu = hl / K = 0.13(GrPr)1/3 for 108 < GrPr < 1012

Characteristic length is the height of the cylinder (l)

K = Thermal conductivity of air

P = Prandtl number of air

Gr= gl3 t / 2

= 1 / Mean temp. of air + 273 K

The properties of air at mean temperature = (T1+T2+T3++T8 )/ 8

Hence h can be evaluated.

NATURAL CONVECTION:V A T1

0c T20c T3

0c T40c T5

0c T60c

85 0.38 30 55 60 65 63 38

= 1/51.8 + 273 = 3 X 10-3

Gr = gl3 t / 2 t = [(T2 + T3 + T4 + T5) / 4 ] [(T1+F6)/2]

= 3 X 10-3 X 9.81 X (0.5)3 X 26.75 / (17.96 X 10-6)2

= 3.05 X 108

where l = length of heater

= Kinematic viscosity of air at mean temp.

Pr = from data book for air mean temp.= 0.698

Hence GrPr = 2.13 X 108

Hence using free convection correlations

Nu = hl / K = 0.13 (GrPr)1/3 where K is the Thermal conductivity of air at mean temp.


23/45

= 72.82

Overall heat transfer co-efficient h = 72.82 X 28.26 X 10-3 / 0.5 = 4.11 w/m2-0c

RESULT:The heat transfer coefficient is found to be -------------- W/m2K


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6. FORCED CONVECTION

AIM:

To find the heat transfer coefficient under forced convection environment.


The important relationship between Reynolds number, Prandtl number and

Nusselt number in heat exchanger design may be investigated in this self contained unit.

The experimental set up (see sketch) consists of a tube through which air is sent in

by a blower. The test section consists of a long electrical surface heater on the tube

which serves as a constant heat flux source on the flowing medium. The inlet and outlet

temperatures of the flowing medium are measured by thermocouples and also the

temperatures at several locations along the surface heater from which an average

temperature can be obtained. An orifice meter in the tube is used to measure the airflow

rate with a U tube water manometer.

An ammeter and a voltmeter is provided to measure the power input to the heater.

A power regulator is provided to vary the power input to heater.

A multi point digital temperature indicator is provided to measure the above

thermocouples input.

A valve is provided to regulate the flow rate of air.


25/45

TABULATION:Sl

No

Inlet temp. of air Outlet temp. of air Temperatures along the duct


26/45

PROCEDURE:1. Switch on the mains.

2. Switch on the blower.

3. Adjust the regulator to any desired power input to heater.

4. Adjust the position of the valve to any desired flow rate of air.

5. Wait till steady state temperature is reached.

6. Note manometer readings h1 and h2.

7. Note temperatures along the tube. Note air inlet and outlet temperatures

8. Note volt meter and ammeter reading.

9. Adjust the position of the valve and vary the flow rate of air and repeat the

experiment.

10. For various valve openings and for various power inputs and readings may be

taken to repeat the experiments. The readings are tabulated

The heat input Q = h A L M T D = m cp (temp. of tube temp. of air)

M = mass of air. cp = specific heat of air.

LMTD=(Avg Temp Of tube outlet air temp) (Avg. temp of tube inlet air temp.)

1n x(Avg. temp of tube outlet temp. of air)

(Avg. temp of tube inlet temp. of air)

H= heat transfer co-efficient. A = area of heat transfer = T1d1

From the above the heat transfer co-efficient h can be calculated. These

experimentally determined values may be compared with theoretical values.

Calculate the velocity of the air in the tube using orifice meter / water manometer.

The volume of air flowing through the tube (Q) = (cd21222gh0 ) / (a12 a22 ) m3 / sec.

ho= heat of air causing the flow.

= (h1 h2)w/a

h1 and h2 are manometer reading in meters.


27/45

a 1= area of the tube.

a2 = area of the orifice.

Hence the velocity of the air in the tube V = Q / a1 m/sec heat transfer rate and

flow rates are expressed in dimension less form of Nusselt number and Reynolds number

which are defined as

Nu = h D/k Re = Dv/

D = Dia. Of the pipe

V = Velocity of air

K = Thermal conductivity of air.

The heat transfer co-efficient can also be calculated from Dittus-Boelter

correlation.

Nu = 0.023 Re 0.8 Pr0.4

Where Pr is the Prandtl number for which air can be taken as 0.7. The Prandtl

number represents the fluid properties. The results may be represented as a plot of Nu

exp/ Nu corr. Vs Re which should be a horizontal line.

FORCED CONVECTION

V A T1 T2 T3 T4 T5 T6 h1cm h2cm50 1 35 42 45 46 47 38 9 19

Avg. Temp. Of heater = (42 + 45 + 46 + 47) / 4 = 45oC

Avg. Temp. of Air = (35 + 38) / 2 = 36.5oC

Vol. Of air flow Q = (Cda1a22gh) / (a12 a22)

Cd = 0.6

A1 = /4 (0.04)2 = 1.256 X 10-3

A2 = /4 (0.02)2 = 3.14 X 10-4

H = water/air (h1 h 2) mtrs

= 1000/1.16 (0.1) = 86.2 mtrs.

Q = [0.6 X 1.256 X 10-3 X 3.14 X 10-4 (2 X 9.81 X 4)] / (1.256X10-3)2

(3.14X10-4)2

= 9.73 X 10-6 / 1.256 X 10-3 = 8.002 x 10-3


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Velocity of air flow = Q / a1 = 6.37 m/sec

Re = D/ r = 15023

R = kinematic viscosity at mean temp.

Using forced convection correlation

Nu = hD /k = 0.023 Re 0.8 Pr0.4

Pr at mean temp = 0.699

= 0.023(15023)0.8 (0.699)0.4

hD/k = 43.7 k= Thermal conductivity of air at mean temp

h = 43.7 X 28.56 X 10-3 / (0.04)

= 31.2 w/mc.

RESULT:

The heat transfer coefficient is found to be ---------------- W/m2K

7. STEFAN BOLTZMAN APPARATUS


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AIM:

To find Stefan-Boltzman constant.


Stefan Boltzman law which establishes the dependence of integral

hemispherical radiation on temperature. We can verify this phenomenon in this unit.

The experimental set up consisting of concentric hemispheres with provision for the hot

water to pass through the annulus. A hot water source is provided. The water flow may

be varied using the control valve provided, there by to control the hot water temperature.

A small disk is placed at the bottom of the hemisphere, which receives the heat radiation

and can be removed (or) refitted while conducting the experiment. A multi point digital

temperature indicator and thermocouples (Fe/Ko) are provided to measure temperature atvarious points on the radiating surface of the hemisphere and on the disc.

SPECIFICATIONS:

1. Mass of the disc = 0.005 kg.

2. Dia. of the disc = 0.020 m.

3. Material of the disc = copper

4. Cp = 381 J/KgK

TABULATION:

Sl.No. T1 T2 T3 Avg.temp. of

hemisphere

Th

T4 Time Steady

temp. of

the disc.


30/45

Td

PROCEDURE:

1. Allow water to flow through the hemisphere. Remove the disc from the bottom of the

hemisphere. Switch on the heater and allow the hemisphere to reach a steady

temperature.

2. Note down the temperatures T1, T2 and T3. The average of these temperatures is thehemisphere temperature (Th).

3. Refit the disc at the bottom of the hemisphere and start the stop clock.

4. The raise in temperature T4 with respect to time is noted. Also note down the disc

temperature at T4 when steady state is reached (Td).


31/45

CALCULATIONS :

Q = (Th4 Td4) A.

= Q / (Th4 Td4) A and =1.

The readings may be tabulated as follows:

T1 T2 T3 T4 Time

40.4 40.1 40.5 33.8 15

33.9 30

34 45

34.1 60

34.2 75

.

34.7 225

Final Temp of the disc 34.7

= Q / b (Th4 Td4) A.

Q = mass of the disc X Cp of disc X d/c

Cp = 381J/Kgo k

Q = 6.35 X 10-3

Avg. Temp. of hemisphere = 40.4 + 40.1 + 40.5 / 3 = 40.33o

C + 273 = 3Td = 34.8o C = 307.8o k -4

A = Area of the disc = / r (0.02)2 = 3.14 X 10-4

dT/dt = 0.1 / 30

J = 6.35 x 10-3 / [(315.3)4 (307.8)4] x 3.14x10-4 = 2.228 x 10-8

RESULT:

Stefan Boltzman constant is found to be------------W/m2 K4

8. THERMAL CONDUCTIVITY OF INSULATING

MATERIAL - LAGGED PIPE

AIM :


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To find the thermal conductivity of different insulating material.

DESCRIPTION OF APPARATTUS :

The insulation defined as a material, which retards the heat flow with reasonableeffectiveness. Heat is transferred through insulation by conduction, convection and

radiation or by the combination of these three. There is no insulation, which is 100%

effective to prevent the flow of heat under temperature gradient.

The experimental set up in which the heat is transferred through insulation by

conduction is under study in the given apparatus.

The apparatus consisting of a rod heater with asbestos lagging. The assembly is inside asMS pipe. Between the asbestos lagging and MS pipe saw dust is filled. The set up as

shown in the figure. Let r1 be the radius of the heater, r2 be the radius of the heater withasbestos lagging and r3 be the inner radius of the outer MS pipe.

Now the heat flow through the lagging materials is given by

Q = K1 2 L(t) / (In(r2)/r1) or

= K2 2 L(t) / (In(r3)/r2)

Where t is the temperature across the lagging.

K1 is the thermal conductivity of asbestos lagging material and

K2 is the thermal conductivity of saw dust.

L is the length of the cylinder.Knowing the thermal conductivity of one lagging material the thermal conductivity of the

other insulating material can be found.

TABULATION :

S.No

Heat temperatureAsbestos

temperature

Sawdust

temperature Volts Amps

1 2 3 avg 4 5 6 avg 7 8 avg


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34/45

SPECIFICATION:

Diameter of heater rod = 20mm

Diameter of heater rod with asbestos lagging = 40mm

Diameter of heater with asbestos lagging and saw dust = 80mm

The effective length of the above set up of cylinders = 500mm

PROCEDURE:

1. Switch on the unit and check if all channels of temperature indicator showing proper

temperature.

2. Switch on the heater using the regulator and keep the power input at some particular

value.

3. Allow the unit to stabilize for about 20 to 30 minutes. Now note down the ammeter,voltmeter reading which given the heat input.

4. Temperatures 1,2 and 3 are the temperature of heater rod, 4,5 and 6 are the temperatures

on the asbestos layer, 7 and 8 are temperatures on the sawdust lagging.

5. The average temperature of each cylinder is taken for calculation. The temperatures are

measured by thermocouple (Fe/Ko) with multi point digital temperature indicator.

6. The experiment may be repeated for different heat inputs.

The readings are tabulated as below:

CALCULATIONS :

Lagged Pipe:

V A T1 T2 T3 T4 T5 T6 T7 T8

90 0.4 108 117 89 51 59 53 41 41

Avg. Temp. of heater = T1 +T2 +T3 / 3 = 104.6 o c

Avg. Temp. of Asbestos lagging = T4 + T5 + T6 / 3 = 54.3 o c

Avg. Temp. of sawdust lagging = T7 + T8 / 2 = 41 o c

The heat flow from heater to outer surface of asbestos lagging =


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q = k1 2 l (t) / ln (r2 / r1)

k1 = Thermal conductivity of asbestos lagging from data look at

= 110.5 X 10-3 w/mo k.

= 54o c

r2 = Radius of the asbestos lagging = 20

r1 = Radius of the heater = 10 mm

l = Length of the heater = 0.5 mtrs.

Substituting these values

q = 110.5 X 10-3 X 2 X X 0.5 X

Substituting this value of q to find the thermal conductivity of sawdust.

25.19 = k2 X 2 X X l X 13.3 / ln (r3/r2)

k2 = 25.19 X ln (40/20) / 2XX13.3X8.

= 0.417

RESULT :

Thermal conductivity of

(i) Asbestas---------------W/mK

(ii) Sawdust----------------W/mK


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9. HEAT EXCHANGER TEST PARALLEL FLOW AND

COUNTERFLOW

Aim:

To find the overall heat transfer co-efficient in parallel flow and counter flow.


Heat exchangers are devices in which heat is transferred from one fluid to

another. Common examples of the heat exchangers are the radiator of a car, condenser at

the back of domestic refrigerator etc. Heat exchangers are classified mainly into three

categories. 1. Transfer type 2. Storage type 3. Direct contact type.

Transfer type of heat exchangers are most widely used. A transfer type of heatexchanger is one in which both fluids pass simultaneously through the device and head is

transferred through separating walls. Transfer type of exchangers are further classifies as

1. Parallel flow type in fluids flow in the same direction.

2. Counter flow type in fluids flow in the opposite direction.

3. Cross flow type in which fluids flow at any angle to each other.

A simple heat exchanger of transfer type can be in the form of a tube

arrangement. One fluid flowing through the inner tube and the other through the annulus

surrounding it. The heat transfer takes place across the walls of the inner tube.


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TABULATION :

FOR PARALLEL FLOW

Sl.No.Time for 1 Lit. of

Hot Water (sec)

Time for 1 Lit. of

cold water (sec)T1 T2 T3 T4

FOR COUNTER FLOW

Sl.No.Time for 1 Lit. of

Hot Water (sec)

Time for 1 Lit. of

cold water (sec)T1 T2 T3 T4


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The apparatus consists of a concentric tube heat exchanger. The hot fluid i.e. hot water isobtained from an electric geyser and flows through the inner tube. The cold fluid i.e.cold water can be admitted at any one of the ends enabling the heat exchanger to run as aparallel flow apparatus or a counter flow apparatus. This can be done by operating thedifferent valves provided. Temperatures of the fluids can be measured using

thermometers. Flow rate can be measured using stop clock and measuring flask. Theouter tube is provided with adequate asbestos rope insulation to minimize the heat loss tothe surroundings.

SPECIFICATIONS:

Length of the heat exchanger

Inner copper tube ID = 12mm

OD = 15mm

Outer GI tube ID = 40mm

PROCEDURE:

1. Connect water supply at the back of the unit. The inlet water flows through

geyser and inner pipe of the heat exchanger and flows out.

Also the inlet water flows through the annulus gap of the heat exchanger and flows out.

2. For parallel flow open valve V2 , V4 and V5.

For counter flow open valve V3, V1 and V5.

3. Control the hot water flow approximately 2 lit./min. and cold water flow approximately 5

lit./min.

4. Switch ON the geyser. Allow the temperature to reach steady state.

5. Note temperatures T1 and T2 (hot water inlet and outlet temperature

respectively).


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6. Under parallel flow condition T3 is the cold-water inlet temperature and T4 is the

cold water outlet temperature.Note the temperatures T3 and T4.Under counter flow

condition T4 is the cold-water inlet temperature T3 is the cold-water outlet

temperature.

7. Note the time for 1 liter flow of the hot and cold water. Calculate mass flow rate

Kg/sec.

8. Change the water flow rates and repeat the experiment.


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U = qh / A L M T D

Where qh = mh cp (Thi Tho)

cp = specific heat of water (j/kgc)

A = Outer area of hot water pipe.

Mh = mass of hot water (kg/sec)

Effectiveness of Heat exchanger

= Actual heat transfer/ Max. possible heat transfer

= (tco tci) / (thi tci)

THEORETICAL METHOD:

The overall Heat transfer co-efficient

1/U = (1/ho) + (1/h1)

Neglecting the thickness of inner tube and film resistance where ho and h1 are the co-efficient of heat transfer of hot and cold side respectively.

h1 = Inside Heat transfer co-efficient (from hot to inner surface of the inner tube)

ho = Out side heat transfer co-efficient (from outer wall of the inner tube to the cold fluid).

Re = hot water flow = D /

= Velocity of hot water.

Knowing the mass flow rates () may be calculated for hot and cold water.

Nu = 0.023 (Re)0.8 (Pr)0.3 = (hiD) /K

K = Thermal conductivity of water.

In a similar manner ho can also be calculated. However for finding ho the

characteristic dia. Is taken as the annulus which is given by the

(ID of the outer pipe OD of outer pipe).

Hence, U the overall Heat transfer co-efficient is evaluated.

Parallel flow / Counter flow Heat exchanger.

Parallel Flow


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hi = 39.79 X 605 X 10-3 / 0.012

= 2006

ho = Volume flow rate of Cold water (1/21) / 1000 m3 / sec.

Qc = 4.76 X 10-5m3 / sec.

Velocity of Cold water flow Vc = Qc / Ac

Ac = Annulus area i.e. /4(0.04)2 - /4 (0.015)2

= 1.08 X 10-5

Vc = 4.76 X 10-5 / 1.08 X 10-3 = 0.044 m/sec

Re = D / = (0.04 0.015) X 0.044 / 0.75 X 10-6

= 1466

Since the flow is not turbulent we can using the following equation.

Nu = 0.37(Re)0.6 (Pr)0.33

hoDc / k = 51.5 Dc = Annulus dia. (0.04 0.015) = 0.025

h o= 1247.

1/U = 1/hi+ 1/ho = 1/2006 + 1/1247

U = 769 W/m2 o c.

This procedure is repeated for counter flow heat exchanger, however care to be

taken while calculating LMTD.


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RESULT :

(i) Parallel flow

Overall heat transfer coeffient by theoretical method ----------- W/ m2 K

Overall heat transfer coeffient by prctical method ----------- W/ m2 K

(i) Counter flowOverall heat transfer coeffient by theoretical method ----------- W/ m2 K

Overall heat transfer coeffient by prctical method ----------- W/ m2 K

te lab - ii manual.doc

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