# 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:

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

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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 = -------------

Temperature at x = 40mm, T40 = -------------

Temperature at x = 60mm, T60 = -------------

Temperature at x = 80mm, T80 = -------------

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 ---------------

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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.

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TABULATION:

Sl.No

.

Voltmeter

Ammeter

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

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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

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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 difference...