thermal lab ii mech manaual ccet

CHENDU

COLLEGE OF ENGINEERING Zamin Endathur, Madurantakam Taluk

Kancheepuram District-603311

Department of Mechanical Engineering

Subject Code/Title: ME 2355 Thermal engineering Lab-II Lab Manual

Name :

Reg No :

Branch :

Year & Semester :

1

TABULATION:

Sl.

No.

Voltmeter

reading

(V)

Ammeter

reading

(A)

T1

0C

T2

0C

Tavg 1

T3

0C

T4

0C

TAvg 2

T5

0C

T6

0C

T7

0C

T8

0C

K

W/m K

KAvg =

2

Ex.No: 1 Date

THERMAL CONDUCTIVITY APPARATUS-GUARDED 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 = mm

Diameter of specimen (d) = mm

THERMAL CONDUCTIVITY APPARATUS

T

T T

T

T

T

T

T

GUARD HEATER MAIN HEATER SPECIMEN PLATES

3

MODEL CALCULATIONS:

FORMULA USED:

Since the guard heater enables the heat flow in uni direction

q = KA dT/dx

Where 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. Watts

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

Q = K1. d 2/4 (Tavg1 T5)/dx

Where dx = mm = m

Diameter of specimen

d = cm = m

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

K avg = (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 at different

locations when the unit reaches steady state. The steady state is defined, as the

temperature gradient across the plate remains same at different 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.

4

Ex.No:2

Date:

Aim:

HEAT TRANSFER THROUGH COMPOSITE WALLS

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.

5

TABULATION:

Sl.No. Voltmeter

reading

Ammeter

reading

T1 T2 T3 T4 T5 T6 T7 T8

COMPOSITE WALLS

WOOD

ASBESTOS

MS

MS

ASBESTOS

WOOD

T8

T7

T6

T5

HEATER

T4

T3

T2

T1

6

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.

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

does not 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

7

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 50.5

Area of the plate / 4 (0.3)2 = 0.07m2

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 = ------------- W

b. Wood section = ------------- W

c. Asbestos section = --------------W

8

TABULATION:

Sl.No. Voltmeter

reading

(V)

Ammeter

reading

(A)

T1 T2 T3 T4 T5 T6

NATURAL CONVECTION

T6

T5

T4

T3

T2

T1

9

Ex.No: 3 HEAT TRANSFER BY 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 or NATURAL CONVECTION.

This apparatus provides students with a sound introduction to the features of free

convection heat transfer from a heated vertical rod. A vertical duct is fitted with a heated

vertical 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 points on the

heated cylinder.

SPECIFICATION:

Length of cylinder = 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 = mm

10

L = Length of heater rod = mm

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

H = Overall heat transfer co-efficient.

THEORETICAL METHOD

Using 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.

11

NATURAL CONVECTION:

V A 0

T1 c 0

T2 c 0

T3 c 0

T4 c 0

T5 c T 0c 6

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

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

=

=

Where l = length of heater

= Kinematic viscosity of air at mean temp.

Pr = from data book for air mean temp.

=

Hence GrPr =

Hence using free convection correlations

Nu = hl / K = 0.13 (GrPr)1/3

where K is the Thermal conductivity of air at mean temp.

=

Overall heat transfer co-efficient h = = W/m2-0c

RESULT:

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

12

Ex.No:4

Date:

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.

TABULATION:

Sl

No

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

13

FORCED CONVECTION

ORIFICE DIA = 20 mm

PIPE DIA. = 40

BLOWER

HEATER

T6 T5 T4 T3 T2 T1

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, for 5min

6. Note manometer readings h1 and h2.

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

8. Note voltmeter 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.

14

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) = (cd a1 a2 2gh0 ) / ( a12 a2

2 ) m

3 /

sec. ho = heat of air causing the flow.

= (h1 h2) w/ a

h1 and h2 are manometer reading in meters.

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.The 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

Pr 0.4

2 )

15

Where Pr is the Prandtl number for air and 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 h2cm

Avg. Temp. Of heater = ( ) / 4 = oC

Avg. Temp. of Air = ( ) / 2 = oC

Vol. Of air flow Q = (Cda1a2 2gh) / ( a12 a 2

Coefficient of discharge

Cd = 0.6

A1 = /4 ( )2

=

A2 = /4 ( )2

=

H = water/air (h1 h 2) mtrs

= 1000/1.16 ( ) = mtrs.

Q =

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

Re = D/ r =

R = kinematic viscosity at mean temp.

Using forced convection correlation

Nu = hD /k = 0.023 Re 0.8

Pr 0.4

Pr at mean temp =

= 0.023( )0.8

( )0.4

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

h = W/mC.

RESULT:

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

16

Ex.No:5

Date:

STEFAN BOLTZMAN APPARATUS

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, thereby 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 at

various points on the radiating surface of the hemisphere and on the disc.

SPECIFICATIONS:

1. Mass of the disc = kg.

2. Dia. of the disc = m.

3. Material of the disc = copper

4. Cp = 381 J/KgK

17

TABULATION:

Sl.No.

T1

T2

T3

Avg.temp. of

hemisphere

Th

T4

Time

Steady

temp. of

the disc.

Td

HEATER

WATER

T2

T1 T3

T4

d

d

18

PROCEDURE:

STEFAN BOLTZMANN APPARATUS

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 the

hemisphere 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).

CALCULATIONS :

Q = (Th4 T 4) A.

= Q / (Th4 T 4) A and =1.

The readings may be tabulated as follows:

T1 T2 T3 T4 Time

Final Temp of the disc

= Q / b (Th4 Td

4) A.

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

19

Cp = 381J/Kgo

K

Q = 6.35 x 10-3

Avg. Temp. of hemisphere = = o

C + 273 =

Td =

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

dT/dt =

20

J =

RESULT:

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

K4

21

Ex.No:6

Date:

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 heat

exchanger 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.

22

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

23

The apparatus consists of a concentric tube heat exchanger. The hot fluid i.e. hot

water is obtained 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 a

parallel flow apparatus or a counter flow apparatus. This can be done by operating the

different valves provided. Temperatures of the fluids can be measured using thermometers.

Flow rate can be measured using stop clock and measuring flask. The outer tube is provided

with adequate asbestos rope insulation to minimize the heat loss to the surroundings.

SPECIFICATIONS:

Length of the heat exchanger

Inner copper tube inner diameter

Outer diameter

=

=

mm

mm

Outer GI tube ID = mm

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 l/min. and cold water flow approximately

5 l/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).

PA

RA

LL

EL

FL

OW

H

EA

T E

XC

HA

NG

ER

C

OL

D

Tci

Thi

Th

o

HO

T

Tco

PA

RA

LL

EL

FL

OW

Thi

Th

o T

o

Tco

T

i -

L

og

e

T

o

T

i

T

o

T

m =

Tci

24

Length of the Exchanger

Ti

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.

CO

UN

TE

R F

LO

W H

EA

T

EX

CH

AN

GE

R

Tco

Thi

Tho

HO

T

Tci

CO

LD

CO

UN

TE

R

FL

OW

T

hi

Tho

T

i

T

o

T

i -

Log

e

T

o

T

i

T

o

T

m =

Tci

T

co

25

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

kg/s.

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

Length of the Exchanger

26

CALCULATIONS:

Refer drawing and find

LMTD (tm) = t1 to / ln (t1 / to)

Please note t1 and to to be calculated as per drawing for Parallel flow and Counter flow.

Qh = A U (L M T D)

Hence the overall Heat transfer co-efficient

U = Qh / A L M T D

Where Qh = mh Cp (Thi Tho)

Cp = Specific heat of water (J/kg0

C)

A = Outer area of hot water pipe.

Mh = Mass of hot water (kg/s)

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)

Neglect the thickness of inner tube and film resistance.

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 for Parallel flow / Counter flow

Heat exchanger.

27

Parallel Flow

Hot Water Temperature Cold Water Temperature

Time taken for

1litre Hot water

flow.

Time taken

for 1 litre

Cold Water

flow.

Inlet

T1

Thi

outlet

T1

Thi

Inlet

T3

Tci

Outlet

T4

Tco

LMTD = (Thi Tci) (Tho Tco) / ln (Thi Tci / Tho Tco)

=

Heat input Qb = A.U LMTD

Hence the overall heat transfer co-efficient, U = Qb / A L M T D

Qb = mb Cb (Thi Tho)

=

Theoretical Method:

1/U = 1/hi + 1/ho

hi = Volume of hot water flow = m3

/ sec.

= m3

/ sec.

Velocity of flow of hot water = m/sec

= m/sec

Re = D / =

=

Using the heat transfer correlation

Nu = 0.023 (Re)0.8

(Pr)0.3

= hiD/k

=

k = Thermal conductivity

of water

Pr = Values from data book

28

hi =

ho = Volume flow rate of Cold water m

3 / s.

Qc = m3

/ sec.

Velocity of Cold water flow Vc = Qc / Ac

Ac = Annulus area i.e. /4(D)2

- /4 (d)2

=

Vc = = m/sec

Re = D / =

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

Nu = 0.37(Re) 0.6

(Pr)0.33

hoDc / k = Dc = Annulus dia. (Dd) =

h o =

1/U = 1/hi + 1/ho =

U = W/m2 o

c.

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

while calculating LMTD.

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 flow

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

K

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

K

29

Ex.No:7

Date:

THERMAL CONDUCTIVITY OF INSULATING

MATERIAL - LAGGED PIPE

AIM :

To find the thermal conductivity of different insulating materials.

DESCRIPTION OF APPARATUS :

The insulator is a material, which retards the heat flow with reasonable effectiveness.

Heat is transferred through insulation by conduction, convection and radiation or by the

combination of these three.

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

conduction is under study.

The apparatus consisting of a rod heater with asbestos lagging. The assembly is

inside an MS 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 with

asbestos 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) / (ln (r2)/r1) or

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

Where t is the temperature difference 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.

30

TABULATION :

S.No

Heater temperatures

Asbestos

temperatures

Sawdust

temperatures

Applied

Voltage

volts

Current

Amps T1 T2 T3 avg T4 T5 T6 avg T7 T8 avg

LAGGED PIPE

SAW DUST

ASBESTOS

HEATER

T1 T3

T4

ASBESTOS

T5 T6

SAW DUST

T7

T8

T4 T1 T7 T5 T8 T6 T3

d1 - HEATER DIA = 20 mm d2 - HEATER WITH ASBESTOS DIA = 40 mm d3 - ASBESTOS & SAW

DUST DIA = 80 mm

LENGTH = 500mm

31

SPECIFICATION:

Diameter of heater rod, d1 =

Diameter of heater rod with asbestos lagging, d2 =

Diameter of heater with asbestos lagging and saw dust, d3 =

The effective length of the cylinder =

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 readings the product of which give 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 saw dust 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

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

C

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

C

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

C

32

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

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

k1 = Thermal conductivity of asbestos lagging, from data look at------------------ o

C

(average temp of asbestos lagging)

= W/m K.

r2 = Radius of the asbestos lagging =

r1 = Radius of the heater = mm

l = Length of the heater = m

Substituting these values

Q = ( ) 2 x l x (t) / (r2 / r1)

Substituting this value of q to find the thermal conductivity of saw dust, K2

Q= K2 x 2 x l x (t) / ln (r3/r2)

K2 = x ln ( )/ 2 x x .

=

RESULT :

Thermal conductivity of

(i) Asbestos---------------W/mK

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

33

Ex.No:8

Date:

HEAT TRANSFER FROM FINS

AIM:

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

FIN efficiency.


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 to the 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 eighth 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 = mm

Duct height

Orifice dia.

w

do

=

=

mm

mm

34

Orifice co-efficient cd =

Fin length L = cm

Fin diameter df = mm

(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 for 10min

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.

35

TABULATION:

Sl.No.

Manometer

readings

Fin surface temp. Amb. temp.

h1 h2 T1 T2 T3 T4 T5 T6

PIN - FIN APPARATUS

T7 T6 T5 T4 T3 T2 T1

HEATER

BRASS PIN - FIN

T8

ORIFICE DIA = 20 mm LENGTH = 145 mm

PIPE DIA = 40 mm DIA = 12 mm

36

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.

B = breadth of the duct.

REYNOLDS NUMBER OF AIRFLOW:

Reynolds number Re = (L x Va x a) / a

Where Va = Velocity of air in the duct.

a = Density of air in the duct.

a = Viscosity of air at to

C.

L = length of fin in m

PRANDTL NUMBER OF AIRFLOW

Prandtl number = (Cpa x a ) / Ka

Where Cpa = Specific heat of air.

a = Viscosity of air

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

37

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 < 40000

Nnu = 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 (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

( D)

D = dia of the fin

A = cross sectional area of the fin.

Kb = thermal conductivity of brass rod.

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

X = distance between thermocouple and heater.

38

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 TD3 / r2

= 1/ (Tav + 273)

T= (Tav Tamb)

Pr = Prandtl Number = ( Cp) / K

2

39

PIN-FIN

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

Mean Temp = o

C

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

- a 2

a1 = /4 =

a2 = /4 =

h = w / a ( )

=

Q =

m

=

m3/s

Velocity of air flow thro duct = Q/A

A = Length X Breadth of the duct

= = m2

Velocity = m/sec

40

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

=

Using the correlation

For 40 > Re

Ex.No:9

Date:

41

TEST ON EMISSIVITY APPARATUS

AIM:

To measure the emissivity of the test plate surface.

DESCRIPTION OF APPARATUS :

An ideal black surface is one, which absorbs the radiation falling on it. Its reflectivity

and transivity is zero. The radiation emitted per unit time per unit area from the surface of

the body is called emissive power.

The emissive power of a body to the emissive power of black body at the same

temperature is known as emissivity of that body. For a black body absorptivity is 1 and by

Kirchhoffs law its emissivity is also 1. Emissivity depends on the surface temperature and

the nature of the surface.

The experimental set up consists of two circular aluminum plates identical in size and

are provided with heating coils at the bottom. The plates or mounted on thick asbestos sheet

and 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. The temperatures of the plates are measured by Ir/Con thermocouples. Each plate

is having three thermocouples; hence an average temperature may be 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 heat dissipation by conduction is same in both cases.

SPECIFICATION:

Diameter of test plate and black surface = mm

PROCEDURE:

a) Connect the three pin plug to the 230V, 50Hz, 15 amps main supply and switch on

the unit.

b) Keep the thermocouple selector switch in first position. Keep the toggle switch in

position 1. By operating the energy regulator 1 power will be fed to black plate.

42

Now 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 off both the energy regulators 1 & 2.

g) For various power inputs repeat the experiment.

TABULATION :

Sl.No.

Black body

temperature

Average

Temp. Tb

Polished body

temperature

Average

Temp. Tp

Chamber

Temp. T4

Emmissivity

T5 T6 T7 T1 T2 T3

EMISSIVITY APPARATUS

CHAMBER

T1 T4 T5

T2 T3 T6 T7

TEST PLATE

DIA. - 150 mm

BLACK PLATE

DIA. - 150 mm

43

CALCULATION

S:

Temperature of the black body in absolute unit T=

Temperature of the polished body in absolute unit T=

Temperature of the chamber in absolute unit T= 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

Avg. temp. of polished plate

= Avg. temp. of Black plate

= Chamber temp. =

Power Input Q = pA (Tp4

- Ta4) = bA (Tb

4 - Ta

4)

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) / (Tp

4 - Ta

4)

=

=

=

RESULT :

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

thermal lab ii mech manaual ccet

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