heat and mass transfer lab

7/25/2019 heat and mass transfer lab

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THERMAL ENGINEERING LAB-II Prepared By: M.VIVEKANANTHAN Page 1

M H L KSHMI

ENGINEERING COLLEGETRICHIRAPPALLI- 621 213

ME2355THERMAL ENGINEERING -II

LABORATORY

LAB MANUAL

MECHANICAL VI SEMESTER

FOR VI SEMESTER B.E. DEGREE COURSE

[REGULATION 2008]

As per the common syllabus prescribed by

ANNA UNIVERSITY, CHENNAI

2012-2013

PREPARED BYMr. M.VIVEKANANTHAN

ASST. PROF / MECHANICAL


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M H L KSHMI

ENGINEERING COLLEGETRICHIRAPPALLI- 621 213

ME2355MECHANICAL ENGINEERING-II

LABORATORY

MASTER RECORD

MECHANICAL -VI SEMESTER

FOR VI SEMESTER B.E. DEGREE COURSE

[REGULATION 2008]

As per the common syllabus prescribed by

ANNA UNIVERSITY, CHENNAI

2012-2013

PREPARED BY

Mr. M.VIVEKANANTHAN

ASST. PROF / MECHANICAL


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CONTENTS

S.NO DATE NAME OF THE EXPERIMENTS PG.NO SIGNATURE

1STEFAN-BOLTZMANN APPARATUS

2TEST ON EMISSIVITY APPARATUS

3

THERMAL CONDUCTIVITY OF POOR

CONDUCTING MATERIAL BY GUARDED

HOT PLATE METHOD

4

HEAT TRANSFER IN NATURAL

CONVECTION

5

HEAT TRANSFER IN FORCED

CONVECTION

6

HEAT TRANSFER FROM A PIN-FIN

APPARATUS BY FORCED CONVECTION

MODE

7

PARALLEL FLOW / COUNTER FLOW HEAT

EXCHANGER

8

PERFORMANCE TEST ON TWO STAGE

RECIPROCATING AIR COMPRESSOR

9

THERMAL CONDUCTIVITY OF

INSULATING

MATERIAL - LAGGED PIPE

10EXPERIMENTS ON AIR-CONDITIONING

SYSTEM

11

PERFORMANCE TEST ON SINGLE/TWO

STAGE RECIPROCATING AIR

COMPRESSOR


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STEFAN-BOLTZMANN APPARATUS

EXP.NO: 01

DATE:

AIM:

To determine the value of Stefan-Boltzmann constant for radiation heat transfer.

THEORY:

Stefan-Boltzmann law states that, the Total emissive power of a perfect black body is

proportional to fourth power of the absolute temperature.

Eb = T4

=5.6697 X 10-8W/m2K

DESCRIPTION OF THE APPARATUS:

The apparatus consists of a flanged copper hemisphere fixed on a flat non-conducting

plate. A test disc made of copper is fixed to the plate. Thus test disc is completely enclosed by

the hemisphere. The outer surface of the hemisphere is enclosed in a vertical water jacket used

to heat the hemisphere to a suitable constant temperature. Three Cr-al thermocouples are

attached at three strategic places on the surface of the hemisphere to obtain the temperature. The

disc is mounted on a Bakelite sleeve which is fitted in a hole drilled at the centre of the base

plate. Other Cr-al thermocouples are fixed to the disc to record its temperature.

SPECIFICATIONS:

Diameter of the disc = 20 mm

Thickness of the disc = 0.5 mm

Mass of the disc (m) = 5 gms

Specific heat of the test disc (Cp) = 380 J/kg-KInner diameter of the hemispherical surface = 200 mm


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OBSERVATION TABLE:

Let TD= T3,Temperature of the disc before inserting into the plate in K = --------------

Sl.no

Hot water temp

T

Sphere top side

tempT1

Sphere left side

tempT2

Sphere right side

tempT4

C K C K C K C K

TEMPERATURE TIME RESPONSES:

SL.NO TIME

Secs

TEMPERATURE

OF DISC in C

T3

TEMPERATURE

OF DISC in K

T3

1

2

3

4

5

6

7

8

9

10


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PROCEDURE:

1. Fill the water in the SS container with immersion heater kept on top of the panel.

2. Remove the test disc before starting the experiment.

3. Heat the water in the SS container to its boiling point.

4. Allow the boiling water into the container kept into the bottom containing copper

hemisphere until it is full and allow sufficient time to attain thermal equilibrium

which is indicated by the three thermocouples provided on the hemisphere.

5. Insert the test disc fixed on the Bakelite sleeve fully in and lock it. Start the

stopwatch simultaneously.

6. Note down the temperature of the disc at an interval of 20 sec for about 5 to 10 min.

CALCULATIONS:

1. Plot the graph of temperature of the disc Vs time and obtain the slope of the line.

2. Average temperature of the hemisphere T avg= (T1+T2+T3 ) / 3 in K

3. Rate of change of heat capacity of the disc = m Cp dT/dt

4. Net energy radiated on the disc = AD( Tavg4TD

4)

Rate of change of heat capacity of the disc = Net energy radiated on the disc

m Cp dT/dt = AD( Tavg4TD

4)

MODEL GRAPH:

Temp in K

Time in sec

RESULT:

Thus the Stefan-Boltzmann constant () of the given specimen is found to be ..

The difference is found to be .


7/41


TEST ON EMISSIVITY APPARATUS

EX.NO: 02

DATE:

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

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

body is called emissive power.

The experimental sets 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. Thetemperatures of the plates are measured by Iron 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 = 150 mm

PROCEDURE:1) Connect the three pin plug to the 230V, 50Hz, 15 amps main supply and switch on the

unit.

2) 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. Now keep the

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

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

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

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

5) For various power inputs repeat the experiment.


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TABULATION:

S.No

Heat input

Black body

Temperature

C

Test Surface

Temperature

C

Chamber

Temperature

T7

Emissivity of

Test Surface

V I Q T1 T2 T3 T4 T5 T6 C

1

2

3

4

EMISSIVITY APPARATUS

CHAMBER

T1 T4 T5

T2 T3 T6 T7

TEST PLATE BLACK PLATE

DIA. - 150 mm DIA. - 150 mm


9/41


CALCULATIONS:

Temperature of Black Surface Tb = (T1+ T2+ T3) / 3 in K

Temperature of Test Surface Tt = (T4+ T5+ T6) / 3 in K

Ambient temperature Ta = (T7+273) in K

Where,

T1, T2, T3, T4, T5 & T6in K

b = Emissivity of the Black Surface = 1

t= Emissivity of the Test Surface

Ab = Area of the Black Surface

At = Area of the Test Surface

Here, At= Ab, Pb= Pt

(Tb4

T74

)

et= ---------------------

(Tt4T7

4)

RESULT:

Emissivity of the Test Surface is found to be ---------------


10/41


THERMAL CONDUCTIVITY OF POOR CONDUCTING MATERIAL BY

GUARDED HOT PLATE METHOD

EX:NO : 03

DATE :

AIM:

To determine the thermal conductivity of a poor conducting material, say asbestos sheet

by guarded hot plate method.

THEORY:

Thermal conductivity is a specific property of any conducting material which is defined

below for a homogeneous solid as the quantity of heat conducted across a unit area normal to the

direction on unit time and for unit temperature gradient along the flow.

K = ( Q dL ) / ( 2A dT )

Where, Q = Heat conducted in watts

dL= Thickness of Plate in m

A = Area of conduction heat transfer in m2

dT = Temperature difference across the length dL in0C

BASIS OF MEASUREMENT:

Experimental measurement of thermal conductivity of solid can be

accomplished by a variety of methods, all based on the observation of the temperature gradient

across a given area of the material conducting heat at a known rate. Each of these methods has

certain unique limitations, and the choice of one over another is governed by the general

temperature level at which K is measured by physical structure of the material in question, andby whether the material is a good or poor conductor.

In measuring the thermal conductivity of poor conductors, the specimens

are taken in the form of sheets in order that the heat flow path be short and the conducting area

large (low dL, higher A)

Guarded hot plate method is generally used to conduct such an

experiment. In this method, electrically heated thermal guards are placed adjacent to the exposed

surface of the source H1, specimen S, and sink H0, these thermal guard plate are independently

maintained at the same temperature as the adjacent surface, to ensure ideally no heat leakage

occurring from source, specimen or the sink boundaries. The enclose drawings given actual

dimensions of various components of the apparatus.


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DESCRIPTION OF APPARATUS:

Enclosed drawing and various specifications associated with all the

components. The test specifications associated with all the components. The test section

assemblies consisting of the asbestos specimen, heaters as well as thermocouples are shown

separately 9 thermocouples are available to measure temperatures of heaters, the two specimens

and the water inlet and outlet. It should be noted that (T3-T4) gives the temperature gradient

across the top asbestos sheet and (T5-T6) gives the corresponding quantity for the bottomspecimen. Provision is also made to measure the cooling water flow rate. The whole assembly is

enclosed in an insulating layer of mineral wool to prevent radiation and connective losses to the

maximum extent possible.

SPECIFICATIONS

Material = Asbestos sheet (commercial grade)

Specimen Diameter, D = 150 mm

Specimen Thickness, L = 12 mm ( 2 pieces, one at the top and one at the bottom )Area of specimen, A = /4 D2Heat input Q = VI watts

Heat input is from two contributions: Main heater and guard heater.

Both these contributions are to be summed up to obtain the total heat that is dissipated by

the heaters. However only half of this heat input will pass (ideally through each of the two

specimens) it may also be noted that since the heaters are sandwiched between 2 layers of

commercial grade asbestos sheet, heat will be dissipated through both the layers. For the purpose

of calculations, average temperature gradient T = [(T3T4) + (T5T6)] / 2is used.

EXPERIMENTAL PROCEDURE:The panels consist of voltmeter, ammeter, and temperature indicator (all digital). Dimmer

controls, volt and ammeter selector (common) switch, thermocouples selector switch and

schematic diagram.

1. Connect the three pin top to 230V, 50 Hz, 5 amps power supply socket, dimmers in OFF

position.

2. Keep the voltmeter and ammeter selector switch in 1 position. Turn the dimmer one in

clockwise and adjust the power input to main heater to any desired value by looking at

the voltmeter and ammeter.

3. Allow the unit to reach steady state. (Anywhere from 20-30 min).

4. Note down the temperature indicated by the digital temperature indicator by turning thethermocouples selector switch clockwise step by step (1,2,3,4,5,6,7,8,9).

5. Repeat the experiment for different power inputs to the heater.

6. Tabulate all the readings and calculate for different conditions.


12/41


OBSERVATION TABLE:

S.No VoltsCurrent

(I)T1 T2 T3 T4 T5 T6 T7 T8 T9

T K

Unit V Amps 0C0

C

0

C

0

C

0

C

0

C

0

C

0

C

0

C

K

W/

mK

1

2

3

4

THERMAL CONDUCTIVITY APPARATUS

T

T T

T T T T

T

GUARD HEATER MAIN HEATER SPECIMEN PLATES


13/41


RESULT:

Thus the thermal conductivity (K) of the asbestos sheet is found to be


14/41


Heat Transfer in Natural ConvectionEX:NO: 04

DATE:

Aim:-

To determine the surface heat transfer co-efficient of a vertical tube losing heat by naturalconvection.

Theory :-

When a hot body is kept in still atmosphere, heat is transferred to the surrounding fluid by natural

convection. The fluid layer in contact with the hot body gets heated; rise up due to the decrease in its

density and the cold fluid rushes in to take place. The process is continuous and the heat transfer takes

place due to the relative motion of hot cold fluid particles.

Applications:-

Transmission lines, pipes, refrigerating coils, hot radiators and buildings etc..

The heat transfer coefficient is given by Newtons law of heating (cooling) :

Experimental Formula , Q = h As ( Ts- T)

Where, h = Average surface heat transfer coefficient (W/ m2 0C)

q = Heat transfer rate (Watts)

As = Area of heat transferring surface = .D.L (m2)

Ts = Average surface temperature

(T1 + T2+ T3 + T4 + T5+ T6+ T7)

TS = ------------------------------------------------0C

7

T8= T= Ambient temperature in the duct = 0C


15/41


The surface heat transfer coefficient, of a system transferring heat by natural convection depends

on the shape, dimensions and orientation of the fluid and the temperature difference between heat

transferring surface and the fluid. The dependence of h on all the above-mentioned parameters is

generally expressed in terms on non-dimensional groups as follow:

h x L

Where, Nu = ----------k

g.L3. . T

Gr = -----------------

v2

Where, L = Length of the pipe (m) T = [Ts T]

K = Thermal conductivity of fluid.

v = Kinematics Viscosity of fluid.

= Coefficient of volumetric expansion for the fluid.

g = Acceleration due to gravity ( 9.81 m/s2)

1

For gases, = ---------------- / 0 k

(Tf+ 273)

(Ts + T)

Tf = ----------------2

For a vertical cylinder losing heat by natural convection,

h x L

Nu = --------- = 0.59(Gr. Pr)0.25

for Gr. Pr < 109, Laminar flow

k

h x L

Nu = --------- = 0.10 (Gr. Pr) 1/3for Gr. Pr > 109 , Turbulent flow

k

L = Length of the cylinder.

All the properties are determined at the mean film temperature (Tf).


16/41


Observations Table:-

S. No. Volt Amp TEMPERATURE,0C

T1 T2 T3 T4 T5 T6 T7 T =T8

1

2

3

4

Schematic Diagram of Free convection heat transfer coefficient apparatus:

Metallic Tube Tc1

2

3

Embedded Heater 4

5

Thermocouples T=T8 6

Rectangular Duct Tc7


17/41


Specification:

1. Diameter of the tube (D) = 40 mm2. Length of tube (L) = 450 mm

Procedure:

1. Note down the specifications of all the instruments provided on the panel and also the

length (L) and diameter (D) of the metallic tube.

2. Switch on the Heater and adjust the heating rate to the suitable level.

3. Wait for some time to ensure the unit to reach steady state.

4. At steady state record the voltage and current readings and the temperatures T1through

T7and T

5. Repeat the experiment for a different heat rate.

Observations

1. O. D. of Cylinder = 40 mm.

2. Length of cylinder = 450 mm.

3. Multichannel Digital Temperature Indicator 0300 0C using Chromel / Alumel thermocouple.

4. Ammeter = Amp. and Voltmeter = Volts.

5. Input to heater Q = V x I Watts.

Result:-

Heat coefficient for a vertical tube losing heat by Natural convection is found to be

h Theoretical = .

hExp = ..


18/41


Heat Transfer in Forced Convection

EX:NO: 05

DATE:

Aim:

To determine the heat transfer co-efficient in forced convection of air in a tube.Introduction:

In many practical situations and equipments, we invariably deal with flow of fluids in

tubes e.g. boiler, super heaters and condensers of a power plant, automobile radiators, water and

air heaters or coolers etc. the knowledge and evolution of forced convection heat transfer

coefficient for fluid flow in tubes is essentially a prerequisite for an optional design of all thermal

system

Convection is the transfer of heat within a fluid by mixing of one portion of fluid with the

other. Convection is possible only in a fluid medium and is directly linked with the transport of

medium itself.

In forced convection, fluid motion is principally produced by some superimposed

velocity field like a fan, blower or a pump; the energy transport is said due toforced convection.

Description:

The apparatus consists of a blower unit fitted with the test pipe. The test section is

surrounded by a Nichrome band heater. Four thermocouples are embedded on the test section

and two thermocouples are placed in the air stream at the entrance and exit of the test section to

measure the air temperature. Test pipe is connected to the delivery side of the blower along with

the orifice to measure flow of air through the pipe. Input to the heater is given through a

dimmerstat and measured by meters.

It is to be noted that only a part of the total heat supplied is utilized in heating the air. A

temperature indicator with cold junction compensation is provided to measure temperatures of

pipe wall at various points in the test section. Airflow is measured with the help of orifice meter

and the water manometer fitted on the board.


19/41


Specification:

Inner diameter (Di) : 25 mm

Length of test section (L) : 400 mm

Blower : 1HP, 3m3/min

Orifice Diameter (D0) : 20 mm

Temperature indicator : Digital type and range 0 - 200 c

Voltmeter : 0 -100 /200v

Ammeter : 02 amp

Heater : Nichrome wire heater wound on

Test Pipe (Band Type) 400 watt

Procedure:

1. Note down all the specifications of the instruments provided on the panel board and the

test section along with the orifice.

2. Start the blower by keeping the valve fully open.

3. Switch on the heater and adjust the heating rate to a suitable level.

4. Allow the system to attain steady state. And note down readings for every 10 min.

5. Record all temperatures, heater input and pressure drop across the orifice.

6. Repeat the experiment at different heat input and air flow rates.

Calculation:

The heat transfer coefficient is given by Newtons law of heating (cooling) :

Experimental Formula, Q = h As ( TbTs)

Where, h = Average surface heat transfer coefficient (W/ m2 0C)

Q = Heat transfer rate, = V x I (Watts)

As = Area of heat transferring surface = .Dt.L (m2)

Ts = Mean surface temperature

Dt= Diameter of the test section = 0.0254 mTb= Bulk Mean temperature

( T2+ T3+ T4 + T5+ T6)

Mean surface temperature, Ts= -----------------------------------0C

5


20/41


(T1+ T7)

Bulk Mean temperature, Tb = ----------------

2

Theoretical Calculation:

Heat input rate = Heat flow rate by air ,

Q = max Cpx ( T1T7)

Where,

Cp=specific heat of air= 1.005 KJ / Kg K

T1= Inlet Temperature of Air in0C

T7= Outlet Temperature of Air in0C

Mass flow rate of air, ma = Cdx Aox a (2gh) / ((Di4D04) / Di4)Where, ma = mass flow rate of air in Kg / sec

Cd = Coefficient of discharge of orifice = 0.68

Ao= Area of Cross Section Orifice in m2= /4 x D0

2

a = density of air at ambient temp. = 1.126 Kg/m3

h = hm[( m / a )1 ]hm= manometer head in meter

m = Density of Mercury = 13600 Kg/m3

ma= U x AiU = Mean Velocity of Flow through tube in m / sec

Ai= Cross Sectional Area of the pipe in m2 = /4 Di

2

Re = Reynolds Number = UxDi/

Where,

= Kinematic Viscosity at Bulk Mean Temperature

Nu = Nusselt Number

h = heat transfer coefficient calculated by using the correlations

Nu = 3.66 For Re 2300 , Turbulant flow

Nu = hxDi/K

Where,

K = thermal conductivity of air at Bulk Mean temp. in w / m k


21/41


Observation Table:

Schematic Diagram:

1) C Channel 2) Motor 3) Blower 4) Adapter6) Orifice 7) Air Inlet Temperature 8) Mica Covered Heater 9) Heater Socket

10) Foam Packing 11) Stainless Still Cladding 12) Monometer

T1:- Air Inlet Temperature ; T6:- Air Outlet Temperature ; T2- T4:- Pipe Wall Temperature

S.

No

Voltage

(V)

(Volts)

Current

(I)

(Amps)

Temperatures

Manometer reading

of water

cm

Air

inlettemp

o

C

Surface temperature

oC

Air

outlettemp

o

CT1 T2 T3 T4 T5 T6 T7 h1 h2 hm


22/41


Result:-

Heat transfer coefficient in forced convection of air in a tube is found to be

h Theoretical = ..

hExp = ..


23/41


Heat Transfer from a Pin-Fin Apparatus by Forced convection mode

EX: NO:06

DATE:

AIM: -

To calculate the heat transfer co-efficient, efficiency and effectiveness of a pin-fin for forced

convection mode.

INTRODUCTION:

Extended surfaces of fins are used to increase the heat transfer rate from a surface to a fluid

wherever it is not possible to increase the value of the surface heat transfer coefficient or the temperature

difference between the surface and the fluid. The use of this is variety of shapes (refer fig. 1).

Circumferential fins around the cylinder of a motor cycle engine and fins attached to condenser tubes of a

refrigerator are a few familiar examples.

It is obvious that a fin surface sticks out from the primary heat transfer surface. The temperature

difference with surrounding fluid will steadily diminish as one move out along the fin. The design of the

fins therefore required knowledge of the temperature distribution in the fin. The main objective of this

experimental set up is to study temperature distribution in a simple pin fin.

APPARATUS:

A brass fin of circular cross section in fitted across a long rectangular duct. The other end of the

duct is connected to the suction side of a blower and the air flows past the fin perpendicular to the axis.

One end of the fin projects outside the duct and is heated by a heater. Temperature at five points along thelength of the fin. The air flow rate is measured by an orifice meter fitted on the delivery side of the

blower.


24/41


SPECIFICATIONS:

1. Duct size (WxB) = 150 mm x 100 mm.

2. Diameter of the fin (Df) = 12 mm.

3. Diameter of the orifice (Do) = 20 mm.

4. Coefficient of discharge Cd = 0.62

5. Centrifugal Blower 1 HP single-phase motor.

6. No. of thermocouples on fin = 5.

(1) to (5) as shown in fig. and indicated on temperature indicator.

7. Thermocouple (6) reads ambient temperature inside of the duct.

8. Thermal conductivity of fin material (Brass) = 110.7 w/m 0C.

9. Temperature indicator = 0300 0C with compensation of ambient temperature up to 500C.

10.Dimmerstat for heat input control 230V, 2 Amps.

11.Heater suitable for mounting at the fin end outside the duct = 400 watts (Band type).

12.Voltmeter = 0100/200 V.

13.Ammeter = 02 Amps.

EXPERIMENTAL PROCEDURE:

Forced Convection: with air circulation (Blower ON)

1. Set the power input to the heater to a desired level through the dimmerstat (VI).

2. Switch on the blower and set the air flow rate to any desired value.

3. Allow the system to attain the steady state.

4. At steady state record the temperatures on the surface (T1, T2, T3, T4, and T5)the ambient

temperature, T6.

5. Note down the difference in level of the two limbs of the manometer.

6. Repeat the experiment by (i) varying the air flow rate and keeping the power input to the

heater constant (ii) varying the power input of the heater and keeping the air flow rate

constant.


25/41


SCHEMATIC DIAGRAM

OBSERVATION TABLE:

S.

No.

V

Volts

I

Amps

Fin Temperatures (0C)

Ambient

Temp.

(0C)

T

Manometer

Reading

in m

T1 T2 T3 T4 T5 T6 h1 h2 hm


26/41


FORMULA USED:

Film Temperature Tf= (Ts+ T)/2 (Property values to be taken at Tf for air)

Average surface temperature Ts= (T1+T2+T3+T4+T5) / 5

Surrounding temperature T=T6

Reynolds number Re=UDeq/

Equivalent diameter Deq= (4 x A)/P

Where, A= Area of duct, WxB (W = Width of the duct in m

B = Breadth of the duct in m)

P = Perimeter of the duct, 2(W+B)

Mean velocity of air in the duct, U = CdAo2g ha and ha= hmm / a

Where , Cd = Coefficient of discharge of orifice=0.62

Ao= Area of orifice, /4 D02

hm= Manometer difference in m

m = Density of Manometer fluid, for mercury=13600 kg/m3

a = Density of Air =1.129 kg/m3

Nusselt Number

NuD= C (ReD)n(Pr)

0.333

Where , NuD= h Df/ K

C and n are taken from the following table


27/41


ReD C n

0.4 - 4.0 0.989 0.33

4 - 40 0.911 0.385

40 - 4000 0.683 0.466

4000 - 40000 0.193 0.618

The fin efficiency is, f = tanh (mL) / mL

The fin effectiveness is, = tanh (mL) / (h Af) / (KbP)

m = h P / kbA

Where, h = heat transfer coefficient W/ m2K

P = perimeter of the fin = D f

Df= Diameter of the fin in m

Af= Area of fin = /4 Df2

Kb= Thermal conductivity of the fin material = 110 W/ m.k

L = Length of the fin in m

T0= Temperature at the base of the fin in0C

T= Surrounding temperature in0C

RESULT :

Thus the value of heat transfer co-efficient under forced condition is determined and also

a) Theoretical value of temperatures along the length of fin.b) Efficiency of the pin-fin for insulated and boundary conditions.


28/41


PARALLEL FLOW / COUNTER FLOW HEAT EXCHANGER

EX.NO: 07

DATE:

AIM:-

To study and compare the heat transfer rate, LMTD, overall heat transfer coefficient and

effectiveness of a heat exchanger working in Parallel flow and counter flow mode.

INTRODUCTION:

Heatexchangersare devices used for effecting the process of heat exchange between two

fluids that are at different temperatures. It is useful in many engineering processes like those in

Refrigeration and Air conditioning system, power system, food processing systems, chemical

reactor and space or aeronautical applications. The necessity for doing this arises in multitude of

industrial applications. Common examples of best exchangers are the radiator of a car, thecondenser at the back of the domestic refrigerator, and the steam boiler of a thermal power plant.

DESCRIPTION & CONSTRUCTION:

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

tube type arrangement as shown in the figure. One fluid flowing through the inner tube and the

other through the annulus surroundings it. The heat transfer takes place across the walls of the

inner tube. The experiments are conducted by keeping the identical flow rates [approx] while

running the unit as a parallel flow heat exchanger and counter flow exchanger.

The temperatures are measured with the help of the temperature sensor. The readings are

recorded when steady state is reached. The outer tube is provided with adequate insulation to

minimize the heat losses.

The total assembly is supported on a main frame. The apparatus consists of a tube in tube

type concentric tube heat exchanger. The hot fluid is water, which is obtained from the hot water

generator it is attached at the bottom of assembly to supply the hot fluid i.e., water with the help

of pump through the inner tube while the cold fluid is flowing through annulus. Pump set isconnected to the hot water generator to suck the water from it & deliver as per requirement.

Different valves are provided in the system to regulate the flow of fluid to the system. The hot

water & cold water admitted at the same end & the opposite end, named parallel & counter flow

heat exchanger accordingly, is done by valve operation.


29/41


SPECIFICATIONS:

Inner Tube Material : copper

Outer Diameter (do) : 40 mm

Inner Diameter (di ) : 12.5 mm

Outer Tube Material : G.I.

PROCEDURE:

1. Switch on the geyser and wait for some time. Start the flow on hot and cold water sides.

2. Adjust the flow rate on the hot water side to a suitable value, say 700 ml/min and that on

the cold side to 800 ml/min.

3. Keep the same flow rates till steady state condition is reached.

4. Note the inlet and outlet temperatures of the cold and hot fluids.

5. Measure the flow rates of hot and cold fluids.

6. Repeat the experiment for different flow rates as well as for parallel flow and counter

flow modes.

7. Note down the I.D and O.D of the inner and outer tubes and also the length of the

exchanger.

CONSTANT:

Cpc&Cph= Specific heat of cold & hot water = 4.178 kJ / kg K


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OBSERVATION TABLE

PARALLEL FLOW: Parallel Flow, in which fluids flow in the same direction.

Flow rate

mh.

(Kg/s)

TIME TAKEN HOT WATER SIDE COLD WATER SIDE

Hot water Cold water

Inlet Temp.

T1=Thi

( C)

Outlet Temp.

T2=Tho

( C)

Inlet Temp.

T3

( C)

Outlet Temp.

T4

( C)

Counter

flow

Parallelflow


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Calculation for Parallel Flow:

1. Heat transfer rate from hot water, Qh= mhx Cphx (ThiTho) in KJ / sec

2. Heat transfer rate from cold water, Qc= mcx Cpcx (TcoTci)in KJ /sec

3. Qavg= (Qh+ Qc) / 2

(Thi - Tci)(Tho - Tco)

4. LMTD =

ln Thi - Tci

Tho - Tco

5.

Overall heat transfer coefficient Uo can be calculated fromQavg

Uo =

AoLMTD

Where Ao = x do x L (m2)

do = outer diameter of the inner tube in m = 0.0125m

L = length of the heat exchanger in m = 1.5m

6. Cmin= smaller of Ccand Ch

Cc ( Tco-Tci)

7. Effectiveness =Cmin ( Thi-Tci)

Ch ( Thi-Tho)

Cmin ( Thi-Tci)

Cc( Tco-Tci)


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RESULT:

PARALLEL FLOW COUNTER FLOW

The heat transfer rate in W

LMTD

Effectiveness

The heat transfer efficiency =

The fin efficiency =

The fin effectiveness =


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PERFORMANCE TEST ON TWO STAGE RECIPROCATING AIR

COMPRESSOR

Ex:No: 08

Date

AIM:

To conduct the performance test on a two stage reciprocating air compressor and to draw

the following graphs:

Delivery pressure Vs Volumetric efficiency

Delivery pressure Vs Isothermal efficiency

Delivery pressure Vs Adiabatic efficiency

APPARATUS REQUIRED:

Stop WatchTachometer

SPECIFICATION:

Make : CEC

Max. Delivery Pressure : 12 bar

Diameter of Low Pressure Cylinder DLP : 70 mm

Stroke Length of Low Pressure Cylinder LLP: 85 mm

Diameter of High Pressure Cylinder DHP : 50 mmStroke Length of High Pressure Cylinder LHp: 85 mm

Coefficient of discharge of orifice : 0.62

Diameter of orifice D0 : 15 mm

Rated speed of compressor : 700 rpm

Energy Meter Constant : 1600 impulse / kW hr

DESCRIPTION:

The two stage air compressor is a reciprocating type (driven by a AC Motor primemover) through a belt. The test rig consists of a base on which the tank (air reservoir) is

mounted. The outlet of the air compressor is connected to the reservoir. The temperature and

pressure of the compressed is indicated by a temperature and pressure gauge. A pressure switchis provided for additional safety. A manometer is provided for measuring pressure difference

across the orifice. The input to the motor is recorded by energy meter.


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PROCEDURE:

1) Check the manometer connections (fill the manometer with water upto half level)

2) Start the compressor.

3) Close the outlet valve and observe the slowly developing pressure.

4) Slightly ahead of the desired pressure, open and adjust the outlet valves so that the

pressure is maintained constant.

5) Note down the readings for different delivery pressures.

CALCULATION:

Theoretical Volume (Vt):

Vs= Vt= ( DLP2LLPN) / 4 x 60

Where, Diameter of Low Pressure Cylinder DLP : 70 mm

Stroke Length of Low Pressure Cylinder LLP: 85 mm

N = Speed of Compressor in rpm

Head (H):

H = h (w/ a)

Where, h = manometer difference (h1h2) in m

w = Density of water = 1000 Kg/m3

a = Density of air = 1.293 Kg/m3

Actual Volume (Va):

Va= CdAo 2 g H

Where, Cd = Coefficient of discharge of orifice = 0.62

Ao= Area of orifice in m2= D0

2/4

g = Acceleration due to gravity (9.81 m/s2)

Volumetric Efficiency (Vol):

Va

Vol = x 100Vt


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Actual Work :

Actual work = (No. of Impulse x 3600 x 0.8) / (Time x E.M.C)Where,

No of Impulse = 10

E.M.C = 1600 imp / kW hr

Isothermal Work :

Isothermal work = PaValn (P3/ P1)Where,

P1= Suction Pressure in N/ m2

P3= Delivery Pressure in N/ m2= Pguage+ Patm

Isothermal Efficiency (Isothermal) :

Isothermal= (Isothermal Work / Actual Work)

Isentropic Work (or) Adiabatic Work :

Isentropic Work = P1V1 + P2V2

Where, = 1.4

V1= VLP= ( DLP2LLP) / 4 ; V2= VHP= ( DHP

2LHP) / 4

P2= P1P3

Adiabatic Efficiency (Adiabatic):

Adiabatic = (Adiabatic Work / Actual Work )


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OBSERVATIN TABLE:

S.

NOReceiver

pressure

Mano

meter

Reading

Speed

Time taken

for 3sec of

energy meter

Temperature in CVol.

Efficiency

%

Isothermal

Efficiency %

Adiabatic

Efficiency %

Bar h1 h2 rpm T1 T2 T3 T4 T5 Vol Isothermal Adiabatic

1

2

3

4

5


37/41


RESULT:

Thus the performance test on a two stage reciprocating air compressor was done and

the following graphs were drawn.

Delivery pressure Vs Volumetric efficiency

Delivery pressure Vs Isothermal efficiencyDelivery pressure Vs Adiabatic efficiency


38/41


THERMAL CONDUCTIVITY OF INSULATING

MATERIAL - LAGGED PIPEEX.NO:09

DATE:

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 r3be the inner radius of the outer MS pipe.

Now the heat flow through the lagging materials is given by

Q = K12 L ( t) / (ln (r2)/r1) or

= K22 L( t) / (ln(r3)/r2)

Where t is the temperature difference across the lagging.

K1is the thermal conductivity of asbestos lagging material and

K2is 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.


39/41


TABULATION :

Heat inputTemperature at Diameter

(D1)0C

Temperature at Diameter

(D2)0C

Temperature atDiameter (D3)

0C

V IVI

Volts

T1T2 T3 T4

T5 T6 T7T8 T9

LAGGED PIPE

SAW DUST

ASBESTOS

T1 HEATER T3

ASBESTOST4 T5 T6

SAW DUSTT7 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


40/41


SPECIFICATION:

Diameter of heater rod, d1 = 50 mm

Diameter of heater rod with asbestos lagging, d2 = 100 mm

Diameter of heater with asbestos lagging and saw dust, d3 = 150 mm

The effective length of the cylinder = 600 mm

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:

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

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

Avg. Temp. of sawdust lagging = T7+ T8+T9/ 3 = oC


41/41

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

Q = k12 l ( t2-t3) / ln (r3/ r2)

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

(average temp of asbestos lagging)

= W/m K.

r2 = Radius of the asbestos lagging = 25 mm

r1 = Radius of the heater = 50 mm

L = Length of the heater = 75 mm

Substituting these values

RESULT :

Thermal conductivity of

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

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

heat and mass transfer lab

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