Ht Lab Finalmanual

7/30/2019 Ht Lab Finalmanual

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HEAT TRANSFER

LAB MANUAL

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LIST OF EXPERIMENTS

1. Composite slab apparatus overall heat transfer co-efficient.2. Stefan Boltzman apparatus.3. Emissivity apparatus.4. Heat transfer through lagged pipe5. Critical heat flux apparatus.6. Heat transfer in pin-fin7. Thermal conductivity of given metal rod.8. Heat transfer in natural convection9. Parallel and counter flow heat exchanger.10.

Heat transfer through a concentric sphere

11.Parallel and counter flow heat exchanger.12.Study of heat pipe and its demonstration.13.Heat transfer in drop and film wise condensation.

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COMPOSITE WALLS APPARATUS

AIM:

The main aim of this experiment is to determine total thermal resistance of composite

wall. And to plot temperature gradient along composite wall structure

SPECIFICATIONS:

1. slabs size:

a) M.S -25cm*25mm.thick

b) Backlite-25cm*10mm.thick

c) Brass-25cm*10mm.thick

2. Nichrome heater wound on mica former and insulator with control unit capacity 200

watt maximum

3. Heater control unit: 230v. 0-2Asingle phase dimmer stat . (1)

4. Voltmeter: 0-250v

5. Ammeter: 0-1amps

6. Multichannel digital temperature indicator

THEORY:

The apparatus consist of a plates of different materials sandwiched between two

aluminum plates. Three types of slabs are provided on the both sides of heater which forms a

composit structure. A small hand press frame is provided to ensure the perfect contact between

the slabs. A dimmer stat is provided for varying the input to the heater and measurement of input

is carried out by a voltmeter and ammeter. The experiment can be conducted at various values of

input slabs, to read the temperature at the surface.

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PROCEDURE:

1. Arrange the plates properly (symmetrically) on both side of heater plate. See that plate

is symmetrically arranged on both sides of heater plate (arranged normally).

2. Operate the hand press properly to achieve the steady environmental conditions.

3. Close the box by cover sheet to achieve the environmental conditions.

4. Start the supply of heater. By varying the dimmer stat, adjust the input (range 30-

70watts) and start water supply.

5. Take the readings of all thermocouples at an interval of 10 minutes until steady state is

reached.

6. Note down the steady state readings in the observation table.

Wall thickness conductivity

1. M. S =2.5 cm 0.46 w/mo K2. Bakelite = 1.0 cm 0.12 w/mo K3. Brass =1.0 cm 110 w/mo K

OBSERVATION

S. NO

Heat supplied (watts) Temperature C

voltmeter Ammeter T1 T2 T3 T4 T5 T6 T7 T8

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CALCULATIONS

MEAN REDING

a. Ta = (T1+T2)/2 0 Cb. Tb = (T3+T4)/2 0 Cc. Tc = (T5+T6)/2 0 Cd. Td = (T7+T8)/2 0 CRate of heat supplied

Q= V*I watts

For calculating the thermal conductivity of composite walls, it is assumed that due to large

dia of the plates, heat flowing through central portion is unidirectional that is axial flow.

Thus for calculations, central half dia. Area where unidirectional flow is assumed is

considered. Accordingly thermocouples are fixed at closed to centre of the plates.

Now,

Heat flux q=Q/a watts/ sq. m

Where A=/4 * d2 (where d= half dia of plates0

A= /4 *0.1252 =0.0122718 m2

1. Total thermal resistance of composite slab R total =(TA-TD)/q sq. m k/watt2. Thermal conductivity of composite slab = K composite = q*b/(TA-TD) w/m/K

Where b= total thickness of the composite slab

= 0.045 m

3. To plot thickness of slab material against temperature gradient

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PRECAUTIONS

1. Heat the dimmer start zero before start.2. Increase voltage slowly.3. Keep all the assembly undisturbed.4. Remove the air gap between plates slowly by moving hand press gently.5. When removing the plates do not disturb the thermocouples.6. Do not increase above 200 V.7. Operate selector switch off temperature indicator slowly.

RESULT AND CONCUSSION:

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STEFAN BOLTZMANN APPARATUS

AIM:

To determine the Stefan Boltzmann constant.

APPARATUS:

The dynamic apparatus of a water heated jacket of hemispherical shape. A copper test

disc is fitted at the center of jacket. The hot water is obtained from a hot water tank, fitted to the

panel, in which water heated by an electric immersion heater. The hot water is taken around the

hemisphere temperature rises. The test disc is inserted at the center. Thermocouples are fitted

inside the hemisphere to average out hemisphere temperature. Another thermocouple fitted at the

center of test disc measures the temperature of test disc.

A timer with a small buzzer is provided to note down disc temperature at the time

intervals of 5 seconds.

THEORY:

All the substances emit thermal radiation. When heat radiation is incident over a body,

part of radiation is absorbed, transmitted through the reflected by the body. A surface whichabsorbs all thermal radiation incidents over it, is called black surface. For black surface,

transmitivity and reflectivity are zero and absorbivity is unity. Stefen boltz men law states that

emissivity of surface is praportionl to fourth poewer of absolute surface temperateure.

PROCEDURE:

1. See that water inlet cock of water jacket is closed and fill up sufficient water in the

heater tank.

2. Put on the heater.

3. Blacken test disc with help of lamp black& let it cool.

4. Put the thermometer and chuck water temperature.

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5. Boil the water and switch off the heater.

6. See that drain cock of water jacket is closed and open water inlet cock.

7. See three is sufficient water above the top of hemisphere (A piezometer tube is fitted to

indicate water level).

8. Note down the hemisphere temperature (i.e. up to channel 1to4).

9. Note down the test disc temperature (i.e. channel no.5).

10. Start the timer. Buzzer will start the ringing. At the start of timer cycle, insert test disc

into the hole at the bottom of hemisphere.

11. Note down the temperature of disc, every time the buzzer rings .Take at least

4to5readings.

OBSERVATIONS:

Hemisphere tempe. Time interval t, sec Test disc temp. T5 C

T1 5

T2 10

T3 15

T4 20

CALCULATIONS:

1) Area of test dist A=3.14*10-4sq.m (d=20mm)

2) Weight of test disc=5.2gms=5.2*10-3

kg

3) Plot a graph. Rise of test disc with time as base and find out its slope at origin

i.e. (dT/dot) at t=0k/see

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4) Hemisphere temp TH= (T1+T2+T3+T4/4+273.15) K

5) Initial test disc temp TD=T5+273.15K

Area of hemisphere is very large as compared to that of test disc, we can put

q=..A (TH4-TD

4)

Where q=heat gained by disc/see

=m.. (dT/dt)

=Stefan Boltzmann constant

m=mass of test disc=5.2*10-3kg

=emissivity test disc=1

A=area of disc

=specific heat copper=381j/kg0 c

= (m.. (dT/dt)/A (TH4-TD

4))

Theoretical valve of is 5.667*10-8W/m2k4

PRECAUTIONS:

1) Never put on heater before putting water in the tank.

2) Put off the heater before draining the water from heater tank.

3) Drain the water after completion of experiment.

4) Operate all the switches and controls gently.

RESULT & CONCLUSION:

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EMISSIVITY MEASUREN\MENT APPARATUS

AIM:

To determine the emissivity of test plate and black plate.

APPARATUS:

The dynamic apparatus uses compactor method for determining the emissivity of test

plate. It consists of two aluminum plates, of equal physical dimensions. Mica heaters are

provided inside the plates. The mounted in an enclose to provide undisturbed surroundings

One of the plates is blackened outside for use a comparator (because black surface has

=1).Another plate having natural surface finish. Input to heaters can be controlled by separatedimmer stats. Heater input is measured on common ammeter and voltmeter. One thermocouple is

fitted on surface on each plate is measure the surface temperature with digital temperature

indicator. By adjusting the input to the heaters, both the plates are brought to same temperature,

so that conduction and convection losses from both the plates are equal and difference in input is

due to different emissivitys.

Holes are provided at back side bottom and the top enclosure for natural circulation of

air over the plates. The plate enclosure is provided with Perspex acrylic sheet at the front

THEORY:

All bodies emits and absorb the thermal radiation to form surroundings the rate of

thermal radiation depends up on temperature of the body. Thermal radiations are electromagnetic

waves and they do not require any medium for propagation.

When thermal radiation strikes a body part of it is reflected, part of it is absorbed an d part of it is

transmitted through the body.

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PROCEDURE:

1. Blacken one of the plates with the help black (normally this is blackened at the works,but if blackening is wiped out, then blackening is necessary.

2. Keep both dimmer knobs at zero position.3. Insert the supply pin-top in the socket (which is properly earthed) and switch on the main

supply.

4. Switch on the mains switch on the panel.5. Keep the meter selector switch (toggle switch) at the black plate side position.6. Adjust dimmer of black plate , so that around 110-120 volts are supplied to black plate7. Now, switch the meter selector switch on other side.8.

Adjust test plate voltage slightly less than that of black plate (say 100-110volts).

9. Check the temperatures (after, say 10minutes) and adjust the dimmers so thattemperatures of both the plates are equal and steady. Normally, very minor adjustments

are required for this.

10.Note down the readings after the plates temperatures reach steady state.OBSERVATIONS:

plateInput

Surface temperature 0 CV I

Test

plate

T1=

Black

plate

T2=

Enclosure temperature T3=0

C

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CALCULATIONS:

1. Enclosure temp TE=T30C= (T3+273.15)

0K

2. Plate surface temp T=T1=T2= 0CTS= (T+273.15)

0K

3. Heat input to black plate, Web=V*I watts4. Heat input to test plate, WT=V*I watts5. Surface area of plates ,A=2*/4D2+(.D.t)

=0.0447 sq. m

Whered=dia of plates=0.16m.

And, t=thickness of plates=0.009m.6. For black plate, Wb= WCVb+Wcdb+WRb.(1)

Where,

WCVb=Convection loss

Wcdb=Conduction loss

WRb=Radiation loss

Similarly for test plate,

WT=WCVT+WCDT+WRT (2)

As the both the plate of same physical dimensions, same material &at same temperature

WCVb=WCVT&WCDb=WCDT

Subtracting the equation (2) from (1) we get,

Webs-WT=WRb-WRT

= (.A.b (TS4-TE

4))-(.A.T (TS4-TE

4))

=.A. (TS4-TE

4)-(b-T)

As emissivity of black plate is 1,

Wb-WT=.A. (TS4

-TE4

)(1-T)Where;

T=emissivity of test plate

=Stefan Boltzmann constant=5.667*10-8w/m2k4

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PRECAUTIONS:

1. Black plate should be blackened.2. Never put your hand or papers over the holes provided at the top of enclosure.3. Keep at least 200mm distance between the back side of unit and the wall.4. Operate all the switches and knobs gently

RESULT&CONCLUTION:

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LAGGED PIPE APPARATUS

AIM:

1. To determine the heat flow rate through the lagged pipe and compare it withthe heater input for known valve of thermal conductivity of lagging material.

2. To determine the approximate thermal conductivity of lagging material byassuming the heater input to be the heat flow rate through lagged pipe.

3. To plot the temperature distribution across the lagging material.APPARATUS:

1. Voltmeter.2. Ammeter.3. Temperature indicator.4. Selector switch.5. Main switch.6. Heater control.7. Assembly.

SPECIFICATIONS:

1. Pipe-a).GI pipe inside 6cm dia (O.D).b).GI pipe middle 8.5cm (mean dia).

c).GI pipe outer 10.7cm (I.D).

d).Length of pipes 1 meter.

2. Heater-Nichrome wire heater (cartridge type) placed centrally having

suitable capacity.

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3. Control panel comprising of-

a) Single panel dimmer stat (0-230).1No.

b) Voltmeter (0-250v)1No.

c) Ammeter (0-2A)1No.

4. Multichannel digital temperature indicator range 0-3000c using cr/al

Thermocouples---1No.

Service required-A.C single phase, 230v electric supply.

THEORY:

The apparatus consist of a concentric pipes mounted on suitable stand. The

hallow space of the inner most pipe consist of the heater. Between first two

cylinders the insulating material with which is to be done filled compactly.

Between second and third cylinders, another material is used for lagging is filled.

The third cylinder is concentric to another outer cylinder. Water flows between

these two cylinders. The thermocouples are attached to the surface of the cylinders

appropriately to measure the temperature.

PROCEDURE:

1. Arrange the pipes in proper fashion with heater assembly. (Arrangednormally).

2. Fill the lagging material in pipes uniformly and by gently pushing press thelagging material (filled normally).

3. See that material gets packed uniformly.4. Cross both ends of pipes and keep the assembly on stands.

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5. Start supply of heater and by varying dimmer stat adjust the input for desiredvalve (range 60-120watts) by using voltmeter and ammeter. Also start water

supply.

6. Take readings of all the 6 thermocouples at an interval of 5 minutes until thesteady state is reached.

7. Note down the steady state reading in the observation table.OBSERVATION TABLE:

1. Inside pipe O.D D1=0.06m.2. Middle pipe mean dia D2=0.085m.3. Outer pipe I.D D3=0.107m.

s.no voltmeter ammeter Thermocouple readings

V I T1 T2 T3 T4 T5 T6

CALCULATIONS:

T inside=(T1+T2/2)0C

Tmiddle=(T3+T4/2)0C

Touter=(T5+T6/2)0C

Ri=inner pipe radius=0.03m.

Ro=outer pipe radius=0.0535m.

Rm=mean radius of middle pipe=0.0425m

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L=length of pipe=0.0425m

K=thermal conductivity w/m.k

q =actual heat input =V*Iwatts

ASSUMPTIONS:

The pipe is so long as compared with that heat flow in radial direction only

middle half length

1. Now first we find out theoretical heat flow rate through composite cylinder.Q= T insideT outside/( (1/k1

K1=0.22 w/m0C and k2=0.13 w/m

oC

Where, actual heat input, qact =VI

2. Now from known value of heat flow rate, value ofcombined thermalconductivity of lagged material can be calculated.

The space between the pipes of dia 6 cm and dia 8.5cm contain commercial

asbestos powder and the space between pipes of dia 8.5 cm and 1.5 cm.

contain saw dust.

K1 (thermal conductivity of asbestos powder)

K2 (thermal conductivity of saw dust.

3. To plot the temperature distribution use formula.

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Where r is the selected radius for corresponding temperature T between the

two pipes of the same lagging material. Thus plot is made for different values of

r.

Note

1. Conductivities of different lagging materials depend upon their density andparticle size. In case of saw dust it depends on the type of wood.

2. These obtained values are apprtoximate because heat flow is not truly inradial direction through the pipe.

Precautions

1. Keep dimmer stat to ZERO position before start.2. Increase voltage gradually.3. Keep the assembly undisturbed while testing.4. While removing or changing the lagging material do not disturb the

thermocouples.

5. Do not increase voltage above 150 volts.6. Operate selector switch of temperature indicator gently.

RESULT & CONCLUSION

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CRITICAL HEAT FLUX APPARATUS

AIM:

To determine the critical heat flux at various bulk temperature water can be

calculated.

APPARATUS:

1. Voltmeter2. Ammeter3. Heater switch4. Lamp switch5. Main switch6. Heater control7. Glass container8. Heater fitting

SPECIFICATIONS:

1. Glass container dia 250mm, height 100mm2. Heater for initial heating - nichrome heater 1kw3. Test heater (r-2) nichrome wire size- 0.18 dia mm4. Length of test heater= 100mm5. Dimmer start6. Voltmeter 0 to 100 volts7. Ammeter 0 to 10 amps8. Thermometer 0 to 1000 C

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THEORY:

When the heat is added to a liquied from a solid submerged substance which

is at a temperature higher the saturatyion temperature of the liquid, it is usual for

the part of the liquid to charge phase. This charge of phase is called boiling.

Boiling is of various types, the type depending upon the temperature

difference between the surface and liquid. The different types are in which a

typically experimental boiling curve obtain in asaturated pool of liquid is drawn

The heat flux supplied to the surface is ploted again(tw-ts) the difference

between the temperature of the surface and the saturation temperation of the liquid.

PROCEDURE:

1. Take sufficient amount of distilled water in the container.2. See that both the haters are completely submerged.3. Connect the heater coil R-1 (1 Kw nichrome coil) and test heater wire across

the studs and make the necessary electrical connections.

4. Switch on the heater R-1 (let varies be at O position)5. Keep it ON till you get the required bulk temperature of water in the

container say 50O

C, 60O

C, 70O

C

6. switch of the heater R-17. Very gradually increase the voltage across test heater by slowly changing the

variac position and stop a while at each position to observe the boiling

phenomenon on wire.

8. Go on increase the voltage till wirw brakes and carefully note the voltageand current at this point.

9. Repeated this experiment by altering the bulk temperature of water.

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Observations

1. Diameter of test heater wire d=2. length of test heater wire l=3. surface area A= d l = m2

Bulk temperature of water

0C

Ammeter reading

I amps

Voltmeter reading

V volts

40oC

50oC

60oC

70oC

CALCULATIONS

The critical heat flux at various bulk temperature water can be calculated by the

following procedure

1. heat input q=vi watts.

2. critical heat flux, hact= q/a watts/m2Peak heat flux in actuated pool boiling=

3. zuber has given following equation for calculating peak heat flux insaturated pool boiling

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q/a=0.18hfg [(lv.g(l-v))]1/4[l-v/ l-v]1/2

Where,

q/a= heat flux watts/m2

hfg= latent heat of vaporization J/kg

lv= liquid vapour surface tension n/m

l=density of liquid kg/m3

v= density of vapour

PRECAUTIONS:

1. keep the variac to zero voltage position before starting the exoeriment2. take sufficient amount of distilled water in the container so that both the

heaters are completely immersed.

3. Connect the test heater wire across the stud.4. Do not touch the water or terminal point when the main switch on5. Operate the variac gently un step and sufficient time in between.6. After the attachment of critical heat flux decreasr slowly the voltage and

bring it to zero position.

RESULT AND CONCLUSION

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PIN FIN APPARATUS

AIM

The aim of the experiment is to study the temperature distribution and the

effectiveness of the fin

THEORY:

Fins are uysed to increase the rate of heat transfer from a surface to the

surrounding. Fluid where evere it is not possible to increase the valve of the

surface heat tyransfer cofficiant rae the temperature difference between the surface

and the fluid.fins are fabricated in varity forms. Fins around the air cooled engines

are a common example.

The aim of the experiment is to study the temperature distribution and the

effectiveness of the fin, which place a impartant role in design.

APPARATUS

1. Pin fin2. Rectangular duct3. Electric heater4. Thermocouples5. Orifice

SPECIFICATIONS

1. Fin12 mm O.D effective length 102 mm with 5 nos. of thermocoupleposition along the length, made of brass, mild steel and alluminium - one

each.

Fin is screwed in heater block wich is heated by a band heater.

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2. Duct150*100 mm cross section, 1000 mm long connected to suction sideof blower.

3. 0.5 HP centrifugal blower with orifice and flow control valve on dischargeside.

4. Orificedia. 22 mm, coefficient of discharge Cd=0.645. Measurements and controls-

a. Dimmerstart to control heater input, 0-230V, 2 ampsb. Voltmeter 0-250V, fort heater supply voltage.c. Ammeter 0-2 amps, for heater current.d. Multichannel digital temperature indicator.e. Water manometer connected to orifice meter.

Procedure

a. Natural conventionOpen the duct cover over the fin. Ensure proper earthing to the unit

and switch on the main supply. Adjust dimmer stat so that about 80

volts are supplied to the heater. The fin will start heating. When the

temperature remain study, note down the temperature of the fin and

duct fluid temperature. Repeat the experiment at different inputs to

heater.

b. Forced conventionClose the duct cover over the fin. Start the blower. Adjust the dimmer

stat so that about 100-110 volts are supplied to the heater. When the

temperature becomes stedy, note down all the temperature and the

manometer difference.

Repeat the experiment at different inputs and at different air flow

rates.

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OBSERVATION TABLE

S.No Manometer

difference

Fin temperature Duct fluid tempe

H m of water T1 T2 T3 T4 T5

CALCULATIONS:

NOMENCLATURE:

tm=average fin temperature= (t1+t2+t3+t4+t5)/5

t=tm-tf

Tmf=mean film temperature=(tm+tf)/2

a= density of air,kgm/m3

w=density of water, kgm/m3

D=dia of pin fin=12*10-3

m

D=dia of orifice=22*10-3

m

Cd=coefficient of discharge of orifice=0.64

=dynamic viscosity-s/m2

Cp=specific heat of air,kj/kg0c

V=kinematic viscosity.m2/s

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Kair=thermal conductivity of air/m0c

=volume expansion coefficient=1/(tmf+273)

H=manometer difference,m of water

V= velocity of air in duct, m/s.

Q= vlume flow rate, m3/s.

Vtmf= velocity of air at men film temperature

All properties are to be evaluated at mean film temperature.

Natural convection

The fin under consideration is horizontal cylinder losing heat by natural

nomvection. For horizontal cylinder, nusselt number,

Nu =1.10(Gr*Pr)1/6

---------for 10-1

< Gr*Pr


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From h, determain m from equation

Using h and m determain temperature distribution in the fin from equation

The rate of heat transfer from the fin can be calculated

Q=(h p k a) *(Tl-Tf) tanh ml

And effectiveness of the fin can be calculated.

= tanh ml/ml

Forced convention

As the natural convention, for horizontal cylinder losing heat by forced convention

Nu =0.615(Re)0.466

---------for 40< Re


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PRECAUTIONS

1. Operate all the switches and controls gently.2. Do not obstruct the suction of the duct are discharge pipe.3. Open the duct cover over the fin for natural convention experiment.4. Fill up water in the manometer and close duct cover for forced convention

experiment.

5. Proper earthing to the unit is necessary.6. While replace the fin, be carefully for fixing thermocouples. Incorrectly

fixed thermocouples may show erratic readings.

RESULTS AND CONCLUSIONS

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THERMAL CONDOCTIVITY OF METALROD

AIM:

To find the thermal conductivity of metal rod.

THEORY:

Thermal conductivity of a material is found to depends on the chemical

composition of the substance which it is composed the phase in which it is exits it

is crystalline structure if a solid the temperature and pressure to which it is

subjected, and weather not it is homogeneous material.

Thermal energy can be conducted in solids by free electrons and by lattice

vibrations. Large number of free electron moves about in the lattice structure of the

materials in good conductures. Energy may also be transfer as vibrationl energy in

the lattice structure of the material.

APPARATUS:

1. Metal Bar-Copper, 25mm O.D., approx, 430 mm long with insulation shellalong the test length and water cooled heat sink at the other end.

2. Test length of bar -240mm3. Thermo couples-Chromel/aluminum, 10 nos.4. Band Nichrome heater to heat the bar.5. Dimmer stat to control the heater input-2A, 230V.6. Voltmeter and Ammeter to measure heater input.

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7. Multichannel Digital temperature indicator,0.10 C least count, 0-2000C withchannel selector switch.

8. Measuring flask to measure water flow.EXPERIMENTAL PROCEDURE

1. Start the electric supply.2. Start heating the bar by adjusting the heater input to, say, 80 volts or 100

bolts.

3.

Start cooling water supply through the heat sink and adjust it to around 350-400 cc per minute.

4. Bar temperature will start rising, Go on checking the temperature at timeintervals of 5 minutes,

5. When all the temperatures remain steady, note down all the observations andcomplete the observation table.

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OBSERVATION

S.No

Test bar Temperature

0

c

Shell

Temp0c

Water

Temp0c

Water

Flow

Rate

Lit/Sec

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T12

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CALCULATIONS-

Heat is flowing through the bar from heater end to water sink. When steady state is

reached, heat passing through the section CC of the bar is heat taken by water.

1)Heat passing through section CC

qcc = m .Cp .T watts

Where,

m= mass flow rate of cooling water,Kg/s,

Cp=Specific heat of water = 4180 J/Kg0C

T= (Water outlet temp)-(water inlet temp)0

C

Now, qcc = -Kcc. [ ]cc .A

A=Cross sectional area of the bar=0.00049 m2

Kcc=--------------- W/m

0

C

2)Heat passing through section BB

qbb = qcc + Radial heat loss between CC & BB

= qcc + / loge(ro/ri)

Where,

k=Thermal conductivity of insulation = 0.35

L1 = Length of insulation cylinder= 0.060m

r0=outer radius=0.105m

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ri=inner radius=0.0125m

qbb = -Kbb . [ ] bb .A

Kbb =------------------ W/ m 0C

3)Similarity,heat passing through section AA,

qaa = qbb + Radial heatloss between B B& AA.

= qbb+ / loge(ro/ri)

Where,

L2= 0.090 m

qaa = -Kaa . [ ] AA .A

Kaa =------------------- W / m0C

RESULTS:

1) Temperature of the bar decreases from hot end to cool end, which satisfiesthe Foururier law of heat conduction.

2) Thermal conductivity of bar at three different sectionsKcc=

Kbb=

Kaa=

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HAET TRANSFER IN FORCED CONVECTION

AIM :

To calculate heat transfer rate and heat transfer coefficient in forced convection.

APPARATUS:

1. Test pipe-33 mm I.D. 50 mm long.2. Band heater for pipe-250W3. Multichannel digital temperature indicator 0-300 0C using Chromel /

Alumel thermocouples.

4. Dimerstat 2 Amps.240 Volts.for heater input control.5. Voltmeter 0-200 volts6. Ammeter 0-2Amps7. Blower to force the air through test pipe8. Orifice meter with water manometer

THEORY:

Whenever a fluid is being fored over the heated surface forced convection heat

transfer occurs the dynamic apparatus consists of circuilarpipe through cold

fluid,i.e. air is being forced. Pipe is heated by a band heater out side the

pipe.temperatyre of the pipe is measured with thermocouples attached to the pipe

surface. Heater input is measured by a voltmeter and ammeter.

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EXPERIMENTAL PROCEDURE:

1. Put ON mains supply.2. Adjust the heater input with the help of dimmer stat.3. Start the blower and adjust the air flow with valve.4. Wait till steady state is reached and note down the reading in the observation

table.

OBSERVATIONS:

S.No

Volt Amp Temperatures0C

Manometer

diference

V I T1 T2 T3 T4 T5 T6 T7 hw

CALCULATIONS-

1. Air inlet temp T1=2. Air outlet temp T2=3. Density of air,

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a = 1.293 x 273 / 273+T1 Kg/m3

4. Diameter of orifice = 22 mmManometer difference=Water head= hw mtrs

Air head, ha =hw(w -a)

where,w= Density of water =1000 Kg/m3

Air volume flow rate, Q=Cd x a0 x 2gha m3/sec

Where, Cd =0.64

a0=c.s area of orifice.

5.Mass flow rate of air,

ma= Q x a Kg/sec

Velocity of air,

V= m/sec

Where ap =Cross sectional area of pipe

= 8.33 x 10-4

m2

6.Heat gained by air, q= ma x Cpa x (T7-T1)

Where,Cpa=Specific heat of air,=1 KJ/Kg K

7.Average inside surface temperature,

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Ts=0C

8.Bulk mean temp.of air

Ts=0C

9.Average surface heat transfer coefficient ,-

Actual Heat loss due to forced convection= q-Heat loss due to radiation

Heat loss due to radiation (q1) =0.4 x A x (Ts

4

-Ta

4

) x (=Stefan BoltzmannConstant)

Actual Heat Loss = qq1

h expt = W/m2k

where A=Inside surface area of the pipe= x di x 1= x 0.033x 0.5=0.0518 m2

10.Reynolds number-

ReD=

=Kinematic viscosity at Tm

D=0.033m

If ReD< 2000,flow is laminar.

For laminar flow= =4.36

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If Reynolds number exceeds 2000,flow is turbulent.

For Turbulent flow,

NuD= (0.023)(ReD)0.8

(Pr)n

Where n=0.4 when fluid is being heated.

n=0.3 when fluid is being cooled

PRECAUTIONS:

1.While putting ONthe supply,keep dimmerstat at zero position and blower

switch OFF.

2.Operate all the switches and controls gently.

3.Donot obstruct the flow of air while experiment is going on.

RESULTS:

htheo from Nu =

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HEAT TRNSFER IN NATURAL CONVECTION

AIM:

To determine the surface heat transfer coefficient for a vertical tube losing heat bynatural convection.

APPARATUS:

1. Tube 38 mm Dia, 500 mm Length.2. Duct size:200mm x 200mm x 800 mm length.3. Multi channel digital temperature indicator 0-3000C using Chromel/Alumel

thermocouple.

4. Ammeter 0-2 Amp.and Voltmeter 0-200 volts.5. Dimerstat 2 Amp,240 volts.

THEORY:

In contrast to the forced convection, natural convection phenomenon is due to the

temperature difference between surface and the fluod and is not created by any

external agency.natural convection flow patterns for some commonly observed

situations are shown in figure

The present experimental set up is designed and fabricated to study the natural

convection phenomenon from a vertical cylinder in terms of varivation of local

heat transfer cofficiant and its comparision its valve obtained by using an

appropriate co relation.

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EXPERIMENTALPROCEDURE

1. Put ON the supply and adjust the dimmer tat to obtain the required heatinput(Say 40W,60W,70W etc)

2. Wait till the steady state is reached ,which is confirmed from temperaturereadings(T1 to T7).

3. Measure surface temperature at the various points i.e. T1 toT7.4. Note the ambient temperature i.e.T85.

Repeat the experiment at different heat inputs (Do not exceed 80W).

OBSERVATIONS

1. O.D. of cylinder(d) =38mm2. Length of cylinder(L) =500mm3. Input to heater =V x I Watts

Sl.No Volt Amp TEMPERATURE,0

C

T1 T2 T3 T4 T5 T6 T7 T8

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CALCULATIONS:

1. h= ------------------------------------------------------ (1)Where, h=Average surface heat transfer coefficient (W/m2 0 C)

q=Heat transfer rate (Watts)

As=Area of the heat transferring surface=.d.1 (m2)

Ts=Average surface temperature=0C

Ta=Ambient temperature in the duct=T80

C

q1=Heat loss by radiation = . A.,(Ts4-Ta4).

Where, = Stefan Boltzman constant = 5.667 x 10-8 W/m2K4

A=Surface area of pipe= 0.05 m2

=Emissivity of pipe material=0.6

Ts & Ta =Surface and ambient temperature in0K respectively

2. = 0.59(Gr,Pr.)0.25 For 104


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5. Compare the experimentally obtained value with the predictions of thecorrelation equations (2) & (3).

PRECAUTIONS:

1. Proper earthing is necessary for the equipment.2. Keep dimmerstat to ZERO volt position before putting on main switch and

increase it slowly.

3. Keep at least 200mm .space behind the equipment.4. Operate the change-over switch of temperature indicator gently from one

position to other ,i.e. from 1 to 8 position.

5. Never exceed input above 80 Watts.

RESULTS & DISCUSSIONS:

The comparison of average heat transfer coefficient is also made with predicted

values are somewhat less than experimental values due to the heat loss by

radiation.

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THERMAL CONDUCTIVITY OF INSULATING POWDER

AIM:

To determine the thermal conductivity of insulating powders using sphere

in sphere method.

APPARATUS & SPECIFICATIONS:

1. Inner sphere-100mm O.D., halved construction.2. Outer sphere-200mm O.D., halved construction.3. Heater-Mica flat heater ,fitted inside inner sphere4. Controls- a) Main Swith-30A , DPDT Switch

b)Dimmerstat-0-230 volts,2A capacity

5. Voltmeter-0-200 volts6. Ammeter- 0-1 Amp7. Multichannel digital temperature indicator ,calibrated for Cr/Al thermo

couples.

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Theory:

Conduction of heat is flow of heatwhich occurs due to exchange of energy

from one molecule to anather with out appreciable motion of molecules. In any

heating process heat is flowing outwords from heat generation point.in order to

reduce losses of heat, various types of insulations are used in practice. Various

powders example asbestos poweder, plaster of faris etc. are used for heat insulation

in order to determine the appropriate thickness of insulation, knowledge of thermal

conductivity of heat insulation material is essential.

EXPERIMENTAL PROCEDURE:

1. Keep dimmerstat knob at ZERO position and switch ON the equipment.2. Slowly rotate the dimmerstat knob,so that voltage is applied across the

heater .Let the temperature rise .

3. Wait until steady state is reached.4. Note down all the temperatures and input of heater interms of volts and

current.

5. Repeat the procedure for different heat inputs.

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OBSERVATIONS;

Sr

No

Temperature0C Heater Input

T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 Volts

V

Amps

I

CALCULATIONS:

1. Heater input = q= V x I Watts2. Average inner sphere surface temperature Ti=(T1+T2+T3+T4) / 4 0C3. Average outer sphere temperature To= (T5+T6+T7+T8+T9+T10) / 6 0C4. Inner sphere radius =50mm=0.05mm5. Outer sphere radius=100mm=0.1mm6. Thermal conductivity K = q (ri-ro) / 4. ri.ro(Ti-To) W/m.K at Ti+To /2

0C

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PRECAUTIONS;

1. Operate the all switches and controls gently.2. Earthing is essential for the unit.RESULT: Thermal conductivity of the insulating powder is ---------------

W/m.K

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PARALLEL AND COUNTERFLOW HEAT EXCHANGER

Aim

To study the experiment on concentric tube hat exchanger

APPARATUS

1. Concentric tubes2. Electric geyser3. Annulus

THEORY:

Heat exchangers are the devices in which the heat s trasfered from one fluid

to another. Exchange of heat is required at many industrial operations as wekk as

chemical process. Common example of of heat exchanger are radiator of car,

condenser of a refregiration unit or colling coil in air conditioner.

Heat exchangers are basically 3 types

1. Transfer type2. Storage type.3. Direct contact type.In transfer type both fluids pass through the exchanger and heat gets transfer

through the separating walls between the fluids.

In the storage type 1st the hot fluid pass through a medium having high heat

capacity and then cold fluid is pass through the medium to collect the heat. Thus

hot and cold fluid are alternately passed through the medium.

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In the direct contact type the fluid are not separated but they mix with each

other and heat passes directly from one fluid to another.

SPECIFICATIONS

1. Haet exchanger a. inlet tube -dia 12.7 mm inner dia, outer dia 11.7 m.copper tube

c. Outertube 25 mm dia N B . GI piped. Length of the heat exchanger is 1 m

2. Electric heater3 KW capacity to supply hot water.3. Valves for flow and direction control5nos4. Thermo meter to messure temperatures10 to 110 o C4 nos5. Measuring flask and stop clock for flow measurement.

PROCEDURE

1. Start the water supply. Adjust the water supply on hot and cold sides. Firstly,keep the valves V2 and V3 clossed and V1- V4 opened so that arrangement

is parallel flow.

2. Put few drops of oil in thermometer pockets. Put the thermometer in thethermometer pockets.

3. Switch on the geyser. Temperature of water will start rising. Aftertemperatures become study, note down the readings and fill up the

observation table.

4. Repeat the experiment by changing the flow.5. Now open the valves V2and V3 and then close the valves V1 and V4 the

arrangement is now counter flow.

6. Wait until the study state is reached and note down the redings.

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OBSERVATION TABLE

Type of

flow

Hot water Cold water

Temperature

Time for 1 lit

water Temperature

Time for 1

lit water

InoC Out

oC Xh SEC In

oC Out

oC Xc SEC

Parallel

flow

Counter

flow

CALCULATIONS

1. Hot water in let temp thi= oCHot water out let temp tho = oC

2. Hot water flow rate mhLet time required for 1 lit of water be Xh sec

Mass of one lit water =1 kg

Mh= 1/Xh kg/s

3. Heat given by the water(in side heat transfer rate).qh = mh*Cp*(thi-tho) watts

4. Similarly for cold waterHeat collect by cold water( out side heat transfer rate)

.qc=mc*cp*(tco-tci)

5. Logarithmic mean temperature difference (LMTD)LMTD= Tm= (Ti-To)/ln(Ti/To)

Where for parallel flow

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Ti=thi-tci, Ti-thi-tci,

To=tho-tco, To=tho-tci

6. Overall heat transfer coefficient Ua. Inside overall heat transfer coefficient Ui

Inside dia of tube =0.011m

Inside surface area of tube, =Ai= di l =*0.011*1=0.03454m2

Now, qh=Ui*Ai*Tm

Ui= qh/(Ai*Tm) w/ m2o C

b. Out side overall heat transfer coefficient UoOut side dia of the tube = 0.012 m

Out side surface area of tube, =Ao= do l =*0.012*1=0.0376m2

Now, qh=Uo*Ao*Tm

Uo= qo/(Ao*Tm) w/ m2o C

7. Effectiveness of the heat exchanger= rate of heat transfer in heatexchanger/max possible heat transfer rate.

=(mh*cp*(thi-tho)/(m* cp*(thi-tci))

PRECAUTIONS

1. Never switch on the geyser unless there is water supply through it.2. If the red indicator on geyser goes of f during operation, increase the

water supply, because it indicates that water temperature exceeds the set

limit.

3. Ensure study water flow rate and temperatures in before noting down thereadings, as fluactuaing water supply can give erratic results.

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RESULT

Heat transfer rate LMTD

inside(W) Outside(W)O

C Ui w/m2o

C Uo w/m2o

C

Parallel flow

Counter flow

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HEAT PIPE APPARATUS

AIM

The main aim of the experiment is to find the thermal conductivity of

heat pipe

SPECIFICATIONS

1. Heat pipe- stain less steel pipe, 25 mm outer dia, 400mm long at bothends, evacuated& filled partially with distilled water- one no

2. Copper and stainless steel pipes of same size as that of pipe- one each.3. Equal capacity heaters at bottom end of each pipe.4. Water filled heat sinks at bottom end of each pipe.5. Measurements and controls

a. Dimmer stat to control heat input to all the heaters 4amps. Capacity-1nos

b. A voltmeters and ammeters to measure input to heaters-one eachc. Multichannel digital temperature indicator to measure temperature

along the pipes. Five thermocouples are provided on each pipe.

d. Thermometers to note down water temperature in heat sinks- 3nos.THEORY:

Heat pipe is an interesting device, which is used to transfer heat from one

location to another. It works with the help of evaporation and condensation of

liquid, which is filled inside heat pipe as working medium.

Heat pipe basically consist of a stain less stell pipe, sealed at both the ends. It is

evacuated & filled partially with distalled water. Stain less stell mesh is provided at

inside periphery of the pipe. When heat is applied at the lower end of the pipe ,

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water inside it evaporates and water passes at upper end of the pipe. The heat is

taken by the medium surrounding upper portion of the heat pipe. The vapour

condenses giving it latent heat of evaporation to the surrounding medium. The

condensed vapor returns to bottom through a mesh packing, thus because of

circulation of vapour, heat pips operates at to isothermal operation and conducts

much heat then conventionl conductors.

PROCEDURE

Fill up sufficient water in heat sinks. Insures proper earthing to the unit put

the thermometers in the grommets provided at the top of heat sinks keep dimmer

stat zero position and start electric supply to unit. Slowly increase the dimmer so

that power is supply to heaters. As same dimmer stat supplies power to all heaters

and all heaters are all same capacity, power in put to all the heater remains same.

This makes the comparison simpler.

Go on noting down the temperature of water in heat sinks every 5 mints (stir

the water before noting down the temperature. After around 30 mints note down

the longitudinal temperature of the pipes, from the temperatures indicator.

Repeat the procedure at different heat inputs, but each time it is necessary to

replace the water. Replace the water when pipes become cool lower them 45oC ,

otherwise removing water at high temperature of pipe may burn the seals at the

bottom of heat sinks.

If experiments in conducted for more time, it is merely to raise the water

temperature &ultimately evaporation of water. Hence it is not recommended to

conduct the experiment for more times than 30 mints

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PRECAUTIONS

1. Proper earthing is necessary2. Stir the water before noting the water temperature in heat sink3. Do not remove water from heat sinks till the pipe become cool4. Operate only one meter selector switch at a time in upward position. Other

to switches must be in down ward position

OBSERVATIONS

1. Heat sink water temperaturesTime, mints S.S.pipes heat

sink

Copper pipe heat

sink

Heat pipe heat

sink

5 min

10mn

2. longitudinal temperature distributionS.S .pipe Copper pipe Heat pipe

T1- T6- T11-

T2- T7- T12-

T3- T8- T13-

T4- T9- T14-

T5- T10- T15-

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GRAPHS

1. plot the graph of heat sink water temperature rise upto 30 min.2. plot longitudinal temperature distribution for pipes.

RESULT AND CONCLUSION

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CONDENSATION IN DROP AND FILM FORM

AIM:

To study the condensation in the drop form and films.

APPARATUS:

1. steam generators2. rot meter3. condensers4. temperature indicator5. selector switch6. heater control7. main switch

SPECIFICATIONS:

1. condensers-made of copper 19mm outer dia, 150mm long, one with naturalsurface and one it chrome plated surface

2. Rotameter-25-250lph for water flow measurement.3. Steam generator with 1.5kw electric and low water level protector.4. Multichannel digital temperature indicator0-3000cusing chromel-alumel

thermocouples.

5. Pressure gauge to measure pressure.6. Necessary valves for water and steam control.

THEORY:

Condensation of vapor is needed in many of the processes, like steam

condensers, refrigeration etc. when vapor comes in contact with the surface having

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temperature lower than saturation temperature, condensation occurs. When the

condensate formed wets the surface, a film is formed over the surface and the

condensation is film wise condensation. When condensate does not wet the

surface, drops are formed over the surface and condensation is drop wise

condensation.

The apparatus consists of two condensers, which are fitted inside a glass

cylinder, which is clamped between two flanges. Steam from steam generator

enters the cylinder through a separator. Water is circulated through the condensers.

One of the condenser is with natural surface finish to promote film wise

condensation and the other is chrome plated to create drop wise condensation.

Water flow is measure by rotameter. Various temperatures are measured by digital

temperature indicator. Steam pressure is measured by a pressure gauge. Thus heat

transfer coefficients in drop wise and film wise condensation can be calculated.

PROCEDURE:

Fill up water in the steam generator and close the water filling valve. Start

water supply through the condensers. close the steam control valve, switch on the

supply and start the heater. After some time, steam will be generated. close water

flow through one of the condensers. Open steam control valve and allow steam to

enter the cylinder and pressure gauge show some reading. Open drain valve and

ensure that air in the cylinder is expelled out. Close the drain valve and observe the

condensers. Depending on the condensers in the operation, drowse or film wise

condensation will be observed. Wait for some time for steady state, and note down

all the readings. Repeat the procedure for the other condenser.

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OBSERVATIONS:

S.No 1 2

Steam pressure, kg/cm2

Water flow rate LPH

Steam temperature-T10C

Drop condes.surface temperature-T20C

Film condens.surface temperature-T30C

Water inlet temperature-T40C

Water outlet temperature form drop

wise condersor-T50C

Water outlet temperature form film wise

condensorT50C

CALCULATION:

(Film wise condensation)

Water flow=Lph= w kg/see

Water inlet temp.T4=0c

Water outlet temp. =0c

(T5 for drop wise condensation T6 for film wise condensation)

Heat transfer rate at condenser wall

Q=W.CP.(T5-T4)watts.

Where cp=specific heat of water=4.2*103j/kgk

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Surface area of the condenser A=9.24*10-3

m3

Heat transfer coefficient,

H1=q/A (TS-TW) W/m2

0c

Where TS=temperature of steam= (T1)

And TW=condenser wall temperature (T2 orT3)

Theoretically, for film wise condensation

Hl= ((.2.g.k3)/ (TS-TW)..L)0.25

Where

= total heat of steam at TS,j/kg

=density of water, kg/m3

g=gravitational acceleration/see2

k=thermal conductivity of water w/m0

c

=viscosity o water-s/m2

L=length of condenser=0.15m

Above valves at mean temperature, Tm= (TS+TW)/20c

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PRECAUTIONS:

1. Operate all the switches and controls gently.2. Never allow the steam to enter the cylinder unless the water is flowing

through condenses.

3. Always ensure that the equipment is earthed properly before switching onthe supply.

RESULT & CONCLUSION:

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Documents

Ht Lab Finalmanual