Heat Fare NEW

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EXPERIMENT NO: 01 DATE: 25-07-12 COMBINED RADIATION AND CONVECTION AIM To determine the heat transfer due to radiation and convection from cylinder. To find emissivity at different temperatures. To find natural convection coefficient. APPARATUS REQUIRED Thermometer Heater Stopwatch Sand paper THEORY If a surface, at a temperature above that of its surroundings, is located in stationary air at the same temperature as the surroundings then heat will be transferred from the surface to the air and surroundings. This transfer of heat will be a combination of natural convection to the air (air heated by contact with the surface becomes less dense and rises) and 12 | Page

Transcript of Heat Fare NEW

Page 1: Heat Fare NEW

EXPERIMENT NO: 01 DATE: 25-07-12

COMBINED RADIATION AND CONVECTION

AIM

To determine the heat transfer due to radiation and convection from cylinder.

To find emissivity at different temperatures.

To find natural convection coefficient.

APPARATUS REQUIRED

Thermometer

Heater

Stopwatch

Sand paper

THEORY

If a surface, at a temperature above that of its surroundings, is located in stationary air at the

same temperature as the surroundings then heat will be transferred from the surface to the air

and surroundings. This transfer of heat will be a combination of natural convection to the air

(air heated by contact with the surface becomes less dense and rises) and radiation to the

surroundings. A horizontal cylinder is used in this exercise to provide a simple shape from

which the heat transfer can be calculated.

Note: Heat loss due to conduction is minimised by the design of the equipment and

measurements mid way along the heated section of the cylinder can be assumed to be unaffected

by conduction at the ends of the cylinder. Heat loss by conduction would normally be included

in the analysis of a real application.

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In the case of natural (free) convection the Nusselt number Nu depends on the Grashof and

Prandtl numbers and the heat transfer correlation can be expressed in the form:

Nusselt number, Nu = f (Gr, Pr)

Rayleigh number, Ra = (Gr Pr)

The average heat transfer coefficient for radiation Hrm can be calculated using the following

relationship:

H rm=ΣζF(T s¿¿ 4−T a

4)(T s−T a)

¿

The average heat transfer coefficient for natural convection Hcm can be calculated using the

following relationship:

T film=T s+T a

2

β= 1T film

N gr=gβ (T s−Ta)D3

υ2

Rad = (Grd × Pr)

Num = b(Rad)n, (c and n can be obtained from the table below)

H fm=kN um

D

Note: k, Pr, and n are physical properties of the air taken at the film temperature Tfilm.

The actual power supplied to the heated cylinder, Qin = V×I

Table for c and n values for natural convection on horizontal cylinder.

Rad c N

10-9 to 10-2 0.675 0.058

10-2 to 102 1.02 0.148

102 to 104 0.850 0.188

104 to 107 0.480 0.250

107 to 1012 0.125 0.333

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Alternatively a simplified equation may be used to calculate the heat transfer coefficient for free

convection : H cm=1.32× ¿¿

The value for Hcm should be calculated using both the original and simplified equations and the

values compared.

EXPERIMENTAL PROCEDURE

The surface of the cylindrical brass specimen is polished using the sand paper.

The dimensions of the cylinder are noted down.

Now with the help of thermometer the room temperature is noted.

The cylinder is now heated till it reaches a temperature of about 200 ̊ C.

Now with the help of tongs the cylinder is taken out of the heater and thermometer is

inserted into the space in the centre of it.

Now the temperature is noted at an interval of 1 minute till it drops down to 90 ̊ C.

OBSERVATIONS

Minutes Fall in temperature

temperature

Minutes Fall in temperature

temperature 0 200 11 128

1 197 12 121

2 190 13 114

3 184 14 110

4 176 15 106

5 169 16 102

6 162 17 98

7 155 18 95

8 148 19 92

9 142 19.40 90

10 135

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Weight of cylinder = 0.995 kg

Diameter of cylinder = 4 cm

Length of cylinder = 125 cm

MODEL CALCULATION

T film=T s+T a

2 =

473+3012 = 387 K

Properties of air at Tfilm, Cp = 0.0294×103 J/kg μ = 13.7 Pas k = 0.03227 W/m2K ρ = 0.03314×103 mol/m3

β= 1T film

¿1

387 = 2.5839×10-3 K-1

N gr=gβ (T s−Ta)D3

υ2 ¿ 9.8 × .0028539 ×172× 0.1253

0.172 = 0.05

N pr=μ Cp

k ¿

13.7× 29.40.03227 = 12.48×106

Nubottom = 0.27(Ngr×Npr)0.25 = 7.591

Nutop = 0.54(Ngr×Npr)0.25 = 14.177

Nulateral = 0.59(Ngr×Npr)0.25 = 15.489

htop=kNutop

D ¿

0.03227 ×7.5910.04 = 12.248 W/m2K

hbottom=kNubottom

D ¿ 0.03227 ×14.177

0.04 = 6.124 W/m2K

hlateral=kNulateral

D ¿

0.03227 ×7.5910.125 = 3.998 W/m2K

Qcon = ((htop×Atop) + (hbottom×Abottom) + (hlateral×Alateral))

= ((12.248×0.001256) + (6.124×0.001256) + (3.998×0.000157))

= 14.75 W

Qtotal=mC

p dTdt

¿0.955 ×502 × (200−90)19 ×60

= 46.25 W

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Qrad = Qtotal – Qcon = 46.25 – 14.75 = 31.25 W

hrad=Qrad

A ∆T ¿ 31.5

0.018× 172 = 10.66 W\m2K

RESULT

Natural Convection coefficient = 10.66 W/m2K

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EXPERIMENT NO: 02 DATE: 01-08-12

SHELL AND TUBE HEAT EXCHANGER

AIM

To study of heat transfer in shell and tube heat exchanger.

INTRODUCTION

To Heat exchanger is deviced in which heat is transferred from one fluid to another. The

necessity for doing this arises in a multitude of industrial applications. Common examples of

heat exchangers are the radiator of a car, the condenser at the back of a domestic refrigerator

and the steam boiler of a thermal power plant.

Heat exchangers are classified in three categories:

Transfer Type

Storage Type

Direct Contact Type

THEORY

A transfer type of heat exchanger is one on which both fluids pass simultaneously trough the

device and heat is transferred through separating walls. In practice, most of the heat exchangers

used are transfer type ones.

The transfer type heat exchangers are further classified according to flow arrangements as –

Single pass

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Multiple pass

A sample example of transfer type heat exchanger can be in the form of a tube type arrangement

in which one of the fluids is flowing through the inner tube and the other through the annulus

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

The heat lost by the hot fluid can be calculated

qh = Heat transfer rate to the hot water.

qh = mh Cph (Thi –Tho)

Heat taken by the cold fluid can also be calculated

qc = Heat transfer rate to the cold water

qc = mc Cpc (Tco – Tci )

Qavg = qc + qh

2

Uo = Qavg

Ao ∆ Tm

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DESCRIPTION

The apparatus consists of 1- 2 pass Shell and Tube heat exchanger. The hot fluid is hot water,

which is attained from an insulating water bath using a magnetic drive pump and it flow

through the inner tube while the cold water flowing through the annuals. For flow measurement

rotameters are provided at inlet of cold water and outlet of hot water line. The hot water bath is

of recycled type with digital temperature controller.

UTILITIES REQUIRED

Water supply 20 lit/min (approx.)

Drain.

Electricity supply: 1 phase, 220 V AC, and 4 Kw.

Floor area of 1.5 m× 0.75 m

EXPERIMENTAL PROCEDURE

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STARTING PROCEDURE

Clean the apparatus and make water bath free from dust. Close all the drain valves provided.

Fill water bath ¾ with clean water and ensure that no foreign particles are there.

Connect cold water supply to the inlet of cold water rotameter line.

Connect outlet of cold water from Shell to Drain.

Ensure that all On/Off switches given on the panel are at OFF position.

Now switch on the main power supply.

Switch on heater by operating rotary switch given on the panel.

Set temperature of the water bath with the help of digital temperature controller.

Open flow control valve and by-pass valve for hot water supply.

Switch on magnetic pump for hot water supply.

Adjust hot water flow rate with the help of flow control valve and rotameter.

Record the temperatures of hot and cold water inlet & outlet when steady state is achieved.

CLOSING PROCEDURE

When experiment is over, switch off heater first.

Switch of magnetic pump for hot water supply.

Switch off power supply to panel.

Stop cold water supply with the help of flow control valve.

Drain cold and hot water from the shell with the help of given drain valves.

Drain water bath with the help of drain valve.

SPECIFICATIONS

Water Shell.

Material = S.S.

Dia. = 220 mm

Length = 500 mm

25% cut baffles at 100 mm distance 4 Nos.

Tube

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material =S.S.

OD = 16 mm

ID = 13 mm

Length of tubes = 500 mm

Nos. of tubes = 24

Temperature Controller = Digital 0- 199.90 C

Temperature Sensors = RTD PT- 100 type (5 nos.)

Temperature Indicator = Digital 0 to 199.90 C with multi-channel switch.

Electric Heater = 2 kW (2 Nos.)

Flow measurement = Rotameter (2 Nos.)

Water bath = Material: SS insulated with ceramic wool and powder coated MS outer shell

fitted with heating element.

Pump = FHP magnetic drive pump (max. operating temp. 850 C).

FORMULAE

Rate of heat transfer from hot water

Qh = MhCph(Thi – Tho ), Watt

Mh = Fh∗10−3∗ρh

3600 , Watt

Rate of heat transfer from hot water

Qc = McCpc(Tco – Tci ), Watt

Mc = Fc∗10−3∗ρc

3600 , Watt

Average heat transfer

Q =Qh+Qc

2 , Watt

LMTD

ΔTm = ΔTi−ΔTo

ln ( ΔTiΔTo

)

Where, ΔTi = Thi - Tc

And, ΔTo = Tho - Tci

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Note that in a special case of counter flow exchanger exists when the heat capacity rates Cc

and Ch are equal, then Thi – Tco = Tho– Tci thereby making ΔTi = ΔTo. In this case LMTD is

of the form 0/0 and so undefined. But it is obvious that since ΔT is constant throughout the

exchanger, hence

ΔTm = ΔTi = ΔTo

Overall Heat Transfer coefficient

Ui =Q

AiΔTm , W/m2°C

Uo =Q

AoΔTm , W/m2°C

OBSERVATIONS

Ai = 3.187×10-3 m2

Ao = 4.827×10-3 m2

Sl NO Hot Water Side Cold Water side

Fh (LPH) Thi °C Tho °C Fc (LPH) Tci °C Tco °C

1 155 50.2 40 115 28 35

2 155 50.8 39 165 28 34

3 115 50.5 37 165 28 33

CALCULATION

Mh = 125× 10−3 ×988.5423600

= 0.034 kg/sec.

Qh = (0.034 × 4180.5 ×(322.5−312)) = 1493.44 W

Mc = 155× 10−3 ×995.673600

= 0.042 kg/sec.

Qc = (0.042×4180.5×(310-301)) = 1580.23 W

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Q = 1493.44+1580.23

2 = 1536.8345 W

ΔTi = Thi – Tco = 12.5

ΔTo = Tho- Tci = 11

ΔTm =ΔTi−ΔTo

ln ( ΔTiΔTo

) = 17.73K

Ai = 0.49m2

Ao = 0.603 m2

Ui = Q

Ai× ΔT m W/m2°C

= 1536.83

.49× 17.73

= 176.89 W/ m2°C

Uo = Q

Ao × ΔT m W/m2°C

= 1536.83

.603× 17.73

= 143 W/m2°C

NOMENCLATURE

Ai =inside area of heat transfer,m2

Ao=outside area of heat transfer,m2

Cpc=specific heat of cold fluid at mean temp. ,J/kgoc

Cph=specific heat of hot fluid at mean temp.,J/kgoc

Do=outer dia.of S.S tube,m

FC=flow rate of cold water,LPH

Fh=flow rate of hot water,LPH

L=length of the tube,m

MC=cold water flow rate,kg/s

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Mh=hot water flow rate,kg/s

Ph=thermal conductivity of hot fluid,W/moC

Q=average heat transfer,W

QC=heat gained by cold water,W

Qh=heat lost by hot water,W

Rh=density of hot fluid, kg/m3

Tc=mean temp. of cold water oC

Tci=cold water inlet temp. oC

Tco=cold water outlet temp. oC

Th=mean temp. of hot water. . oC

Thi=hot water inlet temp. oC

Tho=hot water outlet temp. oC

Ui=inside overall heat transfer coefficient, W/m2

Uo=outside overall heat transfer coefficient, W/m2 oC

ρc=density of cold water at mean temp.kg/m3

ρh=density of hot water at mean temp.kg/m3

∆Tm=logarithm mean temp. difference, oC

PRECAUTIONS

Never switch on main power supply before ensuring that all the on/off switches given on

the panel are at off positions.

Never switch on heaters before filling water bath ¾ with clean water. It may damage heaters

Never run the pump at voltage less than 180 & above 230 volts

Never fully close the delivery and by-pass line valves simultaneously.

Always keep apparatus free from dust.

To prevent cloggig of moving parts, run pump at least onces in a fortnight.

RESULT

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The total heat transfer coefficient for fc: Lph is,

Ui = 176.89 W/m2K Uo = 143.00 W/m2K

EXPERIMENT NO: 03 DATE: 08-08-12

PARALLEL AND COUNTERFLOW HEAT EXCHANGER

OBJECTIVE

To study the heat transfer phenomena in parallel / counter flow arrangements.

AIM

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To To calculate rate of heat transfer, LMTD and overall heat transfer coefficient for both

type of heat exchanger.

To compare the performance of parallel and counter current flow heat exchanger.

INTRODUCTION

To Heat exchanger is deviced in which heat is transferred from one fluid to another. The

necessity for doing this arises in a multitude of industrial applications. Common examples of

heat exchangers are the radiator of a car, the condenser at the back of a domestic refrigerator

and the steam boiler of a thermal power plant.

Heat exchangers are classified in three categories:

Transfer Type

Storage Type

Direct Contact Type

THEORY

A transfer type of heat exchanger is on which both fluids pass simultaneously through the

device and heat is transferred through separating walls. In practice most of the heat exchangers

used are transfer type ones.

The transfer type exchangers are further classified according to flow arrangement as-

Parallel flow in which fluids flow in the same direction.

Counter flow in which they flow in opposite direction and

Cross flow in which they flow at right angles to each other.

A simple example of transfer type of heat exchanger in the form of a tube type arrangement in

which one of the fluids is 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.

DESCRIPTION

The apparatus consists of a tube in tube type concentric tube heat exchanger. The hot fluid is hot

water which is obtained from an insulated water bath using a magnetic drive pump and it flow

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through the inner tube while the cold fluid is cold water flowing through the annuals. The hot

water flows always in one direction and the flow rate of which is controlled by means of a

valve. The coid water can be admitted at one of the end enabling the heat exchanger to run as a

parallel flow apparatus or a counter flow apparatus. This is done by valve operations. For flow

measurement rotameters are provided at inlet of cold water and outlet of hot water line. A

magnetic drive pump is used to calculate the hot water from a recycled type water tank, which is

fitted with heaters and digital temperature controller.

UTILITIES REQUIRED

Electricity supply: single phase, 220 VAC, 50Hz, 5-15Amp socket with earth connection.

Water supply: 10 lit/min (approx.)

Drain

Bench area required: 2m x 0.6 m

EXPERIMENTAL PROCEDURE

Put water in bath and switch on the heaters.

Adjust the required the temp. of hot water using DTC.

Adjust the valve. Allow hot water to recycle in bath through by-pass by switching on the

magnetic pump.

Start the flow through annulus and run the exchanger either as parallel flow or counter flow

unit.

Adjust the flow rate on cold water side by rotameter.

Adjust the flow rate on hot water side by rotameter.

Keeping the flow rates same, wait till the steady state conditions are reached.

Record the temperatures on hot water and cold water side and also the flow rates accurately.

Repeat the experiment with a counter flow under identical flow conditions.

NOMENCLATURE

Ai = inside heat transfer area, m2

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Ao = outside heat transfer area, m2

Cph= specific heat of hot fluid at mean temp., kJ /kg 0C

Cpc = specific heat of cold fluid at mean temp., kJ/kg 0C

Do = outer diameter of tube , m

Di = inner diameter of tube, m

Fh = flow rate of hot water, LPH

Fc = flow rate of cold water, LPH

L = length of tube , m

Mh = mass flow rate of the hot water, kg/s

Mc = mass flow rate of the cold water, kg/s

Q = average heat transfer from the system, W

Qc = heat gained by the cold water,W

Qh = heat loss by the hot water, W

Th = mean temp. of hot water, 0C

Tc = mean temp. of cold water, 0C

Tho = outlet temp. of the hot water, 0C

Thi = inlet temp. of the hot water, 0C

Tco = outlet temp. of the cold water, 0C

Tci = inlet temp. of the cold water, 0C

ΔTm = log mean temp. difference, 0C

Ui = inside overall heat transfer coefficient,W/m2 0C

Uo = outside overall heat transfer coefficient, W/m2 0C

ρc = density of cold water at mean temp., kg/m3

ρh = density of hot water at mean temp., kg/m3

OBSERVATIONS

Sl. No: Flow Fh LPH Thi 0C Tho 0C Flow Fc LPH Tci 0C Tco 0C

1Counter

Parallel

115

115

53.8

53.6

48.2

47.9

Counter

Parallel

135

135

30.1

29.2

34.8

33.8

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2Counter

Parallel

165

165

53.3

53.2

48.5

48.5

Counter

Parallel

135

135

29.5

29.2

34.7

34.1

3Counter

Parallel

165

165

53.3

52.3

47.6

48.1

Counter

Parallel

165

165

29.5

29.3

34.2

33.7

CALCULATIONS

Find the properties of water at

Th = T ho+T hi

2

= 52.4+46.1

2

= 49.2 ◦C

Tc = ( T co+T ci

2 )

= 33+27.1

2

= 30 ◦C

Parallel flow,

Cpc = 4.184 KJ/kg◦ C

ρc = 995 kg/m3

ρh = 988.6 kg/m3

Mh = F h∗ρ h

3600× 1000

= 95∗988.6

3600× 1000

= 0.026 kg/s

Qh = Mh Cph (Thi-Tho)

= 0.026×4.180*103×(52.4-46.1)

= 684.68 W

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Mc = F c∗ρ c

3600× 1000

= 100∗995

3600× 1000

= .0276

Qc = Mc Cpc (Tci-Tco)

= 0.0276×4.184×103×(33 - 27.1)

= 681.3

Q =684.68+681 ,.3

2

= 683 W

ΔTm =

ΔT 1 – ΔT 2

ln ΔT 1ΔT 2

= 19.4 – 19

ln 19.419

= 19.2 ◦C

ΔT1 = Thi-Tci

= 46.1- 27.1

= 19 ◦C

ΔT2 = Tho-Tci

= 52.4 – 33

=19.4 ◦C

Ai = 0.048 m2

Ao = 0.064 m2

Ui = Q

A i ΔT m

= 683

.048× 19.2

= 741.1 W/m2 oC

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Uo = Q

A o ΔT m

= 683

.0642×19.2

= 554.1 W/m2 oC

Similarly for the counter current flow,

Ui = Q

A i ΔT m = 775 W/m2 oC

Uo = Q

A o ΔT m = 579 W/m2 oC

RESULT

For Parallel flow,

Ui = 741.1 W/m2K

Uo = 554.1 W/m2K

For Counter flow,

Ui = 775.0 W/m2K

Uo = 579.0 W/m2K

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EXPERIMENT NO: 04 DATE: 05-09-12

HEAT TRANSFER IN FORCED CONVECTION

AIM

To study the heat transfer in forced convection

INTRODUCTION

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

occurs. The Dynamic apparatus consists of a circular pipe, through which cold fluid ie air is

forced. Pipe is heated by a band heater outside the pipe. Temperature of pipe is measured with

thermocouples attached to pipe surface. Heater input is measured by a voltmeter and ammeter.

Thus, heat transfer rate and heat transfer coefficient can be calculated.

SPECIFICATION

Test pipe-33mm ,1.0 ,500mm long

Band heater for pipe

Multichannel digital temperature indicator 0-3000C using chromel /alumel Thermocouples.

Dimmerstat 2amps,240volts for heater input control.

Volt meter 0-200V

Ammeter 0-2A

Blower to force the air through test pipe.

Orifice meter with water manometer.

EXPERIMENTAL PROCEDURE

Put on mains supply.

Adjust the heater input with the help of dimmerstat.

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Start the blower and adjust the air flow with valve.

Wait till steady state is reached and note down the reading in observation table.

NOTE: The calculated values and actual values may differ appreciably because of heat losses.

The heat loss through natural convection, conduction and heat losses through insulation over the

heater is not considered, but they are present.

OBSERVATION

Sl.NO. Volt(V) I(A)Temperature 0c Manometer

difference,hw(cm of water)

Temperature °C

T2 T2 T2 T2 T2 T2 T2

1. 90 0.40 32 42 45 48 39 34 35 6.5

2. 98 0.44 33 43 46 49 39 36 36 6.5

3. 118 0.53 34 50 54 58 43 38 39 6.5

CALCULATION

Air inlet temperature, T 1 = 33 0C

Air outlet temperature, T7 = 36 0C

Density of air, ρa = 1.299× 273

273+T 1

= 1.149 Kg/m3

Diameter of orifice = 33 mm

Manometer difference = Water head = 0.041 m of water

Air head , ha = hw(ρ wρ a ) =

.041×10001.149

= 35.68 m of air

Air volume flow rate, Q = Cd × a0 × √2gh

= 0.64 × п/4 × (0.033)2 × √(2×9.81×35.68) = 1.4 × 10-2 m3/s

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Mass flow rate of air , ma = Q× ρa

= 1.4 × 10-2 × 1.149

= 1.66 × 10-2 kg/m3

Velocity of air, V = Q

a p

= 1.66× 10−23.14 × .0332

= 16.37 m/sec

Heat gained by air , q = ma ×Cp × a(T7 – T1 )

= 1.66 × 10-2 ×1.047 × 103 × (36-33)

= 52.14 Watt

TS = T 2+T 3+T 4+T 5+T 6

5

= 43.80C

Tm = T 1+T 7

2

= 34.50C

Heat loss by radiation = q1

= 0.4× A × (TS4- Tm

4) σ , A = 0.518m2 , σ = 5.67x10-8

= 0.4 × 0.518 × 5.67 × 10-8 (316.84 – 307.54)

= 13.29 Watt

Actual heat loss = q –q1 = 16.408 Watt

hexp = q-q /A X (TS – Tm) = 38.85 W/m2 k

Reynolds no: , NRe = ρ× V × D

µ

= 16.37 × .03347.85 ×10−6

= 11289.65

For turbulent flow , Nu = 0.23(Re)0.8 (Pr)n = 0.23× 11289.65 0.8 × 0.7010.4

= 34.42, n = 0.4 for heating fluid

hth = 1.865 Watt/m2k

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RESULT

Heat transfer in forced convection has been studied. Experimental and theoretical values of heat transfer coefficient have been calculated.

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Sl.NO. Voltage (V) Hexp w/m2k Hthe. W/m2k

1. 90 3.911 2.827

2. 98 3.858 2.8201

3. 118 4.17 3.15

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EXPERIMENT NO: 05 DATE: 12-09-12

COMPOSITE WALL APPARATUS

AIM

To determine the total thermal resistance of composite wall and to plot the temperature gradient

along composite wall surface

INTRODUCTION

The apparatus consists of plates of different materials sandwiched between two aluminum

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

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. Thermocouples are embedded between interfaces of

input slabs, to read the temperature at the surface. The experiment can be conducted at various

values of input and calculation can be made accordingly.

EXPERIMENTAL PROCEDURE

Arrange the plates properly (symmetrical) on both side of heater plate. See that plates are

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

Operate the hand press properly to ensure perfect contact between the plates.

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

Start the supply of heater. By varying the dimmerstat, adjust the input (range 30-70 watts)

and water supply.

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

reached.

Note down the steady state readings in the observation table.

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OBSERVATIONS

WALL THICKNESS CONDUCTIVITY

M.S. = 2.5 cm 0.46 w/m °K

Bakelite = 1.0 cm 0.12 w/ m °K

Brass = 1.0 cm 110 w/m °K

Sl. no.

Heat supplied (W) Temperature (°C)

Voltmeter (V) Ammeter (A) T1 T2 T3 T4 T5 T6 T7 T8

1 120 0.43 44 44 43 43 32 32 31 31

2 129 0.46 49 49 48 48 33 33 32 32

3 140 0.50 54 54 53 53 32 32 31 31

CALCULATION

For Sl. No. 2:

Mean Readings

T a=T1+T 2

2 =

49+492 = 54 °C

T b=T3+T 4

2 =

48+482 = 53 °C

T c=T 5+T 6

2 =

33+332 = 32 °C

T d=T 7+T8

2 =

32+322 = 31 °C

Rate of heat supplied, Q = VI = 140×0.5 = 70 W

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Heat flux, q ¿QA ¿

700.0122 = 5737.70 N/m2

Total thermal resistance of composite slab, R = Ta+T d

q=¿

54−315737.70 = 4×10-3 m2K/W

Thermal conductivity of composite slab, K composite = qh

Ta−T d =

5737.70× 0.04554−31 = 11.22 W/mK

RESULT

Thermal Conductivity of composite slab = 11.22 W/mK

Total thermal resistance of composite slab = 4×10-3 m2K/W

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EXPERIMENT NO: 06 DATE: 21-09-12

STEFAN BOLTZMAN APPARATUS

AIM

To determine and verify the value of the Stefan – Boltzmann constant.

THEORY

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

radiation is absorbed, transmitted through and reflected by the body. A surface which absorbs

all thermal radiation incident over it, is called black surface. For black surface, transmissivity

and reflectivity are zero and absorptivity is unity. Stefan Boltzmann Law states that emissivity

of a surface is proportional to fourth power of absolute surface temperature ie;

e∝T 4 T4

e = σεT4

Where, e = emissive power of surface.

T = absolute temperature.

Σ = Stefan Boltzmann constant

ε = Emissivity of the surface

σ = 5.667 × 10-8 W/m2K4

For a black body,

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ε = 1, hence above equation reduces to

e = σT4

APPARATUS

The dynamic apparatus consists 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 is heated by an electric immersion heater. The hot water is taken around

the hemisphere, so that hemisphere temperature rises. The test disc is then inserted at the

center. Thermocouples are fitted inside hemisphere to average out hemisphere temperature.

Another thermocouple fitted at the center of test disc measures the temperature of the test disc.

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

of 5 seconds.

EXPERIMENTAL PROCEDURE

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

Put ON the heater.

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

Put the thermometer and check water temperature.

Boil the water and switch OFF the heater.

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

See that there is sufficient water above the top of hemisphere( A piezometer tube is fitted to

indicate water level )

Note down the hemisphere temperatures (i.e up to channel 1 to 4)

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

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

hole at the bottom of hemisphere.

Note down the temperature of disc, every time the buzzer rings. Take atleast 4-5 readings.

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Hemisphere

temperature(0C)Time interval (s) Test disc temperature (0C)

T1 = 67.9 0 29.0T2 = 67.8 5 29.2T3 = 67.7 10 29.4T4 = 68.2 15 29.5

20 29.725 29.930 30.135 30.340 30.545 30.650 30.760 30.865 30.970 31.075 31.180 31.385 31.490 31.595 31.7100 31.9105 32.1110 32.2115 32.3120 32.4125 32.5

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130 32.6135 32.7140 32.8145 32.9150 33.1

OBSERVATIONS

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CALCULATION

Weight of test disc = 5.2×10-3 kg

Hemisphere temperature, T H=T 1+T 2+T 3+T 4

4 =

67.9+67.8+67.7+68.24 = 67.9 °C =

341.05 K

Initial test disc temperature, TD = T5 = 31.725 °C = 304.875 K

As area of hemisphere is very large as compared to that of test disc, we can put, qexp = Aσε(Th

4 – Tb4) = 5.55×10-8 Js-1

qcal = me dTdt

= 5.2×10-3×381× 30.3−30.15

= 0.07294 Js-1

σ=me( dT

dr)

A ¿¿ =

0.072943.14 ×0.0001 × (341.14−304.8754 ) = 5.14×10-8 W/m2K

Theoretical value of σ is 5.667 × 10-8 W/m2 K

In the experiment, this value may deviate due to reasons like convection, temperature drop of

hemisphere, heat losses, etc.

From graph, (dT/dt) = 0.4 Ks-1

PRECAUTIONS

Never turn ‘ON’ the heater before putting water in the tank.

Turn ‘OFF’ the heater before draining the water from heater tank.

Drain the water after completion of experiment.

Operate all the switches and controls gently.

RESULT

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The value of the Stefan – Boltzmann constant is determined = 5.14×10-8 W/m2K

EXPERIMENT NO: 07 DATE: 19-08-12

HEAT TRANSFER NATURAL CONVECTION

AIM

To determine the surface H.T coefficient for a vertical tube losing heat by natural convection.

INTRODUCTION

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

between the surface and the fluid and is not created by any external agency. The present

experiment set up is designed and fabricated to study the natural convection phenomenon from

a vertical cylinder in terms of the variation of local heat transfer coefficient along the length and

also the average heat transfer coefficient and its comparison with the value obtained by using an

appropriate correlation.

APPARATUS

The apparatus consists of a brass tube fitted in a rectangular vertical duct. The duct is open at

the top and bottom and forms an enclosure and serves the purpose of undisturbed surrounding.

One side of the duct is made up of Perspex for visualization. An heating element is kept in the

vertical tube which in turn heats the tube surface. The heat is lost from the tube to the

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surrounding air by natural convection. The temp. of the vertical tube is measured by seven

thermocouples. the heat input to the heater is measured by an ammeter and voltmeter and is

varied by a dimmestat. The vertical cylinder with the thermocouple is shown in fig 2. while the

possible flow pattern and also the expected variation of local heat transfer coefficient is shown

in fig 3. the tube surface is polished to minimize the radiation losses.

SPECIFICATIONS

Diameter of the tube (d) = 38 mm.

length of tube (L) = 500mm.

dust size 200 mm. × 200 mm.×800mm.length

multichannel digital temp. indicator 0-300oc using chromel/ alumel thermocouple.

ammeter 0-2 Amp. And voltmeter 0-200 volts.

dimmersat 2 Amp. 240 volts.

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 get heated , rises 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 and cold fluid particles.

The heat transfer coefficient is given by :

h= (q – q1) / As * (Ts – Ta)

where,

h= average surface heat transfer coefficient (W/ m2 o C)

q= heat transfer rate (watts)

As= area of the heat transferring surface = π .d. l

TS= average surface temp = (T1+T2+T3+T4+T5+T6+T7) / 7 OC

Ta= ambient temp in the duct = T8 o C

q1 = Heat loss by radiation = σ. A.e. (TS4 – TS

4)

Where,

Σ= Stephen boltzman constant= 5.667 × 10-8 W/ m2. k4

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A= surface area of pipe. = 0.0597 m 2

e= emissivity of pipe material

Ts & Ta = surface and ambient temperatures in o K respectively.

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 temp. difference between

heat transferring surface and the fluid. The dependence of h on all the above mentioned

parameters is generally expressed in terms of non-dimensional groups as follow :

h × L / k = A × [g.L3.β.ΔT / v2 × Cp.μ / k]

where,

h × L / K ----------is called the Nusselt no:

g . L3. β.ΔT /v3 ------is called to Grashof no:

c.p.μ / k -------is the prandtl no:

A and n are constant depending on the shape and orientation of the heat transferring surface.

Where,

L = a characteristic dimension of the surface.

K= thermal conductivity of fluid.

V = kinematic viscocity of fluid.

μ= dynamic viscosity of fluid.

Cp= specific heat pf fluid.

β= coefficient of volumetric expansion for the fluid.

g = acceleration due to gravity.

ΔT= [Ts – Ta]

For gases ,

β = 1 / ( Tf + 273) /O K

Tf= (Ts +Ta) / 2

For a vertical cylinder losing heat by natural convection, the constants A and n of equation (2)

have determined and the following empirical correlations obtained.

h × L / K =0.59 (Gr.Pr.) 0.25 for 104 < Gr.Pr. <108 -------(3)

h ×L / K =0.13 (Gr.Pr) <1017 ---------(4)

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L = length of the cylinder.

All the properties of the fluid are determined at the mean film temp.(Tf)

PROCEDURE

put ON the supply and adjust the dimmerstat to obtain the required heat input (say 40W,

60W, 70W, etc)

wait till the steady state is reached , which is confirmed from temp. readings.- (T1-T2)

measure surface temp. at the various points i.e T1 – T7

note the ambient temp. i.e T8

repeat the experiment at different heat inputs (do not exceed 80W)

OBSERVATION

O.D. cylinder = 38 mm.

length of cylinder = 500 mm

input to heater = V × I watts

Sl

No

Volt

(V)

Amp

(A)

TEMPERATURE,ºC

T1 T2 T3 T4 T5 T6 T7 T8

1 106 0.9 155 159 164 167 172 164 165 39

1. 91 0.7 127 130 134 136 136 132 130 38

2. 100 0.8 140 144 148 151 150 146 142 39

PROPERTIES OF AIR

T(ºC) Ρ (kgm/m3)Cp

(kJ/kgm.K)

μ×10-3 (N-

sec /m2)k (w/ m.K) Pr

V×10-6

(m2/ sec)

0 1.293 1.005 17.2 0.0244 0.707 13.2810 1.247 1.005 17.7 0.0251 0.705 14.16

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20 1.205 1.005 18.1 0.0259 0.703 15.0630 1.165 1.005 18.6 0.0267 0.701 16.0040 1.128 1.005 19.1 0.0276 0.699 16.9650 1.093 1.005 19.6 0.0283 0.698 17.9560 1.060 1.005 20.1 0.0290 0.696 18.9770 1.029 1.009 20.6 0.0297 0.694 20.0280 1.000 1.009 21.1 0.0305 0.692 21.0990 0.972 1.009 21.5 0.0313 0.690 22.10100 0.946 1.009 21.9 0.0321 0.688 23.13120 0.898 1.009 22.9 0.0334 0.686 25.45140 0.854 1.013 23.7 0.0349 0.684 27.80

CALCULATION

Model calculation for Sl No. 1:

q = V × I = 106 × .9

= 95.4 W

As = π ×d× l = 3.14 × 38× 500×10-6 = 5.969 ×10-2 m2

Ts = T 1+T 2+T 3+T 4+T 5+T 6+T 7

7

= 163.71 0C

Ta = T8 = 39ºC = 312.15 K

q1 = σ × A × e × (Ts4 – Ta

4)

= 5.667 × 10-8 ×5.969 ×10-2 ×0.4× (436.434 – 312.154) = 36.2W

hexp = (q – q1)

A s×(T s – T a)

= 95.4−36.2

.0596×(436.86−312.15)

= 7.964 W/m2 K

Tf = T s+T a

2

= 436.86+312.15

2

= 374.5 0 K

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Β = 1

T f +273¿¿

= 0.00267 K-1

ΔT = Ts – Ta = 441.86 – 314.15 = 127.71 K

According to correlation,

¿¿= A × [(g× L3 × β . × ΔT

ν2 ) × (Cp×μ / k)]n

Gr = g × L3 × β .× ΔT

ν2

= 2.67 ×10−3 ×5003 ×10−9× 9.81× 124.7122.8 ×10−62

= 785.456 × 106

Pr = 0.690

Gr.Pr = 0.690 × 785.456 ×106 = 541.96×106

hcorr = [k ×0.13 ×(Gr × Pr )1/3

L]

= 6.32 W/m2.K

Local heat transfer coefficient

H = Q

A s× ¿¿

h1 = (q– q1)A s׿¿

= 8.97

Similarly,

h2 = 8.78 W/m2.K

h3 = 8.32 W/m2.K

h4 = 8.02 W/m2.K

h5 = 7.98 W/m2.K

h6 = 8.46 W/m2.K

h7 = 8.41 W/m2.K

PRECAUTIONS

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Proper earthing is necessary for the equipment.

keep dimmerstat to zero volt position before putting on main switch and increase it slowly.

keep at least 200 mm . space behind the equipment.

operate the change over switch of temp. indicator gently from one position to another , i.e

from 1 to 8 positions.

never exceed input above 80 watts.

RESULT

The heat transfer coefficient is having a maximum value at the beginning as expected because

of the just starting of the building of the boundary layer and it decreases as expected in the

upward direction due to thickening of layer and which is laminar one. This trend is maintained

up to half of the lengths (approx.) and beyond that there is little variation in the value of local

heat transfer coefficient because of the transition and turbulent boundary layers. The last point

shows somewhat increase in the value of heat transfer coefficient which is attributed to end loss

causing a temperature drop.

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.

Case 1:

hexp= 7.097W/m2.K

hcorr= 5.86 W/m2.K

Case 2:

hexp= 8.017 W/m2.K

hcorr= 6.067 W/m2.K

Case 3:

hexp= 8.94 W/m2.K

hcorr= 6.94 W/m2.K

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EXPERIMENT NO: 08 DATE: 10-10-12

HEAT PIPE DEMONSTRATOR

AIM

To plot the graph of heat sink water temperature raise upto 30 mins. To plot longitudinal

temperature distribution for pipes

INTRODUCTION

Heat pipe is an interesting device, which is used to transfer heat from one location to another. It

works with the help evaporation and condensation of liquid, which is filled inside the heat pipe

as a working medium.

Heat pipe basically consist of a stainless steel pipe, sealed at both the ends. It is evacuated and

filled partially with distilled water. Stainless mesh is provided at inside periphery of the pipe.

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When heat is applied at lower end of heat pipe, water inside it evaporates and vapour passes to

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

vapour condenses giving its latent heat of evaporation to surrounding medium. The condensed

vapour returns to bottom through the mesh packing thus because circulation of vapor, hat pipe

operates at near to isothermal operation and conducts much heat than conventional conductors.

The dynamic apparatus consist of three pipes, viz, a heat pipe, copper pipe, and a stainless steel

pipe. All the pipes have same physical dimensions.

Copper and stainless steel pipes serve the purpose of comparison of heat pipe performance with

copper pipe as good conductor of heat and with stainless steel pipe as same material of

construction. All pipes are mounted vertically with a band heater at lower end and water filled

heat sink at upper end. When heaters starts heating, the pipes , begin to transfer the heat to heat

sink. Rapid rise of temp .of water in the heat pipe heat sink demonstrates high (apparent)

thermal conductivity of heat pipe.

SPECIFICATIONS

Heat pipe – stainless pipe, 25 mm O.D., 400 mm long at both ends , evacuated and filled

partially with distilled water – 1 no.

copper and stainless steel pipes of same size as that of heat pipe- 1 each.

equal capacity heater at bottom end of each pipe

water filled heat sinks at upper end of each pipe

measurements and controls

dimmerstat to control heat input to all the heaters.-1no.

a volt meter and an ammeter to measure input to heaters one each.

Multichannel digital temperature indicator to measure temp. along length of pipes five

thermo couples are provided on each pipe.

Thermometers to note down water temperatures in heat sinks- 3 nos.

EXPERIMENTAL PROCEDURE

Fill up sufficient water in each sinks.

Ensure proper earthing to the unit put the thermometers at the top of heat sinks.

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Keep dimmarstat zero position and start the electric supply to the unit.

Slowly increase the dimmer so that power is supplied to heaters. As same dimmerstsat

supplies power to all heaters and heaters are of same capacity, power input to all the heaters

remains same. This makes the comparison simpler.

Go on noting down the temperatures of water in heat sinks every 5 minutes(stir the water

before noting down the temp.

After around 30 minutes note down the longitudinal temp. 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 than 45oC , otherwise removing water at

high temp. of pipe may burn the seals at the bottom of heat sinks.

If experiments is conducted for more time, it is merely to raise the water temp. & ultimately

evaporation of water. Hence it is recommended to conduct the experiment for more times

than 30 minutes.

OBSERVATION

Heat Sink Water Temperature

Time (min)Stainless steel sink

(°C)Cu heat sink Heat pipe sink

5

10

15

20

25

30

28

28

29

29

30

30

28

30

33

34

36

37

28

29

30

31

31

32

Longitudinal Temperature Distribution

Stainless steel (°C) Cu pipe (°C) Heat pipe (°C)

T6 = 22 T11 = 123 T1 = 81

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T7 = 21

T8 = 21

T9 = 20

T10 = 20

T12 = 115

T13 = 93

T14 = 76

T15 = 65

T2 = 82

T3 = 82

T4 = 82

T5 = 83

CALCULATION

Voltage, V = 130 V

Current, I = 1.25 A

Length of pipe, x = 0.4 m

Area of cross-section, A = пr2 = 3.14×(1.25×10-3)2 4.91×10-4m2

Heat, q ¿−kA dTdx

For Stainless steel, thermal conductivity, k1 = qx

A ∆ T = 162.5× 0.4

0.000491×(22−20) = 66191.4 W/m2K

For Copper, thermal conductivity, k2 = qx

A ∆ T = 162.5× 0.4

0.000491×(123−65) = 2282.4 W/m2K

For Heat pipe, thermal conductivity, k1 = qx

A ∆ T = 162.5× 0.4

0.000491×(83−81) = 66191.4 W/m2K

PRECAUTIONS

Proper earthing is necessary.

Stir the water before noting the water temp. In heat sink.

Do not remove water from heat sinks till the pipes become cool.

Operate only one meter selector switch at a time in upward position. Other two switches

must be in down ward position.

RESULT

The graph between temperature of heat sink and time was plotted. The longitudinal temperature

distribution for pipes is plotted.

The thermal conductivity values calculated for pipes are:

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For Stainless Steel, k = 66191.4 W/m2K

For Copper, k = 2282.4 W/m2K

For Heat pipe, k = 66191.4 W/m2K

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