heat transfer project-final

7/31/2019 Heat Transfer Project-final

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INDIAN INSTITUTE OF SPACE SCIENCE AND TECHNOLOGY,

TRIVANDRUM

HEAT TRANSFER PROJECT

MADE BY:

Sampathi Divyajyothi (SC09B038)

Maddala Deepthi (SC09B040)

Rupali Sahu (SC09B049)

Prachi Agrawal (SC09B071)


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1. Introduction 32.Specifications 43.Procedure 54.Observations 65.Results and Discussions 76. Inferences 77.References 7


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ABSTRACT

In this work the thermal diffusivity of potato was determined by numerical method

solution of Fouriers second law. The heat transfer model was solved using two

curves of temperature history. The two temperature profiles obtained in two

different positions in biological material under unsteady state heat transfer were

used. The experimental data were adjusted by using the explicit numerical method

of finite differences. The calculated thermal diffusivity for cylindrical and flat plate

geometry presented good results with the average relative deviation less than

0.78%.


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Determination of thermal diffusivities of complex shapes requires high

analytical skills. Here we have used finite difference numerical method

which greatly simplifies the problem. Fouriers second law for

unidirectional transient heat conduction in Cartesian and cylindrical

coordinates was employed. Thermal diffusivity has been assumed

constant over the range of temperatures used. Moisture content was

assumed to be 80%.

Equations used:

a) For cylindrical piece

(Ti+1

-Ti)/t = (Tm-1-2Tm+Tm+1 )/r

2

T(r=0) =T0

T(r=R) =Ts

b) For flat piece

(Ti+1

-Ti)/t=(Tm-1-2Tm+Tm+1)/x

2

T(r=0) =T0

T(r=R) =Ts


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SPECIFICATIONS:

Thermometer 1:

Digital thermometer with range -0 to 10C; least count: 0.1C

Thermometer 2:

Glass and mercury thermometer with range - to 110c; least count:

0.1C

Cylindrical specimen:

Height: 5cm

Radius: 15 mm

Cuboidal specimen:

Height: 5cm;

Width: 1cm;

Thickness: 3cm

Water bath temperature: 0C


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PROCEDURE

1)The specimen was put into a water-bath maintained at a constanttemperature, with its circular surfaces (cylindrical specimen) or

lateral surfaces (cuboidal specimen) insulated with thermocole to

ensure heat transfer in 1 direction.

2)Bores were made at the center and at a distance of R/2 from the centerand thermometers inserted.

3) Temperature readings were taken at equal intervals of time foraround 30 minutes.

4)Above mentioned equations and boundary conditions used tocalculate thermal diffusivity for each set of readings and average

taken.

a. Cuboidal b. Cylindrical


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b.OBSERVATIONSTEMPERATURE PROFILE:

A) For cylindrical specimen:

B) For cuboidal specimen:

X: time(s)

Y: temp(deg C)

X: time(s)

Y: temp(deg C)


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RESULTS AND DICUSSIONS:

For cylinder, the thermal diffusivity has been found out to be2.2x10

7

m

2

/sFor the cuboidal plate, the thermal diffusivity is 2.4 x107m2/s

INFERENCES:

The experiment yielded a greater value of thermal diffusivity for the flat

plate than the cylindrical specimen, but both were quite close to theactual value of 1.3x10

7m

2/s.

Errors might have arisen due to the following reasons:

1.Since the specimen is kept in water bath, its moisture contentincreases.

2.Error in reading temperature values from the thermometer3.The spacing used was too large to apply finite difference numerical

method without much loss in accuracy.

REFERENCES:

[1] Carciofi, B.A.M. et al. Determination of Thermal Diffusivity of MortadellaUsing Actual Cooking Process Data. J. Food Eng. v. 55(1), p. 89-94, 2002

[2] Yunus A. Cengel, Heat and Mass Transfer-a Practical Approach, third

edition, p.286-289, 2010.

heat transfer project-final

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