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Agricultural Energy and Electrification Laboratory Date : March 25th, 2010

Thermodynamics and Heat Transfer

CONDUCTION HEAT TRANSFER THROUGH PLANE WALLS

By :

Gita Pujasari

F14080088

DEPARTEMENT OF AGRICULTURAL ENGINEERING

FACULTY OF AGRICULTURAL TECHNOLOGY

BOGOR AGRICULTURAL UNIVERSITY

2010


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I. INTRODUCTION

A heat transfer situation in which time is not a factor is designed as steady state. Theconsideration of heat transfer where time is not considered affords some simplification in the

analysis. The governing equation for steady state conduction with internal generation is (Welty,

1978 and Holman, 1989)

V2T + = 0 (1)

Where :

VT : temperature gradient in vector form

q : heat flow vector

k : thermal conductivity vector

which is known as the Poisson equation, and for steady state conduction without internal

generation of heat, the Laplace equation applies :

V2T = 0 ...(2)

Both of the above equation apply to an isotropic medium, that is, one whose properties do not

vary with direction, physical properties are also presumed independent of temperature.

The initial consideration is one dimensional steady state conduction without internal

generation of energy. As just discussed. The Laplace equation applies to this case. A general

form of Laplace equation in one dimension is:

= 0 ..(3)

Where

x : critical geometry in the direction of heat transfer, m

: temperature gradient a long x axis

: 0, 1, or 2 in rectangular, cylindrical, and spherical coordinate respectively


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Plane Walls

In the case of he plane wall as shown in figure 1, equation (3) with i = 0 applies.

The equation and boundary conditions to be satisfied are

To

TL

L

x

Figure 1. Steady-state conduction in a plane wall.

= 0(4)

T(x) = T(0) = T0 at x = 0

T(x) = T(L) = TL at x = L

Where

To : temperature at x = 0, K

TL : temperature at x = L, K

Equation (4) may be separated and integrated twice to yield

T(x) = c1x + c2 (5)

And the constants of integration c1 and c2 evaluated, by applying the boundary equation, to be

c1 = and c2 = T0

When c1 and c2 are substituted into equation (5), the final expression for the temperature profile

becomes


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T(x) =

T(x) = .(6)

According to equation (6), the temperature variation in a plane wall under the condition specified

is linear as shown in Figure 1.

The Fourier rate equation may be used to determined heat flux of heat flow rate in this

case. The equation is repeated below, in scalar form in reference

qx = ...(7)

where

qx : heat flux, W/m2

A : cross section area, m2

k : thermal conductivity, W/m K

Since, in the steady- state case, q, is constant, this equation may be separated and integrated

directly as

giving

..(8)

Alternately, the temperature gradient dT/dx could been evaluated from equation (6) and

substituted into equation (7) to achieve the identical result. These two alternate means of

evaluating heat flux, either by direct integration of the Fourier rate equation or by- solving for

temperature and substituting the temperature gradient expression into the rate equation, are both

employed in subsequent examples. One approach may be simpler than another in certain case,

but no general statement can be made in this regard.


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The quantity kA/L, in equation (8), is the thermal conductance for a flat plate or wall.

The reciprocal of this quantity, L/kA, is designated the thermal resistance.

II. OBJECTIVES

1. To asses the characteristic of the steady state heat transfer through the plane wall

2. To determine the rate of steady state heat transfer through the plane wall

III.MATERIALS AND EQUIPMENTS

Material are plane walls made out of alumunumb. The setup of the experiment is

illustrated in fig 2 comprises of Hybrid Recorder Yokogawa DR 130, a water bath at 1000C and

thermocouples. Eleven observation point are inserted into the alumunumb block and the distance

between two observation points are 0.01 m

Water bath

Water 100oC

Hybrid Recorder Aluminum

Block as

Plane wall

Observation

Point

Fig. 2. Experimental set up for heat transfer through a plane wall.

IV.METHODS

1. Record the initial temperature at the observation point located as described in Fig

2

2. Pour boiling water at 100 0C into the water both until the water surface reaches

the height of the alumunium plane wall. Maintain the water at a constant temperature.

3. Record the temperature developed at the observation point in the interval of two

minutes


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V. RESULT AND DISCUSSION

Result from experiment

The experimental data :No x/t T0 T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

1 0 75.3 4535.

229.6

28.

427.2 26

25.

5

25.

425.3 83

2 2 79.755.

3

40.

136.1

34.

131.8 28.5

26.

8

26.

226 86.3

3 4 81.261.

2

46.

542.5

40.

337.6 32.8

29.

7

28.

527.8 89.6

4 6 83.2

64.

7

50.

9 46.9

44.

9 41.9 36.6

32.

8

31.

3 30.2 91.5

5 8 85.167.

5

54.

450.6

48.

745.7 40.2

36.

1

34.

433.1 92.9

6 10 86.970.

5

58.

154.5

52.

849.8 44.4

40.

2

38.

436.9 94.3

7 12 88.472.

7

60.

957.4

55.

952.9 47.5

43.

4

41.

639.9 95.1

8 14 9075.

2

63.

9

60.659.

3

56.4 51.347.

1

45.

3

43.7 95.2

9 16 90.7 7766.

263 62 59.1 54.1

50.

1

48.

446.7 95.2

10 18 92.579.

2

68.

765.6

64.

762 57.2

53.

3

51.

650 97.6

11 20 94.481.

1

70.

867.8 67 64.4 59.7

55.

9

54.

352.6 98.9

12 22 96.183.

6

73.

470.3

69.

766.9 62.3

58.

657 55.3 99.5

13 24 96.184.

6

75.

172.2

71.

669 64.4

60.

9

59.

357.5 99.5

14 26 95.785.

4

76.

674

73.

771.1 66.9

63.

4

61.

960.1 99.6

15 28 96.186.

2

77.

975.3

75.

172.6 68.8

65.

3

63.

862 99.6

16 30 96.286.

8

78.

776.4

77.

373.9 70

66.

8

65.

463.8 99.6

17 32 96 87.1

79.7

77.5 77.6

75.3 71.7 68.6

67.3

65.6 99.6

18 34 96 87. 80. 78.1 78. 76.2 76.8 69. 68. 67 99.5


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6 3 4 9 7

19 36 98.3 8881.

279.5

79.

577.5 74.2

71.

3

70.

268.4 99.5

20 3887.

3

80.

6

78.979.

5

77.3 7471.

4

70.

5

68.5 99.5

21 40 -156.188.

8

82.

881.1

81.

379.5 76.5

73.

8

72.

871.2 99.6

22 42 262.989.

5

83.

682

82.

280.5 77.5 75

74.

172.5 99.6

23 4489.

9

84.

382.8 83 81.3 78.5

76.

4

75.

473.8 99.6

24 46 409.789.

7

84.

382.9

83.

481.5 79.1

76.

876 74.5 99.6

25 4889.

6

84.

382.9

83.

681.9 79.6

77.

2

76.

574.9 99.5

26 5089.

6

84.

482.9

83.

781.9 79.7

77.

4

76.

875.1 99.6

27 52 61.189.

9

84.

883.5 84 82.3 80

77.

777 75.4 99.5

28 54 76.589.

8

84.

683.2

84.

182.3 80.1

77.

8

77.

275.6 99.5

29 56 73 90 84.

883.5 84.

382.6 80.5 78.

2

77.

575.9 99.6

30 58 73.190.

2

84.

983.4

84.

382.5 80.2 78

77.

475.8 99.6

31 60 73.580.

185 83.6

84.

582.8 80.6

78.

376.1 99.5

The calculation data :

No x/t T0 T1 T2 T3 T4 T5 T6 T7 T8 T9 T10

1 0 83 82.23 81.46 80.69 79.92 79.15 78.38 77.61 76.84 76.07 75.3

2 286.

385.64 84.98 84.32 83.66 83 82.34 81.68 81.02 80.36 79.7

3 489.

688.76 87.92 87.08 86.24 85.4 84.56 83.72 82.88 82.04 81.2

4 691.

590.67 89.84 89.01 88.18 87.35 86.52 85.69 84.86 84.03 83.2

5 892.

992.12 91.34 90.56 89.78 89 88.22 87.44 86.66 85.88 85.1

6 10 94.3

93.56 92.82 92.08 91.34 90.6 89.86 89.12 88.38 87.64 86.9

7 12 95. 94.43 93.76 93.09 92.42 91.75 91.08 90.41 89.74 89.07 88.4


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1

8 1495.

294.68 94.16 93.64 93.12 92.6 92.08 91.56 91.04 90.52 90

9 1695.

2

94.75 94.3 93.85 93.4 92.95 92.5 92.05 91.6 91.15 90.7

10 1897.

697.09 96.58 96.07 95.56 95.05 94.54 94.03 93.52 93.01 92.5

11 2098.

998.45 98 97.55 97.1 96.65 96.2 95.75 95.3 94.85 94.4

12 2299.

599.16 98.82 98.48 98.14 97.8 97.46 97.12 96.78 96.44 96.1

13 2499.

599.16 98.82 98.48 98.14 97.8 97.46 97.12 96.78 96.44 96.1

14 2699.

699.21 98.82 98.43 98.04 97.65 97.26 96.87 96.48 96.09 95.7

15 2899.

699.25 98.9 98.55 98.2 97.85 97.5 97.15 96.8 96.45 96.1

16 3099.

699.26 98.92 98.58 98.24 97.9 97.56 97.22 96.88 96.54 96.2

17 3299.

699.24 98.88 98.52 98.16 97.8 97.44 97.08 96.72 96.36 96

18 34 99.5

99.15 98.8 98.45 98.1 97.75 97.4 97.05 96.7 96.35 96

19 3699.

599.38 99.26 99.14 99.02 98.9 98.78 98.66 98.54 98.42 98.3

20 3899.

589.55 79.6 69.65 59.7 49.75 39.8 29.85 19.9 9.95

21 4099.

674.03 48.46 22.89 -2.68 -28.25 -53.82 -79.39

-

104.9

6

-

130.5

3

-

156.1

22 4299.

6

115.9

3

132.2

6

148.5

9

164.9

2

181.2

5

197.5

8

213.9

1

230.2

4

246.5

7262.9

23 4499.

689.64 79.68 69.72 59.76 49.8 39.84 29.88 19.92 9.96

24 4699.

6

130.6

1

161.6

2

192.6

3

223.6

4

254.6

5

285.6

6

316.6

7

347.6

8

378.6

9409.7

25 4899.

589.55 79.6 69.65 59.7 49.75 39.8 29.85 19.9 9.95

26 5099.

689.64 79.68 69.72 59.76 49.8 39.84 29.88 19.92 9.96


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27 5299.

595.66 91.82 87.98 84.14 80.3 76.46 72.62 68.78 64.94 61.1

28 5499.

597.2 94.9 92.6 90.3 88 85.7 83.4 81.1 78.8 76.5

29 5699.

696.94 94.28 91.62 88.96 86.3 83.64 80.98 78.32 75.66 73

30 5899.

696.95 94.3 91.65 89 86.35 83.7 81.05 78.4 75.75 73.1

31 6099.

596.9 94.3 91.7 89.1 86.5 83.9 81.3 78.7 76.1 73.5

Sample of The Calculation :

The data when t = 2 minutes

T(x) = T0 -L

TTL

0

x

T(0) = 86.3 -1.0

7.793.86

x 0.00 = 86.3 W

T(0.01) = 86.3 -1.0

7.793.86

x 0.01 = 85.64 W

T(0.02) = 86.3 -1.0

7.793.86

x 0.02 = 84.98 W

T(0.03) = 86.3 -1.0

7.793.86

x 0.03 = 84.32 W

T(0.04) = 86.3 -1.0

7.793.86

x 0.04 = 83.66 W

T(0.05) = 86.3 -1.0

7.793.86

x 0.05 = 83 W

T(0.06) = 86.3 -1.0

7.793.86

x 0.06 = 82.34 W

T(0.07) = 86.3 -1.0

7.793.86 x 0.07 = 81.68 W

T(0.08) = 86.3 -1.0

7.793.86

x 0.08 = 81.02 W

T(0.09) = 86.3 -1.0

7.793.86

x 0.09 = 80.36 W

T(0.10) = 86.3 -1.0

7.793.86

x 0.10 = 79.7 W

Table of heat flow value


10/14

No TL T0 q/A


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1 75.3 83 19250

2 79.7 86.3 16500

3 81.2 89.6 21000

4 83.2 91.5 20750

5 85.1 92.9 19500

6 86.9 94.3 18500

7 88.4 95.1 16750

8 90 95.2 13000

9 90.7 95.2 11250

10 92.5 97.6 12750

11 94.4 98.9 11250

12 96.1 99.5 8500

13 96.1 99.5 8500

14 95.7 99.6 9750

15 96.1 99.6 8750

16 96.2 99.6 8500

17 96 99.6 9000

18 96 99.5 8750

19 98.3 99.5 3000

20 99.5 248750

21 -156.1 99.6 639250

22 262.9 99.6-

408250

23 99.6 249000

24 409.7 99.6 -

775250

25 99.5 248750

26 99.6 249000

27 61.1 99.5 96000

28 76.5 99.5 57500

29 73 99.6 66500

30 73.1 99.6 66250

31 73.5 99.5 65000

Sample of Calculation

The data when t = 7 minutes, t = 10 minutes and t = 30 minutes

)( Lox TT

L

k

A

q=

Kaluminum = 250 W/m K

t = 7 2/16750)4.881.95(

1.0

250mw

A

qx==

t = 142/9750)7.956.99(

1.0250 mw

Aqx

==


12/14

t = 302/66250)1.736.99(

1.0

250mw

A

qx==

Graph of Time and Temperature from experiment

Graph of Time and Temperature from calculation

Graph of Time and Heat Flux


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VI.DISCUSSION

Acording to the First Law of Thermodynamics, heat transfer changes the internal energy

of both systems involved. As described by the second law of thermodynamics or the Claudius

statement, heat transfer always occurs from a higher-temperature object to a cooler-temperature

one.

In this experiment we will know the characteristic of the steady state heat transfer

through the plane wall and determine the rate of steady heat transfer through the plane wall.

Therere three principles of the heat transfer, they are : conduction, convection, and radiation. In

this experiment, we should discus about conduction heat transfer. The heat transfer by

conduction, occurs via collisions between atoms and molecules in the substance and the

subsequent transfer of kinetic energy. When one end of a metal rod is at a higher temperature,

then energy will be transferred down the rod toward the colder end because the higher speed

particles will collide with the slower ones with a net transfer of energy to the slower ones. For

heat transfer between two plane surfaces, such as heat loss through the wall of a house, the rate

of conduction heat transfer is directly propoetional to the temperature gradient and the cross

sectional area of the path, as the Fourier-Biot Law.

As a result of the experiment, we could see the difference of the temperature in the

different wallas (plane wall). In table 1, we could see the temperature of the wall that near from

the water has the highest temperature (T10), the temperature also increas by the time goes down.

But when the time showed at 20th minute, the machine got error and changes the temperature of

the water that showed. This is could happend because the thermo couple that not in suitable

condition. So, its effect to other walls temperature, mostly in the relation of the temperature with
http://hyperphysics.phy-astr.gsu.edu/HBASE/thermo/firlaw.html#c1http://hyperphysics.phy-astr.gsu.edu/HBASE/thermo/inteng.html#c2http://en.wikipedia.org/wiki/Second_law_of_thermodynamicshttp://en.wikipedia.org/wiki/Clausiushttp://hyperphysics.phy-astr.gsu.edu/HBASE/thermo/temper.html#c1http://hyperphysics.phy-astr.gsu.edu/HBASE/thermo/firlaw.html#c1http://hyperphysics.phy-astr.gsu.edu/HBASE/thermo/inteng.html#c2http://en.wikipedia.org/wiki/Second_law_of_thermodynamicshttp://en.wikipedia.org/wiki/Clausiushttp://hyperphysics.phy-astr.gsu.edu/HBASE/thermo/temper.html#c1


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the time. In table two, we calculate the value Tx when the wallx using To (temperature when x =

11) and TL (temperature when x = 10). The point which nearest the water, have the highest

temperature, so that the point which far from the water, have the lowest temperature. It caused by

conduction heat transfer through plane walls.

The result from the experiment and the calculations is different, because in the

experiment there the error by the human error, the error of the machine, temperature water that

not stady, and the air condition.

VII. CONCLUSION

From the tables and graphic, we could see that the point which nearest the water, have

the highest temperature, and the point that far from the water, have the lowest temperature.

The differences result between experiment and calculation caused by human error, temperature

of water is not stabile and environment factor

VIII. REFERENCES

Holman, J.P. 1997.Heat Transfer. McGraw Hill Book Co., Singapore

Welty, J.R. 1978.Engineering Heat Transfer. John Wiley & Sons, New York, N.Y. USA.

Zemansky, Mark W.1957. Heat and Thermodynamic. McGraw-Hill Book

Company, Inc. New York.

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