Fundamentals Of Finite Element Analysis - Hutton by (Mcgraw-Hill)

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Published: 2014-08-05T15:14:01.996000+00:00

−1

mc

−(1 − s) (1 − s)

 2 2 

˙ mc −1 1

[ k





˙ m ] =

L d s = ˙

mc

L

− s

s



 =

1

2

−1 1

0

−12 2

and note that the matrix is not symmetric.

Hutton: Fundamentals of

7. Applications in Heat

Text

© The McGraw−Hill

Finite Element Analysis

Transfer

Companies, 2004

264

C H A P T E R 7

Applications in Heat Transfer

Using the result of Example 7.8, the stiffness matrix for a one-dimensional

heat transfer element with conduction, convection, and mass transport is given

by

kx A

1

−1

h P L

2

1

˙ mc −1 1

k( e) =

+

+

L

−1

1

6

1

2

2

−1 1

= k( e) + k( e) + k( e)

(7.69)

c

h

˙ m

where the conduction and convection terms are identical to those given in Equa-

tion 7.15. Note that the forcing functions and boundary conditions for the one-

dimensional problem with mass transport are the same as given in Section 7.3,

Equations 7.16 through 7.19.

EXAMPLE 7.9

Figure 7.16a shows a thin-walled tube that is part of an oil cooler. Engine oil enters the

tube at the left end at temperature 50◦C with a flow rate of 0.2 kg/min. The tube is sur-

rounded by air flowing at a constant temperature of 15◦C. The thermal properties of the

oil are as follows:

Thermal conductivity: kx = 0 . 156 W/(m-◦C)

Specific heat: c = 0 . 523 W-hr/(kg-◦C)

The convection coefficient between the thin wall and the flowing air is h =

300 W/(m2-◦C). The tube wall thickness is such that conduction effects in the wall are to

be neglected; that is, the wall temperature is constant through its thickness and the same

as the temperature of the oil in contact with the wall at any position along the length of

the tube. Using four two-node finite elements, obtain an approximate solution for the tem-

perature distribution along the length of the tube and determine the heat removal rate via

convection.

■ Solution

The finite element model is shown schematically in Figure 7.16b, using equal length

elements L = 25 cm = 0 . 025 m. The cross-sectional area is A = (␲ / 4)(20 / 1000)2 =

Air, 15Њ C

T ϭ50Њ C

20 mm

m

ؒ

m

ؒ

1

2

3

4

5

100 cm

1

2

3

4

(a)

(b)

Figure 7.16

(a) Oil cooler tube of Example 7.9. (b) Element and node numbers for a

four-element model.

Hutton: Fundamentals of

7. Applications in Heat

Text

© The McGraw−Hill

Finite Element Analysis

Transfer

Companies, 2004

7.5

Heat Transfer With Mass Transport

265

3 . 14(10−4) m2 . And the peripheral dimension (circumference) of each element is

P = ␲(20 / 1000) = 6 . 28(10−2) m . The stiffness matrix for each element (note that all elements are identical) is computed via Equation 7.69 as follows:

kx A

1

−1

0 . 156(3 . 14)(10−4)

1

−1

k( e) =

=

c

L

−1

1

0 . 025

−1

1

=

1 . 9594

−1 . 9594

−

(10−3)

1 . 9594

1 . 9594

h P L

2

1

300(6 . 28)(10−2)(0 . 025) 2

1

k( e) =

=

=

0 . 157

0 . 0785

h

6

1

2

6

1

2

0 . 0785

0 . 157

˙ mc −1 1

(0 . 2)(60)(0 . 523) −1 1

−3 . 138 3 . 138

k( e)˙ =

=

=

m

2

−1 1

2

−1 1

−3 . 138 3 . 138

−2 . 9810 3 . 2165

k( e) =

−3 . 0595 3 . 2950

At this point, note that the mass transport effects dominate the stiffness matrix and we an-

ticipate that very little heat is dissipated, as most of the heat is carried away with the flow.

Also observe that, owing to the relative magnitudes, the conduction effects have been

neglected.

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