Heat Conduction of a Hollow Cylinder via Generalized ... · PDF fileInternational Research...

International Research Journal of Applied and Basic Sciences. Vol., 3 (4), 758-769, 2012

Available online at http://www. irjabs.com

ISSN 2251-838X ©2012

Heat Conduction of a Hollow Cylinder via Generalized Hankel Transform

A. M. Shafei, S. Rafee Nekoo*

Ph.D. Student of Mechanical Engineering, School of Mechanical Engineering, Iran University of Science

and Technology, P.O. Box 1684613114, Tehran, Iran �

�Correspondent Author, Email: [email protected]

ABSTRACT: In this work, the heat conduction problem of a finite hollow cylinder is

solved as an exact solution method. In order to solve the PDE equation, generalized finite

Hankel and other integral transformations are used. Two case studies for simulations are

presented and verified with simulations which extracted from finite element method. The

results are shown that this approach is proper for solving heat conduction problems in

cylindrical coordinate.

Keywords: Generalized Finite Hankel Transform; Hollow Cylinder; Finite Element; Heat

Conduction

Introduction

Direct measurement of distribution of temperature is difficult and not possible some times. For dealing

with this problem, different methods have been used. In (Sermet, 2012) a new Green’s function in closed

form was introduced for a boundary value problem in thermoelastoestatics for a quadrant domain. Shiah

and Lee (2011) expressed the problem of 3D modeling of anisotropic heat conduction problem with

arbitrary volume heat source. In order for solving the problem, a Boundary Integral Equation and

Multiple Reciprocity Method were applied and three numerical examples were explained to demonstrate

the accuracy of the method. Baldasseroni et al. (2011) demonstrated heat transfer simulation accompanied

by thermal measurements of microfabricated x-ray transparent heater stages. An amorphous silicon

nitride membrane was introduced appropriate for the sample. Thermal specifications in air and vacuum

were analyzed, analytically and with simulation. Hoshan (2009) presented a triple integral equation

method for solving heat conduction equation. A new kind of triple integral was employed to find a

solution of non-stationary heat equation in an axisymmetric cylindrical coordinates under mixed

boundary of the first and second kind conditions, using Laplace transform the triple integral turned into a

singular integral of the second type. Cossali (2009) expressed an analytical solution of the steady periodic

heat conduction in a solid homogenous finite cylinder via Fourier transform, with the sole restriction of

uniformity on the lateral surface and radial symmetry on the bases. A harmonic heating, as an example,

introduced accompanied with simulation results. In (Lambrakos et al., 2008) inverse analysis of heat

conduction problem in hollow cylinder with axisymmetric source distribution was explained with

experimental results. Golytsina (2008) presented mathematical simulation of temperature field in a hollow

rotating cylinder with nonlinear boundary conditions. Volle et al. (2008) introduced a semi-analytical

method for inverse heat conduction of a rotating cylinder using integral transform methods.

In this paper, in order to solve the problem as an exact method, generalized finite Hankel transform is

used. Hankel transform is used for solving problems consist of cylindrical coordinates, but not the hollow

cylinders. Eldabe et al (2004) introduced an extension of the finite Hankel transform, which is capable of

solving hollow cylindrical coordinates, heat equation or wave with mixed boundary values. Povstenko

(2008) expressed the radial diffusion in a cylinder via Laplace and Hankel transform. Akhtar (2001)

presented exact solutions for rotational flow of a generalized Maxwell fluid between two circular

cylinders.

Intl. Res. J. Appl. Basic. Sci. Vol., 3 (4), 758-769, 2012

�

��

Modeling of the Axis Symmetric Problem

In this section, the distribution of temperature in a hollow cylinder is considered, as it is shown in figure

(1). The equation of the problem is in the form of equation (1):

tzz��2rrr

2 uuur

1u

r

1uc =�

�

��

�+++ (1)

Figure (1): Schematic of the hollow cylinder.

Since the case is not dependent onθ , equation (1) can be written as equation (2):

tzzrrr

2 uuur

1uc =�

�

��

�++ (2)

In which:

( )2

2

rrr

tz,r,uu

∂

∂=

( )r

tz,r,uu r

∂

∂=

( )2

2

zzz

tz,r,uu

∂

∂=

( )t

tz,r,uu t

∂

∂=

Where u is the temperature, z the height, r the radius of the cylinder. Also t stands for the time and

c for diffusivity factor. The boundary and initial condition are considered in equations (3-7):

( ) 0tz,a,u = (3)

( ) z

beTtz,b,u = (4)

( ) 0tr,0,u = (5)

( ) 0th,r,u = (6)


�

� ��

( ) 0tz,r,u = (7)

Where bT is a constant temperature. In order to solve the PDE equation, integral transformations are

used. The first transformation is Laplace which changes the domain of time to s , equation (8):

( )z,0r,uusuur

1uc zzrrr

2 −=��

��

�++ (8)

In which:

( ) ( ){ }tz,r,uLsz,r,u =

( )2

2

rrr

sz,r,uu

∂

∂=

( )r

sz,r,uu r

∂

∂=

( )2

2

zzz

sz,r,uu

∂

∂=

Second transformation is Fourier Cosine and therefore equation (8) changes to equation (9):

( ) ( ) ( ) ( )[ ] ( )sn,r,ussh,r,u1sr,0,u2sn,r,u�nur

1uc z

1n

z

22

rrr

2 =��

��

�−+−−+

+ (9)

Where:

( ) ( ) dzz)h

n�cos(sz,r,u

h

2sn,r,u

h

0�=

( )2

2

rrr

sn,r,uu

∂

∂=

( )r

sn,r,uu r

∂

∂=

The boundary conditions of the bottom and top of the cylinder appears in equation (9) which are zeroes.

Forming the equations as following:

( )( ) ( )sn,r,ucn�sur

1uc

2

rrr

2 +=��

��

�+ (10)

Finite Hankel transform can be applied. The general formulation is presented in equation (11) [9]:

( )( )

( )( )

( ) ( )nfkafakJ�a

bk2Jbf

�b

2

x

ml

dx

d

x

2l1

dx

dH 2

i

im

l

im

l2

22

2

2

m −−=�

��

−+

−+= (11)

Selecting l , m equal to zero in equation (11), the left side of equation (10) is appeared and employing

the transformation results the equation (12):

( )( )( )

( ) ( ) ( )( )ucn�ssn,,kuksn,a,uak�J

bk2Jsn,b,u

�

2c

2

i

2

i

i0

i02 +=��

��

�−− (12)

In which:

( ) ( ){ }sn,r,uHsn,,ku 0i =

Simplifying and applying the radial boundary conditions equation (12) turn into equation (13):

( )( ) ( ){ }

( )��

��

�

+

−+−

++=

22

n

b

2

i

2

2

i�n1

1e1

s

T

ckcn�s

1

�

4csn,,ku (13)

We separate equation (13) into two parts 0n = and 0n > . First, for 0n > , inverse finite Hankel

Transform is applied in equation (14):


�

��

( )( )

( ) ( ) ( ) ( ){ }( )

( ) ( ) ( ) ( )[ ]rkYbkJbkYrkJ�n1

1e1

s

T

ckcn�s

1

�

4c

bkJakJ

akJk

2

�sn,r,u

i0i0i0i022

n

b

2

i

2

2

i

2

0i

2

0

i

2

0

2

iinf

1i

2

−��

��

�

+

−+−×

++−=�

=

(14)

Taking inverse Fourier Cosine transform, equation (15) is formed:

( )( )

( ) ( ) ( ) ( ){ }( )

( ) ( ) ( ) ( )( ) ��

��

�−�

��

��

�

+

−+−×

++−=��

= =

zs

n�cosrkYbkJbkYrkJ

�n1

1e1

S

T

ckcn�s

1

�

4c

bkJakJ

akJk

2

�sz,r,u

i0i0i0i022

n

b

inf

1n2

i

2

2

i

2

0i

2

0

i

2

0

2

iinf

1i

2

(15)

Finally, inverse Laplace transform should be taken, equation (16):

( )( )

( ) ( ) ( ) ( ){ }( )

( ) ( ) ( ) ( )( ) ( ) ( ){ }( )tckcn�

i0i0i0i022

n

2

i

2

inf

1n i

2

0i

2

0

i

2

0

2

ib

inf

1i

2

2i

2

e1zh

n�cosrkYbkJbkYrkJ

�n1

1e1

ckcn�

1

bkJakJ

akJkT�2ctz,r,u

+−

= =

−×��

��

�−�

��

��

�

+

−+−×

+−=��

(16)

In order to solve in steady state solution, t is considered infinity, so equation (16) is written in the form of

equation (17):

( )( )

( ) ( ) ( ) ( ){ }( )

( ) ( ) ( ) ( )( ) ��

��

�−�

��

��

�

+

−+−×

+−=��

= =

zh

n�cosrkYbkJbkYrkJ

�n1

1e1

ckcn�

1

bkJakJ

akJkT�2czr,u

i0i0i0i022

n

2

i

2

inf

1n i

2

0i

2

0

i

2

0

2

ib

inf

1i

2

(17)

The second part, 0n = results equation (18):

( )( )

( )1es

T

cks

1

�

2c

2

s,0,ku b

2

i

2

i −+

= (18)

In addition, taking inverse Finite Hankel and Laplace transforms change equation (18) to (19):

( ) ( )( ) ( )

( )

( ) ( ) ( ) ( )[ ]( )

( )( )tck

2

i

i0i0i0i0

i

2

0i

2

0

i

2

0

2

iinf

1i

b

2

2ie1

ck

1rkYbkJbkYrkJ

1ebkJakJ

akJk�Tc

2

tr,0,u

−

=

−−×

−−

=� (19)

As steady-state analytical solution, equation (20) is appeared:

( ) ( )( ) ( )

( ) ( ) ( ) ( ) ( )( )rkYbkJbkYrkJ1ebkJakJ

akJ�T

2

rui0i0i0i0

i

2

0i

2

0

i

2

0inf

1i

b −−−

=�=

(20)

The final steady-state solution is both parts of answers in equation (21):

( )( )

( ) ( )( ) ( ) ( ) ( ) ( )[ ]

( )( ) ( ) ( ) ( ){ }

( )( ) ( ) ( ) ( )[ ] �

�

��

�−�

��

��

�

+

−+−×

+−+

−−−

=

��

�

= =

=

zh

n�cosrkYbkJbkYrkJ

�n1

1e1

ckcn�

1

bkJakJ

akJkT�2c

rkYbkJbkYrkJ1ebkJakJ

akJ�Tzr,u

i0i0i0i022

n

2

i

2

inf

1n i

2

0i

2

0

i

2

0

2

ib

inf

1i

2

i0i0i0i0

i

2

0i

2

0

i

2

0inf

1i

b

(21)


�

� ��

In order to simulate i

k should found by equation (22) [9]:

( ) ( ) ( ) ( ) 0bkYakJakYbkJ i0i0i0i0 =− (22)

In which 0.5a = , 1b = . In table (1) the roots of equation (22) are presented.

Table (1): Roots of the equation (22).

1k 6.24 11k 69.11 21k 131.94 31k 194.77 41k 257.60

2k 12.54 12k 75.39 22k 138.22 32k 201.06 42k 263.89

3k 18.83 13k 81.67 23k 144.51 33k 207.34 43k 270.17

4k 25.12 14k 87.96 24k 150.79 34k 213.62 44k 276.45

5k 31.40 15k 94.24 25k 157.07 35k 219.91 45k 282.74

6k 37.69 16k 100.52 26k 163.36 36k 226.19 46k 289.02

7k 43.97 17k 106.81 27k 169.64 37k 232.47 47k 295.30

8k 50.26 18k 113.09 28k 175.92 38k 238.75 48k 301.59

9k 56.54 19k 119.37 29k 182.21 39k 245.04 49k 307.87

10k 62.82 20k 125.66 30k 188.49 40k 251.32 50k 314.15

Considering C100Tb

�= , 10n = , 1c = , and 50i = at 0.5z = the distribution of temperature is

shown in figure (2). For a better presentation4

3 of the contour is displayed.

Figure (2): The distribution of temperature at 0.5z = .

In order to verify the results, Finite Element Method is used and the temperature field is demonstrated in

figure (3).


�

��

Figure (3): The distribution of temperature FEM.

The details of temperature field at 0.5z = is shown in figure (4) and at 1z = in figure (5).

Figure (4): The distribution of temperature at 0.5z = , Exact and FEM.


�

� ��

Figure (5): The distribution of temperature at 1z = , Exact and FEM.

Finally at 0.75r = and 1z0 << the comparison between methods is shown in figure (6).

Figure (6): The distribution of temperature at 0.75r = and 1z0 << .

Modeling of the General Problem

In previous section the distribution of temperature was solved as an axis symmetric case and because the

problem was independent of � , generalized finite Hankel transform of zero order was used. In this part

the problem and boundary conditions are dependent on � , so the PDE equation (1) is used and boundary

conditions are chosen as equations (23-27):

( ) 0tz,�,a,u = (23)

( ) z

b e�Ttz,�,b,u = (24)

( ) 0t�,0,r,u = (25)

( ) 0th,�,r,u = (26)

( ) 0z,0�,r,u = (27)

Applying the Laplace transformation, equation (1) changes to equation (28):

( )z,0�,r,uusuur

1u

r

1uc zz��2rrr

2 −=��

��

�+++ (28)


�

��

Next step, finite Fourier Cosine transform is employed, so the equation turns into equation (29):

( ) ( ) ( ) ( )[ ]

( )sn,�,r,us

sh,�,r,u1s�,0,r,u2sn,�,r,u�nur

1u

r

1uc z

1n

z

22

��2rrr

2

=

��

��

�−+−−++

+

(29)

Boundary conditions at the bottom and the top of the cylinder are zero so the simplified equation is

appeared as equation (30):

( ) ( )sn,�,r,ussn,�,r,u�nur

1u

r

1uc 22

��2rrr

2 =��

��

�−++ (30)

Taking Periodic Fourier transformation and simplify the equation result:

( )( ) ( )sn,m,r,u~cn�su~

r

mu~

r

1u~c

2

2

2

rrr

2 +=��

��

�−+ (31)

In which:

( ) ( )�−−=

�

�

im�d�esn,�,r,usn,m,r,u~

( )2

2

rrr

sn,�,r,uu~

∂

∂=

( )r

sn,�,r,uu~r

∂

∂=

Considering l equal to zero in equation (11) and applying Hankel Transform the solution forms as

equation (32):

( )( )

( ) ( )2

i

2

2

ickcn�s

sn,m,b,u~

�

2csn,m,,ku

++= (32)

In which:

( )( ) ( )

��

��

� −��

��

�

+

−+−=

m

12I�

�n1

1e1

s

2Tsn,m,b,u~

m

22

n

b

Equation (32) is considered in two parts: 0n = and 0n > . For the first case taking inverse Hankel

transform results equation (33):

( )( )

( ) ( ) ( ) ( ){ }( ) ( )

( ) ( ) ( ) ( )[ ]rkYbkJbkYrkJm

12I�

�n1

1e1

s

2T

ckcn�s

1

�

2c

bkJakJ

akJk

2

�sn,m,r,u~

imimimim

m

22

n

b

2

i

2

2

i

2

mi

2

m

i

2

m

2

iinf

1i

2

−×��

��

� −��

��

�

+

−+−×

++−=�

=

(33)

Using inverse Laplace, Periodic Fourier and Cosine Fourier transforms equation (33) changes to equation

(34):

( )

( )( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )[ ] ( ) ( ) ( ){ }( )tckcn�im�

imimimim

m

22

ninf

1n

infm

infm2

i

2

i

2

mi

2

m

i

2

m

2

iinf

1i

b

2

2i

2

e1zn�coserkYbkJbkYrkJ

m

12I�

�n1

1e1

ckcn�

1

bkJakJ

akJkTc

tz,�,r,u

+−

=

=

−= =

−−×

��

��

� −��

��

�

+

−+−

+−

=

� � � (34)

Moreover, the steady state answer is, equation (35):


�

� ��

( )

( )( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )[ ] ( )zn�coserkYbkJbkYrkJ

m

12I�

�n1

1e1

ckcn�

1

bkJakJ

akJTkc

z�,r,u

im�

imimimim

m

22

ninf

1n

infm

infm

inf

1i2

i

2

i

2

mi

2

m

i

2

mb

2

i

2

−×

��

��

� −��

��

�

+

−+−

+−

=

� � �=

=

−= =

(35)

Also the solution of the second part is, equation (36):

( ) ( )( ) ( ) ( ) ( )

( )( ) ( )

( ) ( ) ( ) ( )[ ] im�

imimimim

m

tckinfm

infm

inf

1i2

i

2

i

2

mi

2

m

i

2

mb

erkYbkJbkYrkJ

m

12I�e1

ckcn�

1

bkJakJ

akJ

2

T

2

t�,0,r,u 2i

−×

��

��

� −−

+−= −

=

−= =

� � (36)

And the steady state answer for this part is, equation (37):

( ) ( )( ) ( ) ( ) ( )

( )( )

( ) ( ) ( ) ( )[ ] im�

imimimim

minfm

infm

inf

1i2

i

2

i

2

mi

2

m

i

2

mb

erkYbkJbkYrkJ

m

12I�1e

ckcn�

1

bkJakJ

akJ

2

T

2

�,0r,u

−×

��

��

� −−

+−= � �

=

−= = (37)

Finally, the analytical answer for steady-state situation is presented in equation (38):

( )

( )( ) ( ) ( ) ( )

( )( )

( ) ( ) ( ) ( )[ ]

( )( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )[ ] ( )zn�coserkYbkJbkYrkJ

m

12I�

�n1

1e1

ckcn�

1

bkJakJ

akJTkc

erkYbkJbkYrkJ

m

12I�1e

ckcn�

1

bkJakJ

akJ

2

T

z�,r,u

im�

imimimim

m

22

ninf

1n

infm

infm

inf

1i2

i

2

i

2

mi

2

m

i

2

mb

2

i

2

im�

imimimim

minfm

infm

inf

1i2

i

2

i

2

mi

2

m

i

2

mb

−×

��

��

� −��

��

�

+

−+−

+−+

−×

��

��

� −−

+−

=

� � �

� �

=

=

−= =

=

−= =

(38)

In order for simulating ik should found by equation (39) [9]:

( ) ( ) ( ) ( ) 0bkYakJakYbkJ imimimim =− (39)

In which 0.5a = , 1b = and 10m10 ≤≤− . In table (2) the roots of equation (39) are presented.

Table (2): Roots of the equation (40).

m -10,10 -9.9 -8,8 -7,7 -6,6 -5,5 -4,4 -3,3 -2,2 -1,1 0

1k 14.5 13.40 12.31 11.23 10.18 9.19 8.26 7.45 6.81 6.93 6.24

2k 18.82 17.80 16.84 15.93 15.10 14.37 13.74 13.23 12.85 12.62 12.54

3k 23.57 22.74 21.97 21.27 20.65 20.11 19.66 19.30 19.04 18.88 18.83

4k 28.84 28.17 27.55 27.00 26.51 26.09 25.74 25.47 25.28 25.16 25.12

5k 34.45 33.89 33.38 32.93 32.53 32.19 31.91 31.69 31.53 31.43 31.40

6k 40.26 39.78 39.35 38.97 38.63 38.34 38.11 37.93 37.79 37.71 37.69

7k 46.2 45.78 45.40 45.07 44.78 44.54 44.33 44.18 44.06 44.00 43.97

8k 52.4 51.84 51.51 51.22 50.97 50.75 50.57 50.43 50.33 50.28 50.26

9k 58.28 57.96 57.66 57.40 57.17 56.98 56.82 56.70 56.61 56.56 56.54

10k 64.40 69.40 63.83 63.60 63.39 63.22 63.08 62.97 62.89 62.84 62.82


�

��

Considering C100Tb

�= and 1c = , simulation results are obtained. In figure (7), the temperature field is

shown and as it was expected; in this case, temperature is dependent on � and with increasing that,

temperature is enlarged. Temperature at ��,� −= has different signs so there is a jump at this point.

Roughness of the figure is caused by selecting small range for i , m , n . If these values are chosen in a

wide range, the result will be smoother.

Figure (7): The temperature field at 0.5z = .

In order to verify the solution, finite element method is used and results the figure (8), the distribution of

temperature at 0.5z = and 2

�� = .

Figure (8): The temperature field at 0.5z = and 2

�� = .


�

� ��

And in figure (9) the variation of temperature at 0.75r = and2

�� = is presented. The verification is not

as precisely as axis symmetric case and the reason is choosing the range of 10m10 ≤≤− .

Figure (9): The temperature field at 0.75r = and2

�� = .

Conclusion

In this paper, the heat conduction of a hollow cylinder was discussed and solved as an exact solution

method. Using Generalized Finite Hankel Transform is systematic, simple and reliable as it was shown in

verifications with Finite Element Method. Two situations, axis symmetric and general case, were

explained and simulated. Finally, the results show that this method is suitable for solving PDE equations

in cylindrical coordinates.

References

Akhtar W, Siddique I, Sohail A (2011) Exact Solutions for the Rotational Flow of a Generalized Maxwell

Fluid Between two Circular Cylinders. Commun Nonlinear SciNumer Simulat 16: 2788-2795

Baldasseroni C, Queen D R, Cooke D W, Maize K, Shakouri A (2011) Heat Transfer Simulation and

Thermal Measurements of Microfabricated x-ray Transparent Heater Stages. Review of Scientific

Instruments 82(9):093904

Cossali G E (2009) Periodic Heat Conduction in a Solid Homogeneous Finite Cylinder. International

Journal of Thermal Sciences 48: 722-732

Eldabe N T, El-Shahed M, Shawkey M (2004) An Extension of the Finite Hankel Transform. Applied

Mathematics and Computation 151: 713-717

Golitsyna E V (2008) Mathematical Simulation of Temperature Field in a Hollow Rotating Cylinder with

Nonlinear Boundary Conditions. Pleiades Publishing, High Temperature 46(6): 835-84

Hoshan N A (2009) The Triple Integral Equations Method for Solving Heat Conduction Equation. Journal

of Engineering, Thermophysics 18(3): 258-262

Lambrakos S G, Michopoulos J G, Jones H N, Boyer C N (2008) Inverse Analysis of Heat Conduction in

Hollow Cylinders with Asymmetric Source Distributions. Journal of Materials Engineering and

Performance 17: 651-661

Povstenko Y Z (2008) Fractional Radial Diffusion in a Cylinder. Journal of Molecular Liquids 137: 46-50

Sermet V (2012) New Closed-form Green Function and Integral Formula for a Thermoelastic Quadrant.

Applied Mathematical Modeling 36: 799-812

Shiah Y C, Lee R F (2011) Boundary Element Modeling of 3D Anisotropic Heat ConductionInvolving

Arbitrary Volume Heat Source. Mathematical and Computer Modeling 54:2392-2402


�

��

Volle F, Maillet D, Gradeck M, Lebouche M (2008) Semi-analytical inverse heat conduction on a rotating

cylinder with Laplace and Fourier transforms. Inverse Problems in Science and Engineering 16(5):

655-674

Date post:	06-Feb-2018
Category:	Documents
Upload:	lynhu
View:	219 times
Download:	2 times

Heat Conduction of a Hollow Cylinder via Generalized ... · PDF fileInternational Research...

Documents