International Conference on Global Trends in Engineering, Technology and Management (ICGTETM-2016)
Computational Modeling of Heating and Cooling Effect of
a Thin Annular Disc
Sunita S. Patil*1 and J.S.V.R.Krishna Prasad*2
*1
SSBT’s College of Engineering & Technology, Bambhori, Jalgaon (M.S.), India
*2
Abstract
M.J. college, North Maharashtra University, Jalgaon (M.S.), India
- The paper is concerned with the inverse
unsteady-state
problem
of
determining
the
2.
A THIN ANNULAR DISC IN THE PLANE
STATE OF STRESS:
Consider a thin annular isotropic disc of
temperature (in heating and cooling process).
Homogeneous boundary conditions of the third kind
thickness h occupying the space D : a
h.
are maintained on curved surfaces of the disc.
The finite Marchi-Zgrablich and Laplace
r
b, 0
z
The differential equation governing the
displacement function U(r, z, t) is
transform techniques are used to find the solutions of
the inverse transient themoelastic problems of a thin
2
U
r2
annular disc.
Keywords Thermoelasticity, Annular Disc, Thermal
Stresses.
problem of determining the temperature (in heating
cooling
process),
displacement
and
functions of the disc occupying the space D: a
0
z
)atT
and at are Poisson's ratio and the linear
coefficient of thermal expansion of the material of the
The inverse unsteady state thermoelastic
and
(1
(2.1)
where
Introduction -
1 U
r r
disc respectively, and T(r, z, t) is the temperature of
the disc satisfying the differential equation
stress
r
2
T
r2
b,
h with the stated boundary conditions.
2
1 T
r r
T
z2
1 T
,
k t
(2.2)
The inverse problem is very important in
view of its relevance to various industrial machines
subjected to heating such as main shaft of the lathe
and turbine and roll of a rolling mill.
subject to the initial condition
T(r,z,0) = 0
(2.3)
the boundary conditions
A related problem of determining the
temperature, displacement and stress functions due to
partially distributed heat supply at z =
a thin annular disc
[2]
(0 <
T r , z, t
< h) in
is reconsidered to study the
T r , z, t
temperature (in heating and cooling process).
Discussion on some similar problems may
be found in

Ozisik
[3]
, Sneddon
k2
r
r
T (r , z , t )
0
(2.4)
r a
T (r , z, t )
0
r b
(2.5)
[8]
, Hetnarski and
Eslami [9].
The corresponding correct expressions are
derived in the present paper.
k1
Also, the numerical
z
T (r , z, t )
0
z 0
(2.6) and the interior condition
results are obtained and presented graphically.
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International Conference on Global Trends in Engineering, Technology and Management (ICGTETM-2016)
Taking Laplace transform of Eqns. (2.10), (2.12) and
z
(2.7)
T (r , z , t )
f (r , t )
using the condition (2.11) gives
z
2
where k is the diffusivity and k1, k2 are the
z
radiation constants on the two curved surfaces of the
2
T*
q 2T *
0; q 2
s
k
2
n
(2.14)
disc.
In order to solve the differential equation
(2.2) subject to the conditions (2.3) – (2.7) we define
the finite Marchi-Zgrablich (FMZ) transform of T as
b
T(
n
, z, t )
rT (r , z, t ) S0 (k1 , k2 ,
n
r )dr
a
(2.8)
z
T*
0;
z
z 0
T*
f *(
n
, s)
z
(2.15)
Solving the differential equation (2.14) and using the
conditions (2.15) yields
T *(
n
, z, s)
f *(
n
, s)
with the inversion formula
cosh( qz )
.
q sinh( q )
(2.16)
T(
T (r , z, t )
n
n 1
, z , t ) S 0 (k1 , k 2 ,
Cn
n
r)
In order to find the inverse Laplace transform of
T*
in (2.16) we need to find the inverse Laplace
transform for G(s) given by
(2.9)
where S0 and Cn are given by Eqns. (19) – (21) and
(25) in the Appendix and
n
are the positive roots of
G ( s)
the transcendental equation (22). Taking the FMZ
transform of Eqns. (2.2)-(2.3), (2.6)-(2.7) and using
cosh(qz )
; q2
q sinh(q )
2
n
s
k
(2.17)
the operational property (18) with p = 0 in the
Appendix and conditions (2.4) –(2.5) we get
2
T
z2
Inverse Laplace transform of G(s)= sum of residues
1 T
k t
2
n
T
According to the method described in [5] we have
est
of
G(s)
at
the
poles
(2.18)
(2.10)
T(
n , z ,0 )
The poles and residues of estG(s) are given by
0
(i) q
(2.11)
T
z
T
z
0;
z 0
f(
2
mn
n ,t)
im
2
n
or
2
m
;
s
2
n
k(
m
m
2
m
)
, m 1,2,...
z
(2.19)
(2.12)
Now we introduce the Laplace transform with respect
to the variable t and define
T *(
e stT (
n , z, s)
(2.20)
n , z , t ) dt.
(ii)
q
0 or
s
k
2
n
0
(2.13)
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(2.21)
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International Conference on Global Trends in Engineering, Technology and Management (ICGTETM-2016)
where
m
and
mn
are given by (2.19) and
n
are the
Now we can find the inverse Laplace transform of
positive roots of the transcendental equation (22) in
G(s) by using (2.18) – (2.21). Then, using (2.16) –
the Appendix.
*
(2.17) the inverse Laplace transform of T (
n
, z, s)
is given by
The mathematical formulation of the problem
described in this section has been studied earlier by
[2]
Pidadi and Khobragade
(2.23)
where we have made use of the convolution theorem
of Laplace transform. Thus the temperature T(r,z,t)
satisfying the conditions (2.2) – (2.7) is given by (2.9),
.
The first term not
involving the infinite series in (3.1) is missing from
the corresponding result of
[2]
. In the second term also
there is an error of a negative sign which may be
corrected by replacing f(r, t) by –f(r, t).
(2.22) – (2.24) and (2.19).
COOLING PROCESS
The temperature change T0(r, z, t) for the
3. THIN ANNULAR DISC SUBJECTED TO
cooling process satisfies the equation
PARTIALLY DISTRIBUTED HEAT SUPPLY
2
T0
r2
In this section an analysis is carried out for
the inverse unsteady state thermoelastic problem of
determining the temperature (in heating and cooling
(3.2)
process), displacement and stress functions of the disc
the boundary conditions
occupying the space D : a
r
b, 0
z
2
1 T0
r r
T0
z2
1 T0
k t
h with the
stated boundary conditions.
T0
HEATING PROCESS
k1
T0
r
0;
T0
k2
r a
T0
r
0
r b
(3.3)
Consider a thin annular disc occupying the
space D. For time t = 0 to t = t0 the disc is subjected
to a partially distributed and axisymmetric heat supply
f(r,t) from the interior point (r, ). After that, the heat
supply is removed and the disc is cooled by the
surrounding medium.
The heating temperature T(r,z,t) of the disc at
T0
z
T0
z
0;
z 0
0
z h
(3.4)
and the continuity condition
T0 (r , z , t0 ) T (r , z , t0 )
(3.5)
time t satisfies the equations (2.2) – (2.7) with u(r,t)
where T(r,z,t) is the heating temperature of the disc at
0. The solution of these equations is given by (2.9)
time t given by (2.9) and (3.1).
Applying FMZ transform with respect to r
together with
in Eqns. (3.2), (3.4), (3.5) and using the operational
property (18) with p = 0 in the Appendix together
with the conditions (3.3) we get
(3.1)
2
T0
z2
2
n 0
T
1 T0
k t
(3.6)
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International Conference on Global Trends in Engineering, Technology and Management (ICGTETM-2016)
h
T0
z
T0
z
0;
z 0
( 1)
(
cos(p z ) cos( m z )dz
0
0
m
2
m
sin(h
p2 )
m
)
.
z h
(3.14)
(3.7)
In view of (3.13) –(3.14), the finite cosine transform
T0 (
n
, z , t0 ) T (
n
of (3.1) is given by
, z , t0 )
(3.8)
(3.15)
With respect to the variable z we define the finite
cosine transform as
ˆ
T0 ( n , p  , t )
Thus
h
T0 ( n , z, t ) cos(p  z)dz ; p 
0
Tˆ0
n
, p , t is given by (3.12) and (3.15).

Inverting the finite cosine transform and noting the
h
remark after Eqn. (3.13) we find
(3.9)
Taking the finite cosine transform of (3.6), (3.8) and
using (3.7) gives
t
k p2
2
n
Tˆ0
Tˆ (
n , p , t0
can be
determined by means of the conditions (3.6) and (3.8).
Indeed, from (3.6) and (3.16) we have
(3.10)
F
t
n , p , t0 ).
k
2
n
F
0
(3.17)
(3.11)
Solving the differential equation (3.10) and using the
n , p , t
Tˆ
n , p , t0
e
k p 2
2
n
t t0
.
(3.12)
In order to find
Before applying (3.8) we note the Fourier series
expansion
condition (3.11) we get
Tˆ0
(3.16)
where the unknown function F( n,t)
Tˆ0
t
Tˆ0
t0
Tˆ
n
m
( 1) m cos( mx)
2
m2
1
1
2
2
cos( x)`
,
2 sin( )
(3.18)
, p , t0 we calculate which yields
the finite cosine transform of the expression in (3.1).
First, we note that

( 1) cos( p z )
( m2 p2 )
1
h cos( m z )
,
sin(
h
)
m
m
1 1
2 m2
h
cos(p  z )dz
0,
(3.19)
0
where
(3.13)
which shows that the finite cosine transform of a term
m
and
p are given by (2.19) and (3.9). Now
using (3.8), (3.1), (3.16) and (3.19) we get
0
independent of z is zero. Therefore, while taking the
T0 (
n
, z , t0 ) T (
n
, z , t0 )
inverse finite cosine transform, a term independent of
z can be added. Also note that
(3.20)
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z
h
International Conference on Global Trends in Engineering, Technology and Management (ICGTETM-2016)
Then
F(
n
parameters are t0 = 0.05, z = 1, a = 0.5, b = 1.0, k
, t ) satisfying (3.17) and (3.20) is given
= 0.38, k1=k2= 0.25, = 0.7501, h = 1 and r = 0.5,
by
0.7, 0.85, 0.95, 1.0.
t
4.
t0
SPECIAL
CASE
OF
PARTIALLY
special
case
DISTRIBUTED
(3.21)
HEAT SUPPLY
The temperature T0(r, z, t) for the cooling process is
Consider
then given by (3.16), (3.21) and the inversion formula
a
wherein
the
prescribed function f(r,t) in (2.7) is given by
S0 (k1 , k 2 ,
T0 (r , z , t )
n 1
n r ) T0 (
Cn
n , z, t )
,
f (r , t ) (1 e t )( h
(3.22)
where S0 and Cn are given by (19) – (21) and (25) in
the Appendix. It may be pointed out that (3.16)
p )
m
that
1
and therefore
Following the simplifications, we find that T(r, z, t)
and h must be such
p for any values of m and  . For (4.2)
m
example, if h = 1 and
m
(4.1)
given by (2.9) and (3.1) reduces to
contains a factor
(
) (r 0.9)
p
= ¾ and then
where
 which vanishes for every
(4 m / 3)
choice of k with m = 3k and
n,
Bmn and Dn are given by
(h
n
)( 0.9) S0 (k1 , k2 ,0.9
n
)
2
 = 4k.
k (1 e t ) (e k mn t 1) /
2
(k mn
1)
Bmn (t )
2
k (1 e t ) (e k mn t 1) /
(k n2 1)
Dn (t )
These
n,
2
mn
2
n
; and
;
Bmn and Dn are given in special case of [1].
Similarly T0 in (3.16) reduces to
(4.3) and F in (3.21) simplifies to
(4.4)
FIG.1: Variation of T
T
T (r , z, t ) (0 t t 0 ) together with
T0 (r , z , t ) (t
t0 )
where T(r, z, t) is
(3.22), (4.3) – (4.4) , where
as
given by (5.2) and T0(r, z, t) is given by (4.22),
(5.3), (5.4).For clarity in the figure, the scales on
the two parts of the t-axis demarcated by
t
t0
are taken to be different. The values of other
ISSN: 2231-5381
Thus the cooling temperature T0(r, z, t) is given by
[1]
n,
Bmn and Dn are given
.Using the series (3.19) it is easily verified that
the expression for T and T0 given by (3.22), (4.3) –
(4.4) and (4.2) satisfy the continuity condition (3.5).
Taking
t0 = 0.05 the variation of heating
temperature T(r, z, t) (0
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t
t0) together with the
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cooling temperature T0(r, z, t)
(t
t0) with t is
The pair of Eqns.(4) – (5) has a nontrivial solution
shown in Fig. 1. The values of other parameters are
only if
taken to be z = 1, a = 0.5, b = 1.0, k = 0.38, k 1=k2 =
J p (k1 , a )Yp (k2 , b)
J p (k2 , b)Yp (k1 , a )
0
0.25, = 0.7501, h = 1 and the curves are drawn for r
(8)
= 0.5, 0.7, 0.85, 0.95, 1.0.
Let
be the nth positive root of the transcendental
n
APPENDIX
equation (8). Using (4) – (5) we can write the solution
FINITE MARCHI-ZGRABLICH TRANSFORM
(3) in the following two forms
Let the Bessel differential equation of order p be given
by
x2 y
(x2
xy
2
p2 ) y
A
J p ( n x)Yp (k1 , n a) Yp ( n x) J p (k1 , n a) ,
Yp (k1 , n a)
yn ( x )
(9)
0
and
(1)
and let the boundary conditions be given by
y(a) k1 y (a) 0 ; y(b) k2 y (b) 0
(2)
yn ( x )
A
J p ( n x)Yp (k2 , nb) Yp ( n x) J p (k2 , nb) .
Y p ( k 2 , n b)
(10)
Define a function S p as follows:
The general solution of Eqn.(1) is
S p ( k1 , k 2 ,
y ( x)
A J p ( x)
B Yp ( x)
n
x)
An J p (
n
x)
Bn Y p (
n
x) ,
(11)
(3)
where Jp( x) and Yp( x) are Bessel functions of the
An Yp (k1 , n a) Yp (k2 , nb) ; Bn
J p (k1 , n a ) J p (k2 , nb).
(12)
first and the second kind, respectively, of order p.
Then the boundary conditions (2) yield
In view of (9) – (10) the function Sp is a solution of
Bessel’s differential equation (1) of order p and
A J p (k1 , a )
B Yp (k1 , a )
0
B Yp ( k 2 , b)
0
satisfies the boundary conditions (2). Because of this
(4)
A J p ( k 2 , b)
fact, the sequence of functions S p ( k1 , k 2 ,
n
x)
n 1
is orthogonal in the interval (a, b). Finite Marchi(5)
Zgrablich (FMZ) integral transform is defined as (see
where, for i =1,2, we have
[1], [2], [4], [7]
)
b
J p ( ki , x )
J p ( x) ki J p ( x),
f(
xf ( x) S p (k1 , k2 ,
n
x)dx
a
(6)
Y p ( ki , x )
n)
Yp ( x )
ki Yp ( x).
(13)
and its inversion formula is given by
(7)
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f(
f ( x)
n
) S p (k1 , k2 ,
n
x)
S0 (k1 , k2 ,
Cn
n 1
n
r)
An J 0 ( n r ) Bn Y0 ( n r ),
(19)
(14)
where the norm Cn may be written
An
Y0 ( n a) k1 nY1 ( n a) Y0 ( nb) k2 nY1 ( nb),
b
x S p2 (k1 , k2 ,
Cn
n
x)dx
(20)
a
(15)
The value of the integral Cn is given by (see Sneddon
Bn
J 0 ( n a) k1 n J1 ( n a) J 0 ( nb) k2
(21)
[8]
,
where
[ Prob. 8 – 14])
Cn
S p (k1 , k2 ,
n
x)
2
1
p2
2
n
x2
S p2 k1 , k2 ,
n
n
is the nth positive root of the transcendental
equation
b
x2
2
J ( nb),
n 1
x
a
J0 ( na
k1
J ( n a) Y0 ( nb) k2 nY1 ( nb)
n 1
(16)
Since S p satisfies the boundary conditions (2) we can
J 0 ( nb
k2
J ( nb) Y0 ( na) k1 nY1 ( na)
n 1
write
(22)
For a = 0.5, b = 1.0 and k1 = k2 = 0.25 the first 20
roots are listed in the Table. With p = 0 the FMZ
(17)
transform may be rewritten
Thus the FMZ integral transform and its inverse are
b
given by (13) – (14) and (17). Also, if we carry out
f(
n
)
rf (r ) S 0 (k1 , k 2 ,
integration by parts twice we can prove the
operational property
n
r )dr
a
(23)
together with the inversion formula
b
2
f 1 f p2
b
f
x 2
f S p (k1, k2 , n x)dx S p (k1, k2 , nb) f k2
2
x x x x
k2
x
a
xb
f(
f (r )
n
n 1
-
a
S p ( k1 , k 2 ,
k1
n
a) f
k1
f
x
2
n
f(
n
).
Cn
SPECIAL CASE p = 0
In our analysis we need the results only for
b2
1
1
2
2
k2
a2
1
1
2
2
k1
the special case p = 0. For this special case we have
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n
r)
(24)
where
x a
(18)
) S0 (k1 , k2 ,
Cn
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2
n
2
n
S 02 (k1 , k 2 ,
S 02 (k1 , k 2 ,
n
n
b)
a)
(25)
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International Conference on Global Trends in Engineering, Technology and Management (ICGTETM-2016)
Finally, the operational property is given by (18) with
p = 0 where f (
n
) is now given by (23).
REFERENCES
[1]
Khobragade N.W. and Durge M.H.,( 2006) An inverse
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10.781796
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62.891112
14.206770
31.532288
48.945081
66.378691
17.654071
35.011765
52.430780
69.866573
unsteady-state thermoelastic problem of a thin annular
disc in Marchi-Zgrablich transform domain.
Bull. Cal. Math. Soc., vol. 98, pp. 255-266.
[2]
Pidadi Sonali S. and Khobragade N.W.,( 2008.) Study
of stress functions in a thin annular disc due to partially
distributed heat supply.
Bull. Cal. Math. Soc. Vol. 100, pp. 691 – 698.

Ozisik
[3]
M.N,( 1968 ) Boundary Value Problems of
Heat
Conduction.
International
Text-Book
Company
,
Stranton,
Pennsylvania.
[4]
Pidadi Sonali S.
and
Khobragade N.W.,( 2008)
Numerical study of inverse transient thermoelastic
problem of a finite length hollow
cylinder.
Bull. Cal. Math. Soc., vol. 100, pp. 599 – 608.
[5]
Churchill R.V. and Brown J.W.,( 1990.) Complex
Variables and Applications.
5th Ed. McGraw Hill, New York.
[6]
Nowacki W., Thermoelasticity, Addison-Wesley Publ.
Co. Inc., 1962.
[7]
Marchi E. and Zgrablich G., (1964) Heat conduction in
hollow cylinders with radiation.
Proc. Edinb. Math.
Soc., vol. 14, pp. 159 – 164.
[8]
Sneddon I.N., ( 1972.) The Use of Integral Transforms.
McGraw Hill, New York.
[9]
Hetnarski R.B.and Eslami M.R.,( 2009)
Thermal
StressesAdvanced Theory and Applications , Springer ,New York.
TABLE CAPTION
TABLE:
First positive 20 roots of equation (22) for
a = 0.5, b = 1.0, k1= k2 = 0.25.
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