106738918
5.1 Unsteady-State Temperature in a Sphere
y
r
0
a
x
Figure 5.1-1 A sphere with temperature w(r,t) dependent on r and t only
The unsteady-state temperature distribution in the solid sphere shown in Figure 5.1-1 is
described by the one-dimensional heat transfer equation in spherical coordinate
w
   2 w 
= 2
r

t
r r  r 
0 < r < a, t > 0
(5.1-1)
Where  is the thermal diffusivity with SI units of [m2/s]. The initial condition is
w(r,0) = wi
The boundary conditions are
w
= 0, at r = 0 for t  0
r
k
w
= h(w  wf) at r = a for t  0
r
The boundary condition at r = 0 indicates that the temperature is symmetric with respect to the
center of the sphere. The boundary condition at r = a is a surface energy balance that requires the
energy transfer to the surface by conduction must be equal to the energy leaving the surface by
convection neglecting radiation. k is the thermal conductivity of the solid [W/moK], h is the heat
transfer coefficient for the convection between the solid surface and the surrounding fluid, and wf
is the temperature of the fluid. It is convenient to introduce the following dimensionless variables
u=
w  wf
wi  w f
 w = u(wi – wf) + wf
=
t
a2


t
=

a2

=
r
 r = a
a
103
106738918
In terms of the dimensionless variables, Eq. (5.1-1) becomes
(wi – wf)
 u
 
u 

= (wi – wf) 2 2
 ( a ) 2

2
a 
( a ) 
a   ( a ) 
u
1   2 u 
= 2



    
(5.2-1)
The initial and boundary conditions become
u(,0) = 1
u
= 0, at  = 0 for   0

 k(w  wf)
u
= h(w  wf)u at r = a for   0
(a )
ha
u
=
u =  Bi u at  = 1 for   0
k

ha
is the ratio of the resistance to heat transfer within the sphere to
k
the resistance to heat transfer by convection outside the sphere.
where the Biot number Bi =
We assume that u(,) can be separated into R(), a function of  alone, and T(), a function of 
alone.
u(,) = R()T()
Differentiating the partial derivative, we get
u
dT
=R
and

d
u
dR
=T

d
In terms of R() and T(), Eq. (5.1-2) becomes
R
dT
T d  2 dR 
= 2


d
 d  d 
(5.1-3)
Divide Eq. (5.1-3) by RT to obtain
1 dT
1 d  2 dR 
=
 
 =  2
2
 d
R d  d 
104
(5.1-4)
106738918
Since the RHS of Eq. (5.1-4) depends on  only, and the LHS depends on  only, they must
equal to a constant  2. The constant must be chosen so that the temperature of the solid is the
same as the temperature of the liquid when time approaches infinity. The ODE on R is
2
1 
dR
2 d R

 =   2R
2



 2  d
d 2 
d 2 R 2 dR
+
=   2R
2
 d
d
(5.1-5)
The above equation can be transformed into a familiar form by a change of variable
 = R  R = --1
dR d --1
=
 - --2
d d
d
d 2R
d 2 --1 --2 d
d 2 --1
--2 d
--3
=



+
2

=
 - 2--2
+ 2--3
2
2
2
d
d
d
d
d
d
In terms of the new variable, Eq. (5.1-5) becomes
--1
d
d
d 2
- 2--2
+ 2--3 + 2--2
- 2--3 =  2--1
2
d
d
d
d 2
=  2
d 2
The solution for R is
 = R = Acos() + Bsin()  R =
1
[ Acos() + Bsin()]

The constant A can be determined from the boundary condition at  = 0.
At  = 0, 0 = A
R=
1
Bsin()

The separation constant  can be determined from the boundary condition at  = 1.
105
106738918
At  = 1,
R=
dR
u
=  Bi u 
=  Bi R

d
1
dR 1
1
Bsin() 
= Bcos() - 2 Bsin()

d


dR
= Bcos() - Bsin() =  BiBsin()
d
At  = 1,
The separation constants n are the roots of
ncos(n) - sin(n) =  Bisin(n)
or
n
cos( n )
+ Bi – 1 = 0
sin( n )
From the ODE for T
1 dT
=  2  T = exp  2n 
 d
The solution for the dimensionless temperature u is
u(,) =


Bn

n 1
sin(n) exp  2n 
The constants Bn are determined from the initial condition
u(,0) = 1 =


Bn
n 1

sin(n)   =


Bn sin(n)
n 1
Multiplying the initial condition by sin(m) and integrating the expression with respect to 
yields
1
  sin(m)d =
0


n 1
1
Bn  sin( n  ) sin(m)d
0
For m  n
1
 sin( 
0
m
x) sin(m)d = [msin(n)cos(m)  ncos(n)sin(m)]/(  2n  2m )
n and m are solutions of n
cos( n )
+ Bi – 1 = 0. Therefore
sin( n )
106
106738918
ncos(n) = (1 – Bi) sin(n) and mcos(m) = (1 – Bi) sin(m)
msin(n)cos(m)  ncos(n)sin(m) = (1 – Bi) sin(n)sin(m) - (1 – Bi) sin(n) sin(m) = 0
1
 sin( 
m
0
x) sin(m)d = 0 for m  n
For m = n
1
 sin
0
2
(n  )d =
1

0
sin( 2n )
1  cos( 2n  )
1
d = [1 –
]
2
2
2n
1
 cos( n  )
+
0 sin( n  )d = – n
0
1
1

0
cos( n  )
n
sin( n )
Bn =
1
1
 sin
2
0
Bn =
1
(n  )d
  sin(   )d
0
n
4[sin( n )  n cos( n )]
n [2n  sin( 2n )]
=

d = –

cos( n )
n
+
sin( n )
2n
cos( n )
n
1  sin( 2n ) 
1
2 
2n 
2
n
In summary, the dimensionless temperature for heat conduction in a solid sphere with convective
boundary conditions and initial constant temperate is given by
u(,) =


n 1
Bn

sin(n) exp  2n , 0    1,  > 0
(5.1-6a)
where n are the roots of
n
cos( n )
+ Bi – 1 = 0
sin( n )
(5.1-6b)
and Bn are given by
Bn =
4[sin( n )  n cos( n )]
n [2n  sin( 2n )]
(5.1-6c)
For a numerical example, let the Biot number Bi = 5. Table 5.1-1 lists the Matlab program used
to plot the dimensionless temperature at the dimensionless time  = 0, .1, .2, .3, .4, .5. The first
ten roots of Eq. (5.1-6b) are computed using Newton’s method. The coefficients Bn are evaluated
next. The temperatures at different times are stored in a matrix with each column represents the
temperature distribution at a given time. At r = 0, the temperature is evaluated using the formula
107
106738918
u(0,) =


Bn n exp  2n , 0    1,  > 0
(5.1-6d)
n 1
Eq. (5.1-6d) is obtained in the limit   0 by using L’Hopital rule.
lim
 0
sin( n  )

=
lim
 0
n cos( n  )
1
= n
Figure 5.1-2 shows the dimensionless temperature distribution at various times. It is clear from
the graph that as time approaches zero, more terms are needed for the partial sum to provide a
smooth distribution.
__________ Table 5.1-1 Matlab program to plot u(,) ___________
Biot=[0 .01 .1 .2 .5 1 2 5 10 inf]';
alfa=[0 .1730 .5423 .7593 1.1656 1.5708 2.0288 2.5704 2.8363 3.1415]';
Bi=input('Bi = ');
z=interp1(Biot,alfa,Bi);
%
lamda=zeros(1,10);
disp('Roots of z*cot(z)+Bi-1 = 0')
for i=1:10
e=1;
while abs(e)>.00001
y=tan(z);
e=(z+(Bi-1)*y)/(1-z*y-z/y);
z=z-e;
end
fprintf('z%g = %g\n',i,z)
lamda(i)=z;
z=z+pi;
end
%
% Evaluate the coefficients for the series
%
Bn=4*(sin(lamda)-lamda.*cos(lamda))./(lamda.*(2*lamda-sin(2*lamda)));
r=[1:25]/25;rp=[0 r];row=length(rp);
t=[0 .1 .2 .3 .4 .5];
col=length(t);u=zeros(row,col);
%
% At r=0, different expression is needed for the temperature calculation
%
for n=1:10
Bnr1=Bn(n)*sin(lamda(n)*r)./r;Bnr0=Bn(n)*lamda(n);Bnr=[Bnr0 Bnr1];
u=u+Bnr'*exp(-lamda(n)^2*t);
end
ax='t=.0t=.1t=.2t=.3t=.4t=.5';
subplot
for i=1:6
ib=1+(i-1)*4;ie=ib+3;
axi=ax(ib:ie);
subplot(3,2,i),plot(rp,u(:,i));axis([0 1 0 1.1]);
108
106738918
xlabel(axi);ylabel('u(r,t)')
end
Bi = 5
Roots of z*cot(z)+Bi-1 = 0
z1 = 2.57043
z2 = 5.35403
z3 = 8.30293
z4 = 11.3348
z5 = 14.408
z6 = 17.5034
z7 = 20.612
z8 = 23.7289
z9 = 26.8514
z10 = 29.9778
1
u(r,t)
u(r,t)
1
0.5
0
0
0.2
0.4
0.6
0.8
0.5
0
1
0
0.2
0.4
t=.0
0.5
0
0.2
0.4
0.6
0.8
0
1
0
0.2
0.4
0.6
0.8
1
0.6
0.8
1
t=.3
1
u(r,t)
1
u(r,t)
1
0.5
t=.2
0.5
0
0.8
1
u(r,t)
u(r,t)
1
0
0.6
t=.1
0
0.2
0.4
0.6
0.8
0.5
0
1
t=.4
0
0.2
0.4
t=.5
Figure 5.1-2 Dimensionless temperature distributions in a solid sphere at various times for Bi = 5
109