Chapter5-3e
Example 5.3-1 Two concentric spherical metallic shells of radii a and b are separated by a solid thermal insulator as shown in Figure 5.3-1.
At r = a, u(a,q) = u1
At r = b, u(b,q) = u2 + u3cosq + u4cos2q
The rate at which solid conducts heat away from the
sphere of radius a is given by
Q = -ka2 sinq dq Figure 5.3-1. Concentric spherical shells
where k is the thermal conductivity of the insulator. How should u1, u2, u3, and u4 be plotted against heat input Q so that k can be determined?
Solution
The Laplace’s equation with axial symmetry in spherical coordinate has the following form
= + = 0 (5.3-7)
We assume that u(r,q) can be separated into R(r), a function of r alone, and T(q), a function of q alone.
u(r,q) = R(r) T(q) (5.3-8)
Differentiating the partial derivative, we get
= R Þ = R
= T Þ = T
In terms of R(r) and T(q), Eq. (5.3-7) becomes
T+ + R= 0 (5.3-9)
Multiply Eq. (5.3-9) by to obtain
+ = - = m = n(n + 1) (5.3-10)
Since the RHS of Eq. (5.3-10) depends on r only, and the LHS depends on q only, they must equal to a constant m. The ODE on r is
+ - n(n + 1) = 0
The equation for R is an Euler equation that has the following solution
R = ra
The constant a can be determined by substituting R = ra and its derivatives into the original ODE to obtain
a(a - 1)ra + 2ara - n(n + 1)ra = 0
a2 + a - n(n + 1) = 0 Þ a = n, and a = - (n + 1)
The solution for R is
R(r) = Arn + Br-(n+1)
From Eq. (5.3-10), the ODE on q is
= - n(n + 1) (5.3-11)
Let s = cos q Þ = = - sin q
= = - cos q- sin q
= - cos q- sin q = - cos q+ sin2 q
Substituting the new independent variable s and the derivative with respect to s yields
- cos q+ sin2 q- cos q+ n(n + 1)T = 0
(1 - s2) - 2s+ n(n + 1)T = 0
The ODE for T is Legendre’s differential equation of order n so that the solution is given as
Tn = Cn Pn(s) + Dn Qn(s)
At q = 0, s = cos q = 1. Since Qn(1) = ¥, Dn must be equal to zero for a finite solution
u(r,q) = (Anrn + Bnr-(n+1)) Pn(cos)
At r = a, u(a,q) = u1
At r = b, u(b,q) = u2 + u3cosq + u4cos2q
P0(x) = 1 P1(x) = x
P2(x) = (3x2 - 1) Þ x2 = (2 P2(x) + 1)
Expressing the boundary condition at r = b in terms of Legendre polynomials yields
u2 + u3cosq + u4cos2q = u2P0(cosq) + u3P1(cosq) + u4(2 P2(cosq) + 1)
Collecting the terms with the same degree of Legendre polynomials yields
u2 + u3cosq + u4cos2q =( u2 + u4)P0(cosq) + u3P1(cosq) + u4P2(cosq)
The temperature gradient can be evaluated from the temperature distribution to give
= [nAnrn-1 - (n - 1)Bnr-n-2] Pn(cos)
= [nAnan-1 - (n - 1)Bna-n-2]Pn(cos)
From the expression for the heat transfer rate Q = -ka2 sinq dq, the constants An and Bn may be determined
Q = -k2p a2 [nAnan-1 - (n - 1)Bna-n-2] Pn(cos)sinq dq
Let x = cos q Þ dx = - sin q dq
q = 0 Þ x = cos q = 1
q = p Þ x = cos q = - 1
Q = -2p k a2 [nAnan-1 - (n - 1)Bna-n-2] dx
Since P0(x) = 1 Þ dx = 0 for n ¹ 0
dx = 2
The expression for the heat rate Q is then
Q = -2p k a2[ - B0a-2](2) = 4p kB0
The constant B0 must now be find in terms of u1, u2, u3, and u4.
At r = a, u(a,q) = u1 = (Anan + Bna-(n+1)) Pn(cosq)
u1 P0(cosq) = (A0 + B0a-1) P0(cosq) (5.3-12)
At r = b, u(b,q) = u2 + u3cosq + u4cos2q
u(b,q) = ( u2 + u4)P0(cosq) + u3P1(cosq) + u4P2(cosq) (5.3-13a)
u(b,q) = (A0 + B0b-1) P0(cosq) + (A1b + B1b-2) P1(cosq) + ... (5.3-13b)
From Eq. (5.3-12)
u1 = A0 + B0a-1 (5.3-14a)
From Eqs. (5.3-13a) and (5.3-13b)
u2 + u4 = A0 + B0b-1 (5.3-14b)
Substracting Eq. (5.3-14a) by (5.3-14b)
u1 - ( u2 + u4) = B0(a-1 - b-1)
B0 = =
The heat rate Q in terms of all the known parameters is
Q = 4p kB0 =
The thermal conductivity k can be obtained from plotting Q versus . The slope of the resulting straight line is that can be used to determined k. For a numerical example, experimental data are given in Table 5.3-1 for the heat transfer rate at various temperatures. Two concentric spherical metallic shells of radii a = 0.15 m and b = 0.20 m are separated by a solid thermal insulator. Use these data to determine the thermal conductivity k.
______Table 5.3-1 Experimental values of Q and temperatures ______
Q(W/s) / u1(oC) / u2(oC) / u4(oC)630
605
576
553
528
503
475
450
429
404
377 / 200
210
220
230
240
250
260
270
280
290
300 / 100
110
120
130
140
150
160
170
180
190
200 / 50
60
70
80
90
100
110
120
130
140
150
The Matlab program used to plot Q versus is listed in Table 5.3-2 and the graph is shown in Figure 5.3-2.
______Table 5.3-2 Matlab program to evaluate k for example 5.3-1 ______
% Least square curve fit to determine the slope and k
%
u1=200:10:300;u2=u1-100;u4=u2-50;
a=.15;b=0.2;
Q=[630 605 576 553 528 503 475 450 429 404 377];
u=u1-u2-u4/3;
c=polyfit(u,Q,1);
slope=c(1);
k=slope*(b-a)/(4*pi*a*b);
fprintf('k(W/m.C) = %5.2f \n',k)
x=[u(1) u(11)];
y=c(1)*x+c(2);
plot(x,y,u,Q,'o')
xlabel('u1 - u2 -u4/3');ylabel('Q')
Figure 5.3-2 Experimental and fitted value for Q versus
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