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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/moK], 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( 2n ) 1 cos( 2n ) 1 d = [1 – ] 2 2 2n 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 [2n sin( 2n )] = d = – cos( n ) n + sin( n ) 2n cos( n ) n 1 sin( 2n ) 1 2 2n 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 [2n sin( 2n )] (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