ME201/MTH281/ME400/CHE400 Newton's Law of Cooling

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1 ME1/MTH281/ME0/CHE0 Newton's Law of Cooling 1. Introduction This notebook uses Mathematica to solve the problem presented in class, in section 5.1 of the notes. The problem is one of transient heat conduction in a slab of finite width L. The boundary condition on the left face of the slab is zero heat flux, and the condition on the right face of the slab is Newton's law of cooling, with an ambient temperature T a. The initial temperature is the function T 0 (x). The relevant material properties are the thermal conductivity k, the thermal diffusivity D f, and the heat transfer coefficient h, all assumed constant. The specific expressions for all functions and parameters are given below in section 2. The mathematical formulation of the problem is given below. T t = D f 2 T, 0 < x < L, t > 0, 2 x with T T H0, tl = 0, k HL, tl + h@t HL, tl - TaD = 0, x x (1) and T Hx, 0L = T 0 HxL. 2. Parameter Values In this section, we define the parameter values and the initial function. This is the only place in the notebook where these quantities are given specifically. This makes it possible to change a value for the entire calculation by just changing it here. The material properties are appropriate for granite. The initial temperature is taken here to be a constant. In[1]:= In[2]:= In[3]:= In[4]:= In[5]:= In[6]:= h = 22.4; H** Wëm 2 ÿk **L L = 0.5; H** m **L k = 2.80; H** WêmÿK **L Df = 1.37 * -6 ;H** m 2 ës **L TA =.0; H** C **L T0@x_D =.0; H** C **L 3. Steady State Solution As in other problems of this sort, we must first find the steady-state solution T s (x), and then reformulate the problem in terms of the transient solution T r. The steady-state solution is obtained by setting the time derivative to zero in the original equation for T. The solution of the resulting equation is a linear function of x, and we require that particular linear function which satisfies the boundary conditions at x = 0 and L. The result (which is intuitively obvious) is that T s HxL is constant and equal to the ambient temperature:

2 2 newtheat.nb In[7]:= := TA 4. Formulation of Problem for Transient Now that we have the steady-state solution, we decompose the full solution into a steady-state part T s (x) and a transient part T r (x,t): T Hx, tl = T s HxL + T r Hx, tl. By substituting this decomposition into the original equation for T, we find the following problem for the transient: (2) T r t 2 T r = D f, 0 < x < L, t > 0, 2 x with T r x H0, tl = 0, k T r x HL, tl + h T r HL, tl = 0, (3) and T r Hx, 0L = T 0 HxL - T s HxL. This is the problem that we solve by separation of variables. 5. Separation of Variables in Transient Problem Following the standard approach in separation of variables, we seek functions which satisfy the equation for T r and the homogeneous boundary conditions at x = 0, L. We assume a form F(x)G(t) for such solutions. The two separated equations then have the form F ² HxL + lf HxL = 0, with F H0L = 0 and kf HLL + hf HLL = 0, and G HtL + ldg HtL = 0. The general solution for the equation for F(x) is a linear combination of sin( l x) and cos( l x). Applying the two homogeneous boundary conditions to the general solution, we get F(x) = cos( resulting eigenvalue equation, we get l x), and simplifying the (4) tan HzL = B i êz, where z = L l, and where B i = hlêk. (5) In[8]:= We define the relevant quantities for Mathematica, denoting the nth value of z by z[n], and the nth eigenvalue by lam[n]. lam@n_d := Hz ê LL^2 ê. z -> z@nd In[9]:= where the values z[1], z[2],.. z[n],.. are the roots of eig1 = eig2, where eig1 = Tan@zD; In[]:= eig2 = Bi ê z; In[11]:= In[12]:= In[13]:= with Bi := Hh * LL ê k The generic eigenfunction is geneig@x_d := Cos@Hz * xl ê LD and the nth eigenfunction is given by Feig@x_, n_d := geneig@xd ê. z -> z@nd

3 newtheat.nb 3 The generic normalization integral is In[14]:= Out[14]= In[15]:= nor = Integrate@Hgeneig@xDL ^ 2, 8x, 0, L<D Sin@2. zd z and the nth normalization integral is norm@n_d := nor ê. z -> z@nd 6. Determination of Eigenvalues We now determine numerically the eigenvalues. We begin by plotting the expressions eig1 and eig2. At each crossing of these, there is an eigenvalue. In[16]:= SetOption@Plot, ImageSize Ø 2D; In[17]:= Plot@8eig1, eig2<, 8z, 0, <, PlotRange -> 8-, <, PlotStyle -> 8RGBColor@1, 0, 0D, RGBColor@0, 0, 1D<, AxesLabel -> 8"z", "eig1 HredL and eig2 HblueL"<, Ticks Ø 880, Pi ê 2, Pi, 3 Pi ê 2, 2 Pi, 5 Pi ê 2, 3 Pi<<D eig1 HredL and eig2 HblueL 5 Out[17]= p 2 p 3 p 2 2 p 5 p 2 3 p z -5 - We see that the first root is between 0 and p/2, and each subsequent root is in an interval of length p. It is also clear that for large n, the nth root is very close to (n-1)p. To find accurate numerical values of the z's, we use the Mathematica command FindRoot. FindRoot requires an initial guess. The nth value of z, as we can see from the graph, is between (n - 1) and (n -0.5 )p. We take as an initial guess (n - 0.5)p The command is then of the form Find- Root[eig1 == eig2,{z, (n - 0.5)p -0.1}]. We now use this command in a Do loop to calculate and display in a table the first roots. This will suffice for any calculation requiring up to terms in the eigenfunction series. We also display the eigenvalue l, and the approximate value z = (n-1)p. In[18]:= In[19]:= In[]:= zroot@n_d := FindRoot@eig1 == eig2, 8z, Hn - 0.5L * Pi - 0.1<D zasymp@n_d := N@Hn - 1L * PiD Do@z@nD = z ê. zroot@nd, 8n, 1, <D;

4 4 newtheat.nb In[21]:= 8, 4<D, 8, 4<D, 8, 4<D<, 8n, 1, <D, TableHeadings -> 8None, 8"n", " " " Out[21]//TableForm= n z@nd zasymp@nd lam@nd We see that the asymptotic formula for z[n] has an error of less than 0.5% for any n equal to or greater than. It is possible to develop even more accurate, but still simple, asymptotic formulas. 7. Representation of the Initial Condition We are finally ready to begin solving the boundary value for the transient T r (x,t). The form of the solution for the transient is

5 newtheat.nb 5 In[22]:= In[23]:= In[24]:= T r Hx, tl = CHnL ExpI-lHnL D f tm cos@hz HnL xêld, n=1 where the coefficients C(n) are determined by the initial conditions satisfied by T r. The formula for the nth coefficient is coeff@n_d := Integrate@Feig@x, nd * HT0@xD - Ts@xDL, 8x, 0, L<D ê norm@nd We now use a Do loop to construct the first coefficients, and then print them out in a table. The nth numerical value is assigned to c[n]. Do@c@nD = coeff@nd, 8n, 1, <D; TableForm@Table@8n, PaddedForm@c@nD, 88, 4<D, PaddedForm@n +, D, PaddedForm@c@n + D, 812, 4<D<, 8n, 1, <D, TableHeadings -> 8None, 8"n", " c@nd", " n", " c@nd"<<d Out[24]//TableForm= n c@nd n c@nd In[25]:= As a check on all that we have done, we see how well our series represents the initial condition, by plotting both the initial condition (in blue) and its representation by the first terms of our series (in red). exactinit@x_d := T0@xD - Ts@xD In[26]:= seriesinit@x_d := Sum@c@nD * Feig@x, nd, 8n, 1, <D (6)

6 6 newtheat.nb In[27]:= seriesgraph = Plot@8exactinit@xD, seriesinit@xd<, 8x, 0, L<, PlotRange -> 80, <, PlotStyle -> 88RGBColor@0, 0, 1D, Thickness@0.004D<, 8RGBColor@1, 0, 0D, Thickness@0.004D<<, AxesLabel -> 8"x", "Exact HblueL, Series HredL"<D Exact HblueL, Series HredL Out[27]= The agreement looks excellent, and is good evidence that our calculations are correct so far. To get a little closer view of the accuracy, we replot with a much narrower range in y. In[28]:= Show@seriesgraph, PlotRange -> 848, 52<D Exact HblueL, Series HredL Out[28]= Now we can see a little overshoot at the right end of the interval. This is not the Gibbs phenomenon, however, because the magnitude of the overshoot decreases to zero as the number of terms is increased. We haven't shown that here, but it can be shown. With some algebraic labor, it also can be shown that the coefficients c[n] diminish like 1/n 2, and not like the 1/n that would be associated with the Gibbs overshoot. 8. Solution of the Initial-Boundary Value Problem In[29]:= Now that we have the eigenvalues and series coefficients, we can construct the series solution of the problem. We define Tran[x,t,n] as the nth partial sum of the series solution, and we define term[x,t,n] as the nth term in the series. As a special case of importance, we define firstterm[x,t] to be the first term in the series. term@x_, t_, n_d := c@nd * Feig@x, nd * Exp@-lam@nD * Df * td In[]:= In[31]:= Tran@x_, t_, n_d := Sum@term@x, t, kkd, 8kk, 1, n<d firstterm@x_, t_d := term@x, t, 1D

7 newtheat.nb 7 In[32]:= Out[32]= In[33]:= Let's look at the first term: firstterm@x, td µ-6 t Cos@ xd We may calculate the e-folding time for this mode -- call it t1 -- as the reciprocal of the coefficient of t in the exponent: t1 = 1 ê Hlam@1D * DfL Out[33]= In[34]:= This time, which is in seconds, equals about 31.7 hours. We compare this with the diffusion time L 2 ëip 2 D f Mwe derived earlier for zero boundary conditions on both sides of the slab. In hours, the time is L 2 ë I Ip 2 DfMM Out[34]= It is interesting that this is less than the 31.7 hours we calculated for the first mode in our problem. There are two reasons for this. The first reason is that our slab is insulated on one end, and this will reduce the cooling rate. One can show that the diffusion time for a slab of width L which has zero temperature on one face and which is insulated on the other face is (4 L 2 MëIp 2 D f ). For our present parameters, the value of this in hours is In[35]:= 4 L 2 ë I Ip 2 DfMM Out[35]=.5436 In[36]:= This is much closer to our calculated value of 31.7 hours, but still smaller. There is an another effect in the present problem, and that is an additional thermal resistance at the boundary (characterized by the reciprocal of the heat transfer coefficient h). This extra resistance increases the decay time to the 31.7 hours that we calculated. Let's look at the decay times of the second and third modes, converting them to hours as we calculate them t2 = 1 ê Hlam@2D * Df * L Out[36]= In[37]:= t3 = 1 ê Hlam@3D * Df * L Out[37]= In[38]:= Knowledge of these decay times will help us choose time values for plotting the solution. To make this a little easier, we construct a table of the first decay times. t@n_d := 1 ê Hlam@nD * DfL

8 8 newtheat.nb In[39]:= 8, 2<D, ê, 8, 5<D<, 8n, 1, <D, TableHeadings -> 8None, 8"Mode", "Decay Time HsL", "Decay Time HhrL"<<D Out[39]//TableForm= Mode Decay Time HsL Decay Time HhrL Graphs of the Solution In[]:= We define here graphs of the solution for T versus x, for various values of time. We test the code with a few graphs, and then we construct a sequence to be shown in a Manipulate panel. The function which produces the graph at time t is named snapshot[t]. We use all the terms we have calculated in the series to compute the solution, although this is somewhat extravagent in that fewer terms would be sufficient for the longer times. Trans@x_, t_d := Tran@x, t, D In[41]:= Temp@x_, t_d := Trans@x, td + Ts@xD Now we are ready to define snapshot[t], which gives us a plot of temperature versus x at time t. From the ambient temperature and the initial distribution, we can get a range of variation of temperature. We extend this range slightly on either side, and call the extended range plrange: In[42]:= plrange = 80, <; In[43]:= In[44]:= Now we define snapshot[t], giving a special definition for snapshot[0] -- namely the initial distribution. For convenience, the function snapshot takes an argument in hours, but converts it internally to seconds, to be consistent with the units of the problem. snapshot@t_d := Module@8time<, time = * t; Plot@Temp@x, timed, 8x, 0, L<, PlotStyle Ø Thickness@0.004D, PlotRange -> plrange, AxesLabel -> 8"x", ""<, PlotLabel Ø Row@8"t = ", PaddedForm@t, 84, 2<D, " hr"<ddd snapshot@0d := Plot@T0@xD, 8x, 0, L<, PlotStyle Ø Thickness@0.004D, PlotRange -> plrange, AxesLabel -> 8"x", ""<, PlotLabel Ø Row@8"t = ", PaddedForm@0, 84, 2<D, " hr"<dd We try it for the initial time and for 1,,, and hr.

9 newtheat.nb 9 In[45]:= snapshot@0d t = 0.00 hr Out[45]= In[46]:= snapshot@1d t = 1.00 hr Out[46]= In[47]:= snapshot@d t =.00 hr Out[47]=

10 newtheat.nb In[48]:= t =.00 hr Out[48]= In[49]:= t =.00 hr Out[49]= As a final task, we produce a sequence of 1 graphs that can be animated to show the dynamics of the heat loss process. The graphs are at 0.25 hr intervals, running from 0 hr to hr. The graphs are output into a Manipulate panel for ease of viewing.

11 newtheat.nb 11 In[]:= i, k<, = Do@mangraph@iD = snapshot@0.25 * id, 8i, 1, 0<D; Manipulate@mangraph@kD, 8k, 0, 0, 1<DD k t = 4.00 hr Out[]=

ME201/MTH281ME400/CHE400 Convergence of Bessel Expansions

ME201/MTH281ME400/CHE400 Convergence of Bessel Expansions 1. Introduction This notebook provides a general framework for the construction and visualization of partial sums of Fourier-Bessel series. The