ࡱ> y%` bjbj"x"x _@@6kNNNd$P,0D&3> (f | | "<,#@# a1c1c1c1c1c1c1$3hD61Y%""%%1xx| | p2///%x8| | a1/%a1///| t :̌_) /0|20&3/6+"6/6/L#v#T/$DZ$}L#L#L#11.dL#L#L#&3%%%%D00xxxxxx  INCLUDEPICTURE "http://www.csun.edu/~pubrels/NewLogoSm.JPG" \* MERGEFORMATINET  College of Engineering and Computer Science Mechanical Engineering Department Mechanical Engineering 501B Seminar in Engineering AnalysisSpring 2009 Class: 14443 Instructor: Larry CarettoFebruary 9 Homework Solutions Kreyszig, page 561 problems 5, 13, 14 and 27 Finding the temperature in a one dimensional solid as a function of time. These problems involve the solution of the diffusion equation with various boundary conditions and initial conditions. All problems start with the same separation of variables process which is described below.  EMBED Equation.3  [1] Here, a = k/c, is the thermal diffusion coefficient with dimensions of length squared over time. The initial condition specifies the value of u at all values of x at t = 0. This condition is written as follows: u(x,0) = f(x) [2] In problem 5 we have f(x) = sin(0.4x) and in problem 13, f(x) = x. The boundary conditions at x = 0 and x = xmax, can be functions of time in general. These are written as follows. u(0,t) = uL(t) u(xmax,t) = uR(t) [3] In problem 5 we have uL = uR = 0 and in problems 13 and 14 we have "u/"x = 0 at x = 0 and x = xmax. We use the basic separation of variables approach in which we assume that the solution, u(x,t), is the product of two functions, T(t) a function of time only and X(x) a function of the distance x only. With this assumption, our solution becomes. u(x,t) = X(x)T(t) [4] We do not know, in advance, if this solution will work. However, we assume that it will and we substitute it for u in equation [1]. Since X(x) is a function of x only and T(t) is a function of t only, we obtain the following result when we substitute equation [4] into equation [1].  EMBED Equation.3  [5] If we divide the final equation through by the product aX(x)T(t), we obtain the following result.  EMBED Equation.3  [6] The left hand side of equation [6] is a function of t only; the right hand side is a function of x only. The only way that this can be correct is if both sides equal a constant. This also shows that the separation of variables solution works. In order to simply the solution, we choose the constant to be equal to -l2. This gives us two ordinary differential equations to solve.  EMBED Equation.3  [7] The first equation becomes EMBED Equation.3 . The general solution to this equation is known to be  EMBED Equation.3 . The second differential equation in [7] can be written as  EMBED Equation.3 . The general solution to this equation is X(x) = Bsin(lx) + Ccos(lx). Thus, our general solution for u(x,t) = X(x)T(t) becomes (after defining AB as C1 and AC as C2)  EMBED Equation.3  [8] At this point we have to consider the initial conditions, boundary conditions and property values for each problem separately. Equation [8] will be the starting point for the analysis of all four problems. Problem 5 Find the temperature, u(x,t) in a bar of silver (length = 10 cm, constant cross section of area 1 cm2, density 10.6 g/cm3, thermal conductivity 1.04 cal/(s cm oC), specific heat 0.056 cal/(g oC) that is perfectly insulated laterally, whose ends are kept at temperature 0oC and whose initial temperature (in oC) is sin(0.4px). In this case the boundary conditions are u(0,t) = 0 and u(xmax,t) = 0. If we substitute the boundary condition at x = 0 into equation [8], get the following result because sin(0) = 0 and cos(0) = 1.  EMBED Equation.3  [9] Equation [9] will be satisfied for all t if C2 = 0. With C2 = 0, we can apply the solution in equation [8] to the boundary condition at x = xmax.  EMBED Equation.3  [10] Equation [10] can only be satisfied if the sine term is zero. This will be true only if lxmax is an integral times p. If n denotes an integer, we must have  EMBED Equation.3  [11] Since any integral value of n gives a solution to the original differential equations, with the boundary conditions that u = 0 at both boundaries, the most general solution is one that is a sum of all possible solutions, each multiplied by a different constant, Cn. This general solution is written as follows  EMBED Equation.3  [12] We still have to satisfy the initial condition that u(x,0) = f(x). Substituting this condition for t = 0 into equation [12] gives  EMBED Equation.3  [13] Since the ordinary differential equation for X(x) forms a Sturm-Liouville problem, the solutions Cnsin(npx/xmax) are eigenfunction solutions to the problem in the interval 0 d" x d" xmax. Thus we can expand any initial condition function, f(x) in terms of these eigenfunctions. We can obtain the usual equation for eigenfunction expansion coefficients of Sturm-Liouville solutions by using the orthogonality relationships for integrals of the sine. If we multiply both sides of equation [13] by sin(mpx/xmax), where m is another integer, and integrate both sides of the resulting equation from a lower limit of zero to an upper limit of xmax, we get the following result.  EMBED Equation.3  [14] In the second row of equation [14] we reverse the order of summation and integration because these operations commute. We then recognize that the integrals in the summation all vanish unless m = n, leaving only this integral to evaluate. Solving for Cm and evaluating the last integral in equation [14] gives the following result.  EMBED Equation.3  [15] For any initial condition, then, we can perform the integral on the right hand side of equation [15] to compute the values of Cm and substitute the result into equation [12] to compute u(x,t). We recognize that equation [15] is the usual equation for eigenfunction expansions in Sturm-Liouville problems. In this problem we have f(x) = sin(0.4px). No dimensions are given for the constant 0.4 in this equation, but we know that it must have units of inverse length to make the argument of the sine a dimensionless quantity. Since other length data are given in units of cm, we assume that the units value multiplying px is 0.4 cm-1; since xmax = 10 cm, the value of 0.4 cm-1 = 4/xmax and we can write this initial condition as f(x) = sin[(4)px/xmax]. If we substitute f(x) = sin[(4)px/xmax] into equation [15] we find Cm as follows.  EMBED Equation.3  [16] Because sin(mpx/xmax) is an orthogonal function in the region 0 d" x d" xmax, the only time that this integral is non zero is when m = 4. We have just shown NOTEREF _Ref63658982 \h  \* MERGEFORMAT 3 that the integral of sin2(mpx/xmax) between x = 0 and x = xmax = xmax/2 so that equation [16] tells us that Cm = 1 when m = 4 and Cm = 0 for all other values of m.  EMBED Equation.3  [17] Substituting this result into equation [12] gives the following solution to the diffusion equation when f(x) sin(px/xmax) and the boundary temperatures are zero.  EMBED Equation.3  [18] For the data in this problem we can compute the thermal diffusivity as follows.  EMBED Equation.3  [19] The solution to the problem uses cm as the unit for length. Problem 13  Adiabatic means no heat exchange with the neighborhood, because the bas is completely insulated, also at the ends. Physical information: The heat flux at the ends of a bar is found to be proportional to "u/"x there. Show that for the completely insulated bar, ux(0,t) = ux(L,t) = 0, u(x,0) = f(x) and separation of variables gives the following solution with An given by (2) in section 11.3.  EMBED Equation.3  Again we use equation [8] as the starting point.  EMBED Equation.3  [8] Taking the first derivative of this equation with respect to x gives.  EMBED Equation.3  [20] The boundary condition that the "u/"x = 0 at x = 0 gives the following result.  EMBED Equation.3  [21] We can satisfy this condition by having C1 = 0. Setting C1 to zero in equation [20] and applying the boundary condition that "u/"x = 0 at x = L gives.  EMBED Equation.3  [22] We can satisfy this condition is the sine term is zero, which it will be if L is an integer multiple of .  EMBED Equation.3  [23] As usual, we take the general solution to be the sum of all solutions that satisfy the differential equation and the boundary conditions. Note that when the eigenfunction is cos(nx/L) we do have a nonzero eigenfunction for n = 0, so we include this in our sum. However, this eigenfunction is simply cos(0) = 1. We will write this term separately outside the sum.  EMBED Equation.3  [24] As usual we find the coefficients, An, by using an eigenfunction expansion. A specific example of this is given in the next problem. The development in the next problem solution shows that the equations for these coefficients match those of equation (2) in section 11.3. The solution gives QTUV   ! 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A0 as t ! ". That is, the steady-state solution is a constant temperature everywhere in the region. This seems correct. For the adiabatic boundary conditions, there is no way for heat to leave the region. The initial temperature profile will head to heat transfer in the x direction so long as there is a temperature difference. At steady state, there is no mechanism to maintain any temperature difference so that the solution approaches a constant. Problem 14  Find the temperature in the bar of problem 13 with L = ,  = c2 = 1 and f(x) = x. Here we use the general solution in equation [24] that satisfies the differential equation and the boundary conditions and apply it to the given initial condition. Setting L =  in equation [24] gives n = n and the following result for our general solution.  EMBED Equation.3  [25] As usual we set t = 0 so that u(x,t = 0) becomes the initial condition, f(x) = x.  EMBED Equation.3  [26] If we multiply both sides of this equation by cos(mx) and integrate the result from 0 to , we obtain.  EMBED Equation.3  [27] We have to consider the case of m = 0, in which case the eigenfunction is 1, separately. This gives.  EMBED Equation.3  [28] The integral  EMBED Equation.3  Thus, all the integrals on the right, except for the integral of A0dx are zero. (This is consistent with the notion that the set of eigenfunctions ym = cos(mx/L) for the cosine solution, which starts at n = 0, gives y0 = 1; since the y0 eigenfunction must be orthogonal to all other eigenfunctions, we expect that the integral of any eigenfunction, other than y0, will be zero.) This gives the following result for A0.  EMBED Equation.3  [29] Returning to equation [27], we find Am for m e" 1, using the result for the orthogonality of the cosine function to give.  EMBED Equation.3  [30] Using integral tables gives the following result for the integral in the numerator.  EMBED Equation.3  [31] The integral in the denominator can also be found using integral tables as  EMBED Equation.3  [32] Using these results and noting that (cos(mp)  1) equals 2 for odd m and zero for even m gives Am as follows.  EMBED Equation.3  [30] Substituting this result for Am (m e" 1) and the previous result for A0 into equation [25] gives.  EMBED Equation.3  [31] Setting  = 1 (given) gives the following result for the first few terms in the series.  EMBED Equation.3  [32] Problem 27  (Non-homogenous heat equation) Show that the problem consisting of ut  uxx = Ne-ax and (2), (3) can be reduced to a problem for the homogenous heat equation by setting u(x,t) = v(x,t) + w(x) and determining w so that v satisfies the homogenous equation and the conditions v(0,t) =v(L,t) = 0, v(x,0) = f(x)  w(x). (The term Ne-ax may represent heat loss due to radioactive decay in the bar.) The differential equation and the boundary conditions given in the problem statement are shown below.  EMBED Equation.3  [33] Substituting the proposed solution, u(x,t) = v(x,t) + w(x) into the diffusion equation gives, after some manipulation, the result shown in equation [34]. In the first step we drop the "w/"t derivative since w is a function of x only. In the second step we drop the v terms since they are exactly the diffusion equation and v satisfies that partial differential equation.  EMBED Equation.3  [34] Since w is a function of x only, the first and last terms in [34] give us an ordinary differential equation that we can solve for w. Writing this equation and integrating it two times gives.  EMBED Equation.3  [35]  EMBED Equation.3  [36]  EMBED Equation.3  [37] We can combine the solution definition, u(x,t) = v(x,t) + w(x), with the initial condition, u(x,t) = f(x), to get an initial condition for v. v(x,0) = u(x,0)  w(x) = f(x)  w(x) [38] We would have to use this initial condition for v(x,0) in developing any eigenfunction expansion to match the initial conditions. The constants of integration in equation [37] can be found by matching boundary temperatures. If u(0,t) = w(0) = uA and u(L,t) = w(L) = uB we have the following results form equation [37].  EMBED Equation.3  [39]  EMBED Equation.3  [40] Solving equation [40] gives C1 as follows.  EMBED Equation.3  [41] Substituting the results of equations [40] and [41] for C1 and C2 into equation [37] gives.  EMBED Equation.3  [42] A rearrangement gives  EMBED Equation.3  [42] The solution for v(x,t) which satisfies the diffusion equation with zero boundary conditions has already been found; it is given by equation [12].  EMBED Equation.3  [12] The coefficients Cn are found by the following formula for eigenfunction expansion coefficients. This is the same formula as equation [15] with f(x) replaced by f(x)  w(x).  EMBED Equation.3  [43] Substituting equation [42] for w(x) gives the equation necessary to compute the expansion coefficients, Cm.  EMBED Equation.3  [44] In this problem we are given uA = uB = 0 which makes only a slight simplification in the final result.  The choice of  l2 for the constant as opposed to just plain l comes from experience. Choosing the constant to have this form now gives a more convenient result later. If we chose the constant to be simply l, we would obtain the same result, but the expression of the constant would be awkward.  As usual, you can confirm that this solution satisfies the differential equation by substituting the solution into the differential equation.  Using a standard integral table, and the fact that the sine of zero and the sine of an integral multiple of p are zero, we find the following result:  EMBED Equation.3      Jacaranda (Engineering) Room 3333 Mail Code Phone: 818.677.6448 Email:  HYPERLINK mailto:lcaretto@csun.edu lcaretto@csun.edu 8348 Fax: 818.677.7062 February 9 homework solutions L. S. 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"2YV%{h0) `!YV%{h0)Φ  dVxڅR=K@]xUHB!`<"lK+A ʟ MN ̛}^YP`j"9j"gʲ*.pfв'k-5aJ]R ]'j, [I~'O2z츂F}B 9U;VG_!A   !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefgijklmnopqrstuvwxyz{|}~_Root Entry  Ff̌Data h?WordDocument_ObjectPool m̌f̌_1111065395F̌̌Ole CompObjfObjInfo  !"%(+.1458;<=>?@CFGHILOPQTWX[^_`adghilopqrstuvwxyz{~ FMicrosoft Equation 3.0 DS Equation Equation.39q`mIyI "u"t=" 2 u"x 2 FMicrosoft Equation 3.0 DS EqEquation Native |_1111066660 F̌̌Ole CompObj fuation Equation.39qԨlInI "u"t="X(x)T(t)[]"t=X(x)"T(t)"t=" 2 u"x 2 =" 2 X(x)T(t)[]"x 2 =T(t)ObjInfo Equation Native  _1111067312F̌̌Ole " 2 X(x)"x 2 FMicrosoft Equation 3.0 DS Equation Equation.39qXI$I 11T(t)"T(t)"t=CompObjfObjInfoEquation Native _1111067331 F̌̌1X(x)" 2 X(x)"x 2 FMicrosoft Equation 3.0 DS Equation Equation.39qHyII 11T(t)dT(t)dt="Ole CompObjfObjInfoEquation Native  2 1X(x)d 2 X(x)dx 2 =" 2l FMicrosoft Equation 3.0 DS Equation Equation.39q\HyII dT(t)_1111067422F̌̌Ole #CompObj$fObjInfo&Equation Native 'x_11110678601F̌̌Ole )CompObj *fdt=" 2 T(t) FMicrosoft Equation 3.0 DS Equation Equation.39qLII T(t)=Ae " 2 tObjInfo!,Equation Native -h_1112179053J$F̌̌Ole /CompObj#%0fObjInfo&2Equation Native 3_1289455079)F̌̌ FMicrosoft Equation 3.0 DS Equation Equation.39qx0. d 2 X(x)dx 2 + 2 X(x)=0 FMicrosoft Equation 3.0 DS EqOle 6CompObj(*7fObjInfo+9Equation Native :uation Equation.39qwXl u(x,t)=T(t)X(x)=Ae " 2 t Bsin(x)+Ccos(x)[]=e " 2 t C 1 sin(x)+C 2 cos(x)[] FMicrosoft Equation 3.0 DS Equation Equation.39qIxoI u(0,t)=0=e " 2 t C 1 sin(0)+C 2 co_1111068776.F̌̌Ole ACompObj-/BfObjInfo0DEquation Native E _1111068823,;3F̌̌Ole JCompObj24Kfs(0)[]=C 2 e " 2 t FMicrosoft Equation 3.0 DS Equation Equation.39qԴ~II u(x max ,t)=0=C 1 e "ObjInfo5MEquation Native N_11121907138F̌̌Ole R 2 t sin(x max ) FMicrosoft Equation 3.0 DS Equation Equation.39q}- x max =nor=nxCompObj79SfObjInfo:UEquation Native V_1111069489=F̌̌ max FMicrosoft Equation 3.0 DS Equation Equation.39qHyII u(x,t)=C n e " n2 t sin( n x) n=1Ole YCompObj<>ZfObjInfo?\Equation Native ]8" " with n =nx max FMicrosoft Equation 3.0 DS Equation Equation.39q¶p'\ f(x)=u(x,0)=C n sin_11373997086EBF̌̌Ole bCompObjACcfObjInfoDeEquation Native f_1137399731GF̌̌Ole jCompObjFHkfnxx max () n=1" " FMicrosoft Equation 3.0 DS Equation Equation.39qHZ< f(x)sinmxx max (ObjInfoImEquation Native nd_1137399754@cLF̌̌Ole |)dx 0x max +" =C n sinmxx max ()sinnxx max () n=1" " 0x max +" dx=C n sinmxx max ()sinnxx max ()dx 0x max +" n=1" " =C m sin 2 mxx max ()dx 0x max +" FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjKM}fObjInfoNEquation Native _1289247459QF̌̌Hk[ C m =f(x)sinmxx max ()dx 0x max +" sin 2 mxx max ()dx 0x max +" =2x max f(x)sinmxx max ()dx 0x max +" FMicrosoft Equation 3.0 DS Equation Equation.39qC`| C m =2x max sin4xx max ()sinOle CompObjPRfObjInfoSEquation Native mxx max ()dx 0x max +" =2x max  4m sin 2 xx max ()dx 0x max +" FMicrosoft Equation 3.0 DS Equation Equation.39q_1289282439VF̌̌Ole CompObjUWfObjInfoXCuhD C m =1m=40m`"4{ FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native _1289455377[F̌̌Ole CompObjZ\fObjInfo]Equation Native _1137401158`F̌̌Ole Xl u(x,t)=e "4x max () 2 t sin4xx max () FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj_afObjInfobEquation Native a_1137409182^heF̌̌EN< =kc=1.04  calcm"s" o C10.6gcm 3 0.056calg" o C=1.752cm 2 s FMicrosoft Equation 3.0 DS EqOle CompObjdffObjInfogEquation Native ruation Equation.39qV u(x,t)=A 0 +A n e " n2 t cos( n x) n=1" "  n =nx max =nLc 2 =_1137409538jF̌̌Ole CompObjikfObjInfol FMicrosoft Equation 3.0 DS Equation Equation.39q8c u(x,t)=e " 2 t C 1 sin(x)+C 2 cos(x)[]Equation Native _1137409552"ToF̌̌Ole CompObjnpf FMicrosoft Equation 3.0 DS Equation Equation.39q/, "u(x,t)"x=e " 2 t C 1 cos(x)"C 2 sin(x)[]ObjInfoqEquation Native _1137409639tF̌̌Ole  FMicrosoft Equation 3.0 DS Equation Equation.39q?l "u(x,t)"x x=0 =0=e " 2 t C 1 cos(0)"C 2 sinCompObjsufObjInfovEquation Native [_1137409760r|yF̌̌(0)[]=e " 2 t C 1 FMicrosoft Equation 3.0 DS Equation Equation.39qH "u(x,t)"x x=L =0=Ole CompObjxzfObjInfo{Equation Native "e " 2 t C 2 sin(L) FMicrosoft Equation 3.0 DS Equation Equation.39q†p L=nor=nLi_1137409869~F̌̌Ole CompObj}fObjInfoEquation Native _1289296127F̌̌Ole CompObjfntegern FMicrosoft Equation 3.0 DS Equation Equation.39qhl u(x,t)=A 0 e "0 cos(0)+A n e " n2 ObjInfoEquation Native _1137411203wF̌̌Ole t cos( n x) n=1" " =A 0 +A n e " n2 t cos( n x) n=1" "  n =nL FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjfObjInfoEquation Native _1289296343F̌̌¶H u(x,t)=A 0 +A n e "n 2 t cos(nx) n=1" " FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjfObjInfoEquation Native ¦l u(x,0)=f(x)=x=A 0 +A n cos(nx) n=1" " FMicrosoft Equation 3.0 DS Equation Equation.39q_1289296361F̌̌Ole CompObjfObjInfo    !$'(),/012589:;<?BCDEFGJMNOPSVWXYZ[^abcdgjklmnoruvwx{~Xo xcosmxdx 0 +" =A 0 +A n cos(nx) n=1" " []cosmxdx 0 +" FMicrosoft Equation 3.0 DS EqEquation Native _1289296389'F̌̌Ole CompObjfuation Equation.39qo x(1)dx 0 +" =A 0 +A n cos(nx) n=1" " [](1)dx 0 +"     Z[ !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYz\]^`abcdegfihkjlmonqprstuwvx{|}~ObjInfo Equation Native  _1137412090F̌̌Ole CompObjfObjInfoEquation Native &_1289296452F̌̌ FMicrosoft Equation 3.0 DS Equation Equation.39q   cosmxdx 0 +" =)sinmxm[] x=0x= =)sinmm[]")sin0m[]=0.Ole CompObjfObjInfoEquation Native s FMicrosoft Equation 3.0 DS Equation Equation.39qW 6 xdx 0 +" =x 2 2() x=0x= = 2 2"0= 2 2=A 0 dx 0 +" =A 0 !A 0 =2 FMicrosoft Equation 3.0 DS Equation Equation.39q­P A m =xcosmxdx 0 +" _1289296535F̌̌Ole "CompObj#fObjInfo%Equation Native &_1295881601F̌:̌Ole *CompObj+fcos 2 mxdx 0 +" FMicrosoft Equation 3.0 DS Equation Equation.39qD xcosmxdx 0 +" =cosmObjInfo-Equation Native ._1137414429F:̌:̌Ole 3xm 2 +xsinmxm[] 0 =cosm"1m 2 FMicrosoft Equation 3.0 DS Equation Equation.39q<D cos 2CompObj4fObjInfo6Equation Native 7X_1295881612F:̌:̌ mxdx 0 +" =cosmxsinmx2m+x2[] 0 =cosmsinm2m+2"0=2 FMicrosoft Equation 3.0 DS Equation Equation.39qOle =CompObj>fObjInfo@Equation Native A„< A m =xcosmxdx 0 +" cos 2 mxdx 0 +" =cosm"1m 2 2="4m 2 modd0meven{_1295879876YF:̌:̌Ole HCompObjIfObjInfoK FMicrosoft Equation 3.0 DS Equation Equation.39q4 u(x,t)=2"4e "n 2 t cos(nx)n 2n=1,3,5,& " "Equation Native L_1295879883F:̌:̌Ole QCompObjRf FMicrosoft Equation 3.0 DS Equation Equation.39quH u(x,t)=2"4e "t cos(x)+e "9t cObjInfoTEquation Native U_1137415900F:̌:̌Ole \os(3x)9+e "25t cos(5x)25+e "49t cos(7x)49+"[] FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj]fObjInfo_Equation Native `_1137416360F:̌:̌/D. 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