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Having trouble with the maths in this question, I realise this is a physics question so I apologise if this isn't allowed, but some mathematicians might be able to solve it well. I asked this in the physics stack exchange and they all freaked out cos its a worked example question and not conceptual of nature so I thought I'd try it here. can anyone help or show me how to solve? please and thank you. A $120\ $mm long metallic rod with a diameter of $11\ $mm is insulated so that heat can only flow in and out of the rod at one end (defined as $x = 0$). This end of the rod is maintained at $18\ ^◦$C while the initial temperature in the rod takes the form of part of a sine wave with a maximum temperature of $65\ ^◦$C.

(a) Show that a temperature profile of the form $$T(x, t) = T_A \sin{2πx\over λ}e^{−t/τ} + T_0$$ is a solution of the temperature diffusion equation when the diffusion constant, $D$, is a function of one or more of the constants $T_A$, $T_0$, λ and τ . Determine this relationship. Note that you are not asked to solve the diffusion equation - just to show that the form given satisfies the equation.

(b) The boundary condition at the ends of the rod must be

Constant temperature end: $\displaystyle {∂T\over ∂t}(0, t) = 0$,

Insulated end: $\displaystyle {∂T\over ∂x}(L, t) = 0$,

where $L$ is the length of the rod. Give brief explanations of why these conditions apply for the current situation.

(c) Using the boundary and initial conditions provided, find numerical values for $T_A$, $T_0$ and λ.

(d) The rod cools over time as heat is lost through the non-insulated end. Suppose it takes $25\ $s for the peak temperature of the rod to drop to $32\ ^◦$C. Calculate a numerical value for τ and hence find $D$.

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    $\begingroup$ plz my mathematical overlords, have mercy on a puny cretin like myself and help me solve this easy question $\endgroup$ – john smith Mar 25 at 8:20
  • $\begingroup$ You are usually expected to provide evidence of your own efforts to solve these problems. What have you tried? $\endgroup$ – asdq Mar 25 at 8:35
  • $\begingroup$ well I know that the diffusion equation is hx = -k(dt/dx) $\endgroup$ – john smith Mar 25 at 9:10
  • $\begingroup$ and dT/dt = D(dT^2/dx^2) $\endgroup$ – john smith Mar 25 at 9:11
  • $\begingroup$ but can't figure out how to make the equation displayed in A in that form $\endgroup$ – john smith Mar 25 at 9:11
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(a) You must check that the given function $T(x,t)$ satisfies the diffusion equation $$ {\partial T\over\partial t}=D{\partial^2 T\over\partial x^2}. $$ We have: $$ {\partial T\over\partial t}=-{1\over\tau}T_A \sin{2\pi x\over\lambda}e^{−t/\tau}, \quad {\partial^2 T\over\partial x^2}=-\left({2\pi\over\lambda}\right)^2T_A \sin{2\pi x\over\lambda}e^{−t/\tau}. $$ Hence: $$ {\partial T\over\partial t}={\lambda^2\over4\pi^2\tau}{\partial^2 T\over\partial x^2}, $$ and this is indeed the diffusion equation, provided: $$ D={\lambda^2\over4\pi^2\tau}. $$ I leave to you now to discuss points (b)-(d).

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