Heat diffusion in an oven

Question

I have this problem where I am preheating an oven to reach $n^\circ$. So lets say within the oven there's a heating element at the bottom, which in turn heats up the rest of the oven. Assuming that the air in the oven behaves like a standard conductor (ignore convection) how can I measure the temperature at a given point $T$ and what will be the equilibrium temperature at a given point?

What I have done so far is make some assumption about the system.

• The walls are perfectly insulated
• Oven is 1-dimensional ($z$ direction)
• Oven has Length $L$ and power output $q$
• Heating Element is placed at $z=0$

From those assumptions I have managed to formulate the equation below using the heat diffusion equation where $\alpha$ is the thermal diffusivity of air, $c$ is the specific heat capacity of the air, $\rho$ is the density and $ \delta $ is the dirac delta function. $$u_t=\alpha u_{zz}+\frac{q}{c\rho}\delta(z)$$ And this equation would have the following initial and boundary conditions where $T_0$ is the initial room temp of the oven. $$u(z,0)=T_0$$ $$ u_z(0,t)=u_z(L,t)=0$$ This is the point at which I'm unsure if I have done the correct approach as feels incorrect to have $u_z(0,t)=0$ whilst also having the heating element at $z=0$. Maybe I could place the heating element at $z=\epsilon$ instead but I'm unsure as how to work this out and would much prefer to have it at $z=0$. I have attempted to solve the equation using the methods I have found from here, but I seem to be having issues with $q$ as it cannot be written as a summation in $X_n(x)$ and $Q_n(t)$ as its just a constant with the only space dependence being the Dirac delta function.

I have also attempted to find the equilibrium temperatures by assuming steady state with $u_t=0$ which results in more issues as I'm unsure on how to find the indefinite integral of the Dirac delta function (uncertain if one even exists). I have attempted to just use the definite integral definition but that gives contradictory results.

If anyone could provide an in-depth answer on how to solve the problem, or even a point in the right direction I'd appreciate it as I've been mulling over this for a couple days but cant think of a solution even still.

Answer 1

A simple model may be fitted for your case. Sovle a simple diffusion equation for thermal energy density $u(z, t)$

$$ \tag{1} \frac{\partial u}{\partial t} = D \frac{\partial^2 u}{\partial z^2}. $$ Where $D$ the diffusion constant. And impose the boundary conditions: $$ D \left[ \frac{\partial u}{\partial z} \right]_L = 0;\\ D \left[ \frac{\partial u}{\partial z} \right]_0 = - P $$ where $P$ is the power generating form heating wire. I will elaborated the details in the following.

effects of boundary condition

To study how the boundary condition related to the heating power, lets integrate Eq.(1) w.r.t $z$. $$ \tag{2} \int_0^L dz \left\{ \frac{\partial u}{\partial t} = D \frac{\partial^2 u}{\partial z^2} \right\} $$

Define $U(t) = \int_0^L u(z, t) dz$. After integration, Eq. (2) renders: $$ \tag{3} \frac{\partial U(t)}{\partial t} = D \left[ \frac{\partial u}{\partial z} \right]_L - D \left[ \frac{\partial u}{\partial z} \right]_0 $$ This equation gives a dynamical meaning for the derivative of $u$ at the boundaries: the derivatives denote the rate of change of total energy.

The $z=L$ is an insulate means to set $\left[ \frac{\partial u}{\partial z} \right]_L = 0$, all heat energies arrive at $z$ are reflected - the slope vanishes there, and energy is preserved.

Remove the boundary at $z=L$, Eq. (3) becomes

$$ \tag{4} \frac{\partial U(t)}{\partial t} = - D \left[ \frac{\partial u}{\partial z} \right]_0 $$ Where $U(t)$ is the total heat capacitance at time $t$. Therefore the slop: $$ \tag{5} \frac{\partial U(t)}{\partial t} = - D \left[ \frac{\partial u}{\partial z} \right]_0 = \text{ rate chage of total heat energy.} $$

In conclusion, what you need to do is to fixed the Neumann boudary conditions:

$$ D \left[ \frac{\partial u}{\partial z} \right]_L = 0 $$

The zero derivative means a reflection boundary. The insulate boudary will reflect all the heat flow into it.

And the Neumann boudary condition at $z=0$: $$ D \left[ \frac{\partial u}{\partial z} \right]_0 = - P $$

P is the power of the heat generator.

Set up these two conditions, then solve the simple diffusion equation. You will get what you expect.