# Heat diffusion in an oven

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.

• You have a problem with your assumptions. Walls are perfectly insulating, meaning any heat that gets in stays in. The heating element adds head forever. This means you will have an oven that gets hotter and hotter without ever reaching an equilibrium temperature. Mar 7 at 20:24
• Also you cannot have air or heat in 1 dimension Mar 7 at 23:04
• Use $-k\frac{\partial u}{\partial z}=q$ at z = 0, where q is the heat flux, and lose the Dirac delta term in the differential equation. Mar 8 at 2:04

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.