# Heat Equation: Solve with Non-Homogeneous Boundary Conditions

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• erobz
It can be found by solving the ODE given by equation (2) and is dependent on the specific heat capacity of the air, the convective heat transfer coefficient, and the heat flux from the fluid.In summary, we have a plane wall with constant thermal conductivity between two thermal reservoirs. The reservoir on the left is kept at a constant temperature and has a high convective coefficient, resulting in a boundary condition at the left wall. At time t=0, the wall and a cold reservoir are brought into thermal contact with a warm reservoir, both initially at the same uniform temperature. The wall and blue reservoir start to heat up, with a boundary condition given at the right wall. However, separation of variables does not work due to the
erobz
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Imagine you have a plane wall with constant thermal conductivity, that is the intermediate between two thermal reservoirs:

The reservoir on the left is being kept at temp ##T_s##, and it is a fluid that has very high convective coefficient ##h##. As a result, the boundary condition at the LHS wall is given by:

$$T(0,t) = T_s$$

at time ##t=0## the wall and the cold reservoir are brought into thermal contact with the warm reservoir. The wall and the blue reservoir are both initially at the same uniform temperature ##T(x,0) = T_o##.

The blue reservoir and the wall start to heat up. The boundary condition on the RHS wall is given by:

$$-k \left. \frac{\partial T}{ \partial x} \right|_{L} = h \left( T(L,t) - T_{air}(t) \right)$$

The heat equation for the plane wall is given by:

$$\frac{\partial^2 T}{\partial x ^2} = \rho c_p \frac{\partial T}{\partial t}$$

So we have three non-homogenous boundary conditions, what is the idea to solve the PDE? Apparently, separation of variables does not work because of the non-zero boundary conditions.

The only thing I know (almost nothing) about solving PDE's I've learned from YouTube videos... Why does having non-zero boundary conditions throw a fork in it, and how big of a fork is it?

Thanks for any guidance!

What do we know about $T_{\mathrm{air}}$?

Have you tried a Laplace transform in time? That gives $$\kappa(p\hat T - T_0) = \frac{\partial^2 \hat T}{\partial x^2}$$ where $\kappa= \rho c_p$ subject to $$\hat T(0,p) = \frac{T_s}{p}, \quad \left.-k\frac{\partial \hat T}{\partial x}\right|_{x=L} = h(\hat T(L,p) - \hat T_{air}(p)).$$ That does leave you with $$\hat T(x,p) = \frac{T_0}{p} + A(p)\cosh(\sqrt{\kappa p}x) + B(p)\sinh(\sqrt{\kappa p}x)$$ which is not easy to invert.

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pasmith said:
What do we know about $T_{\mathrm{air}}$?
It's going to be found via the following ODE:

$$m_{air} c_{air} \frac{dT_{air}}{dt} = h A ( T(L,t) - T_{air}(t) ) + \dot q \tag{2}$$

The first term on the RHS is basically the boundary condition on the right wall. The ##\dot q ## is just a constant term.
pasmith said:
Have you tried a Laplace transform in time? That gives $$\kappa(p\hat T - T_0) = \frac{\partial^2 \hat T}{\partial x^2}$$ where $\kappa= \rho c_p$ subject to $$\hat T(0,p) = \frac{T_s}{p}, \quad \left.-k\frac{\partial \hat T}{\partial x}\right|_{x=L} = h(\hat T(L,p) - \hat T_{air}(p)).$$ That does leave you with $$\hat T(x,p) = \frac{T_0}{p} + A(p)\cosh(\sqrt{\kappa p}x) + B(p)\sinh(\sqrt{\kappa p}x)$$ which is not easy to invert.
I'm not lying when I said the only thing I know about PDE methods is from YouTube. Heck, I have very little formal understanding of ODE's for that matter. Am I understanding correctly that separation of variables simply does not work. Or is it that it takes extra massaging?

I was hoping that if I could find ##T(x,t) = F(x)G(t)##, then I could use:

$$\left. \frac{\partial T}{ \partial x} \right|_{x=L} = \left. \frac{\partial F}{ \partial x} \right|_{x=L} G(t)$$

subbing into (2) to get some kind of system.

I'm likely just getting all tangled up in stuff that won't work.

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pasmith said:
Have you tried a Laplace transform in time? That gives $$\kappa(p\hat T - T_0) = \frac{\partial^2 \hat T}{\partial x^2}$$ where $\kappa= \rho c_p$ subject to [tex]
How does this work? I see you have introduced ##p## and the partial w.r.t. time disappears? I’m obviously oblivious to something.

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pasmith said:
What do we know about $T_{\mathrm{air}}$?
In this context, ##T_{air}## is considered the bulk air temperature outside the thermal boundary layer.

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erobz

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