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enter image description hereI am trying to write down the PDE problem of cooling a square plate from 100 degrees. The air(25 degress) around the plate is going to cool the plate at the surface, but i do not know how to express this. My guess is

\begin{equation} u_t(x,t)-au_{xx}(x,t)=25 \end{equation}

Is this correct? Note that this is a 2d plate but lives in 3d space where the air surrounds it everywhere.

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    $\begingroup$ No. The 25 is a boundary condition, and is not part of the differential equation. In addition, typically, most of the temperature variation is going to be on the air side of the interface, not the solid side. So the surface temperature is going to be much higher than 25 degrees. Your analysis needs to focus on the natural convective heat transfer (and possibly force convective heat transfer) occurring in the air. $\endgroup$ – Chet Miller Apr 21 '18 at 11:29
  • $\begingroup$ The plate is surrounded everywhere by air. So this is a 2d plate in a 3d space. $\endgroup$ – hola Apr 21 '18 at 11:38
  • $\begingroup$ I don't understand. Can you provide a diagram? $\endgroup$ – Chet Miller Apr 21 '18 at 11:50
  • $\begingroup$ So i know the boundary conditions should be 25, and the initial value 100 but i am asking about the PD equation. $\endgroup$ – hola Apr 21 '18 at 11:55
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    $\begingroup$ Like I said, that is accounted for in the boundary conditions, not in the Pde. The way you have it, heat is being generated within the slab at a rate of 25 C per unit time. $\endgroup$ – Chet Miller Apr 21 '18 at 21:21
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There are two ways to write the partial differential equation governing this heat transfer problem depending on whether you wish to consider temperature variations within the plate.

  1. If you do wish to consider temperature variations within the plate, then you can perform an energy balance on a differential element (with volume $a\,dx\,dy$, where $a$ is the thickness) of the plate, which I'll assume lies in the x-y plane. You end up with a conduction term $k\nabla^2T$ corresponding to conductive heat transfer in the x and y directions (where $k$ is the thermal conductivity of the plate and T is the temperature change relative to ambient temperature) and a convective term $-2hT/a$ corresponding to convection to the surrounding air (where $h$ is the convection coefficient). The sum of energy inputs must be equal to the heat storage term, which is $\rho c \frac{\partial T}{\partial t}$, where $\rho$ is the density and $c$ is the heat capacity. Thus, we have $$\rho c \frac{\partial T}{\partial t}=k\nabla^2 T-\frac{2hT}{a}$$ with $T=T(x,y,t)$.

  2. If you don't wish to consider temperature variations within the plate, then you can use a lumped-capacitance approach. Perform an energy balance on the entire plate (of area $A$) to obtain $\rho c Aa\frac{d T}{d t}=-2hAT$, or $$\rho c a\frac{d T}{d t}=-2hT$$ with $T=T(t)$.

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  • $\begingroup$ That convective term should involve a T-25, not just a T. $\endgroup$ – Chet Miller Apr 22 '18 at 2:42
  • $\begingroup$ I defined $T$ relative to 25°C for convenience. But I agree that this is a good point; $T$ is not the absolute temperature or the temperature measured in °C; it is the difference between the beam temperature and 25°C. $\endgroup$ – Chemomechanics Apr 22 '18 at 2:53
  • $\begingroup$ The derivative on the left hand side in case 1 is more correctly written as $\frac{\partial{T}}{\partial{t}}$ or $T_t$, rather than $\dot{T}$. $\endgroup$ – DeltaIV Apr 22 '18 at 7:05
  • $\begingroup$ @DeltaIV Good point; edited. $\endgroup$ – Chemomechanics Apr 22 '18 at 13:01

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