# Heat equation of cooling square plate I 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.

• 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. Apr 21, 2018 at 11:29
• The plate is surrounded everywhere by air. So this is a 2d plate in a 3d space.
– hola
Apr 21, 2018 at 11:38
• I don't understand. Can you provide a diagram? Apr 21, 2018 at 11:50
• So i know the boundary conditions should be 25, and the initial value 100 but i am asking about the PD equation.
– hola
Apr 21, 2018 at 11:55
• 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. Apr 21, 2018 at 21:21

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 $T=T(t)$.
• 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. Apr 22, 2018 at 2:53
• 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 $\dot{T}$. Apr 22, 2018 at 7:05