Conduction through Flat Plate


Consider the system shown above. The top, bottom, front and back of the cube are insulated, so that heat can be conducted through the cube only in the \(x\) direction. In this special case, heat flow is one dimensional. If sides were not insulated, heat flow could be two or three dimensional.

For one dimensional steady heat conduction, \[\frac{d}{d x}\left(\frac{d T}{d x}\right) = 0 \tag*{(9)}\] Boundary conditions: \[\begin{aligned} T &= T_1 \qquad \text{at} \quad x&=0 \\ T &= T_2 \qquad \text{at} \quad x&=L \end{aligned}\] Integrating Eqn.(9) , we get \[\frac{dT}{dx} = C_1 \tag*{(10)}\] Integrating further, we get \[T = C_1x + C_2 \tag*{(11)}\] Using the boundary condition at \(x=0\) gives \[C_2=T_1\] And, from the boundary condition at \(x=L\) gives, \[T_2 = C_1 L + T_1 \qquad \Longrightarrow \quad C_1 = \frac{T_2-T_1}{L}\] Substituting for \(C_1\) in Eqn.(10), we get \[\frac{dT}{dx} = \frac{T_2-T_1}{L}\] From the definition of heat flux (as given by Fourier’s law) \[q = -k\frac{dT}{dx}\] Therefore, \[\begin{align*} q &= -k\frac{T_2-T_1}{L} = k\frac{T_1-T_2}{L} \nonumber \\ Q &= qA =kA\frac{T_1-T_2}{L} = \frac{T_1-T_2}{R} \qquad \text{where  } R =\frac{L}{kA} \tag*{(12)}\end{align*}\] \(R\) is called as the thermal resistance.