Question

A slab of thickness \(L\), as shown in the figure, has cross-sectional area \(A\) and constant thermal conductivity \(k\). \(T_1\) and \(T_2\) are the temperatures at \(x=0\) and \(x=L\), respectively. Which one of the following options is the CORRECT expression of the thermal resistance for steady-state one-dimensional heat conduction?

• A. \(\dfrac{L}{kA}\)
• B. \(\dfrac{k}{LA}\)
• C. \(\dfrac{kA(T_1-T_2)}{L}\)
• D. \(\dfrac{A}{Lk}\)

Hint:

Thermal resistance in conduction: \(R_{cond}=L/(kA)\) (units K/W).
Series/parallel combinations of slabs add like electrical resistances.
If \(k\) varies with \(x\) or \(T\), integrate \(R=\int \frac{dx}{kA}\).

Answer 1

The Correct Option is A

Step 1: For steady one-dimensional conduction with constant \(k\), Fourier’s law gives the heat flux \[ q_x=-k\,\frac{dT}{dx}=\frac{k\,(T_1-T_2)}{L}, \] because the temperature profile is linear between \(x=0\) and \(x=L\). Step 2: The total heat rate is \(Q = q_x A = \dfrac{kA\,(T_1-T_2)}{L}\). Step 3: Define thermal resistance \(R_{th}\) by \(R_{th}=\dfrac{\Delta T}{Q}=\dfrac{T_1-T_2}{Q}\). Substituting \(Q\) from Step 2, \[ R_{th}=\frac{T_1-T_2}{\dfrac{kA\,(T_1-T_2)}{L}}=\frac{L}{kA}. \] Hence, the correct expression is \(R_{th}=\dfrac{L}{kA}\).