GATE 2024 ME – Question 48

Heat Transfer · Modes of heat transfer, one-dimensional conduction, resistance concept · 2 marks · Numerical answer

Consider a slab of 20 mm thickness. There is a uniform heat generation of $\dot{q} = 100$ MW/m$^3$ inside the slab. The left and right faces of the slab are maintained at 150 °C and 110 °C, respectively. The plate has a constant thermal conductivity of 200 W/(m.K). Considering a 1-D steady state heat conduction, the location of the maximum temperature from the left face will be at _____ mm (*answer in integer*).

A slab 20 mm thick with a left face at 150 °C, a right face at 110 °C and a uniform heat generation of 100 MW/m$^3$ inside.

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Correct answer: 6

Explanation

For steady one-dimensional conduction with uniform generation, $\frac{d^2T}{dx^2} = -\frac{\dot{q}}{k}$, so $T = -\frac{\dot{q}}{2k}x^2 + C_1x + C_2$. At $x = 0$, $T = 150$, so $C_2 = 150$. At $x = L = 0.02$ m, $T = 110$: with $\frac{\dot{q}L^2}{2k} = \frac{10^8 \times 0.0004}{400} = 100$, $-100 + 0.02C_1 + 150 = 110$, so $C_1 = 3000$ K/m. The maximum is where $\frac{dT}{dx} = -\frac{\dot{q}}{k}x + C_1 = 0$, so $x = \frac{kC_1}{\dot{q}} = \frac{200 \times 3000}{10^8} = 0.006$ m, which is 6 mm.