Project ForgeConsider steady state, one-dimensional heat conduction in an infinite slab of thickness 2L (L=1 m) as shown in the figure. The conductivity (k) of the material varies with temperature as k=CT, where T is the temperature in K, and C is a constant equal to 2 W·m⁻¹·K⁻². There is a uniform heat generation of 1280 kW/m³ in the slab. If both faces of the slab are maintained at 600 K, then the temperature at x=0 is ________ K (in integer).
Explanation
The governing equation d/dx(k dT/dx)+q̇=0 with k=CT becomes C·d/dx(T dT/dx)+q̇=0, i.e. (C/2)d²(T²)/dx²=−q̇. Substituting φ=T² linearizes this to d²φ/dx²=−2q̇/C=constant. By symmetry (uniform q̇, identical boundary temperatures), integrating gives φ(x)=−(q̇/C)x²+B. At x=L: 600²=−(q̇/C)L²+B, so B=360000+(q̇/C)L²=360000+640000=1,000,000 (using q̇/C=1,280,000/2=640,000). At x=0, φ(0)=B, so T(0)=√1,000,000=1000 K.
Source: GATE ME 2022 Session 2 (ME-2) (2022)
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