Question

The steady-state temperature distribution in a 0.3 m thick plane wall is given by $T_x=600+2500x-12000x^2$ where T is in °C and x is in meter measured from the left surface of the wall. Assume 1D heat flow. Steady-state heat conduction occurs in the wall in x-direction the thermal conductivity of the wall material is 23.5 W/mK. The average temperature of the wall is

• A. 435°C
• B. 400°C
• C. 615°C
• D. 730.2°C

Answer 1

The correct option is C 615°C

$T_{avg}=\frac{1}{L}{\int }_0^L{T}_x{d}_x$

$T_{avg}=\frac{1}{L}{\int }_0^L(600+2500x-12000x^2)dx$

$T_{avg}=\frac{1}{0.3}{\int }_0^{0.3}(600+2500x-12000x^2)dx$

$T_{avg}={615}^{\circ }C$