A sheet of thickness 10 cm has it's one surface maintained at 400

SOLUTION

Concept:

One dimensional heat conduction with temperature dependent thermal conductivity.

The conduction equation is given by

\(\frac{d}{{dx}}\left( {k\frac{{dT}}{{dx}}} \right) = 0\)

Integrating once gives

\(k\frac{{dT}}{{dx}} = {C_1}\)

Now k = a + b T

⇒ (a + b T) dT = C₁ dx

Integrating again →  \(aT + \frac{{b{T^2}}}{2} = {C_1}x + {C_2}\) 

∵ at x = 0; T = T₁ and at x = L; T = T₂;

After solving for boundary conditions,

\({C_2} = a{T_1} + \frac{{bT_1^2}}{2}\;;{C_1} = a\left( {\frac{{{T_2} - {T_1}}}{L}} \right) + b\left( {\frac{{T_2^2 - T_1^2}}{{2L}}} \right)\)

Now heat transfer is given by

\({Q_x} = - kA\frac{{dT}}{{dx}} = - A{C_1}\)

\( \Rightarrow {Q_x} = A\left[ {a\left( {\frac{{{T_1} - {T_2}}}{L}} \right) + b\left( {\frac{{T_1^2 - T_2^2}}{{2L}}} \right)} \right]\)      --- (1)

Average thermal conductivity is given by

\({k_{avg}} = \frac{{k\left( {{T_1}} \right) + k\left( {{T_2}} \right)}}{2} = \frac{{a + b{T_1} + a + b{T_2}}}{2}\)

\(\Rightarrow {k_{avg}} = a + \frac{b}{2}\left( {{T_1} + {T_2}} \right)\)      --- (2)

Schematic:

Calculation:

Given:

L = 10 cm = 0.1 m; A = 2 × 3 = 6 m²; a = 0.2 W/m K; b = 6 × 10^-4 W/m K²;

T₁ = 400 K; T₂ = 300 K;

Heat flow through the sheet is given by equation 1

\({Q_x} = 6\left[ {0.2\left( {\frac{{400 - 300}}{{0.1}}} \right) + 6 \times {{10}^{ - 4}}\left( {\frac{{{{400}^2} - {{300}^2}}}{{2 \times 0.1}}} \right)} \right] = 2460\;W\) (Option 1)

Average thermal conductivity is given by equation 2

\({k_{avg}} = 0.2 + \frac{{6 \times {{10}^{ - 4}}}}{2}\left( {400 + 300} \right) = 0.41\;W/\;m\;K\) (Option 2)

If average thermal conductivity is taken as the constant thermal conductivity throughout, the heat transfer will be

\({Q_x} = {k_{avg}}A\frac{{{\rm{\Delta T}}\;}}{L} = 0.41 \times 6 \times \frac{{400 - 300}}{{0.1}} = 2460\;W\) (Option 3)