A plane slab of 100 mm thickness generates heat. It is observed that the temperature drop between the centre and its surface to be 50°C . If the thickness is increased to 20 cm the temperature difference will be

Concept:

Plane wall with uniform heat generation:

For one dimensional steady state condition

\(T = \frac{{{q_g}}}{{2k}}{x^2} + {C_1}x + {C_2}\)

Here at x = -L, T = Tω ,x = L, T = Tω

\(\left( {T - {T_\omega }} \right) = \frac{{{q_g}}}{{2k}}\left( {{L^2} - {x^2}} \right)\)

\({\left( {\frac{{dT}}{{dx}}} \right)_{x = 0}} = 0 ⇒ {\left( {T - {T_\omega }} \right)_{max}} = \frac{{{q_g}{L^2}}}{{2k}}\)

Calculation:

Initial thickness = 2L₁ = 100 mm;

Final thickness = 2L₂ = 20 cm = 200 mm;

(T - T₁)_max = 50°C;

From the above relation,

\(\frac {{\left( {T - {T_1 }} \right)_{max}}}{{\left( {T - {T_2 }} \right)_{max}}} = \frac {L_1^2}{L_2^2}\)

Substituting the values of L₁ and L₂, we get 

(T - T2)max = 4 × (T - T1)max 

⇒ (T - T2)max = 200°C