A plane wall, 7.5 cm thick, generates heat internally at the rate of 105W/m3. One side of the wall is insulated and the other side is exposed to fluid with T[infinity]= 90 °C and h= 500 W/m2-K. If the thermal conductivity of the wall is 12 W/(m K), calculate the temperature in the center of the wall.

Now if we integrate both sides w.r.t x, one derivative is removed from left side of equation and and a variable is multiplied to the term on the right side and a unknown integrating constant C is added.

[tex]\frac{dT}{dx}[/tex]= - ([tex]q_{s}[/tex]/k)*x + C -------------------- equation (1)

Now if we apply Dirichlet's first boundary equation at x = 0, dT/dx = 0

we get,

[tex]\frac{dT}{dx}[/tex]= - ([tex]q_{s}[/tex]/k)*x + C

0 = - ([tex]q_{s}[/tex]/k)*0 + C

C = 0

Now put C = 0 in equation 1

we get,

k [tex]\frac{dT}{dx}[/tex]= - [tex]q_{s}[/tex]*x ------------------ equation (2)

Now integrate it w.r.t x

T= - ([tex]q_{s}[/tex]/2k)*x² + C₁ ------------------ equation (3)

Now apply Dirichlet's second boundary condition on equation 2, i.e, x=L, k[tex]\frac{dT}{dx}[/tex] = h⁻(T-T₀), equation 2 becomes

h⁻(T-T₀) = L[tex]q_{s}[/tex] ------------------ equation (4)

Put value of T from equation 3 in equation 4

h⁻(- ([tex]q_{s}[/tex]/2k)*L² + C₁-T₀) = L[tex]q_{s}[/tex]

By rearranging it we get,

C₁= T₀ + [tex]q_{s}[/tex]L[(L/2k)+(1/h⁻)] ------------------ equation (5)

Now putting value of C₁ from equation 5 in equation 3, we get

T = T₀ + ([tex]q_{s}[/tex]/2k)[L² + 2kL/h⁻]

Now by putting the given values of T₀ = 90°C, [tex]q_{s}[/tex] = 10 W/m², k = 12 W/mK, L = 0.075m, h⁻ = 500 W/m²K

we get, T= 128°C