A plane wall of thickness 8 cm experiences uniform volumetric heat generation ( q ). One
surface of the wall is insulated, and the other surface is exposed to a fluid at T(infinite) = 20 degree with
convection heat transfer characterized by h=1000W/m2.K. The initial temperature distribution
on the wall is given as: T (x,0) = 300-0.110^4X^2 , where x is in meters. Suddenly, the
volumetric heat generation is deactivated, while convection heat transfer continues to occur at
x=L. If the wall properties are 7000kg/ m^3 ,Cp= 450J/ kg.K, k=108W/ m.K
(a) Determine the magnitude of the volumetric energy generation rate q associated with the
initial condition (t < 0).
(b) Calculate heat flux at the boundary exposed to the convection process at t=0, qx (L,0)
(c) Calculate the amount of energy removed from the wall per unit area (J/m2) by the fluid
stream as the wall cools from its initial steady state condition.

[tex]\dot{q}=-k\frac{d}{dx}(\frac{dT}{dx})\\\dot{q}=-108\frac{d}{dx}(\frac{d(300-0.1\times 10^4 x^2)}{dx})\dot{q}=-108\frac{d}{dx}(2\times (300-0.1\times 10^4 x))\\\dot{q}=-108\times 2\times (300-0.1\times 10^4 )\\\dot{q}=21.6 \times 10^6 W/m^3[/tex]

So the magnitude of the volumetric heat generation rate is [tex]21.6 \times 10^6 W/m^3[/tex]

Part b

[tex]q''_{conv}=h(T(L,0)-T_{\infty})[/tex]

Here

h is given as 1000W/m2K

T(L,0) is found by putting x=L in equation of T. L is given as 8cm=0.08m.

T_inf is the infinite temperature given as 20C or 293K.

So by substituting values in the equation.

[tex]q''_{conv}=h(T(L,0)-T_{\infty})\\q''_{conv}=1000((300-0.1\times 10^4 (L^2)-293)\\q''_{conv}=1000((300-0.1\times 10^4 (0.08^2)-293)\\q''_{conv}=1000((300-0.1\times 10^4 (0.0064)-293)\\q''_{conv}=1000((300-6.4-293))\\q''_{conv}=600 W/m^2[/tex]

So the convective heat flux is [tex]600 W/m^2[/tex].

Part c:

For energy balance

[tex]\dot{E_{in}}-\dot{E_{out}}=\dot{E_{st}}[/tex]

Here as

[tex]\dot{E_{in}}[/tex] is zero

So

[tex]-\dot{E_{out}}=\dot{E_{st}}[/tex]

It is given as

[tex]-\dot{E_{out}}=\dot{E_{st}}=\rho c_p(T_{final}-T_{\infty})-\rho c_p\int\limits^{x=L}_{x=0} ({T_{x,0}-T_{\infty}) \, dx[/tex]

Here as T_final and T_inf will be same so

Also ρ is given as 7000 kg/m3

cp=450 J/kg.K

[tex]\dot{E_{out}}=\rho c_p\int\limits^{x=L}_{x=0} {(T_{x,0}-T_{\infty}) \, dx}\\\dot{E_{out}}=7000 \times 450\int\limits^{x=L}_{x=0} {(300-0.1\times 10^4 x^2-293) \, dx}\\\dot{E_{out}}=3150000[300x-0.1\times 10^4 \frac{x^3}{3}-293x]_0^L\\\dot{E_{out}}=3150000[300(0.08)-0.1\times 10^4 \frac{(0.08)^3}{3}-293(0.08)]\\\dot{E_{out}}=3150000(0.3893)\\\dot{E_{out}}=1.2264 \times 10^7 J/m^2[/tex]

So the amount of energy removed per unit area is [tex]1.2264 \times 10^7 J/m^2[/tex]