Answer:
[tex]h=1.99998\ W/m^2.C[/tex]
[tex]k=33.333\ W/m.C[/tex]
Explanation:
Considering the one dimensional and steady state:
From Heat Conduction equation considering the above assumption:
[tex]\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0[/tex] Eq (1)
Where:
k is thermal Conductivity
[tex]\dot e_{gen}[/tex] is uniform thermal generation
[tex]T(x) = a(L^2-x^2)+b[/tex]
[tex]\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a[/tex]
Putt in Eq (1):
[tex]-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C[/tex]
Energy balance is given by:
[tex]Q_{convection}=Q_{conduction}[/tex]
[tex]h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L[/tex] Eq (2)
[tex]T(x) = a(L^2-x^2)+b[/tex]
Putting x=L
[tex]T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC[/tex]
[tex]\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2[/tex]
From Eq (2)
[tex]h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C[/tex]
