A very large vertical steel plate that is 1 cm thick is inductively heated with internal heat generation rate of 2x10^9 W/m3 . The plate is initially at 20°C, and the left and right surfaces are kept at 20°C throughout the heating process. You are asked to find the steady state temperature distribution through the plate thickness. For steel, lambda=64.6 W/(m°K), C=481 J/(kg°K), and rho=7800 kg/m3 .
a) Simplify the 1D conduction equations with internal heat generation rate for steady state.
b) Solve the differential equation and find an equation for the steady state temperature distribution through the thickness of the plate.
3) Consider the steel plate in Question 2.
a) Write the steady state conduction equation from Question 2a in finite difference form.
b) Divide the plate thickness into 11 nodes (numbered 1 to 11) that are spaced (delta)x=1 mm apart. Using the finite difference equation from Part a write equations for nodes 2 and 3
c) Write in matrix form a system of 11 equations for all of the nodes. For the boundary conditions at nodes 1 and 11, the equations will simply be (Theat)1= 20°C and (theta)11= 20°C.