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ECH0140-HW4 2009 - up The PDE is ∂ T ∂ t = α ∂ 2 T...

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Fall 2008 University of California at Davis Department of Chemical Engineering and Materials Science ECH 140 HW #4 Problems (10 points unless otherwise indicated) #1) 20 points. Solve the ODE: y’’ - y’ - 2y = 10 cos x #2) Relate the term x i / x j to the Kronecker delta. #3) 30 points. Consider a slab ( l thick), covered on one side with perfect thermal insulation. The slab is at uniform temperature T o at times smaller or equal to than zero. At times greater than zero, the other side of the wall is exposed to a fluid at temperature T . The heat transfer from the fluid to the wall is slow enough that the fluid resistance plays a role in the rate that the wall gets heated
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Unformatted text preview: up. The PDE is: ( ∂ T/ ∂ t) = α ( ∂ 2 T/ ∂ x 2 ) where the LHS term accounts for the time dependency, the RHS term for heat diffusion in the slab and α is the thermal diffusivity of the slab material (units cm 2 /s). We have the following BCs, after placing the coordinate system at the insulated side of the slab: IC) t< 0, T=T o , all x BC1) all t, ( ∂ T/ ∂ x)=0, x=0 BC2) t>0, -k ( ∂ T/ ∂ x)= h(T- T ∞ ), x= l Here k is the thermal conductivity of the slab and h is the heat transfer coefficient for the fluid. Using separation of variables, obtain the solution for T....
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