3 A plane wall has constant isotropic properties no internal

3. A plane wall has constant isotropic properties, no internal heat generation, and is initially at a uniform temperature of 25 oC. Suddenly at time t-0 the surface at x Lis exposed to a fluid at 100 °C and the surface at x L is exposed to a fluid at 10 oC, both of which and have convection coefficient h. a) Starting from the general, unsimplified 3D conduction equation, derive the appropriate differential equation which can be solved to determine the steady-state temperature distribution within the wall. Solve the differential equation to find an expression for the steady-

Solution

The eqns of 3 d can be reduced to the 1D steady state eqn, namely Kd²T/dx² =0

The initial temp 25 gets evened out and does not appear in the final aswer. What is important is the flux from the outside wall (RHS) to inside, and the inside to the LHS wall and outward flow to the air.

the flux or thermal gradient at the RHS , assuminhg h is the convection coeff, K the thermal conductivity, and T the temperature:

K( Ts1 - Ts2)/2L = h( 100-Ts1)

RHS equilibriums

h( Ts2-10) = K ( Ts1-Ts2)/2L (LHS equilibrium)

solve for Ts1, Ts2

get 100-Ts1= Ts2-10, so Ts1+Ts2 =90

100-Ts1= K/(2Lh) { Ts1- (90-Ts1)}

Ts1 = [100+90K/(2Lh)]/[ 1+ K/Lh]

Ts2 =90-Ts1

Soln of Kd²T/dx² =0 means T = Ax +B

T (lhs) = Ts2, T(Rhs) =Ts1

get A and B , by adding and subtracting: B = Ts1+Ts2/2 = 90/2 =45

A= (Ts1-45 )/L