THANKS! A plane wall has a thickness 21, and eross-sectional area A. At steady s

Question

THANKS!

A plane wall has a thickness 21, and eross-sectional area A. At steady state conditions, volumetric heat generation occurs within the wall according to:q = q0 cos(ar) Where (IQ is the heat generated per unit volume at the center of the wall (x = 0) and a is a constant. If both sides of the wall are maintained at a constant temperature of Tw, derive and expression for the temperature within the wall as a function of location, x. How much heat is lost through the wall at x = -L, and x = U!

Explanation / Answer

d^2T/dx + q0*Cos(ax)/k = 0

by solving this we'll get, T = -q0*cos(ax)/a^2*k + A*x*q0/k + B*q0/k

boundary conditions are at x = 0 , q = q0 which will get us B, and at x= L , T = Tw which will get us A.

by putting the value of x=0 we can get temperature (at x=0) suppose T comes T0.

Now heat loss at x = L , Q = K*A*(T0-Tw) + q0*Cos(aL)
Heat loss at x = -L, Q = K*A*(Tw-T0) + q0*Cos(a(-L))

I hope you'll be able to solve now.

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