Problem: A liquid is flowing downward along an inclined plane surface, as shown in the figure:
The free liquid surface (y = h) is maintained at temperature Th, and the solid surface (y = 0) is maintained at To. Determine an expression for the temperature distribution in the film, recognizing that the viscous heating effects can be ignored.
Given: non-isothermal film flow, T(0) = T0 and T(h) = Th
Find: An expression for the temperature distribution in the fluid film
Assumptions: steady, laminar, incompressible, Newtonian fluid, ignore viscous heating effects, constant thermal conductivity, assume T = f(y), constant properties
Solution:
Starting with the general form of the differential energy equation for an incompressible Newtonian fluid:
rho*cp*(d/dt(T) + u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -(d/dx(qx) + d/dy(qy) + d/dz(qz)) + mu*Phiv
Steady flow eliminates the first term on the left side; the last term on the right side is eliminated because we are ignoring viscous heating effects.
rho*cp*(u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -(d/dx(qx) + d/dy(qy) + d/dz(qz))
Our assumption of laminar flow means that v = w = 0; the first term on the left side is also zero since T is in not a function of x:
0 = (d/dx(qx) + d/dy(qy) + d/dz(qz))
The flux terms can be related to the temperature gradient using Fourier's law:
qx = k*d/dx(T), qy = k*d/dy(T), and qz = k*d/dz(T)
Furthermore, since the thermal conductivity (k) is constant, our energy equation becomes:
0 = k*(d/dx(d/dx(T)) + d/dy(d/dy(T)) + d/dz(d/dz(T)))
We can divide both sides by k; the first and third terms cancel since T is not a function of x or z:
0 = d/dy(d/dy(T))
If we separate and integrate this equation twice, we end up with the expression:
T(y) = c1*y + c2
We can solve for c1 and c2 by applying the given boundary conditions at 0 and h:
c2 = T0 and c1 = (Th - T0)/h
So we now know the expression that shows us the temperature distribution in the fluid film:
T(y) = (Th - T0)/h*y + T0
Not so bad, huh? Next week we'll begin anew with a fresh subject, but until then, happy studying!
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