Heat Transfer Through a Laminated Round Pipe or Cylinder
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This is treated similar to the flat plate with an important physical exception. The flat plate has equal surface areas on the hot and cool sides.... however each layer of a pipe has a larger outer surface area than its inner surface area. This fact translates into a different equation for the overall heat transfer coefficient.... there are now log terms in the equation. Text section 3.4 treats this problem very well.
The equation for the overall heat transfer coefficient of the
pipe (or cylinder) is slightly different in various textbooks, depending on the way the surface area is referenced. One favorite version for U is
U = 1 / [ ( 1/h1 ) + ( r1/k1 ) *ln ( r2/r1 ) + ( r1/k2 ) *ln ( r3/r2 ) + ( r1/k3 ) *ln ( r4/r3 ) + ( r1/r4 ) * ( 1/h2 ) ]
This example is for a 3-layer pipe. Whatever the number of layers, the last term must always have this form:
( rinner / router ) * ( 1 / houter )
It is important to know that in this setup, the reference area, A, to be used in the heat transfer calculation is the inside surface area of the pipe = 2 * pi * r1 * L.
Also, in practice, most engineers
calculate q/L the heat transfer rate per unit of pipe length, and using the above equations, this becomes...
q/L = U * 2 * pi * r1 * (Tinside - Toutside)
An alternate approach is to define the total thermal resistance for the laminated pipe, Rtotal:
Rtotal= 1 / ( h1* A1 ) + ln ( r2 / r1 ) / (2* Pi* L* k1) + ln ( r3 / r2 ) / (2* Pi* L* k2)+ ln ( r4 / r3 ) / (2* Pi* L* k3) + 1 / ( h2 * A4 )
(And again, this example is for a 3-layer pipe.)
So the heat transfer rate for the total length of pipe becomes...
q = (Tinside - Toutside ) / Rtotal
This approach does not use the inside surface area as the reference area, but keeps the outer aurface area A4 in the equation as well. That's fine .... depends on how you want to set it up..... decide which approach works best for you. It comes down to working with either the overall heat transfer coefficient, U or the total thermal resistance, Rtotal.
Note that if Tinside > Toutside, then q/L
> 0 and heat flows out of the pipe.
If Tinside < Toutside then q/L < 0 and heat flows into the pipe.
The equations for the individual
layer temperatures become...
T1 = Tinner - q/L * [ 1 / (2 * pi * r1 * h1) ]
T2 = T1 - q/L * [ ln(r2/r1) / (2 * pi * k1)]
T3 = T2 - q/L * [ ln(r3/r2) / (2 * pi * k2) ]
T4 = T3 - q/L * [ ln(r4/r3) / (2 * pi * k3) ]
As a check, you should get the same answer for the outer surface T4 from this
equation...
T4 = Toutside + q/L * [ 1 / (2 * pi * r4 *
h2) ]
It is important to note that this approach assumes constant internal and external temperatures and H's along the length of the pipe.