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heat transfer - wetted perimeter vs. heated perimeter

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simchuck

Mechanical
Oct 8, 2003
8
In the design/analysis of a double-pipe heat exchanger, is it correct to use the "heated perimeter" for the annular flow, or should the "wetted perimter" be used?

For example, I have ethylene-glycol flowing in the annular space, with water flow in the inner pipe. I am looking to find the heat transfer coefficient on the outside of the inner pipe (i.e., the ethylene-glycol side).

I have the approach - determine Re, Pr and Nu, and use the Nusselt form (Nu=0.023 x Re^0.8 x Pr^0.3). Just not sure about which perimeter to use for the Re calculation.

Can anyone give me a brief explanation of when the use each is appropriate?
 
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In determining the reynolds number for the annulus, the hydraulic diameter comes from using the perimeter of the inside pipe and flow area between the two pipes. the equivalent diameter is equal to 4 times the hydraulic diameter divided by the wetted perimeter. A good book for heat transfer is "Process Heat Transfer" by Donald Kern.
 
There lies the confusion...

I can understand the logic behind using the perimeter of the inside pipe, aka, the "heated perimeter". But I have been unable to find any definitive and independent verification of that logic.

For example, two references make use of the "hydraulic perimeter", that is, the sum of the OD of the inner tube (ODi)and the ID of the outer tube (IDo)...
"Unit Operations of Chemical Engineering" McCabe and Smith, p.337-339
"Transport Processes and Unit Operations" Geankoplis, p.240-241

IDo is greater than ODi - lets assume equivalent flow areas in the annulus and the center tube, so ODi~1.4IDo. The net effect is a 40% higher result using the "wetted" perimeter versus that calculated using the "heated" perimeter.

So which is correct?
 
As far as I am given to understand, there is no logic behind it.
Substituting the equivalent diameter (=4*hydraulic radius=4*free flow area/heated or cooled wetted-perimeter) for the inner-tube outside diameter is an approximation accepted for heat transfer and pressure-drop calculations in annuli of small equivalent diameters D[sub]e[/sub]. This approach has been substantiated by experiments with annuli.

I don't have the McCabe and Smith or the Geankoplis books with me, thus I cannot comment. I, however, suspect that the use of the combined wetted perimeters or their average, is a confirmed result of experimentation.

Since the HTC is generally proportional to Re[sup]0.5[/sup], from the Nusselt number it would become proportional to D[sub]e[/sub][sup]-0.5[/sup].

For concentric tubes when the inner diameter of the external tube is twice the external diameter of the inner tube d, D[sub]e[/sub]=3*d. This would mean that substituting D[sub]e[/sub] for d provides a conservative approach for estimating the HTC.

Now, let's enlarge the outer tube and make it a S&T HE. When calculating the shell-side convection HTC in S&T baffled HE, the same approach applies. For example, in a square pitch (p) arrangement of tubes with outside diameter d, and pi=3.14, the equivalent diameter is estimated as

4(p[sup]2[/sup]-(3.14)*d[sup]2[/sup]/4)/(3.14d)


On the other hand, both the shell diameter and the equivalent diameter (as above) play a role in estimating the shell-side friction drop. Please comment.
 
Per Applied Chemical Process Design by Aerstin and Street, the wetted perimeter is used for the pressure drop and the heated perimeter is used for the heat transfer using the outer circumference of the inner pipe. See page 128, Section 8.6: Double Pipe Exchangers.
 
Hi simchuck,

In the case of a double pipe heat exchangers, you'll need two distinct equivalent diameters:

- for heat transfer : Deq = (Di^2 - do^2)/do.

- for pressure drop : Deq = Di - do.

I would mention that the correlated given here (sieder and tate) is valid only for turbulent flow.In addition, the value obtained should be multiplied by the viscosity correction term and the ratio (di/do).

where di, do: internal and external tube diameters.
Di : internal pipe diameter

For more in-depth information see, "Process heat transfer" by Kern. There are useful detailed examples on that issue.

Hope this help,

zerokool

process engineer
 
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