!Counter flow concentric tube heat exchger! Incroperia Intro Heat 5th Ex 11.1

[Anonymous]

Since there's not an exact AVE that I could find:

Of interest is the validation of counter flow concentric tube heat exchanger. Just to see if the AD SIM results would yield the problem's heat flow rate. Problem modifications from original due to long mesh times and to eliminate turbulence: Water O.D.: 0.90 meters. Oil O.D.: 1 meter. This gives a length of 37.50 meters.

Data

Water enters internal tube at 30 C at 0.2 kg/s. Oil enters annulus tube from opposite end at 100 C at 0.1 kg/s. Oil leaves at 60 C.

Theoretical Solution

Heat transfer rate: 8524 Watts  (See Incroperia Intro to Heat Transfer 5th Example 11.1 pp. 642-644)

BCs

Design scenario 1: Steady fluid flow (water only) with Fan Surface, external intake at one end, prescribed inlet/outlet other end. Mesh: tets and wedges. LES turb. (Looks fine, no problem there, smoothing off, "SUM" flow rate on inlet/outlet faces, really close)

Design scenario 2: Steady fluid flow (oil only) with Fan Surface, external intake at one end of annulus, prescribed inlet/outlet other end of the annluus. Mesh: tets and wedges. LES turb.  (Looks fine, no problem there, smoothing off, "SUM" flow rate on inlet/outlet faces, really close)

Design scenario 2: Steady heat transfer

Fluid flow loads from Design Scenarios 1. & 2.

Surface applied temperature of 30 C applied to one end (water) and 100 C applied to other end (oil annulus).

Results, with smoothing OFF, there is no combination of surface selecting that yields anything close to 8524 J/s (8524 W)

[John_Holtz]

Hi jrm,

A good starting place is "Help > Autodesk Simulation > Examples > Fluid Flow > Heat Exchanger with Fluid Flow" which compares Simulation to results from a published paper. It sounds very similar to what you are doing.

Secondly, something seems wrong with the example. 3 gal/min (0.2 kg/s) in a 35 inch dia (0.9 m OD) pipe seems to be very slow. Did I do the conversion properly? Perhaps this is molasses instead of water ;-).

And finally, a general suggestion, for you and everyone else. For situations such as this where the results do not match your expectations, save yourself a lot of time. Instead of describing the problem in great detail with images too small to be legible, just attach

1. the expected result (scan of theoretical case, hand calculations, etc)
2. an archive of the model (see "Create, Post, or Provide an Archive of your model").

It is really difficult to determine what is being done without having the model to inspect.

Thanks.

[Anonymous]

John,

You're right about the flow rate being low for such a large diameter. I've adjusted it.

Data (modified from original problem since length:diameter ratio had difficulty meshing)

Counterflow concentric tube heat exchanger. Water on the inside, oil on the outside.

Water enters 400 mm pipe at 0.024 kg/s (381 gpm), 30 C, Cp= 4178  J/kg K, mu = .000803 N s/m^2 K, k = 0.617 W/m K, Pr = 4.95

Oil enters annulus from opposite end, ID = 400 mm, OD = 420 mm at 0.1 kg/s (1.85 gpm), 100 C, leaves at 60 C, Cp= 2131  J/kg K, mu = .0325 N s/m^2 K, k = 0.138 W/m K

The Steady State Heat Transfer is not close to the theoretical Heat Rate of Face at 8524 Watts. The Transient, with smoothing off, and Sum on ALL the internal surfaces of the oil, gives -8523 Watts at load step 4.

Not sure why the Steady State is not returning an answer close to the theoretical. Also, not sure how to retrieve the exit temperature of the oil (should be 60 C). The surface was selected, with smoothing off, temperature activated from results, then subentities (nodes). It is way off in both steady and transient

See attached

file

http://www.mediafire.com/?glpyfnbgq3in8a8

[Anonymous]

See file for hand calcs

[John_Holtz]

Hi Russ,

My co-worker has a previous version of the Incropera and DeWitt book, but it looks like the example is the same. I have attached it in case others are interested in deciphering your calculations.

1. Have you confirmed that all of the equations used are valid over the range of your example? The Reynolds number of 5 for the oil concerns me.
2. Does Simulation work over this range? I did not look at the Help documentation closely. I suggest you start with a simple laminar flow calculation in a single pipe to check whether the governing FEA (Finite Element Analysis) is valid at such low flows.
3. I noticed that the calculated water temperature rise is only 0.09 degrees. Because it is such a small change, I think you will need a very accurate model. You may need a hand created mesh to get to this level of accuracy.
4. Although the outer tube has nodes near the wall (in the boundary layer), there are no nodes near the center of the flow. What I have done in the past is spreadout the boundary layer mesh so that you get nodes approaching the "midplane" of the tube.
5. I suggest using 3 parts along the length (6 parts total), One reason the heat transfer results are inaccurate is because the fluid flow results are inaccurate (mostly at the inlet, but also at the outlet in some cases). The extra parts are to let the flows develop. In the heat transfer analysis, the inlet parts are suppressed and you are left with a section of pipe equal to the calculated length with a good fluid flow result. See below.
6. Is the calculated length 4 m or 2.74 m?
7. Finally, make sure the fluid flow results are converged. When I ran the model with the water, I was getting warnings about stagnation. So it is tricky to say what effect those had.

[Joey.X]

John puts a good heat exchange example and illustrations. I agree to have six parts in simulation and use the result from middle part for validation. However, I don't agree the statement of "heat transfer results are inaccurate is because the fluid flow results are inaccurate". At Reynolds Number=5 (sorry, I directly take this value from above posts and didn't verify it), it's definitely laminar internal flow, fluid simulation results are very accurate even in coarse mesh density for velocity profile and pressure drop along the flow direction, this could be referred from simulation AVEs.

The reason of inaccuracy in this conjugated heat transfer example comes from 

(a) To validate ideal results which are based on fully developed flow, notes that the vicinity of inlet and vicinity of outlet are not fully developed flow because of the enforced inlet and outlet boundary conditions; this is nature for any fluid flow simulation but user needs to know.

So using the middle parts for validation is reasonable if the middle parts are fully developed flow (user needs to specify proper length of inlet parts and outlet parts).

(b) The mesh density of middle part highly impacts the heat transfer coefficient from fluid flow, since the heat transfer coefficient is related to the velocity gradient on the wall. From this point of view, the boundary layer mesh is suggested to catch the proper heat transfer coefficient if user wants accurate benchmark on this kind of examples.

There are possibly other issues, the above are the fundamental ones in model setup phase. 

[Anonymous]

John,

Just to confirm, the problem in Incroperia 3rd is the same as 5th edition. I modified it since the diameter:length ratio was cumbersome to mesh. I started with the original and tweaked it from there. I will address the post concerns regarding Reynolds number and temperature. Meanwhile.

[JH] Although the outer tube has nodes near the wall (in the boundary layer), there are no nodes near the center of the flow. What I have done in the past is spreadout the boundary layer mesh so that you get nodes approaching the "midplane" of the tube. [jrm_1971] Just to confirm, this is done by increasing the number of boundary layers/total extrusion distance for the tet/wedges mesh?

[JH] I suggest using 3 parts along the length (6 parts total), One reason the heat transfer results are inaccurate is because the fluid flow results are inaccurate (mostly at the inlet, but also at the outlet in some cases). The extra parts are to let the flows develop. In the heat transfer analysis, the inlet parts are suppressed and you are left with a section of pipe equal to the calculated length with a good fluid flow result. See below.  [jrm_1971] How long should the temporary parts be? Is this necessary if the one part (2 parts total, 1 water, 1 oil) are sufficiently long enough?

[JH] Is the calculated length 4 m or 2.74 m?  [jrm_1971] The modified problem yields a length of 4 m. This will change (probably longer) as the oil Reynolds no. and water dT problems are addressed.

[JH] Finally, make sure the fluid flow results are converged. When I ran the model with the water, I was getting warnings about stagnation. So it is tricky to say what effect those had.  [jrm_1971] Looking back at the results summary, I get the stagnation warnings too. How do those go away?