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Parallel flow pressure drop theory

Parallel flow pressure drop theory

Parallel flow pressure drop theory

(OP)
Help me with this theoretical question regarding pressure drop in parallel flow. Refer to the JPEG attached for this example. http://files.engineering.com/download.aspx?folder=936838cc-1cfb-45f2-913f-cbc0828f265a&file=ParallelFlowPressureDrop.jpg

I have a smooth duct, 20 mm high by 60 mm wide by 1 m long. I have air flow through the duct at 4 m/s and calculate the pressure drop through that length of duct as being 10.4 Pa.

Now I add two infinitely thin walls in the duct that split the flow equally into 3 channels 20 mm wide. The velocity hasn't changed because the walls are thin. I calculate the pressure drop of each of the 3 channels as 17.5 Pa. When I take the inverse sum of the inverses (parallel flow), I have a total pressure drop of 5.8 Pa.

I'm having difficulty trying to get a practical sense of what is going on here. How is the pressure drop going down just by adding baffles. It seems like the reverse should happen, since I'm adding frictional flow resistance.

Is my assumption of equal velocity between the duct with no baffles wrong? If so, how do I calculate the true velocity in the smaller channels?
 

RE: Parallel flow pressure drop theory

Why the inverse sum of the inverses? The pressure drop is 17.5 Pa and that is all.

Infinitely thin walls don't exist in real life, except in text books.

RE: Parallel flow pressure drop theory

quark is right.

For parallel fluid flow the pressure drop across the parallel runs should be the same. They would even be the same if the new channels were not equal size.

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