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bjarke

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- Thread starter bjarke
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bjarke

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- #2

BvU

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- #3

Arjan82

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Arjan82

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Q = 1.9e+4 m3/s, v = 390e3 m/s...

Ps: I said "Bernoulli *also* takes acceleration into account". Actually, Bernoulli has no friction of course so it *only* takes acceleration into account... What you would actually need is a combination of the two (and then you need to check Reynolds as @BvU suggested, otherwise check Moody's Diagram)

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BvU

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Arjan82

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millimeters, not meters... 🙈... nvm, I cannot brain today...

- #7

bjarke

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Yeah, with these speeds the Reynolds number is turbulant for both scenarios, but it is my understanding that both HP and bernoulli assumes laminar flow.

But the "pipe" in HP is more an orifice. If I halve the L in HP I would get twice the speed, which doesn't seem right either.

Is there some kind of Diameter to length ratio where HP is no longer accurate?

- #8

bjarke

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I use the Darcy-Weisbach equation to figure out what my pressure loss is in my tube, and then I iterate this pressure loss into my bernoulli speed to approximate an exit velocity. v=sqrt((2P-static-2P-headloss)/rho)

Does this exit velocity of 94m/s make any sense?

Another problem is that the pressure loss (P-DW) is very close to, and almost ends up being larger than the initial static pressure. So that if I decrease the tube diamater (D2) the loss will be greater, and the resulting velocity will end up being an imaginary number

- #9

Arjan82

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No, H-P assumes laminar flow indeed. Bernoulli assumes inviscid flow (i.e. no friction at all)Yeah, with these speeds the Reynolds number is turbulant for both scenarios, but it is my understanding that both HP and bernoulli assumes laminar flow.

- #10

Chestermiller

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As best I can estimate, the pressure drop in the 1 mm tube is going to be about 0.4 Bars, so most of the pressure drop is going to be due to the acceleration in the nozzle.

- #11

bjarke

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That sounds much more reasonable than the 20 bar I got. What did you equation look like?

As best I can estimate, the pressure drop in the 1 mm tube is going to be about 0.4 Bars, so most of the pressure drop is going to be due to the acceleration in the nozzle.

- #12

Chestermiller

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Well, I started with the 45 m/s you got using Bernoulli without friction. From that, I estimate a velocity of 45/16 = 2.8 m/s in the 1 mm tube. From that, I got a Re of about 2800 in the 1 mm tube, which is just beyond the HP region. I then used Darcy Weissbach to get the pressure drop in the tube of about 0.47 Bars. So the actual acceleration pressure change would only be about 9.5 Bars, rather than 10 Bars. So then, one could continue to iterate until the sum of the pressure drops due to friction plus acceleration is 10 Bars.That sounds much more reasonable than the 20 bar I got. What did you equation look like?

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