Hagen-Poiseuille vs Bernoulli

In summary: 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.
  • #1
bjarke
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I am trying to calculate the exit velocity of a nozzle (0,25Ø) that is connected to a high pressure syringe (10 BAR), however I cannot understand why Hagen-Poiseuille will have a higher exit velocity then Bernoulli when HP take viscosity into account.

Screenshot-2021-04-12-155018.png
 
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  • #2
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  • #3
Your H-P computation assumes constant velocity. Your Bernoulli computation assumes acceleration from 0 to 45m/s. So your Bernoulli computation also takes the acceleration of mass into account.
 
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  • #4
Actually, I get much more dramatic values for H-P:
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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  • #5
Arjan82 said:
Q = 1.9e+4 m3/s,
seems like the math needs checking :wink:

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  • #6
millimeters, not meters... 🙈... nvm, I cannot brain today...
 
  • #7
BvU said:
Hello @bjarke ,
:welcome: !​

Hagen-Poiseuille is for laminar flow. Check the Reynolds number ...

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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
This is the same problem, but expanded for my entire system.
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

Screenshot 2021-04-13 094913.png


Screenshot 2021-04-13 094933.png
 
  • #9
bjarke said:
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.
No, H-P assumes laminar flow indeed. Bernoulli assumes inviscid flow (i.e. no friction at all)
 
  • #10
This problem calls for the use of the version of the Bernoulli equation which includes the frictional energy dissipation. Both the acceleration and the frictional drag are going to contribute to the overall pressure drop. The main frictional resistance is going to reside in the 1 mm tube rather than the nozzle. And the flow in the tube is going to be turbulent.

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.
 
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  • #11
Chestermiller said:
This problem calls for the use of the version of the Bernoulli equation which includes the frictional energy dissipation. Both the acceleration and the frictional drag are going to contribute to the overall pressure drop. The main frictional resistance is going to reside in the 1 mm tube rather than the nozzle. And the flow in the tube is going to be turbulent.

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.
That sounds much more reasonable than the 20 bar I got. What did you equation look like?
 
  • #12
bjarke said:
That sounds much more reasonable than the 20 bar I got. What did you equation look like?
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.
 

Related to Hagen-Poiseuille vs Bernoulli

What is the difference between Hagen-Poiseuille and Bernoulli's equations?

Hagen-Poiseuille's equation describes the flow of a viscous fluid through a cylindrical pipe, taking into account factors such as viscosity and pipe diameter. Bernoulli's equation, on the other hand, describes the relationship between the velocity, pressure, and height of a fluid in a closed system.

Which equation is used to calculate the flow rate of a fluid in a pipe?

The Hagen-Poiseuille equation is used to calculate the flow rate of a fluid in a pipe, taking into account factors such as viscosity and pipe diameter. Bernoulli's equation is used to describe the relationship between the velocity, pressure, and height of a fluid in a closed system.

Can Hagen-Poiseuille and Bernoulli's equations be used together?

Yes, these equations can be used together to fully describe the flow of a fluid in a pipe. Hagen-Poiseuille's equation can be used to calculate the flow rate, while Bernoulli's equation can be used to calculate the pressure and velocity at different points in the system.

What are the assumptions made in Hagen-Poiseuille and Bernoulli's equations?

Hagen-Poiseuille's equation assumes that the fluid is incompressible, laminar, and flowing through a cylindrical pipe. Bernoulli's equation assumes that the fluid is incompressible, inviscid, and flowing through a closed system with no energy losses.

How do Hagen-Poiseuille and Bernoulli's equations relate to fluid dynamics?

Both Hagen-Poiseuille and Bernoulli's equations are fundamental equations in fluid dynamics, used to describe the behavior of fluids in different systems. They are often used together to fully understand and analyze the flow of fluids in various applications, such as in pipes, pumps, and airplanes.

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