1
$\begingroup$

Hey guys I need some help with a fluid mechanics problem.

I have a capillary with the given dimensions $d=0,15 mm$ and $L=5 cm$ and the operating conditions $p_{1}=4\cdot10^{5} Pa$, $p_{2}=0.25\cdot10^{5} Pa$

I want to calculate the velocity of oxygen flowing from 1 to 2. So I start with the formula for pressure drop with laminar flow

$\Delta p=\zeta \frac{L}{d}\frac{\rho u^{2}}{2}$ with $\zeta =\frac{64}{Re}$

which gets me $u=261 \frac{m}{s}$

After calculating Reynoldsnumber to check wether laminar assumption was right, I see that flow is turbulent ($Re=10175$)

So instead I have to take $\zeta =\frac{0.3164}{\sqrt[4]{Re}}$ (Blasius)

But with a little help of wolfram alpha

solve p=(0.3164/((udrho)/eta)^(1/4))l/d(rho*u^2)/2 for u

I get a velocity of $u=142.9 \frac{m}{s}$.

Does this make any sense? I'm really not sure if I can calulate the velocity like this.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ If those equations are the applicable ones, it should work. You'll have to make sure it's converging though. You need to find a velocity where the friction losses and Reynolds Number make equivalent equations. For example, your Re = 10175 was gotten by assuming u = 261 m/s. When you use that Re in the Blasius equation, then the pressure drop equation, you get a new velocity. That new velocity would have a new Reynolds number, giving you a new velocity from the Blasius equation/pressure drop formula. This becomes iterative; usually you would want a numerical method to solve this AFAIK. $\endgroup$
    – JMac
    Commented Nov 16, 2016 at 17:22
  • $\begingroup$ The Point is: I'm not sure if those equations are the applicable ones. I would like to know if there is a straight forward way to calculate turbulent flow through a pipe like there is with laminar flow. $\endgroup$
    – malleYay
    Commented Nov 16, 2016 at 17:41
  • $\begingroup$ With the information given that seems like the best method. I can't think of any way to solve that with the given info without estimating velocity; and as soon as you do that you're in an iteration loop. $\endgroup$
    – JMac
    Commented Nov 16, 2016 at 20:33
  • $\begingroup$ I don't see why you can't insert the equation for Reynolds number and use the viscosity for oxygen. This way there's no iteration. $\endgroup$
    – jjack
    Commented Nov 19, 2016 at 19:42
  • 1
    $\begingroup$ You have a sixteen fold pressure drop, so your specific volume will drop unless we're talking liauid oxygen ... & thus velocity will increase. There are iterative formulat that take that intop account, I'd jave to hunt them down somewhere. However, that makes the question: Where in your pipe is the velocity you seek? Also, 4 bar pressure, that's less then the inner tube of a racing bike. I don't find 146m/s plausible. $\endgroup$
    – mart
    Commented Apr 24, 2018 at 20:09

1 Answer 1

Reset to default
1
$\begingroup$

One of the assumptions necessary for your pipe-loss formula you used is that the flow is incompressible (M<0.3). You could conceivably break up the problem into smaller segments where the compressible fluid density doesn't vary significantly and approximate those as incompressible individually. The fluid velocity will be directly related to you gasses density and will necessarily be limited to M<1 within the capillary. You'll want to start with the density of oxygen at p1 and calculate the loss going from p1 to a point sufficiently close to p1 where the density does not change significantly, and then recalculate the new pressure, density. You can continue to do this with iterated mass flow until you get convergence with the total delta P.

Improve this answer
$\endgroup$
1
  • $\begingroup$ This makes a clear statement that gases are compressible and that equations for incompressible liquid flow are to be used only after checking the fundamentals. The limitation is not based on a speed value. It is instead based on assumptions that are taken within the energy balance equation as shown by example here. $\endgroup$
    – Jeffrey J Weimer
    Commented Oct 22, 2018 at 2:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.