# Calculating velocity in pipe flow

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 $\zeta =\frac{0.3164}{\sqrt{Re}}$ (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.

• 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.
– JMac
Nov 16, 2016 at 17:22
• 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. Nov 16, 2016 at 17:41
• 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.
– JMac
Nov 16, 2016 at 20:33
• 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. Nov 19, 2016 at 19:42
• 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.
– mart
Apr 24, 2018 at 20:09