# Fluid flow out of pressurized vessel

1. Aug 6, 2014

### koch15

I have a non-rigid vessel holding a fluid with a tube for fluid flow out. It is initially at equilibrium pressure with the atmosphere. (Think a water bladder for a camelback hydration system) A pressure is then applied to the bag to cause fluid to flow out the tube. Is there a way to relate the fluid flow rate to the pressure applied?

Thanks

2. Aug 8, 2014

### jlefevre76

Yes, there is. The pressure drop over the tube is known (pressure in the bladder/vessel minus ambient pressure at the end of the hose). Knowing that pressure drop, you know is has to be equal to the pressure drop due to viscous forces acting inside the tube.

Note, it changes whether you are in the laminar or turbulent regime. It is very difficult to calculate if the length is equal to or shorter than the development length.

$Re=\frac{\rho D V_{avg}}{\mu}$

Laminar: Re < 2300
L_developing = 0.03*Re*D
$Q=\frac{-\pi R^{4}}{8 \mu}\frac{\Delta P}{L}$
(ΔP is pressure drop, L is length of tube, Q is volume flow rate, or V_avg*A_cross sectional).

Source: http://faculty.poly.edu/~rlevicky/Handout12_6333.pdf

Turbulent: Re > 2300
L_developing ≈ 10*D
$\frac{1}{\sqrt{\lambda}}=-2LOG\left(\frac{2.51}{Re\sqrt{\lambda}}+0.269\frac{k}{D}\right)$
$\Delta P=\lambda \frac{L}{D}\frac{\rho}{2}V_{avg}^{2}$
(ΔP is pressure drop, L is length of tube, λ is the friction factor, k is the absolute roughness of the tube where 0 is smooth).

Source: http://www.engineersedge.com/fluid_flow/pressure_drop/pressure_drop.htm

Note: It is iterative, and you will have to guess a value for Re, calculate the friction factor, calculate a new velocity, calculate a new Re, until your solution has converged.

Last edited: Aug 8, 2014