Resistance in a tube

[zebra]

How do I calculate the 'potential?' resistance in a tube, in cm H2O. I'm a physiotherapist and do positive expiratory pressure breathing exercises with lung patients. Problem, the PEP masks are quite expensive, so we make 'blow bottles' - simply a 1 litre bottle, half filled with water (about 10 cm), with a nice thick (about 1 cm diameter) silicone tubing of about 20 cm, through which the patient blows into the water to make it bubble. Another problem - blowing into the water retains the lung pathogens in the water where they can proliferate. I'd like to use the straw as it is to avoid a contamination source, but would like to know what resistance it will give during expiration. If I have a formula, I can use thinner straws or simply make them longer/shorter for +- specific requirements. (I realise the speed of expiration will also play a role in the resistance.) The resistance we're looking for is from 7 - 30 cm H2O.

Thanks!

[insane_alien]

well, its not likely to be that high, and your right, it does vary with the speed of exhalation so i'd suggest putting a small clamp on the tube and adjust as necessary.

[CaptainPanic]

This is easiest done experimentally.

 

The calculations of the resistance ("pressure drop" is a better term) on a straw with a clamp, and most likely some corners aren't very easy. In addition, the pressure drop is dependent on how hard you blow.

 

What's important is the Reynolds number.

Let's see:

Flow = 2 liters / s = 0.002 m3/s - that's just an estimated value, dependent on the patient.

Cross-sectional area of a 1 cm tube = 3.14*0.005^2 = 7.85E-5 m2

Therefore the velocity in the straw is 25 m/s.

 

Reynolds (Re) = density*velocity*diameter / viscosity = 1.2 * 25 * 0.01 / 0.001 = 300

 

That should be a laminar flow.

The thinner straw has a flow 4 times higher since it's about twice as thin. That makes the flow turbulent (flows generally become turbulent at about Re > 1000). That will greatly increase the "resistance" (pressure drop).

 

I am sorry that I am not in the mood to calculate the pressure drop in a theoretical case. Perhaps later (but no guarantees).