How Fast Can Hydrogen Travel Through a 1/16 Tube at 50 PSI?

AI Thread Summary
Hydrogen can be pushed through a 1/16" tube at a maximum pressure difference of 50 PSI, with an initial velocity calculation yielding approximately 108.5 ft/sec. To determine the Reynolds number, the formula RE = (ρVD)/μ is used, resulting in a value of 2292, indicating laminar flow. The friction factor can be approximated as f = 64/RE, leading to a refined friction factor of 0.0279 after iterations. Key assumptions include neglecting compressibility, a tube length of 12 inches, and a smooth surface. The discussion emphasizes the iterative nature of calculating flow characteristics in such scenarios.
xJuggleboy
How fast can Hydrogen be pushed threw a 1/16" tube? we have a maximun pressure diffrence of about 50PSI from one side of the tube to the other.

And how can I calculate the reynolds number for the tube?
 
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How long is the tube?

You're going to have to assume a friction factor and calculate the velocity using

\Delta P = f \frac{L}{D} \frac{1}{2} \rho V^2

Once you calculate a first velocity, calculate the resulting Reynolds number using RE = \frac{\rho V D}{\mu}

Once you have your Reynolds Number, then go to the Moody chart and see what the resultant friction factor is and compare it to your original estimate. If it is close you can stop. If not, get the new friction factor and then go back and redo the entire process with the new friction factor. It may take a few iterations before the friction factors and Reynolds numbers converge to a solution.
 
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I made some assumptions:
- Compressibility neglegted
- T= 70°F
- smooth surface roughness in the tube
- I used the 1/16" as the ID (because I'm too lazy to look up the ID, that's why).
- I assumed a length of 12 inches for the tube
- The tube is straight

Under those conditions, I calculated

\rho_{H2} = .0177 \frac{Lb_m}{ft^3}

and

\mu_{H2} = 4.367x10^{-6} \frac{Lb_m}{ft*s}


I took an initial swag of f=.03 and ran the numbers. If I didn't fat finger the numbers, I get a V = 108.5 ft/sec. That resulted in a RE = 2292 and in the laminar flow regime. That means you could get away with the approximation of f = \frac{64}{RE} which gives you a new f = .0279.

That should get you going in the general area. I am sure someone will let me know if my numbers are off. By feel, I think they're not too unreasonable...
 
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When the pressure difference is P1/P2 > 2 than the line will be chocked and the local velocity will be the velocity of sound.
 
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