I'm afraid that my comment below is so trivial and obvious that all it will prove is that I have not understood the question. Perhaps once you have seen where I have missed the point you can re-ask the question more clearly.
If you put the gas through a pipe line, then all else being equal you...
If the line is liquid-filled the transient will be gone in milliseconds when the block valve opens. The dangerous situation is if you have air (or other non-condensible gas) trapped in the line at 700 psig.
Here is a simple approximation that you could use as a sanity check. It would err on the conservative side so you may slightly oversize your pipes, but it would be a good check on the feasibility of the design.
You said that you looked into the Darcy-Weisbach formula but you don't want to have...
1. It appears that your discharge is at a higher elevation than the source, so I cannot see how a siphon can form.
2. The friction pressure drop from the high point to the discharge almost exactly matches the gain in static head between the same two points. This means that under normal...
In order to be able to offset the pressure loss in the up-leg against the pressure recovery in the down-leg you have to ensure that the pressure at the top of the down-leg remains above the vapor pressure of the liquid. There are basically 2 ways of doing this. You either install a back...
The whole question of entrance and exit pressure losses, resistance coefficients and pressure recovery is generally not well covered in the standard texts. In particular, the texts do not sufficiently emphasize that the pressure changes due to friction and Bernoulli effects must be considered...
Is it possible to avoid the high point by following the contours from the source to the destination? If you bring the highest point down to around 890 m by making the line a bit longer you could make it work.
I agree with LittleInch that you have an error in your Bernoulli calculation. Assuming that you are pumping something similar to water, the pressure recovery due to Bernoulli in going from 1.5" to 3" will be about 6 psi. There would also be a friction loss of about 3.5 psi through the 1.5" to 3"...
@Javier Sanchez - You can use a similar technique to the way you would match the curves for a centrifugal pump system. The difference is that you would use the pressure at the *suction* of the vacuum pump.
The upstream gas pressure is fixed by your pressure regulator. The downstream pressure...
As pointed out by Pierreick and Snickster, the up and down voting system is a bit confusing if you have not seen it on other forums. Here it is even more confusing because we still have the old "Great post!" button as well and it is not immediately clear how the two systems complement each...
@Asisraja D - you have correctly identified the data required for this calculation but you have not got the correct values for that data. See also LittleInch's comment.
@Asisraja D - Your general procedure makes sense, but there are a few items I believe you have got wrong. Bottled nitrogen is unlikely to be at 6.4 Kelvin. For gases the specific gravity is generally defined as the ratio of the molecular mass to that of air, so your value of 19.48 is strange...
To the member(s) who voted down the posts by me and pierreick - you are absolutely entitled to disagree with any post here, but it adds nothing to the discussion to simply vote a post down without explaining why you disagree with the advice given. The free expression of all points of view is...
For the pressures at which you are working, a well designed piping system will have essentially no pressure drop. All the pressure drop is going to be across the membrane. You should get the pressure vs flow characteristic for the membrane from the supplier and check that at the rated flow the...