Gas Release into Volume of Water

[Hamilcar]

Dear Experts,

Would much appriciate any feedback on this hypothetical scenario:
A pipeline is connected to a wellhead which is isolated from the rest of the pipeline by a production valve. The wellhead contains gas in Volume V1 and high Pressure P1. While the pipeline (isolated by valve) contains water of volume V2 (volume of pipeline) and Pressure = 0 barg.

The pipeline is isolted from the wellhead by the production valve and is blanked off with a blind flange on the other side of the line. at time t1, the valve is opened and the pressure within the wellhead equalises with the pipeline to pressure P3.
Given the volume of the wellhead and the pipeline are known, how does one determine the pressure within the pipeline following release of the gas in the wellhead, P3?
I imagine this would be calculated differently to when the pipeline is full of gas rather then water?
If further clarification is required please let me know.

Thank you for your time

 

[zdas04]

In the world where I live, I would put a gauge on it. If this is some sort of academic hypothetical (spelled "homework") then it doesn't belong here. You've been on eng-tips.com for a couple of years so I'll assume it isn't homework, but you haven't been real active so maybe it is.

The thing that makes it look like homework is the P1, V3 etc. If your pipeline is 3 miles of 10-inch pipe or something like that then say so.

This is a first-year chem mixing problem where the wellsite flowline is one tank and the delivery pipeline is another tank, but the second tank has approximately zero volume. In the real world, you would disregard the gas volume that dissolves into the liquid and assume zero compressibility and the final pressure is about the same as the initial wellsite pressure. If you don't disregard the Henry's Law constant then you can remove the dissolved gas volume from the wellsite and recalculate the time zero pressure without that volume.

David