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True, but it can give you a "ballpark figure".

You can make an estimate though. You can start by calculating the flow to obtain a pressure drop of 6 bar over the supply line. You can then take this flow and calculate the mass over 3 minutes.

Hello, I'm working on a project where a Plate Steam Condensor is used to heat district water for urban houses. In short, I have: - superheated steam of 42 bar(g) @ 430 C and condensate of approximately 7 bar(g). - The supply must be 1 MWth to 50 MWth - District water pressure is at minimum...

We at work have a discussion about the gas buffering in a pipeline. The case: We have a 100 km pipeline 30" with no output @ 25 bar(g). What will happen with the pressure at the end of the pipeline if we assume an output of 10 m/s @ 25 bar(g) (=x kg/hour) by opening a valve and an input of x...

Here a little update: For what I now know, they made a calculation for the availability. The background info of the project; the transportation of CO2 to greenhouses. In the night there is no consumption because there is no light and in the winter there is less consumption because the sun...

@zdas04, I know the term availability in the contract is for the whole system. Therefore I am looking for typical values for the availability for compressors. If a typical value is lower I know I need to do something. (Two compressors for instance, redundancy)

Hi All, I have a relative small question, but I can't find the answer! We are going to install a compressor (No vendor selected yet) which will have a various demand and will go off line every night. In the contract we have is a statement that we must meet a technical availability of 99% for...

Thank you all for the fast responses! I did a little workaround for my problem: - I know now that the ramp up/down for the compressor unit is 5 minutes (0 --> 100%). - I have calculated the time the molecules will stay in the pipeline @ maximum consummation. (Which is also around 5 minutes) As...

Of course the gas molecules will flow because it will pressurize the whole pipeline, but I need to give a decent ramp up/down to the consumers and therefore I need to know how much time it will take for the pressure to be stabilized after pressure drop. And since the compressor will rise the...

Edit to background info: Pipe length is approx. 5000 meter with internal diameter of approx. 250 mm --> Volume will be around 230 m3

@ Katmar Thanks for your answer. I think however there is a flaw in your answer. Let us say that we have a 5000 meter pipeline without any flow and just a static pressure. If in one end we put some more gas molecules, i.e. the pressure will get higher, there is no gas velocity (dynamic...

So you say that there is no "delay" in pressure change and a sudden increase in pressure in one end of the pipe is without any delay seen at the other end? My feeling says this is not correct, since it is a gas and not a liquid. I think it takes some time for the pressure change to be seen at...

@ quark, Thanks for the fast response. I understand your explanation, but you don't give an answer of the speed of pressure change over the 5000 meter pipeline. Maybe, I did not made my question correct. What is the time, when at one end of a 5000 meter gas pipeline the pressure instantly...

Hi All, This question is a bit frustrating since I am thinking about it all day! The problem: I have a pipeline of 5000 meter, on one side I have a compressor and on the other side I have a consumer. In one scenario I have an equilibrium and for instance the pressure at the consumer is 4,7...

Hi all, I have a question, does anyone know if the following components can be filtered with an active coal filter? NOx Ethylene Methane If not, is there a way to filter these components? (Catalytic Oxidation for instance?) Thanks in advance. Regards, Eltjo

Yes, I thought so. Thanx for all the replies. If I have news from the manufacturer I'll share the info.

This is correct, I have made some calculations with my old schoolbooks and it uses the same formula (and answers). But when I make that calculation complete with the discharge temperature: T2/T1 = P2/P1^((n-1)/n) The discharge temperatures are lower than the stated 150 C. With your numbers...

Dear people, The answer was indeed as stated above the discharge temperature. They say it is 150 C instead of 180 C. This wil drop the enhalpy from 710 kW to 415 kW. The overall efficiency must then be 92% to get the 450 kW. But I will look into that. Thanx for al the answers!

Yes, thank you for the notification. Ofcourse the R is not correct because there needs to be a correction for the water too. I am going to email the supplier. For information I will post their response here.

@ 25362 You mean that if the outlet temperatures are adjusted downwards the enthalpy will also drops? I just did a "micro-calculation" for an isothermal compression (n=1) (This is the most optimistic calculation right?) Pisoth = Qme * R * T1 * LN (P2/P1) Where: Qme = 10.000 kg/hour --> 2,78...