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Water Content In Natural Gas Calculation
#1
Posted 09 November 2016 - 07:21 AM
#2
Posted 09 November 2016 - 11:39 AM
Is there a specific part of this problem that you are stuck on or don't understand?
Hopefully it is not doing too much for you. You are given a ratio (70 kg water/1E6 Sm3 gas) and are asked to find a total water. My first thought for something like this is to set it up as a proportionality. (If you need an algebra refresher on how to solve proportions, I recommend http://www.purplemat...ules/ratio2.htm).
70 kg water : 1E6 Sm3 gas = x kg water : y Sm3 gas
The problem specifically asks you to find the x, leaving you to figure out what y should be. Once you have y, then the proportion should be easy to solve for x.
#3
Posted 09 November 2016 - 03:53 PM
So, this might be a trick question. So be sure not to leave out anything from the original problem statement. What you have shared above seems to be incomplete information.
Bobby
#4
Posted 10 November 2016 - 11:08 AM
#5
Posted 11 November 2016 - 01:01 AM
#6
Posted 14 November 2016 - 01:12 PM
Romaine,
I think you calculation is only referential, you need take in count a Z (compresibility factor). you are working with 5 MPa, this is 725 psi (aprox. 51 bara) a high pressure, ideal gas equation is not recommendable.
A first aproximation can be Z = 1 - P/500, P in Bar(a), Z = 1 - 51/500 = 0,898
Other point is the water content at 5MPa and 15ºC, this is 350 kg/MMSM3, then a low pressure the value must be higher.
I attached a calculation with Sloan´s formula.
Napo.
Attached Files
#7
Posted 20 December 2017 - 08:55 AM
Hey there, if you need any help with calculations you can hit me up at homework help for college. I'll try to answer all of your questions and help you with any calculations you need.
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