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Re: Jog my memory with simple calc.
Re: Jog my memory with simple calc. -- Newbie	Post Reply	Top of thread	Forum

Posted by: mbeychok ® 09/14/2005, 22:17:47 Author Profile Mail author Edit

Newbie: I realize this is a few months late but I want to emphasize that Zekeman's answer to your posting was correct. However, I would like to make it a bit clearer. (1) You can use the ideal gas law to calculate the density of air, as explained below. Since the number of mols (n) is equal to the mass (w) divided by the gas molecular weight (MW), the ideal gas law can be written as: PV = (w/M)RT Thus the density = w/V =(MW/10.73)(P/R)= 0.154 lb/ft3 where: MW = 28.9 P = absolute pressure, in psia = 14.7 + 15 = 29.7 psia R = universal gas law constant = 10.73 (psia/lb-mols)(ft3/°R) T = absolute temperature, °R = 460 + 60 = 520 (assuming that the compressed air temperature is 60 °F The volume of liquid that must be displace by an equal volume of air = (6000 gal)/(7.48 gal/ft3) = 802.1 ft3 Thus, weight of air required = (802.1 ft3)(0.154 lb/ft3) = 123 lbs Milton Beychok Modified by mbeychok at Fri, Sep 16, 2005, 12:21:33

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Re: Jog my memory with simple calc.
Re: Jog my memory with simple calc. -- Newbie	Post Reply	Top of thread	Forum

Posted by: Cragyon ® 05/27/2005, 08:44:12 Author Profile Mail author Edit

I don't think the ideal gas law is going to help you (PV=nrt). You have to know a lot more than you have given. You will need to know the surface area, pipe size, differential pressure and more. Check out the software at the link below: /product_directory/fluid_flow_software/fluid_flow_calculations_software.htm I bought a copy two weeks ago, works great. They have a free tryout that I think will perform the calculations you need.

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Re: Jog my memory with simple calc.
Re: Re: Jog my memory with simple calc. -- Cragyon	Post Reply	Top of thread	Forum

Posted by: zekeman ® 05/28/2005, 11:31:56 Author Profile Mail author Edit

You have to find the volume of flid expelled,6000 gal or 6000*231/1728 =802 cubic feet Since air at standard pressure has a density of .075 lb per cu ft,at 15 psig it would approximately double, so the amount of compressed ar would be 802*.15=120 lbs

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