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Jog my memory with simple calc.  
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Posted by: Newbie ^{®} 05/25/2005, 13:54:05 Author Profile Mail author Edit 
I am new to this but here goes. If I an off loading a trailer at a rate of 60lb/min with 15 psi of air put on trailer for off loading, how much air will I use? (The trailer size is 6000 gal and the weight of liquid being off load is 8lbs/gal.) I believe that I should use PV=nZRT. Is that right? Modified by Newbie at Wed, May 25, 2005, 13:55:09 
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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/ft^{3} where:
The volume of liquid that must be displace by an equal volume of air = (6000 gal)/(7.48 gal/ft^{3}) = 802.1 ft^{3} Thus, weight of air required = (802.1 ft^{3})(0.154 lb/ft^{3}) = 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.  
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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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