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Saputnik
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Join Date: Sep 2015
Posts: 4

Fuel Vapor Recovery System Efficiency

10/19/2015 6:24 AM

Hi. I'd appreciate if someone knows how to calculate reduced emissions of fuel vapors when using Vapor Recovery System.

I tried to search, but to na avail. I have found only this formula in a presentation:

But, it doesn't say what is S in this formula, and I also don't know if formula is correct at all.

Anyone has better info?

Thanks in advance!
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IdeaSmith
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#1

Re: Fuel Vapor Recovery System efficiency

10/19/2015 8:12 AM

It is maybe worthwhile to explore what the unit of S would be.

Instead of the symbols use the units of each symbol and solve for the unknown.

This way you get an idea of what S might be.

Where this formula came from it should have defined all symbols and given all units.

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PWSlack
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#2

Re: Fuel Vapor Recovery System efficiency

10/19/2015 8:13 AM

<...Anyone has better info?...> Try the individual that made the presentation.

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Saputnik
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#3

Re: Fuel Vapor Recovery System efficiency

10/19/2015 8:23 AM

Thanks for the replies!Problem with presentation is that I don't know who made it, let alone have contact with author! I was just given that presentation by someone who knew I was looking back and forth for this info.

Solving "S" by units is good idea, I have just tried, but I'm afraid it doesn't get me anywhere - if L is 1 (kg of emissions / kg of fuel), VP in Pa (kg/m s^2), MW in kg/mol and T in K, I get S as K*m*mol*s^2 / kg^2, which makes a little sense...
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lyn
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#4

Re: Fuel Vapor Recovery System efficiency

10/19/2015 9:58 AM

Try some of these:

Carbon Adsorption Vapor Recovery Systems - John Zink ...

Emission Calculations and Cost/Benefit Analysis of the ...

50 CAR EFFICIENCY EVALUATION OF - OPW

EPA's Stage II Vapor Recovery Systems Options Paper
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Rixter
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#5

Re: Fuel Vapor Recovery System efficiency

10/19/2015 12:14 PM

Maybe S is fudge factor. You measure L, solve for S, and when you plug S back in, your formula is correct!

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RAMConsult
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#6

Re: Fuel Vapor Recovery System efficiency

10/19/2015 12:35 PM

It's only a guess but I suspect that "S" is the values given for gasoline and diesel vapor release found below the formula. Kinda' makes sense given that gasoline is much more volatile than diesel, but then again you're dealing with the EPA.

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Rixter
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#7

Re: Fuel Vapor Recovery System efficiency

10/19/2015 12:36 PM

http://www3.epa.gov/ttnchie1/ap42/ch05/final/c05s02.pdf

Emissions from loading petroleum liquid can be estimated (with a probable error of ±30 percent)4
using the following expression:

where:
LL = loading loss, pounds per 1000 gallons (lb/103 gal) of liquid loaded
S = a saturation factor (see Table 5.2-1)
P = true vapor pressure of liquid loaded, pounds per square inch absolute (psia)
(see Section 7.1, "Organic Liquid Storage Tanks")
M = molecular weight of vapors, pounds per pound-mole (lb/lb-mole) (see Section 7.1, "Organic
Liquid Storage Tanks")
T = temperature of bulk liquid loaded, °R (°F + 460)
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Mikerho
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#9
In reply to #7

Re: Fuel Vapor Recovery System efficiency

10/20/2015 8:04 PM

How does lb/1000gal = lb/103gal? I would state it as 0.001lb/gal.

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Saputnik
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Join Date: Sep 2015
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#10
In reply to #9

Re: Fuel Vapor Recovery System efficiency

10/21/2015 12:31 AM

It's 10^3, not 103, probably formatting issue as he copied accross documents.
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Saputnik
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#8

Re: Fuel Vapor Recovery System efficiency

10/19/2015 1:42 PM

Rixter, thanks a lot, that was the source I was looking for!
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IdeaSmith (1); lyn (1); Mikerho (1); PWSlack (1); RAMConsult (1); Rixter (2); Saputnik (3)

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