Laserfiche WebLink
Using the Ideal Gas Law P1V1/T1 = P2V2/T2 <br /> Solving for V2 =P1V1T2/P2T1 <br /> ° Assuming P1 = P2 = 14 7 lb/int , P cancels from the equation <br /> leaving V2 = V1T2/Tl . <br /> V1 = Q cf/m x 1440 min/day <br /> T2 = 491 . 58oR T1 = 459 58 + ToF at sight <br /> V2 = Q cf/min x 1440 min/day x 491 58oR/ (459 58o + ToF) <br /> X lb/day = C ug/l x 0 0000000621 lb 1/ug cf x Q cf/min x 1440 <br /> min/day x 491 58oR/ (459 580 + ToF) <br /> Q for the Influent and Effluent samples = The well flow rate + <br /> the air flow <br /> CO2 EMISSIONS <br /> Analysis of the vapor stream has revealed that one of the ma]or <br /> constituent of the extracted vapor is carbon dioxide (CO2) . The <br /> CO2 is found in concentrations well above Its normal <br /> concentration in air (0 033%) and therefore is probably the <br /> product of the oxidation of hydrocarbons by biological and/or <br /> chemical means <br /> Because of the elevated concentrations of CO2 found in the <br /> influent ( . 75-115) , this CO2 represents a significant amount of <br /> hydrocarbon removed. On average in a hydrocarbon molecule there <br /> are 2 Hydrogens (H) for every Carbon (C) therefore each CO2 <br /> represents the oxidation of one CH2 radical This indicates that <br /> for ever mole of CO2 recovered one mole of CH2 has been <br /> destroyed The molecular weight of CH2 is 14 gm/mole . <br /> 14gm/435 6gm/lb= 0 032 lb/mole <br /> The volume of one mole of gas at STP = 22 4 1 <br /> 1 1 = 0353 cf 22 .41 x 0353 cf/1 = 0 . 79cf/mol . <br /> Therefore one cf of pure CO2 represents <br /> 0 . 032 lb/mole / 0 79 cf/mole = 041 lb/cf of CH2 <br /> In gasses 1% = 1/100 of a mole . <br /> 041lb/cf/100%= 00041 lb/cf/% <br /> A sample of the total flow from the vapor extraction system on <br /> 12/09/92 was analyzed for CO2 The sample contained 1" 00% CO2 , <br /> normal air contains 0 . 033% CO2 , the 0 967. in the sample is most <br /> probably the result of oxidation of the hydrocarbons either <br /> chemical or biological At the current rate of vapor extraction, <br /> 79 62 cfm it can be calculated that we are currently removing <br /> 45 45 pounds of hydrocarbon from the site per day as the result <br /> of biological or chemical degradation <br /> 79 62cfm x 0 967% x 0 . 00041lb/cfm/% x 1440min/day = 45 45 lb/day <br /> page 5 of R604 , 12/17/92 <br />