This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: PGE 312 - FALL 2003 Physical and Chemical Behavior of Petroleum Fluids Homework 3- Key 1/9 1. Problem 3.2 The formula of the molecule is C 3 H 6 Because the pressure is low (14.7 psia) the Ideal Gas Law can be used. Thus, calculate the molecular weight as follows. P V n R T m m PV RT M RT RT M V = = = = = = = = = = = = = 3 3 0.103 (100 460) 10.73 42.10l / 14.7 lb psi ft R M RT b lb mol psi ft R lb mol + + + + = = = = = = = = = = = = o o The number of hydrogen atoms (2) for each carbon atom is given. Ttherefore, the general formula of the molecule will have the form C n H 2n . With this information, the molecular weight formula is set up: ( ) 2 ( ) 12 2 14 / 42.1 / 3 molecule M M C n M H n n n nlb lb mole lb lb mole n = + = + = + = + = + = + = + = + = = = = = = = = = = = = Then, the formula of the molecule is C 3 H 6 (Propene) 2. Problem 3.4 In order to calculate the weight fraction, first determine the apparent molecular weight. Multiply the molecular weight by the mole fraction (Columns (1)*(2)). Second, add the result obtained in column (3). This will be the apparent molecular weight of the gas mixture. Third, divide the result of column (3) by the total apparent molecular weight (Column 4). This is the composition of the gas expressed in weight fraction. (1) (2) (3) (4) Component Mole fract. MW MWi*Yi MWi*Yi/ MWi*Yi Methane 0.6904 16.0430 11.0761 0.3025 Ethane 0.0864 30.0700 2.5980 0.0710 Propane 0.0534 44.0970 2.3548 0.0643 i-Butane 0.0115 58.1230 0.6684 0.0183 n-Butane 0.0233 58.1230 1.3543 0.0370 I-Pentane 0.0093 72.1500 0.6710 0.0183 n-Pentane 0.0085 72.1500 0.6133 0.0168 Hexanes 0.0173 86.1770 1.4909 0.0407 Heptane+ 0.0999 158.0000 15.7842 0.4311 1.0000 36.6109 1.0000 Heptane+ properties Specific gravity 0.827 Molecular weigh 158 lb/lb-mol PGE 312 - FALL 2003 Physical and Chemical Behavior of Petroleum Fluids Homework 3- Key 2/9 To obtain the composition of the gas in volume fraction of the gas, assume that the gas mixture behaves as an ideal gas. Therefore, according to Amagats Law of Partial Volumes and using the result given by equation 3-34 (Page 102), the volume fraction of the gas mixture is simply the mole fraction. 3. Problem 3.7 The porosity of the sandstone is 0.18 or 18% Use the following equation in order to calculate porosity: void bulk solid bulk bulk V V V Porosity V V = = = = = = = = The Ideal Gas behavior is used to model the process taking place in the two cells. First, there is gas at certain conditions of P & T in cell 1 occupying a volume V1-Vs. Then, the valve is opened and the gas expands at constant T to a new volume V2-Vs where V2 is the sum of the volumes of the two vessels. Then by difference, the solid volume of the sandstone is calculated and with this information, the porosity is determined....
View Full Document
This note was uploaded on 03/19/2008 for the course PGE 312 taught by Professor Peters during the Spring '08 term at University of Texas at Austin.
- Spring '08