03 Tutorial 1 - Ideal Gas And Real Gas Properties - Solution - 07-06-2017

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  Gas Field Engineering (PDB4013) May 2017 Semester Tutorial -01 Solution Total Marks  –   10 Name of Student:……………………………………Student ID: ………………………   1.   A natural gas has the following composition. Calculate apparent molecular weight, weight fraction, specific gravity, and compressibility factor for the following gas composition at operating pressure 2050 psia and temperature 105ºF. (4 Marks) Component yi Mi Tci ( R ) Pci (psia)  N2 0.03 28.01 239.26 507.5 CO2 0.0175 44.01 547.58 1071 CH4 0.848 16.043 343 666.4 C2H6 0.0591 30.07 549.59 706.5 C3H8 0.023 44.097 665.73 616 i-C4H10 0.0058 58.123 734.13 527.9 n-C4H10 0.005 58.123 765.29 550.6 i-C5H12 0.0035 72.15 828.77 490.4 n-C5H12 0.002 72.15 845.47 488.6 n-C6H14 0.0033 86.177 913.27 436.9 n-C7H16 0.0028 100.204 972.37 396.8 1 Solution:    Component yi Mi Tci ( R ) Pci (psia) yiMi yiTci yiPci N2 0.03 28.01 239.26 507.5 0.8403 7.1778 15.225 CO2 0.0175 44.01 547.58 1071 0.770175 9.58265 18.7425 CH4 0.848 16.043 343 666.4 13.60446 290.864 565.1072 C2H6 0.0591 30.07 549.59 706.5 1.777137 32.48077 41.75415 C3H8 0.023 44.097 665.73 616 1.014231 15.31179 14.168 i-C4H10 0.0058 58.123 734.13 527.9 0.337113 4.257954 3.06182 n-C4H10 0.005 58.123 765.29 550.6 0.290615 3.82645 2.753 i-C5H12 0.0035 72.15 828.77 490.4 0.252525 2.900695 1.7164 n-C5H12 0.002 72.15 845.47 488.6 0.1443 1.69094 0.9772 n-C6H14 0.0033 86.177 913.27 436.9 0.284384 3.013791 1.44177 n-C7H16 0.0028 100.204 972.37 396.8 0.280571 2.722636 1.11104 1 19.59582 373.8295 666.0581 Specific grav 0.675718 P psia Ppr psia T ( F ) T ( R ) Tpr ( R ) Z  i 2050 3.07781 105 565 1.511384 0.775 0.042882 0.039303 0.694254 0.09069 0.051758 0.017203 0.01483 0.012887 0.007364 0.014512 0.014318 1 2.   Five pounds of propane are placed in a vessel at 100°F and 70 psia. Calculate the volume of the gas assuming an ideal gas behaviour. (0.5 Mark) Solution:    mass (lb) T ( F ) T ( R ) P (psia) MW (lb/lb-mol) C3H8 n (lb-mol) R (psi.cuft/(lb-mol.R) V (cuft) 5 100 560 70 44 0.113636 10.73 9.754545 3.   Using the data given in No.2, calculate the density of propane. (0.5 Mark) Solution: mass (lb) T ( F ) T ( R ) P (psia) MW (lb/lb-mol) C3H8 n (lb-mol) R (psi.cuft/(lb-mol.R) V (cuft) z density (lbm/cuft) 5 100 560 70 44 0.113636364 10.73 9.754545455 1 0.512581547 4.   A gas well is producing gas with a specific gravity of 0.63 at a rate of 1.2 MMscf/D. The average reservoir pressure and temperature are 1,400 psi and 160°F. Calculate: a)   Apparent molecular weight of the gas (0.5 Mark)   b)   Flow rate in lb/day (0.5 Mark)  Solution a.     =29∗  =18.27     b.     =   .  −. psi.cuft/(lbmol.R 620 R =3.84 /  Because 1 lb-mol of any ideal gas occupies 379.4 scf at standard conditions, then the daily number of moles that the gas well is producing can be calculated from: =1.2 10  379.4 =3169.57 /  Determine the daily mass of the gas produced: =  =3169.57 18.27=57908.08 /   No.4a specific gravity q (MMscfd) P (psia) T ( F ) Mair (lb/lb-mol) Ma (lb/lb-mol) 0.63 1.2 1400 160 29 18.27 No.4b specific gravity q (MMscfd) P (psia) T ( R ) Mair (lb/lb-mol) Ma (lb/lb-mol) density (lb/cuft) 0.63 1.2 1400 620 29 18.27 3.844813 GFE week 2 part 1 slide 33 (Avogadro law) Because 1 lb-mole of an ideal gas occupies 378.6 cuft at standard condition q (MMscfd) q (scfd) V (cuft) Mole rate (lb-mol/D) Ma (lb/lb-mol) q (lb/D) 1.2 1200000 378.6 3169.572108 18.27 57908.08241 5.   A gas well is producing a natural gas with the following composition: (3 Marks) Component y i  CO 2  0.05 CH 4  0.83 C 2 H 6  0.1 C 3 H 8  0.02