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Solutions Manual for Applied Petroleum Reservoir Engineering Third Edition Revised by Ronald E. Terry
- Brandon Rogers Download Link at the end of this File 1 / 4
CHAPTER 2
2.1 using Equation 2.4: pV nRT=
(a)
()( )110.73 60 460
V379.6SCF
14.7 + == (b)
()( )110.73 32 460
V359.1SCF
14.7 + == (c)
()( )1 10.73 80 460
V378.1SCF
14.7 10 16
+ == +
(d)
()( )110.73 60 460
V371.4SCF
15.025
+ ==
2.2 (a) number of moles of methane 10 16 .625 moles== number of moles of ethane 20 30 .667 moles== total number of moles .625 .667 1.292 moles=+=
(b) from Equation 2.4
()()1.292 10.73 550
p15.25 psia 500 == (c) wt. of mixture 10 20 molecular wt. of mixture23.22 lb/lb-mole moles of mixture 1.292 + === (d) using Equation 2.6 g 23.22
γ0.802
28.97 ==
2.3 using a basis of one mole of gas mixture Component Vol. Fraction Moles Mol. Wt. Weight. lb.Methane .333 .333 16 5.33 Ethane .333 .333 30 10.00 Propane .333 .333 44 14.67 Total weight 30.00 g 30.00 molecular wt. 30 1 30 lb/lb-moleγ1.036 28.97 === = This text is associated with Terry/Applied Petroleum Reservoir Engineering, Third Edition (9780133155587) Copyright 2015, Pearson Education, Inc. Do not redistribute. 2 / 4
2.4 initially the container contains only air but at the end, the container has both air and CO 2 () () 2
14.7 50
moles of air 0.128 moles
10.73 535
moles of CO 10 44 0.227 moles == ==
total moles in the tank at the final state 0.128 0.227 0.355 moles=+=
using Eq. 2.4:
()()0.355 10.73 505
p38.47psia 50 ==
2.5 costofacetylene=
$10.00
20 =$0.50perlb or
$10.00
2026 =$13.00 perlb-mole costofacetyleneperSCF=
$10.00
379.4 =$0.0264 perSCF costofacetyleneperMCF=$0.0264 1000 () =$26.40 perMCF amountofacetyleneusedperday=
1+14.7
( ) 200() 520()
14.7545()
=203.8SCFday costofacetyleneperday=203.8 $0.0264( ) =$5.38 pe r day
2.6 The tank will collapse when the inside pressure reaches the outside pressure minus the pressure that the tank is designed to withstand. This will be used caused by oil being pumped from the tank.() ()
0.75 29.9
collapse pressure 29.1 29.005 inches Hg
16 14.7
=−= the initial volume of the air space, () [ ] 2 i
3.1416 110
V352595,033cuft 4 =−=
the volume of the air space in the tank at the collapse pressure will be:
- )
ii f f 29.1 95,033pV V95,344cuft p29.005 == = This text is associated with Terry/Applied Petroleum Reservoir Engineering, Third Edition (9780133155587) Copyright 2015, Pearson Education, Inc. Do not redistribute. 3 / 4
the volume of oil removed at the time of collapse will be the difference or 311 cu ft the pump removes oil at a rate of ()20,000 5.615 112,300 cu ft day= (a) the time of collapse will be ()() 311 24 60 3.99 minutes
112,300
= (b) total force on roof at time of collapse will be F = pA
- )
() () 2 f
3.1416 11014.7
F 29.1 29.005144 63,920 lb
29.9 4
!" #$ %&=−= () *+ %& ,-
(c) The collapse time would have been less.
2.7 (a) basis of 100 lb of mixture let x = lbof methane
- )
- )
weight 100 moles of mixture = moles of methane + moles ofethane mol.wt. .65 28.97 ==
x100x 100 5.31
16 30 .65 28.97
− += = x67.8lb=
which suggests that mixture is 67.8% by weight methane change the basis to one mole of mixture to calculate the mole or volume fraction let y = methane mole fraction then, ()()()y16 1 y 30 18.83+− = y 0.798=
which suggests that the mixture is 79.8% by volume methane, recognize that mole fraction = volume fraction for the mixture (b) The per cent by volume is greater than the per cent by weight for methane because the methane molecule is lighter than the ethane molecule.
2.8 writing a mole balance on the tanks, we get moles in tank 1 + moles in tank 2 = total moles at the final conditions This text is associated with Terry/Applied Petroleum Reservoir Engineering, Third Edition (9780133155587) Copyright 2015, Pearson Education, Inc. Do not redistribute.
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