PGE 312 – Homework 5
Physical and Chemical
Behavior of Petroleum Fluids I
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.
PV
nRT
mm
PV
RT
M
RT
RT
MV
ρ
=
=→
==
3
3
0.103
(100 460)
10.73
42.10l /
14.7
lb
psi ft
R
MR
T
b
l
b
m
o
psi
ft
R lb mol
⋅+
=
−
⋅−
D
D
l
=
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
MM
C
n
M
H
n
nn
n
l
bl
b m
o
l
e
l
o
l
e n
=⋅
+
⋅
⋅
=+
=−
=
−
→
Then, the formula of the molecule is
C
3
H
6
(Propene)
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
iButane
0.0115
58.1230
0.6684
0.0183
nButane
0.0233
58.1230
1.3543
0.0370
IPentane
0.0093
72.1500
0.6710
0.0183
nPentane
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/lbmol
1/10
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View Full DocumentPGE 312 – Homework 5
Physical and Chemical
Behavior of Petroleum Fluids I
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 Amagat’s Law of Partial Volumes and
using the result given by equation 334 (Page 102), the volume fraction of the gas mixture is simply
the mole fraction.
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
VV
V
Porosity
−
==
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 V1Vs. Then, the valve is opened
and the gas expands at constant T to a new volume V2Vs 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.
First, write the equation of an Ideal Gas for the
initial condition:
11
1
(5
)
PV
nRT
s
n
R
=
−=
T
T
Second, write the Ideal Gas Law for the final
condition after the valve is opened:
22
1
21
(10
)
s
n
R
=
Here, Vs is the solid volume of the sandstone.
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 Spring '08
 Peters

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