Systems and Units
The three systems of units are:
1. The English or the ft-lb-s System
2. The International or the m-kg-s System
3. The Laboratory or the cm-gm-s System
Quantities fall into two main categories:
1. Principal Quantities
Length (L)
Mass (m)
Time (t)
Angle (θ)
Temperature (T)
2. Derived Quantities
Area (A)
Volume (V)
Velocity (ν)
Frequency (Hz.)
Acceleration (a)
Angular Velocity (ω)
Angular Acceleration (â)
Volume Flow Rate (q)
o
Mass Flow Rate ( m )
Density (ρ)
Specific Gravity (SG)
Force (F)
Force due to Inertia (FI)
Force due to Gravity (FG)
Force due to Viscosity (Fµ)
Force due to Elasticity (FE)
Force due to Pressure (FP)
Specific Weight (γ)
Energy (E)
Moment of a Force (M)
Work (W)
Pressure (p)
Stress (τ)
Power (P)
Dynamic Viscosity (µ)
Kinematic Viscosity (ν)
1
1. Principal Quantities
Quantity
English System
International System
mile (mi) = 1760 yd
yard (yd) = 3 ft
foot (ft) = 12 in
inch (in) = 2.54 cm
kilometer (km) = 1000 m
meter (m) = 100 cm
centimeter (cm)
slug (sl) = 32.17404856 lbm
pound mass (lbm) = 453.592370 gm
kilogram (kg) = 1000 gm
gram (gm)
year = 365 d
day = 24 h
hour = 60 m
minute = 60 s
year = 365 d
day = 24 h
hour = 60 m
minute = 60 s
Angle
1 rev. = 2B radians = 360E
1 r = 57.29578E
1 E = 0.017453r
1 rev. = 2B radians = 360E
1 r = 57.29578E
1 E = 0.017453r
Temperature
Ordinary EF= 1.8EC+32
Absolute ER= EF+459.688
ER= 1.8 K
Ordinary EC= (EF-32)/1.8
Absolute K= EC+273.16
K= ER/1.8
Length
Mass
Time
2
2. Derived Quantities
Quantity
Area
Formula
A = L2
Volume
V=L
Velocity
ν=
Frequency
3
dL
dt
Acceleration
Hz.
dν
a=
dt
Angle
2
English Units
SI Units
mi2, ft2, in2, acre
1 mi2 = 640 acres
1 acre = 43,560
ft2
m2
Lab. Units
cm2, darcy, md
1 cm2 =
101,320,790 d
1 d = 1000 md
bbl, ft3, gal, qt
bbl = 5.6146 ft3
bbl = 42 gal
gal = 4 qt = 3.785
ltr
m3, ltr
m = 1000 ltr
ltr = 1000 cc
cm3 = cc
ft/s
m/s
cm/s
s-1
s-1
s-1
ft/s2
m/s2
cm/s2
rad and E
360E = 2B rad
rad and E
360E = 2B rad
rad and E
360E = 2B rad
3
Angular
Velocity
ω=
dθ
dt
rad/s
rad/s
rad/s
Angular
Acceleration
Volume
Flow Rate
aˆ =
dω
dt
rad/s2
rad/s2
rad/s2
bbl/d, ft3/s
m3/s, ltr/s
cc/s
slug/s, lbm/s
kg/s
gm/s
slug/ft3, lb/ft3,
lb/gal
kg/m3
gm/cc
Dw= 103
kg/m3
Dw = 1 gm/cc
Mass Flow
Rate
Density
Specific
Gravity
dV
dt
o
dm
m = dt
m
ρ=
V
q=
ρ
SG = f =
ρw
141.5
131.5+ oAPI
Dw = 1.940
slug/ft3
Dw = 62.428
lbm/ft3
Dw = 8.345
lbm/gal
Dw = 0.433 psi/ft
Dw = 10 EAPI
3
2. Derived Quantities (cont.)
Quantity
Formula
English Units
SI Units
Lab. Units
Force
F=ma
lbf = slug ft/s2
lbf = 444,822 dyne
lbf = 4.44822 N
N = kg m/s2
1 N = 105
dynes
dyne = gm cm/s2
Force due to
Inertia
FI = m a
lbf
N
dyne
Force due to
Gravity=Wt.
FG = m g
lbf
g = 32.17404856
ft/s2
N
g = 9.80665
m/s2
dyne
g = 980.665
cm/s2
lbf
N
dyne
Force due to
Viscosity
Fµ = µ
dν
A
dy
Force due to
Elasticity
FE = E A
lbf
N
dyne
Force due to
Pressure
FP = p A
lbf
N
dyne
Specific Wt.
( = FG/V
lbf/ft3
N/m3
dyne/cc
Energy
E=FL
lbf-ft
J = N-m
erg = dyne cm
Moment of a
Force
M=FL
lbf-ft
J
erg = dyne cm
Work
W=FL
lbf-ft
J
erg = dyne cm
Pressure
p = FN/A
lbf/ft2, lbf/in2 = psi
psc = 14.69594877
psi
Pa = N/m2
bar = 105 Pa
psc = 1.01325
bar
psc = 101325
Pa
dyne/cm2
psc=1013250
d/cm2
psc = 1 atm
psc = 76 cm Hg
Stress
J = FT/A
lbf/ft2, lbf/in2 = psi
bar, Pa
dyne/cm2
Power
P = E/t
lbf-ft/s, hp
1 hp = 550 lbf-ft/s
W = J/s
erg/s
Fµ
lbf-s/ft2 = 47,880
cp
N s/m2= 10
poise
N s/m2= 1000
cp
dyne s/cm2 =
poise
1 poise = 100 cp
ft2/s = 929 stoke
m2/s = 104
stoke
cm2/s = stoke
Dynamic
Viscosity
Kinematic
Viscosity
Fµ =
dν / dyA
< = µ/D
4
EPS-441: Petroleum Development Geology
Units and Conversion
Semester:
Homework #:
Name:
SS#:
Problem #1: Do the following unit conversions:
From
To
38 oAPI
12 oAPI
56 oAPI
40 oAPI
28 oAPI
31 oAPI
0.433 psi/ft
0.433 psi/ft
0.433 psi/ft
0.378 psi/ft
0.394 psi/ft
SG = 0.76
SG = 1.10
SG = 0.74
SG = 1.08
SG = 0.88
48.8 lb/ft3
62.4 lb/ft3
64.3 lb/ft3
48.7 lb/ft3
10.86 lb/gal
8.33 lb/gal
8.33 lb/gal
10.4 lb/gal
0.82 gm/cc
1.02 gm/cc
0.87 gm/cc
0.91 gm/cc
lb/ft3
SG
psi/ft
SG
gm/cc
psi/ft
lb/ft3
lb/gal
gm/cc
o
API
o
API
psi/ft
psi/ft
lb/ft3
lb/gal
o
API
SG
o
API
psi/ft
gm/cc
gm/cc
psi/ft
lb/ft3
gm/cc
SG
psi/ft
psi/ft
lb/ft3
Problem #2: Given a rectangular solid with dimensions 1000 ft x 400 ft x 40 ft. Calculate its
volume in ft3, bbl, acre-ft?.
5
Problem #3: Calculate pressure gradients of the following liquids:
SG
Dw
SG
Dw
SG
EAPI
EAPI
EAPI
Do
= 1.00
= 66.3 lb/ft3
= 1.15
= 9.5 lb/gal
= 0.85
= 42
= 32
= 45
= 58.1 lb/ft3
Problem #4: A well drilled to 3000 ft penetrates a formation containing 28 EAPI oil. If
reservoir pressure is 1300 psia, what is the shut-in surface pressure?. If reservoir pressure is
1000 psia, how many ft of oil will be standing in the wellbore?.
Problem #5: In an area where ambient temperature is 78 EF, two wells, A and B, were drilled.
The depth of well A is 7250 ft and the depth of well B is 8000 ft. The bottom hole temperature
(BHT) in well B is 180 EF. What is the BHT in well A?.
Problem #6: The areal extent of a reservoir as determined with seismic data is 1500 acres.
From logs the following reservoir properties were determined:
Zone
N
h, ft
Sw
1
0.28
4
0.28
2
0.32
7
0.40
3
0.18
3
0.31
4
0.20
11
0.27
a) Calculate the pore volume in the reservoir. Give your answers in acre-ft, ft3, and bbl.
b) Calculate the volume of oil in the reservoir. Give your answers in acre-ft, ft3, and bbl.
c) Calculate the STB of oil if Bo = 1.34 RB/STB.
6
Problem #7: The average reservoir properties of a 300 acre reservoir are: N = 18% and Sw =
36%. Estimated volume of oil originally in place = 410 MM STB. Formation thickness = 2000
ft. Determine the oil formation volume factor.
Problem #8: Consider the sketch below. Given EAPI of oil = 35. water SG = 1.07. If well A
penetrates the oil zone, find the shut-in surface pressure at well A?.
7
EPS-441: Petroleum Development Geology
Units and Conversion
Semester:
Homework #:
Name:
SS#:
Problem #1: A reservoir has an areal extent of 500 acres, an average thickness of 90 ft, and an
average porosity of 20%.
a) What is the reservoir volume available for hydrocarbons?. Answer in acre-ft, bbl, and ft3.
b) If the average water saturation is 35%, what is the reservoir volume available for oil?.
Answer in acre-ft, bbl, and ft3.
c) Same as (b) except reservoir fluid is gas.
d) If Bo = 1.34 RB/STB. What is the oil volume from (b) in surface barrels?. in surface cubic
feet?.
e) If Bg = 310 SCF/CF. What is the gas volume from (c) in surface cubic feet?. in surface
barrels?.
Problem #2: A well is being drilled to a depth of 10,000 ft. The wellbore diameter is 9 inches.
The drill string has an inside diameter of 4.5 inches and an outer diameter of 5 inches. What is
the volume of mud in the hole when it is at total depth (TD) and there is a complete drill string in
the hole?. Ignore collars and bit volume and assume the mud is incompressible. Answer in
cubic feet and barrels.
8
Problem #3: Data for the diagram shown below are:
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
oil gravity = 38 EAPI
water specific gravity = 1.05
average gas gradient = 0.055 psi/ft
only oil flows in oil zone
reservoir B is in water communication with the surface
pressure at oil zone datum in reservoir A = 3820 psig
temperature gradient = 1.5 EF/100 ft.
mean surface temperature = 61 EF
ground level is 3000 ft above sea level
elevations given are sub-sea but answers should be in subsurface
a) Calculate pressure and temperature at each of the 7 zones indicated in the diagram.
b) What is the bubble point pressure for the oil in reservoir A?.
c) Did the oil accumulated before or after faulting in both reservoirs?.
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Problem #4: Seismic data indicates an areal extent of a reservoir as one square mile. Other
reservoir properties are:
Zone
N, %
h, ft
Sw, %
1
15.3
10
36.2
2
26.9
8
13.1
3
18.1
15
24.7
4
22.0
12
19.6
a) Calculate hydrocarbon pore volume in the reservoir. Give your answers in acre-ft, ft3, and
bbl.
b) Calculate the volume of oil in the reservoir. Give your answers in acre-ft, ft3, bbl, and STB if
Bo = 1.18 RB/STB.
c) How many SCF of gas in the reservoir if Bg = 263 SCF/CF.
Problem #5: A well drilled in an area had the following tests run: DST, temperature survey and
a suite of logs. A surface gas and oil samples were taken and recombined at simulated reservoir
conditions and a PVT analysis was done. From these measurements, the following reservoir
rock and fluid properties were determined:
EAPI of oil = 29 EAPI
Bo = 1.38 RB/STB
SGg = 0.78
reservoir depth = 4520 ft
initial reservoir pressure = 2952 psia
bubble point pressure = 1831 psia
reservoir temperature = 151 EF
surface temperature = 70EF
reservoir acreage = 640 acres
Zone
h, ft
N (%)
Sw (%)
1
62
22
31
2
31
18
50
3
5
9
42
4
26
14
21
a) calculate the geothermal gradient in this area?.
b) calculate the original oil in place (OOIP)?.
10
Problem #6: Consider the sketch below. If BHT at well A is 200 EF and BHT at well B is 232
EF.
a) What is the BHT at well C?.
b) What is the ambient temperature in this area?.
11
EPS-441: Petroleum Development Geology
Units and Conversion
Semester:
Homework #:
Name:
SS#:
Problem #1: A well was drilled offshore into an unconsolidated formation. The used sand
control methods were not successful so the well was shut-in. The average annual surface
temperature in that area is 84 EF and the temperature gradient is 1.6 EF/100 ft. The shut-in
surface pressure is 850 psig and the shut-in bottom hole pressure is 2800 psig. It is known that
the wellbore (diameter = 5.5 inches) is filled with gas (SG = 0.78, z = 0.85), oil (API = 32E),
water (SG = 1.08), and water saturated sand (D = 35 gm/cc). Knowing that there is a 2500 ft of
gas, a 2000 ft of sand-free water, and a 500 ft of water-saturated sand, calculate the following:
a) Total depth of the well.
b) Bg in SCF/CF of the gas using average pressure and temperature.
Problem #2: Define the following using equations only - include units:
a) Oil formation volume factor (Bo)
b) Porosity (N)
c) Temperature gradient (GT)
d) Gas saturation (Sg)
e) Oil gravity (EAPI)
f) Fluid pressure gradient (()
12
Problem #3: Using the figure below and the following data:
Dw
Do
Dg
patm
= 1.02 gm/cc
= 0.80 gm/cc
= 0.10 gm/cc
= 14.7 psia
What does gauge A read?.
13