Systems and Units

[Pages:13]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 () Temperature (T)

2. Derived Quantities Area (A) Volume (V) Velocity () Frequency (Hz.) Acceleration (a) Angular Velocity () Angular Acceleration (?) Volume Flow Rate (q)

o

Mass Flow Rate ( m ) Density () Specific Gravity (SG) Force (F)

Force due to Inertia (FI) Force due to Gravity (FG) Force due to Viscosity (F?) Force due to Elasticity (FE) Force due to Pressure (FP) Specific Weight () Energy (E) Moment of a Force (M) Work (W) Pressure (p) Stress () Power (P) Dynamic Viscosity (?) Kinematic Viscosity ()

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Quantity

Length Mass Time

Angle Temperature

1. Principal Quantities

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

kilogram (kg) = 1000 gm

pound mass (lbm) = 453.592370 gm

gram (gm)

year = 365 d day = 24 h hour = 60 m minute = 60 s

1 rev. = 2B radians = 360E 1 r = 57.29578E 1 E = 0.017453r

Ordinary EF= 1.8EC+32 Absolute ER= EF+459.688

ER= 1.8 K

year = 365 d day = 24 h hour = 60 m minute = 60 s

1 rev. = 2B radians = 360E 1 r = 57.29578E 1 E = 0.017453r

Ordinary EC= (EF-32)/1.8 Absolute K= EC+273.16

K= ER/1.8

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2. Derived Quantities

Quantity

Area

Volume

Velocity Frequency Acceleration

Angle

Angular Velocity Angular Acceleration Volume Flow Rate Mass Flow

Rate Density

Specific Gravity

Formula

A = L2

V = L3

=

dL dt

Hz.

a

=

d dt

2

=

d dt

a^

=

d dt

q

=

dV dt

o

m

=

dm dt

=

m V

SG = f = w

141.5 131.5+oAPI

English Units SI Units

mi2, ft2, in2, acre 1 mi2 = 640 acres 1 acre = 43,560

ft2

bbl, ft3, gal, qt bbl = 5.6146 ft3

bbl = 42 gal gal = 4 qt = 3.785

ltr

m2

m3, ltr m3 = 1000 ltr ltr = 1000 cc

ft/s

m/s

s-1

s-1

ft/s2

m/s2

rad and E 360E = 2B rad

rad/s

rad and E 360E = 2B rad

rad/s

rad/s2

rad/s2

bbl/d, ft3/s

m3/s, ltr/s

slug/s, lbm/s

kg/s

slug/ft3, lb/ft3,

lb/gal

Dw = 1.940 slug/ft3

Dw = 62.428 lbm/ft3

Dw = 8.345

lbm/gal Dw = 0.433 psi/ft

Dw = 10 EAPI

kg/m3

Dw= 103 kg/m3

Lab. Units

cm2, darcy, md 1 cm2 =

101,320,790 d 1 d = 1000 md

cm3 = cc

cm/s s-1 cm/s2 rad and E 360E = 2B rad rad/s rad/s2

cc/s gm/s

gm/cc

Dw = 1 gm/cc

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2. Derived Quantities (cont.)

Quantity

Force

Formula

F = m a

English Units

lbf = slug ft/s2 lbf = 444,822 dyne

lbf = 4.44822 N

SI Units

N = kg m/s2 1 N = 105

dynes

Force due to FI = m a

lbf

N

Inertia

Force due to FG = m g Gravity=Wt.

Force due to Viscosity

F?

=

?

d dy

A

lbf g = 32.17404856

ft/s2

lbf

N g = 9.80665

m/s2

N

Lab. Units

dyne = gm cm/s2

dyne

dyne g = 980.665

cm/s2 dyne

Force due to Elasticity

Force due to Pressure

Specific Wt.

Energy

Moment of a Force

Work

Pressure

FE = E A

FP = p A ( = FG/V E = F L M = F L

W = F L p = FN/A

lbf

lbf

lbf/ft3 lbf-ft lbf-ft

lbf-ft lbf/ft2, lbf/in2 = psi psc = 14.69594877

psi

Stress Power

Dynamic Viscosity

J = FT/A lbf/ft2, lbf/in2 = psi

P = E/t

F?

=

F? d / dyA

lbf-ft/s, hp 1 hp = 550 lbf-ft/s

lbf-s/ft2 = 47,880 cp

Kinematic Viscosity

< = ?/D

ft2/s = 929 stoke

N

N

N/m3

J = N-m J

J Pa = N/m2 bar = 105 Pa psc = 1.01325

bar psc = 101325

Pa bar, Pa

W = J/s

N s/m2= 10 poise

N s/m2= 1000 cp

m2/s = 104 stoke

dyne

dyne

dyne/cc erg = dyne cm erg = dyne cm

erg = dyne cm dyne/cm2

psc=1013250 d/cm2

psc = 1 atm psc = 76 cm Hg

dyne/cm2 erg/s

dyne s/cm2 = poise

1 poise = 100 cp

cm2/s = stoke

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Semester:

Name:

EPS-441: Petroleum Development Geology Units and Conversion Homework #:

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 oAPI oAPI

psi/ft

psi/ft lb/ft3

lb/gal oAPI

SG oAPI

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?.

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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.

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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?.

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Semester: Name:

EPS-441: Petroleum Development Geology Units and Conversion Homework #: 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.

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