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Bert Campaneris
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Sample Questions & Worked Out Examples
A.G.T.I(Second Year)
Petroleum Engineering
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Sample Questions for
Chapter 1 Geology and Hydrocarbon Accumulations
Define the following.
(10 marks)
(a) Physical geology (b) Stratigraphy (c) Lithology
Discuss the following completely. (10 marks)
(a) Sedimentation
(b) Sedimentary rocks
(c) Geophysics (d) Geochemistry
3.** Describe the geological time scale and three types of rocks.
(10 marks)
4.** Explain about the structure of the earth.
(10 marks)
5.** How many groups are classified for the rock formation? Describe them completely.
(16 marks)
6.** Write down the notes on the following.
(10 marks)
(a) The origin and habit of oil
(b) Hydrocarbon reservoir
7.** How many traps are forming for the principal classification of petroleum reservoir?
Explain completely. (16 marks)
8.** Draw the representative diagram and write notes on the following.
(a) Domes and anticlines
(b) Structure with unconformity
9.** Describe the following with necessary drawings.
(10 marks)
(a) Salt dome and plug structures
(b) Structures associated with faulting
10.*** Define the following.
(10 marks)
(a) Structural map
(b) Isopacheous map
11.*** Differentiate between normal pressure and abnormal pressure.
(10 marks)
12.** Explain about reservoir temperature and geothermal gradient.
(10 marks)
13.*** Describe the differences between the normal and abnormal pressure.
(10 marks)
Chapter 2 Reservoir Rock Properties
14.** Define the following. (16 marks)
(a) Porosity (b) Primary porosity (c) Secondary porosity (d) Dual porosity
15.* Define permeability of a porous media.
(8 marks)
16.** Describe the practical definition of Darcy. (8 marks)
17.** Discuss the factors affecting the permeability.
(8 marks)
18.** Discuss the Klinkenberg effect completely.
(10 marks)
19.** How the permeability is measured in the laboratory.?
(10 marks)
20.** How the porosity can be measured in the laboratory?
(10 marks)
21.*** Describe the measurement of capillary pressure by mercury injection.
(10 marks)
22.*** Discuss the relation between porosity and permeability.
(10 marks)
Chapter 3 Surface Tension, Wettability, Capillarity, Saturation and Fluid Displacement
23.** What is surface tension? Explain with diagram.
(10 marks)
24.** Describe the measurement of surface tension.
(10 marks)
25.* Explain about the wettability.
(10 marks)
26.** What is the capillarity?
(10 marks)
27.** Discuss about the capillary pressure between spherical particles. (10 marks)
28.*** Explain the repartition of saturation in reservoir rocks.
(10 marks)
29.*** What is the irreducible water saturation?
(10 marks)
30.*** What is the displacement pressure? (10 marks)
31.** Describe the displacement of oil in the pore channels completely. (10 marks)
32.*** Explain about the residual oil with necessary diagram.
(10 marks)
33.** What is the absolute permeability? (5 marks)
34.** What is the effective permeability? (5 marks)
35.*** Explain about the relative permeability.
(10 marks)
36.* Describe the mobility ratio and explain it. (10 marks)
37.** Describe the difference between relative porosity and relative permeability. (10 marks)
Chapter 4 Reservoir Estimates
38.** Describe the volumetric method.
(10 marks)
39.** Explain the geophysical exploration stages. (10 marks)
40.** Explain the field development stages.
(10 marks)
41.** Describe the procedure for determination of the reserve.
(10 marks)
42.** What are the uncertainty in reservoir estimates?
(10 marks)
43.** Explain the field integrated log analysis and reservoir mapping services. (10 marks)
44.** Describe the normalization of log data.
(10 marks)
45.** Explain about the monitoring of fluid interface changes.
(10 marks)
46.* Describe the material balance method for gas reservoirs. (10 marks)
47.* Write down the generalized material balance relation for oil reservoirs.
(10 marks)
48.** Discuss the accuracy of the results for calculation of reserves.
(10 marks)
49.** Describe about the water influx.
(10 marks)
50.*** Compare the methods between volumetric method and material balance method.
(10 marks)
* = Must know, ** = Should know, *** = Could know
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Worked Out Examples for
1. Explain about the structure of the earth.
The internal structure of the earth is revealed by analysis of earthquake waves as they are
reflected and refracted by concentric layers of material with differing density and rigidity.
The core, extending to a radius of 2200 miles from the center of the earth, is composed of
nickel-iron with an average density of 10.7 gm/cc. The inner core is solid while the outer core is
Surrounding the core is an 1800 miles thick mantel, largely composed of iron silicates
whose density, averaging 4.0gm/cc, varies with depth. The mantel is plastic expect for the outer
400miles or so which is solid.
The crust or lithosphere, composed of the more common variety of rocks, varies in
thickness from about 25 miles under the continents to 7miles beneath the ocean basins.
The crust is of insignificant thickness when compared to the core and mantel. It consists of:
Abasaltic layer overlying and floating on the mantel. It is rich in lime, iron and
magnesia with a density of 2.9g/cc.
A second lighter granitic layer, rich in silica and alkalies of 2.7g/cc density,
resting on the basaltic layer and forming the continents.
Driven by density differences due to unequal heating, convection currents circulate in the
plastic. Uplifting magma breaks through the crystal layer forcing a rift into which the upwelling
material solidifies as it surfaces, becoming a mid-ocean ridge.
As new material is continuously added at the rift, the crust spreads, and the continental
land masses are rafted apart.
Compressional forces caused by lateral plate movements cause thickening and folding of
the crust until faulting occurs, and one plate overrides the other, driving it back into the mantel
where it is remelted.
It is now generally accepted that this mechanism caused the break up of the continents
from a single land mass 200 millions years ago.
Mechanical forces act to break down gniess and granite from boulders to cobbles, pebbles,
gravel, and smaller particles. Chemical weathering transforms feldspars into clays, releasing
quartz which become sand. The great abundance of quartz sand is due to the fact that quartz
grains are extremely durable. Sands are also composed of feldspar and carbonate particles.
2. How many traps are forming for the principal classification of petroleum reservoir?
Explain completely.
The principal classifications of petroleum-reservoir-forming traps are as follows:
Domes and Anticlines
Domes and anticlines are formed by uplifting and folding of the strata. When viewed from
above the dome is circular in shape, whereas the anticline is an elongated fold.
Salt Domes and Plug Structures
This commonly occurring geological structure is caused by the intrusion from below of a
salt mass, volcanic material, or serpentine. In pushing up or piercing through the overlying strata
the intrusion may cause the formation of numerous traps in which petroleum may accumulate.
Structure Associated with Faulting
Reservoirs may be formed along the fault plane where the shearing action has caused an
impermeable bed to block the migration of oil and gas through a permeable bed.
Structure with Unconformity
This type of structure can be formed where more recent beds cover older, inclined
formations that have been planed off by erosion. A reservoir may be formed where oil and gas is
trapped by an impermeable overlying layer.
Lenticular Reservoirs
Oil and gas may accumulate in pockets of porous permeable beds, or traps formed by
pinch-outs of the porous beds, within an impermeable bed.
Lens-type reservoirs are formed where sand was deposited along an irregular coastline or
by filling in an ancient river bed or delta. Similar productive zones occur in various porous
sections in thick impermeable limestone beds.
Pinch-outs may occur near the edge of a basin where the sand progressively “shales out”
as the edge of the basin is approached. In river-deposited sand bars, shale-out frequently occurs
within a few hundred feet.
3. Define the following.
(a) Porosity (b) Primary porosity (c) Secondary porosity (d) Dual porosity
(a) Porosity
The void spaces in reservoir rocks are the inter-granular spaces between the sedimentary
particles. Porosity defined as a percentage or fraction of void to the bulk volume of the rock.
While the proportion of void can be calculated for regular arrangements of uniform spheres, the
porosity of reservoir rocks must be determined by direct measurement on core samples in the
laboratory or estimated in situ by well log analysis. In reservoir engineering, only the
interconnected or effective porosity is interest since this is the only capacity which can make a
contribution to flow.
(b) Primary porosity
Primary porosity refers to the void spaces remaining after sedimentation of the granules in
the matrix and hence is a matrix porosity.
( c ) Secondary porosity
Secondary porosity is the contribution from pits, vugs, fractures and other discontinuities
in the bulk volume of the matrix. The contribution of secondary porosity to the overall bulk
porosity is generally small yet it can lead to dramatic increases in bulk permeability.
(d) Dual porosity
The distinguishing factor between primary and secondary porosity is not the mode of
occurrence but the very different flow capacity where an interconnected secondary porosity
system is present. This is called a dual porosity system.
It will normally be possible to distinguish any effect of dual porosity if the “coarse”
system has a flow capacity about two orders of magnitude greater than that of the “fine” system.
4. Define permeability of a porous media.
The permeability of a rock is a measure of its specific flow capacity and can be
determined only by a flow experiment. Since permeability depends upon continuity of pore space,
there is not any unique relation between the porosity of a rock and its permeability, (except that a
rock must have a non-zero porosity if it is to have a non-zero permeability).
Henry Darcy experimented with filtration of water through unconsolidated sand beds, the
results of Darcy’s studies, for horizontal flow, can be written in differential form:
k dp
µ dl
This is the defining equation for the measurement of permeability by flow measurement.
The unit of proportionality k, between velocity and pressure gradient, is the coefficient of
permeability and is usually measured in Darcy.
Note that the coefficient of permeability k, is a rock characteristic independent of the fluid
used for its measurement.
5. What is surface tension? Explain with diagram.
The apparent film which separates two immiscible fluids, such as air-water is caused by
unequal attractive forces of molecules at the interface. It is because of these attractive forces that a
bubble contracts into spherical form. It can be shown theoretically and experimentally that a
pressure difference must exist across any curved fluid interface-the smaller the radius the greater
is the pressure on the concave side.
The pressure in a bubble having a single phase boundary is given by:
in dynes.
P1 − P2 =
where: σ =surface tension, dynes/cm
r = bubble radius, cm
P= pressure, dynes/cm2
Fig. 3.1 & 3.2
6. What is the capillarity?
When liquid wets the surface of a fine bore glass capillary tube, surface tension around the
periphery of the contact pulls the liquid interface up the tube until an equilibrium is reached with
the downward force due to the liquid column height.
The equation expressing this pressure equilibrium is given by:
cosθ σ (2πr ) = πr 2 h( pLl − pV ) g
where: σ = surface tension, dynes/cm
r = radius of tube, cm
h = capillary rise, cm
PL= density of liquid, grams/cu cm
PV= density of vapor, grams/cu cm
g = acceleration due to gravity, 980.6 cm/sec2
θ = contact angle between liquid and solid
cos θ =-1 when θ =180˚, for complete wetting
A pressure P2 applied above the liquid in the capillary sufficient to lower the meniscus to
the same level (pressure P1) as in the reservoir is equal to the capillary pressure, Pc. That is, the
capillary pressure Pc = P2 - P1 in this condition of equilibrium.
Fig. 3.6 & 3.7
7. Explain about the relative permeability.
Relative permeability is the ratio of effective permeability to absolute permeability:
k rw =
k ro =
k rg =
A typical set of oil/water relative permeability curves are shown, together with the
corresponding capillary pressure relations. Both are plotted vs. water saturation to illustrate their
Starting with a 100% water saturated core at point (1) the threshold pressure must be
exceeded before the non-wetting fluid (oil) can displace the wetting phase fluid, water from the
At higher displacement pressures increasing amounts of water are drained from the core
until at (2) the irreducible water saturation is reached. At (2) the relative permeability to water
becomes zero. Note that the irreducible water reduces the amount of pore volume to oil flow.
When water is imbibed into the core which now contains oil and water, the saturation of
water increases up to the point (3), where the residual oil saturation is reached. The relative
permeability to water can not reach a value of 1.0 because of the residual oil remaining in the
8. Describe the volumetric method?
Volumetric methods involve a determination of the bulk reservoir rock volume, average
porosity and fluid saturation, from which the total reservoir hydrocarbon volume is calculated.
Recoverable reserves are then estimated by application of a suitable recovery factor and
the formation/surface volume factor for the produced fluid.
V × φ × (1 − S w ) × R.F
Recoverable oil = b
where: Vb is the bulk reservoir volume
φ is the fractional porosity
(1-Sw) is the hydrocarbon saturation
R.F is the recovery factor
Bo is the formation volume factor (oil)
A recovery factor is approximated considering:
S − S or
- laboratory measurement of oil displacement in cores o
- type of displacement mechanism involved
- correction of sweep efficiency based on similar reservoir
9. Describe the procedure for determination of the reserve.
The gross reservoir rock volume enclosed by the structure above the hydrocarbon/ water contact
is calculated in the following steps:
1. A net sand isopach map, giving the contours of equal thickness of pay with the water contact
assigned zero elevation contour is the most convenient basis for rock volume calculation.
2. The area within each contour is determined by planimetering, and a plot prepared of area
contained in each contour vs. depth.
3. The gross rock volume is A dh which may be found by planimetering again or by application of
a numerical integration rule. In the Schlumberger field studies approach the volumetric
reservoir distribution is calculated numerically and plotted by machine as isopach and
isovolume maps.
An + An +1 + An An +1
- Trapezoidal formula ∆Vb = An + An +1
Vb = A0 + 2 A1 + 2 A2 ........2 An −1 + An + t av × An
Pyramidal formula ∆V =
10. Write down the generalized material balance relation for oil reservoirs.
Generalized material balance relation for oil reservoirs
When hydrocarbons are produced from the reservoir, the attendant pressure drop causes
the remaining fluids to expand and water to encroach, compensating for the removed volume.
The generalized material balance relation below equates the volumes of oil, water and gas
removed with expansion of the oil, free gas, and encroached water.
N ( Bt − Bti ) +
NmBti ( B g − B gi )
+ We = N p Bt + N p ( R p − Rsi ) + BwW p
B gi
Where: Bg = Gas formation volume factor
Bo = Oil formation volume factor
Bt = Total (two phase) formation volume factor Bt=[Bo+(Rsi-Rs) Bg]
Bw = Water formation volume factor
m = Ratio of initial reservoir free gas volume to initial reservoir oil volume
N = Initial oil in place
Np= Cumulative oil produced
Rp = Producing gas-oil ratio
Rs = Solution gas-oil ratio
We = Cumulative water encroached
Wp = Cumulative water produced
I = Initial
Assumptions generally made are that the volume factors for solution gas and free gas from
the cap are identical and that there is negligible release of gas from the produced water.
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Sample Questions for
Chapter 1 Reservoir Fluid Behavior
Classify the composition of reservoir hydrocarbons.
(10 marks)
State the classification process of oil.
(16 marks)
Explain the phase behavior of hydrocarbon fluids. (10 marks)
Explain the phase behavior of a single-component system. (16 marks)
Discuss the phase behavior of multi-component system.
(16 marks)
Describe the phase diagram of a low shrinkage reservoir fluid.
(10 marks)
Discuss the phase diagram of retrograde condensate reservoir fluid. (16 marks)
Explain the diagrams of a dry gas reservoir fluid and wet gas reservoir fluid.(16 marks)
What are the three possible sources of data to get the exact knowledge of a reservoir fluids
physical properties. (16 marks)
10.* Describe the conversion factors between surface and downhole volumes of oil, gas, and
(16 marks)
11.*** Define formation volume factor and explain about the gas formation volume factor briefly.
(16 marks)
12.** Describe the oil formation volume factor completely.
(10 marks)
13.*** Explain about the formation volume factor of water.
(10 marks)
14.*** Define the specific gravity of gas. (10 marks)
Chapter 2 Properties of Gases
15.* What is an ideal gas? Explain completely. (10 marks)
16.** Define the following.
(15 marks)
(a) Boyles' law
(b) Charles' law
(c) Avogadro's law
17.** Define the density of an ideal gas. (5 marks)
18.** Explain the Dalton's law of partial pressures.
(10 marks)
19.* Describe the Amagat's law completely.
(10 marks)
20.** What is the apparent molecular weight of a gas mixture? (10 marks)
21.** Describe the compressibility equation of real gas in mathematical form and explain it.
(10 marks)
22.** Describe the law of corresponding. (10 marks)
23.** Write down the Van der Waals' equation and explain completely. (10 marks)
24.** What is the Beattie-Bridgeman equation. Explain completely.
(10 marks)
25.** Discuss the coefficient of isothermal compressibility of gases using mathematical
expression. (10 marks)
26.** Explain about the coefficient of viscosity of a gas briefly. (10 marks)
27.** Describe the viscosity of pure hydrocarbon gases briefly. (10 marks)
28.** Express the formula for the viscosity of gas mixtures and explain. (10 marks)
Determine the value of gas constant for a system of units in which volume is measured in
cubic inches, pressure in atmosphere absolute, temperature in degrees Kelvin, and quantity
in pound moles.
(16 marks)
Compute the apparent molecular weight, the specific gravity, the composition in weight
fraction and the composition in volume fraction of a gas with the following composition.
(16 marks)
Composition, (mole fraction)
A piece of sand stone with a bulk volume of 1.3 cc is contained in a 5-cc cell filled with
helium at 760 mm Hg. Temperature is maintained constant and the cell is opened to
another evacuated cell of the same volume. The final pressure in the two vessels is 334.7
mm Hg. What is the porosity of sand stone?
(16 marks)
32.** A certain machine used methane as a fuel. The gas is purchased in containers holding 25
lbs of methane and costing of $3.20 exclusive of the container. If the machine used 300 cu
ft per day at 15 oz per sq in. gauge and 100°F, what is the cost of the methane per day?
What will 1.0Mcf at 14,7 psia and 60°F cost?
(16 marks)
Chapter 3 Properties of Reservoir Liquids
Describe the specific gravity of a liquid completely.
(10 marks)
Describe the two-step procedure for the calculation of liquid density at pressures above the
(10 marks)
Define the coefficient of isothermal compressibility of a liquid.
(10 marks)
Explain the pseudocritical properties of hydrocarbon liquid mixtures.
(10 marks)
Describe the coefficient of isobaric thermal expansion of a liquid. (10 marks)
Discuss how to estimate the formation volume factor of oil at pressures above the bubblepoint pressure.
(10 marks)
Explain what is the total formation volume factor. (10 marks)
Give the short notes on the formation volume factor of gas condensate.
(10 marks)
Describe the recombination method for the estimation of formation volume factor of gas
condensate. (10 marks)
Discuss how to estimate the formation volume factor of gas condensate by use of well-gas
gravity correlation. (10 marks)
Describe the estimation of formation volume factor of gas condensate from the total
formation volume factor correlation of Standing.
(10 marks)
Describe the estimation of formation volume factor of gas condensate using the
correlation of Sage and Olds. (10 marks)
Describe the estimation of oil viscosity at bubble-point pressure and below bubble-point
(10 marks)
An oil has a specific gravity 60°/60° of 0.85. What is its density in pounds/cubic foot and
its gravity in °API? (10 marks)
An oil has the following composition.
Mole fraction
What is the API gravity of the oil? (Use ideal-solution principles.) (10 marks)
48.* Calculate the density of the following liquid at its bubble point of 3000 psia at a reservoir
temperature of 200°F. Use the iterative procedure and the data given in Figure 3.1.
(16 marks)
49.* The solubility of a 0.75 gravity gas in an oil of 48°API is 850 scf/STB at 2500 psia and
180°F. What is the density of the liquid at 2250 psia and 180°F?
(16 marks)
50.* What is the density of a crude oil at its bubble point pressure of 2250 psia at 200°F, if Rsb
= 800 scf/STB, the gas gravity is 0.9, the oil gravity is 43°API, and Bob is 1.50?
(16 marks)
* = Must know, ** = Should know, *** = Could know
Ministry of Science and Technology
Department of Technical and Vocational Education
Petroleum Engineering
Worked Out Examples for
1. Classify the composition of reservoir hydrocarbons.
Petroleum consists predominantly of paraffin series hydrocarbons together with lesser
quantities of cyclic hydrocarbons naphthelenes and aromatics mixed together in varying
proportions. There are no definite demarkations between the categories, which often vary in
definition according to local usage. In general oils contain a higher fraction of heavy molecules,
while at the other end of the scale, methane and other light molecules predominate in gases.
Crude oil chemically is quite complex and a typical crude may contain several thousand
different compounds belonging to 18 different hydrocarbon series. A complete chemical analysis
of crude oils, in terms of compounds present, is a difficult, if not impossible task and, less
complete types of analyses are often not useful for determining its physical characteristics.
Difficult in classifying oils by the chemical composition of their constituents has led to
widespread use of simpler, less technical classifications.
2. Describe the phase diagram of a low shrinkage reservoir fluid.
The shape of the two phase envelope and its position on the P-T diagram is determined by
the chemical composition and amount of each constituent present. Each reservoir fluid has a
unique phase diagram.
Fluid at reservoir temperature and pressure at point A exists as undersaturated liquid. If a
sample of this fluid is expanded in a PVT cell at reservoir temperature Tr , from A the
bubble point pressure will be reached at A. This is approximately the path that fluids follow in
moving horizontally through the reservoir to the well bore.
A continued expansion at Tr, yields increasing percentage of gas and decreasing
percentage of oil in the PVT cell until at point B there remains 75% oil.
The path from A to the pressure and temperature the separator is operated at is shown by a
dashed line and indicates the fractions of gas and oil recovered at separator conditions.
3. Discuss the phase diagram of retrograde condensate reservoir fluid.
The phase diagram 1-8 typifies the behavior of a retrograde reservoir. Fluid at point A’ is
above critical temperature and is therefore classified as gas. On reduction of pressure at constant
temperature from point A’, the dew point line is crossed at A and liquid begins to
condense from the reservoir gas.
If the pressure and temperature are reduced from A along the dashed line path to separator
condition, the diagram shows that 25% of liquid is recovered at this point. On further reduction of
pressure to atmospheric pressure about 2% of liquid remains.
The recovered liquid is termed “condensate” or “distilled” which is simply oil, generally
light in color and low in specific gravity.
4. Explain the diagrams of a dry gas reservoir fluid and wet gas reservoir fluid.
Phase diagram of a dry gas reservoir fluid
The phase diagram given in fig. 1-9, on the pressure-temperature plane, typifies the
behavior of a dry gas reservoir. If the pressure and temperature are reduced from the original
reservoir conditions at point A to standard stock tank conditions (60ׁ F and 14.7 psia), there is no
liquid recovery and the reservoir fluid remains completely in the gaseous phase during the
produces a small quantity of liquid condensate on reduction to separator/stock tank conditions,
may be termed “wet gas”. A typical phase diagram for a wet gas is shown in figure 1-10.
5. What is an ideal gas? Explain completely.
As a starting point in the study of the properties of real gases, we will consider a
hypothetical fluid known as an ideal gas. An ideal gas is a fluid in which:
1. The volume occupied by the molecules is insignificant with respect to the volume
occupied by the total fluid.
2. There are no attractive or repulsive forces between the molecules or between the
molecules and the walls of the container.
3. All collisions of molecules are perfectly elastic, that is, there is no loss of internal
energy upon collision
We shall see later why the molecules of an ideal gas must have these properties.
6. Define the density of an ideal gas.
Density is defined as the weight of a gas per unit volume, an equation of state can be used
to calculate the densities of a gas at various temperatures and pressures. The equation for the
density of an ideal gas follows from equation of pv =m/MRT,
ρg = m / v = pM / RT
7. Explain the Dalton's law of partial pressures.
Dalton postulated that each gas in a mixture of gases exerts a pressure equal to the
pressure; it would exert if it alone were present in the volume occupied by the gas mixture. The
total pressure exerted by the mixture of gases is equal to the sum of the pressures exerted by its
components. The pressure exerted by each of the component gases is as known as partial pressure.
This law is valid only when the mixture and each component of the mixture obey the ideal
gas law. It is sometimes called the law of additive pressures.
The partial pressure exerted by each component of the gas mixture can be calculated using
the ideal gas law. Consider a mixture containing nA moles of component, nC moles of component
C, and so on. The partial pressure exerted by each component of the gas mixture may be
determined with the ideal gas equation;
pa = na RT / V, PB = nB RT / V, pC = nC RT / V,....
8. Describe the specific gravity of a liquid completely.
Liquid specific gravity, γo is defined as the ratio of the density of the liquid to the
density of the water, both taken at the same temperature and pressure.
γo = ρo / ρw
Specific gravity appears to be nondimensional since the units of the density of the liquid
are the same as the units of the density of the water; however, this is not strictly true. Actually, in
the English system the units are
γo = £a = ( lb oil / cu ft oil ) / ( lb water / cu ft water )
Sometimes specific gravity is given as sp gr 60 ׁ/ 60ׁ which means that the densities of
both the liquid and water were measured at 60ׁ F and atmospheric pressure.
The petroleum industry also uses another gravity term called API gravity, which is defined
API = [( 141.5 ) / γo ] – 131.5
Where γo is the specific gravity at 60ׁ/ 60ׁ
9. Define the coefficient of isothermal compressibility of a liquid.
The correlation of pseudoreduced compressibility given in Fig. 3-8 was prepared through
the use of calculations like example 3-8 using the reduced density data given in Fig. 3-7. The
values given in Fig. 3-8 are not strictly valid since Fig. 3-7 is a correlation of density data for pure
liquids rather than liquid mixtures. However, values of Co obtained from Fig. 3-8 are as accurate
as those computed with the procedures of example 3-8.
Example 3-9 Rework example 3-8 using Fig. 3-8.
From fig. 3-8 at Tpr = 0.7 and Ppr = 7
Cpr = 0.0046
Co = cpr / Ppc = 0.0046 / 300 psia
Co = 15.3 x 10-6 psia-1
10. Explain the pseudocritical properties of hydrocarbon liquid mixtures.
When the composition of a liquid mixture is known, the pseudo-critical temperature and
pseudocritical pressure can be computed using methods identical to those for a gas mixture given
in equation 2-40. However, the petroleum engineer rarely knows the composition of the very
complex liquid mixtures which are encountered.

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