Chapter 02
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Transcript of Chapter 02
Reservoir Pressures and Temperatures
CONTENTS
1 INTRODUCTION
2 ABNORMAL PRESSURES
3 FLUID PRESSURES IN HYDROCARBON
SYSTEMS
4 PRESSURE GRADIENTS AROUND WATER-
OIL CONTACT
5. TECHNIQUES FOR PRESSURE
MEASUREMENT
6. RESERVOIR TEMPERATURE
2
LEARNING OBJECTIVES
Having worked through this chapter the Student will be able to:
• Having worked through this chapter the student will be able to:
• Define the terms; lithostatic pressure, hydrostatic pressure and hydrodynamic
pressure.
• Draw the normal hydrostatic pressure gradient for water systems.
• Define normal pressured reservoirs, overpressured reservoirs and underpressured
reservoirs
• Describe briefly and sketch the pressure gradients associated with overpressured
and underpressured reservoirs.
• Describe briefly , sketch and present equations for the pressures in a water
supported oil and gas bearing formation.
• Illustrate how a downhole formation pressure device can be used to discriminate
permeability layers after production has commenced.
• Comment briefly what geothermal gradient is in a reservoir where flow
processes occur at constant reservoir temperature.
Reservoir Pressures and Temperatures
Institute of Petroleum Engineering, Heriot-Watt University 3
1. INTRODUCTION
Determining the magnitude and variation of pressures in a reservoir is an important
aspect in understanding various aspects of the reservoir, both during the exploration
phase but also once production has commenced.
Oil and gas accumulations are found at a range of sub-surface depths. At these depths
pressure exists as a result of the depositional process and from the fluids contained
within the prous media. These pressures are called lithostatic pressures and fluid
pressures. These pressures are illustrated in figure 1.
The lithostatic pressure is caused by the pressure of rock which is transmitted through
the sub-surface by grain-to grain contacts. This lithostatic or sometimes called geostatic
or overburden pressure is of the order of 1 psi/ft. The lithostatic pressure gradient
varies according to depth, the density of the overburden, and the extent to which the
rocks are supported by water pressure. If we use this geostatic pressure gradient of
1 psi/ft. then the geostatic pressure Pov
, in psig at a depth of D feet is
pov
= 1.0D (1)
The geostatic pressure is balanced in part by the pressure of the fluid within the pore
space, the pore pressure, and also by the grains of rock under compaction. In un-
consolidated sands, loose sands, the overburden pressure is totally supported by the
fluid and the fluid pressure Pf is equal to the overburden pressure P
ov . In deposited
formations like reservoir rocks the fluid pressure is not supporting the rocks above
but arises from the continuity of the aqueous phase from the surface to the depth D in
the reservoir. This fluid pressure is called the hydrostatic pressure. The hydrostatic
pressure is imposed by a column of fluid at rest. Its value depends on the density of
the water w, which is affected by salinity. In a sedimentary basin, where sediment
has settled in a region of water and hydrocarbons have been generated and trapped,
we can expect a hydrostatic pressure. For a column of fresh water the hydrostatic
pressure is 0.433 psi/ft. For water with 55,000 ppm of dissolved salts the gradient is
0.45 psi/ft; for 88,000 ppm of dissolved salts the gradient is about 0.465 psi/ft.
Its variation with depth is given by the equation.
Pf =
wDg (2)
where g is the acceleration due to gravity.
There is another fluid pressure which arises as a result of fluid movement and that
is called the hydrodynamic pressure. This is the fluid potential pressure gradient
which is caused by fluid flow. This however does not contribute to in-situ pressures
at rest.
4
Pressure (psia)
De
pth
(F
t.) FP GP
Overburden
Pressure (OP)
Underpressure
Overpressure
Normal
14.7
(FP = Fluid Pressure, GP = Grain Pressure)
0
Figure 1 Gives the relationship between the lithostatic pressure and the hydrostatic
pressure. 1
Fluid pressure in hydrocarbon accumulations are dictated by the prevailing water
pressure in the vicinity of the reservoir. In a normal situation the water pressure at
any depth is:
PdP
dDx D 14.7psiaw
water (3)
where dP/dD is the hydrostatic pressure gradient
This equation assumes continuity of water pressure from the surface and constant
salinity. In most cases even though the water bearing sands are divided between
impermeable shales, any break of such sealing systems will lead to hydrostatic pres-
sure continuity, but the salinity can vary with depth.
Reservoirs whose water pressure gradient when extrapolated to zero depth give an
absolute pressure equivalent to atmospheric pressure are called normal pressured
reservoirs.
EXERCISE 1If the average pressure gradient in a region is 0.47 psi/ft, calculate the pore
pressure in a normally pressurised formation at 7400ft. Convert the pressure from psi to KPa, then express the pressure in MPa. What is the pressure gradient in
KPa/m?
Reservoir Pressures and Temperatures
Institute of Petroleum Engineering, Heriot-Watt University 5
2. ABNORMAL PRESSURE
Under certain conditions, fluid pressures may depart substantially from the normal
pressure. Overpressured reservoirs are those where the hydrostatic pressure is greater
than the normal pressure and underpressured reservoirs are below normal pressure.
Figure 1. They are called abnormal pressured reservoirs and can be defined by the
equation:
P = dP
dDx D + 14.7 psia + Cw
water (4)
where C is a constant, being positive for overpressured and negative for an under-
pressured system.
For abnormally pressured reservoirs, the sand is sealed off from the surrounding strata
so that there is not hydrostatic pressure continuity to the surface.
Conditions which cause abnormal fluid pressure in water bearing sands have been
identified by Bradley 2 and include (Figure 2):
Original Deposition
Dense ShaleShale deposited too
quickly to allow
fluid equilbrium
FP-Too High
Upthrust
Reservoir
North Sea
Glacier
Greenland 3 km thick
1300 psi/1000 m ice
Normal Surface
(a)
(b)
(c)
Figure 2 Causes of overpressurring
• Thermal effects, causing expansion or contraction of water which is unable
to escape ; an increase in temperature of 1˚F can cause an increase of 125 psi
in a sealed fresh water system.
6
• Rapid burial of sediments consisting of layers of sand and clay. Speed of burial
does not allow fluids to escape from pore space.
• Geological changes such as uplifting of the reservoir, or surface erosion both
of which result in the water pressure being too high for the depth of the burial.
The opposite occurs in a down thrown reservoir.
• Osmosis between waters having different salinity, the sealing shale acting as a
semi-permeable membrane. If the water within the seal is more saline than the
surrounding water, the osmosis will cause a high pressure and vice versa.
Overpressured reservoirs are common in Tertiary deltaic deposits such as the North
Sea, Niger delta and the Gulf Coast of Texas. In the North Sea one mechanism for
overpressure is the inability to expel water from a system of rapidly compacted
shales.
With abnormally pressured reservoirs a permeability barrier must exist, which inhibit
pressure release. These may be lithological or structural. Common lithological
barriers are evaporates and shales. Less common are the impermeable carbonates
and sandstones. Structure permeability barriers may result from faults which, in
some cases, seal. The subject on of abnormal pressures is covered more fully in
the Geology Module
If reservoirs are all normal pressured systems then the pressure gradient for these
reservoirs would be virtually all the same, other than from the influence of salinity.
The figure below shows the water pressure gradients for a number of reservoirs in
the North Sea and indicates the significant overpressuring in this region. Often these
overpressuring show regional trends. For example the fields depicted in figure 3
show an increase in abnormal pressure in the south east direction. Clearly if all these
reservoirs were normally pressured then the pressure depths values would lie on the
same gradient line with a zero depth pressure value of atmospheric pressure.
Reservoir Pressures and Temperatures
Institute of Petroleum Engineering, Heriot-Watt University 7
4
21
3
5
Note: Water gradient lines drawn
through known or projected
oil/water contacts
Alwyn
Lyell
Ninian
OWC
Heather
OWC
Cormorant
OWC
S.W> Ninian
N.W. Alwyn
Thistle OWC
Brent OWC
Statfjord OWC
5000 6000 7000 8000 9000 10,000
13,000
12,000
11,000
10,000
9,000
8,000
Pressure, psig
Su
bse
a D
ep
th (
Fe
et)
Figure 3 Examples of overpressured reservoirs in the North Sea3
3. FLUID PRESSURES IN HYDROCARBON SYSTEMS
Pressure gradients in hydrocarbon systems are different from those of water systems
and are determined by the oil and gas phase in-situ specific gravities, o and
g of
each fluid.
The pressure gradients are a function of gas and oil composition but typically are:
dP
dD = (0.45 psi / ft)
water (5)
dP
dD = (0.35 psi / ft)
oil
(6)
dP
dD = (0.08 psi / ft)
gas (7)
8
For a reservoir containing both oil and a free gas cap a pressure distribution results,
as in the Figure 4 As can be seen, the composition of the respective fluids gives rise
to different pressure gradients indicated above. These gradients will be determined by
the density of the fluids which result from the specific composition of the fluids.
Depth (Ft.)
Formation Pressure (PSI)
Gas-Oil Contact
0.17 psi/ft
f = 0.39 gm/cc
0.29 psi/ft
f = 0.67 gm/cc
0.47 psi/ft
f = 1.09 gm/cc
Oil-Water Contact
1
4000
8800
8700
8600
8500
4050 4100 4150
2
3
4
5
67
89
1011
12
13
De
pth
(F
t.)
Figure 4 Pressure distribution for an oil reservoir with a gas cap and an oil-water contact.
The nature of the pressure regime and the position and recognition of fluid contacts
are very important to the reservoir engineer in evaluating reserves, and determining
depletion policy.
The data used for these fluid contacts comes from:
(i) Pressure surveys
(ii) Equilibrium pressures from well tests
(iii) Flow of fluid from particular minimum and maximum depth
(iv) Fluid densities from reservoir samples
(v) Saturation data from wireline logs
(vi) Capillary pressure data from cores
(vii) Fluid saturation from cores
EXERCISE 2If the pressure in a reservoir at the OWC is 3625 psi, calculate the pressure at the top if there is a 600ft continuous oil column. If a normal pressure gradient exists outwith the reservoir, calculate the pressure differential at the top of the reservoir. Redo the calculations for a similar field, but this time containing gas.
Reservoir Pressures and Temperatures
Institute of Petroleum Engineering, Heriot-Watt University 9
4. PRESSURE GRADIENTS AROUND THE WATER-OIL CON-
TACT
Water is always present in reservoir rocks and the pressure in the water phase Pw
and the pressure in the hyrocarbon phase Po are different . If P is the pressure at the
oil/water contact where the water saturation is 100%, then the pressure above this
contact for the hydrocarbon and water are :
Po = P -
ogh (8)
Pw
= P - wgh (9)
The difference between these two pressures is the capillary pressure Pc: see Chapter
8. In a homogenous water-wet reservoir with an oil-water contact the variation of
saturation and phase pressure from the water zone through the capillary transition
zone into the oil is shown in Figure 5). In the transition zone the phase pressure
difference is given by the capillary pressure which is a function of the wetting phase
saturation. (Chapter 8).
Oil Phase Pressure
Oil Zone
Capilliary
Transition
Zone
Water Zone
Water Saturation, Sw Pressure, P
Vertical
Depth
D
WOC
Oil Gradient
Water Gradient
Water Phase Pressure
po = pFWL -
ogh
pw = pFWL -
wgh
(pc
= o)
Swc
Sw
pc
FWL
pFWL0 1
pc
(Sw)
gh =
Figure 5 Pressure Gradients around the Water-Oil Contact
Pc = P
o - P
w (10)
at hydrostatic equilibrium
Pc(Sw)
= ∆ gh
∆ = w-
o
h = height above free water level
10
The free water level, FWL, is not coincident with the oil-water contact OWC. The
water contact corresponds to the depth at which the oil saturation starts to increase
from water zone. The free water level is the depth at which the capillary pressure
is zero.
The difference in depth between the oil-water contact and the free water level depends
on the capillary pressure which in turn is a function of permeability, grain size etc.
Providing the phase is continuous the pressures in the respective phases are:
Po = P
FWL -
ogh (11)
Pw = P
FWL -
wgh (12)
On the depth-pressure diagram the intersection of the continuous phase pressure line
occurs at the free water level.
5. TECHNIQUES FOR PRESSURE MEASUREMENT
Earlier tests for vertical pressure logging have been replaced by open-hole testing
devices that measure the vertical pressure distribution in the well, and recover for-
mation samples.
One such device which was introduced in the mid seventies which has established
itself in reservoir evaluation is the repeat formation tester RFT (Schlumberger trade
name). It was initially developed as a device to take samples. Over the years however
its main application is to provide pressure -depth profiles over reservoir intervals. The
device places a probe through the well mud cake and allows small volumes of fluid
to be taken and pressure measurements to be made (Figure 6). It can only be operated
therefore in an open hole environment. The unit can be set at different locations in
the well and the pressure gradient thereby obtained. This device has been superseded
by different tools provided by a number of wireline service providers. The principle
is the same of measuring with a probe in open hole the pressure depth profile.
Reservoir Pressures and Temperatures
Institute of Petroleum Engineering, Heriot-Watt University 11
Pressure Guage
Seal Valve
to Upper Chamber
Seal Valve
to Upper Chamber
Piston
Filter
Flow Line
Formation
Probe Closed
Probe Open and
Sampling
Packer
Chamber 1
Chamber 2
Flow Line
Equalising Valve
(To Mud Column)
Packer Mud Cake
Figure 6 Original Schematic of the RFT Tool
These open hole pressure measurements have proved valuable at both the appraisal
stage and can be used to establish fluid contacts. It has also proved particularly valu-
able during the development stage in accessing some of the dynamic characteristics
of the reservoir. The pressure changes in different reservoir layers resulting from
production reveal the amount of interlayer communication and these pressure meas-
urements can be a powerful tool in understanding the characteristics of the reservoir
formation.
By comparing current pressure information with those obtained prior to production,
important reservoir description can be obtained which will aid reservoir depletion,
completion decisions and reservoir simulation.
In 1980 Amoco3 published a paper with respect to the Montrose Field in The North
Sea which illustrates the application of pressure-depth surveys. Figure 7 shows the
pressure depth survey in 1978 of a well after production since mid 1976. Only the
top 45ft of the 75ft oil column had been perforated. The initial pressure gradient in-
dicates the oil and water gradients at the condition of hydrostatic equilibrium. The
second survey shows a survey after a period of high production rate, and reveals the
reservoir behaviour under dynamic conditions. The various changes in slope in the
pressure profile reveal the partial restricted flow in certain layers. Similar surveys
in each new development wells (Figure 8) show the similar profiles and enable the
detailed layered structure of the reservoir to be characterised which is important for
reservoir simulation purposes.
12
4000
8100
8200
8300
8400
8500
8600
8700
8800
2500
2550
2600
2650
Layer 4
Layer 1
Layer 2
Layer 3
Layer 5
2500 3000 3500
14 16 18 20 22 24 26
Tru
e v
ert
ica
l su
bse
a d
ep
th -
me
tre
s
Tru
e v
ert
ica
l su
bse
a d
ep
th -
fe
et
Reservoir pressure - MPa
Reservoir pressure - psig
Perforations Original
pressure
gradient
Top paleocene
Gr%
0 100Sw%
100 0
%
0 50
Figure 7 RFT Pressure Survey in Development Well of Montrose Field 3.
Reservoir pressure - psig
8000
8100
8200
8300
8400
8500
8600
8700
8800
8900
9000
2450
2500
2550
2600
2650
2700
3000 3200 3400
18 20 22 24 26 28
Reservoir pressure - MPa
Tru
e v
ert
ica
l su
bse
a d
ep
th -
fe
et
Tru
e v
ert
ica
l su
bse
a d
ep
th -
me
tre
s
symbol ?Well number Date
22/17-A6 05/04/77
A8 27/01/78
A11 20/12/77
A15 15/08/78
A17 02/11/78
A18 28/03/79
A6A8
A11A15
A17A18
Original
pressure
gradient
Figure 8 RFT Pressure Syrveys on a number of Montrose Wells3.
Reservoir Pressures and Temperatures
Institute of Petroleum Engineering, Heriot-Watt University 13
6. RESERVOIR TEMPERATUREThe temperature of the earth increases from the surface to centre. The heat flow out-
wards through the Earthʼs crust generates a geothermal gradient, gc. This temperature
variation conforms to both a local and regional geothermal gradient, resulting from
the thermal characteristics of the lithology and more massive phenomenon associated
with the thickness of the earthʼs crust along ridges, rifts and plate boundaries.
In most petroleum basins the geothermal gradient is of the order of 1.6˚F/100 ft.
(0.029 K/m) The thermal characteristics of the reservoir rock and overburden give
rise to large thermal capacity and with a large surface area in the porous reservoir
one can assume that flow processes in a reservoir occur at constant reservoir tem-
perature. The local geothermal gradient will be influenced by associated geological
features like volcanic intrusions etc. The local geothermal gradient can be deduced
from wellbore temperature surveys . However they have to be made under stabilised
conditions since they can be influenced by transient cooling effects of circulating
and injected fluids.
During drilling the local thermal gradient can be disturbed and by analysis of the
variation of temperature with time using a bottom hole temperature (BHT) gauge
the local undisturbed temperature can be obtained.
Without temperature surveys the temperature at a vertical depth can be estimated
using a surface temperature of 15 oC (60 oF) at a depth D.
T(D)
= 288.2 + gcD (K)
Solutions to Exercises
EXERCISE 1
If the average pressure gradient in a region is 0.47 psi/ft, calculate the pore pressure
in a normally pressurised formation at 7400ft. Convert the pressure from psi to KPa,
then express the pressure in MPa. What is the pressure gradient in KPa/m?
Multiply KPa by 0.145 to get psi.
1 US foot = 0.3048m.
SOLUTION
Pressure in formation = 0.47 * 7400 = 3478 psi
Converting to KPa = 3478 / 0.145 = 23986 Kpa
Converting to MPa = 23986 / 1000 = 23.99 MPa
Pressure gradient = 0.47 psi/ft = (0.47 / 0.145) KPa/ft = 3.2414 KPa/ft
= (3.2414 /0.3048) KPa/m
= 10.63 KPa/M
14
EXERCISE 2
If the pressure in a reservoir at the OWC is 3625 psi, calculate the pressure at the top
if there is a 600ft continuous oil column. If a normal pressure gradient exists outwith
the reservoir, calculate the pressure differential at the top of the reservoir. Redo the
calculations for a similar field, but this time containing gas.
SOLUTION
Typical pressure gradients are (psi/ft):
Water – 0.45
Oil – 0.35
Gas – 0.08
Pressure at seal = 3625 - (600*0.35) = 3415 psi
To calculate the pressure differential across seal, look at fluid gradient differential
from OWC to seal 600ft above…
Differential = (0.45-0.35) * 600 = 60 psi
If the reservoir is gas then the differential becomes…
(0.45 – 0.08) * 600 = 222 psi higher in the reservoir than surrounding area
REFERENCES
1. Dake,L.P. Fundamentals of Reservoir Engineering. Elsevier 1986
2. Bradley,J.S. Abnormal Formation Pressure. The American Association of
Petroleum Geologists Bulletin. Vol 59, No6, June 1975
3. Bishlawi,M and Moore,RL: Montrose Field Reservoir Management. SPE
Europec Conference, London,(EUR166) Oct.1980