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.


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?



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.



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.



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.



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.



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

Documents

Chapter 02