Invaded Zone Water Saturation From
SHALLOW
Resistivity
Calculate invaded zone water saturation (Sxo) using the same method
as used to calculate water saturation in the uninvaded zone (Sw).
To do this, replace the following terms in any water saturation
equation given earlier in this Chapter:
RESD by RESS
RW@FT by RMF@FT
RSH by RSHS
Sw by Sxo
Where:
RESD = deep resistivity (ohmm)
RESS = shallow resistivity (ohmm)
RMF@FT = mud filtrate at form temperature (ohmm)
RSH = resistivity of shale on deep log (ohmm)
RSHS = resistivity of shale on shallow log (ohmm)
RW@FT = water resistivity at form temperature (ohmm)
Sw = water saturation in the invaded zone (fractional)
Sxo = water saturation in the invaded zone (fractional)
COMMENTS:
Most analysts assume RSHS = RSH.
Invaded
zone water saturation is used to find the amount of hydrocarbon
flushed by the invasion process. The hydrocarbon thus moved is
called the moveable hydrocarbon saturation, and that left behind
is the residual hydrocarbon saturation. A simple empirical formula
relating water saturation and invaded zone water saturation is:
1: Sxo = Sw ^ (1/5)
This
is a traditional relationship and local knowledge may provide
different results.
Invaded
Zone Water Saturation from Electromagnetic Logs
The
electromagnetic propagation log measures the travel time and attenuation of
microwaves propagated along the borehole wall. The analysis concept is
similar to a sonic log, but the frequencies involved are microwave
(electromagnetic) in the gigahertz range instead of acoustic in
the kilohertz range. The travel time is largely influenced by
the amount of water in the formation and does not depend too greatly
on other components, such as hydrocarbon or matrix rock. Because
of its shallow investigation, the log measures the water volume
of the flushed zones.
The
response equation for the electromagnetic propagation log follows
the classical form:
1:
TP = PHIe * Sxo * TPw (water term)
+ PHIe * (1  Sxo) * TPh (hydrocarbon term)
+ Vsh * TPsh (shale term)
+ (1  Vsh  PHIe) * Sum (Vi * TPi) (matrix term)
Where:
TPh = log reading in 100% hydrocarbon
TPi = log reading in 100% of the ith component of matrix rock
TP = log reading
TPsh = log reading in 100% shale
TPw = log reading in 100% water
PHIe = effective porosity (fractional)
Sxo = water saturation in invaded zone (fractional)
Vi = volume of ith component of matrix rock
Vsh = volume of shale (fractional)
The
hydrocarbon does not contribute any signal, so the equation is
solved directly for water filled porosity, assuming Vsh = 0.0.
This is then compared to the total porosity, after shale volume
corrections, based on data derived from the dual water method
described in Section 8.11.
Water Satuation from EPT Method
Calculate
loss free propagation time from measured propagation time.
1: Tpo = (TPL ^ 2  ((ATTN  50)) ^ 2 / 3604) ^ 0.5
The
value of Tpw varies with temperature and is given by:
2: FT1 = SUFT + (BHT  SUFT) / BHTDEP * DEPTH)
3: IF LOGUNITS$ = "METRIC"
4: THEN FT1 = 9 / 5 * FT1 + 32
5: Tpw = 20 * (710  FT1 / 3) / (444  FT1 / 3)
Calculate
water filled porosity from propagation time.
6: PHIept = (Tpo  TPM) / (Tpw  TPM)
Calculate
water saturation of invaded zone.
7: IF Vsh < 1.0
8: THEN SWept = (PHIept  BVWSH * Vsh) / (PHIt  BVWSH * Vsh)
9: OTHERWISE SWept = 1.0
Where:
ATTN = measured attenuation time of formation (db/m)
BHT = bottom hole temperature (degrees Fahrenheit or Celsius)
BHTDEP = depth at which BHT was measured (feet or meters)
BVWsh = bulk volume water in 100% shale (fractional)
FT1 = formation temperature (degrees Fahrenheit or Celsius)
PHIept = total porosity from electromagnetic log (fractional)
PHIt = total porosity from any total porosity (fractional)
SUFT = surface temperature (degrees Fahrenheit or Celsius)
SXOept = invaded zone saturation (fractional)
TPL = measured propagation time of formation (nsec/m)
TPM = loss free propagation time of matrix (nsec/m)
Tpo = loss free propagation time of formation (nsec/m)
Tpw = loss free propagation time of water (nsec/m)
Vsh = volume of shale (fractional)
COMMENTS:
In tar
sands and heavy oil, SXOept will equal SW since there is no
invasion in these reservoirs. This is also true in oil
reservoirs drilled with genuine oil base mud. There may be a bit
of invasion, but it does not change SW very much.
Note that this value of
SXOept does not depend on any knowledge
of resistivity log data or assumed fluid resistivities.
This
is a thin bed tool as it sees zones of 34 inches in thickness
Shale
volume can be calculated from the EPT attenuation curve in a fashion
similar to the gamma ray. This helps to resolve laminated shaly
sands.
A
graphical solution exists but is more complicated to use than
these simple formulae.
RECOMMENDED
PARAMETERS: 
Material 
Relative
Dielectric Permitivity 
TPM
LossFree Propagation Time ns/m 



Gas
or Air 
1.0 
3.3 
Oil 
2.2 
4.9 
Water 
56
 80 
25
 28 
Quartz 
4.7 
7.2 
Limestone 
7.5 
9.6 
Dolomite 
6.9 
8.7 
Anhydrite 
6.5 
8.4 
Dry
Clay 
5.7 
8.0 
Halite 
5.6
 6.3 
7.9
 8.4 
Gypsum 
4.2 
6.8 
Shale 
5.0
 25 
7.5 
The
value for water varies with temperature  see Step 5 for more
precise values.
NUMERICAL
EXAMPLE:
1. Given a zone with:
ATTN = 200 db/m
FT = 43 degrees C = 109 degrees F
TPL = 15 nsec/m
TPM = 7.2 nsec/m (sandstone)
BVWSH = 0.30
Vsh = 0.33
PHIe = 0.11
Tpo = (15 ^ 2  ((200  50) ^ 2) / 3604) ^ 0.5 = 11.9
Tpw = 20 * ( 710  109 / 3) / (444  109 / 3) = 33.0
PHIept = (11.9  7.2) / (33.0  7.2) = 0.182
SWept = (0.182  0.30 * 0.33) / (0.11) = 0.75
This
is a reasonable number for Sxo in a sand such as Sand D.
