# saturation height function

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Transition Zone Analysis

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procedure and example of manual calcuation of SHF

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• Transition Zone Analysis

• Porosity Cross-Plots

End-point values for porosity logs are determined from

cross-plots of log response versus core porosity (NOB)

Density Log Calibration with Core Data

y = -1.7702x + 2.7103

0

0.5

1

1.5

2

2.5

3

0 0.1 0.2 0.3 0.4

Core Porosity at NOB

Bu

lk D

en

sit

y, g

/cc

Density Log

Sonic Log Calibration with Core Data

y = 126.52x + 49.905

0

20

40

60

80

100

120

140

160

0 0.1 0.2 0.3 0.4

Core Porosity at NOB

Tra

ns

it T

ime

, m

icro

-se

co

nd

Sonic Log

• Shale Content Cross-Plots

min

minlog

SPSP

SPSPI

sh

SP

min

minlog

sh

I

In this case, use

Larionov Older Rocks correlation

In this case,

Make a new correlation

• Calibration of Well Logs

with Core Data

• OWC

Oil Rim

Bottom Water

Cap Rock

If Capillary Pressure is Negligible

10

Sw

Swc

Initial Saturation Distribution

No transition zone

• OWC

Oil Rim

Bottom Water

Cap Rock

If Capillary Pressure is Significant

10

Sw

Swc

Initial Saturation Distribution

Top transition zone O/w

Transition Zone

• Transition Zones

Shape of (Sw vs depth) curve is similar to capillary pressure curve

Oil-water capillary pressure Pcow = Po - Pw

At equilibrium:

Forceup = Forcedown

PcowA = Ah(w - o)g

Hence;

h = Pcow/ (w - o)g

OWCh

A

Pcow

SwSwc0 1

h

SwSwc 10

• Example Calculation

Given:

Water density = 995 kg/m3

Oil density = 689 kg/m3

Acc. of gravity g = 9.8 m/s2

Oil

WaterWater

(A) h = 5 m

(B) h = 15 m

(C) h = 30 m

OWC

Oil

WaterWater

(A) h = 5 m

(B) h = 15 m

(C) h = 30 m

OWC

Pcow psi

Sw

0 10

16

8

24

0.40.2 0.80.6

For point A: Pcow = 5x9.8x(995-689)/6900 = 2.2 psi

Swi = 0.88

For point A: Pcow = 15x9.8x(995-689)/6900 = 6.5 psi

Swi = 0.39

For point A: Pcow = 30x9.8x(995-689)/6900 = 13 psi

Swi = 0.24

• Transition Zone Analysis

Defining reservoir rock facies

Objectives

Checking validity of Sw values calculated from well logs

Checking validity of OWC level

Calibration of well logs with core data

Estimating ( cos) for various reservoir rock facies

Calculating capillary pressure curves for reservoir conditions

• Transition Zone Analysis

Procedure

For each rock facies, formulate suitable transforms for:

k in terms of and Vcl

k

Vcl

Swc in terms of k

log k

Swc

• Transition Zone Analysis

Procedure, continued

Formulate suitable J-Functions from core data

J-Function vs Sw

J

Sw0 1

kPJ c

cos

• Transition Zone Analysis

Procedure, continued

Convert J-Function to normalized J-Function

wc

wcww

S

SSS

1

*

J-Function vs Sw*

J

Sw*0 1

Jmax

• Transition Zone Analysis

TVDSS Vcl Sw k Swc Sw*

3422.0 0.232 0.091 0.200 621 0.199 0.001

3422.5 0.241 0.077 0.213 487 0.207 0.007

--- --- --- --- --- --- ---

--- --- --- --- --- --- ---

3501.0 0.162 0.175 0.598 25 0.301 0.425

3501.5 0.225 0.107 0.601 412 0.212 0.494

--- --- --- --- --- --- ---

--- --- --- --- --- --- ---

3533.0 0.208 0.096 0.967 270 0.225 0.957

3533.5 0.182 0.124 0.989 85 0.262 0.985

OWC 0.197 0.115 1.000 152 0.244 1.000

Procedure, continued

Calculate k, Swc and Sw* for every point above OWC

from log analysis results

wc

wcww

S

SSS

1

*

• Transition Zone Analysis

TVDSS k Sw* h J cos

3422.0 0.232 621 0.001 112.0 0.4040

3422.5 0.241 487 0.007 111.5 0.3461

--- --- --- --- --- ---

--- --- --- --- --- ---

3501.0 0.162 25 0.425 33.0 0.0284

3501.5 0.225 412 0.494 32.5 0.0970

--- --- --- --- --- ---

--- --- --- --- --- ---

3533.0 0.208 270 0.957 1.0 0.0025

3533.5 0.182 85 0.985 0.5 0.0008

OWC 0.197 152 1.000 0 0

Procedure, continued

Calculate h and (J cos) for every point above OWC

from log analysis results

h = height above OWC

kghJ ow )(cos

• Transition Zone Analysis

Procedure, continued

Plot (J cos) versus Sw* and fit the best J-Function Curve

(J cos) vs Sw*

J cos

Sw*0 1-0.2

Jmax cos

• Transition Zone Analysis

Procedure, continued

Calculate cos for each reservoir facies at several

values of Sw*

( cos)res = (J cos) / J

Calculate the average value of ( cos)res

Calculate the reservoir J-function for each reservoir

facies using the average value of ( cos)res

Jres = Jlab (J cos) /( cos)res for each Sw* value

• Transition Zone Analysis

Procedure, continued

Compare the curves of laboratory and reservoir J-

functions versus Sw*

Estimate the value of for each reservoir rock facies

if is known

Calculate the coefficient of J-function for use in

Petrel model

Use the reservoir J-function to formulate a transform

relating Sw* to Jres or a selected function of Jres

• Reservoir Capillary Pressure Curves

Use the value ( cos)res to calculate required capillary pressure

curves for various facies from their normalized J-Functions

Pc

Sw0 1Swc

kJPc

cos

Determine average and k for various

reservoir rock facies

Estimate Swc from transforms

Calculate Pc values from J, cos, and k

Plot Pc versus Sw

Assume several values for Sw* between 0 and 1

Calculate corresponding values of Sw

Determine J values from Sw*

• Transition Zone Analysis Example

Given capillary pressure data

Sw 1 0.871 0.729 0.612 0.521 0.453 0.389 0.342 0.301 0.273 0.257 0.244 0.241 0.240

Pc, psi 0 5 10 15 20 25 30 40 50 60 70 80 90 100

0

5

10

15

20

25

30

35

40

45

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Water Saturation Sw

Capill

ary

Pre

ssure

, psi

Sample A

Porosity = 18%

Permeability = 236 md

Swc = 24%

• Transition Zone Analysis Example

Given k and Swc transforms

log k = - 28.56por2 + 21.14por - 4.6Vsh - 0.38

-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

0 0.05 0.1 0.15 0.2 0.25 0.3

Porosity

Lo

g k

Vsh

0

0.1

0.2

0.3

0.4

y = 0.0192x2 - 0.1434x + 0.4555

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

-1 -0.5 0 0.5 1 1.5 2 2.5 3

Log kS

wc

Oil-water contact elevation = -2866 ftss

Reservoir oil density o = 764 kg/m3

Reservoir water density w = 982 kg/m3

Acceleration of gravity g = 9.8 kg/m2

Other data

• Transition Zone Analysis Example

Calculated Jlab versus Sw*

Sw* 1 0.83 0.643 0.489 0.37 0.28 0.196 0.134 0.08 0.043 0.022 0.005 0.001 0

J 0 0.55 1.10 1.65 2.19 2.74 3.29 4.39 5.49 6.58 7.68 8.78 9.88 10.97

y = -38.494x5 + 105.36x4 - 108.03x3 + 53.181x2 - 15.649x + 3.63

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Sw*

J

• Transition Zone Analysis Example

Calculated Jcos versus Sw* from log dataDepth Porosity Vsh Sw k Swc Sw* h J cos

ft subsea md 0 ft

2758 0.155 0.271 0.696 9.2 0.335 0.542 108 0.01713

2758.5 0.12 0.017 0.430 46.5 0.270 0.219 107.5 0.04357

2759 0.166 0.11 0.422 68.6 0.257 0.223 107 0.04479

2759.5 0.223 0.287 0.597 39.2 0.276 0.444 106.5 0.02907

2760 0.168 0.035 0.310 160.1 0.233 0.101 106 0.06738

2760.5 0.201 0.241 0.589 40.4 0.275 0.433 105.5 0.03080

2761 0.218 0.333 0.646 21.8 0.298 0.496 105 0.02162

2761.5 0.241 0.175 0.276 178.2 0.230 0.060 104.5 0.05851

2762 0.164 0.19 0.577 27.8 0.288 0.405 104 0.02788

2762.5 0.148 0.277 0.797 7.1 0.347 0.689 103.5 0.01476

2763 0.128 0.089 0.567 28.1 0.288 0.392 103 0.03142

2763.5 0.161 0.256 0.748 12.8 0.320 0.630 102.5 0.01882

2764 0.173 0.074 0.323 120.8 0.240 0.109 102 0.05550

2764.5 0.233 0.139 0.256 226.8 0.224 0.042 101.5 0.06521

2765 0.255 0.069 0.230 685.8 0.203 0.034 101 0.10785

2765.5 0.252 0.256 0.438 90.4 0.248 0.253 100.5 0.03919

2766 0.224 0.209 0.465 91.4 0.248 0.288 100 0.04159

2766.5 0.106 0.355 0.950 0.8 0.470 0.905 99.5 0.00563

2767 0.105 0.19 0.795 4.5 0.370 0.674 99 0.01335

2767.5 0.101 0.161 0.773 5.3 0.362 0.644 98.5 0.01469

2768 0.17 0.275 0.751 13.3 0.319 0.634 98 0.01785

2768.5 0.103 0.354 0.913 0.7 0.478 0.834 97.5 0.00523

2769 0.147 0.145 0.585 27.8 0.288 0.418 97 0.02747

2769.5 0.25 0.067 0.218 648.3 0.204 0.018 96.5 0.10119

2770 0.258 0.095 0.215 544.6 0.207 0.010 96 0.09082

2770.5 0.144 0.07 0.448 56.2 0.263 0.251 95.5 0.03885

2771 0.106 0.128 0.680 8.9 0.337 0.517 95 0.01792

2771.5 0.21 0.291 0.679 28.9 0.287 0.550 94.5 0.02283

2772 0.176 0.253 0.722 19.6 0.302 0.602 94 0.02043

2772.5 0.152 0.351 0.885 3.6 0.382 0.813 93.5 0.00937

2773 0.215 0.171 0.402 114.4 0.242 0.212 93 0.04417

2773.5 0.16 0.295 0.845 8.2 0.340 0.765 92.5 0.01364

2774 0.11 0.233 0.803 3.4 0.385 0.679 92 0.01053

2774.5 0.106 0.293 0.847 1.6 0.427 0.733 91.5 0.00732

2775 0.146 0.12 0.577 35.1 0.280 0.413 91 0.02905

2775.5 0.221 0.196 0.447 99 0.246 0.267 90.5 0.03944

2776 0.195 0.217 0.585 45.5 0.271 0.430 90 0.02831

2776.5 0.195 0.309 0.765 17.2 0.308 0.660 89.5 0.01731

2777 0.191 0.395 0.836 6.3 0.353 0.747 89 0.01052

2777.5 0.162 0.273 0.803 10.9 0.327 0.708 88.5 0.01495

2778 0.11 0.34 0.886 1.1 0.450 0.792 88 0.00573

2778.5 0.168 0.383 0.926 4 0.376 0.881 87.5 0.00879

2779 0.194 0.045 0.310 275 0.220 0.116 87 0.06745

2779.5 0.257 0.178 0.348 222.8 0.225 0.159 86.5 0.05244

2780 0.185 0.086 0.413 143.8 0.235 0.232 86 0.04937

2780.5 0.199 0.291 0.693 22.8 0.296 0.564 85.5 0.01884

2781 0.131 0.055 0.481 44.3 0.271 0.289 85 0.03219

2781.5 0.165 0.008 0.282 196.7 0.228 0.070 84.5 0.06007

• Transition Zone Analysis Example

Transition zone plot

• Transition Zone Analysis Example

Calculated (cos)res for various Sw* values

Sw* J cos J cos

0 0.0853 3.630 0.0235

0.1 0.0602 2.499 0.0241

0.2 0.0453 1.919 0.0236

0.3 0.0371 1.565 0.0237

0.4 0.0296 1.268 0.0233

0.6 0.0172 0.713 0.0242

0.8 0.0090 0.377 0.0239

1 0 0

0.0238Average

Using a value of 0.025 N/m for res:

res = 18

• Calculating Initial Water Saturation

From Capillary Equilibrium

Required items are:

( cos)res Sw* as function of J k and Swc transforms

kghJ ow

cos

)(

For every point, determine height above

OWC (h)

Estimate k and Swc from transforms

Calculate corresponding values of Sw*

Calculate corresponding values of Sw Sw = Swc + Sw*(1 - Swc)

Oil

Water

OWC

h

• Calculating Initial Water Saturation

From Capillary Equilibrium

Example

Density difference = 205 kg/m3

g = 9.8 m/s2

= 0.215

Vcl = 0.141

OWC at 1833 mss

( cos)res = 0.037 N/m

Transforms

Swc = 0.388 0.055 log k

log k = 14.5 - 5.8 Vcl

y = -1,2764x5 + 5,9212x4 - 10,565x3 + 9,1674x2 - 4,179x + 1

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1

0 0,4 0,8 1,2 1,6

J-Function

Sw

*

Estimate Swi at depth = 1821 mss

• Calculating Initial Water Saturation

From Capillary Equilibrium

Example, continued

From transforms: k = 199 md and Swc = 0.262

Height above OWC h = 12 m

J = 0.611

Hence; Sw* = 0.176

Sw = 0.262 + 0.176(1 0.262) = 0.392