Page 48 - Basic Well Log Analysis for Geologist

THE SPONTANEOUS POTENTIAL LOG

Figure 11. Determination of formation water resistivity (Ry) from an SP log. This example is an exercise involving the charts on
Figures 12 through 14.
Given:
Ryar = 0.51 at 135° (BHT)
= 0.91 at 135° (BHT)
Rn
Surface temperature = 60°F
Total depth = 8,007 ft
Bottom Hole Temperature (BHT) = 135°F
From the Log track:
l. SP = —40mv (spontaneous potential measured from log at a formation depth of 7.446 ft and uncorrected for
bed thickness). It is measured here as two 20mv divisions from the shale baseline. The deflection is negative.
so the value (—40mv) is negative.
WIV Bed thickness equals 8 ft (7,442 to 7.450 ft).
Resistivity short normal (R,) equals 28 ohm-meters.
& Formation depth equals 7,446 ft.

Procedure:
l. Determine T,—Use Figure 8 to determine the temperature of the formation (Tp). Use BHT = 135°F, TD =
8.007 ft. surface temperature = 60°F, and formation depth = 7,446 ft. (Your answer should be 130°P).
to Correct R,, and R,,yto Tp—Use Figure 9 to correct the values for the resistivity of mud and of mud filtrate,
using T;(130°F) from step 1. Use R,, = 0.91 at 135°F and Ray = 0.51 at 135°F. (Your answers should be: Ry,
= 0.94 at 130°F and Ry = 0.53 at 130°F).

. Determine SP—Read directly from the SP curve in Fig. 11. [t measures two units (at a scale of 20 my per
division) fron: the shale baseline. The deflection is negative. so your answer is also (--40 mv) negative.
. Correct SP to SSP—Correcting SP for the thin-bed effect will give a value for SSP; use the chart in Figure 12
to find the SP Correction Factor.
Given RR, (or R,,/Ry,) = 28/0.94 = 30. Bed thickness (read from SP log) Fig. 11 equals 8 ft. Correction
factor (from Fig. 12) = 1.3.
SSP = SP x SP Correction factor (Fig. 12)
SSP = (—40 mv) x 1.3
SSP = —52 mv (Your answer)
. Determine Ry/Rye ratio—Use the chart in Figure 13 (Your answer should be 5.0).

. Determine Ry.—Divide the corrected value for Ry by the ratio Ryyy/R,, value.
Rye = Rin (Rn Rwe)
Rye = 0.53/5.0
Rye = 0.106

. Correct Ry. to Ry—Use the chart in Figure 14, and the Ry. value in step 6 (Your answer should be R,, = 0.11
at Ts).

NOTE: The term short normal describes a log used to measure the shallow formation resistivity, or the resistivity
of the invaded zone (R;). Short normal resistivity (R,,) was used in Procedure Step 4 above, and its use as a
logging/resistivity term is common.