Exercise SP Log

Problem0 100

1375TVDm

1400TVDm

1425TVDm

From Logging data :• Temperature @ Surface = 25 C• Temperature @ 1500 TVDm = 75 C• SP curve as shown picture on the left side

From Mud Logging data :• Rmf @ 25 C = 0.09 ohm-m

Questions :• Find Rw@ depth 2000 MDm (1390 TVDm) ?• Find Rw@ depth 2050 MDm (1420 TVDm) ?• Do they use Fresh mud, Salt mud or Oil based mud ?

Notes :TVD m = meter True Vertical DepthMD m = meter Measured Depth

Solution• Determine the temperature at the depth were asked

Temperature gradient25 C ?? 75 C

Surface

calculation is always using TVD !!

1390 m

1420 m

1500 m

Temperature gradient = (75-25)/(1500-0) = 0.033 C/m

Temperature@ 1390 TVDm = 25+(0.033*1390)

= 71.33 C

Temperature@ 1420 TVDm = 25+(0.033*1420)

= 72.33 C

1390 TVDm

Rmf @ 25 C = 0.09 ohm-m

Temperature@ 1390 TVDm = 71.33 C

Rmf @ 71.3 C = 0.045 ohm-m

Halliburton Solution

1390 TVDmRmf @ 25 C = 0.09 ohm-m

Temperature@ 1390 TVDm = 71.33

C Rmf @ 71.3 C = 0.045 ohm-m

SSP = -25 mV

Rw @ 1390 TVDm

= 0.028 ohm-m

Schlumberger Solution

@ 1390 TVDm

• SSP = -K log (Rmfe/Rwe)• K = 60+0.133T (F)• T @ 1390 TVDm = 71.3 C = 160.4 F• K = 60+(0.133x160.4) = 81.33

We need Rmfe and Rwe data !!!!!!!!!!!

Determination of RmfeT @ 1390 TVDm = 71.3 C

= 160.4 F

Rmf @ 71.3 C = 0.045 ohm-m

Rmfe @ 160.4 F = 0.035 ohm-m

Determination of RweT @ 1390 TVDm = 71.3 C

= 160.4 F

Rmfe @ 160.4 F = 0.035 ohm-m

SSP = -25 mV

Rwe @ 160.4 F = 0.017 ohm-m

Determination of RweT @ 1390 TVDm = 71.3 C

= 160.4 F

Rwe @ 160.4 F = 0.017 ohm-m

Rw @ 160.4 F = 0.03 ohm-m

Exercise SP Log

• From log header, Rmf = 0.09 Ω-m @ 75 °F

• From interpolation, formation temperature = 150 °F

• From log, SSP = −50 mV

• Determine : Rmf @ 150 F and Rw @ depth