Corrected D exponent

The original “d” exponent is good for constant mud weight but in reality several drilling operations drill with various mud weights in hole due to weight up. In order to account for mud weight variation, so modification of d exponent, called “corrected d exponent”, has been made to correct for mud weight changes.

The corrected d-exponent is listed below.

dc = log (R ÷ 60N) ÷ log (12W ÷ 1000D) x (MW1 ÷ MW2)

Where;

dc = corrected “d” exponent

R = penetration rate in feet per hour

d = exponent in drilling equation, dimensionless

N = rotary speed in rpm

W = weight on bit in kilo pound

D = bit size in inch

MW1 = initial mud weight in ppg

MW2 = actual mud weight in ppg

 

Example: Determine the corrected d-exponent from following information.

Rate of penetration (R) = 90 ft/hr

Rotary drilling speed (N) = 110 rpm

Weight on bit (W) = 20 klb

Bit Diameter (D) = 8.5 in

MW1 = 9.0 ppg

MW2 = 12.0 ppg

Solution: dc = log [90÷ (60 x 110)] ÷ log [(12 x 20) ÷ (1000 x 8.5)] x (9.0 ÷ 12.0)

dc = 1.20 x 0.75

dc = 0.9

** Please remember that single d exponent or corrected d exponent valve does not help identify abnormal pressure. The trend of d exponent will help drilling personnel detect high formation pressure zones while drilling.

Please find the excel sheet for calculating the corrected D Exponent

Ref book: Drilling Formula Book Formulas and Calculations for Drilling, Production and Workover, Second Edition

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