Question

A vertical well is drilled up to a depth of 4000 ft. Further drilling starts with 10 ppg of fresh mud and \(50000 \, lbf\) weight on bit (WOB). An equivalent circulation density (ECD) of \(10.75 \, ppg\) was recorded. The total circulation pressure loss is estimated to be 110 psi. The still density is \(65.5 \, ppg\). The decrease in hook load is ________ lbf (rounded off to one decimal place). (Note: 1 ppg mud is equivalent to \(0.052 \, psi/ft\)).

Hint:

In drilling calculations, always account for buoyancy using the buoyancy factor \((1 - \rho_{mud}/\rho_{steel})\). Equivalent circulation density (ECD) must be considered during circulation.

Answer 1

Step 1: Calculate buoyancy factor.
Buoyancy factor = \(\dfrac{\rho_{mud}}{\rho_{steel}}\). Here, \[ \rho_{mud} = 10 \, ppg, \quad \rho_{steel} = 65.5 \, ppg \] \[ BF = 1 - \frac{\rho_{mud}}{\rho_{steel}} = 1 - \frac{10}{65.5} = 0.847 \] Step 2: Hook load decrease.
Decrease in hook load due to buoyancy: \[ \Delta HL = WOB \times \frac{\rho_{mud}}{\rho_{steel}} \] \[ \Delta HL = 50000 \times \frac{10}{65.5} = 50000 \times 0.1527 \] \[ \Delta HL \approx 7635 \, lbf \] But considering dynamic ECD = 10.75 ppg: \[ \Delta HL = 50000 \times \frac{10.75}{65.5} = 50000 \times 0.1641 = 8205 \, lbf \] Step 3: Additional correction for circulation loss.
Given circulation pressure loss = 110 psi. Equivalent height: \[ h = \frac{110}{0.052 \times 65.5} \approx 32.9 \, ft \] This adds ~570 lbf buoyancy adjustment. Final Answer: \[ \boxed{8205 \, lbf} \]