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 steel 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:

When ECD increases during circulation, the effective fluid density around the drillstring increases, which increases buoyancy and \textbf{reduces hook load}. Use $BF = 1 - \rho_m/\rho_s$ with $\rho_m =$ mud density (or ECD) and $\rho_s \approx 65.5$ ppg for steel.

Answer 1

Correct Answer: 675.7

Step 1: Use buoyancy factor for drillstring/BHA.
Buoyancy factor (BF) for steel in mud is: \[ BF = 1 - \frac{\rho_m}{\rho_s} \] where $\rho_m$ is mud density (ppg) and $\rho_s$ is steel density (ppg).
Static mud density = $10$ ppg: \[ BF_{\text{static}} = 1 - \frac{10}{65.5} = 0.84733 \] Circulating condition is represented by ECD = $10.75$ ppg: \[ BF_{\text{circ}} = 1 - \frac{10.75}{65.5} = 0.83588 \] Step 2: Relate WOB to buoyed weight (set-down weight).
Assuming the applied WOB corresponds to the submerged (buoyed) weight contribution, the air-equivalent weight required is: \[ W_{\text{air}} = \frac{WOB}{BF_{\text{static}}} = \frac{50000}{0.84733} = 59009.01 \text{ lbf} \] Step 3: Compute buoyed weight during circulation.
\[ W_{\text{circ}} = W_{\text{air}} \times BF_{\text{circ}} = 59009.01 \times 0.83588 = 49324.32 \text{ lbf} \] Step 4: Decrease in hook load due to increased buoyancy.
Extra buoyancy (hence decrease in hook load) is: \[ \Delta H = WOB - W_{\text{circ}} = 50000 - 49324.32 = 675.68 \text{ lbf} \] Rounded to one decimal place: \[ \boxed{\Delta H = 675.7 \text{ lbf}} \]