Question:medium

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.)

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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.
Updated On: Feb 16, 2026
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Correct Answer: 675.7

Solution and Explanation

Step 1: Calculate the buoyancy factor for the drillstring. 
The buoyancy factor for steel immersed in drilling fluid is given by:

\[ BF = 1 - \frac{\rho_m}{\rho_s} \]

where $\rho_m$ is the mud density and $\rho_s$ is the density of steel, both expressed in ppg.

For static conditions, the mud density is $10$ ppg:

\[ BF_{\text{static}} = 1 - \frac{10}{65.5} = 0.84733 \]

During circulation, the effective mud density increases to an ECD of $10.75$ ppg:

\[ BF_{\text{circ}} = 1 - \frac{10.75}{65.5} = 0.83588 \]

Step 2: Convert applied WOB to air-equivalent weight.
The applied weight on bit represents the submerged (buoyed) weight of the drillstring under static conditions.
To find the corresponding weight in air:

\[ W_{\text{air}} = \frac{WOB}{BF_{\text{static}}} = \frac{50000}{0.84733} = 59009.01 \text{ lbf} \]

Step 3: Determine the buoyed weight during circulation.
When circulation starts, the increased mud density reduces the buoyancy factor, which changes the submerged weight:

\[ W_{\text{circ}} = W_{\text{air}} \times BF_{\text{circ}} = 59009.01 \times 0.83588 = 49324.32 \text{ lbf} \]

Step 4: Evaluate the change in hook load.
The additional buoyancy caused by circulation results in a reduction in hook load.
This reduction is calculated as:

\[ \Delta H = WOB - W_{\text{circ}} = 50000 - 49324.32 = 675.68 \text{ lbf} \]

Rounding to one decimal place:

\[ \boxed{675.7 \text{ lbf}} \]

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