Question

A drill string in a wellbore is composed of 6000 ft of drill pipe with an internal diameter of 4.67 inches. Drilling fluid is pumped at a rate of 80 cycles per minute, with a pump factor of 0.21 bbl/cycle. The amount of drilling fluid held in the drill collar and drill bit assembly may be assumed negligible. The time required to circulate the drilling fluid from the surface to the drill bit (in minutes, rounded to two decimal places) is ______. [1 bbl = 5.61 ft\(^3\)]

Hint:

Find the drill pipe internal capacity in barrels, then divide by the pump output rate in barrels per minute.

Answer 1

Correct Answer: 7.57

Step 1: Compute the internal cross sectional area of the drill pipe:
The internal diameter is 4.67 inches. Converting to feet: \[ ID = \frac{4.67}{12} = 0.38917 \text{ ft} \] The radius is half of this: \[ r = \frac{0.38917}{2} = 0.19458 \text{ ft} \] The cross sectional area is: \[ A = \pi r^2 = \pi \times (0.19458)^2 = \pi \times 0.037863 = 0.11895 \text{ ft}^2 \]
Step 2: Compute the internal volume, capacity, of the drill pipe:
Multiplying the cross sectional area by the length of the drill pipe: \[ V = A \times L = 0.11895 \times 6000 = 713.7 \text{ ft}^3 \] Converting this volume to barrels, using 1 bbl = 5.61 ft\(^3\): \[ V = \frac{713.7}{5.61} = 127.22 \text{ bbl} \]
Step 3: Compute the pump output rate:
The pump delivers 80 cycles per minute, and each cycle delivers 0.21 bbl of fluid, so the pump output rate is: \[ Q = 80 \times 0.21 = 16.8 \text{ bbl/min} \]
Step 4: Compute the circulation time from surface to bit:
Since the drill collar and bit hold negligible fluid volume, the time to circulate fluid down to the bit is simply the drill pipe capacity divided by the pump rate: \[ t = \frac{127.22}{16.8} = 7.5728 \text{ minutes} \] Rounding to two decimal places: \[ t \approx 7.57 \text{ minutes} \]
Final Answer:
\[ \boxed{7.57 \text{ minutes}} \]