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: Derive the pipe capacity per foot of length: 
Instead of finding the total volume first, find the capacity of the drill pipe per foot of its length. The volume per foot, in ft$^3$/ft, is $\pi/4$ times the internal diameter squared, with the diameter in feet: $ID = 4.67/12 = 0.38917$ ft, so $ID^2 = 0.15145$ ft$^2$. The volume per foot is $\dfrac{\pi}{4} \times 0.15145 = 0.7854 \times 0.15145 = 0.11895$ ft$^3$/ft. 

Step 2: Convert the per foot capacity to barrels per foot: 
Dividing by 5.61 ft$^3$ per barrel gives the capacity in bbl/ft: $0.11895 / 5.61 = 0.021203$ bbl/ft. 

Step 3: Multiply by the total pipe length to get total capacity: 
Multiplying the capacity per foot by the 6000 ft of drill pipe: $0.021203 \times 6000 = 127.22$ bbl. This matches the total volume found by computing the full pipe volume first and converting afterward, confirming the capacity is correct. 

Step 4: Divide by the pump output rate to get the circulation time: 
The pump output rate is $80$ cycles per minute times $0.21$ bbl per cycle, which is $16.8$ bbl/min. Since the drill collar and bit assembly hold a negligible volume, the surface to bit circulation time is: $t = 127.22 / 16.8 = 7.5728$ minutes, which rounds to $7.57$ minutes. 

Final Answer: 
\[ \boxed{7.57 \text{ minutes}} \]