Question

A drilling fluid following the power law model is circulated in a wellbore at a rate of 600 gal/min. The internal diameter (Di) of the drill pipe is 4.276 inches, and the power law flow behaviour index n for the fluid is 0.67.
The wall shear rate is given by \[ \dot{\gamma}_w = \left(\frac{3n+1}{4n}\right)\frac{8V}{D_i} \] where V is the average velocity of the drilling fluid inside the drill pipe.
[Given: 1 gallon = 3785.4 cm3, 1 inch = 2.54 cm]
The wall shear rate (in s^-1) is __________ (rounded off to one decimal place).

Hint:

First find the mean pipe velocity V from Q/A in consistent units, then plug into the given wall shear rate formula.

Answer 1

Correct Answer: 338

Step 1: List the given data: 
Flow rate Q = 600 gal/min 
Internal diameter Di = 4.276 in 
Power law index n = 0.67

Step 2: Convert the internal diameter into centimetres: 
Di = 4.276 x 2.54 
Di = 10.86104 cm, so the radius is 5.43052 cm

Step 3: Convert the flow rate into cm3/s: 
Q = 600 gal/min x 3785.4 cm3/gal 
Q = 2271240 cm3/min 
Q = 2271240 / 60 
Q = 37854 cm3/s

Step 4: Compute the cross-sectional flow area inside the drill pipe: 
\[ A = \pi \left(\frac{D_i}{2}\right)^2 = \pi (5.43052)^2 \]
(5.43052)^2 = 29.4905 
A = 3.14159 x 29.4905 
A = 92.647 cm2

Step 5: Compute the average velocity V: 
V = Q / A 
V = 37854 / 92.647 
V = 408.58 cm/s
Step 6: Compute the power law coefficient (3n+1)/(4n): 
3n + 1 = 3 x 0.67 + 1 = 2.01 + 1 = 3.01 
4n = 4 x 0.67 = 2.68 
(3n+1)/(4n) = 3.01 / 2.68 = 1.12313

Step 7: Compute the term 8V/Di: 
8V = 8 x 408.58 = 3268.66 
8V/Di = 3268.66 / 10.86104 = 300.95 s^-1

Step 8: Multiply the power law coefficient by this term to get the wall shear rate: 
\[ \dot{\gamma}_w = 1.12313 \times 300.95 \]
\[ \dot{\gamma}_w = 338.0 \text{ s}^{-1} \]

Final Answer: 
\[ \boxed{\dot{\gamma}_w = 338.0 \text{ s}^{-1}} \]