Question:
A double acting duplex pump with a rod diameter of 2.5 inch and a stroke of 20 inch is to be operated at 60 strokes per minute for drilling down to 10000 ft. The flow rate is 600 gpm. If the volumetric efficiency of the pump is 80%, the liner size is \(\underline{\hspace{2cm}}\) inch (round off to one decimal place).
A double acting duplex pump with a rod diameter of 2.5 inch and a stroke of 20 inch is to be operated at 60 strokes per minute for drilling down to 10000 ft. The flow rate is 600 gpm. If the volumetric efficiency of the pump is 80%, the liner size is \(\underline{\hspace{2cm}}\) inch (round off to one decimal place).
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For pump calculations, ensure to use the correct conversion factors and check the efficiency of the pump.
Updated On: Jan 9, 2026
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Verified By Collegedunia
Correct Answer: 6.5
Solution and Explanation
The flow rate can be calculated using the following formula for a duplex pump:
\[
Q = \frac{2 \cdot A \cdot S \cdot N}{231},
\]
where:
- \( A \) is the cross-sectional area of the pump liner,
- \( S \) is the stroke length,
- \( N \) is the number of strokes per minute,
- 231 is the conversion factor for gallons to cubic inches.
Rearranging the formula to find \( A \), the cross-sectional area: \[ A = \frac{Q \cdot 231}{2 \cdot S \cdot N}. \] Substituting the values: \[ A = \frac{600 \cdot 231}{2 \cdot 20 \cdot 60} = 3.85 \, \text{in}^2. \] Now, the area of the pump liner is related to its diameter \( D \) by: \[ A = \frac{\pi D^2}{4}. \] Rearranging to solve for \( D \): \[ D = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \cdot 3.85}{3.14}} \approx 2.8 \, \text{in}. \] Thus, the liner size is approximately 2.8 inches.
- \( A \) is the cross-sectional area of the pump liner,
- \( S \) is the stroke length,
- \( N \) is the number of strokes per minute,
- 231 is the conversion factor for gallons to cubic inches.
Rearranging the formula to find \( A \), the cross-sectional area: \[ A = \frac{Q \cdot 231}{2 \cdot S \cdot N}. \] Substituting the values: \[ A = \frac{600 \cdot 231}{2 \cdot 20 \cdot 60} = 3.85 \, \text{in}^2. \] Now, the area of the pump liner is related to its diameter \( D \) by: \[ A = \frac{\pi D^2}{4}. \] Rearranging to solve for \( D \): \[ D = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \cdot 3.85}{3.14}} \approx 2.8 \, \text{in}. \] Thus, the liner size is approximately 2.8 inches.
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