Question 6 part d 6. While drilling, the mud is being circulated at 2.5 bbl/minute. The surface pressure seen at thepumps is 2,500 psi.Hydraulic horsepower = flow rate * Pressure = q* P7.a.b.d.What is the hydraulic power required for the mud pump in horsepower (HP)?Assuming the mud pumps run on diesel, and assuming the pumps are 45% efficient(only 45% of the power of the fuel creates pumping power), how much fuel would thisone mud pump consume per week if operating 24 hrs/day under these conditions?Assume pumps are 45% efficient, 55% of the power goes to waste, so the power neededis 1/efficiency, orEstimate the bottomhole pressure at 7,000 ft assuming the mud weight is 11 ppgand a surface pressure of 2,500 psi. (the bottomhole pressure is the sum of a) the surfacepressure and b) the pressure caused by the column of mud in the wellbore)Assuming we wanted to keep the same bottomhole pressure, we could reduce thesurface pressure by increasing the mud weight. What mud weight would be required toreduce the surface pressure to 500 psi but still keep the same bottomhole pressure as(c)?A pump is pumping water at 50 gal/min; P= 70 psi. The efficiency of the pump is 0.73.Compute the hydraulic horsepower (hp) of the pump.

Mud circulated = 2.5 bbl/min Surface Pressure = 2500 Psi Now, Hydraulic Horsepower is given by,…

Assuming we wanted to keep the same bottomhole pressure, we could reduce the surface pressure by increasing the mud weight. What mud weight would be required to reduce the surface pressure to 500 psi but still keep the same bottomhole pressure as (c)?

Assuming we wanted to keep the same bottomhole pressure, we could reduce the surface pressure by increasing the mud weight. What mud weight would be required to reduce the surface pressure to 500 psi but still keep the same bottomhole pressure as (c)?

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Radiation

Radiation can be described as the energy in the form of a particular matter or electromagnetic waves traveling in the air and can be in the form of light, sound, and heat. Radiation can be detected through special instruments like X-rays. The radiation from a cooler object is less intense as compared to the radiation from hot objects.

Question

Question 6 part d

6. While drilling, the mud is being circulated at 2.5 bbl/minute. The surface pressure seen at the
pumps is 2,500 psi.
Hydraulic horsepower = flow rate * Pressure = q* P
7.
a.
b.
d.
What is the hydraulic power required for the mud pump in horsepower (HP)?
Assuming the mud pumps run on diesel, and assuming the pumps are 45% efficient
(only 45% of the power of the fuel creates pumping power), how much fuel would this
one mud pump consume per week if operating 24 hrs/day under these conditions?
Assume pumps are 45% efficient, 55% of the power goes to waste, so the power needed
is 1/efficiency, or
Estimate the bottomhole pressure at 7,000 ft assuming the mud weight is 11 ppg
and a surface pressure of 2,500 psi. (the bottomhole pressure is the sum of a) the surface
pressure and b) the pressure caused by the column of mud in the wellbore)
Assuming we wanted to keep the same bottomhole pressure, we could reduce the
surface pressure by increasing the mud weight. What mud weight would be required to
reduce the surface pressure to 500 psi but still keep the same bottomhole pressure as
(c)?
A pump is pumping water at 50 gal/min; P= 70 psi. The efficiency of the pump is 0.73.
Compute the hydraulic horsepower (hp) of the pump.

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