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 a. What is the hydraulic power required for the mud pump in horsepower (HP)? b. 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 c. 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 weiibore) d. 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. Show every step of your calculations. Note: Each conver5ion can build on the previous. Conversions 1. Do the following conversions/calculations: a. How much does a 1 kg mass weigh in Newtons? b. Convert that 1 kg mass to lbm. And how much does that mass weigh in lbf? c. Convert watts to horsepower. 2. Convert fresh water density (1.0 g/cc) into the following units: a. kg/m3 b. lbm/ft3 c. lbm/gal 3. A drilling mud has a density of 15 ppg (Ibm/gal). What is the pressure gradient in psi/ft? 4. The frac gradient for a reservoir is 0.7 psi/ft. What is the frac gradient in ppg? Power calculations also show your work. 5. Assuming the following data, calculate the required drill rig hoisting power in horsepower and in watts. 0 30,000 ft maximum drilling depth capability o 25 Ib/ft drill pipe 0 125 lb/ft drill collars (assume using 500 ft of collars) 0 25,000 lb traveling block 0 5 ft/s maximum velocity 0 10 ppg drilling mud 0 steel density= 8000ki m The following formulas will be needed. fluid density object density Bouyant weight = (dry weight * bouyaricy factor) Power = Bouyant weight * velocity Bouyancy factor = 1