HW#5_Olivas_Sadie

docx

School

University of Texas, El Paso *

*We aren’t endorsed by this school

Course

301

Subject

Mechanical Engineering

Date

Dec 6, 2023

Type

docx

Pages

Uploaded by ConstableWrenMaster997

Show every step of your calculations. If you use a conversions calculator and write down the answer without all the detail covered in class, you won’t get any credit. The same criterion will be used in tests. Note: Each conversion can build on the previous. Conversions 1 Do the following conversions/calculations: 1. (6 pts) How much does a 1 kg mass weigh in Newtons? kg * 9.8 m/s 2 = 9.8 Newtons (1 Newton = 1 kg * m/s 2 ) 2. (6 pts) Convert that 1 kg mass to lb m . And how much does that mass weigh in lb f ? kg * * 32.174049 ft/s 2 = 70.9315519 lbf 3. (6 pts) Convert watts to horsepower. 1 watt ÷ 745.7 watts/horsepower = 0.00134102 horsepower 2 Convert freshwater density (1.0 g/cc) into the following units: 1. (3 pt) kg/m 3 1 g/cc ÷ 1000 g/kg * (100 cm) 3 /m 3 = 1000 kg/m 3 2. (3 pt) lb m /ft 3 g/cc ÷ 453.592 g/lbm * (30.48 cm) 3 /ft 3 = 62.428 lbm/f 3 3. (3 pt) lb m /gal 1 g/cc ÷ 453.592 g/lbm * (30.48 cm) 3 /ft 3 ÷ 7.48 gal/ft 3 = 8.3499lbm/gal 3 (6 pts) A drilling mud has a density of 15 ppg (lb m /gal). What is the pressure gradient in psi/ft? 15 lbm/gal * 7.48 gal/ft 3 * 32.18 ft/s 2 ÷ 32.18 lbm/ft*s 2 ÷ (12 in) 2 /ft 2 = 0.7791667 psi/f 4 (6 pts) The frac gradient for a reservoir is 0.7 psi/ft. What is the frac gradient in ppg? 0.7 psi/ft * (12 in) 2 /ft 2 * 32.18 lbm/ft*s 2 ÷ 32.18 ft/s 2 ÷ 7.48 gal/ft 3 = 13.47593583 ppg

Power calculations – also show your work. 5. (20 pts) Assuming the following data, calculate the required drill rig hoisting power in horsepower and in watts.  30,000 ft maximum drilling depth capability  25 lb/ft drill pipe  125 lb/ft drill collars (assume using 500 ft of collars)  25,000 lb traveling block  5 ft/s maximum velocity  10 ppg drilling mud • steel density= 8000 𝑘? ? 3 Buoyancy factor = 1 – (10 ppg ÷ 66.76 ppg) = 0.85 Dry weight = (30000 ft – 500 ft) * 25 lb/ft + 500 ft * 125 lb/ft + 25000 lb = 825000 lb Buoyant weight = 825000 lb * 0.85 = 701250 lb Power = 701250 lb * 5 ft/s = 3506250 lb*ft/s Horsepower = 3506250 lb*ft/s ÷ 550 (lb*ft/s)/hp = 6375 horsepower Watts = 6375 hp * 745.7 watts/hp = 4753837.5 watts The following formulas will be needed. 𝐵𝑜𝑢𝑦𝑎𝑛𝑐𝑦 𝑓𝑎𝑐𝑡𝑜? = 1 − 𝑓𝑙𝑢𝑖? ?𝑒𝑛𝑠𝑖𝑡𝑦 / 𝑜𝑏𝑗𝑒𝑐𝑡 ?𝑒𝑛𝑠𝑖𝑡𝑦 𝐵𝑜𝑢𝑦𝑎𝑛𝑡 𝑤𝑒𝑖?h𝑡 = (??𝑦 𝑤𝑒𝑖?h𝑡 ∗ 𝑏𝑜𝑢𝑦𝑎𝑛𝑐𝑦 𝑓𝑎𝑐𝑡𝑜?) 𝑃𝑜𝑤𝑒? = 𝐵𝑜𝑢𝑦𝑎𝑛𝑡 𝑤𝑒𝑖?h𝑡 ∗ 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 6. While drilling, the mud is being circulated at 2.5 bbl/minute. The surface pressure seen at the pumps is 2,500 psi. 𝐻𝑦??𝑎𝑢𝑙𝑖𝑐 h𝑜?𝑠𝑒𝑝𝑜𝑤𝑒? = 𝑓𝑙𝑜𝑤 ?𝑎𝑡𝑒 ∗ 𝑃?𝑒𝑠𝑠𝑢?𝑒 = ? ∗ 𝑃 1. (9 pts) What is the hydraulic power required for the mud pump in horsepower (HP)? 2.5 * 42 gal/bbl * 2500 psi ÷ 7.48 gal/ft 3 * (12 in) 2 /ft 2 ÷ 33000 ft*lbf/min = HP 2. (9 pts) 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

153.1356 hp * 60 min/hr * 24 hours/day * 7 days/week * (1/0.45) * 0.746 kW/hp = 2558957.7054 kW/week 2558957.7054 kW/week * 1 week/7 days * 1 day/24 hours * 1 gal/40.7 kW/hour = 374.2479 gal/hour *168 hours/week = 62873.65369 gal/week 3. (7 pts) 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) 3999.72 psi + 2500 psi = 6499.72 psi 4. (7 pts) 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)? 6499.72 psi – 500 psi = 5999.72 psi 5999.72 psi ÷ 7000 ft * (12in) 2 /ft 2 * 32.18 lbm/ft*s 2 ÷ 32.18 ft/s 2 ÷ 7.48 gal/ft 3 = 16.5 ppg 7. (9 pts) 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. 50 gal/min* 70 psi ÷ 7.48 gal/ft 3 * (12 in) 2 /ft 2 ÷ 33000 ft*lbf/min * (1/0.73) = 2.79699788 HP

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version