Please consider a pump that was located at the bottom of an oil well. The pump f

Question

Please consider a pump that was located at the bottom of an oil well. The pump forced a mixture of water and oil to the surface 435 m above. The well produces 800 barrels per day of the oil/water mixture. We will ignore inefficiencies and viscous dissipation as we answer the questions below. a. What was the energy required (in kW-hr) to pump 1 lbm of the oil/water mixture to the surface? b. What was the power draw of the pump in MW? c. What is the daily cost of electricity? DATA Density of the oil/water mixture = 0.930 gm/ml. Temperature of the oil = 45 oC Diameter of the vertical pipe = 0.43 ft Cost of electricity = $0.12/kW-hr

Explanation / Answer

given,

       mass of mixture pumped,m = 1 lbm = 0.4535 kg

     height to which mxture has to be pumped, h = 435m

Energy required is, E = weight*h = (m*g)*h = (0.4535*9.8)*435 = 1933.27 J

                                                                                           = 1933.27*2.78*10^{-6} KW-hr

                                                                                           = 0.005374 KW-hr

Since 1 barrel = 119.24 kg. Therefore 0.4535 kg of mixture = 0.0038 barrels

800 barrels are produced per day(=24 hrs). 0.0038 barrels or 0.4535 kg of mixture is produced in

                                                       (24*0.0038)/800 = 1.14*10^{-4} hrs

Power = Energy /time = 0.005374 /(1.14*10^{-4}) = 47.14 KW = 47.14*10^{-3} MW

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.