Given the following data: Bingham Plastic mud with: - Density (rho_m) =10.2 lb/g

Question

Given the following data: Bingham Plastic mud with: - Density (rho_m) =10.2 lb/gal Theta_600 = 65, Theta_300 = 40 Drill pipe: - ID = 4.0 in - OD=4.5 in - Length=14693 ft Drill Collar: - ID=4.0 in - OD=7.5 in - Length=1000 ft Maximum flow rate (Q) = 400 gpm Surface connections length = 500 ft Bit: tri-cone roller-cone jet bit (12-12-12) Mud pump volumetric efficiency (eta_vol -pump) = 85% Determine the pump hydraulic horse power used if the friction factor (f) is taken to be: 1/Squareroot f = 2.28 - 4 log(24.25/Re^0.9)

Explanation / Answer

Power (W) = [ Flow (m3/h) * Static Pressure (liquid column meters) * Spc.gravity ] / [ Pump Efficiency * 270 ] * 735,48

Static Pressure (liquid column meters <> mcl <> m) = [Static Pressure (kg/cm2) / Density (kg/cm3)] *10

* 270 and 735,48 are conversion factors to correct units.

* This Power is absorved power for the pump, it's not considered driver efficiency.

3) If you want to calculate Hydraulic power you have not to consider the pump efficiency

1. W=Pa x m3/s
2. Jose's formula Power (W) = [ Flow (m3/h) * Static Pressure (liquid column meters) * Spc.gravity ] / [ Pump Efficiency * 270 ] * 735,48

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