🗊Презентация Drilling Engineering

Категория: Физика
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Слайды и текст этой презентации

Слайд 1


Drilling Engineering – PE 311 Hydraulics of Drilling Fluids
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Drilling Engineering – PE 311 Hydraulics of Drilling Fluids

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Drilling Engineering, слайд №2
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For incompressible fluids, the specific weight of the liquid in field unit is given by If P0 = 0 then The fluid density
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For incompressible fluids, the specific weight of the liquid in field unit is given by If P0 = 0 then The fluid density

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Example: Calculate the static mud density required to prevent flow from a permeable stratum at 12,200ft if the pore pressure of the formation fluid is 8500psig. Solution: The mud density must be at least 13.4 lbm/gal
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Example: Calculate the static mud density required to prevent flow from a permeable stratum at 12,200ft if the pore pressure of the formation fluid is 8500psig. Solution: The mud density must be at least 13.4 lbm/gal

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EOS of gas:
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EOS of gas:

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A well contains tubing filled with methane gas (MW = 16) to a vertical depth of 10000ft. The annular space is filled with a 9.0 lbm/gal brine. Assuming ideal gas behavior, compute the amount by which the exterior pressure on the tubing exceeds the interior tubing pressure at 10,000ft if the surface tubing pressure is 1000 psia and the mean gas temperature is 140F. If the collapse resistance of the tubing is 8330 psi, will the tubing collapse due to the high external pressure?
Описание слайда:
A well contains tubing filled with methane gas (MW = 16) to a vertical depth of 10000ft. The annular space is filled with a 9.0 lbm/gal brine. Assuming ideal gas behavior, compute the amount by which the exterior pressure on the tubing exceeds the interior tubing pressure at 10,000ft if the surface tubing pressure is 1000 psia and the mean gas temperature is 140F. If the collapse resistance of the tubing is 8330 psi, will the tubing collapse due to the high external pressure?

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The pressure in the annulus (external pressure) at D = 10,000 ft is P2 = 0.052 * 9.0 * 10,000 + 14.7 = 4,695 psia The pressure in the tubing (internal pressure) at D = 10,000ft Pressure difference = p2 – p = 4695 – 1188 = 3507 < 8330 psia The tubing will withstand the high external pressure
Описание слайда:
The pressure in the annulus (external pressure) at D = 10,000 ft is P2 = 0.052 * 9.0 * 10,000 + 14.7 = 4,695 psia The pressure in the tubing (internal pressure) at D = 10,000ft Pressure difference = p2 – p = 4695 – 1188 = 3507 < 8330 psia The tubing will withstand the high external pressure

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Drilling Engineering, слайд №8
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Drilling Engineering, слайд №9
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Слайд 10


The effective density exerted by a circulating fluid against the formation that takes into account the pressure drop in the annulus above the point being considered. The ECD is calculated as:  – mud density, ppg P – Sum of the hydrostatic pressure and the frictional pressure drop in the annulus between the depth D and surface, Psig D – the true vertical depth, ft
Описание слайда:
The effective density exerted by a circulating fluid against the formation that takes into account the pressure drop in the annulus above the point being considered. The ECD is calculated as:  – mud density, ppg P – Sum of the hydrostatic pressure and the frictional pressure drop in the annulus between the depth D and surface, Psig D – the true vertical depth, ft

Слайд 11


Example: A 9.5-PPG drilling fluid is circulated through the drill pipe and the annulus. The frictional pressure losses gradient in the annulus is 0.15. Calculate the equivalent circulating density in PPG. Solution:  = 9.5 + P/0.052 = 9.5 + 0.15 / 0.052 = 12.4 PPG
Описание слайда:
Example: A 9.5-PPG drilling fluid is circulated through the drill pipe and the annulus. The frictional pressure losses gradient in the annulus is 0.15. Calculate the equivalent circulating density in PPG. Solution:  = 9.5 + P/0.052 = 9.5 + 0.15 / 0.052 = 12.4 PPG

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Drilling Engineering, слайд №12
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Слайд 13


10,000 ft of 19.5-lbm/ft drillpipe and 600 ft of 147 lbm/ft drill collars are suspended off bottom in a 15-lbm/gal mud. Calculate the effective hook load that must be supported by the derrick. Density of steel is 65.5 lbm/gal Solution: W = 19.5 * 10000 + 147 * 600 = 283200 lbm We = W(1 - f/s) = 283200 * (1 - 15/65.5) = 218300 lbm (density of steel = 65.5 lbm/gal = 490lbm/cu ft)
Описание слайда:
10,000 ft of 19.5-lbm/ft drillpipe and 600 ft of 147 lbm/ft drill collars are suspended off bottom in a 15-lbm/gal mud. Calculate the effective hook load that must be supported by the derrick. Density of steel is 65.5 lbm/gal Solution: W = 19.5 * 10000 + 147 * 600 = 283200 lbm We = W(1 - f/s) = 283200 * (1 - 15/65.5) = 218300 lbm (density of steel = 65.5 lbm/gal = 490lbm/cu ft)

Слайд 14


Energy balance: Pp is heat entering the system Pf is heat loss due to friction
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Energy balance: Pp is heat entering the system Pf is heat loss due to friction

Слайд 15


Applying the energy equation for a flow through a nozzle with neglecting: effects of elevation: D2 - D1 = 0 effects of uptream velocity vo = 0 Heat entering the system Pp = 0 and friction loss Pf = 0
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Applying the energy equation for a flow through a nozzle with neglecting: effects of elevation: D2 - D1 = 0 effects of uptream velocity vo = 0 Heat entering the system Pp = 0 and friction loss Pf = 0

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Drilling Engineering, слайд №16
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Assuming a constant Pb through all the nozzles Pressure drop across the bit  – lbm/gal ; q – gpm ; At - in2
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Assuming a constant Pb through all the nozzles Pressure drop across the bit  – lbm/gal ; q – gpm ; At - in2

Слайд 18


The purpose of the jet nozzles is to improve the cleaning action of the drilling fluid at the bottom of the hole. Since the fluid is traveling at a vertical velocity vn , before reaching to the hole and traveling at zero vertical velocity after striking the hole bottom hence all the fluid momentum is transferred to the hole bottom. Force is time rate of change of momentum, hence: Substitute vn to the equation above gives Where Fj is the hydraulic impact force given in pounds.
Описание слайда:
The purpose of the jet nozzles is to improve the cleaning action of the drilling fluid at the bottom of the hole. Since the fluid is traveling at a vertical velocity vn , before reaching to the hole and traveling at zero vertical velocity after striking the hole bottom hence all the fluid momentum is transferred to the hole bottom. Force is time rate of change of momentum, hence: Substitute vn to the equation above gives Where Fj is the hydraulic impact force given in pounds.

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Example: A 12.0 lbm/gal drilling fluid is flowing through a bit containing three 13/32 in nozzles at a rate of 400 gal/min. Calculate the pressure drop across the bit and the impact force developed by the bit. Solution: Assume Cd = 0.95 Hydraulic impact force:
Описание слайда:
Example: A 12.0 lbm/gal drilling fluid is flowing through a bit containing three 13/32 in nozzles at a rate of 400 gal/min. Calculate the pressure drop across the bit and the impact force developed by the bit. Solution: Assume Cd = 0.95 Hydraulic impact force:

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Drilling Engineering, слайд №20
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Newtonian Model Non-Newtonian Model Bingham-plastic model Power Law model: Yield power law model:
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Newtonian Model Non-Newtonian Model Bingham-plastic model Power Law model: Yield power law model:

Слайд 22


Pseudoplastic (Time-independent shear thinning fluids) If the apparent viscosity decreases with increasing shear rate Dilatant (Time-independent shear thickening fluids) If the apparent viscosity increases with increasing shear rate
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Pseudoplastic (Time-independent shear thinning fluids) If the apparent viscosity decreases with increasing shear rate Dilatant (Time-independent shear thickening fluids) If the apparent viscosity increases with increasing shear rate

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Thixotropic (Time-dependent shear thinning fluids): If the apparent viscosity decreases with time after the shear rate is increased to a new constant value Rheopectic (Time-dependent shear thickening fluids): If the apparent viscosity increases with time after the shear rate is increased to a new constant value Drilling fluids and cement slurries are generally thixotropic
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Thixotropic (Time-dependent shear thinning fluids): If the apparent viscosity decreases with time after the shear rate is increased to a new constant value Rheopectic (Time-dependent shear thickening fluids): If the apparent viscosity increases with time after the shear rate is increased to a new constant value Drilling fluids and cement slurries are generally thixotropic

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Drilling Engineering, слайд №24
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A rotational viscometer is used to determine type of the fluid and the rheological model of the fluid. This can be done by varying the speed of the rotor (varying the shear rate) and reading the dial reading (shear stress). To convert the speed to shear rate and dial reading to shear stress, simply use these corellations:  = 1.703 x rpm, 1/s  = 1.06 x Dial Reading
Описание слайда:
A rotational viscometer is used to determine type of the fluid and the rheological model of the fluid. This can be done by varying the speed of the rotor (varying the shear rate) and reading the dial reading (shear stress). To convert the speed to shear rate and dial reading to shear stress, simply use these corellations:  = 1.703 x rpm, 1/s  = 1.06 x Dial Reading

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Drilling Engineering, слайд №26
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Drilling Engineering, слайд №27
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Drilling Engineering, слайд №28
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Drilling Engineering, слайд №29
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