Презентация Drilling Engineering онлайн
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- Тип файла:ppt / pptx (powerpoint)
- Всего слайдов:29 слайдов
- Для класса:1,2,3,4,5,6,7,8,9,10,11
- Размер файла:862.50 kB
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- Автор:неизвестен
Слайды и текст к этой презентации:
№1 слайд
Содержание слайда: Drilling Engineering – PE 311
Hydraulics of Drilling Fluids
№2 слайд
Содержание слайда:
№3 слайд
Содержание слайда: For incompressible fluids, the specific weight of the liquid in field unit is given by
If P0 = 0 then
The fluid density
№4 слайд
Содержание слайда: 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
№5 слайд
Содержание слайда: EOS of gas:
№6 слайд
Содержание 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?
№7 слайд
Содержание 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
№8 слайд
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№9 слайд
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№10 слайд
Содержание 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
№11 слайд
Содержание 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
№12 слайд
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№13 слайд
Содержание 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)
№14 слайд
Содержание слайда: Energy balance:
Pp is heat entering the system
Pf is heat loss due to friction
№15 слайд
Содержание слайда: 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
№16 слайд
Содержание слайда:
№17 слайд
Содержание слайда: Assuming a constant Pb
through all the nozzles
Pressure drop across the bit
– lbm/gal ; q – gpm ; At - in2
№18 слайд
Содержание 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.
№19 слайд
Содержание 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:
№20 слайд
Содержание слайда:
№21 слайд
Содержание слайда: Newtonian Model
Non-Newtonian Model
Bingham-plastic model
Power Law model:
Yield power law model:
№22 слайд
Содержание слайда: 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
№23 слайд
Содержание слайда: 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
№24 слайд
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№25 слайд
Содержание 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
№26 слайд
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№27 слайд
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№28 слайд
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№29 слайд
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