Lesson 5 Air, Gas, Mist cont

# Lesson 5 Air, Gas, Mist cont - PETE 689 Underbalanced...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PETE 689 Underbalanced Drilling (UBD) Lesson 5 Lesson 5 Air, Gas and Mist Drilling Read: UDM Chapter 2.1­2.4, Pages 2.1­2.74 Harold Vance Department of Petroleum Engineering Air, Gas, and Mist Drilling Air, Gas, and Mist Drilling Circulating Pressures. Equipment design. Operating procedures. Limitations of Dry Air Drilling. Natural Gas Drilling. Mist Drilling. Harold Vance Department of Petroleum Engineering Circulating Pressures Circulating Pressures Calculating standpipe pressure starts with predicting the pressure just below the bit, and working your way back to the surface. 4. Pi 3. Pa 1. PS 2. Pb Pressure Calculation Steps for Pure Gas Fluid Harold Vance Department of Petroleum Engineering Bit Pressure Drop Bit Pressure Drop As air flows through the jets, it expands in response to the decrease in pressure and it’s velocity increases. Once the pressure drop exceeds a certain level, the air velocity reaches the prevailing speed of sound. Harold Vance Department of Petroleum Engineering Bit Pressure Drop Bit Pressure Drop At this point, the air cannot expand any faster and the upstream pressure becomes independent of the downstream pressure. This implies that under sonic discharge conditions the standpipe pressure is independent of the annular pressure. Harold Vance Department of Petroleum Engineering Bit Pressure Drop Bit Pressure Drop The condition for the onset of sonic flow in ideal gases is: Pa 2 = Pb k+1 ­k K­1 Pa = upstream pressure at the onset of sonic flow, psia. Pb = downstream pressure. k =ratio of specific heat at constant pressure to that at constant volume. For air, k = 1.4 and Pa = 1.89Pb Harold Vance Department of Petroleum Engineering Bit Pressure Drop Bit Pressure Drop If the upstream pressure is more than 1.89 times the annulus pressure beneath the bit, flow through the bit will be sonic. Harold Vance Department of Petroleum Engineering Upstream Bit Pressure Upstream Bit Pressure Sonic Flow G TaR 2 k+1 0.5 Pa= 1­k An Sgk k+1 G = mass flow rate of air in lbm/s An = total area of the bit nozzles, sq.in. Ta = air temperature above the bit, 0R R = the universal gas constant, 53.3 ft – lbf/lbm 0R for air. S = gas gravity ( 1 for air). g = gravitational constant, 32.17 ft/s2 Harold Vance Department of Petroleum Engineering Upstream Bit Pressure Upstream Bit Pressure Sonic Flow Noting that the density of air under standard conditions is 0.0764 lbm/cu.ft. the above equation reduces to: GTa0.5 Pa = 1.88 =0.00239 An Q = air flow rate (scfm). Harold Vance Department of Petroleum Engineering QTa0.5 An Upstream Bit Pressure Upstream Bit Pressure Sub­sonic Flow If the air flow velocity through the jets remains sub­sonic, the pressure above the bit is related to the mass flow rate and the annulus pressure beneath the bit by: Pa = Pb 1+ R ( k­1) G2Tb 2gSAn2Pb2 Harold Vance Department of Petroleum Engineering K K­1 Upstream Bit Pressure Sub­sonic Flow For air this becomes: For air this becomes: Pa = Pb 1+ G= 0.236 G2Tb 3.5 An2Pb2 Qρg 60 Tb = Temperature below the bit, 0R ρg = Gas density at STP, lbm/cu.ft. ρair = 0.0764 at STP, lbm/cu.ft. Harold Vance Department of Petroleum Engineering Upstream Bit Pressure Sub­sonic Flow The circulating air cools as itexpands through the The circulating air cools as it expands through the bit. Assuming ideal behavior, the temperature decrease can be estimated from: Tb = Ta Pb K­1 k Pa Indicating that the absolute air temperature below the bit will be approximately 17% lower than that above the bit if flow through the jets is sonic. Harold Vance Department of Petroleum Engineering Standpipe Pressure Standpipe Pressure Ps = √ Pa2 + β Tav2 (e 2αh/Tav – 1) e 2αh/Tav α = β = S 53.3 1.625 x 10­6Q2 Di5.333 Di = internal diameter of the drillstring, ft. Harold Vance Department of Petroleum Engineering Steps To Steps To Predicting Standpipe Pressure Assess whether flow through bit is sonic or sub­sonic. If sonic, the pressure above the bit is determined with equation 2.21 This value is used in equation 2.25 to predict standpipe pressure. Harold Vance Department of Petroleum Engineering Steps To Predicting Standpipe Pressure If flow through the bit is sub­sonic, the annulus pressure below the bit must be first predicted (Angel’s analysis, etc) using equation 2.12. The pressure above the bit is determined by equation 2.23. This value is used in Equation 2.25 to determine standpipe pressure. Harold Vance Department of Petroleum Engineering Important Point Important Point When air drilling, large changes in annulus pressure may result in smaller changes in standpipe pressure, or in the case of sonic flow through the bit, no change in standpipe pressure at all. Hole problems that lead to an increase in annulus pressure may be indicated by small or no changes in standpipe pressure. Harold Vance Department of Petroleum Engineering Important Point It is very important to monitor the standpipe pressure closely and react promptly to unanticipated changes. It is important to know if flow through the bit is sonic or not. If flow is sonic, standpipe pressure will not change with changes in annulus pressure. Harold Vance Department of Petroleum Engineering Example Example 8.1/2” hole at 6000’ drilled with 4.1/2” drillpipe air rate is 1,400 scfm. Penetration rate ranges up to 300 ft/hr. Bit has no nozzles in one example and three 14/32’s in the other. Harold Vance Department of Petroleum Engineering Predicted bottomhole annular and standpipe pressure at various penetration rates in a 6,000 foot dry, air drilled hole. Harold Vance Department of Petroleum Engineering Predicted standpipe pressure as functions of penetration rate, for a bit with and without nozzles, in a 6,000 foot deep, air drilled hole. Harold Vance Department of Petroleum Engineering Equipment Design Equipment Design Compressor output is often expressed in standard cubic feet per minute, scfm. Common output ratings are 750 to 1,000 cfm. Altitude (and corresponding atmospheric pressure) has an effect on the actual output of compressors. Harold Vance Department of Petroleum Engineering Compressor Design Compressor Design (Assuming that air behaves as an ideal gas) Vt = Vo 14.7 ( T1 + 460) 520 P1 The air delivery rate, Qo, expressed in scfm, can be found from: Qo = Q 520 P1 14.7 (T1+460) Appendix A includes a table showing atmospheric pressure at differing elevations. Harold Vance Department of Petroleum Engineering Normal Atmospheric Pressure at Different Altitudes Harold Vance Department of Petroleum Engineering Effect Of Elevation On Effect Of Elevation On Compressor Output At 6,000’ above sea level the ambient pressure is 11.8 psia. At this elevation, a compressor rated at 1,000 scfm free air delivery will deliver only 803 scfm if the ambient temperature is 600F. At ambient temperature 6,000’ elevation, the delivery rate will drop to 745 scfm. Harold Vance Department of Petroleum Engineering Compressor Rating Compressor Rating Single stage compressors typically have a maximum discharge pressure of ~135 psi. Multi­stage compressors have discharge pressures from 250 ­ 350 psi. Boosters will be required for standpipe pressures up to 1,500 psi. Harold Vance Department of Petroleum Engineering Mist And Formers Pump Harold Vance Department of Petroleum Engineering Blooie Lines Blooie Lines Diameter too large will not carry cuttings efficiently to the reserve pit. Too small, additional pressure imposed downhole. Blooie line cross­section should be at least equal to that in the annulus over the longest section of the hole to be drilled with air. Harold Vance Department of Petroleum Engineering Frictional pressure losses down two different 150 foot long blooie lines Harold Vance Department of Petroleum Engineering Blooie Line Harold Vance Department of Petroleum Engineering Measurement of Air Injection Rate An orifice meter should be installed between the compressor An orifice meter should be installed between the compressor and the mist injector to measure the air injection rate. Q = Fb Fg √ 520 hwPf T Fb= orifice flow factor, (Appendix B) Tf Fg= (1/s) 0.5 S = Gas gravity, (1 for air). Remember to add the prevailing atmospheric pressure to the gauge pressure to obtain the absolute pressure. Harold Vance Department of Petroleum Engineering Meter chart from a well in the Arkoma Basin. The drillstring became stuck at 3:45 a.m. while drilling at 10,845 feet. Harold Vance Department of Petroleum Engineering Operating Procedures Operating Procedures Unloading a well with air. Dmax = Pmax/0.433 . Dmax = maximum water interval to be unloaded by air compressors. Pmax = delivery pressure of the air system. It is, however unusual to unload more than 2,000’ of water at any one time. Harold Vance Department of Petroleum Engineering Limitations of Dry Air Drilling Limitations of Dry Air Drilling Water inflows. Downhole fires. Wellbore instability. Harold Vance Department of Petroleum Engineering Water Inflows Water Inflows Mud ring formed from formation water wetting the cuttings. Cuttings stick together and accumulate at the shoulder on the top of the BHA, Drillstring and the walls of the hole. Cuttings mixed with a small amount of water will form a mud ring at the top of the drill collars where hole cleaning is critical. Harold Vance Department of Petroleum Engineering Downhole Fires Downhole Fires Effect of pressure on combustible concentrations of natural gas in air Harold Vance Department of Petroleum Engineering Reverse Circulation Air Drilling Reverse Circulation Air Drilling Advantages Reduced Damage to Permeable formations. Quality and size of drill cuttings is improved. Wellbore integrity is improved Less air volume required. Harold Vance Department of Petroleum Engineering Reverse Circulation Air Drilling Reverse Circulation Air Drilling Disadvantages Greater likelihood of cuttings plugging the bit. Surface equipment needs improvement. Large inflows above the bit may cause problems circulating down the annulus. Harold Vance Department of Petroleum Engineering Natural Gas Drilling √ Vt = 4gdc ρc – ρf 3Cdρf Vt = Terminal Velocity (ft/s). g = Gravitational acceleration, 32.17 ft/sec2 dc = Characteristic particle diameter, ft. Cd = Drag coefficient. ρc = Density of cuttings, lbm/ft3 ρf = Density of fluid, lbm/ft3 Harold Vance Department of Petroleum Engineering Terminal Velocity Of Terminal Velocity Of Natural Gas At Atmospheric Pressure Vtg = Vtair (1/S)0.5 Vtg = terminal velocity in natural gas. Vtair = Terminal velocity in air. S = Specific gravity gas. Harold Vance Department of Petroleum Engineering Natural Gas Drilling Natural Gas Drilling Lower density of natural gas than air results in: • • • • Lower BHP. Lower drag forces. Higher required circulation rates. Non­ideal behavior of natural gas is not usually a problem since operating pressures are low (< 2,200 psi) and ideal behavior can be assumed. Harold Vance Department of Petroleum Engineering Natural Gas Injection Rate Natural Gas Injection Rate A first order estimate of the minimum injection rates can be derived by taking Angel’s figures for air drilling at the appropriate depth and penetration rate and dividing these by the square root of the gas’s specific gravity. Usually acceptable in practice. Harold Vance Department of Petroleum Engineering Mist Drilling Mist Drilling Liquid volumes are only 1 to 2 percent at the prevailing temperature and pressure. Harold Vance Department of Petroleum Engineering For A Lightened Drilling Fluid: For A Lightened Drilling Fluid: VgP VgP FgP = = VmP VgP + VfP + VsP g = gas VfP VfP FfP = = VmP VgP + VfP + VsP f = liquid VsP VsP FsP = = VmP VgP + VfP + VsP P = solids FgP + FfP + FsP = 1 Harold Vance Department of Petroleum Engineering Eq. 2.35 For A Lightened Drilling For A Lightened Drilling Fluid: We can assume that the gaseous phase acts as an ideal gas. Solid and liquid phases are incompressible. Harold Vance Department of Petroleum Engineering For A Lightened Drilling Fluid: For A Lightened Drilling Fluid: VgP = Vgo Po P VƒP = Vƒo + Vƒ VsP = Vso=Vs Harold Vance Department of Petroleum Engineering Eq. 2.37 Substituting Eq. 2.37 into 2.35 Substituting Vgo FgP = Po P Po Vgo + Vf + Vs P From this, the gas fractional at pressure is… Harold Vance Department of Petroleum Engineering Substituting Eq. 2.37 into 2.35 FgP = = Fgo P Fgo + (Fƒo+ Fso) Po Fgo Fgo + (1­ Fgo) P Po Harold Vance Department of Petroleum Engineering For The Liquid And Solid For The Liquid And Solid Volume Fractions: FfP = FsP= Fƒo P (1­Fgo)+ (Fgo) Po Fgo (1­Fgo)+ (Fgo) P Po Harold Vance Department of Petroleum Engineering Mixture Density (mist) Mixture Density (mist) ρmP = ρmo Vmo VmP Assuming that the gaseous phase obeys the ideal gas law and the solid and liquid are incompressible, Harold Vance Department of Petroleum Engineering Mixture Density (mist) ρmP = ρmo ( Vgo + Vƒ +Vs) Po Vgo + Vƒ + Vs P = ρmoVmo VmoFƒo + VmoFso + VmoFgo Harold Vance Department of Petroleum Engineering Po P Mixture Density (mist) Mixture Density (mist) Finally ρmP = ρmo Po 1­Fgo 1 ­ P Harold Vance Department of Petroleum Engineering Example Mist Drilling Example Mist Drilling Liquid is metered at 10.7 BPH (1cfm) Dry air injection rate = 2,000 scfm Standpipe pressure = 210 psig Liquid volume fraction at atmospheric pressure is: 1cfm/(2,000 scfm + 1cfm)~5 x 10­4 = .05% Harold Vance Department of Petroleum Engineering Example con’t Example con’t Assume atmospheric pressure = 15 Psia (P/Po) = (210+15)/15 = 15 VgP = 133 cfm Substituting into equation 2.40 FfP = 7.5 x 10­3 Harold Vance Department of Petroleum Engineering Example con’t Example con’t Assume that there is a water inflow of Assume 96.3 BWPH at BHP = 135 psig 96.3 Assume volume of cuttings is Assume negligible. negligible. Total fluid volume is now 10 cfm. At the blooie line exit, the liquid volume At fraction will be 10/2,000 x 100 = 0.5 % fraction At bottom hole conditions the liquid At fraction would be 4.78% - foam will probably form. probably Harold Vance Department of Petroleum Engineering Effect Of Temperature Effect Of Temperature On Mist Density Equation 2.37 can be re­written: VgP,T = Vgo Po T P To Harold Vance Department of Petroleum Engineering Effect Of Temperature Effect Of Temperature On Mist Density BHT, in reality has little effect on bottom hole flow rates. Harold Vance Department of Petroleum Engineering Effect Of Temperature Effect Of Temperature On Mist Density Depth 5000’ 10,000 Temp 140 0F 220 0F Flow rate 231 cfm 262 cfm % Liquid 4.15 % 3.68 % Pressure changes can result in up to 10 fold changes in volumetric flow rates. Harold Vance Department of Petroleum Engineering Hole Cleaning, Mist Hole Cleaning, Mist Water droplets act similarly to cuttings with slip velocity of near zero ­ mists do not clean the wellbore more efficiently than dry gas. Therefore annular velocities are high. Circulating fluid density is increased however and may add to the frictional pressure losses. Harold Vance Department of Petroleum Engineering Hole Cleaning, Mist Hole Cleaning, Mist The increased density will lower the terminal velocity of the cuttings, but will increase the BHP reducing the volumetric flow rate at the bottom of the hole. Higher air injection rates are usually required when misting than with dry air. Harold Vance Department of Petroleum Engineering Application Of Angel’s Application Of Angel’s Method To Mist Drilling Determine the penetration rate that would generate the same mass of cuttings as the mass of liquid entering the well over a time period. This includes any base liquid, foamer, and water influx. Harold Vance Department of Petroleum Engineering Apparent Equivalent ROP Apparent Equivalent ROP If the total liquid rate is L (BPH) the mass flow of liquid entering the well will be 350.5L assuming that the liquid is water ( 62.4 lbm/ft.cu.) 350.5L 380L = ROPе = π Db 2 169 Db2 4 12 This is added to the actual anticipated penetration rate. Harold Vance Department of Petroleum Engineering Angel’s Method For Mist Angel’s Method For Mist The minimum air injection rate, required for good hole cleaning during mist drilling, is determined; either from Angel’s charts or from the approximation in equation 2.17. Harold Vance Department of Petroleum Engineering Example Example Hole size = 7 7/8” depth = 5,000’ Drillpipe size = 4 1/2” Anticipated ROP = 30 feet/hr Qo = 671, N = 65, H = 5,000/1,000 = 5 Minimum air rate for dry air = Qa = Qa +NH = 670 + 65x5 = 995 scfm Harold Vance Department of Petroleum Engineering Example Example Liquid injection rate is 6 BPH. Water influx is 3.8 BPH. Total liquid rate is 9.8 BPH. Penetration rate that would give this mass cuttings per hour is 60 ft/hr. Harold Vance Department of Petroleum Engineering Example Example The minimum air rate required for dry air drilling at a penetration rate of 90 ft/hr using the value of N for 90 ft/hr, N = 98.3 would be 1,162 scfm. Harold Vance Department of Petroleum Engineering Limitations To Mist Drilling Limitations To Mist Drilling Higher air injection rates than dry gas (up to 40%). Waste water disposal. Wellbore instability. Corrosion. Harold Vance Department of Petroleum Engineering ...
View Full Document

## This note was uploaded on 10/30/2011 for the course PETROLEUM 689 taught by Professor Jeromej.schubert during the Fall '11 term at Texas A&M University-Galveston.

Ask a homework question - tutors are online