Reservoir Estimates One of the important functions of the reservoir engineer is the periodic calculation of reservoir oil and gas in place and the recovery anticipated under the prevailing reservoir drive mechanisms. Reverse estimation methods are usually categorized into three types: 1. Analogy 2. Volumetric methods 3.performance based techniques Material balance calculations Decline curve analysis Pressure transient analysis Numerical simulation techniques. 1.Analogy: During this period , before any wells are drilled on the property , any estimates will be of a very general nature based on experience from similar pools or wells in the same area. i.e by analogy. 2. Volumetric methods The volumetric methods involve a determination of the bulk reservoir rock volume , average porosity , fluid saturations , formation volume factors from which the total reservoir hydrocarbon volume is calculated. Recoverable reserves are then estimated by application of a suitable recovery factor and the formation /surface volume factor for the produced fluid. Recoverable oil = Where : Vb x Ø x (1-Sw) x R.F. Bo Vb is the bulk reservoir volume Ø is the fractional porosity (1-Sw) is the hydrocarbon saturation R.F. is the recovery factor Bo is the oil formation volume factor A recovery factor is approximated considering : - laboratory measurement of oil displacement in cores So –Sor So -type of displacement mechanism involved -correlation of sweep efficiency based on a similar reservoir Reservoir estimates are needed at various stages of a project 1)Geophysical exploration stages Same order –of- magnitude estimate of the reserves which a structure might contain is necessary to rank various projects probability of economic success prior to making bids or possibly to relinquish undrilled acreage. The first estimate is based on the volume of the structure determined from seismic maps supplemented by information on local geological trends which may indicate the thickness of porous beds which may be encountered. By applying the common range of rock parameters , porosity (7 to 30%) , water saturation (8 to 40%) and recovery factor (10 to 50%) a possible range of reserves that the structure might contain is estimated. 2)Exploration stage With the drilling of a discovery well the uncertainty of encountering hydrocarbons is removed , and measured values for porosity , and water saturation became available for the section of pay traversed. Assuming that well log data corroborate the prior seismic data , now only the contour of the hydrocarbon /water contact (O.W.C.) is required to make reasonable estimate for this stage. 3)Field development stage: As new wells are drilled the volume and geometrical distribution of the reservoir become even more accurately defined as well as the average reservoir porosity and saturation values . On the other hand , fluid withdrawals and injections into the reservoir and the corresponding changes in fluid interfaces must be accounted for as the inventory of reserves is continuously upgraded . Whether the accounting of hydrocarbon reserves is made by computer or manually , the procedures are the same in principle. Obviously , the trend in reservoir studies is toward numerical simulation on which not only the static inventory of reserves is kept , but which can predict the future behaviour of a field. Calculation of the reserve The gross reservoir rock volume enclosed by the structure above the hydrocarbon /water contact is calculated in the following steps: 1) A net isopach map , giving the contour of equal thickness of pay with the water contact assigned zero elevation contour is the most convenient basis for rock volume calculations. 2) The area within each contour is determined by planimetering , and a plot prepared of area contained in each contour versus depth: 3) The gross rock volume is A dh which may be found by planimetering again or by application of a numerical integration rule . In the Schlumberger field studies approach the volumetric reservoir distribution is calculated numerically and plotted by machine as isopach and isovolume maps. V =h An + A n+1 + An A n+1 3 - Trapezoidal formula V = h An +An+1 2 Vb = h Ao +2A1 +2A 2 …2A n-1 +A n + t av x An 2 - Pyramidal formula 3.Performance based techniques Material balance calculations In many cases , porosity , saturations , and reservoir bulk volume are not known with any reasonable accuracy , and emphasis on volumetric calculations for reserve estimates is not advisable. The material balance is a useful auxiliary tool for confirming reservoir estimates. The material balance equation allow dependable estimates of the initial hydrocarbons in place as well as prediction of the future reservoir performance. The material balance equation (MBE) relates the volumes of fluids withdrawn and encroached to the resulting reservoir pressures. Its principal utility , however , lies in predicting reservoir behavior and not in the estimation of initial hydrocarbon in place. The generalized material balance equation: The general form of the oil material balance , first presented by Schilthuis is derived as a volume balance which equates : (a) cumulative observed production , expressed as an underground withdrawal , to (b) the expansion of the fluids in the reservoir resulting from a finite pressure drop. For the general case, the material balance has the form: underground fluid expansion of oil zone expansion of any reduction of withdrawal of oil , = (includes liquid oil+dissolved gas)+ gas cap + HCPV due to ,water, gas(Bbls) (Bbls) (Bbls) connate water expansion and rock compaction and water encroachment (Bbls) Symbols: cumulative oil production, Np ,STB cumulative gas production ,Gp ,SCF cumulative water production , Wp ,STB original oil in place ,N ,STB original gas in place ,G ,SCF cumulative water influx ,res. bbl , We m = gas cap size , G X Bgi oil zone size , N X Boi Bo,Bg , Bw , Bt are PVT properties of the different reservoir , s fluids Bt = Bo+(Rsi-Rs)Bg Ri = GOR = Kg/Ko x Mo/Mg x Bo/Bg +Rs (instantaneous gas oil ratio) Average GOR ,Rp = Cumulative gas production = Gp SCF Np STB Cumulative oil production Bgc = gas cap Bg Bgs = solution gas Bg Gpc = cumulative gas production from gas cap Gps = cumulative gas production from solution gas. Derivation of the oil MBE Let us define the following quantities: N= the initial oil in place(STB) = VBO (*1-Swi) /Boi (!) Gas cap expansion = New gas cap volume - original gas cap volume New gas cap volume = (G-Gpc) B gc Original gas cap volume = G x Bgci (1)Gas cap expansion = (G-Gpc) B gc - G x Bgci (2) Remaining release gas = original soluble gas –remaining soluble gas – - cumulative produced gas. = [N Rsi -(N-Np)Rs -Gps] Bgs (3) Remaining oil volume = (N-Np) Bo (4) Net water influx = (W e -W p Bw) (5) Rock and water expansion is neglected in the presence of gas. Gas Cap Gas Cap Gas cap expansion Release gas Oil Remaining Oil (N Boi ) Connate water Pi Connate Water P Rock & water expansion Net water influx Aquifer ( water bearing zone) Aquifer ( water bearing zone) Condition I Condition II Pressure = Pi NP = Zero GP = Zero WP = Zero P<Pi NP = +ive GP = +ive WP = +ive Now, by equating the initial conditions , to the final conditions resulting from a finite pressure drop. N Boi = (N-NP) Bo + [(G-GPC) Bgc -G Bgci] +[NRsi –(N-NP)Rs–GPS ]Bgs+(We -WPBw) NBo – NPBo + G Bgc- GPC Bgc - G Bgci + N RSi Bgs- N RS Bgs+ NP Rs Bgs - N Boi = -GPS Bgs + (We-WP Bw) NBoi-NBo-NRsi Bgs +NRsBgs=-NPBo+NPRsBgs-GPSBgs+GBgc-GPcBgc-GBgci+(We-WPBw) N[Boi-Bo-RsiBgs+RsBgs]=Np[-Bo+RsBgs]-Gps Bgs+GBgc-Gpc Bgc-GBgci+(We-WpBw) But : Bgc = Bgs = Bg and Bgci = Bgsi Bgi N[Boi-Bo+(-Rsi+Rs)Bg] = Np[-Bo+RsBg]-Gps Bg+GBg –GpcBg-GBgi+(We-WpBw) Np[Bo-RsBg]+(Gps+Gpc)Bg =N[Bo-Boi+(Rsi-Rs)Bg]+ G(Bg-Bgi)+(We-WpBw) But : (Gpc+Gps)Bg = GpBg Therefore , and Rp = Gp/Np and Bt = Bo+(Rsi-Rs)Bg Np [ Bo - RsBg]+GpBg =N[(Bo-Boi)+(Rsi-Rs)Bg]+ G(Bg-Bgi)+(We-WpBw) By adding and subtracting RsiBg -RsiBg Np[Bo-(Rsi-Rs)Bg]-RsiBg]+GpBg =N[(Bo-Boi)+(Rsi-Rs)Bg]+ G(Bg-Bgi)+(WeWpBw) Np[Bt-RsiBg]+Np Rp Bg =N[(Bo-Boi)+(Rsi-Rs)Bg]+ G(Bg-Bgi)+(We-WpBw) Np [Bt – Rsi Bg + Rp Bg] = N [(Bo - Boi)+(Rsi-Rs)Bg] + G(Bg-Bgi) +(We-WpBw) And finally : Np [Bt + (Rp - Rsi) Bg] = N (Bt – Bti ) + mNBoi(Bg - Bgi) + (We-WpBw) …..(1) Bgi Where : m = GBgi Nboi , G = mNBoi Bgi N (Bt – Bti ) + Np [Bt + (Rp - Rsi) Bg] = Cumulative oil withdrawal Depletion Drive mechanism mNBoi(Bg - Bgi) Bgi Gas cap drive mechanism + (We-WpBw) Water drive mechanism and this is the generalized material balance equation for combination drive reservoir neglecting the rock and connate water expansion. N ( B t – B t I ) = D.D.I Np[Bt+(Rp-Rsi)Bg] m N Boi/ Bgi (Bg - Bgi) = GCDI BgiNp [Bt + (Rp - Rsi) Bg] ( We- Wp Bw) = Np[Bt+(Rp-Rsi)Bg] WDI Driving index of any mechanism: The driving index of any mechanism represents the fractional contribution of the total oil withdrawal produced by that mechanism. Driving Index GCDI DDI WDI Time or Np or Pressure It is clear from the previous figure that the DDI and GCDI decrease both of them with time. This is because that both of the gas cap gas and solution gas are limited volume wise and pressure wise. However , the water aquifer is a huge water bearing zone (unlimited volume in case of infinite aquifer), which means that the volume of water available in the aquifer is very great. Normally , the reservoir pressure in the aquifer will remain more or less constant , while the reservoir pressure is decreasing with time. Therefore, the pressure draw down between the aquifer and the reservoir increases with time which means increasing in water influx and consequently an increasing water drive index(WDI). Rock and water expansion: Definitions: 1. Compressibility : The change in volume per unit volume per unit pressure. 2. Compressibility factor (z): It is a measure of the deviation of natural gas with respect to ideal gas. C = - V . 1 V P Note that formation compressibility = pore volume compressibility. Cf = - V . 1 V (Pi-P) Now , let us consider the rock expansion = Cf Vp (Pi-P) …….(A) Saturation = fluid volume Pore volume Soi = (1-swi) = OOIP = Vp NBoi Vp And thus: Vp = (NBoi ) (1-Swi) substitute in (A) Rock expansion = Cf (NBoi ) (Pi-P) (1-Swi) Wate expansion : Cw = - Vw . 1 Vw Vw = Vw Cw SW = Vw (Pi-P) then : Vw = Vp. Sw Vp (Pi-P) Therefore Water expansion = Vw = Cw x Vp Sw(Pi-P) Water expansion = Vw = Cw x Sw NBoi 1-Swi also , (Pi-P) Therefore , Rock & water expansion = Cf (NBoi ) (Pi-P) + (1-Swi) + Cw x Sw ( NBoi ) (Pi-P) (1-Swi ) = (Cf +CwSw) ( NBoi ) (Pi-P) (1-Swi ) The most general form of Material Balance Equation is: Np [Bt + (Rp - Rsi) Bg] = N (Bt – Bti ) + mNBoi(Bg - Bgi) Bgi + (We-WpBw) + (Cf +CwSw) (NBoi ) (Pi-P) ……(2) (1-Swi ) Case (1): Water drive reservoir: a) Below the bubble point pressure: The driving mechanisms involved are : 1.Water drive mechanism 2.Depletion drive mechanism The material balance equation is : Np [Bt + (Rp - Rsi) Bg] = b) N (Bt – Bti ) + (We-WpBw) …………………..(3) Above the bubble point pressure: The driving mechanisms involved are : 1.Water drive mechanism 2.Depletion drive mechanism and 3.Rock and water expansion mechanism. The material balance equation is : Np Bo = N (Bo – Boi ) (Since Rp = Rsi + ( We –Wp Bw ) + (Cf +CwSw) (NBoi ) (Pi-P )......(4) (1-Swi ) = Rs = Constant) . Effective oil compressibility : Co = - Vo Vo Bo - Boi = 1 P Co Boi = Bo-Boi Boi 1 (Pi-P) (Pi-P) Substitute this value in equation (4): Np Bo = N Co Boi (Pi-P) + ( We –Wp Bw ) + (Cf +CwSw) (NBoi ) (Pi-P ) ..(5) (1-Swi ) + Np Bo + Wp Bw = N Boi (Pi-P) Cf +CwSw (1-Swi ) Np Bo + Wp Bw = N Boi (Pi-P) Cf +CwSw +Co So (1-Swi ) Np Bo + Wp Bw = N Boi Ce (Pi-P) + Co We + We ………..(6) + We …....…..(7) …………………….(8) Where : Ce = Cf + Cw Sw +So So = effective oil compressibility. 1-Swi Case (2): Gas cap drive reservoir: The driving mechanisms involved are : 1.Gas cap drive mechanism and 2.Depletion (solution gas) drive mechanism Np [Bt + (Rp - Rsi) Bg] = N (Bt – Bti ) + mNBoi (Bg - Bgi) Bgi Case (3): Depletion drive reservoir: a) Below the bubble point pressure: The driving mechanism involved is : Depletion (solution gas) drive mechanism only. Np [Bt + (Rp - Rsi) Bg] = c) N (Bt – Bti ) Above the bubble point pressure (Under-saturated reservoir) : The driving mechanism involved is : Rock and fluid expansion only. The material balance equation is: Np Bo = N(Bo-Boi) + Cf +CwSw +Co So (1-Swi ) Bo-Boi = Co Boi (Pi –P ) , finally: Np Bo = N Boi Ce P NBoi ( Pi - P ) + We If there is water production :, the equation form becomes : Np Bo + Wp Bw = N Boi Ce P This is the material balance equation for depletion drive reservoir (DDR) producing above the bubble point pressure (under- saturated reservoir) , taking into account the rock and water compressibilities ( expansion mechanism ). However, the rock and water compressibilities are small relative to the oil compressibility and thus , the rock and water expansion mechanism can be cancelled. Therefore , the equation will be as follows: Np Bo = N (Bo - Boi ) Np Bo = N Bo - N Boi and therefore : (N – Np ) = N Boi this is the simplest form of the material balance equation which represents a depletion drive reservoir (DDR) producing above the bubble point pressure (under- saturated reservoir) neglecting the rock and water expansion mechanism. The last equation can be driven simply by considering the initial and remaining oil in-place only ( of course , in addition to the connate water). Pressure = Pi Original Oil in-Place Remaining oil N Boi (N-Np) Bo Connate water 11 Pressure = P < Pi Connate water Reservoir performance curves Typical performance curves of primary recovery mechanisms: It is very rare to find a reservoir driven by single mechanism. In most cases , the reservoirs are driven by more than one mechanism. For example, if you have a water drive reservoir producing above bubble point pressure, the mechanisms will be water influx and fluid and rock expansion. Once the reservoir pressure introduce to become lower than bubble point pressure , the driving mechanism of the same reservoir will be water drive and solution gas drive. Sharp reduction in pressure GCDR P Fluid &Rock Expansion Mechanism DDR WDR 0 Np/N or R.F GOR 1.0 Bg is very high GCDR Pb WDR F& Rock Expansion 0.0 DDR NP/N or R.F 1.0 The following table represents the different recovery mechanisms and the recovery factors associated with each of them: Reservoir mechanism Ultimate recovery factor,% Fluid &rock expansion(F&R Exp.M) 5% of the OOIP Depletion drive mechanism(D.D.M) 15-30% of the OOIP Gas cap drive mechanism(G.C.D.M) 40-50% of the OOIP Water drive mechanism (W.D.M.) 50% or more of the OOIP 1- For water drive reservoirs: A rapid pressure reduction is occurred and then the water influx can compensate this reduction causes gradual pressure reduction. Gas oil ratio ( GOR) : Almost constant = solubility of gas 2-For Gas cap drive reservoirs: The reduction in reservoir pressure is slow due to the high expansibility of the gas , then reduction of the reservoir pressure is compensated by the gas cap expansion . Near the end of the reservoir life , reservoir pressure is considerably decline , the gas cap gas invades the oil zone and you can,t avoid the production of the gas cap gas . thus you will concern a rapid increase in GOR , which is corresponding to rapid decline in the reservoir pressure. 3-For depletion drive reservoirs: The reservoir pressure declines slowly at the beginning and then rapidly. The GOR remains more or less constant for the period from the initial reservoir pressure up to the Pb ( bubble point pressure). Then it decreases for a short period and then increases again to reach a maximum value after which it starts to decrease again. To explain this, it is well known that the gas solubility above the bubble point pressure is constant and therefore the surface GOR is constant as well. Once the pressure is reduced below bubble point pressure , the solution gas start to release from oil , however , this gas can not be move (immobile) until its saturation exceeds the critical gas saturation. So , the observed surface GOR in this short period ( from Pb to Sgc) decreases. After that , the released free gas becomes mobile and can be produced to be added to the solution gas which results in increasing the GOR. Near the end of the reservoir life , the reservoir pressure declines to a low value which means an increase in the gas formation volume factor (Bg) . since the surface gas oil ratio ,GOR is expressed as : GOR = K g . o .o + Rs Kg g g Thus , surface GOR starts to decreases due to increasing g Rock and fluid expansion : As for the rock and fluids expansion which dominates for a very short period of reservoir life , the reservoir pressure declines very rapidly while thesurface GOR remains constant. How to determine the reservoir driving mechanism: The worst conditions is the rock and fluid expansion only. This exists in the case of DDR above Pb. Np o + N W p w Np o + = = N oi Ce P W p w oi Ce P i.e Y Y = Constant. X Past performance: 1-production data 2-pressure data and from production commencement so far. 3-PVT data. Gas cap or We X X Y X X Correct assumption O O O O O O Lost production or Thief zone P Where : Y = Apparent N = Np o + W p w oi Ce P The driving mechanism of any reservoir can be determined as follows :(1) plot the past performance (GOR & pressure Vs Np/N ) and compare the same with the typical performance curves of the different driving mechanisms to guess which mechanism (2) to check if your guessing is correct or not ,do the following: (3) assume the lowest efficient driving mechanism i.e (that is to say) a fluid & rock expansion only (D.D.R above Pb ) i.e. under saturated DDR (4) The M.B. equation for this reservoir can be written as: NPBo +WPBW= N BOICe P (5) Use the past performance data of your reservoir (P, NP. and PVT data) (6) Use the past performance to solve the M.B. equation at different time intervals and plot apparent N versus pressure . If this plot represent a horizontal line , this means that your assumption is good and your driving mechanism is really fluid & rock expansion mechanism. (7)If not , there are two possibilities : a) The apparent (N) increases with pressure . This indicate that there is an other external force rather than DDR mechanism which might be gas cap or water influx (We) or both . To check which of which go to the logs. b)The apparent (N) is decreases , this means that your reservoir connected to a thief zone . General material balance can be also derived as follows : EXPASNSION TERMS Oil : N(o – oI) Gas cap : m N oi gc-gci gci Liberated gas (liberated .solutionn gas) water : Vp Sw (w-wi) = N oi ( 1+m ) Sw Cw P wi 1-Sw Vp (1-) . m-mi rock: : N (Rsi –Rs) gs = N oi (1+m) .Cf mi P 1-Sw VOIDAGE TERMS Npo OIL : Librated gas (librated solution gas) : (Gps- NpRs) gs Gas cap : Gpc gs Water : –W e + Wpw Injection : –Gi gi –Wi w Equating the expansion terms to the voidage terms , and solving for the original oil in place , we get General M.B Equation for oil in – place : N = Np o + [ Gps – NpRs ] gs +Gpc gc+ W pw – W e -Gi gi –W i w (o-oi)+(Rsi-Rs) gs+moi(gc-gci)+ oi(1+m)(CwSw+Cf) P gci (1-Swi ) IF no distinction is made between solution gas , gas cap gas and injection gas , the general material balance equation can be written as: N = Np o + [ Gp– NpRs ] g – ( W p +Wp w ) -Gi g – W i w (o-oi)+(Rsi-Rs) g+moi(g-gi)+ oi (1+m)(CwSw+Cf) P gi Notice that : (1-Swi ) G= m Noi gi Although this equation is general and includes all terms yet special cases may be considered in which certain of the terms become zero and allow considerable simplifications . some of these special cases are given in the following pages. When there is no gas injection or water injection , the principal unknowns in the MB equation are N , m and We. any one of these unknowns can be calculated if the other two are satisfactorily known. also, it is some times possible to solve simultaneously to find two of these unknowns. one of the important uses of the MB equation is predicting the effect of production rate and/or injection rates (for gas or water)on the reservoir pressure. Special Cases of the MB Equation: 1-Under saturated oil reservoir above the bubble point pressure: In this case : m = zero (no gas cap ) the producing GOR R = Rs = constant , namely Rsi , so that Gp = Np Rsi , Also , with no gas injection , Gi = reduced to : N = Np o + ( W p –Wi ) w - W e (o-oi)+ oi (CwSw+Cf) P (1-Swi ) or N = Np o + ( Wp –Wi ) w - W e oi Ce P zero , the MB equation where: Co = Bo-Boi . 1 P Boi Ce = = Compressibility of single phase oil Co So +Cw Sw +Cf 1 -Swi Also , if Wp = zero , Wi = zero and We = zero, Therefore , Np o = N oi Ce P 2- Solution gas drive reservoir ( initially at bubble point pressure): (under saturated oil reservoir below Pb): In this case , Pi = Pb , m = zero , We = zero , Wp = zero or negligible , and Wi = zero , ( no water or gas injections) . Also , usually , rock and connate water expansions are negligible compared to gas and hydrocarbons expansion , so that , Cf = zero and Cw = zero . Hence the MB Equation reduces to: N = Np o + [ Gps – NpRs ] gs (o-oi)+(Rsi-Rs) g Where : N = the oil in place at the bubble point pressure. It is customary to base solution gas drive calculations on an initial (N) value representing the oil in place at the bubble point pressure Pb . Thus the original oil in place at the initial pressure Pi would be this oil in place at the bubble point pressure plus the cumulative oil production obtained until the Pb. Is reached. (case 1 obtained before). That is : Ni = Nbubble + N I- b Gas reservoirs For these gas reservoirs, the reservoir fluid is always a single phase gas for the life of the reservoir because in these systems the formation temperature is greater than the cricondentherm. Phase diagram : These diagrams are drawn from PVT date: P single phase liquid 2 Retrograde condensation C Bubble point line 1 Two phase region 1 Cricondentherm Single phase (gas) Dew point line Temperature Point C, the critical point: the point at which the liquid and gas are in equilibrium or has the same properties. Point 1, the cricondentherm : the maximum temperature where the two phase or liquid may exist. Point 2 the cricondenbar : the maximum pressure where the two phase or gas may exist. Point 3 retrograde condensation: at which , although the pressure is reduced , the gas is condensed to the liquid state. How you can confirm the type of reservoir: According to the phase diagram: as shown in the attached figure: C B A P C1 C B1 2 C2 A3 B2 A2 A1 Separator Temperature Reservoir A is a gas reservoir A –A1 A – A2 Dry gas reservoir at the surface A-A3 Wet gas reservoir Reservoir B B –B1 Gas reservoir B- B2 Gas Condensate reservoir Reservoir C C–C1 Under saturated reservoir Contains no free gas C- C2 Saturated reservoir contains both oil and free gas Note that : -the reservoir temperature is constant -reservoir pressure declines with production Determination of the original gas in place(OGIP): As with oil reservoirs, there are two main ways to estimate the original gas in place : the volumetric method and Material balance method. 1-Volumetric method : this approach is used early in the life of the reservoir ( for instance , before 5% of the reserves have been produced). The standard cubic feet of gas in a reservoir which has a gas pore volume of Vg ( cu.ft) is simply : G = Vg g Where: g =gas formation volume factor ( Scf/ft 3) . As (g) changes with pressure , the gas in place also changes as pressure declines. Also, the gas pore volume (Vg) may also be changing owing to water influx into the reservoir. The simplest form of the equation for the volumetric or pore volume method that assumes a homogenous , isotropic reservoir is: G = 43,560 A h (1-Sw) gi (1) where : G = original gas in place ,scf 43,560 = conversion factor , sq.feet per acre. A = reservoir productive area , acres h = net thickness , feet = porosity,% Swi = average water saturation ,% and gi = initial gas formation volume factor, cu.ft/scf. The significance and derivation for gi is as follows: (g) is used to signify gas formation volume factor which is equal to the volume of gas at reservoir temperature and pressure divided by the volume of the same amount of gas at standard conditions of temperature and pressure. With this factor, we can relate gas reservoir volume to its surface volume using the real gas law: Z n R Tf gi = Vreservoir = Pi = Zi Tf Psc Vstandard Zsc n R Tsc Zsc Tsc P Psc gi = Psc . Tf Zi cu.ft /scf Pi Tsc Normally , with field units , Tsc = 520 oR , Psc = 14.7 psia , Zsc =1.0 g = 0.0283 Z T cu.ft/ scf P gi = 0.00504 Z T bbl/scf P During the early life of the reservoir (such as zero , one or two wells drilled) , only one or two estimates for each of the parameters in equation (1) are available . however, the geologist may have provide a structure map primarily based on geophysical data from which a rough net- pay isopach can be generated. In this case , it would be suitable to use equation (1) together with estimates of average values over the reservoir for all parameters except ( A) and (h) . Net hydrocarbon (gas) , volume for this situation is determined by numerical integration of the net pay isopach. This result in acre-ft is substituted into the previous equation (1) in place of the product (A h) or VB. The volumetric method makes use of subsurface and isopach maps based on the data from electric logs , cores , and drill –stem and production tests . - A subsurface contour map is a map showing lines connecting points of equal elevations on the top of a marker bed , and it therefore , represents a map showing geologic structure. -A net isopach map is a map showing lines connecting points of equal net formation thickness. Two equations are used to estimate reservoir bulk volume (VB): - Trapezoidal formula : VB = h ( An + - Pyramidal formula : 2 VB = h (( An + 3 An+1 ) An+1 ) + (An x An+1)1/2 Calculation of unit recovery from volumetric gas reservoirs: During the development period , VB is not known , so, it is better to place the reservoir calculations on a unit basis (acre-ft). Therefore , Connate water = 43,560 Sw ft 3/ ac-ft Reservoir gas volume = 43,560 (1- Sw) Reservoir pore volume = 43,560 ft 3/ ac-ft ft 3/ ac-ft Calculation of the initial gas in place (G) : G= 43,560 (1- Sw) gi Ga = 43,560 (1- Sw) ga scf/ ac-ft scf/ ac-ft The unit recovery = the difference between the initial gas in-place and the gas remaining at abandonment pressure. Therefore , the unit recovery ,U.R = 43,560 (1- Sw) (gi - ga ) And the recovery factor , R.F.= ( G-Ga ) X100 G scf/ ac-ft =100 (gi - ga ) gi % Calculation of unit recovery from water drive gas reservoirs: In many reservoirs under water drive , the pressure suffers an initial decline , after which water enters the reservoir at a rate equal the production and the pressure stabilizes. In this case , the stabilize pressure equals abandonment pressure.(Pb). Then , under abandonment conditions , water volume = 43,560 (1- Swi) ,and Reservoir gas volume = 43,560 Sgr Surface units of gas = 43,560 Sgr x ga The Unit recovery ,U.R= 43,560 [ (1-Swi) gi - Sgr x ga] scf/ac-ft The Recovery factor ,R.F = 100 [ (1-Swi) gi - Sgr x ga] % (1- Swi) gi Material balance equation of gas reservoirs If enough production –pressure history is available for a gas reservoir , the initial gas in –place (G) , the initial reservoir pressure (Pi) , and the gas reserves can be calculated without knowing A , h, , or Sw . This is accomplished by forming a mass or mole balance on the gas. That is : Moles produced = initial moles in – place =remaining moles in equation form: nf = ni - np np = ni - nf applying the gas law , PV = n Z R T gives : Psc Gp = Tsc Zst PI VI - Ti Zi Pf Vf Tf Zf where : vi = initial gas pore volume ,ft3 pf = final pressure vf = final gas volume after producing Gp (separator) of gas Vf We = = Vi - We + WpBw cubic feet of water encroached into the reservoir \Wp = cubic feet of water produced Bw = water formation volume factor , bbl/STB Psc Gp Tsc = PI VI - Pf (Vi - We+Wp Bw) Ti Zi Tf Zi Gp = Standard cubic feets (SCF) of produced gas measured at standard temperature and pressure. in case of volumetric reservoirs , there is no water influx ( We) , Wp generally negligible , then : Psc Gp = PI VI Tsc - Pf Vi Ti Zi (2) Tf Z where : Tf = formation temperature Vi = reservoir gas volume Pi = initial pressure and P = reservoir pressure after producing scf (Gp ), the reservoir gas volume can be put in units of scf by use of Bgi , that is Vi = G Bgi combining equation (2) P (3) and (3) and solving for P gives Z = Pi - Tf Psc x Gp Tsc Bgi G from which it is obvious that a plot of P/Z versus Gp will result a straight line of slope (m) equals m = Tf Psc and intercept at Gp = zero. of Pi/Zi Tsc Bgi G thus , both (G) and (Pi) can be obtained graphically. the slope m = Gp (P/Z) Equation (2) may be written in terms of gas volume factors gi and gf by solving it for (Gp) : Psc Gp = PI V I Tsc Gp - Pf (Vi - We+Wp Bw) T Zi = Pi Tsc T Vi - gi = Psc Zi T gf scf/ft3 = Pf Tsc Psc Zf T Gp = gi Vi = G gi Vi - gf ( Psc Psc Zf T , scf/ft3 Pi Tsc Zf Pf Tsc ( Vi – W e +W p Bw) Zi T Psc x Tsc Vi –W e +W p w) Therefore , Gp = G - gf ( G – W e +W p w) Dividing by Gp = gf Gp G = g g G Replace Gp - G gf : gf gf gi We + Wp + gi 1 - 1 gf by ft3/scf = G ( w gf + We - W p w gi - instead of scf/ft3 , gi ) + W e -W p w therefore , production term = expansion term +Water influx term - water production term For volumetric reservoir: production volume = expansion volume Gp g = G ( gf - gi ) P/Z Pi Zi Water drive Volumetric Reservoir Pa Za 0 Reserve G P/Z Vs Gp graph for gas reservoir Gp The steady state radial flow equation Well bore damage and improvement effects: Ways to quantify damage or improvement : 1- skin factor (S): The pressure drop at a damaged or improved well differs from that at undamaged well by the additive amount : Ps = 141.2 q . S k h From Darcy, s equation : Q = 0.00708 k h (pe -pwf ) o ln re/rw pe = pwf + 141.2 q o ln re/rw kh pat r = pwf + 141.2 q o ln r / rw kh if there is a skin ( altered zone) around the well : Pe P Ka P1 K Pwf Pwf1 R1 rw Ra Re Re P1 – Pwf = 141.2 q o ln re/rw K.h P1 - Pwf1 = 141.2 q o ln re/rw Ka . h Subtract equation (2) from equation (1) Pwf – Pwf1 = Pwf- Pwf1 141.2 q o ln ra/rw 1 h Ka (1) (2) 1 K = pressure drop due to skin . Multiply by K and divide by K Gas reservoirs For these gas reservoirs, the reservoir fluid is always a single phase gas for the life of the reservoir because in these systems the formation temperature is greater than the cricondentherm. Phase diagram : These diagrams are drawn from PVT date: P single phase liquid 2 Retrograde condensation C Bubble point line 1 Two phase region 1 Cricondentherm Single phase (gas) Dew point line Temperature Point C, the critical point: the point at which the liquid and gas are in equilibrium or has the same properties. Point 1, the cricondentherm : the maximum temperature where the two phase or liquid may exist. Point 2 the cricondenbar : the maximum pressure where the two phase or gas may exist. Point 3 retrograde condensation: at which , although the pressure is reduced , the gas is condensed to the liquid state. How you can confirm the type of reservoir: According to the phase diagram: as shown in the attached figure: C B A P C1 C B1 2 C2 A3 B2 A2 A1 Separator Temperature Reservoir A is a gas reservoir A –A1 A – A2 A-A3 Wet gas reservoir Dry gas reservoir at the surface Reservoir B B –B1 Gas reservoir B- B2 Gas Condensate reservoir Reservoir C C–C1 Under saturated reservoir Contains no free gas C- C2 Saturated reservoir contains both oil and free gas Note that : -the reservoir temperature is constant -reservoir pressure declines with production Determination of the original gas in place(OGIP): As with oil reservoirs, there are two main ways to estimate the original gas in place : the volumetric method and Material balance method. 1-Volumetric method : this approach is used early in the life of the reservoir ( for instance , before 5% of the reserves have been produced). The standard cubic feet of gas in a reservoir which has a gas pore volume of Vg ( cu.ft) is simply : G = Vg g Where: g =gas formation volume factor ( Scf/ft 3) . As (g) changes with pressure , the gas in place also changes as pressure declines. Also, the gas pore volume (Vg) may also be changing owing to water influx into the reservoir. The simplest form of the equation for the volumetric or pore volume method that assumes a homogenous , isotropic reservoir is: G = 43,560 A h (1-Sw) gi where : G = original gas in place ,scf 43,560 = conversion factor , sq.feet per acre. A = reservoir productive area , acres h = net thickness , feet = porosity,% Swi = average water saturation ,% and gi = initial gas formation volume factor, cu.ft/scf. (1) The significance and derivation for gi is as follows: (g) is used to signify gas formation volume factor which is equal to the volume of gas at reservoir temperature and pressure divided by the volume of the same amount of gas at standard conditions of temperature and pressure. With this factor, we can relate gas reservoir volume to its surface volume using the real gas law: Z n R Tf gi = Vreservoir = Pi = Zi Tf Psc Vstandard Zsc n R Tsc Zsc Tsc P Psc gi = Psc . Tf Zi cu.ft /scf Pi Tsc Normally , with field units , Tsc = 520 oR , Psc = 14.7 psia , Zsc =1.0 g = 0.0283 Z T cu.ft/ scf P gi = 0.00504 Z T bbl/scf P During the early life of the reservoir (such as zero , one or two wells drilled) , only one or two estimates for each of the parameters in equation (1) are available . however, the geologist may have provide a structure map primarily based on geophysical data from which a rough net- pay isopach can be generated. In this case , it would be suitable to use equation (1) together with estimates of average values over the reservoir for all parameters except ( A) and (h) . Net hydrocarbon (gas) , volume for this situation is determined by numerical integration of the net pay isopach. This result in acre-ft is substituted into the previous equation (1) in place of the product (A h) or VB. The volumetric method makes use of subsurface and isopach maps based on the data from electric logs , cores , and drill –stem and production tests . - A subsurface contour map is a map showing lines connecting points of equal elevations on the top of a marker bed , and it therefore , represents a map showing geologic structure. -A net isopach map is a map showing lines connecting points of equal net formation thickness. Two equations are used to estimate reservoir bulk volume (VB): - Trapezoidal formula : VB = h ( An + - Pyramidal formula : 2 VB = h (( An + 3 An+1 ) An+1 ) + (An x An+1)1/2 Calculation of unit recovery from volumetric gas reservoirs: During the development period , VB is not known , so, it is better to place the reservoir calculations on a unit basis (acre-ft). Therefore , Connate water = 43,560 Sw ft 3/ ac-ft Reservoir gas volume = 43,560 (1- Sw) Reservoir pore volume = 43,560 ft 3/ ac-ft ft 3/ ac-ft Calculation of the initial gas in place (G) : G= 43,560 (1- Sw) gi Ga = 43,560 (1- Sw) ga scf/ ac-ft scf/ ac-ft The unit recovery = the difference between the initial gas in-place and the gas remaining at abandonment pressure. Therefore , the unit recovery ,U.R = 43,560 (1- Sw) (gi - ga ) And the recovery factor , R.F.= ( G-Ga ) X100 G scf/ ac-ft =100 (gi - ga ) gi % Calculation of unit recovery from water drive gas reservoirs: In many reservoirs under water drive , the pressure suffers an initial decline , after which water enters the reservoir at a rate equal the production and the pressure stabilizes. In this case , the stabilize pressure equals abandonment pressure.(Pb). Then , under abandonment conditions , water volume = 43,560 (1- Swi) ,and Reservoir gas volume = 43,560 Sgr Surface units of gas = 43,560 Sgr x ga The Unit recovery ,U.R= 43,560 [ (1-Swi) gi - Sgr x ga] scf/ac-ft The Recovery factor ,R.F = 100 [ (1-Swi) gi - Sgr x ga] (1- Swi) gi % 3 PETROLEUM RESERVOIR ENGINEERING Lectures Selected-Compiled –Edited By Dr.Eng.Mostafa Mahmoud A. Kinawy CHAPTER II FLUIDS FLOW IN POROUS MEDIA -Steady State Flow -Pseudo- Steady State Flow -Unsteady State Flow Dimensionless Quantities: Td = 0.0002637 K t Ct r w2 ct = total system compressibility ,psi-1 ct =cf + co so +cw sw +cf (for single phase oil flow) tdA = (rw2) , A = total drainage area A rD = r rw PD = P 141.2 q Kh with these dimensionless quantities , the diffusivity equation becomes : 2 PD + 1 PD = PD 2 rD rD rD tD 1-dimensionless quantities provide a convenient way to summarize the increasing number of solutions being developed to depict well or reservoir pressure behavior over a broad range of time , reservoir properties , boundary , geometry conditions. 2- an unfortunate consequence of the generalized dimensionless solution approach is that the dimensionless parameters do not provide the engineer with the physical feel available when normal dimensional parameters are used. 3- Note that 141.2 q = 0.8686 m , where m is the slope of the semi – kh log plot , = 162.6 q kh Solutions to the diffusivity equation: 1- for infinite reservoirs: 1-1-Constant Terminal Rate Solution: initial and boundary conditions: initial conditions : p = pi at t 0 every where boundary conditions q = constant for t 0 ( or r p ) = constant r p pi as r for all t the solution is : pD = fn (rD , tD) note that : = -1/2 Ei ( - rd2 ) 4td pD = pi – p 141.2 q kh 1/2 [ ln ( td ) rD2 + 0.80907] for td 100 This is the "exponential integral "solution , also known as the line source or" Theis" solution , where the exponential integral is defined as : Ei (-x) = - x e –u du u and its values may be taken from tables , graphs or approximated: Ei (-x) ln (x) + 0.5772 ( for x 0.0025) for rd 20 , , td 0.5 rD2 or for td 25 , the exponential integral solution rD2 is sufficiently accurate. At the operating well , rD = 1.0 and so , td = td rD2 1-2 Constant Terminal Pressure Solution: The initial and boundary conditions are : initial conditions : p = pi at t 0 every where Boundary conditions : p = constant at r = rw for t 0 p pi as r for all t A solution was found by Van Everdingen & Hurst , to give the dimensionless cumulative production ( water influx ) Q(t) as a function of dimensionless time tD . A set of curves relating the dimensionless rate qD against dimensionless time tD for a constant pressure test , where : q D = 141.2 k h (pi –pwf) q the following figures , after Craft& Hawkins , are useful to show the difference between the constant terminal rate and constant terminal pressure cases. t=0 time t=1 t=10 t= 100 P=10 P =100 The steady state radial flow equation Well bore damage and improvement effects: Ways to quantify damage or improvement : 2- skin factor (S): The pressure drop at a damaged or improved well differs from that at undamaged well by the additive amount : Ps = 141.2 q . S k h From Darcy, s equation : Q = 0.00708 k h (pe -pwf ) o ln re/rw pe = pwf + 141.2 q o ln re/rw kh pat r = pwf + 141.2 q o ln r / rw kh if there is a skin ( altered zone) around the well : Pe P Ka P1 K Pwf Pwf1 R1 rw Ra Re Re 141.2 q o ln re/rw K.h P1 – Pwf = (1) P1 - Pwf1 = 141.2 q o ln re/rw Ka . h Subtract equation (2) from equation (1) 141.2 q o ln ra/rw 1 h Ka Pwf – Pwf1 = Pwf- Pwf1 (2) 1 K = pressure drop due to skin . Multiply by K and divide by K Pwf – Pwf1 = PS PS 141.2 q o K h K Ka 1 ln ra/rw skin factor(s) = 141.2 q o K h . S II) if the skin is viewed as a zone of finite thickness with permeability Ka: s = III) K - 1 ln ra/rw Ka An apparent well bore radius ,rwa may be defined so that the correct pressure drop results: rwa = rw e-s notice that : ln re/rwa = ln relrw + s since re/rwa =re/rw x rwlrwa for damaged zone: ka<k which means +ive skin factor for a simulated zone , ka>k which means –ive skin factor the value of skin factor ,s is an indication about how much is the severity of damage or improvement , ra can be obtained from resistively logs. s can vary from about -5 to + for too badly well, s is +ive for damage , and negative for improvement> hydraulically fractured wells often show values of s from -3 to -5. introducing the skin factor ,s: pd = pi-pwf 141.2 qo kh = [ 1/2 {ln td +0.80907} +s ] Build –Up Tests Pressure Introduction: Pressure build-up testing is probably the most familiar transient well testing technique. the test requires shutting –in a producing well . it should be noticed that: (1) the well is produced at a constant rate ,q , either from the start-up , or for a long enough period to establish a pressure distribution before shutting-in ; (2) the pressure is measured immediately before shut-in, and is recorded as a function of time during shut-in period. the resulting pressure buildup curve is analyzed for reservoir properties , well bore conditions as follows: figure (i) schematically shows rate and pressure behavior for an ideal pressure build-up test. in that figure, tp is the production time and t is the running shut-in time. flow rate q idealId case 0 actualIi case ( history case) tp t shut –in time Pressure Pwf ( t =0) 0 tp t figure (i) represents the idealized pressure history for a pressure build-up test as in transient well tests , knowledge of surface and subsurface mechanical conditions is important in build-up test data interpretation . therefore , it is recommended that tubing and casing size , well depth , packer locations , etc , be determined before data interpretation starts . stabilizing the well at a constant rate before testing is important for reliable test. to determine the degree and adequacy of the stabilization , one way is to check the length of the pre-shut-in constant rate period against the time required for stabilization, as follows: ct A ts = 380 k where : ts =the minimum shut-in time ,hrs for a well in the center of a symmetrical drainage area(A) , and tR = 946 ct A k where : tR = readjustment time , the time required for a short lived transient to die out, hrs, for a single well in the center of a constant pressure square. Basic Analysis Method: (Horner ,s method ) it is applicable to an infinite , homogeneous , one –well reservoir containing a fluid of small and constant compressibility , and so, it applies well to newly completed well in oil reservoirs above bubble point pressure(Pb). derivation: the exponential integtal solution given before equation 12 gives the pressure drop for a well flowing for a time tp dimensionless) form: Pi-Pwf = -q 4k h -q 4k h Ei ( - cf rw2 ) 4 k tp ln ( ct rw2 ) 4 k tp where : = Euler,s constant = 1.78 as follows in dimensional (not if now , the well is shut-in for a time t , after having produced for a time t p , then pressure drop at t can be obtained by the principle of superposition as follows: Pi-Pws = (pressure drop caused by rate q for time (tp+t) ) + (pressure drop caused by rate change -q for time t) or: Pi - Pws = -q ln ( ct rw2 ) + q 4k h 4 k (tp+t) ln ( ct rw2) 4k h 4 k (t) and Pws = - q pi ln ( tp + t) t 4k h Where : Pws = the well pressure after shut-in Pi = the well flowing pressure before shut-in In the usual practical oilfield units , the last equation became: Pws = Pi - 162.6 q ln ( tp + t) t kh Pws = Pi - m log ( tp + t) t This is Horner,s equation , applicable for ideal Build-Up test. this equation indicates that a plot of observed shut-in bottom hole pressure , Pws Vs log [tp + t] should have a straight –line portion with slope = -m t that can be used to estimate reservoir permeability, k = 162.6 q m h As indicated by the Horner , s equation , the straight line portion of the Horner plot , see Figure(ii) , may be extrapolated to tp + t = 1.0 , log tp + t = 0 t t , the equivalent to infinite shut-in time , to obtain an estimate of (Pi ) . That is an accurate estimate only for short production periods. It is also noticed that , as a result of using the principle of superposition , the skin factor does not appear in Horner , s equation. Pi Pwf, Psi slope =slope =-m = slope per cycle. P1hr (tp +t)/ t However, the skin factor may be estimated from the build-up test data plus the flowing pressure immediately before the build-up test (Pwf) , since the skin does affect this flowing pressure. S = 1.153 P1hr - Pwf t = zero - log ( K ) + 3.227 S 2 m ct rw The first part of the build-up plot is usually non-linear resulting from the combined effect of the skin factor and well bore storage. the latter is due to the normal practice of closing in the well at surface rather than down hole , which results in the flow rate not being reduced to zero instantaneously . Use of Horner,s method for analysis of pressure build-up tests during the infinite acting period: During the infinite acting period of time, after : - Well bore storage effects have diminished , and - tD > 100 ( which occurs after few minutes for most unfractured systems Pws = Pi - m log (tp +t)/ t slope = -m and intercept = Pi which is a straight line semi-log plot , From this plot , we make the following estimates: 1- K = 162.6 q m.d (undamaged zone permeability) m h (tp +t)/ t = 1 ( this is accurate only for 2- Pi : by extrapolation to short production periods , otherwise, extrapolated values give P*= false pressure. 3- S = 1.153 P1hr - Pwf t = zero - log ( K ) + 3.227 S 2 m ct rw This equation is good as long as tp >> 1 hr , otherwise , for example in case of drill stem tests (DST). S = 1.153 P1hr - Pwf t = zero + log (tp+1 ) - log ( K ) + 3.227 S 2 m tp ct rw The value of tp may be approximated from : tp = 24 x Vp q Where : Vp = cumulative volume produced since the last pressure equalization, or the Np 4- The average permeability , Kavg = q ln (re/rw) 7.08 h (Pe=Pwf) Here , the Pwf is the stabilized well pressure corresponding to the stabilized flow rate (q). 5-Productivity ratio (PR) = Kaverage = 0.868 m Kundamaged ln (re/rw) Pe -Pwf = 2 m log re/rw Pe -Pwf 6- Damage factor = 1 - PR Also , S = Ka Kund - ln ( ra/rw) Ka and PR = ln re/rw ln (re/rw) +S If the productivity ratio ,PR 1 this means stimulation , sinc3e Kavg > Kun. If PR 1 , This means damage , also , Damage factor = 1- PR , Positive + Damage : may be Negative - Stimulation Also , S may be Positive + Kun > Kavg Negative - Kun < Kavg Damage Stimulation Remarks: In all pressure build-up test analysis , the log-log data plot ( log (Pws- Pwf) Vs log t ) should be made . when well bore storage dominates , the plot will have a unit slop straight line. This helps choosing the data for the straight line semi-log plot ; use the 1-1.5 cycles in time rule of thumb or the following equation to estimate the line for the beginning of the straight line semi-log plot: t > 170000 C e0.145 (Kh / ) Actual build up tests , definition of the infinite –acting region of the test , (middle times region , or infinite –acting period or semi-log straight linen region) Pws E Early Time Region ETR True formation Well interference MiddleTime Region MT R Skin or well storage Log to (tp +t)/ t Late y Time Region LTR We can divide a build-up curve into three regions , as shown in the above figure. 1- An early time region : during which a pressure transient is moving through the formation nearest the well bore 2- A middle –time region ,: during which , the pressure transient has moved away from the well bore into the bulk formation and 3- A late –time region : in which the radius of investigation has reached the well,s drainage boundaries. well storage: -well bore storage affects short-time transient pressure behavior , also called after f low ,after production , after injection , and well bore unloading or loading. if it is not considered in transient test design , analysis , wrong conclusions may be made. Well –bore storage constant(coefficient or factor) "C": C= V or V = C P P where : V = change in volume of fluid in the well bore , bbls at well bore conditions. P = change in bottom hole pressure , psi C = well –bore storage coefficient , bbl/psi for a well –bore with changing liquid level: iC = Vu ( = g) 144 gc Vu ( if g =gc) 144 where : = the well –bore volume per unit length , bbl/ft Vu = density , g = acceleration of gravity , gc= conversion factor ii- for a well bore completely full of a single phase fluid: C = Vw . c where : Vw = total well –bore volume , bbls c = compressibility of the well bore fluid at well bore conditions psi-1 Dimensionless well bore storage coefficient (CD): By definition: CD = 5.6146 C 2 ct h rw2 where : , ct , h are relevant to the reservoir. Surface flow rate (q) and sand face flow rate ( qsf ) : Well bore storage causes qsf to change more slowly than q . The following figure shows ( qsf/q ) as a function of time when the well is opened and surface rate changed from 0 to q at time = zero. Note that : When C = 1.0 zero , qsf /q = 1.0 at time = 0.0 C1 C2 C3 qsf/q 0 0 td Effect of well bore storage on sand –face flow rate C3 > C2 > C1 The sand face flow rate may be calculated from: Qsf = q + 24 C . = q [ dP dt 1-CD d d td ( td , CD ……)] where: PD is a special function which accounts for well bore storage , and is shown in the given figure: Notice that , by definition: C . P = V or C dP dt , the additional flow rate , The slope of PD Vs = dV dt t , in hours and q in bbls per day. tD on log –log graph is 1.0 , during well bore storage domination . The dimensional values P and t are proportional to PD and tD The location of the log-log unit slope line does not give any information about the reservoir , but can be used to estimate the apparent well bore storage coefficient from : C = q 24 where : t t P , P are values read for a point on the log-log unit slope line . In the above figure , C D = tD at PD = 1.0 The log –log plot of data is valuable , when early pressure data is available to recognize well bore storage effect. Make this plot is a part of the transient test analysis. It helps delineate important periods of the analysis. 1- slope = 1.0 ------------- well bore storage dominant 2- after the end of that period by 1 to 1.5 cycles in time , the standard semi-log plot starts. Also , time for the beginning of standard semi-log plot technique may be estimated from: Td > (60+3.5 S) CD or t > (200,000+12000S) C ( k h/ ) approximately for drawdown and injection tests , and from: tD > 50 CD e 0.14s or t > 170000 C e 0.14S ( k h / ) for pressure build-up and fall off tests. 3- In between , type –curve matching techniques apply. approximately The Pressure Build-Up Procedure: 1-Estimate the probable shut –in time for accurate data . Minimum shut-in time , ts 190 co re2 ko where : ts = ko required shut-in time , hrs = permeability to oil ,md o = oil viscosity ,cp = porosity ,% co = oil compressibility, psi-1 2-put the well on a particular production rate and hold until the well is stabilized with a pressure bomb in place at the mid point of the perforation . 3-shut the well in and measure the build-up in pressure vs time 4-plot pressure versus log 10 { tp + t } , tp is the production time of t the well prior to shut-in , and t is the time since the well was shut-in. 5-determine the slope per cycle (m) from the straight line portion of the curve. 6-determine the flow capacity , k h = 162.6 q m 7-determine the static reservoir pressure , the skin effect , flow efficiency , the productivity index , and damage ratio. the flow efficiency , FE : also called " the condition ratio , productivity ratio ,PR , or completion factor. FE = Jactual = P - Pwf - Ps J ideal Where : J = productivity index. P - Pwf = K actual k (avg) Draw Down Testing Often the first significant event at a production well is the initial production period that results in a pressure drawdown at the formation face. Thus, it seems logical to investigate what can be learned about the well and reservoir from pressure drawdown data. Here , we will consider the drawdown test analysis for infinite acting and pseudo-steady state periods, and we will deal only with constant rate drawdown testing. Drawdown testing may provide information about: 1-formation permeability 2-skin factor and 3-the reservoir volume communicating with the well. The attached figure schematically illustrates the production and pressurehistory during a drawdown test. Producing Rate,q Shut-in 0 0 time , t Bottom hole flowing pressure, Pws Pwf Pws 0 time , t Ideally , the well is shut-in until it reaches static pressure of the reservoir before the test. That requirement is met in new reservoirs , it is less met in old reservoirs. The drawdown test is run by producing the well at a constant flow rate wile continuously recording bottom –hole pressure. While most reservoir information obtained from a drawdown test also can be obtained from a pressure build-up test , there is an economic advantage to drawdown test , that the well is produced during the test. Advantages of D/D Test: 1-the well is producing during the test 2-the possibility for estimating reservoir volume Disadvantages of D/D Test: 1. the difficulty of maintaining a constant production rate. Pressure drawdown analysis in infinite –acting reservoirs: The pressure at a well producing at a constant rate in an infinite acting reservoir is given by: pi - pwf = 141.2 q [ Pd (td …….)+ S]………..(1) k h If the reservoir is at Pi initially , the dimensionless pressure at the well ( rD=1.0) is: PD = 1/2 [ ln (td) +0.80907] ……………………(2) When (td/rD2) > 100 and after well bore storage effects have diminished: td = t , hours 0.002637 k t ……….(3) Cf rw2 By combining equation (1) and (3) and rearranged , we get the familiar form of the pressure drawdown equation: Pwf = Pi – 162.6 q [ log t + log ( k ) - 3.2275 +0.86858 S] Cf rw2 k h Equation (4) describes a straight line relationship between ( P wf ) and log(t). By grouping the intercept and slope terms , together , it may be written as: Pwf = m log t + P1hr m = Pwf -P1hr log t Theoretically , a plot of Pwf vs log t , on a semi-log paper should be a straight line with a slope , m and intercept , P 1hr as shown in the following figure: pwf , psi Deviation from straight line caused by damage and well bore storage effects slope = m = 162.6 q kh P1hr 0.1 1 10 time , t ,hours Figure (2) Pwf, psi ET R 0 ETR = Early M T R Time , t , hours time region , affected by : 1- damage zone 2- well storage LTR = Late time region , affected by: 1- No flow boundary 2- well interference L T R A plot of Pwf Vs log t yields a straight line portion appears after well bore damage and storage effects have diminished. The slope of the straight line in figure (2) m = -162.6 q k h The intercept at log t = 0 , which occurs at t = 1 hr is also determined : p 1 hr = pi + m [ log ( k ) - 3.2275 + 0.86859 S] Cf rw2 Formation permeability ; k = 162.6 q k h S =1.1513 [ P1hr –Pi - log ( The skin factor, k ) - 3.2275 ] Cf rw2 m Example: Estimate the oil permeability and skin factor from the drawdown data: Known reservoir data are: h =130 ft , = 20% , rw = 0.25 ft , Pi = 1154 psi , qo = 348 stb/D , m = -22 psi/cycle, o = 1.14 bbl/stb , o= 3.93 cp , P1hr = 954 psi , Cf = 8.74 x 10 -6 psi-1 Solution ko = 162.6 x 348 x 3.93 x 1.14 = 89 md (-22) x( 130) S= 1.1513 [ (954-1154) - log ( 89 ) + 3.2275] 0.2 X 3.93X8.74X10 -6 X(0.25)2 -22 S = 4.6 Two graphs of pressure drawdown data are required: 1. log-log data plot [ log (Pi-Pwf) Vs log (t) ] to estimate when well bore storage effects are no longer important. Well bore storage C = q p 24 t Log P slope /cycle log t ,hrs We can estimate when the semi-log straight line begin: t > (200,000 + 12000 S)C ( k h / ) Reservoir limit test: A drawdown test run specifically to determine the reservoir volume communicating with the well is called a reservoir limit test. The dimensionless pressure during pseudo-steady state flow is a linear function of dimensionless time. The following equation is obtained: Pwf = m* t + Pint…………………..(5) where : m* = -0.23395 q ……………..(6) Cf h A Flowing m Pressure Pwf,psi P1hr 2060 2040 2020 Flow time , hours 2000 0.1 1 10 102 103 Pint = Pi - 70.6 q [ ln( A ) +ln ( 2.2458 ) + 2S] …..(7) and rw2 k h Equation (5) Pwf = m* t CA + Pint of bottom hole flowing pressure (Pwf) indicates that a Cartesian plot Vs time (t) should be a straight line during pseudo-steady state flow , with slope m* given by equation (6). m* = -0.23395 q , and intercept Pint given by equation (7). Ct h A Pwf m* Pint 0 5 10 15 20 Flowing 25 30 35 40 time ,hours Note that : cf = dv . 1 Vp dp q = dv /dt cf = q dt o Vp . dp Vp = q o cf (dp/dt) The slope may be used to estimate the connected reservoir drainage volume: h A = - 0.23395 q where the volume in cubic feet. ct x m* If ( h) is known , the drainage area may be estimated. Another technique has been proposed for analyzing pseudo-steady state data , but this one appears to be the simplest . If pressure data are available during both the infinite acting period , and PSS period , it is possible to estimate the drainage shape for the test well. The semi-log plot is used to estimate (m) and (P 1hr) ; the Cartesian plot is used to get (m*) and (Pint). The system shape factor is estimated from: CA = 5.456 m exp [ 2.303 (P1hr-Pint) ] m* m By knowing the shape factor , use table (C-1) , to determine the reservoir configuration with the shape factor closest to that calculated. This process may by refined by computing ( tda)pss = 0.1833 m* tpss m and using the exact value for (tDA ) , from the table (C-1) , the time tpss is when the Cartesian straight line starts. Drill Stem Testing (DST) Different methods are used for evaluating formation productivity includes core analysis , well logging and drill stem testing. Despite the tremendous value of core analysis and logging , some shadow of doubt always remains concerning the potential productivity of an exploratory well , and this doubt is not dispelled until a sizable sample of oil has been delivered to the surface. This drawback is not inherent in drill stem testing. The decision to run a drill stem test on a zone is often based on shows of oil in the cuttings which , in the opinion of the geologist or engineer in charge , deserve detailed investigation . This may happen many times in the course of drilling a wildcat , with as many as 20 or 30 tests being conducted on a single well. Although the cost of such detailed testing is quite high , it is much better to test and be sure , rather than miss a productive zone. General procedure: A drill stem test is a temporary completion whereby the desired section of the open hole is isolated , relieved of the mud column pressure , and allowed to produce through the drill pipe (drill stem). The basic test tool assembly consists of: 1- a rubber packing element or packer which can be expanded against the hole to segregate the annular sections above and below the element. 2- A tester valve to control flow into the drill pipe , that is to exclude mud during entry into the hole and to allow formation fluids to enter during the test , and 3- By –pass valve to allow pressure equalization across the packer after completion of the flow test. 4- Anchor : This is merely the extension below the tool which supports the weight applied to set the packer. 5- Pressure recorders: These furnish a complete record of all events which may occur during a particular test. This record is in the form of a graph of pressure versus time. Two pressure recorders are usually desirable and should be located so that one will measure the pressure inside, and the other the pressure outside the anchor. These two measurements allow accurate determination of weather or not the perforations have become plugged during the test. 6- Safety Joints: These merely afford a means of unscrewing the drill string at a point convenient for fishing operations , should the packers become stuck. Figure (1) illustrates the procedure for testing the bottom section of a hole. While going in the hole, the packer collapsed , allowing the displace mud to rise as shown by the arrows. After the pipe reaches bottom and the necessary surface preparations have made , the packer is set (compressed and expanded) ; this isolates the lower zone from the rest of the open hole. The compressive load is furnished by a slacking off the desired amount of drill string weight which is transferred to the anchor pipe below the packer. The tester valve is then opened and thus the isolated section is exposed to the low pressure inside the empty or nearly empty drill pipe . Formation fluids can then inter the pipe as shown in the second picture. At the end of the test , the tester valve is closed , trapping any fluid above it and the by-pass valve is opened to equalize the pressure across the packer. Finally the setting weight is taken off and the packer is pulled free. The pipe is then pulled from the hole until the fluid containing section reaches surface. As each successive stand is then broken (unscrewed) , its fluid content may be examined. Frequently such stand –by-stand sampling is neither necessary nor desirable , and the test is reversed (circulated opposite to the normal direction ) as shown. This reversal is performed by closing the blow out preventers and pumping mud down the annulus, the mud then enters the drill pipe through the reversing ports, thereby displacing any formations fluid in the pipe. The recovered fluid samples as they are discharged at the surface. Although the above is a very common type of test , there are many other variations of procedure as indicated in Figure (2). These are: 1-Straddle Packer Test: is necessary when isolation from formations both above and below the test zone is necessary. Such a situation commonly arises when evidence (electric logs, radioactivity logs, detailed sample analysis etc.) indicates that a zone previously passed by has productive possibilities. Straddle testing is less desirable than conventional testing , from both a cost and an operational hazard stand point. Two packers are more apt to become stuck than one , since any material which sloughs or caves from the test zone may accumulate between the packers. 2-The Cone Packer or Rat- Hole Method: Is used when the test section is smaller in diameter than the hole above. This situation commonly occurs as a consequence of coring operations in which the corehead used was smaller than the regular bits. The coneshaped packer is compressed against the shoulder , forming the necessary seal. 3-Wall Over Cone Packer Test: if the formation opposite the shoulder is soft, an additional conventional wall packer placed above the cone packer may be necessary to provide the desired seal. 7- Testing Through Perforations in the Casing: Zones behind casing may be tested through perforations by the same basic procedures except that the packer used has slips which engage or grab the casing wall. These are commonly called hook-wall packers. Since the slips support the compressive load used to expand the packer element , no anchor pipe is required, and the packer may be reset several times at different depths if necessary. Testing inside casing is widely used in soft rock areas where open hole testing is particularly hazardous. Factors affecting drill stem testing: 1-Condition of the hole: the close tolerance between the hole and the tool assembly requires a full gage, clean , well bore if the tool is to reach bottom in an undamaged , unplugged condition. The cake and cavings shoved ahead of the packer may plug the perforations and/or choke when the valve is opened. It is common practice to circulate for some time prior to testing, so that all cuttings are removed from the hole. The drilling mud should be conditioned to the desired density and viscosity before the test is started. 2-Pressure surges: the drill stem test conditions represent a severe case of pressure surge, because the lower end of the pipe is closed, necessitating displacement of the total drill pipe volume. Special consideration should be givn to pipe running and pulling speeds to avoid undue bottom hole pressure variations. 3-operating conditions: (a) The length and location of test section govern the amount of tail pipe required and the choice of a conventional or straddle test. The testing of short sections is more conclusive , and is generally preferable. (b) The packer seat location , while of no particular importance for tests run inside casing, is critical for a successful open hole test. The seat should be placed in a true gage section of the hole opposite as dense and consolidated formation as possible. As a general rule, electric or caliper logs are not taken prior to open hole testing and the main guides to packer seat location are the drilling time and sample logs. (c)Size and number of packers: the pressure differential which a packer can stand depends on the amount it must expand to furnish the desired seal. The service companies which supply these tools have a standard range of sizes for various hole diameters. The ratio of hole diameter to unexpanded packer diameter is kept as low as possible, and commonly ranges from 1.1 to 1.2. In deep , open hole tests , two closely spaced packers are often run as a precautionary measure. This often eliminates test failure due to by-passing in fractured zones; also, if one packer fails , the other may carry the load. The cost of the second packing element is quite nominal compared to the cost of misrun. (d)Choke sizes: the size of the bottom hole and surface orifices selected depends on the anticipated test conditions. The bottom choke is of primary importance and is used to govern the flow rate. The top choke is used primarily as a safety measure and should be considerably larger than the bottom choke in order to minimize surface pressure in case a flowing test is obtained. (e) Use of cushions: this refers to the practice of placing a certain length or head of liquid inside the drill pipe , rather than running it dry. This is commonly done for two reasons: 1-to reduce the collapse (external )pressure on the drill pipe in deep holes and/or 2- to reduce the pressure drop on the formation and across the packer (s) when the tool is first opened. (f) length of test: this is difficult if not impossible to predict until after the test has commenced and some observations are available. While this is a safe and necessary practice in many cases, it should not be carried to the extreme that back pressure becomes large enough to prevent formation fluid flow. The Material Balance Equation as a Straight Line (The Havlena and Odeh Method ) Normally , when using the material balance equation, an engineer considers each pressure and the corresponding production data as being separate points from other pressure values. From each separate point, a calculation for a dependent variable is made . the results of the calculations are sometimes averaged. The Havlena –Odeh method uses all the data points , with the further requirement that these points must yield solutions to the material balance equation that behave linearly to obtain values of the independent variable. The straight-line method begins with the material balance equation written as: Np [Bt+ (Rp-Rsi)Bg ] + Wp Bw – Wi – GiBgi = N (Bt – Bti ) + (1+m) Bti { cw Sw + cf } P+ mBti (Bg-Bgi) 1-Swi + We Bgi Havlena –Odeh defined the following terms and rewrote the equation as follows: F = Np [Bt+ (Rp-Rsi)Bg ] + Wp Bw – Wi – Gi Bgi Eo = Bt - Bti Ef,w = { cw Swi + cf } P 1-Swi Eg = Bg –Bgi And finally : F = N Eo +N (1+m) Bti Ef,w + { N m Bti } Eg + W e -----(1) Bgi In the last equation , F represents the net production from the reservoir. Eo , Ef,w , and Eg represent the expansion of oil, formation and water and gas respectively. Havlena and Odeh examined several cases of varying reservoir types with this equation and found that the equation can be rearranged into the form of a straight line. For instance, consider the case of no original gas cap , no water influx , and negligible formation and water compressibilities. The equation can be reduced to F=N Eo ----------------(2) This would suggest that a plot of F as the the X Y coordinate and Eo as coordinate would yield a straight line with slope N and intercept equal to zero. For the case of a saturated reservoir with an initial gas cap and neglecting the compressibility term, Ef,w, equation (1) becomes F = NEo + mNBti Eg + We Bgi If N is factored out of the first two terms on the right –hand side and both sides of the equation are divided by the expression remaining after factoring, we get : F Eo+ m Bti = Eg N + We Eo + m Bti Eg Bgi Bgi F/[Eo+(mBti/Bgi) Eg] MM STB Slope =1.0 N We/[Eo+(mBti/Bgi) Eg] , MMSTB

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