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Well Performance Analysis Adam Muhammad approved by Mr. Akhmad Sofyan, M.T. Day – 4 (Part 2) eq Day - 5 Reference 1. Brown, Kermit E., Beggs, D. H. 1977.“The Technology of Artificial Lift Methods”. Vol 1. Tulsa, Oklahoma. PennWell Publishing Company. Outline – Day 4 (Part 2) Eq Day 5 ❖ Limited Correlations ❖Development and Utilization of The Best Correlations in Predicting Pressure Loss 1 Limited Correlations Poettmann and Carpenter Method • Assumptions : 1. The correlations is applicable to the pipe sizes used in field study (2, 2 ½ and 3 in.) and extension of the correlation to other size should be done with caution. 2. The total output flow rate was used to calculate the density at any one pressure point 3. Flow pattern were ignored 4. The effect of viscosity were assumed to be negligible 5. The acceleration term of the general energy equation was considered to be negligible. This will be true for most cases but the term is significant at very high velocities and was included in the friction factor 6. Although the energy loss factor is assumed to be an average value over the length of vertical tube it was actually found to vary from the bottom to the top of the tubing string Poettmann and Carpenter Method Cont’d • The Poettmann and Carpenter correlation is still used quite extensively in the field. It can be used with confidence for the following conditions : 1. Tubing sizes 2 in., 2 ½ in., 3 in. 2. For viscosities less than 5 cp 3. For GLR less than 1500 SCF/Bbl 4. For rates greater than 400 Bbl/d • Extension of Poettmann and Carpenter Method is Fancher and Brown Method and Hagedorn and Brown Method to Account for Viscous Effects Poettmann and Carpenter Method Cont’d • Poettmann and Carpenter started with the general energy eq : 1 ∆ + ∆ℎ + ∆ 2 2 + W + Iw = 0 Assume : W = 0 and kinetic energy to be negligible and that Τ = 1, then: 2 ∆ + ∆ℎ+ Iw = 0 Poettmann and Carpenter Method Cont’d Assume : Total losses du to irreversibilities can be correlated by the Fanning Eq : Iw = ∆ℎ = 4 ∆ℎ 2 2 1 4 2 1+ 2 − 2 ∆ ; where : f = correlating energy loss factor, then : 2 1 ∆ + ∆ℎ+ ;V= ,v= 4 ∆ℎ 2 2 =0 ; rewrite the abbove eq as follows : Poettmann and Carpenter Method Cont’d ∆ℎ = 2 4 2 2 1+ 2 2 1 ∆ℎ = 1 ∆ ; A = ¼ 2 and converting to field units 1 1 ∆ 2 2 2 1+ 7.413 1010 2 ; = ; Then , ∆ℎ = 1 1 ∆ 2 2 2 1+ 7.413 1010 2 Where : w = qm = mass flow rate, or : ∆ℎ = 1 1 ∆ 2 2 1+ 7.413 1010 −2 2 ; Ass : 2 7.413 1010 −2 2 = K ; Then, ∆ℎ = 1 1 2 1+ ∆ Poettmann and Carpenter Method Cont’d Dividing the above eq into ∆ : ∆ ∆ = 1 1 ∆ℎ ∆ 2 1+ ∆ ∆ (1 + ) = 1 1 ∆ℎ ∆ 2 ∆ ∆ℎ = (1 − 2 ) 1 1 2 ∆ + 1 − 2 1 1 2 ∆ ; It is noted that : = 1 1 2 ∆ (1 − 2 ) Poettmann and Carpenter Method Cont’d Then, ∆ ∆ℎ = + ∆ ∆ℎ ; It is know that = =+ 2 7.413 1010 2 Poettmann and Carpenter developed their energy loss factor correlation by knowing ∆ ∆ℎ in psi/ft and solving the above eq for the energy loss factor as follows : ∆ = 7.413 1010 2 144∆ℎ − 2 Example Problem Worked By the Poettman and Carpenter Method Given : d = 1.995 in ID P1 = 500 psig P2 = 1000 psig T1 = 120 oF T2 = 150 oF SGg = 0.65 SGw = 1.07 Example Problem Worked By the Poettman and Carpenter Method Cont’d Given : Density oil = 22 OAPI qo = 400 STB/d qw = 600 STB/d Viscosity Gas = 0.018 cp SFTo = 30 dynes/cm SFTw = 70 dynes/cm GLR = 500 SCF/STB Solution 1. SGo = 0.9218241042 2. qg = 500,000 SCF/d 3. GOR = 1250 cu ft/Bbl 4. m (massa per STB oil) = 350 (Sgo) + 0.0764 (GOR) (SGg) + 350 (Sgw) (WOR) = 946,4634365 lbm/STB oil 5. Total mass flow rate/d = qo x m = 378585,3746 lbm/day Solution Cont’d 7. P = 500 psig 1. Rs = 56,50240974 Where : Frick utilizes the following formula for calculating Rs for a particular set of conditions : = 18 100.0125 1Τ 0.83 100.00091 2. Bo = 1,045192931 Where : Frick has used the following for calculated Bo : = 0.972 + 0.000147 1.175 0.5 + 1.25 Solution Cont’d 3. Vm = 5.61 Bo + 5.61 (WOR) + vol free gas ( 14.7 )( )( ) 520 1 Vm = 49.82700728 cu ft 4. = 19,00575715 lbm cu ft 5. Flowing pressure gradient @ 500 psig = 0.176 psi/ft (From Fig 2.37) 8. P = 1000 psig Bo = 1.083, Rs = 120 SCF/Bbl, z = 0.9, Vm = 31.8 cu ft, = 29.8 lbm/cu ft, flowing pressure gradient = 0.236 psi/ft → Kermit E Brown Vol 1 Page 317 Solution Cont’d 9. Average pressure gradient = 0.206 psi/ft 10. The distance between 500 and 1000 psig = 2427,184466 ft 11. d v = 33.55917392 ; f = 0.008 12. ∆ @ ∆ℎ ∆ @ ∆ℎ 500 = 1 144 1000 = 1 144 + 2 7.413 1010 2 + 2 7.413 1010 2 = 0.177 psi/ft = 0.235 psi/ft 13. The average gradient is 0.206 psi/ft → giving a distance of 2427, 18 ft between 500 and 1000 psig 2 Development and Utilization of The Best Correlations in Predicting Pressure Loss Generalized correlation of Hagedorn and Brown • An effort was made by Hagedorn and Brown to determine a generalized which would include all practical ranges of flow rates, a wide range of GLR, all ordinarilly used tubing sizes and the effects of fluid properties. • Starting with the basic flow equation in symbolic differential form based on one pound mass of flowing fluid : 144 + ℎ + + + = 0 Generalized correlation of Hagedorn and Brown Cont’d • Assumes steady state flow and its units are in ft. • Hagedorn and Brown chose to treat the mixture of gas and liquid as a homogeneous mixture of combined properties. • Assuming that no work is done by the flowing fluid, the equation becomes : 144 + ℎ + + = 0 • Assuming that the frictional loss can be defined by the Darcy Weisbach Eq : ℎ = 2 Generalized correlation of Hagedorn and Brown Cont’d • The following eq for the mixture results : ℎ 144 + ℎ + + =0 2 Where ∶ vm is an average of the mixture • Writing this eq in integral form to solve: 2 ℎ2 2 ℎ2 ℎ 144 න + න ℎ + න + න =0 2 1 ℎ1 1 ℎ1 Generalized correlation of Hagedorn and Brown Cont’d • Evaluating three of terms : 2 2 2 − 2 (ℎ2 − ℎ1) 2 144 න + ℎ2 − ℎ1 + + =0 2 2 1 • Solving h2 – h1, we have : 1 2 2 − 2 144 2 − 2 ℎ2 − ℎ1 = 1+ 2 Generalized correlation of Hagedorn and Brown Cont’d • Similar to the method of Poettmann and Carpenter, 1 2 = 1 − 2 • Substiuting Vm (P1 – 1 P2) for 2 ℎ2 − ℎ1 = 144 , then : Vm ∆ −∆ 1+ 2 2 2 ; vm = vsl + vsg Generalized correlation of Hagedorn and Brown Cont’d ∆ℎ = 144 Vm ∆ −∆ 1+ ( ( vsl + vsg)2 2 vsl + vsg)2 2 • The average integrated density of the mixture is: 1 = • Solving for the average pressure gradient : ∆ ( + ) ∆ ( + )2 144 = + + ∆ℎ 2 ∆ℎ Generalized correlation of Hagedorn and Brown Cont’d • If g = gc and the conversion to units commonly used in the field is made: • Based on 1 bbl of liquid, the mass is : • Where, w = qm Generalized correlation of Hagedorn and Brown Cont’d • In this procedure we need to detemine the average integrated density. It cannot be calculated directly because of the slippage or fallback of the liquid phase. • If Ve = volume of a section of pipe, then Hl Ve = (1-Hl) Ve, is that portion occupied by gas Generalized correlation of Hagedorn and Brown Cont’d • By substituting than : • Hl → 1, vsl → 0, vsl → 0 (This eq will reduce to single phase liquid flow) Generalized correlation of Hagedorn and Brown Cont’d • Hl → 0, vsl → 0, vsl → 0 (This eq will reduce to single phase gas flow) Example Problem By Generalized Method of Hagedorn and Brown Given : d = 1.995 in ID P1 = 500 psig P2 = 1000 psig T1 = 120 oF T2 = 150 oF SGg = 0.65 SGw = 1.07 Example Problem By Generalized Method of Hagedorn and Brown Cont’d Given : Density oil = 22 OAPI qo = 400 STB/d qw = 600 STB/d Viscosity Gas = 0.018 cp SFTo = 30 dynes/cm SFTw = 70 dynes/cm GLR = 500 SCF/STB Solution Solution Cont’d Solution Cont’d Solution Cont’d Example Problem By Duns and Ros Method Given : d = 1.38 in ID P1 = 500 psig P2 = 1000 psig T1 = 120 oF T2 = 150 oF SGg = 0.65 SGw = 1.07 Example Problem By Duns and Ros Method Cont’d Given : Density oil = 22 OAPI qo = 400 STB/d qw = 600 STB/d Viscosity Gas = 0.02 cp SFTo = 30 dynes/cm SFTw = 70 dynes/cm GLR = 500 SCF/STB Solution Solution Cont’d Solution Cont’d Solution Cont’d Homework # 1 Calculate with Poettmann and Carpenter Method and Hagedorn and Brown Method Given : d = 1.380 in ID @ 1 ¼ in Tubing P1 = 500 psig P2 = 1000 psig T1 = 120 oF T2 = 150 oF SGg = 0.65 SGw = 1.07 Homework # 1 Calculate with Poettmann and Carpenter Method and Hagedorn and Brown Method Cont’d Given : Density oil = 22 OAPI qo = 400 STB/d qw = 600 STB/d Viscosity Gas = 0.018 cp SFTo = 30 dynes/cm SFTw = 70 dynes/cm GLR = 500 SCF/STB Homework # 2 Calculate with Duns and Ros Method Given : d = 1.995 in ID @ 2 in Tubing P1 = 500 psig P2 = 1000 psig T1 = 120 oF T2 = 150 oF SGg = 0.65 SGw = 1.07 Homework # 2 Calculate with Duns and Ros Method Given : Density oil = 22 OAPI qo = 400 STB/d qw = 600 STB/d Viscosity Gas = 0.02 cp SFTo = 30 dynes/cm SFTw = 70 dynes/cm GLR = 500 SCF/STB Thank You END OF SLIDES – FOR DAY 4 (PART 2) eq DAY 5