Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
ESP Design Examples 10.1. Introduction In order to design ESP procedures have to be followed:
system,
the
following
1. Collect data (well, production, fluid, electrical). 2. Determine the production capacity of the well. 3. Determine tubing size. 4. Calculate Total Dynamic Head (TDH). 5. Select pump stage type. 6. Calculate the number of pump stages required. 7. Check pump shaft loading and pump housing
pressure.
8. Calculate motor suitable motor.
and
horsepower
requirements
select
a
9. Calculate the protector thrust bearing load and select a suitable protector. 10. Determine the correct cable size and select a suitable cable type. 11.
Calculate the surface voltage and KVA requirements and select a suitable switchboard and transformer.
In current stage, we will calculate only the number of stages; the other calculations will be involved in the next chapters. To calculate the number of stages, we have to calculate first the Total Dynamic Head (TDH) that the pump has to deal with. The TDH is the sum of three basic components: 1. The Net Vertical Lift or net distance which the fluid must be lifted (Dynamic Fluid Level, DFL). 2. The friction loss in the tubing string. 3. The wellhead pressure which the unit must pump against. See the drawings below.
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Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
TDH = Net vertical lift in ft + friction loss in ft + WHP in ft
Pump setting depth does not affect net vertical lift 2
Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
Direction angle of the well does not affect the net vertical lift
10.2. Example 1: (water well) Casing size: Perforation depth: S.G. of produced fluid: Tubing sizes available: Pump setting depth: Static Fluid Level: Productivity Index: Bottom Hole Temperature: Wellhead Pressure required: Desired Flow rate: Hz: Viscosity:
5.5 inch 6000 ft. 1.0 2 3/8 inch 5500 ft 1820 ft 0.9 bpd/psi 260 deg F 100 psi 1300 bpd 60 1 cp
Solution: 1. Determine the production capacity of the well Flowing fluid Drawdown
level
(DFL)
=
Static
fluid
level
+
Max Drawdown possible = 6000 ft – 1820 = 4180 ft. Max Drawdown (psi) = 4180 x 0.433 psi/ft = 1810 psi Max possible flow (drawing fluid right down to the perforations = 1810 psi x 0.9 BPD/psi = 1629 BPD 2. Calculate TDH @ 1300 bpd Flowing Fluid Level @ 1300 BPD = SFL + DD Draw Down (psi)
= 1300 BPD/ 0.9 BPD/psi = 1444 psi
Draw Down (feet) = 1444 psi/ 0.433 psi/ft = 3335 ft. Flowing Fluid Level = 1820 ft + 3335 ft = 5155 ft 3
Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
Friction Loss (refer to Chapter 2) = 171 ft WHP = 100 psi / 0.433 psi/ft = 231 ft. Total Dynamic Head = DFL + WHP + Friction loss TDH =
5155
+
231
+ 285 = 5671 Ft
3. Select pump stages type Select a pump suitable for installation in 5.5 inch casing and also capable of producing 1300 BPD, Select for example Centrilift pump type FC1200 (from the manufacturer catalogue)
4. Calculate the number of stsges At 1300 bpd the FC1200 generates 22.5 ft per stage Number of stages required = TDH / Head per stage Number of stages required = 5671 ft/ 22.5 ft per stage = 252 stages From the table of the pump housings of the manufacturer, see below; select the suitable housing/s which contains number of stages greater than the calculated ones. Select either housing 15 (262 stages) or hosing 12 (209 stages) + housing 3 (51 stages) of total 260 stages.
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Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
5. Check shaft loading and Housing Pressure limitations Pump Hp = 0.36 HP/stage x 260 x 1.0 (SG) = 93.6 HP Maximum shaft HP rating = 125 HP (see Catalog) Therefore the standard shaft is O.K. Maximum head in operating range =31.2 X 260 = 8112'. Maximum pressure in range = 8112 ft x 0.433 = 3513 psi Housing pressure rating = 5000 psi (see Catalogue) Therefore the housing is O.K 6. Select Motor Motor hp required = 94 hp Select a motor suitable for installation in 5.5 inch well. Therefore select a 450 series motor (4.50 inch O.D.) Select a 450 series 100 HP, 2080 volt, 31 amp Note: Remember BHT = 260 deg F 7. Calculate Protector protector type
thrust
bearing
load
and
choose
Maximum pressure in operating range = 3513 psi Cross sectional area of Pump shaft = 0.371 sq. in. TB load for floater pumps = 3321 psi x 0.371 sq. in = 1232 lbs Select Solid Shoe Bearing (standard) rated at 1427 lbs (information in protector section of catalog) 5
Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
FMB 450 SERIES MOTORS 450 SERIES MOTOR MODEL FMB 4.50 INCH (114.3 MM) O.D. SIZE HP
VOLT/AMPS
LENGTH*
WEIGHT
60 HZ 60
50 HZ 50
60 HZ
50 HZ
FT.
M.
LBS.
KG.
735/51
612/51
17
5.2
843
383
60
50
840/44
700/44
17
5.2
843
383
60
50
945/40
787/40
17
5.2
843
383
60
50
1270/30
1058/30
17
5.2
843
383
75
62.5
644/77
537/77
21
6.3
1030
467
75
62.5
924/52
770/52
21
6.3
1030
467
75
62.5
1129/43
941/43
21
6.3
1030
467
85
71
1290/43
1075/43
23
7.1
1161
527
85
71
2080/27
1733/27
23
7.1
1161
527
100
83
1000/64
833/64
27
8.3
1353
614
100
83
1150/56
958/56
27
8.3
1353
614
100
83
2080/31
1733/31
27
8.3
1353
614
120
100
1200/64
1000/64
32
9.8
1605
728
120
100
2080/37
1733/37
32
9.8
1605
728
8. Determine Cable size and voltage drop in cable Motor is rated 100 HP, 60 HZ. , 31 Amp Pump Load is 87 HP Motor operating current = 94/100 x 31= 29 amp. Choose a cable size with a volts drop < 30v/1000 ft Choose No 4 AWG cable
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Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
Voltage drop at 29 amps = 13 volts/1000 ft VD corrected @ 260 DF = 13 x 1.4 = 18.2 volts We have 5600 ft of cable allowing for 100 ft at surface Voltage drop = 5.6 x 18.2 = 102 volts 9. Calculate Surface transformer
voltage
and
size
switchboard
and
Surface Voltage required = 2080 (motor) +102 = 2182 v.
Switchboard MODEL
KVA
VOLTS
AMPS
1500-SSC 2400-SSC 3300-SSC 4800-SSC 33MR-SSC
415 664 913 1660 913
1500 2400 3300 4800 3300
160 160 160 200 160
Therefore use a 2400 volt rated Switchboard.
Therefore use a 125 KVA 3 phase dual wound with Multi tapped secondary
transformer
10.3. Example 2: (oil well) • Casing size: • Perforation depth: • Tubing sizes available:
7.0 inch 6800 ft. 3.5 inch OD EUE, 9.3 lb. 7
Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9 • • • • • • • • • • • • • • • • • •
Pump setting depth: 5500 ft Static Bottom Hole Pressure: 3000 psi Oil Cut: 25 % Water S.G.: 1.085 Gas S.G: 0.70 Oil API Gravity: 32 (0.866) GOR: 300 scf/bbl Productivity Index: 1.42 bpd/psi Bottom Hole Temperature: 160 deg F Wellhead Temperature: 100 deg F Wellhead Pressure require: 100 psi Desired Flow rate: 2300 STbfpd Viscosity of oil @ WHT: 15 cp Viscosity of oil @ BHT: 5 cp Viscosity of water @ WHT: 0.9 cp Viscosity of water @ BHT: 0.6 cp Hz: 50 Casing is unvented, i.e., string has unvented paker.
Solution: 1. Calculate the average specific gravity Ave S.G. = (0.75 x 1.085 + 0.25 x 0.866) = 1.03 2. Calculate the fluid gradient
Fluid Gradient = 0.433*1.03 = 0.446 psi/ft 3. Calculate the Static Fluid Level (SFL) SFL = 6800-3000/0.446 = 73 ft 4. Calculate bottom perforations
hole
flowing
pressure
at
mid-
Pwf @ Mid Perfs = 3000 – 2300 / 1.42 = 1380 psi 5. Calculate bottom hole flowing pressure at Pump Setting Depth (PSD), which is pump intake pressure (PIP). Pwf @ PSD (PIP) = 1380 –(6800-5500)*0.446 = 800 6. Calculate the drawdown Drawdown = 2300/1.42 = 1620 psi 8
psi
Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
Drawdown in ft = 1620 / 0.446 = 3632 ft 7. Calculate Dynamic Fluid Level (DFL) DFL = SFL +DD = 73 + 3632 = 3705 FT Or Drawdown = 6800 – 1380 / 0.446
= 3705
ft
Fluid column above the pump = 800 / 0.446 = 1794 8. Calculate Rs @ PIP reservoir
ft
or get it from PVT data of the
Where: Rs = Gas in solution = Gas Specific gravity (0.7) PIP = Pump Intake Pressure (800 psi) BHT = Bottom Hole Temperature (160 OF) Rs = 0.7x[(800/18)*(100.0125*32 scf/bbl
/100.00091*160)]1.2048
=
137
9. Calculate the total produced gas (QGT) QGT = GOR*(1-wc)* Q = 300*(1-0.75)*2300/1000 = 172.5 mscf 10. Calculate the total gas in solution @ PIP (QGIS) in mscf QGIS @ PIP = 137*(1-0.75)*2300 /1000 = 78.8 mscf 11. Calculate Total free gas @ PIP (QGF) in mscf QGF = QGT Unvented)
-
QGIS
=
172.5–78.8
=
94
mscf
(Casing
12. Calculate g (gas formation volume factor) @ PIP it from PVT data of the reservoir:
is
or get
Where Z is the gas compressibility factor If Z is not available, it can be calculated from the following equations:
g = 5.04*0.886 *(160 +460)/(800+14.7) = 3.4 9
bbl/mscf
Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
13. Calculate the volume of free gas under reservoir condition (QRG), this will enter the pump as the system in unvented. QRG = QGF x g = 94*3.4 = 319
bbl/day
14. Calculate o (oil formation volume factor) @ PIP it from PVT data of the reservoir:
or get
As Rs < GOR, so, F = 137 x (0.7/0.866)0.5 + 1.25 x 160 = 323 o = 0.972 + 0.000147 x (323)1.175 = 1.1 Rbbl/STbbl 15. Calculate the volume of oil at reservoir condition (QRO). QRO = QO x o = 1.1*(1-0.75)*2300 = 634 Rbblpd QRW = volume of water (w~1) = 0.75 * 2300 = 1725 Rbblpd 16. Calculate the total volume of fluid entering the pump under reservoir condition (QRT). QRT = QRO + QRW + QRG = 634+1725+319 = 2678 Rbblpd 17. Calculate free gas % at the pump intake % of free gas = 319 / 2678 *100 = 11.9 % 18. In order to recalculate the real average specific gravity including gas and accordingly, the real fluid gradient, we have to calculate the Total Weight of the Fluid Oil Mass = Qo x OSG x 62.4 #/cuft x 5.6146 cuft/bbl = 2300 x 0.25 x 0.866 x 62.4 x 5.6146
=
174,457 Lb
Water Mass = Qw x WSG x 62.4 #/cuft x 5.6146 cuft/bbl = 2300 x 0.75 x 1.085 x 62.4 x 5.6146
=
655,726 Lb
Gas Mass = Qo x GOR x GSG x 0.0752 #/cuft x 5.6146 cuft/ bbl = 2300 * 0.25 * 0.7 * 0.0752 = 9,080 Lb Total Mass = 174,457 + 655,726 + 9,080 = 839263 Lb 19. Calculate the density, specific gravity and gradient Density = Mass Lb/ volume cuft = 839263 / (2678 * 5.6146) = 55.82 lb / cuft
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Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9 S.G = 55.82 / 62.4 = 0.8945
Gradient (G) = 0.8945 * 0.433 = 0.3872
psi/ft
20. Recalculate the TDH based on the new gradient Dynamic fluid level = PSD–PIP/G = 5500–800/0.3872 = 3434 ft Friction loss in 3-1/2” Tubing based on average viscosity of 10 cp and rate 2678 Rbpd (refer to section 2.20) = 166’ WHP in ft = 100 / 0.3872 = 258 ft TDH = 3434 + 166 + 258 = 3858 ft 21. Select a pump suitable for installation in 7” casing and also capable of producing 3678 Rbfpd Select SN2600 of Reda
Head per stage = 46 ft/stage 22. Calculate the number of stages required for TDH and motor HP # of stages = TDH /ft per stage =
3858/46 = 84 stages
From pump SN2600 catalog select Hsg # 70 (88 stages) Note The average viscosity of the fluid does not have serious effect on either head or rate of the pump. That why we do not take into consideration. HP required = # of stages * hp per stage * fluid SG HP required = 88 * 1.32 * 0.8945 = 104 hp
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Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
From motors catalog select Series 540, 117 hp motor, 1751 V, 40.5 A
Series 540 Motors - MK Type HP 60 HZ 80
50 HZ 67
VOLTS 60 HZ 50 HZ
AMPS
Type S
1235
1029
40
UT S
2241 100
1868
22
83
UT S
1305
1088
51.5
UT S
2313 120
1928
27
100
UT S UT
1105
921
69.5
CT S
2270 140
1892
32.5
117
UT S UT
1022
852
85
CT S UT
1299
1083
69.5
CT S UT
2101 1751
12
40.5
CT
Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
23. Check shaft loading and Housing Pressure limitations Check pump shaft hp (Monel, 256 hp) with motor hp, if the motor hp is more than pump shaft hp, ask for high strength shaft (inconel, 410 hp) Pump head at shut in condition = 58’ * 88 = 5104‘ Pump discharge pressure at shut in = 5104 * 0.3872 = 1976 psi Housing pressure rating = 5000 psi (from catalog) Therefore the housing is O.K 24. Calculate Protector thrust bearing load Maximum pressure in operating range = 1976 psi Cross sectional area of Pump shaft = 0.601 sq.in TB load for floater pumps = 1976 * 0.601 = 1188 lb Select Solid Shoe Bearing (standard) rated at 2637 lbs (information in protector section of catalog) 25. Cable Size Selection and voltage drop in cable Motor is rated 120 hp, 60 HZ. 2270 v, 32.5 Amp Pump Load is 104 hp Motor operating current = (104/ 117) * 40.5 = 36 amp. Choose a cable size with a volts drop < 30v/1000 ft Choose AWG #6 cable Voltage drop at 40.5 amps = 24 volts/1000 ft Correct for Temp. [(160+100)/2]= 130 Deg F) = 24 * 1.12 = 26.88 volts We have 5500 ft of cable allowing for 100 ft at surface Voltage drop = 5.6 * 26.88 = 151 volts 26. Calculate Surface voltage Surface Voltage required = 1751(motor) +151= 1902 v.
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Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
27. Calculate Size of Switchboard and Transformer Surface KVA = (1.732 * Volt * Amps) / 1000 KVA = (1.732 * 2902 * 40.5) / 1000 = 134 KVA Therefore Selecta SB 1512, 182 KVA, 70 A Select Transformer with minimum 200 KVA, 3 φ, and tapped secondary
multi
MDFH 1500 V Maximum 14 20 45 90
Amp Load To To To To
20 45 90 160
36 52 117 234
KVA To To To To
52 117 234 416
1512 3900 V Maximum, 1900 V Minimum 35 70 115
To To To
70 115 165
91 182 299
To To To
182 299 429
10.4. Example 3 (viscosity correction) In this example will take the same data and some results of example 2 except the average viscosity of the oil is 140 cs @ average temperature (140oF).
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Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
Average water viscosity = (0.9 + 0.6) / 2 = 0.75 CS Average fluid = WC x W + (1-WC) x O Average fluid = 0.25 x 0.75 + 300 x 0.75 = 105 cs So, the required fluid rate is 2678 downhole bfpd (78.12 US gpm) with viscosity 225 cs. At QW = 2678 bpd the head per stage HW = 46 ft (from pump curve as water) and W = 68 % Pseudocapacity (P)=1.95 x 0.5 x [0.04739 x H0.25746 x Q0.5]-0.5 P = 1.95*105^0.5*(0.04739*46^0.25746*78.12^0.5)^-0.5 = 18.86 CH1Q=1.0045-0.002664*18.86-0.00068292*18.86^2 +0.000049706*18.86^3-0.0000016522*18.86^4 +0.000000019172*18.86^5 = 0.8815 CQ=0.9873+0.009019*18.86-0.0016233*18.86^2 +0.000077233*18.86^3-0.0000020528*18.86^4 +0.000000021009*18.86^5 = 0.88851 C=1.0522-0.03512*18.86-0.00090394*18.86^2 +0.00022218*18.86^3-0.000011986*18.86^4 +0.00000019895*18.86^5 = 0.517 HVis = 46 x 0.8815 = 40.6 ft Qvis = 2678 x 0.88851 = 2380 bfpd (downhole) = 69.4 gpm vis = 68%*0.517 = 35.2 % BHPvis = 60 * 40.6 * 0.943 / (3960 x 0.352) = 1.65 hp/stg (compared with 1.32 hp/stg of example 2) The friction loss at 108 cs and 2380 bfpd = 236 ft New TDH = 3434 + 166 + 236 = 3838 ft Required No of stages = 3838 / 40.6 = 95 stages (compared with 84 stgs of example 2) Required HP = 95 x 1.65 = 157 hp (compared with 104 hp of example 2) t = 2678/2300 = 1.164 (as per example 2) New STBPD = 2380/1.164 = 2045 STBPD Conclusion In case if the viscosity of the oil becomes 140 cs: Production rate will be 2045 bpd instead of 2300 bpd. The number of stages will be 95 instead of 84 stages. The horsepower required will be 157 hp instead of 104 hp.
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Mohamed Dewidar 2013
Electric Submersible Pumps Chapter 9
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