reservoir and production well test data

ESP Design - Well Data Sheet

Design Specifications For ESP Pump

Reservoir And Production Well Test Data

Note - Water Cut

Mid Pt Perfs @ 7488’

see problems

5-1/2” O.D.

Note - APIo

© 2015 PetroSkills, LLC. All rights reserved.

Total Dynamic Head RequiredTotal dynamic head required to pump the desired capacity refers to the feet (meters) of liquid being pumped (pressure required to pump liquid expressed as height of liquid) and is the sum of the (1) net well lift, (2) well tubing friction loss, and (3) wellhead discharge pressure. Refer to the figure.

H = Hd + Ft + Pd

Head

(Ph

) Calcu

lation

Ph =

Ph =

Ph =

Ph =

Ph = Hd + Ft + PdWhere Ph is the:

FIELD UNITS

METRIC UNITS

or

or

and

and

Total Dynamic Head


Module Exercise Problem

Subject: Basic ESP Pump Design Example ________________________________________________________________________________

1) A field with several producing wells on natural flow has been forecasted to require future artificiallift facilities. In the near future, wells will no longer have the reservoir energy for primaryproduction on natural flow. Detailed engineering studies have concluded that an electricsubmersible pump (ESP) will be installed for each well. Sufficient power will be available for theprojected life of the ESP project. ESP Design Target Rate is 800 BFD / 127.2 m3D

From Well Data (for well with Pwf below the bubble point pressure)

Depth to mid point of perforations: 7488 ft / 1673 m Previous test data: 40o API gravity oil (specific gravity of 0.825):

Pres = 925 psig / 6377.9 kPa Qo = 750 BFD / 119.3 m3/D Pwf = 550 psig / 3792.3 kPa Pump Operating Conditions: 60 Hz frequency / sand free wellbore / very low GLR

For the above field data description: a) Vogel analysis determined that a Pwf = 517 psig / 3564.5 kPa is required to produce

800 BFD / 127.2 m3/D b) Determine the total dynamic head (TDH = HL + HF + HD) required for the pump if:

- friction head HF in the tubing is 25 ft / 4 m - discharge head at surface HD is 25 ft / 4 m

c) Determine the number of pump stages required for producing 800 BFD / 127 m3/dd) Determine the total motor horsepower required (at 60 Hz) for the pump design

……………………………………………………………………………………………

Solution: Fluid Gravity (25% oil x 0.825 s.g.) + (75% water x 1.05 s.g.) = 1.00 fluid s.g. Fluid Gradient fluid s.g. x gradient of water = 1.00 x 0.433 psi/ft = 0.433 psi/ft

= 1.00 x 9.795 kPa/m = 9.795 kPa/m From Vogel Analysis Pwf = 517 psi / 3565 kPa at 800 BFD / 127 m3/D Determine Lift Head 517 / (0.433 x 1.00) = 1194 ft of fluid column with pump operating

3565 / (9.795 x 1.00) = 364 m of fluid column with pump operating 7488 ft (surface to mid pt perfs) - 1194 ft = 6294 ft to the surface 2283 m (surface to mid pt perfs) – 364 m = 1919 m to the surface

HL to be lifted x fluid s.g. of 1.00 = 6294 ft / 1919 m Total Dynamic Head 6294 ft HL + 25 ft HF + 25 ft HD = 6344 ft TDH Required From Pump

1919 m HL + 4 m HF + 4 m HD = 1927 m TDH Required From Pump

Pump Stages 6344 ft / (24 ft/stage) = 264 stages required 1927 m / (7.3 m/stage) = 264 stages

Motor Power Reqt. 264 stages x (0.22 BHP per stage) = 58 HP Motor (43.3 kw)


ESP PUMP Curve For 5-1/2” Well Casing

H-27 ESP

Pump Curve


ESP Design - Well Data Sheet

Design Specifications For ESP Pump

Reservoir And Production Well Test Data

Note - Water Cut

Mid Pt Perfs @ 1830 m / 6000 ft

see problems

5-1/2” O.D.

Note - APIo65%40o

50


Total Dynamic Head RequiredTotal dynamic head required to pump the desired capacity refers to the feet (meters) of liquid being pumped (pressure required to pump liquid expressed as height of liquid) and is the sum of the (1) net well lift, (2) well tubing friction loss, and (3) wellhead discharge pressure. Refer to the figure.

H = Hd + Ft + Pd

Head

(Ph

) Calcu

lation

Ph =

Ph =

Ph =

Ph =

Ph = Hd + Ft + PdWhere Ph is the:

FIELD UNITS

METRIC UNITS

or

or

and

and

Total Dynamic Head


Module Exercise Problem

Subject: ESP Pump Design 85 m3/ day / 530 BFD Problem ________________________________________________________________________________ 1) A field with several producing wells on natural flow has been forecasted to require future artificial

lift facilities. In the near future, wells will no longer have the reservoir energy for primary production on natural flow. Detailed engineering studies have concluded that an electric submersible pump (ESP) will be installed for each well. Sufficient power will be available for the projected life of the ESP project. The Design Target Rate for well # 29 is 85m3D / 530 BFD.

Well Data (for well with Pwf below the bubble point pressure)

Depth to mid point of perforations: 1830 m / 6000 ft Previous test data: 40o API gravity oil (specific gravity of 0.825): Water Cut is 65%

Pres = 710 psig / 4896 kPa Qo = 660 BFD / 105 m3/D Pwf = 273 psig / 1880 kPa

For the above field data description: a) Vogel analysis determined that a Pwf = ____ psig / ____ kPa is required to produce

85 m3/D / 530 BFD b) Determine the total dynamic head (TDH = HL + HF + HD) required for the pump if:

- friction head HF in the tubing is: 4 m / 25 ft - discharge head at surface HD is: 4 m / 25 ft

c) Determine the number of pump stages required for producing 85 m3/d / 530 BFD d) Determine the total pump design motor HP required when operating at 50 Hz

………………………………………………………………………………………… Solution: Fluid Gravity (35% oil x ____ s.g.) + (65% water x ____ s.g.) = _____ fluid s.g. Fluid Gradient fluid s.g. x gradient of water = ____ x 0.433 psi/ft = _____ psi/ft = ____ x 9.792 kPa/m = _____ kPa/m From Vogel Analysis Pwf = ____ kPa / ____ psi at 85 m3/D / 530 BFD Determine Lift Head ____ / (0.433 x ____) = ____ ft of fluid column with pump operating ____ / (9.795 x ____) = ____ m of fluid column with pump operating 6000 ft (surface to mid pt perfs) - ____ ft = ____ ft to the surface 1830 m (surface to mid pt perfs) – ____ m = ____ m to the surface HL to be lifted x fluid s.g. of ______

HL => ____ ft x ____ = ____ ft ____ m x _____ = ____ m

Total Dynamic Head ____ ft HL + 25 ft HF + 25 ft HD = ____ ft TDH Required From Pump ____ m HL + 4 m HF + 4 m HD = ____m TDH Required From Pump Pump Stages ____ ft / (___ ft/stage) = ____ stages required ____ m / (___ m/stage) = ____ stages required Motor Power Reqt. ____ stages x (_____ BHP per stage) = ____ HP Motor (_____ kw)

reservoir and production well test data

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