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SOLUTION:
SUBJECT TPG 4245 PRODUCTION WELLS
EXAM DECEMBER 6. 2004
Premisses for the exam
This exam is deviating from previously exams. Previously there were strictly defined and limited tasks. In this case a field scenario with some guideline questions will be presented. The questions may be answered in many different ways.
The teacher does not expect that all questions are answered equally thorough as the exam is only limited to 4 hours. It is accepted if one of the questions is not answered.
A good solution will reflect:
-The skill to perceive the totality in the problem
-The skill to make reasonable assumptions regarding the number of wells, pipe size and other design parameters, and verify/controll the assumptions by calculations.
-the ability to do critical resoning about the assumptions based on the calculated results
Background for the proposed solution
Considering the above premises many good solutions may be presented. The difference will depend on what assumptions and priorities are made. (Only after extensive try and failure one may hope that the different alternatives will converge to a somewhat uniform “optimized” solution.) Solutions sketched below are teachers recommendation, but far from “the only right”.
Answers connected to the single questions
a)Oil, gas, or both?
(In the exercise text there was a typing error, so the gas rate was given as 1200 Sm3/d at an oil rate of 600 Sm3/d. The correct is 12000 Sm3/d at an oil rate of 600 Sm3/d.)
Saturation pressure, at wellhead temprature
Average Rt = 19.86 Sm3/Sm3 ≈ 20 Sm3
At saturation pressure:
The flow in the top of the production tubing will be one phase oil when the pressure is above 57.4 bar. There is a reason to assume that the pressure in the top of the production tubing must be above the saturation pressure to be able to drive the oil through to the processing unit. The reservoir pressure is way above.
b)Skin
Productivity index, adjusted from the production test, see attached figure
We may compare this with the productivity index from a formula. But the we must first estimate the formation volume factor from the given fluid data.
Formation volume factor at saturation pressure
At reservoir pressure, the formation volume factor will be slightly reduced due to compression. The reduction will be relatively small. We may for example assume Bo= 1.085 at reservoir conditions.
Estimated productivity index from equivalent drainage radius: 500 m
This is somewhat larger than the measured productivity index (29.4). This indicates a relatively small skin. If we want, we may estimate the skin factor through the inflow performance relationship.
Expressed as skin factor:
c)Number andthe location of well
We choose to compare the vertical and horizontal wells, for a pressure difference of 20 bar between the reservoir and well. A too large pressure difference will give an unnecessary large pressure loss.
C-1. Vertical
Measured productivity index: J = 29.4 Sm3/d/bar
Production per well: Q=J/(PR – PW) = (29.4)(20) = 588 Sm3/d
Number of wells to fulfill the available processing capacity (Q = 10000 Sm3/d):
It will be reasonably to place these equally through the reservoir, possibly with a certain distance from the original water – oil contact.
C-2. Horizontal
We will here consider two horizontal wells, located higher up in the reservoir, for example as sketched. This will make it possible to produce over a longer time before water breakthrough.
To estimate the productivity index from such wells we will assume
-some anisotropy, for example:
-well placement4 m beneaththe hang:b = 4 m, this will give a geometric skin, estimated as
-formation damage as for the vertical test well: Sf = 1.6
-total skin factor
S = Sb + Sf = 1.4 + 1.6 = 3
-Productivity index, estimated as for stationary inflow. (We may expect a certain water drive. Otherwise the assumption about stationary inflow is conservative compared to the pseudo stationary.)
If we as previously assume a 20 bar drawdown, this will give a capacity per well:(216)(20) = 4310 Sm3/d. For two wells 8620 Sm3/d.
This seems to be in accordance with the available processing capacity: 10000 Sm3/d. If the well pressure is somewhat lower, or the productivity index is somewhat larger than estimated, we may fill the available processing capacity with two horizontal wells.
We therefore recommend:
-2 horizontalwells (alternatively 3)
-located high up in the reservoir, as sketched
d)Production tubing
If we assume a flowing velocity of 2 m/s in the tubing:
Maximum limit: 200 mm
We may as a basis assume a maximum pipe size, 200 mm, then v = 1.7 m/s
Oil density
Reynolds number
Friction factor
Initial well pressure: pw = 226 – 20 = 206 bar.
Length of the vertical section 2100 – 70 = 2030 m.
Estimated top pressure
This pressure is most likely not sufficient to make the liquids flow to the processing unit 14 km away. It is therefore most likely necessary to implement artificial lift from day one.
e)Lift methods
With rates as assumed it is most likely that the following two lift methods are applicable:
-Continuous gas lift
-Electric submersible pump
A down –hole pump should give significant pressure to drive the produced fluids through pipelines on the seafloor and up the riser. This may be estimated (if time).
Figure In flow performance relationship based on production data
TPG4245 Prod.br.løsn.
06.12.2018
Iptadmin/372/HAA