Solution: Exam 12.13.06

Solution: Exam 12.13.06

Exercise 1

a)Sand production

Reservoir A:

Upper limit for inflow pressure loss:pA =pR-pw = 30 bar

-Gives the upper limit for production: qA =JApA= 20.30 = 600 Sm3/d

Reservoir B

Upper limit for inflow pressure loss:pB=33 bar

- gives upper limit for production: qB =JBpB= 10.33 = 330 Sm3/d

b)Pipe diameter

As estimated above, the reservoir A will cope a maximum inflow pressure loss: pR – pw = 30 bar, which will be reached at a production of 600 Sm3/d. We may use the following as a basis:

Volume flow, down hole:

At a wellbore pressure of:

Static pressure loss in the production tubing, independent of the pipe dimension:

Pressure loss due to friction, assume close to vertical well; measured depth = true vertical depth

Well head pressure estimate:

Well head pressure estimated for alternative production tubings:

Diameter 0.08 og 0.1 m synes begge akseptable ut fra gitte opplyninger. Sandproduksjon kan muligens bli et problem i framtida. Valg av diameter: 0.08 metervilopprettholde høgere strømningsfart når produksjonen synker, at sand ikkje så lett akkumuleres i brønnen. Så vi velger her røyrdiameter 80 millimeter og gjør videre berekninger ut fra dette.

A diameter of 0.08 and 0.1 seem both to be acceptable from the given information. Sand production will might be a future problem. By choosing a diameter of 0.08 m one is able to sustain a larger flowing velocity when the production is declining, i.e. sand will not easily accumulate in the well

(Under other circumstances, a diameter of 0.1 m might bethe ”correct choice”. The important thing is that you make a rational selection and clarify the premises)

c)Choke

With our selection of pipe, we expect a wellhead pressure of 54 bar, resulting in a pressure difference between x-mas tree and separator: pc= 54-20 =34 bar. Based on the formula for a one phase liquid flow:

nozzle diameter: 1.2 cm

(If we choose the larger pipe diameter 0.1 m, so that the well head pressure estimate will be 58 bar, the nozzle opening will be:1.0 .10-4 m2 og 1.1 cm.)

One phase liquid flow assumes that the wellhead pressure is above the saturation pressure. This is the situation in our case, but should be commented, maybe clarified by calculations.

d)Total production, limited by sand criteria

When both reservoirs are producing together, one of the reservoirs will most likely reach the limit for sand production before the other. The total production will therefore be limited by the reservoir first reaching the sand production limit.

When both reservoirs are producing, the pipe between A and B will be filled with liquid from reservoir B, with a density: 900 kg/m3. This results in a static difference in well pressure:

Assumption 1: Reservoir A i limiting the total production

Well-pressure at reservoir A : pwA = 200-30 = 170 bar

Well-pressure at reservoir B :

Inflow pressure loss for reservoir B: pRB-pwB= 219-178.8 = 40.2 bar.

Reservoir B is overridingthelimit for sand-production: 33 bar.

Assumption 2: Reservoir B is limiting the total production:

Well-pressure at reservoir B : pwB = 219-33 = 186 bar

Well-pressure at reservoir A :

Inflow pressure loss for reservoir A:pRA-pwA= 200-172.2 = 27.8 bar.

The inflow pressure loss in reservoir A is lower than the limit for sand production: 30 bar.

Production under assumption 2:

Reservoir A:

Reservoir B:

Total: qt = qA + qB =786 Sm3/d

e)Initial production capacity

The production capacity limited by sand production is estimated above. We will examine if it is possible to produce at this rate with the given pressures and flowing conditions.

Since two reservoirs are producing fluid with a different density, the density in the pipe will be an average value. We assume volumetric proportion of mixture so that:

QA = qABoA=456 . 1.5 = 684 m3/d

QB = qBBob=330 . 1.2 = 396 m3/d

Qt=QA + QB = 684 + 396 = 1080 m3/d

Average density:

Flowing velocity (when: d= 0.08):

Pressure loss along the production tubing, if we assume a vertical well above the reservoir layers:

Well head pressure:

pth = pwA- pt=177.2 - 130.9=46.3 bar

The separator pressure is 20 bar. The well head pressure is larger than this, so we have to choke back the well not to override the limit for sand production.

The consideration to sand production is limiting the initial production rate.

f)Cross flow

Shut main valve, zero total production, we get

Well pressure at reservoir A is then:

The flow from both reservoirs, when assuming that the injectivity is equal to the productivity

Reservoir B is producing 68 Sm3/d, which will flow into A.

g)Total production to avoid cross flow

If the well-pressure at A does not override the reservoir pressure: 200 bar, fluid will not flow into reservoir A. With a well pressure of 200 bar. at A, the well-pressure at reservoir B is then:

This results in zero production from reservoir A, so the total production to avoid cross flow is then:

Exercise 2

a) Productivity index

Effective drainage radius:

Below the productivity index isestimated by

1. Cinco-Ley’ correlation

2. Semi analytic formula from the compendium

a-1 :Cinco-Ley’s correlation

Produktivity index for ”deviated, vertical wells” :

Inflow efficiency:

Effective productivity index:

a-2. Semi analytical formula fromthe compendium

Effective productivity index:

We may observe that the difference between the estimates is: 6%. Without making a standpoint of view of what estimate is the most credible, further calculations are based on the estimate: J= 60.8 Sm3/d.

b) Well pressure at a production of 1000 Sm3/d

c) Liquid fraction

Formation volume factor for gas

Superficial velocity, gas:

Superficial velocity, liquid:

Liquid fraction:

We are predicting the liquid fraction 30% (29.4%) just after the gas injection.

d) Change in liquid fraction, velocity and gradient

Up the pipe, the pressure will decline. At a constant temperature, the reduction in pressure gives:

-Gas expansion and evaporation of gas from the liquid; therefore: increasing gas volume

-A certain reduction in liquid volume, due to the evaporation

The increase in gas volume will then always be larger than the reduction in liquid volume (molecules occupy larger volume in gas phase than in the liquid phase). Since the mass flow along the pipe is constant, the increase in gas volume and total volume means:

-decreasing flowing density

-increasing flowing velocity

-decreasing liquid fraction (flux fraction and in-situ)

Pressure gradient:

Decreasingthe averagedensity, implies:

- the pseudo static contribution: is decreasing up along the pipe.

- the contribution due to friction is increasing when the velocity is increasing up the pipe.

At moderate flowing velocity (3 – 4 m/s), the decreasing density will dominate, so the pressure gradient (Pa/m) is decreasing upwards the pipe. When the flowing velocity is sufficiently large, the friction term will dominate, so a further increase in velocity will increase the pressure gradient.

The temperature has been assumed to be constant. In the real world the temperature will always decline upwards along the pipe. The drop in temperature will contribute to a reduction in the evaporation and reduction in expansion. It will contribute in “the opposite direction” to the pressure loss. Normally the contribution from pressure loss will dominate, so that the total effect will be as described above. I certain cases (large pressure, large drop in temperature) the drop in temperature may dominate, so the total effect will be opposite of what is described above.

e) Production under certain circumstances

We want to investigate if it still is possible to produce by drawdown between reservoir and separator

Average pressure for the production tubing:

Average temperature: reservoir temperature (333 K), or a little lower since we have some cooling: we may assume: 320 K

Gas density:

Two phase density:

Two phase friction factor:

Liquid flux fraction:

Well-head pressure

The estimated well-head pressure is larger than the separator pressure.

It is therefore possible to produce at the predicted rate.