Exercise

1. How can you differentiate between a wellbore screenout and a tip screenout?

A wellbore screenout causes a rapid pressure increase because the blockage

occurs in the wellbore (not too far from the pumps) and there is no place for the

fluid to escape. On a log pressure versus log time plot the wellbore screen-out

shows a unit or even higher slope. A tip screenout takes place a long way from

the wellbore, and the entire fracture volume may be expanded—particularly the

width and/or the height before the pressure within the wellbore and at the surface

increases to the preset limit. On a log-log plot, the slope is less than one.

2.

In planning a fracturing treatment for a well, we know or have calculated the following

information: Well parameters:

rw = 0.328 ft

re = 3000 ft.

h = 45 ft (pay layer thickness)

k = 0.5 md.

formation fluid: oil

= 0.8 cp

B = 1.2 RB/STB

 

Design calculations:

Total mass of proppant that can be placed into the pay layer:

100,000 lbm

Proppant specific gravity: = 2.65 (water = 1.0)

Proppant pack porosity p = 0.38

Proppant pack permeability kf =60,000 md,

Volume of 100,000 lbm proppant = 975 ft3

One-wing propped volume is Vf  =  487.5 ft3.

Max. prod. Rate = 135.6 STB/D (xf 1= 901.4 ft)

Optimal areal proppant concentration = 50,000 lbm / (901.4 ft 45

ft) = 1.23 lbm/ft2

Optimal width = 0.144 in.

Assume a slurry injection rate of (2qi) = 30 bpm for the two wings and a created fracture

height of 67.5 ft. E =  2.4106 psi,  = 0.2, the Power Law rheology is K' = 0.022 lbf/(ft2 sec0.63)

and n' = 0.63; the leakoff coefficient with respect to the permeable zone is CL = 0.004 ft/min1/2

and the spurt loss is negligible.

Determine the following:

a) How much proppant is to be injected?

b) What maximum proppant concentration is necessary to achieve the

optimum proppant placement?

c) What will be the apparent leakoff coefficient?

d) Calculate the necessary injection time using the PKN model. What will

be the fluid efficiency?

e) Determine the pad volume and the ramped proppant schedule.

f) Give the proppant schedule in terms of added proppant to clean liquid

(assume 2.7 specific gravity of proppant.)

solution:As with any other complex algorithm it is best to use consistent system of units (i.e., SI.)

(Therefore, rp = 0.6667)

a) The fracture height is 1.5 times the pay layer thickness. Therefore, approximately 150,000

lbm proppant needs to be injected into the formation. The mass of proppant in one wing will be

M = 22,680 kg/rp = 34,020 kg from which , (22,680 kg) will be placed into the target layer.

b) If we know the volume of one wing at the end of pumping, the maximum

proppant concentration is calculated by dividing the mass by the volume. To

calculate the volume of the fracture at the end of pumping, we need to know the

fracture width when the target length is reached. From the PKN width equation

(incorporating power law rheology)

The average width at the end of pumping is

The maximum proppant concentration is the injected proppant mass divided by the volume of

the dynamically created fracture wing at the end of pumping:

Notice that this concentration is mass proppant per volume of slurry. This should be the

proppant concentration in the last injected slurry stage. Also this will be the proppant

concentration everywhere in the fracture at the end of pumping. In terms of added proppant to

clean liquid this is 1425 kg/m3 that is 11.9 lbm/gal.

c) The apparent leakoff coefficient with respect to the fracture area is less than

the value with respect to the permeable area because part of the surface are is

"idle":

d) The injection time is calculated solving the following equation:

 

The positive root of the quadratic equation is  , therefore te = 3378 s = 56.31

min.

Once the injection time is known, we calculate the injected slurry volume (for one

wing)

Since the dynamically created volume of one wing at the end of pumping is

the fluid efficiency is e = 27.34 %.

e) Knowing the fluid efficiency the Nolte parameter is calculated as

from which the pad volume (for one wing) is calculated as

and the pad injection time is 

The proppant concentration of the injected slurry should be zero for the first 27.4

min, and then it should be increased according to:

f) The above function can be approximated by "stairs." The proppant concentration is

understood here as mass per slurry volume. To calculate the mass of proppant added to unit

volume of clean liquid we use (assuming 2.7 specific gravity of proppant) :

For instance, the maximum proppant concentration ce = 926.5 kg/m3 requires cadded = 1425 kg added proppant per m3 clean liquid (11.89 ppga)

When the proppant schedule is converted into stages, we take the average concentration for the interval. Table 1 is a possible stage design:  

Stage Start, End, Slurry Prop. Add. Added Prop

  min min Vol., m3 conc,

kg/m3

Prop.

kg/m3

Prop.,

ppg

In

Stage

lbm

          neat neat  

          fluid fluid  

Pad 0 36.7 87.8 0 0 0 0

1 36.7 38.0 7.6 272 303 2.5 16,800

2 38.0 42.0 12.7 484 592 4.9 20,360

3 42.0 46.0 12.7 614 799 6.7 25,830

4 46.0 50.0 12.7 726 1000 8.3 30,540

5 52.0 54.0 12.7 826 1200 10.0 34,740

6 54.0 56.3 12.7 898 1358 11.3 21,770

Sum: 268.6       150,000    

Table 1: Stage design (Injected volume and proppant for two wings)

3. How much horsepower should be ordered for a fracturing treatment if the

calculated surface treating pressure is 7,000 psi and the two-wing flow rate is 20

BPM? How much energy is consumed for pumping (assuming 60 % overall

efficiency)?

The power requirement is N = pst (2qi) = 4.826107 Pa 0.0530 m3/s = 2.56106 W (3433 HHP )

The energy consumed is W = N te / 0.6 = 2.56106 W 3951 s / 0.6 = 1.691010 J (4680 kWhr )

4. In a mini-frac test, 39.75 m3 (10,500 gal) of fluid was injected into one fracture wing over a

period of 20 minutes. Estimate the leakoff coefficient, if E’ = 16.9 GPa, the closure pressure is

pC =  22.1 MPa (3200 psi), the permeable height is 9.75 m (32 ft) and the analysis of the

bottomhole pressure fall-off curve resulted in the following straight line:

Use the Radial model for analysis

The created fracture radius is

The apparent leakoff coefficient with respect to the total fracture surface is

Since only hp =  9.75 m is permeable , the ratio of permeable to total surface is

Therefore, the leakoff coefficient with respect to the permeable surface is

5. A bilinear flow analysis of a pressure drawdown test with production rate

qo = 124.3 STB/D indicates a straight line on the plot of pressure versus t1/4 with

slope mbf = 84.3 psi/hr 0.25. Determine the equivalent fracture half-length if the

following data are available: h = 45 ft, k = 2 md, B = 1.2 RB/STB,  = 0.8 cp,

= 0.15, ct = 105 1/psi and kf = 60,000 md.

In SI units: H = 45 ft = 13.72 m

K = 2 md = 1.9710-15 m2

B = 1.2

 = 0.810-3 Pas

 = 0.15

ct = 10-5 1/psi =  1.4510-9 1/Pa

kf = 5.9210-11 m2

qo = 2.28710-4 m3/s

mbf =  7.5104 Pa/s0.25

since

therefore

In other words, the created fracture is equivalent to an optimal fracture of half-length 387 ft.

Also, from kf1/2w1/2 = 6.7 10-7 m15, we obtain w = 0.0063 m = 0.248 in. Therefore

the volume of the effectively placed one fracture wing is 1180.006313.72 = 10.2

m3 = 360 ft3.

Assuming a specific gravity of 2.7 and a pack porosity of 0.38, the amount of

proppant corresponding to this packed volume is 17,100 kg (37,700 lbm) for one

wing, that is, 75,400 lbm for two wings.

If, for instance, the injected amount was twice as much, the overall effectiveness

of the proppant placement is about 50 % (which is a very good value for a

fracture treatment.)

The approximate end of the bilinear flow period is

therefore the specialized plot (from which mbf has been obtained) should contain time points

below three and a half hours.