5. Hydraulic Fracturing 2

8/6/2019 5. Hydraulic Fracturing 2

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5. HYDRAULIC FRACTURING 2

1TUNIO, May' 2011,,, Courtesy AP Aung


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FRACTURING HYDRAULICS

The hydraulic power required to pump qt barrels of fluidper minute into a well with a surface injection pressure of

ps psi is given by the product ps qt or, converting tohydraulic horsepower,



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Surface Injection Pressure

The pump pressure or surface injection pressure, ps, isequal to the sum of the bottom-hole fracture treating

pressure pt, the frictional pressure drop in the pipe pf,and the pressure drop through the perforations p ,minus the hydrostatic pressure ps,

(4.20)

If, during the treatment, the pumps are stopped, thesurface pressure will drop, as there is no flow.



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In the momentary absence of frictional pressure losses,

the bottom-hole fracture treating pressure can be

calculated with greater precision from the expression

Where pi is the instantaneous surface shut-downpressure.

In addition, the frictional pressure losses in the system

(pt + pp) can be measured indirectly, by use of theequation

.

(4.22)4TUNIO, May' 2011,,, Courtesy AP Aung


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Since the pressure drop across the perforations is usually

small in comparison with the other pressure terms, one

can assume that it is negligible and obtain the surface

injection pressure as

Since pf and pp in Eq (4.20) are influenced by the flowrate, the surface injection pressure will also be dependent

upon rate.

(4.23)



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Bottom-hole Treating Pressure

The bottom-hole treating pressure ptis determined fromthe fracture gradient Gf and the depth D to the fracture,since

(4.25)

(4.24)



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Hydrostatic Pressure

The hydrostatic pressure ps is obtained from the densityof the fluid including the propping agent.

(4.26)

7

where

is the specific gravity of the fracture fluid,

x is the concentration of sand in pounds per gallon of fluid,

2.63 is the specific gravity of sand.

TUNIO, May' 2011,,, Courtesy AP Aung


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If the specific gravity of the fracturing fluid is given at

60oF, the specific gravity at the average well temperature

Tis calculated from


(4.27)



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Where is the thermal coefficient of expansion of thefluid.

If the fracture fluid is a crude oil, the value of is

obtained from volume correction tables.


The hydrostatic pressure ps is equal to the hydrostaticgradient times the depth,

(4.28)



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Pressure drop across the perforations is obtained directly

from Bernoullis equation and converting necessary units.

Pressure Drop Across Perforations

.

Where in pounds per gallon, q in gallons per minute, Ain square inches, and pressure in pounds per square inch.



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This type of calculation involves consideration of the

following variables:

Fracturing fluid coefficient C

Injection rate q

FRACTURE DESIGN CALCULATION

Total injection volume V

Area of the fractureA

Weight of propping agentS

Surface injection pressure ps Horsepower required Hh

Productivity ratio of the well PR



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Example (4.5)

Design a fracture treatment assuming an injection rate,

size of treatment, and fracture gradient, given:



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Example (4.5) the frictional pressure drop is 1826 psi

volume per unit area of the fracture is

0.00833 cu ft/sq ft porosity of most fracturing sand is of the

order of 35 per cent

The specific gravity of sand is 2.63 and the

density is 2.63 x 62.4 or 164 lb/cu ft

equivalent diameter (de) of the circular

pipe is 3.62 in

Solution

A review of past treatments in the area shows that a

treatment of 40,000 gal of lease oil at an injection rate of30 bbl per minute has given effective results in this

reservoir, also, it is noted that the fracture gradients have

been approximately 1.0 psi per ft. 14

Fracture pressure is 1826 psi

TUNIO, May' 2011,,, Courtesy AP Aung


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Example (4.5)

With this fracture gradient, the bottom-hole treating

pressure is

Then the differential pressure across the fracture face is

Since lease oil is to be used as the fracturing fluid, the

fracturing fluid coefficient is Cc



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Example (4.5)

The pumping time for 40,000 gal of fluid at a rate of 30

bbl/min is

Solving forx

And the efficiency from Fig (7.11) is 31 per cent. Then from

Eq (4.18) the fracture area is



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Example (4.5)

Then the weight of sand, S, necessary to pack one squarefoot of fracture is

The maximum amount of sand needed is



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Example (4.5)

And the sand concentration in pounds per gallon of oil is

176,600/40,000 or 4.42 lb/gal.

The actual flow rate (including sand) is the total volume

(oil plus sand) divided by the total fracture time, or



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Example (4.5)

The coefficient of thermal expansion for oil approximately

0.0005 per degree. The specific gravity of the oil at the

average well temperature is

And the density is

The hydrostatic pressure developed at 2000 ft is



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Example (4.5)

The average flow velocity in the equivalent pipe is

20

The Reynolds number of the flow in the equivalent pipe is

found from



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Example (4.5)

Then the surface injection pressure, neglecting the

pressure drop across the perforations is

Since for 5 in, J-55 casing 60 per cent of the bursting

pressure (5320 psi) is 3724 psi, the injection pressure used

21

s w n e sa e m . e y rau c orsepower scalculated as

On the basis of 600 hydraulic horsepower per pump truck,

five pumpers are required for the job.

To calculate the productivity ratio, several assumptions

must be made.TUNIO, May' 2011,,, Courtesy AP Aung


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Example (4.5)

Since the gradient is 1.0 psi per ft and the well depth is

2000 ft, it is assumed that the fracture plane is horizontal.

2

= =, ,ft. the ratio of fracture radius to external radius is rf/re =252/660 = 0.381.



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Example (4.5)

From Fig (4.7) the permeability of 10-20 mesh sand for a

fracture treating pressure of 2000 psig is 60,000 md, the

value ofkfW/ k h is

From Fig (4.6) the PR is 5.0, so that, if the initial production

was 10 bbl per day, the production after fracturing is 50 bbl

per day.



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