How a fracture geometry affects production:

The goal of determining a fracture dimensions is so that we can find the optimal

width and length of a fracture that would allow us to maximize the production, as

that’s the whole purpose of the petroleum industry.

First we look at the Dimensionless productivity index, JDpss

It’s formula is

1

ln re+0.75−0.5 lnk fV f

2k h f+0.5 lnC fD+ ln

x frw

+sf

𝑟𝑒, 𝑘𝑓, 𝑉𝑓, 𝑘 ℎ𝑓 are usually given, and thus can be considered constant. We need

to concentrate on Cfd, sf, xf, and rw. We want to make those values small, since

they’re in the denominator, and that will make the whole equation bigger, which is

what we want. For that we are going to plot ln ( x frw )+sf vsC fD. On the graph we need

to see at which point the function reaches its minimum, and that will tell us the

maximum for the Jd.

On the following graph you can see that Cfd is 1.6 when the abovementioned

function is at its minimum.

1.6

Now that we know the optimal fracture conductivity, Cfd, it tells us that fractures

whose conductivity is more or less than 1.6 will correspond to a different

Productivity Index, which means it will perform less. Knowing the Cfd is very

important as it will allow us to determine the width and length of a fracture. Those

can be calculated with the following formulas:

• x f opt=√ k f V f

2C fDopt khwopt=√C fDopt kV f

2k f h

Proppant Number:

• For a given proppant volume, well drainage area, proppant and

reservioir permeabilities, there exists a proppant number, Np

N p=2k fV f

kV r

It is also important to know it as it helps us to identify Cfd value. On the next

2 pictures there are graphs that represent Cfd vs Jd that correspond to a

particular Proppant Number.

The 1st graph is for Np values less than 0.1. As we can see from the graph,

the maximum of those function occurs when Cfd is 1.6. That again tells us

the Maximum Productivity index will be at Cfd=1.6.

The next graph is for Np>0.1. As Np gets bigger, due to increased propped

volume or lower reservoir permeability, Cfd also increases. It is governed by

the formula

1.6+exp−.583+1.48 ln (Np )1+.142 ln (Np )

, for 0.1≤Np≤10.

¿Cfd=Np for Np≥10.

Example on how to calculate optimal dimentions and maximum productivity

index.

Solution:

Given Data: k = 15 mdB

o= 1.1 rsb/stb

μ= 1 cpMass of proppant (Mp) = 150,000 lbProppant specific gravity = 2.65Φ(Porosity) = 0.38K

f = 60000md

h(reservoir) = 50ft h(fracture) = 100ft

Area = 4*106

sq.ft.JD

of unfractured well=0.12

Find optimal dimensions, pseudosteady-state productivity index and folds of increase from fracturing.

1st

we find the propped volume in the pay (Vf) and reservoir volume (Vr)

V f=

−hh f

(Mp )

62.4∗(1−φ )∗ρ p=

50100

∗150,000

2.65∗62.4∗(1−0.38)=732 ft

3

Vr= A*h= 4*10^6*50=2*10^8 ft3Proppant number: N p=

2∗k f∗V f

k∗V r

=2∗60000∗73215∗2∗108

=0.0293

N p is smaller than 0.1, therefore the maximum fracture conductivity is 1.6(¿C ¿¿ fDopt)¿, and

the formula to calculate the maximum dimensionless pss Productivity index is:

JDmax pss(Np)=1

0.99−0.5∗ln(Np)= 10.99−0.5∗ln (0.0293)

=0.36

x f opt=√ k f V f

2C fDopt kh=√ 60000∗732

2∗1.6∗15∗50=¿135 ft ¿

wopt=√C fDopt k V f

2k f h=√ 1.6∗15∗7322∗60000∗50

=0.054 ft=0.65 inch.

Finally, to calculate maximum pss PI:

J= kh141.2B μ

JDmax pss=15∗50

141.2∗1.1∗10.36=1.75 STB/d/psi

Initial JDwas 0.12. To find folds of increase: 0.360.12

=3.

That means that productivity increased 3 times.