units and conversion factors used in reservoir technonlogy. in this module you will learn about...

Units and Conversion FactorsUsed in reservoir technonlogy.

In this module you will learn about

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References Summary

1 Introduction

2 Standard unit systems

3 Converting equations to fit data units

4 Summary

Topic overview

2.1 Gravitational conversion factors

3.1 Capillar pressure over F.W.L

3.2 Darcy´s law

3.3 The line source solution

UnitsUnits

Topic overview

1 Introduction

2 Standard Units


3.2 Darcy´s law 3.3 The line source solution

3.1 Capillar pressureover free water level

4 Summary

Convert data units to fit equation units

Use converted units in equation

Convert equationto fit data units

Data fits directlyinto equation

SingleApproach

MultipleApproach

3 Converting equations to fit input units

SI System International

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3.2 Darcy´s law


Section 1: Introduction

You are probably familiar with the international SI-system of units. However, it is not used in many everyday situations. Take an example: You measure the time mostly in days, hours , and minutes. But according to the SI-system, time should be measured in seconds.Take another example: You measure the speed of your car in kilometers per hour, not in metre per second.

How come that the fundamental units of the SI-system are not used in everyday situations? Some of the answer leads to habits, an some of the answer leads to the visual impact of the units used.

If you where to measure all time in seconds, you would get an awful lot of digits to manage So what do you do ? You bundle up some seconds and give it a name: 1 hour, 1 day ..... 1 Year. The advantage is that you get few numbers to keep up with. You could call this technique "downscaling" or "packaging" of the data, to make it more easy for people to understand the dimensions of the count. When you make a bread at home, the "units" are spoons and cups rather than cubic metre.

Many of the units in use to day are only plain old habits and perhaps not particularly practical in egineering! As an engineer, perhaps in the petroleum industry, you have to be able to cope with the different units in use.

In this module we will show you some of the common units you will encounter when facing the petroleum industry in everyday situations.More special, and rewarding, you will learn a smart technique to make equations fit your data units. This way you can put the numbers directly into the equation without any prior conversions, more about this in section 3.

Section 2 will start giving you some of the more common units in the oil industry. The value of these units in SI system will be shown in examples.

Go to section 2

Units pageon W3

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3.2 Darcy´s law


The oil industry makes use of at least four different system of units.These are:

•The U K (Imperial) System of Measurements [absolute units]- This was earlier the cgs system (centi, gram, second)

•The Metric System of Measurements [SI] [absolute units]•The U S System of Measurements [absolute units]•Oil Field Units [ OFU] [hybrid units]

Actually in practical diciplines these units are often mixed to the big gold-medallion and are called field units. Here are shown some selective units cropped from the systems above.

Section 2: Standard unit systems

Table fromL.P.Dake [1]

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3.2 Darcy´s law


Section 2.1: Gravitational conversion factors

Sometimes one has defined a number to have the value 1, just as one defines an hour to be 60 minutes. This can be confusing when you are familiar with the SI system with consistent units.

Example: According to Newtons´s 2 law, F = ma , a mass of one kilogram would exert a force of 9,81 Newton on the earth at normal conditions. Let´s say we defined this 9,81N to be one kilogram force, this way we could say that a one kilogram mass, would have a one kilogram force on it. This is actually the case when we in everyday language says that "my weight" is 72 kilograms. Because weight is synonymous with force, you are actually saying: "i have a force of 72 kilograms working on my body".Now someone would protest and say "you cannot use the same symbol for both mass and force!", but you can if you have defined it that way!So a one kilogram mass would have a one kilogram force acting on it.As you can see, you can not use these two relationships when calculating in SI units, because the force are measured in Newtons not kilograms.So to get the "72 kilogram" force concistent with SI units you would have to multiply it with 9,81 and get 706,32 Newtons.

In this case g=9,81 is an example of an gravitational conversion factor.

The purpose of the example above, is that in the British system, the units for mass and force are the same, and are called Pound. It´s often referred to as Pound mass and Pound force ( lbm, lbf ).While this approach works fine in the British system, one has to use a gravitational conversion factor when converting to SI or other concistent units.

Next page show an example of converting pressure in PSI ( Pounds per square inch) to Pascal

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Section 2.2: Gravitational conversion factor

To convert a pressure measured in PSI to Pascal, one has to make use of a gravitational coversion factor which expresses that 1 lbf is equivalent to 32,2 lbm ft/s2. Also, we use that 1ft is equal to 30,48 cm and that 1 N = 1 kg m/s2.

]in

)ft/s(lb2,32[]

in

lb[psi][

2

2

2mf ppp

]m

N[00668,0]

m0,0254

m/s4,452kg[]

in

m0,0254*in

]s

ft

m0,3048ft

[lb

kg4536,0lb 2,32

[222

2

2

222

2

ppp mm

)lbft/(s lb 32,2 2fmcg

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3.2 Darcy´s law


Section 3: Converting equations to fit data units

Introduction:

Take an example: You are a lab-assistant measuring to sides of a rectangle, the one side side L1 you measure with a centimeter scale, the other side L2 you measure with a millimeter scale. There are hundred rectangles to be measured, and you deliver the results to an engineer needing the data for further calculations. The problem is that the enginer calculates the rectangle by the formula A[m2] = L1[m]• L2[m]. So every time you give him two measurements, L1, L2 , he has to convert the centimetres and millimetres to metres, so he can put the data in to his equation. With this method he has to convert 100 pairs of data to metre units. So, is there any easier way to do this ?

Think of an arbitrary number L1 measured in centimeters. Let’s convert L1 to metre: L1[cm•10-2m/cm]. Now you can put L1 data into the equation. We do the same with L2 but this is millimetre:

L2[cm•10-3m/cm]. Now if you put ”data-numbers” into L1 and L2, they will convert to meters.

The trick is now that you put the converted L1 and L2 into the equation, remove the [m], and only keep indicated which units to put into the equation: A[m2] = L1 10-2[cm]• L2 10-3[mm] =

10-5 L1 [cm]• L2 [mm], and voila! Every time you put data of cm and mm into the equation, you find the area in m2.

Oops!: When using this technique you must always tell the user of the equation wich input data the equation is valid for or else he will bite the dust. You will see in section 3.2, 3.3, 3.4 how we

concistently convert the equations to fit the data, and how we report what kind of data the equations are valid for.

The big trickin converting equationsis to learn the sentence:Start with the equation in your selected units and convert it to the original units

Click here for an example!

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Example of converting an equation

We start with a simple linear SI unit equation for speed:

]s[]m/s[]m/s[]m/s[ 20 tavv

We want to be in mm/s , to be miles/hour^2 , and t in weeks! But still we want the output (v) to be in m/s

We start by putting the input units into the equation:

0v a

]weeks[]hr

miles[]

s

mm[]m/s[

20 tavv

This equation is not valid until we have converted the units to the original ones, this way we get the appropriate conversion factors:

]Week

3600S127Week[]

hr

s)3600(hrftmile

0,3048mft5280miles

[]sm

mm10m

[]m/s[

2

22

3

0

tavv

Now we collect all the factors and write the equation in the new units:

]weeks[]hr

miles[]

s

mm[106,37]m/s[

203 tavv

And this equation is only valid for the given conversion factor, and units.

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3.2 Darcy´s law


Section 3.1: Capillar pressure over Free Water LevelThe expression for capillary pressure over free water level (fwl) is given by:

ghp c

And the default units are SI: , [pascal]

To use the equation in field units (PSI, Square-inch, Foot) one has to convert it using a gravitational conversion factor ( the same factor as mentioned in section 2.2 ):

]kg/m3 ]m/s[81,9 2g cp

]kg4536,0

c2in

mlb

c2in

2ft/sm

lb2,32

2in

flb223cccc m)1054,2(

[][][][

ggpppp

cg

]ft

0,3048mft[]m/s[]

3ft

m3048,0ft

kg4536,0m

lb]

kg4536,0c 2

333

ftin12

m3048,02ft2inc

222

22[

hgg

p

144

ft][]3ft

mlb[

PSIc ][h

p

We used a intermediate step her to get 12^2 because that

way we get a fractional number 1/144 instead of 0,006944…

As you can see her, we use pound force and pound mass, leaving the g´s out

We cancel out the g´s onBoth sides, because convertingOne of them to the otherLeads to canceling.

Look here if you need to check up the conversion factors used

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3.2 Darcy´s law


Section 3.2: Darcy´s law

Darcy´s lawL

pkAq

Valid for: k darcy, cp, A cm^2, q cm^2/s, L cm, p atm

Shall be used with Oil Field Units: k md, cp, A ft^2, q bbl/d, L ft, p psiWe use the general rule: Start with the new units, convert to the old:

]ft

cm30,48ft[

]psiatm

106,8046psi[]ft

cm102903,9ft[]

mdD

10md[]

daysec

86400d

bblcm

10159bbl[

ft][

psi][]ft[md][]

d

bbl[

2-

2

2323-

33

2

L

pAkq

L

pAkq

Arranging factors gives us: L

pkAq

001127,0

If we dont write the new units, into the equation like this:ft][

]psi[

]cp[

]ft[]md[001127,0]bbl/d[

2

L

pAkq

We have to tell explicitly in the text what units the equation are valid for, this will be done in the next section when converting the Line Source Solution

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3.2 Darcy´s law


Section 3.3: The line source solution

The line source solution for well testing:

)4

ei(),(2

kt

crqptrppi

And it´s associated units before and after conversion:

Before After

Darcy units Practical units

p atm psi

q cm^3/s bbl/d

k D md

r,h cm ft

c 1/atm 1/psi

t s hr

)psi

atm(

14,696

1psi:p

ds

6060

bblcm

10159

day

bbl:

33

Q

ft

30,48cmft: h

md

d0,001md: k

psi/atm)14,696psi/(1: c

hr

s6060hr: t

)00105,0

ei(6,702

kt

crqp

This equation valid for:P: psi, q: bbl/d, k: md, (r,h): ft, c: 1/psi, t: hr

conversion

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3.2 Darcy´s law


Section 4: Summary

The oil industry is a multinational business. The units used in a particular company may well express their country of origin. Some oil company´s have implemented the SI units as their internal units language, but they still have to cope with the units used by other company´s when exchanging data and statistics.

The abillity to convert equations to fit any kind of input data, is a necessitywhen working with data from different suppliers.Also when reading litterature, different units systems are used.

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3.2 Darcy´s law


Developers

Module made by

StudentOdd Egil OverskeidPetroleum Technology Dept.Stavanger University CollegeNorway

Topic author and coordinatorProfessorSvein M. SkjævelandPetroleum Technology Dept.Stavanger University CollegeNorwayhttp://www.ux.his.no/~s-skj/

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References

Units table section2 taken from:[1] L.P. Dake [fundamentals of petroleum reservoir engineering]

units and conversion factors used in reservoir technonlogy. in this module you will learn about...

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