Units and Conversion Factors Used in reservoir technonlogy.

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Units and Conversion FactorsUsed in reservoir
technonlogy.
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In this module you will learn about
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Topic overview
1 Introduction
Units
2 Standard Units
4 Summary
3 Converting equations to fit input units
2.1 Gravitational conversion factors
3.1 Capillar pressure over free water level
3.3 The line source solution
3.2 Darcys law
Convert data units to fit equation units
Use converted units in equation
Single Approach
Convert equation to fit data units
Data fits directly into equation
Multiple Approach
SI System International
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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.

Units page on W3
Go to section 2
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Section 2 Standard unit systems

• 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.

Table from L.P.Dake 1
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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 Newtonss 2 law, F ma , a
  mass of one kilogram would exert a force of 9,81
  Newton on the earth at normal conditions. Lets
  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 g9,81 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. Its 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.

Back
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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 Am2
  L1m L2m. 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. Lets convert L1 to metre
  L1cm10-2m/cm. Now you can put
  L1 data into the equation. We do the same with L2
  but this is millimetre
• L2cm10-3m/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 Am2 L1 10-2cm L2 10-3mm
• 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 trick in converting equations is 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

We want to be in mm/s , to be
miles/hour2 , 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
This equation is not valid until we have
converted the units to the original ones, this
way we get the appropriate conversion factors
Back
Now we collect all the factors and write the
equation in the new units
And this equation is only valid for the given
conversion factor, and units.
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Section 3.1 Capillar pressure over Free Water
Level

• The expression for capillary pressure over free
  water level (fwl) is given by

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 )
We cancel out the gs on Both sides, because
converting One of them to the other Leads to
canceling.
We used a intermediate step her to get 122
because that way we get a fractional number 1/144
instead of 0,006944
Look here if you need to check up the conversion
factors used
As you can see her, we use pound force and pound
mass, leaving the gs out
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Section 3.2 Darcys law

• Darcys law

Valid for k darcy, cp, A cm2, q cm2/s, L
cm, p atm
Shall be used with Oil Field Units k md, cp,
A ft2, q bbl/d, L ft, p psi
We use the general rule Start with the new
units, convert to the old
Arranging factors gives us
If we dont write the new units, into the equation
like this
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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Section 3.3 The line source solution

• The line source solution for well testing

Before After
Darcy units Practical units
p atm psi
q cm3/s bbl/d
k D md
r,h cm ft
c 1/atm 1/psi
t s hr
And its associated units before and after
conversion
conversion
This equation valid for P psi, q bbl/d, k md,
(r,h) ft, c 1/psi, t hr
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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
  companys have implemented the SI units as their
  internal units language, but they still have to
  cope with the units used by other companys when
  exchanging data and statistics.
• The abillity to convert equations to fit any kind
  of input data, is a necessity
• when working with data from different suppliers.
• Also when reading litterature, different units
  systems are used.

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Developers
Module made by
Student Odd Egil Overskeid Petroleum Technology Dept. Stavanger University College Norway
Topic author and coordinator Professor Svein M. Skjæveland Petroleum Technology Dept. Stavanger University College Norway http//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