tech bulletins cable

7/25/2019 Tech Bulletins Cable

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Technical Bulletins

"Quality You Can Pull On..."


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Calculating Cable Length by Measuring Conductor Resistance

L1 = Length of new cable installed on the Drum in feet (ft).

R1 = Total resistance of center conductor of installed cable in Ohms ()T1 = Temperature of cable when R1 is measured in degrees Fahrenheit (F)r = Resistance of this cable per foot at 65F. This factor should be calculated and recorded in the cable log book.

To convert from metres to feet: 1 foot = 0.3048 metresTo convert from Celsius to Fahrenheit: TF = (TC x 1.8) + 32

r = (R1 / L1) x [(302.5) / (234.5 + T1)]

Example:L1 = 25,500 FeetR1 = 265.4 T1 = 75Fr = (R1 / L1) x [(302.5) / (234.5 + T1)]

r = 0.01017 /ft

This value of r, should be recorded in the Cable Log Book.

As sections of cable are cut off, the remaining length, L, of cable on the drum can be calculated using the recordedvalue of r, the Resistance Rof the centre conductor of the remaining length, and the Temperature, Tof theremaining length of cable.

L = (R / r) x [(302.5) / (234.5 + T)]

Example:r = 0.01017245 (/ft)R = 195.4 T = 92F

L = (R / r) x [(302.5) / (234.5 + T)]L = 17796.7

The length of the cable is now 17,796 feet.

To obtain the best results, it is recommended to use a Fluke or equal quality 4 or 5 digit 1% type Ohmmeter with goodleads. It is Important to use the same Ohmmeter to establish the value of rand later measurements of R.

Technical Bulletin Number 001

05/2005


3/25


Locating Electrical Leaks (Part One)

Electrical leaks, or the break down of the insulation of the conductor, are typically caused by one of the following reasons:

Physical Damage: The cable, jumped the sheave wheel, over-run in the hole, drum crush, kick back from a large gun, cut-inwhile spooling cable, or other accidental mechanical damage.Excessive Temperature: Operating the cable at bottom hole temperatures in excess of the maximum temperature rating of

the cable. High tensions at maximum temperature.Excessive Tension: Repeated tensions over 60% of rated breaking strength or a little as one pull in excess of 75% of rated

strength of the cable.Manufacturing Defects: Inner armor coverage less than 97%, Non uniform spacing of inner armor wires, eccentricity in

conductor insulation, Crossed inner armor wires, Tape lap joints or string filler knots in multi-conductor cable.

There are basically three different types of electrical leaks:Dead short: The resistance , or leak, is less than 100 Ohms.Hi resistance: The leak is as high as 20 Meg-Ohms.Intermittent leak: This is the worst type as the leak at times disappears.Wet leak: Any of the above 3 types of leaks can have moisture present which can complicate the location of the leak.

Moisture in the leak can generate a small voltage between the copper conductor and the zinc of the armor wire, which will

give misleading resistance measurements.

The fastest way of locating an electrical leak is to burn it outwith a high voltage, high current source* and in doing thisyou dry the leak and reduce it to a dead short. There are several methods of easily locating a dead short. When it isknown that the cable has been abused in some way, tension or temperature, then burning out the leak is the bestprocedures. If, however you have a fairly new cable and a factory defect is suspected, the burn out method should not beused as it completely destroys the area around the leak and the exact cause of the leak can not be determined. There aremethods of locating leaks within a few inches without first burning out the leak as will be covered in later technical bulletins.

Part-1, Method for locating a dead shortleak. A dead short leak, as described above, is when the copper conductor is indirect contact with the armor or the resistance between the conductor and armor is less than 100 ohms. The only instrumentrequired is an accurate digital ohmmeter that reads to at least 0.1 Ohm. Before any leak location process is started be surethat both ends of the cable conductor are completely disconnected from any tools , collector, etc. and clean!

After both ends of the conductor are cleaned, using the digital Ohmmeter, measure and record, the following resistances.

R: Total conductor resistance, end to end, Ohms.Rt: Resistance between conductor & armor measured at the Truck end, Ohms.Rw: Resistance between conductor & armor measured at the whip end, Ohms.L: The total length of the cable, feet.

To be sure your problem can be classified as a dead short, make the following calculation:((Rw + Rt) - R) < 300 Ohms

If the this calculation is greater than 300 Ohms, you do not have a dead shortand you should use another method oflocating the leak, (Part 2 & 3), or burn outthe leak to obtain a more direct short. If the above calculation results with avalue less than 300 Ohms then the leak location can be calculated as follows:

Lw: Distance of the leak from the whip end of the cable, feet.Lt: Distance of the leak from the truck end of the cable, feet.

Lw = [ R + Rw Rt ] x [ L / 2R ]Lt = [ R + Rt Rw ] x [ L / 2R ]

Example:Cable length:

L = 20,500 feetR = 220.5 OhmsRw = 165.0 OhmsRt = 325.5 Ohms

Leak Location:Lw = [ 220.5 + 165.0 325.5 ] x [ 20,500 / 2x220.5 ] = 2,789 feet from whip endLt = [ 220.5 + 325.5 165.0 ] x [ 20,500 / 2x220.5 ] = 17,711 feet from truck end

Methods for locating high resistance electrical leaks and eliminating the effects of a wet leak will be covered in later Technical Bulletins. 05/2



4/25


Locating Electrical Leaks (Part Two): Leak Locator Bridge

The most common locations of cable electrical leaks are:

1. Within 1000 feet of the whip end due to physical damage.2. A drum crush, two or three layers down on the drum.3. Failure at the dog knotwhere the drum end of cable

passes through the flange.4. Damage on the bed layer due to installation problems.

The simple leak locator bridge is one of the fastest methodsof locating the approximate location of a leak. This isimportant as it will quickly indicate whether the leak can becut off the whip end or the cable will need to be strung upand pulled down to a location on the drum. The simple leakbridge circuit, shown below, can locate leaks within about +/-200 feet if the leak is less than 10,000 Ohms. Moresophisticated leak bridges can locate a leak as high as 1 Meg-Ohm to an accuracy of +/- 50 feet.

Remember before any type of leak location procedure is startedbe sure that both ends of the cable are completely free andclean. When the Micro-ammeter reads zero, the percentageindicated on the potentiometer will be the same as thepercentage of the cable length to the leak. Just how this workscan be further understood by referring to the set up andoperation of a prototype leak locator instrument.

The leak bridge shown here is available from CSR in Rosenberg,Texas. Web site: www.csrusa.net

Operating Instructions Leak Locator Bridge

Setup: Check that both ends of the cable are free, clean and NOT

connected to anything. Turn power switch OFF.

Check and adjust the ZERO on the meter with the small

screw on the meter face. Plug one test lead into the RED WHIP socket and clip thislead to the WHIP end of the cable.

Plug one test lead into the BLACK TRUCK socket and clipthis lead to the TRUCK end of the cable,

Plug one test lead into the GREEN ARMOR socket and clipthis lead to the cable ARMOR.

Rotate knob A to read 50. Connect the AC power cord and turn power ON.

Operation: With the power on, the meter needle should deflect off

of zero. If a very small or no deflection occurs, starting with button

1 hold down and observe the deflection of the meter.

If the meter deflects to the left, WHIP, rotate knob Acounter clockwise, until the meter reads zero. If the meter deflects to the right, TRUCK, rotate knob A

clockwise, until the meter reads zero. Hold down button 2 until the meter again reads zero.

Repeat this with button 3 held down. Record the final dial reading, 0.252 in the example

shown below.

Leak LocationMultiply the dial reading, 0.252 by the length of the cable. Forexample with a cable length of 20,000 feet, the leak is located at:

0.252 X 20000 = 5040 feet from WHIP end

Accuracy This type of instrument will only locate a leak to within +/-

200 feet. For maximum accuracy:

1. Burn out the leak to the lowest resistance possible2. Be sure all test leads fit firmly in sockets3. The lengths of the WHIP & TRUCK test leads must be equal.4. The test leads should be as short as possible

Check for a wet leak. This is done by measuring theresistance of the leak with an Ohm meter and then reversingthe leads of the Ohm meter and see if the leak resistance isthe same. If the leak resistance is the same you are ok. Ifthere is a significant difference between the two readings,you need to dry out the leak with a burn out box.

05/2



5/25


Precise Method for Locating Electrical Leaks

The most common method of locating electrical leaks in a

wire-line is to burn outthe leak until the conductor isshorted to the armor and then use a digital ohmmeter todetermine an approximate location of the leak. This is aquick and easy method of finding a leak; but, there is adisadvantage. When burning out the leak, it can melt thecopper, plastic and sometimes burns the inner armor wires.If this occurs it can be impossible to determine the cause ofthe original leak.

If it is important to determine exactly what caused the leakor if the high voltage equipment is not available to burnout the leak, then a alternate leak locating procedure isrequired. This procedure will locate leaks that are as highas 10 Meg-Ohms and locate the leaks within +/-1 inch. To

use this method, the following shop conditions must exist:

The shop is setup to reel the cable from a metal pay offor truck drum to a metal take up shop drum.

The pay off frame is electrically insulated from the takeup stand.

A POWER SOURCE, such as a 6 or 12 volt car battery or abattery charger is available, with leads long enough toconnect the power between the pay off and take up stands.

A collector is mounted on the pay off, truck, drum and iscalled TRUCK DRUM.

A DC voltmeter that will indicate plus and minusvoltages and has 200mv sensitivity and 10 Meg Ohmsinput impedance is required.

A set of well insulated test leads to reach between themeter, the collector on TRUCK DRUM and one test lead,fixed with a copper hook, to reach the cable.

1. If the meter indicates a POSITIVE voltage, the leak istowards the SHOP DRUM.

2. If the meter indicates a NEGATIVE voltage, the leak istowards the TRUCK DRUM.

Setup Procedure String the cable between the pay off, truck, and shop

drums. Going around a sheave wheel is ok as long as thissheave wheel is electrically insulated from both pay offand take up drums.

Connect the leaking conductor to the collector on

TRUCK DRUM.

Be sure the insulation on the conductor is clean.

The other end of the leaking conductor, on the SHOPDRUM, must be free and clean.

Turn the meter on, set it at the lowest dc voltage range,200 mv.

Connect the Meter NEGATIVE, long test lead to thecollector, mounted on the TRUCK DRUM.

The test lead with the copper hook is connected to theMeter POSITIVE, and the copper hook is hung on thecable between the pay off and take up.

Connect the POSITIVE lead of the POWER SOURCE to theframe of TRUCK DRUM.

Connect the NEGATIVE lead of the POWER SOURCE tothe frame of SHOP DRUM.

DO NOT CONNECT THE POWER SOURCE DIRECTLY TO

THE CABLE ARMOR, ONLY TO THE FRAME. The powersource can arc when connected and burn an armor wire.

Leak Location With all of the connections made in accordance to the

SETUP PROCEDURE listed above, place the copper hookfrom the meter positive in contact with the cable armor.

The meter will indicate a positive voltage if the leak ison the SHOP DRUM

The meter will indicate a negative voltage if the leak ison the TRUCK DRUM.

The cable is then spooled in a direction to move the leakoff of the drum it is located on.

The copper hook, connected to the meter positive

terminal, is held in contact with the cable armor as it isbeing spooled.

When the leak comes off the drum, the meter voltagereading will start to decrease.

When the meter reads zero, the leak is located directlyunder the copper hook location.

If the meter voltage changes polarity, you have passedthe leak.

Connection Summary TRUCK DRUM has the collector Connect leaking conductor to the collector Connect NEGATIVE METER lead to the Collector TRUCK Connect the test lead with the copper hook to the

METER POSITIVE Connect POSITIVE power source to frame of

TRUCK DRUM. Connect NEGATIVE power source to frame of

SHOP DRUM

Miscellaneous Be sure the plastic insulation is CLEAN on both ends. The pay off and take up stands must not be connected

mechanically. Do not connect the power source leads directly to the

cable armor. This connection can spark when connected,burning cable armor.

05/2

(Pay-off) (Take-up)



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Problems in Locating Electrical Leaks

Bulletins 002, 003, 004 have described three methods of locating electrical leaks in electromechanical cable. There are

four problems which when encountered increase the difficulty in locating the leaks.

1. Wet Leaks: An electrical leak occurs when there is a rupture in the plastic insulation surrounding the copper conductor.Sometimes there is direct contact between the copper and the armor wires, (dead short), other times there is burntplastic from the break down process that leaves a carbon trail between the copper and armor, and in other cases thereis moisture in the cable between the copper and armor forming a wet leak.

The armor wires are covered with a Zinc coating. Zinc is above and Copper is below Hydrogen in the electromotiveseries of metals, so when they are in a conductive medium, such as salt water, there will be a voltage generatedbetween them. In the case of Zinc and Copper the voltage is about 0.83 volts. If you would like to run a little physicsexperiment, simply clip a short piece of armor wire to the negative lead of a digital voltmeter and the positive lead to apiece of copper wire and place them in a glass of salt water. The meter will indicate a voltage in the range of 0.83 volts.This voltage in a leak can significantly distort the location of the leak.

To determine if you have a wet leak, first measure the resistance of the leak and then reverse the leads of theohmmeter. If you have a wet leak there will be a significant difference between the 2 resistance measurements. Thewet leak can be dried out by repeated application of voltage from a Hi-pot or a burn out box.

If you are not equipped to burn out the leak, then using the resistance method, (Technical Bulletin 002), calculate thedistance to the leak, Lw, and then reverse the polarity of the Ohmmeter and again calculate the distance to the leakLw. Take the average of these two values, (Lw +Lw)/2 and this will be the correct location.

Some leak locator bridges have a polarity switch built in to obtain the values of Lw and Lw. Again the correct locationwill be the average of the two readings. If a reversing switch is not included, then swap the TRUCK and WHIPconnections to get the value of Lw. Again the correct location will be the average of the two values.

2. Very High Resistance Leaks: are leaks of greater than 10 Meg Ohms and require the repeated application of Hi-potvoltages of several thousand volts. Once the leak is less than one Meg Ohm, the burn out box can be used to reduce the

leak to several hundred or less Ohms. As mentioned in previous Bulletins, the burn out boxis a DC power supplywith an adjustable output up to 600 or 800 volts with a current capacity of 1 to 3 amps. This type of power supply isLETHAL, so must be used very cautiously!!!

3. Multiple Leaks: There is no straight forward way of locating multiple leaks. Generally when multiple leaks occur theyare all fairly close together. The best approach is to use any of the described leak locating methods and cut the cableand check both pieces of cable using the same locating methods and cut again. After all the cutting, be sure to Hi-potboth lengths of cable, to ensure they are clear of leaks before they are spliced back together.

4. Intermittent Leaks: are leaks that are not always present. These type of leaks typically are noticed during a job, andlater when the cable is brought to the cable shop, the cable tests clear. If the leak does not appear when the cable istested at 1,000 VDC, then to get the leak to reappear the cable is spooled back and forth in a number of ways whilewatching the ohmmeter to see when the leak occurs. Some shops pass the cable around a capstan, through a postformer or around several sheave wheels. Spooling is done at very low and then at very high tensions. Once the leakreappears the spooling is stopped and the leak is located using one of the standard methods.

05/2005



7/25


Increase in Cable Resistance with Wellbore Depth

The temperature of the earth increases with depth. The rate of temperature increase can vary from 10 to 25 degrees

Fahrenheit for each one thousand feet increase in depth. With the increase in temperature there is an increase in theresistance of copper conductors. The resistance of copper, in fact, will double at a temperature of 370F. This higherconductor resistance will directly increase the attenuation of signals and increase the power requirements to operatedown hole tools.

The plot shown below shows the overall increase in conductor resistance as a tool is lowered in the hole. For thisexample the total conductor resistance of a 25,000 foot Dakota Cable, type 7K-464-FTD, is shown as the tool is loweredin a 20,000 foot well with bottom hole temperatures of 500F and 300F.

Conductor Resistance Versus Depth

Graph is based on the following formula

For the above graph the following values were usedR = Total conductor resistance Ohms.r = Resistance per 1000 feet at 68F, (20C) Ohms/Mft, (9.8Ohms/Mft).L = Total length of cable Mft., (25Mft).d = Depth of cable in bore hole Mft.Hd = Hole depth to bottom Mft., (20Mft).Ts = Temperature at the surface degrees F, (75F).Tb = Bottom hole temperature degrees F, (300F & 500F).

To convert from metres to feet: 1 foot = 0.3048 metres

To convert from Celsius to Fahrenheit: TF = ( TC x 1.8 ) + 32

ConductorResistanceatDepth(Ohms)

Depth 1,000s of Feet

400

350

300

250

200

R = r (L - d)(234.5 + Ts)

d

HdTs + (Tb - Ts)

234.5 +

2

+ d(234.5 + 68) (234.5 + 68)

300F

0 5 10 15 20

500F

12/2005



8/25


Stuck Point Location

When a cable becomes stuck and will not move at the recommended maximum allowable tension, then the first step in

deciding what action to take is to determine where the cable is stuck. In cased hole work it is most commonly, but notalways, the tool that is stuck. In open hole operations, there is always the problem of the cable becoming key-seated inthe bore hole wall. In any situation it is best to make a quick check of the depth to the stuck point before deciding on thebest action to take.

The quick procedure for locating the approximate depth of the stuck point, (Ds) is as follows:

Pull on cable to remove all slack and put the cable under strain. Note and record the indicated depth from the measuring device, (D1). Note and record the tension in the cable. Increase the tension exactly 1000 pounds (4.44 kN) & record the indicated depth, (D2). Calculate the depth of the stuck point: Ds = (D1 - D2) / K (1000 feet). K is the stretch coefficient of the cable, which is listed in the Wireline Works Catalogue, ft/Kft/Klbs. To convert from metres to feet: 1 foot = 0.3048 metres

Nominal Values of Cable Stretch Coefficients:

Cable OD-Inches 3/16 7/32 1 / 4 9/32 5/16 3/8 7/16 15/32 0.49

K ft / Kft / Klbs 3.0 2.2 1.9 1.6 1.2 1.0 0.70 0.77 0.60

ExampleCable type-Dakota 1-R-322-FAH, 5/16Mono-cable ; K = 1.2Cable becomes stuck at an indicate depth of, D1 = 16500 ft.With the cable under strain the line tension is = 3,300 lbs.The tension is then increased to 4,300lbs and the indicated depth is D2= 16480 ft.

Ds = (D1 - D2)/K = (16500-16480) / 1.2 ( 1000 ft) = 16,600 feetIn this example the stuck point depth is close to the indicated tool depth, so it is the tool that has become stuck.

Depth CorrectionsIf a more accurate stuck point is important, then the following factors can be considered: The stretch, ( D2-D1) when measured at the truck includes the stretch in the cable from the truck to the well head. A

more accurate method of measuring the stretch is to mark the cable at the well head an then measure the stretchwhen the tension is increased.

If the rig-up distance is known, it can be subtracted from the calculated depth based on measurements of stretch atthe truck.

For well seasoned cables the stretch coefficient should be reduced by 5%. In very deep hot holes the effective value of K can increase by 10%. If there are reasons not to increase the tension by 1000 lbs, then just increase the tension by 500 pounds and then

take the value of Ds. calculated using the above formula.

05/2005



9/25


Cable Breaking Strength

New and properly maintained Dakota cables have been

designed and manufactured to have a breaking strength thatyou can depended on. This bulletin will discuss these cables

and how field operating conditions effect breaking strength.

A later bulletin will discuss the mechanical failure or reduced

breaking strength of cable due to effects such as fatigue, acid,

H2S, corrosion and wear.

The breaking strength of cable, listed in the Wireline Works

catalog, is the guaranteed minimum strength at which the

cable will break when the ends of the cable are prevented

from rotating. When a cable is loaded, with no rotation

allowed, the outer armor wires are stressed slightly more than

the inner armor wires. For this reason when a cable breaks

with ends fixed, the outer armor wires will always break first

and the inner armor wires stretch out before they break.

Oilfield cables are constructed with two layers of contra-

helically applied armor wires. Under load each layer of wires

develop torque. The torque developed by the inner armor is in

opposition to the torque of the outer armor. The torque

developed by each layer of armor wires is determined

primarily by the total area of steel in each layer and the

distance of the wires from the cable center. The outer armor

wires are always further from the cable center than the inner

armor and for practical reasons the outer armor layer has a

greater area of steel. The outer armor layer, therefore,

develops much greater torque than the than the inner armor

layer. This imbalance in torque can be partially but not

completely offset by adjusting the lay angles of the inner andouter armor wires.

If a cable under load is free to rotate, such as a cable hanging

in a vertical cased hole, the dominant torque of the outer

armor wires will cause the cable to rotate in such a direction

as to unwind the outer armor and reduce its stress. As the

outer armor wires unwind, the inner armor wires are forced to

wind tighter and this increases the stress in the inner armor. If

allowed, this unwinding will continue until the torque

between the layers is equal and when this occurs the stress in

the inner armor is much higher than in the outer armor. When

a cable is free to rotate or is forced to unwind by improper

operating conditions the breaking strength is significantly

reduced and when it does break, the inner armor will breakfirst. and then the outer armor wires will stretch out before

they break.

In normal operations, with proper tensions going in and out

of the well, the lower portion of the cable, if free to rotate

will unwind in proportion to the tension but due to friction in

the borehole, there is less unwinding near the surface, so the

cable breaking strength at the surface is close to ends fixed

strength. The breaking strength will be reduced further by

field operations that force the cable to unwind. This includes:

trying to control pressure with a tight pack-off instead of

using more flow tubes; wide cable tension variations that

result from allowing the cable to free fallinto the hole and

coming out of the hole at speeds that cause excessive hightensions; improper sheave grove size or sheave alignment can

also contribute to loosening the outer armor. When the outer

armor has become loose it is important to have a cable shop

normalizeand post-form the cable to tighten the outer

armor and restore its normal breaking strength.

Dakota cables are designed to exceed the catalog breaking

strength. All incoming armor wire has certified tensile

strength. In addition Wireline Works routinely tests the wire

and finished cables to verify the strength.

Calculating Cable Breaking StrengthEXAMPLE - Dakota Cable type 1-R-322-PH

( units-- inches, square inches, psi, pounds, degrees)D = 0.322 - Cable diameter

do=0.0445 - Outer armor wire diameter

Do= D - do = 0.2775 - Pitch diameter outer armor layer

di = 0.0445 - Inner armor wire diameter

Di = Do - do - di = 0.1885 - Pitch diameter inner armor layer

Dc= Di - di = 0.144 - Effective core diameter after compression

Ni = 12 - Number of inner armor wires

No= 18 - Number of outer armor wires

Ai = Ni(/4)di2 = 0.018663 - Total cross sectional area of all inner armor w

Ao= N0(/4)do2 = 0.027995 - Total cross sectional area of all outer armor w

Li = 1.50 - Inner armor lay distanceLo= 2.50 - Outer armor lay distance

sini = Di / [ (Di )2 + ( Li )

2 ]1/2 = 0.3672

sino = Do / [ (Do )2 + ( Lo )

2 ]1/2 = 0.3293

cosi = Li / [ (Di )2 + ( Li )

2 ]1/2 = 0.9301

coso = Lo / [ (Do )2 + ( Lo )

2 ]1/2 = 0.9442

i = 21.54 - Inner armor lay angle

o = 19.22 - Outer armor lay angle

P = 0.33 - Poissons ratio for plastic

S = 270,000 - Wire tensile strength

Breaking Strength - Ends Fixed

F = [ cos2 i - P(Dc / Di )sin2 i] / [ cos2o - P(Dc / Do)sin2 o ] =0.9- armor stress ratio, ends fixed

BF = S[ F ( Aicos i ) + ( Ao Cos o ) ]

BF = 11,370 - Calculated minimum

BF = 11,200 - Catalog minimum

Breaking Strength - Free RotationR= [ AiDisini ] / [ AoDosino] = 0.50496 - armor stress ratio, ends f

BR= S[ ( Aicos i ) + R ( Ao Cos o ) ]

BR= 8,120 - Calculated minimum

BR= 7,900 - Catalog minimum

05/2005



10/25


Drum Crush

The term drum crushrefers to a cable electrical failures that occur as a result of the cable being crushed, smashed or

distorted to such an extent that the armor wires press and distort the plastic insulation and in some cases cut through theinsulation and contact the conductor. It is possible to have the conductor insulation distorted to such an extent that itaffects the signal transmission characteristics of the cable with out an actual electrical short. These failures are caused bycables at high tension being spooled over cable spooled at abnormally low tension, cable installed with an incorrecttension profile or cable that is operated in a manner to cause excessive rotation. For these reasons a more correct term forthis type of failure would be Cable Crush Failure, ( CCF).

CCF, never occurs on the whip end of the cable. Typically the failures are found at a minimum of 4 or 5 layers down andmore commonly deeper than that. Failures can and often do occur in layers of cable that have never or not recently beenoff the drum.

An important characteristic of CCF is that the failure frequently does not occur immediately. For a failure to occur theplastic insulation must cold flow under pressure and this can be a slow process. A CCF condition in some cases may havebeen setup several jobs, days or even weeks before the actual failure does occur. It is this time lag that makes it difficult to

always identify the actual cause of the failure.

Factors that can contribute to CCF include:1. Poor drum cable entry hole.2. Irregular drum core.3. Spreading of drum flanges.4. Incorrect tension profile on initial cable installation.5. Single break cable installation.6. Loss of normal cable tension in field operations.7. Excessive cable rotation.8. Low or non uniform cable inner armor coverage.

A further explanation of these failures:1. A rough or bad angle of the cable entry hole in the drum can result in a CCF from the pressure of all the wraps on the

drum. This is a special case and is easily identified.2. Irregularities in the diameter of the drum core results in irregularities in the spooling pattern which causes pressure

points on the cable and distorts its shape. When the cable shape is distorted, it generates gaps in the inner armorpermitting easier cold flow of the plastic.

3. When the drum flanges spread there is as much as half of the diameter of the cable on each side of the top layer thenthere is a situation where the cable can cut in. When this occurs, the cable shape is distorted resulting in easier plasticcold flow.

4. There is no one size fits allwhen it comes to installing a cable on the drum. The correct tension profile that should beused depends on the type of cable, the cable length and expected depth of operations. In general after the bed layer isestablished the spooling tension is increased each layer for 3 or 4 more layers up to a tension of about 1/3 of the cablebreaking strength. This tension is maintained for half the cable length after which the tension is reduced eachsuccessive layer. This is just a very general rule and experienced cable service men in each area know how to adjustthese tensions for best spooling. If too much installation tension is used in shallow hole areas, then the cable will notspool properly at the low tensions in shallow operations. If the installation tension is too low, when installing a cable tobe used in deep hole operations, then the problem is more serious as CCF can result. When the installation tension islow, and the spooling tension coming out of the hole is high it can result in a CCF. The failure will typically occur severallayers down on the drum and at a cable cross over between wraps or at the flange where the cable moves from onelayer to the next.

5. When a cable is spooled on a drum the cable must move over one diameter distance for each wrap. If this move isaccomplished at one point in the wrap it is called a single break spooling. All qualified cable spoolers now use thedouble break spooling method, which moves the cable half a cable diameter midway around each wrap. The singlebreak results in a more sever distortion of the cable armor making it more susceptible to a CCF by over laying layers.

6. One frequent causes of a CCF resulting from field operations is re-spooling cable back on the drum after loss of normalcable tension. A common cause of loss of cable tension is over running the hole bottom. When this occurs it results inseveral wraps of the cable going back on the drum at low tension, followed by the high tension of cable when pulledoff of TD. This problem has become more frequent because of deviated holes, where it is necessary to approach TD very

Continued on next page05/2005



11/25


slowly to avoid over run. The cable can also loose normal tension when any restriction is encountered going in the hole.When this over run occurs the cable will start back on the drum at lower tension than the subsequent layers. If thistension differential is too great, it sets up a condition for a CCF. As explained above this failure is usually not immediate

and therefore if a cable has been over run it might be saved by bringing it to a cable shop and have a normal tensionprofile reestablished or by operating the cable in a deeper hole very soon after the previous problem. Any othersituation that causes a loss of normal spooling tension coming out of the hole, such as clamping off the cable to correcta spooling problem or clamping off at a side entry sub, can also lead to a CCF. There are some radical operatingconditions where the friction in the hole and tool are such that a wide variation of the in and out tension can not beavoid. In these cases special powered sheave or capstans must be used to reduce the tension differential and to avoidcable crush.

7. One factor in the ability of a cable to withstand cable crush is the hoop strengthof the armor wires. This is greatlyreduced when the outer armor unwinds and becomes loose. If the end of the cable is free to rotate then the outerarmor will try to unwind in proportion to the tension on the cable. This is the most frequent cause of failures inpressure cable.a. The cable should never be spooled out of the hole at a speed greater than a speed that results in a tension greater

than 125% of the static tension at that depth. Higher tensions result in excessive unwinding of the outer armor.Further, going into the hole the speed should never be less than that speed that maintains a tension greater than

75% of static tension. Failure to follow these rules will cause the cable to progressively unwind the outer armorsetting up a condition for CCF. When cables are operated outside these limits, the cable should be brought into acable service center for normalizing, (tightening the outer armor), and post forming. when the outer armor becomesloose. Examining the whip end of the cable, if you can easily move the outer armor with your finger nail or a smallscrew driver, then it is time for service.

b. Coming out of the hole the cable unwinds and going back in the hole the cable attempts to tighten back up. If aspring centralizer is used going into the hole a swivel head must be used to avoid sever unwinding of the outer armor.

c. The amount a cable unwinds under free conditions, depends not only on the tension in the cable but also thefriction between the armor layers. For this reason seasoned cables with mud and corrosion products between thearmor layers will rotate less and are less subject to CCF. New cables, however rotate very easily and new cable used inhigh pressure wells, with grease in flow tubes, must be carefully checked for loose outer armor. It would be goodpractice to have a new pressure cable brought to a service center to have the armor tighten after the first 20 or 30operations and thereafter when the outer armor becomes loose.

d. Alloy cables used in sour gas operations, represent a special problem. The alloy armor does not rust or produce

corrosion byproducts and these cables are normally used with high pressure grease in the flow tubes, so these cablesrotate more freely throughout the life of the cables. These cables MUST be brought into the cable shop regularly fornormalizing and post-forming. Good practice for these alloy cables would be to have the armor tightened every 20operations throughout the life of the cable.

e. Cables can be forced to unwind when pulled through a tight packer, dragging on any fixed object, run over a sheavewheel that does not have the proper grove or the sheave and truck are not properly aligned. Any forced unwindingof the cable further reduces its resistance to CCF.

f. Dakota cables are manufactured with a special, patent pending, compound called TCI, (torque, compressioninhibitor), that is placed between the armor layers. The material results in new cables that have sufficient frictionbetween the armor layers that a new Dakota Cable will closely act like a seasoned cable and will rotate 50% lessthan a typical new cable under similar conditions.

8. In the manufacture of Dakota cables careful attention is paid to the inner armor coverage. Coverage is how completelythe surrounding inner armor wires cover the plastic core. This coverage is maintained between 98% and 99%. This

range of coverage provides the maximum protection to the core to withstand CCF and still provides the necessary gapsto allow the cable to bend. It is also important that the inner armor wires are spaced evenly around the core. AtWireline Works the inner armor spacing is carefully controlled and inner armor wires are pressed into the core so thatthis spacing is maintained through the final armoring operation. When a CCF occurs the inner armor coverage shouldbe checked to determine if it was a contributing factor. The inner armor coverage should be checked on a new sectionof cable as once the cable shape is distorted the coverage can not be judged.

The cost of cable and more importantly the cost of a cable failure on a job are so great, that the cost of bringing a cableto a service center to have the armor tightened and the proper tension profile reestablished, is minor by comparison.When ever a cable has encountered any of the conditions described above where a CCF could result, get your truck to aservice center before there is a failure.

05/2005

Technical Bulletin Number 009ii


12/25


New Cable Care & Treatment

To get the maximum trouble free service from a cable, it very important to give special consideration to how a new cable

is treated on its first few runs in the field. During the manufacturing of a cable the tension is only a few hundred pounds,the cable is passed from one reel to another so no rotation is possible and the temperature is always moderate. In fieldoperations the cable is under very high tension, is free to rotate, and subjected to very high temperatures.

In field operations the higher tensions and temperature cause several important changes in the cable: When a new cable is first lowered in a well, the tension on the cable generates a torque and the cable end needs to

rotate to relieve this torque. If the end is carrying a tool that can easily rotate the cable will spin out hundreds of turnsto relieve this initial torque and become normalized. The amount of rotation depends on depth, tension and typeof cable.

During manufacturing, the inner armor wires are partially embedded into the plastic core by means of pinch rollers orpre-form rollers. With the high tensions in field operations the armor wires exert an increased radial pressure on thecore causing further embedment in the core resulting in a reduction in diameter. Higher down hole temperaturessoften the plastic which will cause this diameter reduction to occur more rapidly. When the inner armor wires are fully

embedded, the diameter of the cable will stabilize, which normally occurs in 20 or 30 operations. In the case of mono-cables the diameter reduction is only a few thousandths but unless the cable is allowed to rotate it will result in looseouter armor wires which can accumulate into a bird cage.

When the effective core diameter is reduced due to embedment, the armor wires wrap around this smaller diameter,resulting in an increase in cable length.

Cable can also be forced to excessively unwind as a result of using a hydraulic pack-off to control pressure, not enoughclearance in flow tubes, poor truck and sheave alignment or an incorrect sheave groove. Cables forced to unwind havea reduced breaking strength, are more susceptible to drum crush and loose outer armor wires that can be milkedintoa bird cage. A typical new cable is more susceptible to all of the problems associated with cable torque and rotationas there is minimal friction between the inner and outer armor wires. After a number of field operations the spacesbetween the armor wires become filled with mud and corrosion byproducts which increases the friction between thearmor layers reducing the problem caused by forced cable rotation.

During the manufacture of DAKOTA cables a special material called TCI is applied on the inner armor layer to fill theinterstitial spaces between the inner and outer armor wires. This compound not only controls the fluid and gas passagebetween armor layers but by the inclusion of sharp particles the friction between the armor layers is significantlyincreased, reducing cable rotation. A new DAKOTA cable with TCI will act more like a seasoned cable.

Based on the above explanations, here are a few DOs and DONTs that should be observed when breaking in aNEW cable:

DO: Run in and out of the well at half the normal speed. Run .004clearance in flow tubes. Use sheave wheels with the proper grove size.

DONT: Run tools that restrict cable rotation. Run in deviated holes. Apply any pressure with a hydraulic packer.

05/2005



13/25


Alternate method of leak location

It should be noted that no single method of leakdetection is fool-proof. Therefore it is highly

recommended that more than one technique be

used to provide confidence in the location of

the leak.

The method described below is quick, easy and will

locate a low resistance leak within +/-50 feet. This is

usually close enough for service and repair. To use

this method accurately, you will require a digital

multi-meter (DMM) with a 4+ digit display and an

input impedance of 10 Meg-Ohms or greater. Mostquality DMM's meet these requirements but there

are some lower quality DMM's with only a 3 digit

display and a 1 Meg-Ohm input resistance.

To get started the total length of the cable, L, must

be known. (See Tech Bulletin - 001). This method for

locating the leak assumes the entire length of cable

is on the truck drum.

The procedure is as follows:

Disconnect the leaking conductor from thecollector and connect a 6 or 12 volt car battery

between the conductor at the whip end and

collector end.

Measure the battery voltage, at the cable ends,

Vb, not at the battery terminals.

05/2005

Collector End

Whip End*

BatteryLeads

Vb

DMM

+

Battery

*Measure at cable ends on the conductor

Measure the voltage between the armor and the

conductor at the Collector end, Vc.

Measure the voltage between armor and the

conductor at the whip end, Vw.

If all measurements have been done carefully, and

the leak is "stable" the following formulae will

validate: Vc + Vw = Vb. If this checks out, or is close,then you can have confidence in the procedure.

The location of the leak from the whip end, Lw, is:

Lw = L(Vw / Vb)

Example:

L = 18,000 feet

Vb = 12.635 volts, measured at the cable ends

Vw = 2.456

Vc = 10.179

CHECK: Vw + Vc = 2.456 + 10.179 = 12.635 = Vb

Lw = L(Vw / Vb) = 18,000( 2.456 / 12.635) = 3,498 Ft.

The leak is located +/- 50 feet from 3,498 feet from

whip end.

Whip End

Measure on cable ground

Vc

+

Battery

Collector End

Measure on cable ground

Vw

+

Battery

Collector End

Whip End



14/25


Breaking Strength

The breaking strength of any Dakota cable can be found on both the Cable Specification Sheet andin the Catalogue. These can either be found on the website www.wirelineworks.com or call your

Wireline Works Representative to send you physical copies. These values of breaking strength are

theoretical values assuming the cable is in new physical condition, and the cable is pulled straight

without rotating. Dakota cables are regularly tested to verify that the breaking strength exceeds the

catalogue values.

There are many factors which can effect the breaking strength of a cable after it has been in the

field which include:

Physical wear on the cable which reduces the diameter of the outer armor wires; hence,

reducing the breaking strength of the cable Corrosion of the cable will reduce the effective diameter of both the inner and outer armor

wires and again reduce the breaking strength

H2S exposure can embrittle the steel and drastically reduce its breaking strength, as it bends

over the sheave wheel.

CO2 exposure will also cause accelerated corrosion

Excessive rotation of the cable, caused by improper operating tensions or hydraulic packers

can reduce the breaking strength by as much as 30%

Splices if done properly can withstand loads over 90% of the cables breaking strength.

However they loose much of their strength if put into compression (spudding), and tend to

deteriorate quickly when run over sheaves frequently. Shims used in splicing need to be

inspected regularly for wear. Fatigue of armor wires occurs when the cable is yo-yoedat high tension. When it is

necessary to yo-yoa cable, then at every 10 or 20 cycles the upper sheave wheel or truck

should be moved so that a fresh section of cable is passing under the measuring head and

over the sheave wheels.

Physical Damage such as kinks, armor scratches, dents, etc. to the cable can result in a much

reduced breaking strength

Operating Strength of a cable is expressed as the percent of ends fixed breaking strength (BS) of the

cable. For GIPS cables Wireline Works recommends an Operating Strength of 60% of the breaking

strength. Sour service cables should not normally be operated over 50% of there breaking strength.

The cable will operate an unlimited number of tension cycles to its Operating Strength without

permanent damage to the cable. When the cable is stressed to above the operating strength, there

may be permanent irreversible damage. Above the recommended operating strength there can be

plastic forced out of the gaps in the inner armor resulting in less electrical insulation between the

conductor and armor. There may also be additional elongation of the cable and when tension is

released Zkinks may begin to form in the copper conductor. If these high tensions are repeated it

will lead to electrical failure.

05/2005



15/25


Hydrogen Sulfide Standard Cables

Hydrogen Sulfide (H2S), is lethal to breath, very corrosive, and it can embrittle the standard GIPS (Galvanized Improved

Plow Steel) armor wire used on oilfield electro-mechanical cables. When water is present and galvanized steel armorwires come in contact with H2S there is a chemical reaction. The first and fastest reaction is with the Zinc resulting in theformation of Zinc Sulfide, which is black, in addition to the release of nascent or atomic Hydrogen (H). The second and ongoing reaction is with Iron forming Iron Sulfide and again atomic Hydrogen

The more stable state of Hydrogen is H2. However the reaction we are discussing results in a large percentage of atomichydrogen which is extremely small. The molecules are so small they can diffuse and accumulate within the crystalstructure of steel. As the accumulation continues the atomic Hydrogen seeks a more stable state and combines withanother H forming H2, which is twice as large as H. In this larger state it does not diffuse back out of the structure aseasily as it went in. This packing of Hydrogen in the steel crystal structure generates an internal stress and in time canlead to micro stress cracks in the steel. Even before there is advanced stress cracking, the accumulation of Hydrogen in thesteel crystals results in the crystals elements being unable to move internally, causing the steel to become extremelybrittle. A strand of GIPS wire exposed to sufficient Hydrogen Sulfide can result in the steel wire breaking like a glass rodwhen bent.

After the Zinc has been used up in the chemical reaction the H2S continues to react with the Iron. This action can takesplace faster in the presence of water. If the well fluid is mostly oil, then the reaction of dissolved H2S on the cable isslower but there are no safe or unsafe standards.

The presence of CO2 where there is water present results in the formation of Carbonic acid. This acid environment seemsto accelerate the action of H2S on iron, but again the published data is not complete enough for any standard guidelinesto be complete. Another catalyst occurs when the carbonic acid etches the steel surface providing additional surface areaexposed to the effects of H2S. Field experience has shown that when CO2 and H2S are both present in bore hole fluidsthat include water, the embrittlement of steel is much faster and more severe. In addition to CO2, the well pressure,temperature and the total time of exposure, are factors that can radically effect the degree of embrittlement.

The National Association of Corrosion Engineers has published a guide line for the use of GIPS in wells containing H2S:Maximum H2S in ppm for GIPS = (50.000 / Well pressure, psi)

Example: Well pressure = 1000 psi , GIPS can be used with H 2S concentrations up to 50 ppm.

The above guidelines are very general and what is safe depends on a number of other factors. The nature of H2Sembrittlement is that up to a point the embrittlement is reversible without permanent damage to the cable. Over timethe H2 will diffuse out of the wires and the cable will return to normal.

If a standard cable has been exposed to H2S and has successfully come out of the hole, you need to make a quick check ofthe armor wire by bending a wire around a rod (2 to 3 times the wire diameter) 5 complete wraps. Unwrap the wire, if itdoes not break then it is likely there has been no permanent damage by micro fracture, and the cable can be saved. TheH2 in the cable armor will ultimately diffuse out of the armor. If a wire breaks in this wrap test but there were no outerarmor wires broken coming out of the hole, then it is best to let the cable sit for a few days to allow the H 2 to diffuse outof the wires. Do not use this cable in an H2S well again until it has made several trips in normal wells and the wrap testhas passed.

Although not recommended, if you are considering running a standard, plow steel, cable in a well containing H2S, thenhere are several pointers:

Run an older cable, less Zinc. Use plenty of the pressure control grease, Liquid ORing 4-I, ( or equal). Use the National Association of Corrosion Engineers guide lines, for allowable H 2S. Use larger diameter sheave wheels No hydraulic pack off pressure . Use more flow tubes with greater clearance, 0.004 Get in and out of the hole as quickly as possible, within correct operating speeds.

If your operating conditions do not fall within these guidelines, then an alloy cable should be used. H2S and alloyarmored, MP35 & stainless steel, cables will be covered in another technical bulletin.

05/2005



16/25


Voltage and Current Ratings

Voltage

Voltage ratings are determined by the thickness of primaryplastic insulation. The published dielectric strength for FEP and

PTFE are as high as 500 & 350 volts/mil under ideal laboratory

conditions. The Voltage Rating used for oil field cables is a

conservative 50 Volts DC /mil of insulation As an example,

Dakota Cable 1-R-224-PH has 24.5 mils of insulation. At 50

Volts / mil this would indicate a rating of 1,250 Volts DC. The

actual catalog rating is rounded off at 1,200VDC.

The sixty cycle AC RMS voltage rating of a cable is less than

the DC rating. The peak voltage of the sine wave AC voltage is

1.4 times the RMS value, so this would make the AC voltage

rating of the Dakota Cable 1-R-224-PH only 700VAC.

With AC voltages there is always the threat of corona

discharge that can deteriorate the plastic insulation. A

complete treatment of corona problems is very complex and

includes effects of temperature, pressure, conductor size,

frequency as well as insulation thickness. A simplified

conservative formula, ( NTIS), for calculating the volts per mil

for the onset of corona for this cable is:

E=0. 868 / d[ log(D / d)]

Where: EVolts / mil for the onset of corona

d diameter of conductor inches

D Diameter over insulation inches

E =0. 868 / 0.059[log(0.108 / 0.059)] = 56 Volts / mil, is where

corona could start to be a problem.By rating Dakota Cables at 50 Volts / mil, corona should never

be a problem.

Current

The maximum current in a cable is determined by the

allowable voltage drop and the heat generated by the current

in the cable that is on the drum. Unless the maximum current

is on continuously for several hours, the maximum current will

normally be limited by the maximum voltage.

Using a 25,000 ft. 1-R-224-FTH cable as an example, the

maximum allowable current can be calculated using the values

of resistance and voltage rating listed in the catalog.

Ld = cable length on drum (kft);

Lh = cable length in bore hole (kft);

The conductor resistance, of the cable on the drum is:

(4.0 x Ld) Ohms (from Wireline Works Catalogue)

The armor of the cable on the drum has no resistance as it

is all shorted on its self and the drum.

The conductor resistance, of the cable in the hole is

(4.0 x Lh) Ohms .

The armor resistance, Ohms, of cable in the hole is

(4.4 x Lh) Ohms (from Wireline Works Catalogue)

Example #1:

Length of cable in the hole is 20,000 feet; therefore, Lh = 20Length of cable on the drum is 5,000 feet; therefore, Ld = 5;

The Voltage required at the tool Vb = 700;

The cable voltage rating Vmax= 1200, (from Wireline Works

Catalogue)

Total cable loop resistance, Rc = (4.0 x 5) + (4.0 x 20) +

(4.4 x 20) = 188 Ohms

Total allowable Voltage drop, Vd = Vmax Vb = 1200 70

= 500 Volts

Current = Voltage /Resistance

Maximum current that can be supplied is Imax = Vd / Rc =

500 / 188 = 2.6 amps.

Now consider the heating effect of the cable on the drum.Power = (Current)2 x Resistance = Current x Current x

Resistance

Power (watts) dissipated in drum cable, Pd = (Imax x Imax) x

(4.0 x Ld) = (2.6 x 2.6) x (4.0 x 5) = 135 Watts. In this example

the heat from 135 watts, a typical light bulb, dissipated in th

500 pounds of cable on the drum plus the steel drum will ha

little effect on the cable temperature.

Example #2:

Lh =5; Ld = 20; Vb = 700; Vmax = 1200; Calculate Imax

Rc = (4.0 x 20) + (4.0 x 5) + (4.4 x 5) = 122 Ohms

Vd = 1200 700 = 500 VoltsImax = 500 / 122 = 4.1 amps

Now consider the heating effect of the cable on the drum

Pd = (Imax x Imax) x Rd

Rd = 4.0 x Ld

Pd = (4.1 x 4.1) x (4.0 x 20) = (16.8 x 80) = 1,344 watts.

This is nearly 10 times the wattage of the other example but

still not a serious problem for short periods. 1,344 Watts is

about the power of a kitchen hot plate. It would take a ve

long time to heat up a 2,000 pound cable on the drum plus a

steel drum with a kitchen hot plate. This example does

however, illustrate that the problem of maximum current

becomes more serious when most of the cable is on the drumThere are too many variables to calculate the maximum

allowable time limit, including: ambient temperature, layers

cable on the drum, air circulation, spooling tensions, etc.

Experience has indicated that cable on the drum can tolerate

without damage, 1/10 watt per foot for periods of 24 hours.

this example that would be 1,250 watts.

08/2005



17/25


15/15 Versus 12/18 Armor Packages for 7/32Cables

Choosing between a 15 X 15 and 12 X 18 armor package on

7/32cables is always a topic of discussion among operators.The electrical characteristics of both type cables is essentially

the same. The finished diameter, weight and rated breaking

strength of both constructions are the same. There are,

however, several factors that influence the choice for general

operations with cables using standard galvanized steel (GIPS),

armor wires and there are other special factors in the choice

for cables using Alloy armor wires. This technical bulletin

outlines factors affecting which 7/32" cables to choose when

selecting from GIPS or Alloy armor wires with either 12 inner

and 18 outer armor wires or 15 inner and 15 outer armor

wires. These factors have been reported by some wireline

operators and are presented to assist you in your decision

making. Please note that your operational practices and

environment must be considered in making any decision, and

that your own experience may vary from that reported by

these wireline operators.

Armor Specifications

Armor Package 12 inner X 18 outer 15 inner X 15 ou

Cable Diameter - inches 0.224 +.005/-.002 0.224 +.005/-.0

Diameter Inner Wires inches 0.0310 0.02

Diameter Outer Wires inches 0.0310 0.03

Steel Area Inner inches square 0.009056 0.0070

Steel Area Outer inches square 0.013585 0.0150

Total Steel Area inches square 0.022641 0.0221

Rated Breaking Strength pounds 5,600 5,6

Torque Factor** 2.2

** Torque Factor = (Area of outer armor)( Pitch diameter of outer

armor) / ( Area of inner armor)( Pitch diameter of inner armor).

09/2005

Comparing 7/32Standard GIPS

15 X15 Construction

Larger outer armor wires wear longer.

Larger outer wires are stiffer and therefore easier to

thread through flow tubes.

Larger outer wires do not become crossed overas

easily during re-heading.

Smaller inner wires will corrode to brittleness faster,

reducing cable life.

The larger Torque Factor means this type of cable,

especially when new, will try to unwind more , which

can result in loose outer armor wires.

The outer armor of the 15 X 15 construction will

require more frequent trips to a service center for

normalizationand post forming, to tighten the

outer armor.

12 X 18 Construction

Larger inner armor wires will not corrode and become

brittle as fast.

The smaller Torque Factor means the cable will not

unwind as easily, so the outer armor will stay tight

longer, requiring less service.

The outer and inner armor wires are the same

diameter making a better head termination.

Comparing 7/32Alloy, Stainless & MP35

The very high costs of alloy cables and their resistance to

corrosion makes the decision on the best armor package

different. The primary consideration is on obtaining maximu

cable life. With these armor materials the cost of the frequen

cable service is small compared to the cost of the cables.


Larger outer armor wires will wear longer.

Alloy wires do not corrode, so smaller inner armor

wires are not a problem.

With no corrosion of the armor wires, the normal

corrosion products, that inhibit cable rotation, are not

present between the armor wires, so the cable under

load will unwind more, loosening the outer armor.

This cable construction, with a high torque factor and

low rotational resistance, requires frequent trips to the

service center to have the outer armor tightened and

post formed.


The lower torque factor means this construction will

unwind less than the 15X15 construction and will

require less frequent service.

The same diameter armor wires make a better

head termination.

This construction would be favored for use in

extremely remote locations where cable service is not

readily available.



18/25


Length Stability: Compound Pre-StressingTM (patent pending)

The cable length is the primary method of determining the depth of

a tool in open hole operations. In using either magnetic marks ormeasuring wheels, accurate depth measurements depend on knownstretch characteristics of the cable. There are two types of cablestretch; elastic and inelastic.

Elastic stretch is an elongation of the cable that is directlyproportional to the tension applied and when tension is removedthe cable returns to its original length. This is a characteristic of aseasonedcable. Since elastic elongation or stretch is proportionalto tension, the elastic stretch can be calculated from the measuredtension on the cable.

Inelastic stretch is a permanent elongation of a cable that occurswhen sufficient tension is applied to a new cable and the cableremains elongated after the tension is removed. In the manufacture

of new cables when the inner armor is applied to the core there arevoids between the underside of the inner armor and the core,figure 1. After adequate tension is applied to the cable the innerarmor will embed into the plastic core material, figure 2. When theinner armor wires are fully embed, essentially all inelastic stretchwill have occurred and the cable is seasoned. A seasoned cablewill have predictable elastic stretch characteristics as long as it is notsubjected to tensions over 60% of rated breaking strength.

With full embedment there is also a permanent change in theeffective cable core diameter. In manufacturing a cable the amountof this diameter change is important, so that after full embedmenthas occurred, the finished cable, will have the correct outsidediameter. The desired final effective diameter, Dc, of the core iseasily calculated: Dc = D 2do 2 di

D = required finished cable diameter do & di = outer and inner armor wire diameters

The initial diameter of the core required for full embedment can beclosely approximated by: Dc= [ (Dc+di)^2 (N/2Cos Ai )(di)^2]^1/2

N = number of inner armor wires Ai = lay angle of inner armor wires

Using the Dakota slammer cable type 7-Y-484 as an example, therequired finished effective core diameter is: Dc = 0.4842 X 0.06702 X 0.0535= 0.243

To allow for full embedment the initial core diameter required is: Dc= [ (0.243+ 0.0535)^2 (16 / 2Cos22 )( 0.0535)^2]^1/2

= 0.251 Dc- Dc = 0.251 - 0.008is the core compression.

This shows that the diameter of a newly assembled cable of thistype will decrease 8 thousandths of a inch when the inner armor isfully embedded in the core. Since the inner armor wires arehelically wrapped around this smaller effective core diameter thelength of the embedded inner armor layer will be longer. Thisincrease in length, li, can be calculated : li= L[ (Dc- Dc ) /( Dc + di)] X [ Tan Ai]^2 = L[ 0.008 /

(0.243 + 0.0535)] X [Tan22]^2 = L[0.0044]

If the original length of the cable, L is 25,000 feetthen the increa

of the inner armor length, liis: li= ( 25,000 )(0.0044) = 110 feet

The outer armor in contact with the inner armor layer experiencethe same reduction in diameter but this is a smaller fraction of itsoriginal diameter, so the elongation, lo, of the outer armor is lessthan the inner armor: lo = L[(Dc-Dc)/( Dc+2di +do)]X[Tan Ao]^2 =

L[(0.008)/(0.243+2*0.0535+0.067)](Tan 19)^2 =L[ 0.0023] lo = (25,000)(0.0023) = 58 feet.

It is conventional cable manufacturing practice to embed the innearmor in the core to seasonit by subjecting the finished cable ta prestressing operation. This operation typically applies a tensionof about 1/3 of the cable rated breaking strength as it passes

between two capstans. The tension in the cable is lost as it leavesthe final capstan and goes on the shipping reel.

If the tension in this standard prestressing operation has embeddthe inner armor in the core, then when the tension is removed, thinner armor, now wound around a smaller core, would like to belonger than the outer armor as the above calculations demonstraBecause of the friction between the inner and outer layers theouter armor can not shrink back over the inner armor . Since theshorter outer armor has much greater strength, it will push back the inner armor, which will reduce the pressure on the core.

The nature of plastics is that they have a memory and though theprestressing operation may have initially fully embedded the innearmor into the plastic core, when the inner armor pressure isreduced the core will start to recover its original shape. When thioccurs, the cable is not fully seasoned and will have excessiveinelastic stretch.

Wireline Works Inc. has a proprietary cable seasoning processknown as Compound Pre-StressingTM (patent pending). With thisprocedure the inner armor, core assembly is first pre-stressed withsufficient tension to fully embed the inner armor wires into thecore. After pre-stressing the inner armor, the outer armor is thenapplied to the already embedded and elongated inner armor. Thipermits the outer armor to add rather than reduce the pressure othe core ensuring a fully seasoned cable.

02/2006

Voids

Figure 1 Figure 2

Dc D



19/25


Armor Coverage

Armor coverage is a very important property of

electromechanical wireline cables. Proper design and armorcoverage permits cables to operate under tough operating

conditions of high temperatures and high tensile loads. The

term armor coveragerefers to how close the armor wires

are together. If a layer of armor wires were to have 100%

coverage, it would mean that all of the armor wires in that

layer were touching their adjacent wires.

There are a number of very important reasons why the armor

coverage of both the inner and outer armor layers must be

carefully controlled. If the coverage on either layer were

100%, the cable would be so stiff it would not be able to

bend around a sheave wheel without forcing one of the wires

out of the layer, creating a high wire. If the armor coverage is

too low a premature electrical short could result under high

temperature and high load conditions.

The range and requirements for the coverage of the inner and

outer armor layers is quite different. To calculate coverage

there are at least 4 good formulas. In the case of oil field

electro mechanical cables (wirelines), there is little difference

in the calculated coverage values using any of these formulas.

The formula that has been accepted by the major oil field

service companies and Wireline Works is:

% Ci = x100

% Co = x100

Ci = Percent coverage of the inner armor layers

Co = Percent coverage of the outer armor layers

di = Diameter of the individual inner armor wires

do = Diameter of the individual outer armor wires

Ni = The number of wires in the inner armor layers

No = The number of wires in the outer armor layers

i = the lay angle of the inner armor wires

o = the lay angle of the outer armor wires

Dc = Finished cable outside diameter

Inner Armor Coverage acceptable range: 97.5% to 99.5%,

ideal is 98.5%.

The importance of keeping the inner armor coverage as high

as possible is to contain the plastic insulation covering the

conductor. During cable manufacturing the equal spacing

between the inner armor wires is carefully controlled and the

inner armor wires are partially embedded in the plastic

insulation to preserve this equal spacing. This equal spacing is

important to spread the coverage equally between each wire

thus minimizing the gap at any one location.

A cable under load generates a pressure on the core and if t

inner armor wires are not close enough (low coverage %), thplastic insulation can be squeezed out between the armor

wires. With an inner armor coverage over 98% cables can

operate under rated operating conditions of temperature an

tension without the plastic insulation being squeezed out in

the gap between the inner armor wires. When a cable is

subjected to high downhole temperatures and excessive

tension some plastic insulation may be forced out between

the inner armor wires, even when the armor coverage is in a

acceptable range. Excessive operating conditions, stuck tools

and pulling out of the weak point can often create this

phenomena. In these cases it is good operational practice

to cut back on the cable end to be assured of full

electrical insulation.

Outer Armor Coverage acceptable range: 96.5% to 98.5%,

ideal is 97.5%.

The importance of allowing a lower coverage on the outer

armor is to give the cable sufficient flexibility to wrap aroun

standard sheave wheels. On cables used in high pressure

operations it is important to keep the outer armor coverage

on the high side to better control pressure in the flow tubes

The outer armor being applied over the inner armor can not

be embedded to control spacing. For this reason, new cables

when they are first spooled may cause the outer armor wire

to shift around resulting in what appears to be an excessivel

wide gap. This is perfectly normal. When a new cable has a

dark protective grease applied, this grease will collect in this

wider gap giving an appearance that the cable is gappy.

After a few runs in the hole the outer armor will equalize th

gap between the adjacent armor wires and the appearance

a gappycable will disappear.

Sample inner armor coverage calculation:

Wireline Works 1-R-288-TH

Ci = (100);

di = 0.0405;

do = 0.040;Dc = 0.288;

Ni = 12;

i = 19.5

Ci = 99.1%

06/2006

di

(Dc - 2 do - di) Sin[ ] Cos [i]

Ni

do

(Dc - do) Sin[ ] Cos [o]

Ni

di

(Dc - 2 do - di) Sin[ ] Cos [i]

Ni

180



20/25


Minimizing Cable Torque During Cable Design

All electromechanical Wireline cables that are used in oil well

service operations are designed with two layers of armor wiresaround a core of insulated conductors. By design, wirelinecables develop torque when subjected to load. The inner layerof armor wires is normally wrapped around the core in a righthand direction while the outer layer of armor wires arewrapped over the inner armor wires in a left hand direction. Bywrapping the layers in opposite directions the torque from theinner armor opposes the torque from the outer armor. Theresult is that the net torque in the cable is the differencebetween the torque generated by each armor layer. In theory itis possible to design a cable in which the torque generated byeach layer is exactly equal, resulting in a cable that has notorque under load and therefore would not rotate under loadbut there are a number of reasons that this is not a practicaldesign for oilfield service operations. To better understand the

impact of cable torque see Technical Bulletin Wireline Torque.

The torque in the cable is thedifference between the torque in theouter armor, Qo, and the innerarmor, Qi, The factors thatdetermine the torque in eachlayer are:

Qo No do2 to Sin [o]

Qi Ni di2 ti Sin [i]

Di = the pitch diameter of each inner layerDo = Is the pitch diameter of each outer layerNi = Is the number of wires in each inner layerNo = Is the number of wires in each outer layerdi = Is the diameter of the wires in each inner layerdo = Is the diameter of the wires in each outer layeri = Is the lay angle of each inner layero = Is the lay angle of each outer layerti = Is the wire tension in each inner layerto = Is the wire tension in each outer layer

The net torque in the cable, Qc= Qo Qi

Looking at the picture of a standard cable type with 12 innerand 18 outer armor wires and the factors that determine thetorque in each layer , the outer armor torque will be greaterthan the inner armor torque because:

Do > Di the outer armor is always over the inner armor No > Ni in any cable with equal diameter armor wires do = di With the standard 12 / 18 armor package

These 3 factors give the dominate torque to the outer armorlayer. When designing a cable the torque can be minimized byadjusting the lay angles of each layer. To decrease the torquegenerated by the outer armor wires the lay angle of the outerarmor is reduced to as small of an angle that does notcompromise the spooling characteristics of the cable . This anglein most cases is 19 degrees. Experience has shown that whenthe lay angle is less than 19 degrees the outer armor wires areeasily crossed if the cable gets slack during operations.

To off set the dominant outer armor torque, the lay angle of

the inner armor is increased. From cable design we knowthere is an angle of maximum torque for the inner armor. This because the portion of cable tension carried by the innerarmor decreases as the inner armor lay angle is increased.Therefore; even though the larger lay angle will result inincreasing the component of tension that generates torque,the tension is decreased excessively, the torque will alsodecrease. The result is an angle of maximum torque for theinner armor, which in turn is the angle that results in theminimum cable torque. ( Q =Qo-Qi ).

The minimum torque angle for a cable with a 12/18 armorpackage is about 25 to 26 degrees. Again in cable design theare compromises to be made. As the lay angle of the innerarmor is increased to reduce cable net torque, it also reduce

the breaking strength. The best compromise betweenbreaking strength and cable torque is an inner armor layangle of 23 degrees.

The actual computation of cable torque under load is acomplicated problem as when the angles of inner and outerarmor are changed, the tension carried by each layer and thtorque are changing.

The full calculation of cable torque, qcc, for the WirelineWorks cable 1-R-322 looks like this:

cd = 0.322 dia = 0.0445 doa = 0.0445 nia = 12noa = 18 aia = 23 aoa = 19 pr = 0.47

qcc = 0.0151162 inch-pounds of torque / pound of tension

At a working tension of 5000 pounds, the cable torque,Qc would be:Qc = 75.6 inch pounds of torque

06/2006

Do

2

Di

2CableTorque,

inch-pounds/pound

0.017

20

0.0165

0.016

0.0155

0.015

15 25

Inner Armor Lay Angle ia

30 35

qcc =

(cd doa) doa2noa sin (oa) cos2(oa) (cd 2 dia 2 doa) pr sin2(oa)

cd doa

dia2(cd dia 2 doa) nia sin (ia) cos2(ia) (cd 2 dia 2 doa) pr sin2(ia)

cd dia 2 doa

doa2noa cos (oa) cos2(oa) (cd 2 dia 2 doa) pr sin2(oa)

cd doa

2 nia cos (ia) cos2(ia) dia2 +(cd 2 dia 2 doa) pr sin2(ia)

cd dia 2 doa



21/25


Wireline Torque

Electromechanical Wireline Cables are designed and

manufactured to minimize the inherent torque in the cable,

see Technical Bulletin Minimizing Cable Torque during

Design". However, all cables inherently have some torque

and will develop a need to rotate relative to the tension

applied during operations. This is generally not a problem

as long as the cable is allowed to rotate freely. In todays

complex oilfield there are a lot of variables that affect and

restrict cable rotation. If the cable is not allowed to rotate

in proportion to tension, torque build up will begin to occur

in certain areas of the cable depending on what is

restricting it from rotating properly. For example, the pack-

off will restrict the cable from rotating and the cable willaccumulate torque as the cable passes through pack-off.

This results in torque build up and loose outer armor.

Cable rotation can be restricted and torque imbalance may

occur from the following operations:

Deviated or crooked well bores.

Going in and out too fast and not observing the 80/120

tension rule, (see Technical Bulletin #9).

Pulling out of a well at high speeds that result in

excessive tension.

Centralized and decentralizing tools.

Heavy and viscous drilling mud and completion fluidsaffect the tension of the cable.

Grease heads or pack-offs used to wipe or

control pressure.

Pulling out of a rope socket under high load conditions.

Low fluid bypass conditions.

Field experience has shown that almost always loose outer

armor is caused from the torque imbalance resulting from

improper running conditions. During the seasoning or

breaking in period for new cables there will generally be

some areas in the cable that become loose. These areas do

not cause problems under everyday use; however, it wouldbe good insurance to normalize (tighten loose areas) a

standard GIPS cable after 20 to 30 runs. This would tighten

any loose outer armor that may have occurred due to the

core embedment of a new cable.

If a cable has been run into a well bore with any condition

that may prevent free rotation or cause torque imbalance,

the cable will need attention to keep it from failing. The

standard approach is to normalize the cable to be sure the

outer armor is tight. If you feel or see your cable trying to

curl up while laying on the ground during rig ups it has

excessive torque. Running the cable in this condition will

risk breaking, or getting a strand cross-over which can cau

the cable to strand at deeper depths. Remember every bir

cage you see is caused by getting too much slack in one

location of the cable. It is a good idea to rehead, when

possible, with inner armor strands on the cables that are

using grease heads because they are lubricated and cantorque up relatively easily.

Lack of tension means low rotation is required, and high

tensions means a lot of rotations required to prevent torq

build up. If you come out of a well with very high tension

and torque in the cable, the next time you go into a well

with very little tension, there will be a lot of torque in the

cable wanting to be released. Armor separation, high

strands, or bird caging are not the only issues to worry

about with torque build up, you may also experience early

pullouts, cable breaks, and excessive compression on the

conductor which can short out the cable. The more youunderstand the affects of torque the better off you are in

preventing cable failures and/or well site disasters.

The torque generated at maximum working load for

standard cables has been calculated as follows:

TYPE Z-224 R-224 R-288 R-322 R-380 R-42

LOAD

(pounds) 3360 3360 6000 6720 8600 1170

TORQUE

(Inch-pounds) 55.5 40.7 93.5 116.7 176.5 268

06/2006



22/25


Cable Rotation

By design, wireline cables develop torque when subjected to

load, see Technical Bulletin Minimizing Cable Torque DuringCable design. The load on the wireline cable is a result of theweight of the tool, the weight of the cable and any dynamicfriction due to running conditions. If the tool end of the cable isfree to rotate, the cable will try to rotate to reduce this torque.

All cables used in oil field service operations are built with thetorque of the outer armor dominant over the opposing torqueof the inner armor. To balance the torque the cable will unwindin a direction to loosen the outer armor, which will tighten theinner armor. If the cable is free to rotate, this unwinding orrotation will continue until the torque in the inner armor equalsthe torque of the outer armor. The number of revolutions, Nf(per 1000 ft per 1000 lb), that the end of the cable will make toequalize the torque can be calculated as follows:

Example of Cable Rotation with Tool End Free to RotateUsing Wireline Works Cable # 1-R-322 as an example:Cd=0.322, doa=0.0445, dia=0.0445, noa=18,nia=12, oa =19.22, ia=21.54, pr*=0.47**

*pr = Poisson Ratio

**Testing has shown that 0.47 is the best value for EM cables

Nf = 17.2 revolutions / 1000 ft / 1000 pounds tension.

With 10,000 feet of new 1-R-322-PH cable lowered into astraight dry hole, with a 500 lb tool, the total revolutions,N, the cable would make to equalize torque would be:

Cable weight lb / kft = 188 ;Average tension due to cable = 188 X 10 / 2 = 940 lbEffective tension is 940 + 500=1440=1.44 klb;Nf x 1.44 x 10 = 17.2 x 1.44 x 10 = 247 revolutions

There are very few straight, dry holes but this calculationindicates the amount of rotation a new cable will try to maketo equalize the torque. With fluid in the hole the tension would

increase with cable speed coming out of the hole resulting in

additional unwinding revolutions.

Standard Wireline Works Cable RotationNfNumber of Revolutions per 1000 feet per 1000 poundtension cable end free to rotate

TYPE Z-224 R-224 R-288 R-322 R-380 R-42

Nf 83 48 22 17 10

The above calculations represent the possible rotation of atypical new cable. As a cable becomes seasoned it will rotaless with tension changes. A seasonedcable is one in whichthe outer armor wires have dug into the Zinc of the innerarmor wires and mud plus corrosion by products have collectebetween the armor layers add to the friction between layersreducing the amount of cable rotation. The amount of thisinitial new cable rotation has been reduced in Wireline Workscables by including a material called TCI (Torque CompressionInhibitor) Technical Bulletin #10, between the armor layers. TCcontains materials that increase the friction between the armolayers, reducing the rotation of new cables, so new cables wilperform more like seasonedcables.

Cables armored with alloy wires like MP-35 or 31MO are anextreme case in cable rotation. The reason is that the alloyarmor does not have a soft Zinc coating and it does not corrodcreating friction between armor layers that reduces the rotatiin cables armored with galvanized wire. For these reasons allocables will continue to rotate in use and must be given extra

care in field operations and periodically the outer armor needto be tightened.

In operations, keep in mind that when ever the tension on thecable changes it will try to rotate. When cable tension isincreased above the static tension by frictional drag on the cabthe increase in cable torque will try to unwind the outer armorwires. Frictional drag comes from bore hole friction and tightpressure control equipment. This frictional drag increases withthe speed of the cable spooling. Coming out of the hole too facan result in excessive frictional tension on the cable forcing thcable to rotate excessively , further loosening the outer armor.Going into the hole the tension in the cable is reduced by thefrictional drag and the cable will try to rotate to tighten theouter armor. Going into the hole too fast will not give the cab

time to rotate to tighten the armor. Experience has shown thatfor standard GIPS armored cables if the tension going into thehole is not less than 80% of static tension at that depth and thtension coming out of the hole is never more than 120% ofstatic tension, the cable armor will remain tight. This rule doenot apply to alloy cables, which require special care.

Cable rotation can cause the stress in the outer armor wires tobe reduced, which not only leads to loose outer armor wires balso significantly reduces the cable breaking strength. Thereduction in cable breaking strength with the cable free torotate will be covered in a later Technical Bulletin. Foradditional effects of cable rotation and cable torque seeTechnical Bulletin Wireline Torque.

06/2006

Nf = 48 x 106(cd doa) (cd + dia + 2 doa)

dia2(cd 2 (dia + doa)) nia pr sin3(ia) + dia2(cd + dia + 2 doa) nia cos 2(ia) sin (ia) +

dia2doa2) nia noa 2ym ((doa cd) cos (ia) sin (oa) (cd dia 2 doa) cos (oa) sin (ia))

doa2noa sin (oa) (cd doa) cos2(oa) + (2 (dia + doa) cd) pr sin2(oa)

(cd dia 2 doa) (cd doa)2sin(2 oa) cos2(ia) +

sin(ia) ((cd doa) (cd + dia + 2 doa)2cos2(oa) +

prcd (cd

2

4 doa cd + doa (4 dia + 5 doa) cos (

oa) sin (

oa) sin2

(

ia) +

12

cd3+ (4 dia + 6 doa) cd25 dia2cd + 8 doa3) pr sin2(oa) cos (ia) +

dia cd2+ doa2(dia + doa) sin (2 oa) sin2(ia) +

dia3+ 5 doa dia2+ 8 doa (doa cd) dia 6 cd doa2 sin (2 ia) sin2(oa)



23/25


Temperature Rating of Cables

The maximum temperature rating

The maximum temperature rating of Wireline Works cables,

as listed in the catalog, is based on the following

operating conditions:

The maximum bore hole temperature is not greater than

the cable rated temperature.

The cable is operated under a normal tension profile.

There are no unusual hole conditions or restrictions causing

excessive tension.

There are three factors that will influence the temperature

rating of a cable.

The nominal melting pointof the plastic used for

insulating the conductor. The pressure exerted by armor wires on the core by normal

loads

The inner armor coverage

The nominal melting point

Plastics are called amorphous materials and as such do not

have a specific melting point. Crystalline materials, such as

metals and water, are characterized by the fact that they do

have a very specific temperature at which they change from a

solid state to a liquid state. Amorphous materials, like plastics,

do not have a specific temperature at which they change

from a solid material to a liquid st

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