Dynamometer Cards

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Dynamometer Cards (Surface and Downhole)

Surface Dynamometer Cards

Webster defines a “dynamometer” as an instrument used to measure force. In

terms of its use in the “oil patch”, a dynamometer records polished rod load in relation to polished rod position. The result is a plot of “load versus position” – commonly called the surface “dynamometer card” that measure the amount of work being done by the pumping system. Dynamometer systems allow the user to record the necessary information to generate a “surface card” – i.e., load and position data. This data can then be analyzed by any “wave equation” driven diagnostic program. Analysis results include downhole cards, load/stress calculations, counterbalance information, pump displacement, calculated fluid levels, estimated electrical costs, etc. Through the years, the polished rod dynamometer has been the principal tool for analyzing the operation of rod pumped wells. The shape of the surface dynamometer card is determined by changing downhole conditions. Ideally, these conditions would be apparent from the surface dynamometer card by visual interpretation. However, because of the complex behavior of the rod string and the great diversity of card shapes, visual diagnosis is not always possible. Though much information can be gained from visual interpretation of surface cards, success is directly linked to the skill and experience of the analyst – and even the most experienced analysts are often misled into an incorrect diagnosis. The graphic shown in the next slide illustrates how “surface card” shape relates to pump depth/size, SPM, stroke length, and rod diameter.

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Polished Rod Load Points vs. Polished Rod Position

Points =

Dynamometer

Card

The Way It

Used To Be

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Position Input Devices

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Raw Position and Load Values for Dynamometer

Card Generation

If these

“

position points

”

are plotted against the

“

load points

”

–

what is the result?

A surface dynamometer card

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The following several slides are a library of “API 11L2”

generated surface dynamometer cards and actual cards

gathered in the field.

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Calculated Downhole Pump Card

Bottom Hole Analysis

In the studies of surface dynamometer cards, it is understood that one of the primary things that affect the shape of the actual dynamometer card is the load condition at the bottom of the hole. Therefore, bottomhole conditions such as crooked hole, paraffin, scale, sand, and solids all affect the loads and the shape of the card.In an effort to find out exactly what was happening at the bottom of the hole, W. E. Gilbert and others designed a

bottomhole dynagraph that measured loads and displacements at the bottom of the hole. His work, and a number of bottomhole dynagraphs, was published in 1936. In 1967, Sam Gibbs received a patent on a mathematical method for simulating the sucker rod pumping system. His work and the work of others made it possible to use a surface dynamometer card as a basis for a simulated bottom hole card. In 1986 G. Albert designed an electronic bottomhole analyzer which measured the bottom hole conditions electronically in the same manner that W. E. Gilbert had measured them mechanically. This tool confirmed that the mathematical simulation from surface dynamometer cards does indeed give accurate bottomhole loads and displacements.

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¾ The mysterious "black magic" of an experienced

surface card “dynamometer” analyst was and continues to

be based on a lifetime of observation and experience.

Today, with a thorough understanding of the principles

involved and aided by computer simulation, an accurate

analysis of the problem is possible in nearly every case.

¾ To bridge the gap between interpretation of the

surface card (“black magic”) and quantitative downhole

data, today’s diagnostic programs make use of a

mathematical solution based on a model of the rod

pumping system, known throughout the industry as the

“wave equation”. The resulting subsurface or downhole

card removes personal judgment and experience from the

diagnosis of downhole conditions.

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Rod Pumping Visualization

For Four of the Basic Downhole Pump Card Shapes

Net Pump Stroke (in.)

Load (L

bs)

Full Pump

Load (L

bs)

Gas

Interference

or

Compression

Load (L

bs)

_Tubing

Movement

Load (L

bs)

Pump-Off

Click Anywhere to Continue

0.2

Net Pump Stroke (in.)

Rod Pumping Visualization

Traveling Valve

Standing Valve

Downhole Pump card

Pump Stroke (Inches)

Load (L

bs)

Pressure Above Plunger Pressure Below Plunger

Click Anywhere to Continue

Tubing Anchor

Full Pump

Fluid Load

On Plunger

Ball Seat Pump Plunger Pump Barrel

Maximum Fluid Load

Minimum Fluid Load

Top of Stroke F1-03 Fluid Level Pump Hold-Down Rod String Tubing Casing Polished rod in the hole – bottom of stroke Ball

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Start of Upstroke

Net Stroke (in.)

Load (L

bs)

SV Open – Rods Support Fluid Column (load increases)

SV Closed – Tubing Supports Fluid Column (load)

Pressure Above Plunger Pressure Below Plunger Tubing Anchor

TV Closed (loads increase to maximum)

Click Anywhere to Continue

F2-04

Upstroke Completed

Net Stroke (in.)

Load (L

bs)

SV Open – Rods Support Fluid Column (load)

Click Anywhere to Continue

Tubing

Anchor

TV Closed (load @ maximum)

F3-05

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Downstroke Begins

SV Open – Rods Support Fluid Column

SV Will Close – Tubing Supports Fluid Column (load)

Net Stroke (in.)

Load (L

bs)

Click Anywhere to Continue

Tubing

Anchor

TV Closed

TV Will Open (loads decrease)

F4-06

Pump Downstroke Completes

SV Closed – Tubing Supports Fluid Column (load)

Net Stroke (in.)

Load (L

bs)

TV Open

(loads @ minimum)

Click Anywhere to Continue

Tubing

Anchor

F5-07

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Pump Stroke Complete

SV Closed – Tubing Supports Fluid Column (load)

Net Stroke (in.)

Load (L

bs)

TV Open

TV Will Close

Click Anywhere to Continue

Tubing

Anchor

F6-08

Full Downhole Pump Card

Rod Pumping Visualization

Traveling Valve

Standing Valve

Net Stroke (in.)

Load (L

bs)

Click Anywhere to Continue

Tubing

Anchor

_Gas

Interference

(high pressure gas)

Downhole Pump Card

G1-09

Fluid Level

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Start of Upstroke

TV Closed

(loads increase to maximum)

Net Stroke (in.)

Load (L

bs)

SV Open – Rods Support Fluid Column (load increases)

SV Closed – Tubing Supports Fluid Column (load)

Click Anywhere to Continue

Tubing

Anchor

G2-10

Fluid Level

Pump Upstroke Completes

Net Stroke (in.)

Load (L

bs)

SV Open – Rods Support Fluid Column (load)

TV Closed

Click Anywhere to Continue

Tubing

Anchor

G3-11

Fluid Level

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Start of Downstroke

TV Closed

SV Open – Rods Support Fluid Column (load)

SV Will Close – Tubing Supports Fluid Column (load)

Net Stroke (in.)

Load (L

bs)

Click Anywhere to Continue

Tubing

Anchor

G4-12

Gas below pump must be compressed before traveling valve can open

Fluid Level

Downstroke (cont.)

TV Closed

TV Will Open

(loads decrease)

SV Closed – Tubing Supports Fluid Column (load)

Net Stroke (in.)

Load (L

bs)

Click Anywhere to Continue

Tubing Anchor G5-13 Fluid Level Gas Compression Complete

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Pump Downstroke Completes

Standing Valve Closed – Tubing Supports Fluid

Column (load)

Net Stroke (in.)

Load (L

bs)

Traveling Valve Open

(loads @ minimum)

Click Anywhere to Continue

Tubing

Anchor

G6-14

Fluid Level

Pump moves down

Bottom of Pump Stroke

Traveling Valve Closed

Standing Valve Closed – Tubing Supports

Fluid Column (load)

Net Stroke (in.)

Load (L

bs)

Traveling Valve Open

Click Anywhere to Continue

Tubing Anchor G7-15 Fluid Level

Gas Interference

Card

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Rod Pump Visualization

Traveling Valve

Standing Valve

Net Stroke (in.)

Load (L

bs)

Click Anywhere to Continue

Tubing

Anchor

Pump-Off

(low pressure gas)

Downhole Pump Card

O1-16

Fluid Level

Start of Upstroke

Traveling Valve Closed (load increases to maximum)

Standing Valve Closed – Tubing Supports Fluid

Column (load)

Net Stroke (in.)

Load (L

bs)

Standing Valve Will Open – Rods Support Fluid

Column

Click Anywhere to Continue

Tubing

Anchor

O2-17

Fluid Level

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Low Pressure

Gas

Upstroke Continues

Net Stroke (in.)

Load (L

bs)

Standing Valve Open – Rods Support Fluid Column

Traveling Valve Closed

(loads @ maximum)

Top of Pump Stroke

Click Anywhere to Continue

Tubing

Anchor

O3-18

Fluid Level

Pump moves up the hole

Start Downstroke

Traveling Valve Closed

Standing Valve Open – Rods Support Fluid Column

Standing Valve Closed – Tubing Supports Fluid Column

Net Stroke (in.)

Load (L

bs)

Click Anywhere to Continue

Tubing Anchor O4-19 Low Pressure Gas

Traveling valve must contact liquid before it can open

Fluid Level

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Downstroke (cont.)

Traveling Valve Closed

Standing Valve Closed – Tubing Supports Fluid Column

Net Stroke (in.)

Load (L

bs)

Traveling Valve Open

(loads to minimum)

Click Anywhere to Continue

Tubing

Anchor

O5-20

Traveling valve opens late in the downstroke

Fluid Level

Pump Downstroke Completes

Standing Valve Closed – Tubing Supports Fluid Column

Net Stroke (in.)

Load (L

bs)

Traveling Valve Open

(loads @ minimum)

Click Anywhere to Continue

Tubing

Anchor

O6-21

Fluid Level

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Bottom of Pump Stroke

Traveling Valve Closed

Standing Valve Closed – Tubing Supports Fluid Column

Net Stroke (in.)

Load (L

bs)

Traveling Valve Open

Click Anywhere to Continue

Tubing

Anchor

O7-22

Fluid Level

Pump Off Card

(low pressure gas)

Rod Pump Visualization

Traveling Valve

Standing Valve

Net Stroke (in.)

Load (L

bs)

Click Anywhere to Continue

No Tubing Anchor

Tubing

Movement

Downhole Pump Card

T1-23

Tubing movement reduces effective pump stroke

Fluid Level

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Start of Upstroke

Traveling Valve Closed

(loads to maximum)

Standing Valve Closed – Tubing Supports Fluid Column

Net Stroke (in.)

Load (L

bs)

Click Anywhere to Continue

T2-24

Tubing moves up due to rods now supporting the fluid load Lost effective pump stroke

Upstroke Continues

Net Stroke (in.)

Load (L

bs)

Standing Valve Open – Rods Support Fluid Column

Traveling Valve Closed

(loads @ maximum)

Top of Pump Stroke

Click Anywhere to Continue

Standing Valve Closed – Tubing Supports Fluid Column

T3-25

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Start Downstroke

Traveling Valve Closed

Standing Valve Open – Rods Support Fluid Column

Standing Valve Closed – Tubing Supports Fluid Column

Net Stroke (in.)

Load (L

bs)

Click Anywhere to Continue

T4-26

Downstroke (cont.)

Traveling Valve Closed

Standing Valve Open – Rods Support Fluid Column

Standing Valve Closed – Tubing Supports Fluid Column

Net Stroke (in.)

Load (L

bs)

Click Anywhere to Continue

Traveling Valve Open

(loads decrease to minimum)

T5-27

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Downstroke Completes

Standing Valve Closed – Tubing Supports Fluid Column

Net Stroke (in.)

Load (L

bs)

Traveling Valve Open

(loads @ minimum)

Click Anywhere to Continue

T6-28

Pump moves down

Bottom of pump Stroke

Traveling Valve Closed

Standing Valve Closed – Tubing Supports Fluid Column

Net Stroke (in.)

Load (L

bs)

Traveling Valve Open

Click Anywhere to Continue

No Tubing Anchor T7-29

Tubing Movement

Card

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The following several slides show example calculated

downhole pump cards and explanations that are the

result of the “wave equation” diagnostic solution.

The shape of a downhole pump card showing full liquid fillage (with anchored tubing) is approximately rectangular.

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1. At point “A”, the traveling valve closes and the load begins to be transferred from the tubing to the rods.

2. Between points “A” and “B”, tension in the pull rod is increasing as the rods are picking up the fluid.

3. At point “B”, the entire fluid load is borne by the rods and the standing valve opens.

4. Between points “B” and “C”, fluid is being lifted toward the surface. At the same time, the pump chamber below the traveling valve is filling completely with liquid through the open standing valve.

5. At point “C”, the top of the stroke has been reached and the downward tendency of the pump motion causes the standing valve to close.

6. Between points “C” and “D”, the fluid load is being transferred back to the tubing. Because the pump chamber has filled completely with liquid (nearly incompressible) the pump cannot move downward until the entire fluid load has been released. This is one of the reasons for the rectangular card shape. The pump remains stationary (if the tubing is anchored) while the load is being transferred back to the tubing from the rods.

7. At point “D”, the traveling valve opens and the pump begins to descend. 8. Between points “D” and “A”, the pump descends with the traveling valve

open (standing valve closed) through the fluid that entered the pump chamber during the upstroke.

9. At point “A”, the traveling valve is closed by the tendency of the pump to move upward. This action begins another pumping cycle.

Important Conclusion

When a pump fills completely with liquid (with anchored tubing), traveling and standing valve actuation occurs at the top and bottom of the stroke with little movement of the pump. This gives the downhole pump card a characteristic rectangular appearance.

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A downhole pump card with unanchored tubing (full liquid fillage) has a parallelogram shape. The amount of tubing movement (in inches) can be scaled off from the downhole card to determine the amount of pump displacement being lost to unanchored tubing.

Detailed Description

2. Between points “A” and “B”, tension in the pull rod is increasing as the rods are picking up the fluid. The pump is moving relative to the casing as the fluid load is being picked up. The pump is stationary relative to the tubing. Since the tubing is not anchored, the tubing shortens as the load is removed from it. Because the pump is riding along with the tubing, the pump moves relative to the casing. This movement can be detected with the use of the “wave equation” diagnostic solution. The horizontal distance between points “A” and “B” is the amount of tubing “stretch” in inches.

4. Between points “B” and “C”, fluid is being lifted toward the surface. At the same time, the pump chamber below the traveling valve is filling completely with liquid through the open standing valve.

5. At point “C”, the top of the stroke has been reached and the downward tendency of the pump motion causes the standing valve to close.

6. Between points “C” and “D”, the pump load is transferred from the rods to the tubing. As the load shifts to the tubing, the tubing stretches downward relative to the casing. Thus, pump movement relative to the casing can be detected by the “wave equation” diagnostic solution.

7. At point “C”, the traveling valve opens and the pump begins to descend. 8. Between points “D” and “A”, the pump descends with the traveling valve open

(standing valve closed) through the fluid that entered the pump chamber during the upstroke.

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The shape of a downhole pump card with gas interference shows a gas compression curve in the upper portion of the downstroke (points “C” - “D”).

Detailed Description

2. Between points “A” and “B”, tension in the pull rod is increasing as the rods are picking up the fluid. If the fluid in the lower portion of the pump chamber is compressible (very gassy), a slight upward movement of the pump may occur during the load pick-up.

4. Between points “B” and “C”, fluid is being lifted to the surface. At the same time, the pump chamber below the traveling valve is filling with a mixture of liquid and high-pressure gas through the open standing valve.

5. At point “C”, the top of the stroke has been reached and the downward tendency of the pump causes the standing valve to close.

6. Between points “C” and “D”, the fluid load is being transferred back to the tubing. Because of the compressible gas that entered the pump during the charging cycle, the load transfer takes place along a “compression curve”. The pump moves downward during load transfer – which compresses the gas in the chamber below the closed traveling valve. As the pressure in the gas below the traveling valve increases, the load is removed from the rods.

7. At point “D”, the pressure in the compressed gas in the pump chamber is high enough to offset the pressure in the tubing at which point the traveling valve opens. The pump continues to descend.

8. Between points “D” and “A”, the pump descends with the traveling valve open (standing valve closed) through the fluid that entered the pump chamber during the upstroke.

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Fluid pound is a limiting case of gas interference. Pump intake pressure is low and the incompletely filled (with liquid) pump contains almost incompressible fluid. The load release thus takes place more abruptly than the gradual transfer that occurs with gas interference does.

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Detailed Description

4. Between points “B” and “C”, fluid is being lifted to the surface. At the same time, the pump chamber below the traveling valve is filling with a mixture of liquid and low-pressure gas. Liquid fillage can be much less than the volume of the pump chamber.

6. Between points “C” and “D”, the fluid load is being transferred from the rods to the tubing. Because the gas is under low pressure, little compression takes place as the pump descends. Eventually, the plunger encounters nearly incompressible liquid in the incompletely filled (with liquid) pump chamber. At this point, the load is rapidly released from the rods and the pump is said to “pound fluid”.

7. The traveling valve opens at point “D” and the pump continues to descend.

8. Between points “D” and “A”, the pump descends with the traveling valve open (standing valve closed) through the fluid that entered the pump chamber during the upstroke.

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A worn traveling valve or plunger causes the pump to pick up the fluid load slowly at the bottom of the stroke and to release it prematurely at the top of the stroke.

Detailed Description

2. Between points “A” and “B”, tension in the pull rod is increasing as the rods are picking up the fluid. The pump is moving slowly during this part of the cycle – thus its displacement rate is low. The pump slippage rate is a sizeable portion of the displacement rate. This causes the fluid load pick-up to be more gradual than usual.

4. Between points “B” and “C”, fluid is being lifted toward the surface. At the same time, the pump chamber below the traveling valve is filling completely with liquid through the open standing valve. In addition to this, fluids are slipping back around the worn traveling valve or plunger into the chamber below. This subtracts from the volume available for entry of new fluids from the reservoir.

5. At point “C”, the pump speed has again slowed down enough so that the slippage rate exceeds the displacement rate of the pump. This closes the standing valve. In a worn pump of this type, the load release begins prematurely near the top of the stroke.

6. Between points “C” and “D”, the fluid load is being transferred back to the tubing. The load is released with the pump still moving upward because of slippage. This happens because the slippage rate exceeds the pump displacement rate in this portion of the stroke.

7. At point “D”, the traveling valve opens and the pump begins to descend. 8. Between points “D” and “A”, the pump descends with the traveling valve

open (standing valve closed) through the fluid that entered the pump chamber during the upstroke.

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A worn standing valve causes the pump to prematurely pick-up load near the bottom of the stroke. It also causes a delayed release near the top of the stroke.

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1. At point “A”, the traveling valve closes and the load begins to be transferred from the tubing to the rods. The load transfer begins with the pump still on the downstroke. This happens because the slippage rate past the standing valve exceeds the displacement rate of the slowly moving pump as it approaches the bottom of the stroke. This closes the traveling valve while the pump is still moving downward.

3. At point “B”, the entire fluid load is borne by the rods and the standing valve opens. When the standing valve opens, slippage ceases.

4. Between points “B” and “C”, fluid is being lifted toward the surface.

6. Between points “C” and “D”, the fluid load is being transferred back to the tubing. The load can be released with the pump moving down – even with complete liquid fillage. This can happen because slippage past the standing valve exceeds the displacement rate of the slowly moving pump.

7. At point “D”, the displacement rate of the pump exceeds the slippage rate of the standing valve and the traveling valve opens. The pump continues downward. 8. Between points “D” and “A”, the pump descends with the traveling valve open

(standing valve closed) through the fluid that entered the pump chamber during the upstroke. Slippage past the standing valve is occurring – which decreases volumetric efficiency.

9. At point “A”, the pump has slowed down enough so that the slippage rate past the standing valve exceeds the displacement rate of the pump. This closes the traveling valve and a new pump cycle begins.

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Pumps can “hit” up or down or both. This “hitting” or “bumping” condition can occur for any downhole situation – gas interference, fluid pound, unanchored tubing, etc. The case shown above has full liquid fillage.

¾ If the pump “hits” down, a load loss (compression) will be

shown at the lower left of the downhole card (at the bottom

of the stroke on the downhole pump card).

¾ If the pump “hits” up, a load increase (tension) will be

shown at the top of the stroke (upper right on the downhole

pump card).

9 Pumps that are identified as “hitting” up or down should

be “re-spaced” to prevent equipment damage.

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Detailed Description

1. At point “A”, the plunger is below the bent section and the load

on the pull rod is the same as for a full pump.

2. At point “B”, as the plunger reaches the “bend”, the load on the

pull rod increases because the plunger must “squeeze” by this

portion of the pump barrel.

3. At point “C”, the load reaches a maximum value and then

decreases as the plunger moves away from the bend.

4. On the downstroke, the load on the pull rod is normal until the

plunger reaches the “bad” spot in the barrel at point “E”.

5. The load on the pull rod decreases until the plunger reaches

point “F”.

6. The pull rod load returns to normal after the plunger moves

away from the bent portion of the pump barrel.

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Detailed Description

1.

At point “A”, the plunger is below the bent section and the load on the pull rod is the same as for a full pump.

2. At point “B”, as the plunger reaches the “bend”, the load on the pull rod increases because the plunger must “squeeze” by this portion of the pump barrel.

3. At point “C”, the load reaches a maximum value and then decreases as the plunger moves away from the bend.

4. On the downstroke, the load on the pull rod is normal until the plunger reaches the “bad” spot in the barrel at point “E”.

5. The load on the pull rod decreases until the plunger reaches point “F”.

6. The pull rod load returns to normal after the plunger moves away from the bent portion of the pump barrel.

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Detailed Description

1.

From point “A” to the worn place in the barrel, the load on the pull rod is normal.

2. At point “C”, as the plunger moves through the bad portion of the barrel, fluid leaks by the plunger and causes the load to decrease on the pull rod. 3. At point “D”, a good seal is again established between the plunger and the

pump barrel.

4. On the downstroke, as long as the traveling valve remains open, the load on the pull rod appears to be normal.

5. If, however, the pump barrel is worn, it is possible that a corresponding load increase will occur at the same place on the downstroke – if the worn spot causes enough pressure loss in the pump barrel for the traveling valve to start picking up the fluid load.

6. The pull rod load returns to normal after the plunger moves away from the worn portion of the pump barrel.

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The graphic shown above helps to explain the downhole pump card shape for a full pump experiencing upstroke fluid inertia effects. This card shape is representative of wells with large plungers, of shallow depth (usually less than 4000’), and a high water cut.

Detailed Description

1. From points “A” to “B”, the inertia of the fluid in the tubing causes the load on the pull rod to increase as the plunger moves on the upstroke and “accelerates” the fluid above it.

2. At point “B”, the load on the pull rod reaches a maximum value. 3. Between points “B” and “C”, the pressure “pulse” travels up the fluid

column and the pull rod drops – until the pressure “pulse” travels up the tubing and reflects back down. When this “reflected” wave reaches the plunger, it increases the pull rod load, but not as much as before. 4. The pull rod load returns to normal, assuming no further reflected

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More Downhole Pump Card Examples

Combinations of the following conditions can occur simultaneously.

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Things That Affect Calculated Downhole Pump Cards

¾ Rod String Information:

9 Too short -

9 Too long -

9 Wrong length per rod -

9 Wrong taper length –

9 Wrong weight per foot –

9 Wrong “Modulus of Elasticity” –

9 Wrong “Speed of Sound”

-The examples shown in this slide and the next show what can happen to the calculated downhole pump card when input data is incorrect.

This example shows the downhole pump card shifted up and with a different shape as a result of an incorrect number of rods in each taper of a three taper rod string.

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The example above shows what happens to the downhole pump card when the “weight per foot” of each rod size is entered incorrectly in the database.

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Things That Affect Calculated Downhole Pump Cards

¾ Incorrect Surface Loads

9 Strain gauge

9 Bad calibrated load cell, load cable / connections

¾ Poor Position Data

9 Position switch (TOS / Simulated Data Input)

9 Bad “real position” device /cable

9 Improper position device installation

• Damping Factor values control the amount of work from the surface dynamometer card that is applied to the calculated downhole pump card. 9 Increasing the “Damping Factors” results in smaller downhole cards/fluid

loads.

9 Decreasing the “Damping Factors” results in larger downhole cards/fluid loads.

Note: Changing the “Damping Factors” is one way to help

“model” conditions such as excessive friction or entrained gas.

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Class Exercise – Dynamometer Cards

Refer to the exercise handout. Record your observations in the

space provided below each card or card pair.

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Extreme pump off (“gut pumped”) … Bad pump off set-point? … Restricted pump intake? ...

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Pumping unit “rolling” ...

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No pump valve action … Tubing leak ?

TV not closing properly … Possible “ring valve” pump ?

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Load cable short / connection? ...

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TV hung open – then slams shut ...

Parted rod string … Inoperative pump valves ...

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Worn pump - upstroke leakage (TV or bbl./plunger fit) ...

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Rod part ...

Stuck pump valves - trash, well not pumping ...

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Load cable - possible radio antenna problem ...

No tubing anchor or not enough TAC tension ...

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Upstroke pump wear ...

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Upstroke pump wear ...

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Position input problem: Proximity switch / simulated position, bad inclinometer / potentiometer, bad real position device ...

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Timing – RPC SPM too high … Calibration / Data input ...

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Sticking SV, load cable problem, load cell problem ...

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Load cable … Bad position input …

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Position input - device shaking from fluid pound or unstable foundation? ...

Bumping bottom - with reflections on upstroke ...

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RPC problem with “stroke” closure ? Unit rolling ?

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Hitting on the upstroke …

Comments? Questions?

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Pumping Unit Counterbalance

The most overlooked component of rod pumping systems?

Why Worry About Pumping Unit Counterbalance?

Practical field-based experience indicates that 75% of all pumping units are improperly counterbalanced. Incorrect counterbalance has two main drawbacks:

o Reduced gearbox life o Excessive energy usage

Remember that “balancing” a pumping unit is primarily aimed at the reduction of loading on the gearbox – and thus will not always reduce energy requirements. In some cases, energy usage may actually increase slightly. However, in cases where the unit is badly “out of balance”, energy usage will usually be reduced. One case study has shown an average energy use decrease of “7%-9%” per unit “balanced”. Perhaps the most important aspect of gearbox overloading caused by improper counterbalance is the effect on the life of the gearbox. A “rule of thumb” is that the life of a gearbox is reduced as the cube of the gearbox overload amount.

Example: A 320,000 lbs. gearbox has a calculated maximum torque of 456,000 in-lbs. (43% overload). Thus, 320000 / 456000 = (.7) cubed = 0.33 = (1/3) of the rated life of the gearbox (with ideal maintenance).

A “10%” overload reduces expected gearbox life by “25%” A “20%” overload reduces expected gearbox life by “42%”

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Essentially, counterbalance stores energy on the downstroke and then gives it up on the upstroke – all to reduce the required gearbox torque.

A pumping unit is said to be “weight heavy” if it has more counterbalance than is required – the weights are too big, are positioned too far out on the crank arms, or in some cases the crank arms themselves may be too large. When referred to as “rod heavy”, a pumping unit does not have enough counterbalance – just the opposite of the “weight heavy” description.

Remember that these “out of balance” conditions or a “balanced” situation will remain constant only if the well conditions do not change because of fluid level, pump wear, downhole friction, etc. If the pumping unit is located in a waterflood or steam injection response area, changing downhole conditions mean more frequent checks and adjustments – best accomplished through the use of RPCs and central site software programs.

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rotated. (See below.)

The unit of “torque” measurement is pounds of force times inches of

distance – resulting in “inch-pounds of torque”.

What is “torque” as it

relates to pumping unit

gearboxes?

When considering a pumping unit gearbox, the gear reducer is subjected

to two simultaneous torques. One is the torque applied by well loads at

the polished rod transmitted through the walking beam and pittman arms.

The other is the torque applied by the counterweights. (See below.)

The net torque is the algebraic

sum of these two torques because

they act in different directions. To

complicate matters, these applied

torques are continually changing.

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The angle of the walking beam at the tail bearing changes continuously

(R1) and the direction that the pittman arms “pull” on the walking beam

changes (R2), together causing both the lever arm length and the applied

forces to change at the tail bearing. Compare “

B

” in the graphics above.

because these forces and lever arms are changing continually, complicated

geometrical relationships are necessary to solve them – today’s wave

equation based torque analysis software programs handle this situation

easily.

Pumping unit “structural unbalance” also plays a part

in determining the torque load on the gearbox.

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Net Gearbox Torque (in-lbs) = TF * (W - SUB) – M SIN 0

Where –

W = Polished Rod Load (lbs.)

SUB = Structural Unbalance (lbs.)

TF = Torque Factor (at current crank position)

M = Maximum Counterbalance (CBT, in-lbs)

0 = Crank Angle (degrees)

Pumping Unit Counterbalance

“Counterbalance” can be defined as the measurement of the torque applied to a pumping unit gearbox by the unit cranks and counterweights. Basically, the term “CBE” (counterbalance effect) refers to the value of this torque as measured in the field in “pounds”. “CBT” (counterbalance torque, sometimes referred to as counterbalance “moment”) refers to this value when calculated in “in-lbs of torque”.

If a pumping unit gearbox had to supply all the force or torque necessary to operate the entire pumping system, it would have to be extremely large. The purpose of counterbalance (cranks and weights) is to help reduce the amount of work or torque that the gearbox has to provide to operate the system. Counterbalance helps the gearbox to handle the polished rod load on the upstroke and on the downstroke the gearbox moves the counterbalance using the rod load to help.

The Next Slide Illustrates Crank and Counterweight

Counterbalance Effects

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A pumping unit is said to be “balanced” when gearbox torque on the upstroke and on the downstroke is equal because the proper amount of counterbalance (cranks and counterweights) has been applied to the system.

Improperly “balanced” pumping units can and often do result in overloading the gearbox or prime mover – leading to excessive energy usage or expensive failures.

To determine if a pumping unit installation is “balanced”, use a diagnostic program that is capable of doing a torque analysis of the unit gearbox.

A less desirable method is to record the amps being using by the pumping unit on the upstroke and downstroke. Because amps are also a direct measurement of the torque requirements on the gearbox, unit “balance” or “unbalance” can be easily seen from such a measurement of amps for conventional pumping units.

Peak amp measurements are very difficult to accurately determine from “improved geometry” units (multiple peak amp readings in a single stroke).

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Using data taken from an API torque analysis calculation has major advantages, especially if the information is generated by a good computer analysis software program:

Counterbalance effect is easily calculated using data stored in the analysis software database - no potentially dangerous field

measurements are necessary.

Counterweight placement based on calculated “optimum”

counterbalance requirements eliminates "trial and error" positioning of weights when using amps as a guide. Weights are simply positioned according to analysis software calculations, greatly reducing time and cost considerations.

The risk of injury is reduced because the counterweights are “handled” only one time.

Using Measured Motor Amps vs. API Torque Analysis

Balancing beam pumped wells has traditionally been done using an amp meter or amp clamp, which requires a "trail and error" approach. This method is based on the fact that the current drawn by the electric motor is directly proportional to the work being done by the motor. The well is assumed to be balanced when the peak amps drawn on the upstroke, by the electric prime mover, is equal to the peak amps drawn on the downstroke. This method has several disadvantages:

It can be very time consuming (Remember, time is money) - weights may have to be adjusted several times and/or weights may have to be added or removed.

It likely will be inaccurate - the nature of using amp measurements to balance a pumping unit requires that the well be shut down for an extended period of time (likely more than once). This means a fluid level rise will occur in the casing, thus changing conditions from normal and likely causing “incorrect” amp readings. Also, improved geometry units are very difficult to balance because of multiple amp peaks during a single stroke.

It is dangerous - weights are heavy and not easy to handle, increasing the possibility of injury - especially if weights have to be removed or added to the pumping unit cranks.

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When Does It Become Necessary to “Balance” a Pumping Unit?

One “rule of thumb” suggests a “10%” difference between the calculated upstroke and downstroke peak gearbox torque. This percentage should be “refined” based on torque reduction and decreased energy use for a particular operating area.

Whenever necessary to reduce unacceptable torque levels on the gearbox.

Counterbalance Measurement:

The most accurate method to measure counterbalance in the field is to use a calibrated load cell to record the polished rod load (lbs.) when a conventional pumping unit is stopped as close to “90” or “270” degrees (wellhead at the observer’s right) as possible – with the brake off. For a Mark II unit, stop the unit at the “six o’clock” position.

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In most cases, however, a pumping unit will not stop in these positions and remain motionless (without using the brake) because of “unit unbalance” or other well conditions. In these situations, a chain must be used to “tie off” the polished rod to the wellhead and the unit

counterbalance “propped up” or “tied off”, as appropriate, before a measurement of counterbalance can be recorded.

Note: This method can be dangerous and should be undertaken only if absolutely required and only under the most rigorous of safe practices.

Field calculation of counterbalance is much safer, relatively simple to accomplish and provides acceptable accuracy - if the user is properly trained, and if the needed crank and weight information or counterbalance charts have been requested from the pumping unit manufacturer. There are three common scenarios that are usually encountered when “determining” counterbalance in the field.

Method One - “Formula or calculation” based counterbalance

measurement:

CBT (inlbs) = CBTC (inlbs) (cranks) + CBTW (inlbs) (weights)

CBTC (cranks) value is available from manufacturer information based on the individual crank identifier as observed from the cranks on each pumping unit. If this value is not provided in “inlbs.”, but in “lbs.”, use this formula to convert to “inlbs”: W (lbs.) x C.G = CBTC (inlbs.), where “W” is the weight of both cranks in pounds and “C.G” is the distance to the center of gravity of the cranks (in inches). The pumping unit manufacturer provides the “W” and “C.G” values.

CBTW (weights) = (M - X) (NW + nZ) – where …

“M” = Maximum distance from the center of the gearbox crankshaft to the center of gravity of the weight (inches)

“X” = Distance the weights are located from the long end of the cranks (inches) “N” = Number of master weights

“W” = Weight of each master weight (pounds) “n” = Number of auxiliary weights

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1. At each pumping unit, note the “ID” of each master and each auxiliary weight (if installed) and physically measure the distance from the end of the crank to the leading edge of each master weight (“X”).

2. Then, based on the “ID” of each weight – refer to the proper

manufacturer’s data for the “M” distance and for the weight in pounds of each master (“W”) or auxiliary weight (“Z”).

3. Use the formula to calculate the “CBT” value of the counterweights. Add the “CBTC” (cranks) value to the “CBTW” (weights) value.

This is the “CBT” value to be entered into whatever rod pumping analysis program is being used.

Important: If more than one “size” of counterweight is used on a pumping unit, repeat the entire calculation for each different size counterweight.

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Counterbalance Class Exercise

1

A Lufkin conventional pumping unit has 5456B cranks. Installed on the cranks are four # 5ARO master weights and two # 5A auxiliary weights, all measured at 6” (average distance) from the end of the cranks. From the Lufkin pumping unit catalog: “CBTC” = 70,303 inlbs., “M” = 40.1”, “W” = 913 lbs., and “Z” = 366 lbs. Calculate the “CBT” for this unit.

________________________________________

______________________________

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Counterbalance Class Exercise 1

A Lufkin conventional pumping unit has 5456B cranks. Located

on the cranks are four # 5ARO master weights and two # 5A

auxiliary weights, all measured at 6” from the end of the cranks.

From the Lufkin catalog: “CBTC” = 70,303 inlbs., “M” = 40.1”,

“W” = 913 lbs., and “Z” = 366 lbs. Calculate the “CBT” for this

unit.

CBT (inlbs) = CBTC (inlbs) (cranks) + CBTW (inlbs) (weights)

CBTW (weights) = (M - X) (NW + nZ)

CBTW = (40.1-6) * {4(913) + 2(366)}

CBTW = (34.1) * {(3652) + (732)}

CBTW = (34.1) * (4384)

CBTW = 149,494 inlbs.

CBT = 70,303 + 149,494

CBT = 219,797 inlbs.

Counterbalance Class Exercise 2

Using the same information available from exercise 1, calculate where to position the counterweights calculate (“X”) if the “optimal CBT” is 260,000 inlbs.

Was there a problem with the result?

“X” = M – {(OCBT-CBTC) / (NW + nZ)}

“X” = 40.1 – {(260,000-70,303) / (4*913 + 2*366)}

“X” = 40.1 – {(189,697) / (3652 + 732)} = 40.1 - {189,687 / 4384} “X” = 40.1 – 43.3

“X” = - 3.2”

Yes – the negative calculation of “X” indicates that additional counterbalance needs to be added to this pumping unit in order to achieve the goal of 260,000 inlbs. of counterbalance. What is the solution?

Re-calculate “X” by adding an third auxiliary weight

--“X” = M – {(OCBT-CBTC) / (NW + nZ)}

“X” = 40.1 – {(260,000-70,303) / (4*913 + 3*366)} “X” = 40.1 – {(189,697) / 3652 + 1098

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Method Two – Manufacturer’s Chart-based Counterbalance Determination

Some pumping unit manufacturers provide counterbalance tables to enable the user to determine “CBT” (inlbs.) or “CBE” (pounds), rather than calculating CBT as explained and “exercised” in method one.

Crank and weight information still must be physically gathered from each pumping unit. Under this scenario, the cranks will have some type of “position scale” marked off on each crank.

o The counterweights will have a pointer or arrow that “points” to their position on the crank. Record the “ID” of the cranks and of each weight and weight “position” as indicated by the “pointer”.

o Determine the “average” value of the position of all installed counterweights. o Then, use the appropriate manufacturer’s chart to retrieve the CBT values for the cranks and counterweights.

This is the value to be entered into an available analysis program. If the

manufacturer provides only CBE (pound) charts, determine this value in the same way as described above. Analysis programs may require the conversion of CBE to CBT before calculating gearbox torque.

Once the “average” position of all installed counterweights is calculated, charts similar to those shown in the following slides can be used to determine the correct “CBT” (Maximum Moment) or “CBE” (Effective Counterbalance) value.

3.5 + 3.5 + 9.5

+ 9.5 = 26 / 4 =

6.5

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CBT (in.lbs.)

Counterbalance

Chart

CBE (lbs.)

Counterbalance

Chart

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An American pumping unit with KA-117-53 cranks has two “RJ” master weights at position “2” and two “RJ’s” at position “10”. From the table provided on the next slide, calculate the existing “CBT”. (Note: The column labeled “Position of Counterweights” refers to the average position of all counterweights on the cranks.)

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An American pumping unit with KA-117-53 cranks has two “RJ” master weights at position “2” and two “RJ’s” at position “10”. From the table provided on the next slide, determine the existing “CBT”. (Note: The table column labeled “Position of Counterweights” refers to the average position of all counterweights on the cranks.)

2 + 2 + 10 + 10 = 24 / 4 = “6.0” average position From the counterbalance table: “CBT” is 1,145,480 inlbs.

Note that the “CBT” and “CBE” values shown in

manufacturer’s counterbalance sheets apply only to situations where there are four installed counterweights. If there are one – two – or three counterweights, these values will require adjustment based on the number of counterweights actually installed on the pumping unit.

Counterbalance Class Exercise 4

Normally, “Position of Counterweights” will very seldom be calculated as an “even” number. What changes would be required in the calculations if “average position” were calculated to be a decimal? (Ex: Not “6.0”, but “6.4”)

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_______________________________________

From the Counterbalance Chart:

“Position 6.0” = 1,145,480 & “Position 7.0” = 1,189,400

1,189,400 – 1,145,480 = 43,920 / 10 = 4,392 per decimal fraction. Thus, an average “Position 6.4” = 1,145,480 + (4 * 4,392)

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Visit the wellsite and note the number and type of beam weights attached to the walking beam.

Using the correct manufacturer’s API data sheets, refer to the effective

counterbalance portion of this information to determine the “CBE” value in pounds based on the number of beam weights installed.

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Converting “CBE” (pounds) to “CBT” (inlbs.)

Sources of “CBE” values:

From a recent wellsite dynamometer survey measurement From the pumping unit manufacturer’s counterbalance sheets From “CBE” values determined by using wellsite RPC functionality

Use this formula to manually convert “CBE” to “CBT” – then enter the “CBT” value directly into an analysis program:

CBT = (CBE – SU) x TF, where

-CBT = Counterbalance torque in “inlbs.”

CBE = Counterbalance effect at the polished rod in “lbs.”

SU = Structural unbalance in “lbs.” - defined as the force required at the polished rod to hold the unit walking beam in a horizontal position with the pitman arms disconnected from the crank arms. This force is “positive” when acting up on the polished rod or “negative” when exerting a downward pull. This value is included in the manufacturer’s API dimensional data sheets.

TF = Torque factor (ins.) can be defined as the number that, when multiplied by the load at the polished rod, results in the gearbox torque from that polished rod load. The pumping unit manufacturers can supply “torque factor tables” for each size unit. In addition, the number can be calculated from the unit’s geometric dimensions, but the calculations are complex and best done by a computer application.

Note: Use the “torque factor” at “90” degrees for “clockwise” unit rotation or the “torque factor” at “270” degrees for “counter-clockwise” unit rotation.

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Counterbalance Class Exercise 5

Given: CBE = 17,000 lbs. TF = 68.450” SU = 480 lbs. Calculate: CBT in “inlbs.” CBT = (CBE – SU) x TF CBT = (17,000 – 480) * 68.450 CBT = 16,520 * 68.450 CBT = 1,130,794 inlbs.

This would be the value to be entered into a rod pumping analysis software program.

Field Determination of Rotaflex Counterbalance

Total CBE = Standard Minimum CBE + Auxiliary CBE Standard Rotaflex CBE:

800DX and 900 Series = 9400 lbs. 1100 Series = 9800 lbs.

Auxiliary Rotaflex CBE:

The auxiliary counterweights are supplied in properly dimensioned metal blocks of varying thickness.

o To calculate the weight (lbs.) of the installed auxiliary counterweights, measure each stack of auxiliary

counterweights (four). Measure the height, width, and depth of each stack. (See pictures on the next slide.)

Use this formula to calculate the CBE value for each counterweight stack: (H” x W” x D” X 0.2833).

Add these four values together plus the “standard CBE” value for the installed Rotaflex unit size to obtain total installed CBE (lbs.).

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Manual Calculations for Balancing a Pumping Unit Using

Optimum Counterbalance Torque Calculations or a

Counterbalance Chart

The best and most accurate method of properly balancing a pumping unit is accomplished with a diagnostic torque analysis evaluation from a “wave equation” driven analysis program. Torque analysis programs calculate the degree of “unbalance” and the increased gearbox load caused by any unit “unbalance”. Also calculated is the “optimum counterbalance torque” needed to “balance” the pumping unit, the correct weights needed, and the proper position of the counterweights.

Once “optimum counterbalance torque” value is available, the position of the weights (“X”) to achieve the “optimum counterbalance torque” can be manually calculated and thus “balance” the pumping unit.

“X” = M – {(OCBT-CBTC) / (NW + nZ)} }, where “X” is defined as the overall average distance the weights are from the end of the cranks when all weights are the same size (Refer to the previous explanation of terms – other than “OCBT” is “optimum CBT”) To solve for “X” when one or more of the counterweights are a different size:

“Q” = Required counterbalance torque To move one weight:

“X” = (CBTC – Q) + {M1*(W1+Z1)} + {M1*(W1+Z1)} + {M1*(W1+Z1)} ) + {M1*(W1+Z1)} (W1 + Z1) To move two weights:

“X” = (CBTC – Q) + {M1*(W1+Z1)} + {M1*(W1+Z1)} + {M1*(W1+Z1)} ) + {M1*(W1+Z1)} (W1 + Z1) + (W2 +Z2)

To move three weights:

“X” = (CBTC – Q) + {M1*(W1+Z1)} + {M1*(W1+Z1)} + {M1*(W1+Z1)} ) + {M1*(W1+Z1)} (W1 + Z1) + (W2 +Z2) + (W3 + X3)

To move four weights:

“X” = (CBTC – Q) + {M1*(W1+Z1)} + {M1*(W1+Z1)} + {M1*(W1+Z1)} ) + {M1*(W1+Z1)} (W1 + Z1) + (W2 +Z2) + (W3 + X3) + (W4 + Z4)

If “X” is calculated to be a negative value, add more counterweight. If “X” is calculated to be more the value of “M”, reduce the amount of counterweight.

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Important:

Always move “lead” and “lag” weights to the same

distance from the end of the crank arms to prevent

changing the “phase angle” of the cranks – which could

affect gearbox torque calculation and correct

counterbalance determination.

In addition, it will be necessary to use the manufacturer’s counterbalance charts when the “X” position cannot be “calculated” as indicated previously by using a formula solution.

For example:

An American pumping unit has KA-117-53 cranks with three “RJ” counterweights located at position “5” on the cranks. It has been

determined that in order to “balance” the pumping unit - 1,100,000 in-lbs. of counterbalance is needed.

Where must the three counterweights be “re-positioned” to achieve this amount of counterbalance and thus “balance” the pumping unit?

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From the chart above, note that the two KA-117-53 cranks and four “RJ” counterweights at position “5” have a value of 1,101,560 in-lbs. The value of the cranks alone is 551,200 in-lbs. To calculate the value of a single counterweight, subtract the value of the cranks from the total value and divide by four:

(1,101,560 – 551,200) / 4 = 137,590 in-lbs. per counterweight. This means that the maximum value of the cranks and three counterweights at position “5” is: 551,200 + 3(137,590) = 963,970 in-lbs. By subtracting this number from the calculated optimum value, it can be determined how much addition

counterbalance is needed: 1,100,000 – 963,970 = 136,030 in-lbs.

Note that the counterbalance value difference between each position (“1” to “2”, “2” to “3”, “3” to “4”, etc.) is 43,920 in-lbs. To find the counterbalance value for a single counterweight, divide this number by four: 43,920 / 4 = 10,980 in-lbs.

Knowing that an additional 136,030 in-lbs. of “CBT” is needed to “balance” the unit, calculate how much “position” to add the current counterweight position of “5” for the three “RJ” counterweights: “X” = 136,030 / (3 X 10,890) = “4.1”. Therefore, “5” + “4.1” = “9.1” is the final calculated counterweight position for the three “RJ”