© Executive Confidantes, 2018

CASE STUDY ENGINEERING

BACKGROUND:
 

The transporting of tube and pipe throughout a mill is generally done by a conveyor system. The main goal is to get the material from point A to point B to maintain production flow. This is generally accomplished by belt conveyors (flat) or “V” roll conveyors (angled). For example, to transport rectangular material, either a flat belt or 90 degree “V” roll could be used. The belt carrying method is simple, but does not provide side to side support. However, it can work with any material with at least one flat side, and flat material tends to require very little side guidance. The “V” drive roll solution provides both transportation as well as side to side support. However, it requires that the angle of the material being transported approaches 90 degrees to be most efficient.
 

If a round pipe was transported on a flat conveyor, there would be a tendency for the pipe to roll sideways. To maintain the position on the transportation system, side forces must be applied to the pipe to prevent this from happening. Thus a compromise is desired, and achieved with a conveyor “V” drive roll. As the roll turns, forward motion is achieved by the friction between the pipe and roll. Rolls that are motor driven are called drive rolls.
 

If we assume that a 150 degree angle is chosen to compensate for the variation in material straightness along a 20’ length, then Figure 1 below represents the 150 degree drive rolls present on a typical conveyor system supporting a pipe with an OD of 365mm. The pipe is made of carbon steel and has a wall thickness of 20mm.


Referring to the diagram, the weight (W) of the pipe will create a horizontal and vertical force at each point of contact with the drive roll. The drive roll must supply an equal and opposite force at each point of contact or the pipe will break the roll. To understand the manner in which the forces interact, we shall initially assume that the pipe is symmetrical, causing the weight to be evenly distributed at each contact point. The following equations apply to conveyor drive rolls where the drive roll angle is indicated by 2Θ:


W = SG x A
A = (OD - WT) x WT x
π
W = SG x (OD - WT) x WT x π
W = 0.02466 x (OD - WT) x WT
Where W = theoretical weight per meter (kg/m)

SG = Specific Gravity of carbon steel (7.85 kg/m³ )
A = cross sectional area of pipe (m² )
OD = nominal outside diameter in mm
WT = nominal wall thickness in mm


For a pipe with an OD of 365mm and a WT of 20mm, the weight per meter is:

W = 0.02466 x (365 – 20) x 20 = 170.154 kg/m, which is divided in half by the two contacts points with the roll.

FIGURE 1

As the perfectly round pipe rolls down the perfectly aligned conveyor, the horizontal force applied by the weight of the pipe to each contact point on the conveyor rolls is: 


SF = cos(Θ) x W


Where SF = Side Force (pipe exerts on drive roll)

              Θ = ½ the roll angle


SF = cos(75) x 85.1 = 0.259 x 85.1 = 22.0 kg


Applying the same formulas to Figure 2, this represents a 120 degree conveyor drive roll:

SF = cos(60) x 85.1 = 0.5 x 85.1 = 42.55 kg at each contact point on the drive roll.

FIGURE 2

ANALYSIS:
 

The importance of these last two calculations is that they identify the amount of horizontal force that would be required to overcome the weight of the pipe and roll it up the drive roll, resulting in a combination of horizontal and vertical movement. The 150 degree drive roll requires approximately ½ the horizontal force, making it twice as susceptible to being moved out of center.


Consider now the imperfect world. The pipe is neither straight nor round. As it progresses down the conveyors, it moves from side to side. By viewing the relative position of the pipe within the roll, it is apparent that the larger roll angle provides less horizontal control as the pipe advances.


A test system meeting the specifications of API 5CT requires precise positioning of the pipe within the tester (alignment and centering). To achieve the positioning and transportation (linear motion through the mechanics) pinch assemblies may be used. The pinch assemblies, together with the slide and elevate mechanisms and weldments, comprise the center section upon which the testers are located. In a pinch assembly, the bottom roll is referred to as the drive roll, while the upper roll is referred to as the pinch roll. The drive roll is driven by a motor whose speed is varied to match the conveyor line. The pinch roll applies downward pressure so that there is adequate friction to advance the product as well as stabilize it within the drive roll. To achieve repeatability, the center section needs to consistently hold the pipe within a fraction of a millimeter of true center.

To meet API 5CT, let us consider the condition where a magnetic sensor is spinning around the pipe at a nominal air gap of 2.0mm. Let the nominal center of the pipe coincide with the center of rotation of the sensor. Let the pipe have a straightness of 2mm /m and an out of round condition of 0.5mm. A magnetic flux leakage (MFL) tester operates by introducing the pipe to a magnetic field and measuring the disturbances in the field caused by discontinuities in the pipe. The strength of the magnetic field
produced by the disturbance will be inversely proportional to the square of the distance between the sensor and the pipe. To put this is simple terms, as the distance between the sensor and the pipe varies from 1.0mm to 3.0mm from the nominal 2.0mm, there will be an effect on the signal amplitude:


A /A  = (d 2 )² / (d 1 )²
Let A = Amplitude at 2.0mm referenced as A = 100%
Let A  = Amplitude at 1.0mm then A₂  = 400%
Let A  = Amplitude at 3.0mm then A₂  = 44.4%
The difference in Amplitude caused by a change of +/-1mm would be:
400% - 44.4% = 355.5% or 11dB.


This makes it quite clear that centering the pipe within the rotary, minimizing the air gap and keeping the pipe stable are critical to repeatability.


The preferred method of adding stability to the test is to add additional force to the pipe to prevent it from moving side to side or up and down. Figure 3 shows the configuration with a 150 degree pinch roll and 150 degree drive roll where the drive roll makes contact with the pipe at two points: the bottom left and bottom right of the pipe. Each point of contact must resolve forces in three directions: the “X” direction for horizontal control, the “Y” direction for vertical control and the “Z” direction for linear motion through the center section.


The “X” direction will be the horizontal forces applied by the drive roll to balance the horizontal component of the pipe’s weight on the drive roll plus the horizontal component of the downward force applied from the pinch roll.


The “Y” direction is the upward force applied by the drive roll to counter the weight of the pipe on the roll plus the vertical component of the downward force applied from the pinch roll.


The “Z” direction is the forward force applied by the drive roll to overcome the friction developed by the weight of the pipe on the drive roll plus the vertical component of the downward force applied from the
pinch roll.

FIGURE 3

As explained earlier, the horizontal force required to move the pipe off center in the 150 degree roll is approximately ½ that of the 120 degree roll. By adding additional downward force with the pinch roll, it is possible to increase the effective downward force, but not change the ratio.


As seen from this configuration, the effective downward force applied by the top pinch roll is:


EDF = sin(Θ) x DF

Where DF = Downward Force (roll exerts on product)

              Θ = ½ the roll angle
EDF = sin(75) x DF = 0.966 x DF
To simplify the example, we shall approximate EDF = DF.


Following this line of analysis, the top pinch roll will need to deliver 22 kg/m to each contact point on the 150 degree drive roll. Therefore, DF = 44 kg/m, just to match a 120 degree drive roll without a pinch roll.


Therefore the use of 120 degree drive rolls with 150 degree pinch rolls is a better alternative than using 150 degree drive rolls for OCTG material handling systems, as show in Figure 4.

FIGURE 4