Tag: friction

Three Ways Static Electricity is Generated

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EXAIR published a white paper, Basics of Static Electricity, explaining what causes static electricity; how it is generated; and steps to eliminate it. Download it now by clicking this Link, and begin to remove the static issues in your plant or processes.

In this blog, I would like to expand on the subject about how static can be generated.  On a molecular scale, the outer electrons that are orbiting the nucleus can be “stripped” and redistributed from one atom to another.  This will cause an electrical charge imbalance called static.  An additional electron will create negatively charged static while atoms losing an electron will create a positively charged static.  With non-conductive materials like plastic, paper, rubber, glass, etc, the electrons cannot move back to the original atom. There are three common methods of static generation that will cause this phenomenon to occur.  I will explain each one in brief detail below:

Contact

Contact – Whenever objects hit each other, electrons can be passed to or received from the surface of another object. The number of electrons being transferred is based on the type of triboelectric material.  But, with plastic bottles or trays bumping into each other on conveyor belts, static can be generated relatively easy.

Detachment

Detachment – when one material is being separated from another material by peeling, electrons may not able to return back to the original molecule. Adhesive tape and protective films are prevalent in generating static charges by detachment because of the larger surface areas.  As an example; when the backing material is being removed from labels, the static will cause the labels to be misaligned or cause jams.

Frictional – This is one of the most common reasons for generating large static forces. It is caused by two non-conductive surfaces being rubbed together.  The amount of force being applied to the material as it slides back and forth will create higher static charges.   As an example, it is noticed when you rub a balloon on your hair.  The more times that you rub the balloon against your hair, the stronger the static forces, allowing the balloon to “stick” to the wall.  It is also noticed as sheets of material are stacked or running over rollers.

Static tends to propagate.  The more contact, detachment, and friction that occurs; the higher the static charges.  Even when the static is removed from the surface, static charges can still regenerate by the same mechanisms above.  So, controlling the static can be determined by the type of treatment as well as the location for removal.

Another variable that affects static generation is humidity.  Most process problems are noticed during the winter months as the ambient air is drier.  With a lower relative humidity, static can develop easier and with greater strength.  We always refer to winter as static season.  You may even notice this when you walk across the carpet and get zapped by touching a door handle. 

Production problems can occur like dirty surfaces, tearing, alignment, jamming and shock to staff with static.  EXAIR has a number of Static Eliminators to remove these process snags that can cost your company money.  You can contact an Application Engineer at EXAIR to discuss any static issues that are occurring.

John Ball
Application Engineer
Email: johnball@exair.com
Twitter: @EXAIR_jb

How To Choose The Right Style & Size Vacuum Cup

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When you are using vacuum to pick and place different objects, how do you know which style vacuum cup and size you need?  EXAIR offer’s (3) different styles of vacuum cups in a variety of sizes to cover a wide variety of applications.

The first is the round cup, they are suited to smooth, flat surfaces.  They grip and release quickly, hold their shape with frequent use, and grip well when used with vertical loads.  Also, round cups are offered with cleats for a better gripping power when the load is heavy.  Vacuum Cups with cleats provide extra rigidity that lends itself to heavy loads.  The extra rigidity prevents the cup from peeling away or deforming when a heavy load is required.

Large Round Cleated Vacuum Cup
Round Cup With Cleats

Next we have the Oval Cup, they provide the most gripping power due to their larger surface area.  They naturally lend themselves to heavier loads.  They are ideal for flat rigid sheet materials such as wood, glass, cardboard and composites.

Oval cup
Oval Cups Provide The Most Vacuum Which Make Them Ideal For Heavy Loads Due To Their Larger Contact Area

Last but not least we have the Bellows Cup, they are best suited for curved, uneven or textured surfaces.  The bellows or more accurately called “convolutions”, provide a collapsible area that allows the cup to quickly compress when it contacts the surface of item to be moved.  Please know that the grip/release time is greater due the the larger internal volume of the cup.

bellows cup
Bellows Cup Are Excellent For Uneven Or Textured Surfaces, However They Longer Pick-Up & Release Times

Here is an example: If we want to pick up a sheet of glass that weighs 20 lbs, what size vacuum cup(s) do we need?  Fortunately we have published charts in our catalog with the math already completed, but this is how we arrived at the answer.  Using the formula below:

Cup Diameter Formula

D = Cup Diameter
W = Weight of item
N = Factor of safety ( We recommend a safety factor of 2 for horizontal lifting and 4 for vertical lifting).  For this exercise we will use a safety factor of 4.
U = # of cups (Remember it takes a minimum of 3 points to make a plane)
V = Vacuum level (psi) – Exair Porous E-Vac’s are rated to 21″ HG, Non – Porous 27″ HG and the Adjustable E-Vac 25″ HG.   (Each inch of HG equals .491154 PSI)
π = 3.14159
μ = Vertical surface friction coefficient (.5 for typical non-porous materials such as metal, glass, stone, etc..)

Cup Diameter Formula Solved

The calculated diameter is 2.26″ so we would choose the next standard size which is 2.5″ diameter.

When you are looking for Vacuum Cup application assistance or expert advice on safe, quiet and efficient point of use compressed air products give us a call.   We would enjoy hearing from you!

Russ Bowman, CCASS

Application Engineer
EXAIR Corporation
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Considerations for Ejecting Parts with an Air Nozzle: Weight and Friction

Published on by johnball2014Leave a comment

I had a customer wanting to reject a container off a conveyor belt.  The container held yogurt, and when an optic detected a reject, they wanted to operate a solenoid to have a nozzle blow the container into a reject bin.  They had a range that went from 4 oz. (113 grams) for the small containers to 27 oz (766 grams) for the large.  He wanted me to suggest one nozzle for all sizes, as they would automatically regulate the pressure for the full range of products.  In looking at the largest size, this container will need the most force to blow off the conveyor.  The two factors that affects the force in this type of application is weight and friction.  When it comes to friction, it is generally an unknown for customers.  So, I was able to help with a couple of things to determine the friction force.

Friction is a dimensionless number that represents the resistance created between two surfaces.  We have two types; static friction, ms, and kinetic friction, mk.  Static friction is the maximum amount of resistance before the object begins to move or slide.  Kinetic friction is the amount of resistance that is created when the object is moving or sliding.  So, Static friction is always greater than kinetic friction, ms > mk.  For this application, we will use an air nozzle to “shoot” horizontally to hit the rejected product.

Let’s take look at our customer’s application.  We have a system to reject a non-conforming part with air.  The conveyor has a urethane belt.  The container is plastic.  For the largest container, they have a weight of 27 oz. (766 grams).  Being that the conveyor belt is only 12” (30.5 cm) wide, we can determine that if we get the part moving, it will continue off the belt and into the reject bin.  The equation for the maximum amount of force required to move a container is below as Equation 1.

Equation 1

Fs = ms * W

Fs – Static Force in ounces (grams)

ms  – Static Friction

W – Weight in ounces (grams)

One way to determine the amount of force is to use a scale similar to a fish scale.  The scale should have a maximum indicator to help capture the maximum amount of force.  You will have to place the object on the same belt material because different types of materials will create different static forces. Keep the scale perpendicular to the object, and slowly pull on the scale.  Once the part begins to move, record the scale reading.  For the exercise above, it showed 9.6 oz. (271 grams) of force to move the 27 oz. (766 gram) object.

Another way would be to calculate the static friction, ms.  Static friction can be found by the angle at which an object starts to move.  By placing the container on a section of supported urethane conveyor belt, you can lift one end until the object starts to slide.  The height of the lift can be measured as an angle.  As an example, we take 3 feet (0.9 meter) of supported urethane conveyor belt, and we lifted one end to a height of 1 foot (0.3 meters) before the 27 oz (766 gram) container moved.  To determine static friction, it is the tangent of that angle that you lifted.  With some right triangle trigonometry equations, we get an angle of 19.5o.  Thus, ms = tanq or ms = tan(19.5o) = 0.354.  If we plug this into Equation 1, we get the following:

Imperial Units                                                    SI Units

Fs = ms * W                                                         Fs = ms * W

= 0.354 * 27 oz.                                                = 0.354 * 766 grams

= 9.6 oz. of force                                              = 271 grams of force

1″ Flat Super Air Nozzles

Now that we have the static force, we want to be slightly higher than that.  In looking at the force requirements that are published in the EXAIR catalog, it shows that the model 1126 1” Super Flat Air Nozzle has a 9.8 oz. (278 grams) of force at 80 PSIG (5.5 Bar).  This force is measured at a 12” (30.5 cm) distance with a patented .015” (0.38mm) shim.  So, this nozzle will be able to slide the largest container into the reject bin.

1″ Flat Super Air Nozzle shims

To expand on the benefits in using the EXAIR Flat Super Air Nozzles, the force can be changed easily with a regulator or with a Shim Set.  This is a unique feature as most competitive flat nozzles do not allow you to do this.  The patented shims control the force rating in a wide range with lower air consumption and lower noise levels; making them safe and efficient.  So, if this manufacturer decided to produce other sizes in the future, then they could change the shim to target even larger containers.  The flexibility of using the EXAIR Flat Super Air Nozzles allow you to increase or decrease the force by just removing two screws and changing the thickness of the shim inside.  EXAIR does offer a pack of shims with different thicknesses which are called a Shim Set.

With air pressure or shim manipulation, the customer could use the same nozzle for the yogurt containers.  If you have any applications that need products to be rejected quickly, an Application Engineers at EXAIR will be happy to help you with a solution.

John Ball
Application Engineer
Email: johnball@exair.com
Twitter: @EXAIR_jb

Photo: Yogurt by BUMIPUTRA.  Pixabay Licence

Calculating Static Friction To Eject Parts with Air

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2″ Flat Super Air Nozzle

In today’s fast-paced world, companies are always looking for ways to do things faster, cheaper, more efficiently without sacrificing safety.

A cereal company had a high-speed system to check the quality of each box of cereal.  When a box did not meet the quality criteria for visual and/or weight, the box would be rejected.  The rejection system that they used was a quick blast of compressed air to remove the box from the conveyor line into a non-conforming bin.  For their first attempt, they tried to use a ¼” copper tube with a solenoid valve attached to a reservoir tank.  When a “bad” box was detected, the solenoid would be triggered, and compressed air would “shoot” the box off the rubber conveyor belt.  The ¼” copper tube can be an inexpensive, common, and easy-to-use device; but they found that the copper tube was very loud (above OSHA limits for noise exposure) and not very effective.  As a note, this company had a safety committee, and they wanted to keep all blowing devices below 80 dBA in this department.  The ¼” copper tube was around 100 dBA.  So, they contacted EXAIR to get our expertise on this type of application.

The cereal company gave me some additional details of the operation.  The box weighed 26 oz. (740 grams) with a dimension of 7.5″  wide by  11″ tall by  2 3/4″ deep (19 cm X 28 cm X 7 cm respectively).  The issue with the ¼” copper tube was the small target area compared to the area of the box.  With any slight variation in the timing sequence, the force would miss the center of mass of the box.  The box could then spin and remain on the conveyor belt.  This would cause stoppage and disruption in the system.  They asked if EXAIR had a better way to remove the defective boxes.

I recommended a model 1122, 2” Flat Super Air Nozzle.  The reason for this style of nozzle was for a variety of reasons.  First, we needed a larger area to “hit” the box.  This Flat Super Air Nozzle has a width of 2” versus the ¼” copper tube.  This increased the target area by 8 times.  So, any small variations in time, we could still hit the center of mass and remove the box.  The second reason was the force rating.  The model 1122 has a force of 22 oz. (624 grams) at 80 PSIG (5.5 bar).  This is slightly under the 26 oz. (740 grams) weight of the cereal box, but we are just sliding the box and not lifting it.  If we can overcome the static friction, then the box can be easily removed.  With Equation 1, we can calculate the required force.

Equation 1:

Fs = ms * W

Fs – Static Force (grams)

ms  – Static Friction

W – Weight (grams)

From the “Engineering Toolbox”, the static friction between rubber and cardboard is between 0.5 to 0.8.  If I take the worse case condition, I can calculate the static force between the belt and cereal box using Equation 1:

Fs = 0.8 * 740 grams

Fs = 592 grams

The model 1122 has a force of 22 oz. (624 grams), so plenty enough force to move the box from the rubber conveyor belt.

The third reason for this nozzle is the noise level.  The noise level of the model 1122 is 77 dBA, well below the safety requirement for this company.  Noise levels are very important in industries to protect operators from hearing loss, and the model 1122 was able to easily meet that requirement.  I added an additional reason for recommending the 2” Flat Super Air Nozzle; compressed air savings.  Companies sometimes overlook the cost when using compressed air for blow-off devices.  In this comparison, the ¼” copper tube will use 33 SCFM (934 SLPM) at 80 PSIG (5.5 bar) while the model 1122 will only use 21.8 SCFM (622 SLPM).  This is a 33% reduction in compressed air; saving them money.

At the intro, I mentioned that companies are looking to do things faster, cheaper, more effective without sacrificing safety.  For this company, we were able to increase production rates by removing every cereal box from the conveyor belt.  We also saved them money by reducing the compressed air requirement as well as keeping it safe by reducing noise.

If you have an application that needs products to be moved by air, you can contact an Application Engineer at EXAIR to help you with a solution.

John Ball
Application Engineer
Email: johnball@exair.com
Twitter: @EXAIR_jb