Calculating Force and Pressure For Air Nozzles

I assisted with an application where logs were being shaved to make thin laminate.  Because the logs were non-concentric or entirely smooth, the beginning of the sheet was riddled with scrapes and defects until it was about 8 foot (2.4 meters) long.  This was a very quick process, and once good product was coming from a shaved log, the machine would divert the material from the scrap bin to the production feed line with a nip roll.  At the speeds that the material was traveling, they needed to kept pressure on the leading edge of the sheet so that it would not “curl” up before the nip roll closed and grabbed the sheet. The drive rolls were pushing the laminate product toward the nip roll and they needed to keep the curl pushed flat along a plate and wondered if we had a product that could accomplish this.

We suggested a series of 2” flat air nozzles, model 1122, to keep the product pressed down on the plate with the force from the airflow.  In their trial runs, they tried to find the correct amount of air pressure to keep the product flat.  Once they found the pressure required, they noticed that the thin and delicate laminate was getting damaged.  Of course, it was just at the beginning length when it was being held in place as it slid into the nip roll, approximately 3 feet (0.9 meters).  Like any company, they did not want to waste any more product and wondered if we had anything else that we could recommend.

Thus a question was presented, and a solution was needed.  In thinking about this, it took me to my Michigan days where snow was abundant.  When walking on snow, you would fall through, but if you had snow shoes, you could stay on top of the snow.  This brought me to the factors of Pressure and Force.  Like with the laminate, if a smaller area does damage to the product (boots through the snow), can we expand the area to keep it from being damaged (snow shoes on top of the snow).

Snow Shoes

Snow Shoes

With the application, we needed to apply the same force on the material.  The equation for force is F = P *A (Equation 1), where F – Force, P – Pressure, and A – Area.

We can do an equality statement from Equation 1 which shows F = P1 * A1 = P2 * A2 (Equation 2).  The amount of pressure required from other EXAIR products can be determined, i.e. if I can double the surface area, then I can reduce the pressure by ½.  For model 1122, we can determine the pressure that was generated from Equation 1 and from the catalog data:

Imperial Units of Model 1122                                                      S.I. Units of Model 1122

F = 1.4 lbf (catalog)                                                                       F = 0.624 Kg (catalog)

A1 = Length X Width                                                                    A1 = Length X Width

= 5 inches X 2 inches (catalog)                                                   = 12.7 cm X 5.1 cm (catalog)

= 10 in^2                                                                                         = 64.8 cm^2

P1 = F/A1 (Rearranging Equation 1)                                         P1 = F/A1 (Rearranging Equation 1)

= 1.4 lbf/10 in^2                                                                            = 0.624 Kg/64.8 cm^2

= 0.14 PSI (pounds per in^2)                                                     = 0.0096 Kg/cm^2

Super Air Amplifier

Super Air Amplifier

Now that we have all the information from model 1122, we can determine the pressure required for a different product to keep the force the same.  With the 2” Super Air Amplifier, model 120022, it has a much larger footprint than the 2” flat air nozzle, model 1122.  So, with Equation 2, we can determine the amount of pressure required.  We will use model 1122 for our P1 and A1, and we will use model 120022 for P2 and A2.  From the catalog data for model 120022, we get a target area as follows:


Imperial Units for Model 120022                                               S.I. Units for Model 120022

A2 = pi * (diameter/2)^2                                                              A2 = pi * (diameter/2)^2

= 3.14 * (5.15 in/2)^2                                                                    = 3.14 * (13.1 cm/2)^2

= 20.8 in^2                                                                                      = 134.7 cm^2


When we apply the information to Equation 2, we get the following information:


Imperial Units                                                                                  S.I. Units

P2 = P1 * A1 / A2                                                                              P2 = P1 * A1 / A2

=(0.14 PSI * 10 in^2) / 20.8 In^2                                               =(0.0096 Kg/cm^2 * 64.8cm^2) / 134.7 cm^2

= 0.067 PSI                                                                                       =0.0046 Kg/cm^2


Now that the area was increased like the snow shoes above, the pressure was reduced and no additional waste was incurred.  Sometimes you have to think outside the igloo.  As with any application or product, you can always contact us at EXAIR for help.


John Ball
Application Engineer


Image courtesy of VasenkaPhotography. Creative Comment License

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