Pressure – The Inner Working of the Basic Pressure Gauge

Everyday here at EXAIR we talk about pressure, specifically compressed air pressure. The other day I was looking up our model 9011, 1/4″ NPT Pressure Gauge , and it got me to wondering just how does this small piece of industrial equipment work. The best way to find out is to tear it apart.


Most mechanical gauges utilize a Bourdon-tube. The Bourdon-tube was invented in 1849 by a French watchmaker, Eugéne Bourdon.  The movable end of the Bourdon-tube is connected via a pivot pin/link to the lever.  The lever is an extension of the sector gear, and movement of the lever results in rotation of the sector gear. The sector gear meshes with a spur gear (not visible) on the indicator needle axle which passes through the gauge face and holds the indicator needle.  Lastly, there is a small hair spring in place to put tension on the gear system to eliminate gear lash and hysteresis.

When the pressure inside the Bourdon-tube increases, the Bourdon-tube will straighten. The amount of straightening that occurs is proportional to the pressure inside the tube. As the tube straightens, the movement engages the link, lever and gear system that results in the indicator needle sweeping across the gauge.

Pressure Gauge Top

The video below shows the application of air pressure to the Bourdon-tube and how it straightens, resulting in movement of the link/lever system, and rotation of the sector gear –  resulting in the needle movement.

If you need a pressure gauge or any of the EXAIR Intelligent Compressed Air® Products, feel free to contact EXAIR and myself or one of our Application Engineers can help you determine the best solution.

Brian Bergmann
Application Engineer

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Compressed Air Regulators: The Design and Function


Compressed air regulators are a pressure reducing valve that are used to maintain a proper downstream pressure for pneumatic systems.  There are a variety of styles but the concept is very similar; “maintain a downstream pressure regardless of the variations in flow”.  Regulators are very important in protecting downstream pneumatic systems as well as a useful tool in saving compressed air in blow-off applications.

The basic design of a regulator includes a diaphragm, a stem, a poppet valve, an orifice, compression springs and an adjusting screw.  I will break down the function of each item as follows:

  1. Diaphragm – it separates the internal air pressure from the ambient pressure. They are typically made of a rubber material so that it can stretch and deflect.  They come in two different styles, relieving and non-relieving.  Relieving style has a small hole in the diaphragm to allow the downstream pressure to escape to atmosphere when you need to decrease the output pressure.  The non-relieving style does not allow this, and they are mainly used for gases that are expensive or dangerous.
  2. Stem – It connects the poppet valve to the diaphragm. This is the “linkage” to move the poppet valve to allow compressed air to pass.  As the diaphragm flexes up and down, the stem will close and open the poppet valve.
  3. Poppet valve – it is used to block the orifice inside the regulator. It has a sealing surface to stop the flowing of compressed air during zero-flow conditions.  The poppet valve is assisted by a spring to help “squeeze” the seal against the orifice face.
  4. Orifice – it is an opening that determines the maximum amount of air flow that can be supplied by the regulator. The bigger the orifice, the more air that can pass and be supplied to downstream equipment.
  5. Compression springs – they create the forces to balance between zero pressure to maximum downstream pressure. One spring is below the poppet valve to keep it closed and sealed. The other spring sits on top of the diaphragm and is called the adjusting spring.  This spring is much larger than the poppet valve spring, and it is the main component to determine the downstream pressure ranges.  The higher the spring force, the higher the downstream pressure.
  6. Adjusting screw – it is the mechanism that “squeezes” the adjusting spring. To increase downstream pressure, the adjusting screw decreases the overall length of the adjusting spring.  The compression force increases, allowing for the poppet valve to stay open for a higher pressure.  It works in the opposite direction to decrease the downstream pressure.

With the above items working together, the regulator is designed to keep the downstream pressure at a constant rate.  This constant rate is maintained during zero flow to max flow demands.  But, it does have some inefficiencies.  One of those issues is called “droop”.  Droop is the amount of loss in downstream pressure when air starts flowing through a regulator.  At steady state (the downstream system is not requiring any air flow), the regulator will produce the adjusted pressure (If you have a gage on the regulator, it will show you the downstream pressure).  Once the regulator starts flowing, the downstream pressure will fall.  The amount that it falls is dependent on the size of the orifice inside the regulator and the stem diameter.  Charts are created to show the amount of droop at different set pressures and flow ranges (reference chart below).  This is very important in sizing the correct regulator.  If the regulator is too small, it will affect the performance of the pneumatic system.

The basic ideology on how a regulator works can be explained by the forces created by the springs and the downstream air pressures.  The downstream air pressure is acting against the surface area of the diaphragm creating a force.  (Force is pressure times area).  The adjusting spring force is working against the diaphragm and the spring force under the poppet valve.  A simple balanced force equation can be written as:

Fa  ≡ Fp + (P2 * SA)

Fa – Adjusting Spring Force

Fp – Poppet Valve Spring Force

P2 – Downstream pressure

SA – Surface Area of diaphragm

If we look at the forces as a vector, the left side of the Equation 1 will indicate a positive force vector.  This indicates that the poppet valve is open and compressed air is allowed to pass through the regulator.  The right side of Equation 1 will show a negative vector.  With a negative force vector, the poppet valve is closed, and the compressed air is unable to pass through the regulator (zero flow).

Let’s start at an initial condition where the force of the adjusting spring is at zero (the adjusting screw is not compressing the spring), the downstream pressure will be zero.  Then the equation above will show a value of only Fp.  This is a negative force vector and the poppet valve is closed. To increase the downstream pressure, the adjusting screw is turned to compress the adjusting spring.  The additional spring force pushes down on the diaphragm.  The diaphragm will deflect to push the stem and open the poppet valve.  This will allow the compressed air to flow through the regulator.  The equation will show a positive force vector: Fa > Fp + (P2 * SA).  As the pressure downstream builds, the force under the diaphragm will build, counteracting the force of the adjusting spring.  The diaphragm will start to close the poppet valve.  When a pneumatic system calls for compressed air, the downstream pressure will begin to drop.  The adjusting spring force will become dominant, and it will push the diaphragm again into a positive force vector.  The poppet valve will open, allowing the air to flow to the pneumatic device.  If we want to decrease the downstream air pressure, the adjusting screw is turned to reduce the adjusting spring force.  This now becomes a negative force vector; Fa < Fp + (P2 * SA).  The diaphragm will deflect in the opposite direction.  This is important for relieving style diaphragms.  This deflection will open a small hole in the diaphragm to allow the downstream air pressure to escape until it reaches an equal force vector, Fa = Fp + (P2 * SA).  As the pneumatic system operates, the components of the regulator work together to open and close the poppet valve to supply pressurized air downstream.

Compressed air is expensive to make; and for a system that is unregulated, the inefficiencies are much greater, wasting money in your company.  For blow-off applications, you can over-use the amount of compressed air required to “do the job”.  EXAIR offers a line of regulators to control the amount of compressed air to our products.  EXAIR is a leader in manufacturing very efficient products for compressed air use, but in conjunction with a regulator, you will be able to save even more money.  Also, to make it easy for you to purchase, EXAIR offer kits with our products which will include a regulator.  The regulators are already properly sized to provide the correct amount of compressed air with very little droop.   If you need help in finding the correct kit for your blow-off application, an Application Engineer at EXAIR will be able to help you.

John Ball
Application Engineer
Twitter: @EXAIR_jb

People of Interest: Robert Boyle – January 25, 1627 – December 31, 1691

Robert Boyle was born on January 25, 1627 in Lismore Castle, County of Waterford, Ireland.  He was an Anglo-Irish natural philosopher, chemist, physicist and dabbled in many other areas of study. He published the book The Sceptical Chymist in 1661, and many consider him and his work as the foundation of modern chemistry.  He was a very devout Anglican, and published numerous works in this area as well.

Robert Boyle

One of Boyle’s most famous discoveries was to become the first of the gas laws, relating the pressure of a gas to its volume. With Robert Hooke, a young university student as his laboratory assistant, Boyle began experimenting with air.  Together they made their first great discovery, now known as Boyle’s Law.

J-Tube 2
Boyle used a ‘J’ Tube – Sealed on the Short End, and Open at the Long End

The experiment was performed using a ‘J’ shaped glass tube sealed on the shorter leg, and open to atmosphere on the longer leg.  Quicksilver (mercury) was poured into the tube, such that the level was equal on each side. The volume of the trapped air was noted. Additional mercury was poured into the tube and it was observed that the mercury did not stay level, and measurements of the heights on each tube leg were recorded.  The height difference of the mercury is effectively a measure of the pressure of the trapped air. Boyle, through the experiment and the data,  discovered a relationship between the volume and the pressure of air.  The data as published, is shown below.

Boyle's Data

Boyle noticed the pressure times the volume of air for the initial condition equaled the pressure times the volume at any other mercury height.

Known as Boyle’s Law – P ∝ 1/V,      pressure is proportional to the inverse of the volume

Alternately, PV = k,       pressure times volume is equal to a constant

For comparing the same substance under two different sets of conditions, the law can be expressed as P1V1 = P2V2

Of note is that Boyle’s Law, combined with Charles’s law and Gay-Lussac’s Law formed the combined gas law, and in combination with Avogadro’s law is the basis for the ideal gas law – PV=nRT, which include temperature, the amount of the substance, and the ideal gas constant to the mix.

It is noted that Boyle credited fellow scientist Richard Towneley for making the connection between the pressure of a gas and volume, but Boyle’s experiments and observations using the ‘J’ tube confirmed Towneley’s predictions, and the rest as they say is history.

If you would like to talk about compressed air or any of the EXAIR Intelligent Compressed Air® Products, feel free to contact EXAIR and myself or one of our Application Engineers can help you determine the best solution.

Brian Bergmann
Application Engineer

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Experiment Data from the book New Experiments Physico-Mechanicall, Touching the Spring of the Air, and Its Effects (1660)


Intelligent Compressed Air: How to Develop a Pressure Profile

An important part of operating and maintaining a compressed air system is taking accurate pressure measurements at various points in the compressed air distribution system, and establishing a baseline and monitoring with data logging.  A Pressure Profile is a useful tool to understand and analyze the compressed air system and how it is functioning.

Pressure Profile 1
Sample Pressure Profile

The profile is generated by taking pressure measurements at the various key locations in the system.  The graph begins with the compressor and its range of operating pressures, and continues through the system down to the regulated points of use, such as Air Knives or Safety Air Guns.  It is important to take the measurements simultaneously to get the most accurate data, and typically, the most valuable data is collected during peak usage periods.

By reviewing the Pressure Profile, the areas of greatest drop can be determined and the impact on any potential low pressure issues at the point of use.  As the above example shows, to get a reliable 75 PSIG supply pressure for a device or tool, 105-115 PSIG must be generated, (30-40 PSIG above the required point of use pressure.)  As a rule of thumb, for every 10 PSIG of compressed air generation increase the energy costs increase 5-7.5%

By developing a total understanding of the compressed air system, including the use of tools such as the Pressure Profile, steps to best maximize the performance while reducing costs can be performed.

If you have questions about getting the most from your compressed air system, or would like to talk about any EXAIR Intelligent Compressed Air® Product, feel free to contact EXAIR and myself or one of our Application Engineers can help you determine the best solution.

Brian Bergmann
Application Engineer

Send me an email
Find us on the Web 
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Video Blog: How To Rebuild Pressure Regulators

Today’s video blog is a how-to on rebuilding EXAIR pressure regulators.   Regulators can wear out over time and extensive adjustment as well as if they are not used on a clean compressed air supply.  If you have any questions on an EXAIR product, please contact an Application Engineer.

Thanks for watching!



Brian Farno
Application Engineer Manager

Consider these Variables When Choosing Compressed Air Pipe Size

Here on the EXAIR blog we discuss pressure drops, correct plumbing, pipe sizing, and friction losses within your piping system from time to time.   We will generally even give recommendations on what size piping to use.  These are the variables that you will want to consider when selecting a piping size that will suit your need and give the ability to expand if needed.

The variables to know for a new piping run are as follows.

  • Flow Rate (SCFM) of demand side (products needing the supplied compressed air)
  • System Pressure (psig) – Safe operating pressure that will account for pressure drops.
  • Minimum Operating Pressure Allowed (psig) – Lowest pressure permitted by any demand side point of use product.
  • Total Length of Piping System (feet)
  • Piping Cost ($)
  • Installation Cost ($)
  • Operational Hours ( hr.)
  • Electical Costs ($/kwh)
  • Project Life (years) – Is there a planned expansion?

An equation can be used to calculate the diameter of pipe required for a known flow rate and allowable pressure drop.   The equation is shown below.

A = (144 x Q x Pa) / (V x 60 x (Pd + Pa)
A = Cross-Sectional are of the pipe bore. (sq. in.).
Q = Flow rate (cubic ft. / min of free air)
Pa = Prevailing atmospheric absolute pressure (psia)
Pd  = Compressor discharge gauge pressure (psig)
V = Design pipe velocity ( ft/sec)

If all of these variables are not known, there are also reference charts which will eliminate the variables needed to total flow rate required for the system, as well as the total length of the piping. The chart shown below was taken from EXAIR’s Knowledge Base.

Airflow Through 1/4″ Shed. 40 Pipe

Once the piping size is selected to meet the needs of the system the future potential of expansion should be taken into account and anticipated for.   If no expansion is planned, simply take your length of pipe and start looking at your cost per foot and installation costs.    If expansions are planned and known, consider supplying the equipment now and accounting for it if the additional capital expenditure is acceptable at this point.

The benefits to having properly sized compressed air lines for the entire facility and for the long term expansion goals makes life easier.   When production is increased, or when new machinery is added there is not a need to re-engineer the entire system in order to get enough capacity to that last machine.   If the main compressed air system is undersized then optimal performance for the facility will never be achieved.   By not taking the above variables into consideration or just using what is cheapest is simply setting the system up for failure and inefficiencies.   All of these considerations lead to an optimized compressed air system which leads to a sustainable utility.

Brian Farno
Application Engineer Manager

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