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# Fluid Power - Introduction to Pressure: The Basics

Submitted by Russ Rochambeau, Engineering Manager, IFPS Hydraulic Specialist, Valin Corporation
When discussing fluid power, pressure is the basis for producing any kind of work.  Work cannot be achieved without pressure.  In order to understand the finer points of fluid power, one must first understand the concept of pressure.  Pressure is defined as the measure of force acting perpendicular to a unit area.  In mathematics, pressure is stated as P=F/A where P equals the Pressure, F equals force and A equals the area we are discussing.  To put the discussion into relatable terms, right now you are experiencing approximately 14.7 psi of pressure exerting force on all areas of your body.  Pressure is applied in all directions regardless of shape or size.  Pressure can act both outward and inward, depending on the circumstances.  Additionally, pressure will always act perpendicular to the surface of the body upon which it is acting.

French mathematician Blaise Pascal was the inventor of the hydraulic press and found many theories surrounding fluid power. He’s best known for Pascal’s Law, which is a principle in fluid mechanics that states that pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions through the fluid such that the pressure variations remain the same.  As we’ve already discussed Pressure is equivalent to force per area.  Therefore, we can see that if pressure equals force divided by area, then Force would equal pressure multiplied by area, or F = PA.

Here is an example of Pascal’s law with fluid power:

In this example, we see that by applying 1 newton of force to the small piston creates equal pressure acting on the larger piston. Since the larger piston is 1,000 times greater surface area, the pressure creates 1000 Newtons of Force in the large piston.

In a fluid power system, the pressure is typically transferred to some type of actuator used to perform work.   Actuators can be rotary, linear or a combination of the two.  Linear actuators are often referred to as cylinders or rams while rotary actuators are called motors.

In addition to the formula F = P x A, the formula for the area of a circle can be expressed as A = D2 x 0.7854.  Using these two equations we can find the minimum bore size required to lift a specific load at a given pressure. The term bore is used to refer to the diameter measurement of the piston.  For example, what would the minimum bore diameter be to lift a load of 6000 lbs at an operating pressure of 1500 psi?  Using the formula  F = P x A, we find that the Area is equivalent to 4 square inches.  Then by plugging this number for A into the formula A = D2 x 0.7854, we find that the minimum bore diameter cylinder would need to be 2.26 inches.

When discussing fluid power and pressure, it’s critical to understand how Force, Pressure and area are all dependent on one another to ensure a system runs properly.  It is also important to understand the concept of “Effective Area.”   Effective area is the area in which the pressurized fluid acts to create force. In a double acting cylinder the effective area of the rod side is different from the effective area of the bore side.  When calculating the effective area for the rod side (retraction force), the area of the rod must be removed from the total area (bore area), as the pressure only acts upon this area in the retracting direction of movement.

In the above diagram we see that the effective area in retract mode can only be calculated by subtracting the rod area from the bore area. Understanding the relation between bore area, rod area, Force and Pressure, let’s take a look at another example.

Let’s say you have a cylinder with a 5” bore diameter, a 2” rod diameter in retract mode.  How much pressure would it take to lift an 8000 lb load?

We start with our formula F = P x A.  We’re looking for P.  We know that F = 8000.  To Determine A, we need to find the Effective area.  So we start with the Bore Area = D2 x 0.7854.  Thus A = (5)2 x 0.7854 or A = 19.64.  The Rod Area = (2)2 x 0.7854 or A = 3.14.  In this scenario the Effective Area is 19.64 – 3.14 or 16.5.

We now plug 16.5 into our original equation F = P x A to determine 8000 = P x 16.5.  Thus P = 484.85 psi

This is a brief introduction into the world of fluid power and pressure.  In our next installment, we’ll further explore fluid power, introducing concepts of Velocity, Flow, Work and Torque.