1.3: Definition of Hydraulic Pressure and Its Units of Measure
- Page ID
- 116995
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)One of the fundamental concepts in hydraulics is pressure. Pressure is the intensity of force applied over a given area. It is created when a force is exerted by one object onto the surface of another.
Imagine a box weighing 32 pounds resting on a surface with an area of 4 square feet. Since the force is evenly distributed, each square foot supports a portion of the total force. In this case, the pressure exerted on the surface is:
32 pounds / 4 square feet =8 lbs/ft²
This means the pressure exerted by the box is 8 pounds per square foot (8 lbs/ft²).
Calculating Pressure
Pressure is determined using the following formula:
P = F x A
Where:
- P = Pressure
- F = Force (measured in Newtons or pounds)
- A = Area over which the force is applied (measured in square meters or square inches)
Units of Measurement
The units used for pressure depend on the measurement system:
- In the English system, pressure is commonly measured in pounds per square inch (psi), which is written as lb/in².
- In the SI (International System) of units, pressure is measured in Pascals (Pa), where 1 Pascal (Pa) = 1 N/m².
Understanding the Relationship Between Force, Area, and Pressure
Although force and pressure are closely related, they are not the same. A single force can produce different pressures depending on the area over which it is applied.
If the same 5 lbs weight is placed on a 2.5 ft² surface, the pressure exerted is:
5 x 2.5 = 2 lbs/ft²
If the same weight is placed on a 1 ft² surface, the pressure increases:
5 x 1=5 lbs/ft²
This principle explains why a person wearing high heels exerts more pressure on the ground than someone wearing flat-soled shoes. The smaller the contact area, the greater the pressure applied.
Fluid Pressure and Pascal’s Law
In hydraulics, pressure is also created within a fluid. One way to generate fluid pressure is by placing a weight on a container filled with liquid. The fluid must exert pressure against the container walls to support the weight.
If the liquid inside the container is at rest, the pressure at every point within the fluid remains equal. This concept was first described in the seventeenth century by the French scientist Blaise Pascal and is known as Pascal’s Law.
Using the same P = F/A formula, the pressure in a hydraulic system can be calculated.
If a 10-pound force is applied to an area of 0.1 ft², the pressure inside the system is:
10 pounds / 0.1 ft² = 100 lbs/ft²
Understanding these principles is essential for working with hydraulic systems, as pressure determines how effectively hydraulic machinery can operate.