Pascal’s LawRecall from chapter 1 that the foundation ofmodern hydraulics was established when Pascaldiscovered that pressure in a fluid acts equally inall directions. This pressure acts at right anglesto the containing surfaces. If some type ofpressure gauge, with an exposed face, is placedbeneath the surface of a liquid (fig. 2-6) at aspecific depth and pointed in different directions,the pressure will read the same. Thus, we can saythat pressure in a liquid is independent ofdirection.Pressure due to the weight of a liquid, at anylevel, depends on the depth of the fluid from thesurface. If the exposed face of the pressure gauges,figure 2-6, are moved closer to the surface of theliquid, the indicated pressure will be less. Whenthe depth is doubled, the indicated pressure isdoubled. Thus the pressure in a liquid is directlyproportional to the depth.Consider a container with vertical sides(fig. 2-7) that is 1 foot long and 1 foot wide. Letit be filled with water 1 foot deep, providing 1cubic foot of water. We learned earlier in thischapter that 1 cubic foot of water weighs 62.4pounds. Using this information and equation 2-2,P = F/A, we can calculate the pressure on thebottom of the container.Since there are 144 square inches in 1 square foot,This can be stated as follows: the weight of acolumn of water 1 foot high, having a cross-sectional area of 1 square inch, is 0.433 pound.If the depth of the column is tripled, theweight of the column will be 3 x 0.433, or 1.299pounds, and the pressure at the bottom will be1.299 lb/in2 (psi), since pressure equals the forcedivided by the area. Thus, the pressure at anydepth in a liquid is equal to the weight of thecolumn of liquid at that depth divided by theFigure 2-6.—Pressure of a liquid is independent of direction.cross-sectional area of the column at that depth.The volume of a liquid that produces the pressureis referred to as the fluid head of the liquid. Thepressure of a liquid due to its fluid head is alsodependent on the density of the liquid.If we let A equal any cross-sectional area ofa liquid column and h equal the depth of thecolumn, the volume becomes Ah. Using equation2-4, D = W/V, the weight of the liquid above areaA is equal to AhD.Figure 2-7.—Water pressure in a 1-cubic-foot container.2-5
Integrated Publishing, Inc. - A (SDVOSB) Service Disabled Veteran Owned Small Business