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CE 150 Fluid Mechanics G.A. Kallio Dept. of Mechanical Engineering, Mechatronic Engineering & Manufacturing Technology California State University, Chico CE 150 1 Fluid Statics Reading: Munson, et al., Chapter 2 CE 150 2 Fluid Statics Problems • Fluid statics refers to the study of fluids at rest or moving in such a manner that no shearing stresses exist in the fluid • These are relatively simple problems since no velocity gradients exist thus, viscosity does not play a role • Applications include the hydraulic press, manometry, dams, and fluid containment (tanks) CE 150 3 Pressure at a Point • Pressure is a scalar quantity that is defined at every point within a fluid • Force analysis on a wedge-shaped fluid element is presented in the text (Figure 2.1) CE 150 4 Pressure at a Point • The result shows that pressure at a point is independent of direction as long as there are no shearing stresses (or velocity gradients) present in the fluid • This result is known as Pascal’s law • For fluids in motion with shearing stresses, this result is not exactly true, but is still a very good approximation for most flows CE 150 5 The Pressure Field • Now, we ask: how does pressure vary from point to point in a fluid w/o shearing stresses? • Consider a small rectangular element of fluid (Figure 2.2): CE 150 6 The Pressure Field • Two types of forces acting on element: – surface forces (due to pressure) – body forces (due to external fields such as gravity, electric, magnetic, etc.) • Newton’s second law states that F ma – where F Fs Wk̂ pxyz xyz k̂ m xyz CE 150 7 The Pressure Field • The differential volume terms cancel, leaving: p k̂ a • This is the general equation of motion for a fluid w/o shearing stresses • Recall: p p p p î ĵ k̂ x y z – known as the pressure gradient CE 150 8 Pressure Field for a Fluid at Rest • For a fluid at rest, a 0 : p k̂ 0 • This vector equation can be broken down into component form: p 0, x p 0, y p z • This shows that p only depends upon z, the direction in which gravity acts: dp dz CE 150 9 Pressure Field for a Fluid at Rest • This equation can be used to determine how pressure varies with elevation within a fluid; integrating yields: p2 z2 dp dz p1 z1 • If the fluid is incompressible (e.g., a liquid), then p1 p2 ( z2 z1 ) • For a liquid with a free surface exposed to pressure p0 : p p0 h CE 150 10 Pressure Field for a Fluid at Rest • If the fluid is compressible, then (or ) is not a constant; this is true for gases, however, the effect on pressure is not significant unless the elevation change is very large • The pressure variation in our atmosphere is such an exception • To integrate the pressure field equation, we need to know how atmospheric air density varies with elevation CE 150 11 Atmospheric Pressure Variation • From the pressure field equation, dp g dz • Atmospheric air can be regarded an an ideal gas, P = RT, so: dp pg dz RT • Separating variables & integrating: p2 p1 dp p2 g z2 dz ln p p1 R z1 T CE 150 12 Atmospheric Pressure Variation • In the troposphere (sea level to 11 km), temperature varies as: T Ta z – where Ta is the temperature at sea level and is the lapse rate • Completing the integration yields: z p pa 1 Ta g R – where the lapse rate for a standard atmosphere is = 0.00650 K/m CE 150 13 Pressure Measurement • Manometer – gravimetric device based upon liquid level deflection in a tube • Bourdon tube – elliptical cross-section tube coil that straightens under under influence of gas pressure • Mercury barometer – evacuated glass tube with open end submerged in mercury to measure atmospheric pressure • Pressure transducer – converts pressure to electrical signal; i) flexible diaphragm w/strain gage ii) piezoelectric quartz crystal CE 150 14 The Manometer • Simple, accurate device for measuring small to moderate pressure differences • Rules of manometry: – pressure change across a fluid column of height h is gh – pressure increases in the direction of gravity, decreases in the direction opposing gravity – two points at the same elevation in a continuous static fluid have the same pressure CE 150 15 Hydrostatic Force on a Plane Surface • The forces on a plane surface submerged in a static fluid are due to pressure and are always perpendicular to that surface • These forces can be resolved into a single resultant force FR , acting at a particular location (xR, yR) along the surface • For a horizontal surface: FR = pA xR, yR is at the centroid of the surface CE 150 16 Hydrostatic Force on a Plane Surface • Resultant force on an inclined plane surface defined by angle : CE 150 17 Hydrostatic Force on a Plane Surface • The resultant force is found by integrating the differential forces over the entire surface: FR dF hdA h A A c – where hc is the vertical distance to the centroid of the area, which can be found from the y-direction distance to the centroid: hc yc sin CE 150 18 Hydrostatic Force on a Plane Surface • The location of the resultant force is found by integrating the differential moments over the area; this yields expressions which contain moments of inertia: I xc yR yc yc A xR I xyc yc A xc – refer to Figure 2.18 for the centroids and moments of inertia of common shapes CE 150 19 Hydrostatic Force on a Plane Surface • The resultant force on vertical, rectangular surfaces can be found using a graphical interpretation known as the pressure prism: CE 150 20 Hydrostatic Force on a Plane Surface • The resultant force is h FR pav A A 2 • Graphically, this can be interpreted as the volume of the pressure prism: h 1 FR A h bh volume 2 2 • This force passes through the centroid of the pressure prism, located a distance h/3 above the base CE 150 21 Hydrostatic Force on a Plane Surface • The pressure prism can also be used for vertical surfaces that do not extend to the free surface of the fluid; here, the cross section is trapezoidal and the resultant force is: 1 FR h1 h2 A 2 – located at: 1 h2 h1 y A h1 3 h2 h1 2 CE 150 22 Hydrostatic Force on a Curved Surface • See section 2.10 CE 150 23 Buoyancy • The resultant buoyant force on a submerged or partially submerged object in a static fluid is given by Archimedes’ principle: FB V – where V is the submerged volume of the object “The buoyant force is equal to the weight of the fluid displaced by the object and is in a direction opposite the gravitational force” • This force arises from the net pressure force acting on the object’s surface CE 150 24 Buoyancy • The line of action of the buoyant force passes through the centroid of the displaced volume, often called the center of buoyancy (COB) • The stability of submerged objects is determined by the center of gravity (COG): – Stable: COG is below COB – Unstable: COG is above COB • For floating objects, stability is complicated by the fact that the COB changes with rotation CE 150 25 Pressure in a Fluid with Rigid-Body Motion • From before, p k̂ a • Consider an example of linear motion - an open container of liquid accelerating along a straight line in the y-z plane (Figure 2.29); we then have: p 0, x p a y , y CE 150 p ( g az ) z 26 Pressure in a Fluid with Rigid-Body Motion • Due to the imbalance of forces on the liquid in the y and z directions, the slope of the liquid surface will change to produce a pressure gradient that offsets the acceleration; it can be shown that this slope is: ay dz dy g az CE 150 27