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Unformatted text preview: .S 20803 kPa, inlet diameter, which cavita- it a water tern- ilet velocity is rsibilities such P R E s s U R E A N D Nithcomputa~ the effects of :hed__tube. Set then vary the sure to predict his chapter deals with forces applied by fluids at rest or in rigid—body m is likely to motion. The fluid property responsible for those forces is pressure, “ n 2—128 and which is a normal force exerted by a fluid per unit area. We start this ‘ "' i " ‘ es denser than chapter with a detailed discussion of pressure, including absolute and gage ' ' V H ‘ ' made to float pressures, the pressure at a point, the variation of pressure with depth in a 3‘ ‘ ' ‘ r - 1 ! ould happen if gravitational field, the barometer, the manometer, and other pressure mea— ' ' ‘ 7‘7 ' ‘ ‘ surement devices. This is followed by a discussion of the hydrostatic forces ~ H v ‘ ' ‘ ‘ : ~ applied on submerged bodies with plane or curved surfaces. We then con- ' ‘7 ' A ' sider the buoyant force applied by fluids on submerged or floating bodies, ‘ ‘ ‘ ‘ ' ' ‘ ' “ and discuss the stability of such bodies. Finally, we apply Newton’s second i ‘ ' ' ‘l a ' law of motion to a body of fluid in motion that acts as a rigid body and ana- ‘ ' n: ' lyze the variation of pressure in fluids that undergo linear acceleration and ‘ I “1. H in rotating containers. This chapter makes extensive use of force balances for bodies in static equilibrium, and it would be helpful if the relevant topics 7 ‘ from statics are first reviewed. John Ninomiya flying a cluster of 72 helium-filled balloons over Temecula, California in April Of 2003. The helium balloons displace approximately 230 m3 of air, providing the necessary buoyant force. ' Don’t try this at home! Photograph by Susan Dawson. Used by permission. , , 74 . . . .. PRESSURE AND FLUID STATICS 3—1 I PRESSURE Pressure is defined as a normal force exerted by a fluid per unit area. We speak of pressure only when we deal with a gas or a liquid. The counterpart of pressure in solids is normal stress. Since pressure is defined as force per unit area, it has the unit of newtons per square meter (N/mz), which is Called a pascal (Pa). That is, ‘ 1Pa = lN/m2 The pressure unit pascal is too small for most pressures encountered in practice. Therefore, its multiples kilopascal (1 kPa = 103 Pa) and megapas- cal (1 "MPa = 106 Pa) are commonly used. Three other pressure units com- monly used in practice, especially in Europe, are bat: standard atmosphere, and kilogram—force per square centimeter: 1 bar =105 Pa = 0.1 MPa = 100 kPa 1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bars 1kgf/cm2 : 9.807 N/cmz = 9.807 x 104N/m2 = 9.807 X 104 Pa 150 pounds ‘ 300 pounds AW: 50 “’2 = 0.9807 bar , = 0.9679 atm Note the pressure units bar, atm, and kgf/cm2 are almost equivalent to each P=3psi P:6psi other. In the English system, the pressure unit is pound—force per square W 150M . inch (lbf/in2, or psi), and 1 atm = 14.696 psi. The pressure units kgf/cm2 P =‘7n= Afeet = 50in2 = 3 P51 and lbf/in2 are also denoted by kg/cm2 and lb/inz, respectively, and they are commonly used in tire gages. It can be shown that l kgf/cm2 = 14.223 psi. FIGURE 3—1 Pressure is also used on solid surfaces as synonymous to normal stress, The normal stress (or “pressure”) which is force acting perpendicular to the surface per unit area. For exam- on the feet of achubby person is ple, a lSO-pound person with a total foot imprint area of 50 in2 exerts a much greater than on the feet of pressure of 150 lbf/50 in2 = 3.0 psi on the floor (Fig. 3—1). If the person aslim person. stands on one foot, the pressure doubles. If the person gains excessive weight, he or she is likely to encounter foot discomfort because of the increased pressure on the foot (the size of the bottom of the foot does not change with weight gain). This also explains how a person can walk on fresh snow without sinking by wearing large snowshoes, and how a person cuts with little effort when using a sharp knife. The actual pressure at a given position is called the absolute pressure, and it is measured relative to absolute vacuum (i.e., absolute zero pressure). Most pressure—measuring devices, however, are calibrated to read zero in the atmosphere (Fig. 3—2), and so they indicate the difference between the absolute pressure and the local atmospheric pressure. This difference is called the gage pressure. Pgage can be positive or negative, but pressures below atmospheric pressure are sometimes called vacuum pressures and are measured by vacuum gages that indicate the difference between the atmospheric pressure and the absolute pressure. Absolute, gage, and vacuum pressures are related to each other by 3—2 Pgage Pubs _ Patm Some basic pressure gages. " - pvac = Pm“ _ pabs (3-2) l l ‘ DresserInstruments, Dresser, Inc. Used by I _ _ ‘ _ permission. This 13 1llustrated 1n F1g. 3—3. :m't area. We 3 counterpart as force per iich is called countered in nd megapas- 'e units com- ‘ atmosphere, )4 Pa alent to each e per square inits kgf/cm2 and they are = 14.223 psi. ormal stress, a. For exam- l in2 exerts a If" the person .ns excessive cause of the foot does not can walk on row a person lte pressure, 3ro pressure). 1d zero in the between the difference is but pressures ressures and between the , and vacuum (3—1 ) (3—2) atm Absolute Absolute vacuum 7 I abs vacuum Like other pressure gages, the gage used to measure the air pressure in an automobile tire reads the gage pressure. Therefore, the common reading of 32.0 psi (2.25 kgf/cmz) indicates a pressure of 32.0 psi above the atmos- pheric pressure. At a location where the atmospheric pressure is 14.3 psi, for example, the absolute pressure in the tire is 32.0 + 14.3 = 46.3 psi. In thermodynamic relations and tables, absolute pressure is almost always used. Throughout this text, the pressure P will denote absolute pressure unless specified otherwise. Often the letters “a” (for absolute pressure) and “g” (for gage pressure) are added to pressure units (such as psia and psig) to clarify what is meant. XAMPLE 3—1 ' * 'AbshlUte Pressure of a Vacuum Chamber a I _ ‘acuum gage connected to archamber' reads, ata lecation .w_here;__-_ atmospheric pressure is 14.53 psi.‘ Determine the abSolute pressure inithe I mber. ' ssure in thechamber-is to be determined. halysis The absOIUte pressure is easily determined from Eq. 3—‘2 to'be Pa... = Pam, — Pvac = 14.5 — 5.8 = 8.7 psi éiscussian Note that the local value of the atmospheric pressure is used Then determining the absolute pressure. _ ,OLUTIDN The gage pressure Of air‘vacuum chamber is git/en. absolute Pressure at a Point Pressure is the compressive force per unit area, and it gives the impression 0f being a vector. However, pressure at any point in a fluid is the same in all directions (Fig. 3—4). That is, it has magnitude but not a specific direction, and thus it is a scalar quantity. This can be demonstrated by considering a Small wedge—shaped fluid element of unit length (Ay = 1 into the page) in €quflibrium, as shown in Fig. 3—5. The mean pressures at the three surfaces are P1, P2, and P3, and the force acting on a surface is the product of mean P—-> 75 CHAPTER‘ 3 FIGURE 3—3 Absolute, gage, and vacuum pressures. FIGURE 3—4 Pressure is a scalar quantity, not a vector; the pressure at a point in a fluid is the same in all directions. : - 76 PRESSURE AND FLUID STATICS pressure and the surface area. From Newton’s second law, a force balance in the x— and z-directions gives 2 Fx = max = 0: P1 AyAz * P3 Ayl sin 6 = 0 (3—33) >N 1 BF, = maZ = 0: P2 AyAx —~P3 Ayl cos 6 ~ Epg Ax Ay Az = 0 (3—3h) where p is the density and W = mg = pg Ax Ay Az/Z is the weight of the fluid element. Noting that the wedge is a right triangle, we have Ax = l cos 6 and Az = 1 sin 6. Substituting these geometric relations and dividing Eq. 3—3a by Ay Az and Eq. 3—3b by Ax Ay gives 157 P 1 _ P 3 = 0 (3—4a) l PZ—P3—EpgAz=0 (3—4h) p2 Ax Ay The last term in Eq. 3—4b drops out as Az —> 0 and the wedge becomes (Ay : 1) , , infinitesimal, and thus the fluid element shrinks to a point. Then combining the results of these two relations gives FIGURE 3—5 m _ _ Forces acting on a wedge-shaped fluid P1 " P2 _ P3 _ P (3‘5) Element in equmbfium- regardless of the angle 6. We can repeat the analysis for an element in the yz—plane and obtain a similar result. Thus we conclude that the pressure at a point in a fluid has the same magnitude in all directions. This result is applicable to fluids in motion as well as fluids at rest since pressure is a scalar, not a vector. Variation of Pressure with Depth It will come as no surprise to you that pressure in a fluid at rest does not change in the horizontal direction. This can be shown easily by considering a thin horizontal layer of fluid and doing a force balance in any horizontal direction. However, this is not the case in the vertical direction in a gravity field. Pressure in a fluid increases with depth because more fluid rests on deeper layers, and the effect of this “extra weight” on a deeper layer is bal— anced by an increase in pressure (Fig. 3—6). To obtain a relation for the variation of pressure with depth, consider a rectangular fluid element of height Az, length Ax, and unit depth (Ay = 1 into the page) in equilibrium, as shown in Fig. 3—7. Assuming the density of the fluid p to be constant, a force balance in the vertical z—direction gives 2Fz=maz=0z PleAy—PzAxAy—pgAxAyAz=O where W = mg = pg Ax Ay Az is the weight of the fluid element and AZ = Z2 — zl. Dividing by Ax Ay and rearranging gives AP = P2 —’P1 = — pg Az = —ys Az (3—6) FIGURE 3‘6 ' . where ys = pg is the specific weight of the fluid. Thus, we conclude that the The Press/11f? Of a flmd at FeSt pressure difference between two points in a constant density fluid is propor- mcreases Wlth depth (as a res‘flt tional to the vertical distanceAz between the points and the density p of the 0f adde‘i Weight)- fluid. Noting the negative Sign, pressure in a static fluid increases linearly with depth. This is what a diver experiences when diving deeper in a lake. a balance in (3—3a) z = 0 (3—3b) eight of the Ax = 1 cos iividing Eq. (3—4a) (3—4b) ge becomes 1 combining (3—5) ament in the 7F6SSM7‘6 at a his result is )ressure is a 'est does not " considering 1y horizontal l in a gravity luid rests on layer is bal— 1, consider a .pth (Ay = 1 he density of :tion gives =0 element and (3—6) clude that the lid is propor- nsity p of the zases linearly er in a lake. An easier equation to remember and apply between any two points in the same fluid under hydrostatic conditions is Pbelow : Pabove + : Pabove + ’yslAZI «3—7) where “below” refers to the point at lower elevation (deeper in the fluid) and “above” refers to the point at higher elevation. If you use this equation consistently, you should avoid sign errors. For a given fluid, the vertical distance AZ is sometimes used as a measure of pressure, and it is called the pressure head. We also conclude from Eq. 3—6 that for small to moderate distances, the variation of pressure with height is negligible for gases because of their low density. The pressure in a tank containing a gas, for example, can be consid— ered to be uniform since the weight of the gas is too small to make a signif— icant difference. Also, the pressure in a room filled with air can be assumed to be constant (Fig. 3—8). If we take the “above” point to be at the free surface of a liquid open to the atmosphere (Fig. 3—9), where the pressure is the atmospheric pressure Pam, then from Eq. 3~7 the pressure at a depth it below the free surface becomes P = Patm + pgh or Pgage = pgh (3—8) Liquids are essentially incompressible substances, and thus the variation of density with depth is negligible. This is also the case for gases when the elevation change is not very large. The variation of density of liquids or gases with temperature can be significant, however, and may need to be considered when high accuracy is desired. Also, at great depths such as those encountered in oceans, the change in the density of a liquid can be significant because of the compression by the tremendous amount of liquid weight above. The gravitational acceleration g varies from 9.807 m/s2 at sea level to 9.764 m/s2 at an elevation of 14,000 m where large passenger planes cruise. This is a change of just 0.4 percent in this extreme case. Therefore, g can be assumed to be constant with negligible error. For fluids whose density changes significantly with elevation, a relation for the variation of pressure with elevation can be obtained by dividing Eq. 3—6 by Az, and taking the limit as Az —9 0. This yields LP _ _ (3 9) dz pg Note that dP is negative when dz is positive since pressure decreases in an upward direction. When the variation of density with elevation is known, the pressure difference between any two points 1 and 2 can be determined by integration to be 2 . AP=P2—P1= —Jpgdz (3310) 1 . For constant density and constant gravitational acceleration, this relation reduces to Eq. 3—6, as expected. Pressure in a fluid at rest is independent of the shape or cross section of the container. It changes with the vertical distance, but remains constant in ii; i 77 u ' CHAPTER ,3 . Z >1 FIGURE 3—7 Free—body diagram of a rectangular fluid element in equilibrium. FIGURE 3—8 In a room filled with a gas, the variation of pressure with height is negligible. : 73 ,. , PRESSURE AND FLUID STATICS other directions. Therefore, the pressure is the same at all points on a hori~ zontal plane in a given fluid. The Dutch mathematician Simon Stevin (1548—1620) published in 1586 the principle illustrated in Fig. 3—10. Note Pabm: Pam that the pressures at points A, B, C, D, E, F, and G are the same since they V are at the same depth, and they are interconnected by the same static fluid. ‘ i i However, the pressures at points H and I are not the same since these two points cannot be interconnected by the same fluid (i.e., we cannot draw a curve from point I to point H while remaining in the same fluid at all times), although they are at the same depth. (Can you tell at which point the pressure is higher?) Also notice that the pressure force exerted by the fluid is always normal to the surface at the specified points. A consequence of the pressure in a fluid remaining constant in the hori- zontal direction is that the pressure applied to a confined fluid increases the FIGURE 3—9 pressure throughout by the same amount. This is called Pascal’s law, after Pressure in aliquid at rest increases Blaise Pascal (1623-1662). Pascal also knew that the force applied by a linearly with distance from the free fluid is proportional to the surface area. He realized that two hydraulic surface. cylinders of different areas could be connected, and the larger could be used to exert a proportionally greater force than that applied to the smaller. “Pas~ cal’s machine” has been the source of many inventions that are a part of our daily lives such as hydraulic brakes and lifts. This is what enables us to lift a car easily by one arm, as shown in Fig. 3—11. Noting that P1 = P2 since both pistons are at the same level (the effect of small height differences is negligible, especially at high pressures), the ratio of output force to input force is determined to be 5-2 fl_fl P=P —9 — a — 3—11 1 2 A1 A2 F1 A1 ‘ ) PA—PB~PC—PD—PE=PF—PG-Patm l 9871 PH¢PI HGURE3—1O Under hydrostatic conditions, the pressure is the same at all points on a horizontal plane in a given fluid regardless of geometry, provided that the points are interconnected by the same fluid. :s on a hori— mon Stevin 3—10. Note l6 since they static fluid. :e these two nnot draw a fluid at all ich point the by the fluid , in the hori- increases the PS law, after applied by a V0 hydraulic ould be used maller. “Pas— a part of our )les us to lift 1 = P2 since 1ifferences is orce to input (3—1 1) irdless of The area ratio Az/Al is called the ideal mechanical advantage of the hydraulic lift Using a hydraulic car jack with a piston area ratio of Az/A1 = 100, for example, a person can lift a lOOO-kg car by applying a force of just 10 kgf (: 90.8 N). 3-2 — PRESSURE MEASUREMENT DEVICES The Barometer AtmOSpheric pressure is measured by a device called a barometer; thus, the atmospheric pressure is often referred to as the barometric pressure. The Italian Evangelista Torricelli (1608—1647) was the first to conclu- sively prove that the atmospheric pressure can be measured by inverting a mercury—filled tube into a mercury container that is open to the atmosphere, as shown in Fig. 3—12. The pressure at point B is equal to the atmospheric pressure, and the pressure at C can be taken to be zero since there is only mercury vapor above point C and the pressure is very low relative to Patm and can be neglected to an excellent approximation. Writing a force balance in the vertical direction gives Patm = pgh (3—12) where p is the density of mercury, g is the local gravitational acceleration, and h is the height of the mercury column above the free surface. Note that the length and the cross-sectional area of the tube have no effect on the height of the fluid column of a barometer (Fig. 3—13). A frequently used pressure unit is the standard atmosphere, which is defined as the pressure produced by a column of mercury 760 mm in height at 0°C (pHg = 13,595 kg/m3) under standard gravitational acceleration (g = 9.807 m/sZ). If water instead of mercury were used to measure the standard atmospheric pressure, a water column of about 10.3 m would be needed. Pressure is sometimes expressed (especially by weather forecasters) in terms of the height of the mercury column. The standard atmospheric pressure, for example, is 760 mmHg (29.92 ian) at 0°C. The unit mmHg is also called the torr in honor of Torricelli. Therefore, 1 atm = 760 torr and l torr = 133.3 Pa. Atmospheric pressure Patm changes from 101.325 kPa at sea level to 89.88, 79.50, 54.05, 26.5, and 5.53 kPa at altitudes of 1000, 2000, 5000, 10,000, and 20,000 meters, respectively. The typical atmospheric pressure in Denver (elevation = 1610 m), for example, is 83.4 kPa. Remember that the atmospheric pressure at a location is simply the weight of the air above that location per unit surface area. Therefore, it lchanges not only with elevation but also with weather conditions. The decline of atmospheric pressure with elevation has far—reaching rami— fications in daily life. For example, cooking takes longer at high altitudes Since water boils at a lower temperature at lower atmospheric pressures. Nose bleeding is a common experience at high altitudes since the difference betWeen the blood pressure and the atmospheric pressure is larger in this Case, and the delicate walls of veins in the nose are often unable to with- stand this extra stress. : ' For a given temperature, the density of air is lower at high altitudes, and thus a given volume contains less air and less oxygen. So it is no surprise 79 CHAPTERS W: rghA FIGURE 3—11 Lifting of a large weight by a small force by the application of Pascal’s law. [1 P atm FIGURE 3—12 The basic barometer. v . £50,, PRESSURE AND FLUID‘STATICS FIGURE 3—1 3 The length or the cross—sectional area of the tube has no effect on the height of the fluid column of a barometer, provided that the tube diameter is large enough to avoid surface tension (capillary) effects. FIGURE 3—14 At high altitudes, a car engine generates less power and a person gets less oxygen because of the lower density of air. that we tire more easily and experience breathing problems at high altitudes. To compensate for this effect, people living at higher altitudes develop more efficient lungs. Similarly, a 2.0—L car engine will act like a 1.7—L car engine at 1500 III altitude (unless it is turbocharged) because of the 15 percent drop in pressure and thus 15 percent drop in the density of air (Fig. 3—14). A fan or compressor will displace 15 percent less air at that altitude for the same volume displacement rate. Therefore, larger cooling fans may need to be selected for operation at high altitudes to ensure the specified mass flow rate. The lower pressure and thus lower density also affects lift and drag: airplanes need a longer runway at high altitudes to develop the required lift, and they climb to very high altitudes for cruising in order to reduce drag and thus achieve better fuel efficiency. Measuring Atmospheric Pressure with a Barometer EXAMPLE 3—2 Determine the atmospheric pressure at a location where the barometric read- " ing is 740 mm Hg and the gravitational acceleration is g = 9.805 m/sz. ' Assume the temperature of mercury to be 10°C, at which its density is d. 13,570 kg/m3. SOLUTION The barometric reading at a location in height of mercury col- umn is given. The atmospheric pressure is to be determined. r Assumptions The temperature of mercury is assumed to be 10°C. _' Properties The density of mercury is given to be 13,570 kg/m3. ‘Analysis From Eq. 3—12, the atmospheric pressure isdetermined to be Palm = pgh , 1 N 1 kPa = 13,570k / 3 9.805 m/ 2 0.740 r ( gmx S X In)- 1kg - nvs2 1000 N/m2 - '= 98.5 kPa ' Discussion Note that density changes with temperature, and thus this effect :should be considered in calculations. ' EXAMPLE 3—3 Intravenous infusions usually are driven by gravity by hanging the fluid bottle of at sufficient height to counteract the blood pressure in the vein and to force the fILIid into the body (Fig. 3~15). The higher the bottle is raised, the higher the flow rate of the fluid will be. (a) If it is observed that the fluid and the blood pressures balance each other when the bottle is 1.2 m above ‘7 the arm level, determine the gage pressure of the blood. (b) if the gage pres- sure of the fluid at the arm level needs to be 20 kPa for sufficient flow rate, determine how high the bottle nfust be placed. Take the density of the fluid to be 1020 kg/m3. Gravity Driven Flow from an IV Bottle SOLUTION It is given that an IV fluid and the blood pressures balance each other when the bottle is at a certain height. The gage pressure of the blood and elevation of the bottle required to maintain flow at the desired rate are to be determined. a- igh altitudes. levelop more L car engine percent drop 3—14). A fan for the same r need to be d mass flow ift and drag: required lift, reduce drag , 81 _ CHAPTER3 Asgumptians 1 The IV fluid is incompressible. 2 The N bottle is open tofy’ffg ‘ théiatmosphere- ‘ Properties The density of the IV fluid is given to be p = 1020 kg/m3. ' Analysis (a) Noting that the IV fluid and the blood pressures balance each other when the bottle is 1.2 m above the arm level, the gage pressure of the blow in the arm is simply equal to the gage pressure of the N fluid at a", depth of 1.2 m. _ PgageJl—m : Pabs — Patm 2’ pgharm~bottle z (1020 kg/m3)(9.81 m/szxrzo “(Pa 2) 1 FIGURE 3—15 1000 kg ' W8 1 kN/m Schematic for Example 3—3. = 12.0 kPa 3(5) To provide a gage pressure of 20 kPa at the arm level, the height of the surface of the W fluid in the bottle from the arm level is again determined Pgage, arm : pghaIm—bottle to be h = Pgage, arm netric read— ' “meOt‘fle pg )805 W5?- , I 20 kPa (1000 kg - m/s2)<1kN/m2) ’ density ‘3 ‘ (1020 kg/m3)(9.81 m/sz) lkN lkPa = 2.00 m nercury col— ‘ _‘ss,i_on Note that the height of the reservoir can be used to control flow ~ , ‘ gravity—driven flows. When there is flow, the pressure drop in the tube- {o frictional effects also should be considered. For a specified flow rate,‘ equires raising the bottle a little higher to overcome the pressure drop. d to be 7' _ _ , , _ , MPLE3—4 Hydrostatic Pressure in a Solar Pond kPa ) i with Variable Density 0 Mint ponds are small artificial lakes of a few meters deep that are used to - solar energy. The rise of heated (and thus less dense) water to the sur- _is prevented by adding salt at the pond bottom. In a typical salt gradi— olar pond, the density of water increases in the gradient zone, as shown ,: . 3—16, and the density can be expressed as l 77 S p = po 1 + tan2(Z ere p0 is the density on the water surface, s is the vertical distance mea— ed downward from the top of the gradient zone (5 = —z), and H is the 3 this effect :fluid bottle and to force raised, the at the fluid l.2 m above e gage pres- nt flow rate, of the fluid Sun Increasing salinity and density p0 = 1040 kg/m3 Surface zone ialance each of the blood ired rate are FIGURE 3—1 6 Schematic for Example 3—4. 82 P, kPa FIGURE 3—1 7 The variation of gage pressure with depth in the gradient zone of the solar pond. PRESSURE AND FLUID'SVTATICS ' ' thickness of the gradient zone. For H = 4 m, p0 = 1040 kg/m3, and a thick— ness of O. *m for the surface zone, calculate the gage pressure at the bot- tom of the gradient zone. SOLUTION The variation of density of saline water in the gradient zone of a solar pond with depth is given. The gage pressure at the bottom of the gradi- ent zone is to be determined. Assumptions The density in the surface zone of the pond is constant. Properties The density of brine on the surface is given to be 1040 kg/m3. Analysis We label the top and the bottom of the gradient zone as 1 and 2, respectively. Noting that the density of the surface zone is constant, the gage pressure at the bottom of the surface zone (which is the top of the gradient zone) is 1 kN P1 = pgh1 = (1040 kg/m3)(9.81m/s2)(0.8 m)( > = 8.16 kPa 1000 kg -' m/s2 since 1 kN/m2 = 1 kPa. Since 5 = -z, the differential change in hydrostatic pressure across a vertical distance of ds is given by dP = pg ds integrating from the top of the gradient zone (point 1 where s = O) to any location 5 in the gradient zone (no subscript) gives P-P1=Jpgds —-> P=P1+mell+tan2(11)gds 0 0 4H Performing the integration gives the variation of gage pressure in the gradi— ent zone to be P—P + 4:]{s‘ h“1<ta 77-3) 1 908 W 1H 4H Then the pressure at the bottom of the gradient zone (5 = H = 4 m) becomes ’ ' 44 P2=8.16kPa+(1O4-0 kg/m3)(9.81m/s2) ( m) sinh—1<tan35l—><———1——kI—\I———2> 7T 44 1000kg-m/s = 54.0 kPa (gage) V Discussion The variation of gage pressure in the gradient zone with depth is plotted in Fig. 3—17. The dashed line indicates the hydrostatic pressure for the case of constant density at 1040 kg/m3 and is given for reference. Note that the variation of pressure with depth is not linear when density varies with depth. That is why integration was required. The Manometer ' We notice from Eq. 3—6 that an elevation change of —Az in a fluid at rest corresponds to AP/pg, which suggests that a fluid column can be used to measure pressure differences. A device based on this principle is called a manometer, and it is commonly used to measure small and moderate pres— sure differences. A manometer consists of a glass or plastic U—tube contain— ing one or more fluids such as mercury, water, alcohol, or oil. To keep the size of the manometer to a manageable level, heavy fluids such as mercury are used if large pressure differences are anticipated. ‘ V Pi 1d a thick- at the bot- t zone of a ithe gradi— am. .0 kg/m3. s 1 and 2, t, the gage ie gradient 8.16 kPa hydrostatic = O) to any )g ds 1 the gradi— H = 4 m) lkN > lOkg-m/s2 'ith depth is pressure for :rence. Note ansity varies 1 fluid at rest in be used to le is called a ioderate pres— -tube contain— . To keep the :h as mercury Consider the manometer shown in Fig. 3—18 that is used to measure the pressure in the tank. Since the gravitational effects of gases are negligible, the pressure anywhere in the tank and at position 1 has the same value. Fur— thermore, since pressure in a fluid does not vary in the horizontal direction within a fluid, the pressure at point 2 is the same as the pressure at point 1, P 2 z P 1‘ . . . . . . . . . . . The differential fluld column of height h is in static equilibrium, and it is open to the atmosphere. Then the pressure at point 2 is determined directly from Eq. 3—7 to be P2 = Palm + pgh (3—13) where p is the density of the manometer fluid in the tube. Note that the cross—sectional area of the tube has no effect on the differential height h, and thus the pressure exerted by the fluid. However, the diameter of the tube should be large enough (more than several millimeters) to ensure that the surface tension effect and thus the capillary rise is negligible. EXAMPLE 3—5 Measuring Pressure with a Manometer :anometer is used to measure the pressure of a gas in a tank. The fluid sedhas a specific gravity of 0.85, and the manometer column height is 55 as shown in Fig. 3—19. if the local atmospheric pressure is 96 kPa, rmine the absolute pressure within the tank. OLUTIDN The reading of a manometer attached to a tank and the I r heric pressure are given. The absolute pressure in the tank is to be ~ 'umptians The density of the gas in the tank is much lower than the den»; fthe manometer fluid. rties The specific gravity of the manometer fluid is given to be 0.85._ ke the standard density of water to be 1000 kg/m3. _ ‘ ' ' . is The density of the fluid is obtained by multiplying its specific y by the density of water, p = so (pHZO) = (0.85)(1000 kg/mS) = 850 kg/m? hen from Eq. 3—13, {Pam + pgh ' 1N lkPa =96kP + 850k / 3 9.8111/2 0.55 a ( gm)( 1 S)( m) 1kg.m/Sz NOON/m2 _ = 100.6 kPa (Dismissian Note that the gage pressure in the tank is 4.6 kPa. Some manometers use a slanted or inclined tube in order to increase the res— olution (precision) when reading the fluid height. Such devices are called inclined manometers. I Many engineering problems and some manometers involve multiple 1mmiscible fluids of different densities stacked on top of each other: Such Systems can be analyzed easily by remembering that (1) the pressure change. across a fluid column of height h is AP =‘ pgh, (2) pressure increases ’\ , 83 . CHAPTER 3 FIGURE 3—18 The basic manometer. FIGURE 3—1 9 Schematic for Example 3—5. “.__.__Tv.. ‘ downward in a given fluid and decreases upward (i.e., Pbottom > Pmp), and (3) two points at the same elevation in a continuous fluid at rest are at the same pressure. ’ The last principle, which is a result of Pascal’s law, allows us to “jump” from one fluid column to the next in manometers without worrying about pressure change as long as we stay in the same continuous fluid and the fluid is at rest. Then the pressure at any point can be determined by starting with a point of known pressure and adding or subtracting pgh terms as we advance toward the point of interest. For example, the pressure at the bot— tom of the tank in Fig. 3—20 can be determined by starting at the free sur— face where the pressure is Patm, moving downward until we reach point 1 at the bottom, and setting the result equal to P1. It gives 84-. ‘ i D FLUID STATICS PRESSURE AN FIGURE 3—20 , Patm + plghl + Pzghz + Paghs 2 P1 i In stacked—up flu1d layers at rest, the 3; pressure change across each fluid In the special case of all fluids having the same density, this relation reduces if layer of density p and height h is pgh. t0 Patm + pg(h1 + hz ‘i‘ h3) = P]_- r Manometers are particularly well-suited to measure pressure drops across ‘i a horizontal flow section between two specified points due to the presence A flow section of a device such as a valve or heat exchanger or any resistance to flow. This or flow device is done by connecting the two legs of the manometer to these two points, as shown in Fig. 3—21. The working fluid can be either a gas or a liquid whose density is p1. The density of the manometer fluid is p2, and the differential fluid height is h. The two fluids must be immiscible, and p2 must be greater than p1. A relation for the pressure difference P1 — P2 can be obtained by starting at point 1 with P1, moving along the tube by adding or subtracting the pgh terms until we reach point 2, and setting the result equal to P2: P1 + p1g(a + h) - ngh — plea = P2 (344) Note that we jumped from point A horizontally to point B and ignored the part underneath since the pressure at both points is the same.._Simplifying, FIGURE 3—21 P1_ P2 2 (p2 _ pagh (345) Measuring the pressure drop across a . _ ' _ . flow section or a flow device by a Note that the d1stance a must be included in the analys1s even though it has differential manometer. no effect on the result. Also, when the fluid flowing in the pipe is a gas, then p1 << p2 and the relation in Eq. 3—15 simplifies to P1 — PZ E ngh. EXAMPLE3—6 Measuring Pressure with a Multifiuid Manometer The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Fig. 3—22. The tank is located on a moun— tain at an altitude of 1400 m where the atmospheric pressure is 85.6 kPa. Determine the air pressure in the tank if f71 = 0.1 m, h2 = 0.2 m, and h3 0.35 m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively. SOLUTION The pressure in” a pressurized water tank is measured by a multi— . fluid manometer. The air pressure in the tank is to be determined. > P top), and est are at the us to “jump” )rrying about fluid and the :d by starting terms as We .“6 at the bot— the free sur- [Ch point 1 at ation reduces drops across the presence to flow. This :wo points, as liquid whose 1e differential ust be greater ed by starting .cting the pgh (3—14) .d ignored the Simplifying, (3—15) though it has pipe is a gas, P 2 E Pzgh- Vlanometer easured by a :l on a moun— is 85.6 kPa. m, and n3 ' 1000 kg/m3, w d by a multi—- ed. ipres$ure at the air—water interface. 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively. Aha/ysis Starting with the pressure at point 1 at the air—water interface, moving along the tube by adding or subtracting the pgh terms until we reach point 2, and setting the result equal to Patm since the tube is open to the atmosphere gives A 0 Pl + pwaterghl + poiighz * pmmurygh3 = P2 = Pm Solving for P1 and substituting, E:P1:: Pann “- Pwaterghl T Poiighz + Pmmuryghg “1 Palm + g(pmercuryh3 "T pwaterhl _ poilhz) L = 85.6 kPa + (9.81 m/s2)[(13,600 kg/m3)(0.35 m) - (1000 kg/m3)(0.1 m) l. N l kPa _ 5 k/ 3 0.2 (8 0 ng I10] 1kg . m/sz 1000 N/m2 _ [130 kPa ause‘of the risk of exposure to mercury vapor during an accident. MPLE 3—7 Analyzing a Multifluid Manometer with EES pressure in the tank using EES. Also determine what the differential Were replaced by seawater with a density of 1030 kg/m3. TION The pressure in a water tank is measured by a muIt-ifluid ‘cury is replaced by seawater are to be determined using EES. *is We start the EES program by double—clicking on its icon, open a mospheric pressure in Pa for unit consistency): g=9.81 Patm= 85600 h1=0.1; h2=0.2; h3=0.35 » rw=1000; roil= 850; rm: 13600 Pl+rw*g*h1+ r0i1*g*h2—1‘m*g*h3 : Patm Here P1 is the only unknown, and it is determined by EES to be Assamptiafl The air pressUre in the tank is uniform (i.e., its variation with “““a_tio-n is negligible due to its low density), and thus we can determine the; properties The densities of water, oil, and mercury are given to be I Gu35ian Note that jumping horizontally from one tube to the next and L ng that pressure remains the same in the same fluid simplifies the 7 considerably. Also note that mercury is a toxic fluid, and mercury‘_'_fj meters and thermometers are being replaced by ones with safer fluids L, i." der the multifluid manometer discussed in Example 3—6. Determine ‘5 height h3 would be for the same air pressure if the mercury in the last eter. The air pressure in the tank and the differential fluid height h3 “ I-e, and type the following on the blank screen that appears (we express I 85 CHAPTER‘ 3 ' Mercury FIGURE 3—22 Schematic for Example 3~3; drawing not to scale. ' i 86 PRESSURE AND FLUID STATICS FIGURE 3—23 Various types of Bourdon tubes used to measure pressure. They work on the same principle as party noise—makers (bottom photo) due to the flat tube cross section. Photo by John M. Cimbala. P1 = 129647 Pa '=“ 130 kPa which is identical to the result obtained in Example 3—6.'The height of the fluid column h3 when mercury is replaced by seawater is determined easily by replacing “hi-3:035” by “P12129647” and “rm=13600” by “rm=1030,” and clicking on the calculator symbol. It gives [13 = 4.62 m Discussion Note that we used the screen like a paper pad and wrote down the relevant information together with the applicable relations in an orga— nized manner. EES did the rest. Equations can be written on separate lines or on the same line by separating them by semicolons, and blank or com— ment linesican be inserted for readability. EES makes it very easy to ask “what if” (questions and to perform parametric studies, as explained in Appendix 3 on the DVD. Other Pressure Measurement. Devices Another type of commonly used mechanical pressure measurement device is the Bourdon tube, named after the French engineer and inventor Eugene Bourdon (1808—1884), which consists of a bent, coiled, or twisted hollow metal tube whose end is closed and connected to a dial indicator needle (Fig. 3—23). When the tube is open to the atmosphere, the tube is unde- flected, and the needle on the dial at this state is calibrated to read zero (gage pressure). When the fluid inside the tube is pressurized, the tube stretches and moves the needle in proportion to the applied pressure. Electronics have made their way into every aspect of life, including pres— sure measurement devices. Modern pressure sensors, called pressure trans- ducers, use various techniques to convert the pressure effect to an electrical effect such as a change in voltage, resistance, or capacitance. Pressure trans- ducers are smaller and faster, and they can be more sensitive, reliable, and precise than their mechanical counterparts. They can measure pressures from less than a millionth of 1 atm to several thousands of atm. A wide variety of pressure transducers is available to measure gage, absolute, and differential pressures in a wide range of applications. Gage pressure transducers use the atmospheric pressure as a reference by venting the back side of the pressure-sensing diaphragm to the atmosphere, and they give a zero signal output at atmospheric pressure regardless of altitude. Absolute pressure transducers are calibrated to have a zero signal output at full vacuum. Difierential pressure transducers measure the pressure difference between two locations directly instead of using two pressure transducers and taking their difference. Strain-gage pressure transducers work by having a diaphragm deflect between two chambers open to” the pressure inputs. As the diaphragm stretches in response to a change in pressure difference across it, the strain gage stretches and a Wheatstone bridge circuit amplifies the output. A capacitance transducer works similarly, but capacitance change is measured instead of resistance change as the diaphragm stretches. a t» eight of the ad easily by rm=1030,” wrote down in an orga- Jarate lines nk or com— aasy to ask (plained in :ment device :ntor Eugene 'isted hollow icator needle ube is unde— to read zero .ed, the tube ssure. eluding pres— :ssure trans- an electrical :essure trans- reliable, and HS pressures easure gage, ations. Gage :e by venting ere, and they 9 of altitude. ;nal output at ire difference 3 transducers iragm deflect e diaphragm it, the strain 1e output. A 3 is measured Piezoelectric transducers, also called solid-state pressure transducers, work on the principle that an electric potential is generated in a crystalline Substance when it is subjected to mechanical pressure. This phenomenon, first discovered by brothers Pierre and Jacques Curie in 1880, is called the piezoelectric (or press-electric) effect. Piezoelectric pressure transducers Very suitable for high-pressure applications, but they are generally not as sensitiVe as diaphragm—type transducers, especially at low pressures. Another type of mechanical pressure gage called a deadweight tester is used primarily for calibration and can measure extremely high pres- sures (Fig. 3—24). As its name implies, a deadweight tester measures pres- sure directly through application of a weight that provides a force per unit area~the fundamental definition of pressure. It is constructed with an inter— nal chamber filled with a fluid (usually oil), along with a tight-fitting piston, cylinder, and plunger. Weights are applied to the top of the piston, which exerts a force on the oil in the chamber. The total force F acting on the oil at the piston—oil interface is the sum of the weight of the piston plus the applied weights. Since the piston cross—sectional area A6 is known, the pres— sure is calculated as P = F/AE. The only significant source of error is that due to static friction along the interface between the piston and cylinder, but even this error is usually negligibly small. The reference pressure port is connected to either an unknown pressure that is to be measured or to a pres— sure sensor that is to be calibrated. 3—3 1‘ INTRODUCTION TO FLUID STATICS Fluid statics deals with problems associated with fluids‘at rest. The fluid can be either gaseous or liquid. Fluid statics is generally referred to as hydrostatics when the fluid is a liquid and as aerostatics when the fluid is a gas. In fluid statics, there is no relative motion between adjacent fluid lay— ers, and thus there are no shear (tangential) stresses in the fluid trying to deform it. The only stress‘we deal with in fluid statics is the normal stress, which is the pressure, and the variation of pressure is due only to the weight of the fluid. Therefore, the topic of fluid statics has significance only in gravity fields, and the force relations developed naturally involve the gravi- tational acceleration g, The force exerted on a surface by a fluid at rest is normal to the surface at the point of contact since there is no relative motion between the fluid and the solid surface, and thus there are no shear forces acting parallel to the surface. Fluid statics is used to determine the forces acting on floating or sub- merged bodies and the forces developed by devices like hydraulic presses and car jacks. The design of many engineering systems such as water dams and liquid storage tanks requires the determination of the forces acting on their surfaces using fluid statics. The complete description of the resultant hydrostatic force acting on a submerged surface requires the determination 0f the magnitude, the direction, and the line of action of the force. In the following two sections, weconsider the forces acting on both plane and Curved surfaces of submerged bodies due to pressure. have a much faster frequency response compared to diaphragm units and are \ . . 87 . , CHAPTER3‘ ' u Weights Oil reserv Reference pressure port oir Adjustab e plunger Crank FIGURE 3—24 A deadweight tester is able to measure extremely high pressures (up to 10,000 psi in some applications). 88 ‘ " a t ’ PRESSURE’ANDFLUID STATICS' FIGURE 3—25 . Hoover Dam. Courtesy United States Department of the Interim; Bureau ofReclamation-Lower Colorado Region. FIGURE 3—26 When analyzing hydrostatic forces on submerged surfaces, the atmospheric pressure can be subtracted for simplicity when it acts on both sides of the structure. 3—4 “ HYDRDSTATIC FORCES 0N SUBMERGED PLANE SURFACES A plate (such as a gate valve in a dam, the wall of a liquid storage tank, or the hull of a ship at rest) is subjected to fluid pressure distributed over its surface when exposed to a liquid (Fig. 3—25). On a plane surface, the hydro- static forces form a system of parallel forces, and we often need to determine the magnitude of the force and its point of application, which is called the center of pressure. In most cases, the other side of the plate is open to the atmosphere (such as the dry side of a gate), and thus atmospheric pressure acts on both sides of the plate, yielding a zero resultant. In such cases, it is convenient to subtract atmospheric pressure and work with the gage pressure only (Fig. 3—26). For example, Pgage = pgh at the bottom of the lake. Consider the top surface of a flat plate of arbitrary shape completely sub- merged in a liquid, as shown in Fig. 3—27 together with its normal View. The plane of this surface (normal to the page) intersects the horizontal free surface at angle 6, and we take the line of intersection to be the x—axis (out of the page). The absolute pressure above the liquid is P0, which is the local atmospheric pressure Patm if the liquid is open to the atmosphere (but P0 may be different than Patm if the space above the liquid is evacuated or pres- surized). Then the absolute pressure at any point on the plate is P = P0 + pgh = P0 + pgy sin 6 (3—16) where h is the vertical distance of the point from the free surface and y is the distance of the point from the x—axis (from point 0 in Fig. 3—27). The resultant hydrostatic force F R acting on the surface is determined by inte- grating the force P dA acting on a differential area dA over the entire sur- face area, ‘ FR: JPdA= J(PO-Fpgysin0)dA=P0A+pgsin6JydA (3—17) A A A But the first moment of area I y dA is related to the y-coordinate of the A centroid (or center) of the surface by Ill pgh Patm + pgh (a) P considered (17) PaLm subtracted atm :orage tank, or lbuted over its 1C6, the hydro d to determine h is called the is open to the iheric pressure lCh cases, it is i gage pressure ie lake. )mpletely sub- normal View. horizontal free the x—aXis (out ich is the local ;phere (but PO :uated or pres- is (3—1 6) 1rface and y is ig. 3—27). The mined by inte- the entire sur- J y dA (3—17) A )rdinate of the Pressure distribution :P avg P=P0+pgysin0 Center of pressine FIGURE 3—27 : Hydrostatic force on an inclined plane surface completely submerged in a liquid. 1 yc = — Jy dA (3—18) A A Substituting, FR = (P0 + pgyC sin 0)A = (P0 + pghC)A = PCA = P..ng (3—19) where PC = P0 + pghC is the pressure at the centroid of the surface, which is equivalent to the average pressure Pavg on the surface, and 110 = yC sin 0 is the vertical distance of the centroid from the free surface of the liquid (Fig. 3—28). Thus we conclude that: The magnitude of the resultant force acting on a plane surface of a completely submerged plate in a homogeneous (constant density) fluid is equal to the product of the pressure PC at the centroid of the surface and the area A of the surface (Fig. 3~29). The pressure P0 is usually atmospheric pressure, which can be ignored in most force calculations since it acts on both sides of the plate. When this is not the case, a practical way of accounting for the contribution of PO to the resultant force is simply to add an equivalent depth hequiv = PO/pg to hC; that is, to assume the presence of an additional liquid layer of thickness hequiv on top of the liquid with absolute vacuum above. Next we need to determine the line of action of the resultant force F R. Two parallel force systems are equivalent if they have the same magnitude and the same moment about any point. The line of action of the resultant hydrostatic force, in general, does not pass through the centroid of the sur— face—it lies underneath Where the pressure is higher. The point of intersec— non of the line of action of the resultant force and the surface is the center 0f pressure. The vertical location of the line of action is determinedgby equating the moment of the resultant force to the moment of the distributed pressure force about the x-aXis: M = JdeA= Jy(P§+pgysin0)dA=P0JydA+pgsin6Jy2dA A A A A Pressure prism Plane surface 89 .‘CHAPTER3’ i, ‘ ofsurface. A : FIGURE 3—28 The pressure at the centroid of a plane surface is equivalent to the average pressure on the surface. 9.0 PRESSURE AND‘FLUID STATICS /Center of pressure Centroid of area FIGURE 3—29 The resultant force acting on a plane surface is equal to the product of the pressure at the centroid of the surface and the surface area, and its line of action passes through the center of pressure. yPFR =Po'yCA + Pg Sin 0 [mo (3—20) where yP is the distance of the center of pressure from the x—axis (point 0 in Fig. 3—29) and [mo = J y2 dA isthe second moment of area (also called A the area moment of inertia) about the x—axis. The second moments of area are widely available for common shapes in engineering handbooks, but they are usually given about the axes passing through the centroid of the area. Fortunately, the second moments of area about two parallel axes are related to each other by the parallel axis theorem, which in this case is expressed as Ixx,0 = Ixx,C + where [m C is the second moment of area about the x—axis passing through the centroid of the area and 3/0 (the y—coordinate of the centroid) is the distance between the two parallel axes. Substituting the F R relation from Eq. 3—19 and the [my 0 relation from Eq. 3—21 into Eq. 3—20 and solving for yP yields I XX, C [3’0 + Po/(Pg Sin 6)lA For P0 = 0, which is usually the case when the atmospheric pressure is ignored, it simplifies to yp = yc + (3—22a) Ixx, C y C Knowing yP, the vertical distance of the center of pressure from the free sur- face is determined from hp = )2], sin 6. The I“, C values for some common areas are given in Fig. 3—30. For areas that possess symmetry about the y—axis, the center of pressure lies on the y- axis directly below the centroid. The location of the center of pressure in such cases is simply the point on the surface of the vertical plane of symme— try at a distance hp from the free surface. Pressure acts normal to the surface, and the hydrostatic forces acting on a flat plate of any shape form a volume whose base is the plate area and whose length is the linearly varying pressure, as shown in Fig. 3—31. This virtual pressure prism has an interesting physical interpretation: its volume is equal to the magnitude of the resultant hydrostatic force acting on the plate since F R = f P dA, and the line of action of this force passes through the centroid of this homogeneous prism. The projection of the centroid on the plate is the pressure center Therefore, with the concept of pressure prism, the problem of describing the resultant hydrostatic force on a plane surface reduces to finding the volume and the two coordinates of the cen— troid of this pressure prism. yp = yc + (3—2211) Special Case: Subme’rged Rectangular Plate Consider a completely submerged rectangular flat plate of height [9 and width a tilted at an angle 6 from the horizontal and whose top edge is hori— zontal and is at a distance 's from the free surface along the plane of the plate, as shown in Fig. 3—32a. The resultant hydrostatic force on the upper surface is equal to the average pressure, which is the pressure at the mid— point of the surface, times the surface area A. That is, 'u. x (3—20) ;is (point 0 in :1 (also called ments of area ooks, but they :1 of the area. Les are related 9 expressed as (3-21) 1g through the .s the distance Eq. 3—19 and P yields (3—22a) ‘ic pressure is (3—22h) n the free sur- —30. For areas i lies on the y— of pressure in [1’16 of symme- :es acting on a )late area and ig. 3—31. This on: its volume acting on the )asses through 1e centroid on Qt of pressure toe on a plane ,es of the cen— te height [9 and p edge is hori— e plane of the 3 on the upper re at the mid“ .fljfs”..., A = ab, 1m C = ab3/12 A = 7R2, [m C = 77-R4/4 (a) Rectangle ([7) Circle a/Z a/2 A = ab/2, I“ C = ab3/36 A = «122/2, 1% C = 0.109757R4 (d) Triangle (e) Semicircle FIGURE 3—30 The centroid and the centroidal moments of inertia for some common geometries. Tilted rectangular plate: FR = PCA = [P0 + pg(s -l- b/Z) sin 6]ab (3—23) The force acts at a vertical distance of hp = yP sin 6 from the free surface directly beneath the centroid of the plate Where, from Eq. 3—22a, I b ab3/12 ' 2 [s + b/Z + PO/(pg sin 6)]ab ‘ b u b2 ' 2 12[s + 12/2 + PO/(pg sin 0)] When the upper edge of the plate is at the free surface and thus s = 0, Eq. 3—23 reduces to yin—S (3—24) Tilted rectangular plate (3 = 0): FR = [P0 + pg(b sin 6)/2]ab For a completely submerged vertical plate (6 é 90°) Whose top edge is hori— zontal, the hydrostatic force can be obtained by setting sin 6 = 1 (Fig. 3—3219) Vertical rectangular plate: FR = [P0 + pg(s + b/ 2)]ab FR = (P0 + pgb/2)ab i3—27) hWhen the effect of P0 is ignored since it acts on both sides of the plate, the Ydrostatic force on a vertical rectangular surface of height [2 whose top (3—25) (3—26) Vertical rectangular plate (5 : O): 91 A = flab, In, C = 77ab3/4 (c) Ellipse A = 7mm, 1“, C = 0109757111;3 (f) Semiellipse Pressure prism FIGURE 3—31 The hydrostatic forces acting on a plane surface form a pressure prism whose base (left face) is the surface and whose length is the pressure. ' , ,92 PRESSURE AND FLUID STATICS‘ ; FR = [P0 + pg(s + 22/2) sin 9142: y g FR = [P0 + pg(s + b/2)]ab (a) Tilted plate (17) Vertical plate (0) Horizontal plate FIGURE 3—32 Hydrostatic force acting on the top surface of a submerged rectangular plate for tilted, vertical, and horizontal cases. edge is horizontal and at the free surface is F R = pgabZ/Z acting at a dis- tance of 2b/3 from the free surface directly beneath the centroid of the plate. ‘ ‘ The pressure distribution on a submerged horizontal surface is uniform, f and its magnitude is P = P0 + pgh, Where h is the distance of the surface 3 from the free surface. Therefore, the hydrostatic force acting on a horizontal rectangular surface is Horizontal rectangular plate: FR = (P0 + pgh)ab (3—28) and it acts through the midpoint of the plate (Fig. 3—32c). EXAMPLE 3—8 Hydrostatic Force Acting on the Door of a Submerged Car A heavy car plunges into a lake during an accident and lands at the bottom a of the lake on its wheels (Fig. 3—33). The door is 1.2 m high and 1 m wide, and the top edge of the door is 8 m below the free surface of the water. Determine the hydrostatic force on the door and the location of the pressure center, and discuss if the driver can open the door. SOLUTION A car is submerged in water. The hydrostatic force on the door is to be determined, and the likelihood of the driver opening the door is to be assessed. Assumptions 1 The bottom surface of the lake is horizontal. 2 The passen— ger cabin is well—sealed so that no water leaks inside. 3 The door can be approximated as a vertical rectangular plate. 4 The pressure in the passenger cabin remains at atmospheric value since there is no water leaking in, and thus no compression of the air inside. Therefore, atmospheric pressure can— cels out in the calculations since it acts on both sides of the door. 5 The weight of the car is larger than the buoyant force acting on it. . l i l l FIGURE 3—33 l Schematic for Example 3—8. y "0 + pgh)ab l plate tal cases. acting at a dis- )id of the plate. 2; ace is uniform, 3 of the surface on a horizontal if (3—28) at the botto :e on the do the door is to: 2 The passen-Tg’ a door can be: the passenget aaking in, and pressure can-f e door. 5 The" praparties We take the density of lake water to be 1000 kg/m-3 throughout. if Analysis - The average (gage) pressure on the door is the pressure value at the‘Ce‘nt-mid (midpoint) of the door and is determined to be Pavg = Pc = pghc : pg(s + M) ' 1kN = 000k/ 3 9.811m2 8+ 1.2/2 (1 gm)( 8X 2 m) lOOOkg - m/s2 = 84.4 kN/m2 Then'the resultant hydrostatic force on the door becomes L FR = PflVgA = (84.4 kN/m2)(1 m X 1.2 m) x 101.3 kN The pressure center is directly under the midpoint of the door, and its dis- tance from the surface of the lake is determined from Eq. 3—24 by setting ‘po‘; 0, yielding “ __ + + [,2 8+1.2+ 1.22 W” 2 12(s+b/2) 2 12(8+1.2/2) sign A strong person can lift 100 kg, whose weight is 981 N or about ‘ ‘Also, the person can apply the force at a point farthest from they2 -(_1 m farther) for maximum effect and generate a moment of 1 kN - m. ' sultant hydrostatic force acts under the midpoint of the door, and thus a :e of 0.5 m from the hinges. This generates a moment of 50.6 kN ‘ m, I about 50 times the moment the driver can possibly generate. There— _:‘ . impossible for the driver to open the door of the car. The driver’s _ L to let some water in (by rolling the window down a little, for ‘ nd to keep his or her head close to the ceiling. The driver should [ open the door shortly before the car is filled with water since at _ the pressures on both sides of the door are nearly the same and the door in water is almost as easy as opening it in air. -— 8.61 In 3—5 r HYDROSTATIC FORCES 0N SUBMERGED CURVED SURFACES For a submerged curved surface, the determination of the resultant hydrosta— tic force is more involved since it typically requires integration of the pres— sure forces that change direction along the curved surface. The concept of the pressure prism in this case is not much help either because of the com- plicated shapes involved. The easiest way to determine the resultant hydrostatic force F R acting on a two—dimensional curved surface is to determine the horizontal and vertical Components FH and F V separately. This is done by considering the free—body dlagram of the liquid block enclosed by the curved surface and the two plane surfaces (one horizontal and one vertical) passing through the two ends of the curved surface, as shown in Fig. 3—34. Note that the vertical sur— face of the liquid block considered is simply the projection of the curved surface on a vertical plane, and the horizontal surface is the projection of the Curved surface on a horizontal plane. The resultant force acting on the Curved solid surface is then equal and opposite to the force acting on the Curved liquid surface (Newton’s third law). CHAPTER 3 l . l ...
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