This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Roosevelt Dam in Arizona. Hydrostatic pressure, due to the weight of a standing fluid, can causeenormous forces and moments on large-scale structures such as a dam. Hydrostatic fluid analysis is the subject of the present chapter. (Courtesy of Dr. E.R. Degginger/Color-Pic Inc.)|vv58|e-Text Main Menu|Textbook Table of Contents|Study GuideChapter 2Pressure Distributionin a FluidMotivation. Many fluid problems do not involve motion. They concern the pressuredistribution in a static fluid and its effect on solid surfaces and on floating and submerged bodies.When the fluid velocity is zero, denoted as the hydrostatic condition, the pressurevariation is due only to the weight of the fluid. Assuming a known fluid in a givengravity field, the pressure may easily be calculated by integration. Important applications in this chapter are (1) pressure distribution in the atmosphere and the oceans, (2)the design of manometer pressure instruments, (3) forces on submerged flat and curvedsurfaces, (4) buoyancy on a submerged body, and (5) the behavior of floating bodies.The last two result in Archimedes principles.If the fluid is moving in rigid-body motion, such as a tank of liquid which has beenspinning for a long time, the pressure also can be easily calculated, because the fluidis free of shear stress. We apply this idea here to simple rigid-body accelerations inSec. 2.9. Pressure measurement instruments are discussed in Sec. 2.10. As a matter offact, pressure also can be easily analyzed in arbitrary (nonrigid-body) motions V(x, y,z, t), but we defer that subject to Chap. 4.2.1 Pressure and PressureGradientIn Fig. 1.1 we saw that a fluid at rest cannot support shear stress and thus Mohrs circle reduces to a point. In other words, the normal stress on any plane through a fluidelement at rest is equal to a unique value called the fluid pressure p, taken positive forcompression by common convention. This is such an important concept that we shallreview it with another approach.Figure 2.1 shows a small wedge of fluid at rest of size x by z by s and depth binto the paper. There is no shear by definition, but we postulate that the pressures px, pz ,and pn may be different on each face. The weight of the element also may be important.Summation of forces must equal zero (no acceleration) in both the x and z directions.FxFz00pzbpxb zxpnb s sinpnb s cos12(2.1)bxz|vv59|e-Text Main Menu|Textbook Table of Contents|Study Guide60Chapter 2 Pressure Distribution in a Fluidz (up)pnsElement weight:d W = g ( 1 b x z)2zpxxxOFig. 2.1 Equilibrium of a smallwedge of fluid at rest.Width b into paperpzbut the geometry of the wedge is such thats sinzs cosx(2.2)Substitution into Eq. (2.1) and rearrangement givepxpnpz12pnz(2.3)These relations illustrate two important principles of the hydrostatic, or shear-free, condition: (1) There is no pressure change in the horizontal direction, and (2) there is avertical change in pressure proportional to the density, gravity, and depth change. Weshall exploit these results to the fullest, starting in Sec. 2.3.In the limit as the fluid wedge shrinks to a point, z 0 and Eqs. (2.3) becomepxpzpnp(2.4)Since is arbitrary, we conclude that the pressure p at a point in a static fluid is independent of orientation.What about the pressure at a point in a moving fluid? If there are strain rates in amoving fluid, there will be viscous stresses, both shear and normal in general (Sec.4.3). In that case (Chap. 4) the pressure is defined as the average of the three normalstresses ii on the element13p(xxyyzz)(2.5)The minus sign occurs because a compression stress is considered to be negativewhereas p is positive. Equation (2.5) is subtle and rarely needed since the great majority of viscous flows have negligible viscous normal stresses (Chap. 4).Pressure Force on a FluidElementPressure (or any other stress, for that matter) causes no net force on a fluid elementunless it varies spatially.1 To see this, consider the pressure acting on the two x facesin Fig. 2.2. Let the pressure vary arbitrarilypp(x, y, z, t)(2.6)1|vvAn interesting application for a large element is in Fig. 3.7.|e-Text Main Menu|Textbook Table of Contents|Study Guide2.2 Equilibrium of a Fluid Element61ydzp dy dz(p+dypd x) dy dzxxFig. 2.2 Net x force on an elementdue to pressure variation.dxzThe net force in the x direction on the element in Fig. 2.2 is given bydFxp dy dzpdx dy dzxppdx dy dzx(2.7)In like manner the net force dFy involvesp/ y, and the net force dFz concernsp/ z. The total net-force vector on the element due to pressure isdFpresspxipyjkpdx dy dzz(2.8)We recognize the term in parentheses as the negative vector gradient of p. Denoting fas the net force per unit element volume, we rewrite Eq. (2.8) aspfpress(2.9)Thus it is not the pressure but the pressure gradient causing a net force which must bebalanced by gravity or acceleration or some other effect in the fluid.2.2 Equilibrium of a FluidElementThe pressure gradient is a surface force which acts on the sides of the element. Theremay also be a body force, due to electromagnetic or gravitational potentials, acting onthe entire mass of the element. Here we consider only the gravity force, or weight ofthe elementdFgravorg dx dy dzfgrav(2.10)gIn general, there may also be a surface force due to the gradient, if any, of the viscous stresses. For completeness, we write this term here without derivation and consider it more thoroughly in Chap. 4. For an incompressible fluid with constant viscosity, the net viscous force is2fVSVx222Vy2Vz22V(2.11)|vvwhere VS stands for viscous stresses and is the coefficient of viscosity from Chap.1. Note that the term g in Eq. (2.10) denotes the acceleration of gravity, a vector act-|e-Text Main Menu|Textbook Table of Contents|Study GuideChapter 2 Pressure Distribution in a Fluiding toward the center of the earth. On earth the average magnitude of g is 32.174 ft/s29.807 m/s2.The total vector resultant of these three forces pressure, gravity, and viscousstress must either keep the element in equilibrium or cause it to move with acceleration a. From Newtons law, Eq. (1.2), we haveaffpressfgravpfvisc2Vg(2.12)This is one form of the differential momentum equation for a fluid element, and it isstudied further in Chap. 4. Vector addition is implied by Eq. (2.12): The accelerationreflects the local balance of forces and is not necessarily parallel to the local-velocityvector, which reflects the direction of motion at that instant.This chapter is concerned with cases where the velocity and acceleration are known,leaving one to solve for the pressure variation in the fluid. Later chapters will take upthe more general problem where pressure, velocity, and acceleration are all unknown.Rewrite Eq. (2.12) asp(ga)2VB(x, y, z, t)(2.13)where B is a short notation for the vector sum on the right-hand side. If V and adV/dt are known functions of space and time and the density and viscosity are known,we can solve Eq. (2.13) for p(x, y, z, t) by direct integration. By components, Eq. (2.13)is equivalent to three simultaneous first-order differential equationspxBx(x, y, z, t)pyBy(x, y, z, t)pzBz(x, y, z, t)(2.14)Since the right-hand sides are known functions, they can be integrated systematicallyto obtain the distribution p(x, y, z, t) except for an unknown function of time, whichremains because we have no relation for p/ t. This extra function is found from a condition of known time variation p0(t) at some point (x0, y0, z0). If the flow is steady (independent of time), the unknown function is a constant and is found from knowledgeof a single known pressure p0 at a point (x0, y0, z0). If this sounds complicated, it isnot; we shall illustrate with many examples. Finding the pressure distribution from aknown velocity distribution is one of the easiest problems in fluid mechanics, whichis why we put it in Chap. 2.Examining Eq. (2.13), we can single out at least four special cases:1. Flow at rest or at constant velocity: The acceleration and viscous terms vanishidentically, and p depends only upon gravity and density. This is the hydrostaticcondition. See Sec. 2.3.2. Rigid-body translation and rotation: The viscous term vanishes identically,and p depends only upon the term (g a). See Sec. 2.9.3. Irrotational motion (V 0): The viscous term vanishes identically, andan exact integral called Bernoullis equation can be found for the pressure distribution. See Sec. 4.9.4. Arbitrary viscous motion: Nothing helpful happens, no general rules apply, butstill the integration is quite straightforward. See Sec. 6.4.Let us consider cases 1 and 2 here.|vv62|e-Text Main Menu|Textbook Table of Contents|Study Guide2.3 Hydrostatic Pressure Distributions63p (Pascals)High pressure:p = 120,000 Pa abs = 30,000 Pa gage120,00030,000Local atmosphere:p = 90,000 Pa abs = 0 Pa gage = 0 Pa vacuum90,00040,000Vacuum pressure:p = 50,000 Pa abs = 40,000 Pa vacuum50,00050,000Fig. 2.3 Illustration of absolute,gage, and vacuum pressure readings.Gage Pressure and VacuumPressure: Relative TermsAbsolute zero reference:p = 0 Pa abs = 90,000 Pa vacuum0(Tension)Before embarking on examples, we should note that engineers are apt to specify pressures as (1) the absolute or total magnitude or (2) the value relative to the local ambient atmosphere. The second case occurs because many pressure instruments are ofdifferential type and record, not an absolute magnitude, but the difference between thefluid pressure and the atmosphere. The measured pressure may be either higher or lowerthan the local atmosphere, and each case is given a name:1. p2. ppapaGage pressure:Vacuum pressure:p(gage)p(vacuum)p papa pThis is a convenient shorthand, and one later adds (or subtracts) atmospheric pressureto determine the absolute fluid pressure.A typical situation is shown in Fig. 2.3. The local atmosphere is at, say, 90,000 Pa,which might reflect a storm condition in a sea-level location or normal conditions atan altitude of 1000 m. Thus, on this day, pa 90,000 Pa absolute 0 Pa gage 0 Pavacuum. Suppose gage 1 in a laboratory reads p1 120,000 Pa absolute. This valuemay be reported as a gage pressure, p1 120,000 90,000 30,000 Pa gage. (Onemust also record the atmospheric pressure in the laboratory, since pa changes gradually.) Suppose gage 2 reads p2 50,000 Pa absolute. Locally, this is a vacuum pressure and might be reported as p2 90,000 50,000 40,000 Pa vacuum. Occasionally, in the Problems section, we will specify gage or vacuum pressure to keep youalert to this common engineering practice.2.3 Hydrostatic PressureDistributions0 and 2VIf the fluid is at rest or at constant velocity, athe pressure distribution reduces top0. Equation (2.13) forg(2.15)|vvThis is a hydrostatic distribution and is correct for all fluids at rest, regardless of theirviscosity, because the viscous term vanishes identically.Recall from vector analysis that the vector p expresses the magnitude and direction of the maximum spatial rate of increase of the scalar property p. As a result, p|e-Text Main Menu|Textbook Table of Contents|Study Guide64Chapter 2 Pressure Distribution in a Fluidis perpendicular everywhere to surfaces of constant p. Thus Eq. (2.15) states that a fluidin hydrostatic equilibrium will align its constant-pressure surfaces everywhere normalto the local-gravity vector. The maximum pressure increase will be in the direction ofgravity, i.e., down. If the fluid is a liquid, its free surface, being at atmospheric pressure, will be normal to local gravity, or horizontal. You probably knew all this before, but Eq. (2.15) is the proof of it.In our customary coordinate system z is up. Thus the local-gravity vector for smallscale problems isggk(2.16)where g is the magnitude of local gravity, for example, 9.807 m/s2. For these coordinates Eq. (2.15) has the componentspxpy0pz0g(2.17)the first two of which tell us that p is independent of x and y. Hence p/ z can be replaced by the total derivative dp/dz, and the hydrostatic condition reduces todpdz2orp2p1dz(2.18)1Equation (2.18) is the solution to the hydrostatic problem. The integration requires anassumption about the density and gravity distribution. Gases and liquids are usuallytreated differently.We state the following conclusions about a hydrostatic condition:Pressure in a continuously distributed uniform static fluid varies only with verticaldistance and is independent of the shape of the container. The pressure is the sameat all points on a given horizontal plane in the fluid. The pressure increases withdepth in the fluid.An illustration of this is shown in Fig. 2.4. The free surface of the container is atmospheric and forms a horizontal plane. Points a, b, c, and d are at equal depth in a horizonAtmospheric pressure:Free surface|vvFig. 2.4 Hydrostatic-pressure distribution. Points a, b, c, and d are atequal depths in water and thereforehave identical pressures. Points A,B, and C are also at equal depths inwater and have identical pressureshigher than a, b, c, and d. Point Dhas a different pressure from A, B,and C because it is not connectedto them by a water path.WaterDepth 1abcdMercuryDepth 2|Ae-Text Main MenuB|CTextbook Table of ContentsD|Study Guide2.3 Hydrostatic Pressure Distributions65tal plane and are interconnected by the same fluid, water; therefore all points have thesame pressure. The same is true of points A, B, and C on the bottom, which all have thesame higher pressure than at a, b, c, and d. However, point D, although at the same depthas A, B, and C, has a different pressure because it lies beneath a different fluid, mercury.Effect of Variable GravityFor a spherical planet of uniform density, the acceleration of gravity varies inverselyas the square of the radius from its centergg0r0r2(2.19)where r0 is the planet radius and g0 is the surface value of g. For earth, r0 3960statute mi 6400 km. In typical engineering problems the deviation from r0 extendsfrom the deepest ocean, about 11 km, to the atmospheric height of supersonic transportoperation, about 20 km. This gives a maximum variation in g of (6400/6420)2, or 0.6percent. We therefore neglect the variation of g in most problems.Hydrostatic Pressure in LiquidsLiquids are so nearly incompressible that we can neglect their density variation in hydrostatics. In Example 1.7 we saw that water density increases only 4.6 percent at thedeepest part of the ocean. Its effect on hydrostatics would be about half of this, or 2.3percent. Thus we assume constant density in liquid hydrostatic calculations, for whichEq. (2.18) integrates toLiquids:p2orp1z1z2(z2p2z1)(2.20)p1We use the first form in most problems. The quantity is called the specific weight ofthe fluid, with dimensions of weight per unit volume; some values are tabulated inTable 2.1. The quantity p/ is a length called the pressure head of the fluid.For lakes and oceans, the coordinate system is usually chosen as in Fig. 2.5, withz 0 at the free surface, where p equals the surface atmospheric pressure pa. WhenTable 2.1 Specific Weight of SomeCommon FluidsSpecific weightat 68F 20Cv||lbf/ft3N/m3Air (at 1 atm)Ethyl alcoholSAE 30 oilWaterSeawaterGlycerinCarbon tetrachlorideMercuryvFluid000.0752049.2055.5062.4064.0078.7099.1846000,011.8007,733008,720009,790010,050012,360015,570133,100e-Text Main Menu|Textbook Table of Contents|Study Guide66Chapter 2 Pressure Distribution in a FluidZp pa b air+bAirFree surface: Z = 0, p = pa0WatergFig. 2.5 Hydrostatic-pressure distribution in oceans and atmospheres.p pa + h waterhwe introduce the reference value ( p1, z1)(negative) depth z,Lakes and oceans:p( pa, 0), Eq. (2.20) becomes, for p at anypaz(2.21)where is the average specific weight of the lake or ocean. As we shall see, Eq. (2.21)holds in the atmosphere also with an accuracy of 2 percent for heights z up to 1000 m.EXAMPLE 2.1Newfound Lake, a freshwater lake near Bristol, New Hampshire, has a maximum depth of 60m, and the mean atmospheric pressure is 91 kPa. Estimate the absolute pressure in kPa at thismaximum depth.SolutionFrom Table 2.1, take9790 N/m3. With pathe pressure at this depth will be91 kN/m2p91 kPa(9790 N/m3)( 60 m)587 kN/m2By omitting pa we could state the result as p60 m, Eq. (2.21) predicts that1 kN1000 N678 kPaAns.587 kPa (gage).The simplest practical application of the hydrostatic formula (2.20) is the barometer(Fig. 2.6), which measures atmospheric pressure. A tube is filled with mercury and inverted while submerged in a reservoir. This causes a near vacuum in the closed upperend because mercury has an extremely small vapor pressure at room temperatures (0.16Pa at 20C). Since atmospheric pressure forces a mercury column to rise a distance hinto the tube, the upper mercury surface is at zero pressure.|vvThe Mercury Barometer91 kPa and z|e-Text Main Menu|Textbook Table of Contents|Study Guide2.3 Hydrostatic Pressure Distributions67p1 0( Mercury has a verylow vapor pressure.)z1 = hp2 pa( The mercury is incontact with theatmosphere.)ph= aMzpaz2 = 0MMercury(b)(a)Fig. 2.6 A barometer measures local absolute atmospheric pressure: (a) the height of a mercury column is proportional to patm; (b) a modern portable barometer, with digital readout, uses the resonating silicon element ofFig. 2.28c. (Courtesy of Paul Lupke, Druck Inc.)From Fig. 2.6, Eq. (2.20) applies with p1paor00 at z1M(0h and p20:h)pahpa at z2(2.22)MAt sea-level standard, with pa 101,350 Pa and M 133,100 N/m3 from Table 2.1,the barometric height is h 101,350/133,100 0.761 m or 761 mm. In the UnitedStates the weather service reports this as an atmospheric pressure of 29.96 inHg(inches of mercury). Mercury is used because it is the heaviest common liquid. A water barometer would be 34 ft high.Hydrostatic Pressure in GasesGases are compressible, with density nearly proportional to pressure. Thus density mustbe considered as a variable in Eq. (2.18) if the integration carries over large pressurechanges. It is sufficiently accurate to introduce the perfect-gas law pRT in Eq.(2.18)|vvdpdz|e-Text Main Menu|pgRTgTextbook Table of Contents|Study GuideChapter 2 Pressure Distribution in a FluidSeparate the variables and integrate between points 1 and 2:2dpp1lnp2p1gR21dzT(2.23)The integral over z requires an assumption about the temperature variation T(z). Onecommon approximation is the isothermal atmosphere, where T T0:p2g(z2 z1)RT0p1 exp(2.24)The quantity in brackets is dimensionless. (Think that over; it must be dimensionless,right?) Equation (2.24) is a fair approximation for earth, but actually the earths meanatmospheric temperature drops off nearly linearly with z up to an altitude of about36,000 ft (11,000 m):TT0Bz(2.25)Here T0 is sea-level temperature (absolute) and B is the lapse rate, both of which varysomewhat from day to day. By international agreement  the following standard values are assumed to apply from 0 to 36,000 ft:T0518.69RB288.16 K0.003566R/ft15C0.00650 K/m(2.26)This lower portion of the atmosphere is called the troposphere. Introducing Eq. (2.25)into (2.23) and integrating, we obtain the more accurate relationppa 1BzT0g/(RB)wheregRB5.26 (air)(2.27)in the troposphere, with z 0 at sea level. The exponent g/(RB) is dimensionless (againit must be) and has the standard value of 5.26 for air, with R 287 m2/(s2 K).The U.S. standard atmosphere  is sketched in Fig. 2.7. The pressure is seen to benearly zero at z 30 km. For tabulated properties see Table A.6.EXAMPLE 2.2If sea-level pressure is 101,350 Pa, compute the standard pressure at an altitude of 5000 m, using (a) the exact formula and (b) an isothermal assumption at a standard sea-level temperatureof 15C. Is the isothermal approximation adequate?SolutionPart (a)Use absolute temperature in the exact formula, Eq. (2.27):ppa 1(0.00650 K/m)(5000 m)288.16 K101,350(0.52388)v||e-Text Main Menu|5.26(101,350 Pa)(0.8872)5.2654,000 PaThis is the standard-pressure result given at zv68Ans. (a)5000 m in Table A.6.Textbook Table of Contents|Study Guide2.3 Hydrostatic Pressure Distributions504040Altitude z, km6050302020.1 km56.5CAltitude z, km60Part (b)30Eq. (2.24)Eq. (2.27)10Eq. (2.26)Troposphere01.20 kPa2011.0 km10Fig. 2.7 Temperature and pressuredistribution in the U.S. standard atmosphere. (From Ref. 1.)69 60 40101.33 kPa15C 20Temperature, C0+204080Pressure, kPa0120If the atmosphere were isothermal at 288.16 K, Eq. (2.24) would apply:pa exppgzRT(9.807 m/s2)(5000 m)[287 m2/(s2 K)](288.16 K)(101,350 Pa) exp(101,350 Pa) exp(0.5929)60,100 PaAns. (b)This is 11 percent higher than the exact result. The isothermal formula is inaccurate in the troposphere.Is the Linear Formula Adequatefor Gases?The linear approximation from Eq. (2.20) or (2.21), pz, is satisfactory for liquids, which are nearly incompressible. It may be used even over great depths in theocean. For gases, which are highly compressible, it is valid only over moderate changesin altitude.The error involved in using the linear approximation (2.21) can be evaluated by expanding the exact formula (2.27) into a series1BzT0n1nBzT0n(n 1) Bz2!T02(2.28)where n g/(RB). Introducing these first three terms of the series into Eq. (2.27) andrearranging, we obtain|vvp|e-Text Main Menu|paaz11 BzT0nTextbook Table of Contents2|Study Guide(2.29)70Chapter 2 Pressure Distribution in a FluidThus the error in using the linear formula (2.21) is small if the second term in parentheses in (2.29) is small compared with unity. This is true ifz(n2T01)B20,800 m(2.30)We thus expect errors of less than 5 percent if z or z is less than 1000 m.2.4 Application to ManometryFrom the hydrostatic formula (2.20), a change in elevation z2 z1 of a liquid is equivalent to a change in pressure (p2 p1)/ . Thus a static column of one or more liquidsor gases can be used to measure pressure differences between two points. Such a device is called a manometer. If multiple fluids are used, we must change the density inthe formula as we move from one fluid to another. Figure 2.8 illustrates the use of theformula with a column of multiple fluids. The pressure change through each fluid iscalculated separately. If we wish to know the total change p5 p1, we add the successive changes p2 p1, p3 p2, p4 p3, and p5 p4. The intermediate values of pcancel, and we have, for the example of Fig. 2.8,p5p10(z2z1)w(z3z2)G(z4z3)M(z5z4)(2.31)No additional simplification is possible on the right-hand side because of the different densities. Notice that we have placed the fluids in order from the lighteston top to the heaviest at bottom. This is the only stable configuration. If we attemptto layer them in any other manner, the fluids will overturn and seek the stablearrangement.A Memory Device: Up VersusDownThe basic hydrostatic relation, Eq. (2.20), is mathematically correct but vexing to engineers, because it combines two negative signs to have the pressure increase downward. When calculating hydrostatic pressure changes, engineers work instinctively bysimply having the pressure increase downward and decrease upward. Thus they use thefollowing mnemonic, or memory, device, first suggested to the writer by Professor Johnz = z1z2zz3z4|vvFig. 2.8 Evaluating pressurechanges through a column of multiple fluids.z5|Known pressure p1Oil, op2 p1 = og(z 2 z1)Water, wp3 p2 = w g(z 3 z 2)Glycerin, GMercury, Me-Text Main Menu|p4 p3 = G g(z 4 z 3)p5 p4 = M g(z 5 z 4)Sum = p5 p1Textbook Table of Contents|Study Guide2.4 Application to Manometry71Open, pazA, pAAJump acrossz1, p1Fig. 2.9 Simple open manometerfor measuring pA relative to atmospheric pressure.z 2 , p2 pa1p = p1 at z = z1 in fluid 22Foss of Michigan State University:pdownpup z(2.32)Thus, without worrying too much about which point is z1 and which is z2, the formula simply increases or decreases the pressure according to whether one is movingdown or up. For example, Eq. (2.31) could be rewritten in the following multiple increase mode:p5p10z1z2wz2z3Gz3z4Mz4z5That is, keep adding on pressure increments as you move down through the layeredfluid. A different application is a manometer, which involves both up and downcalculations.Figure 2.9 shows a simple open manometer for measuring pA in a closed chamberrelative to atmospheric pressure pa, in other words, measuring the gage pressure. Thechamber fluid 1 is combined with a second fluid 2, perhaps for two reasons: (1) toprotect the environment from a corrosive chamber fluid or (2) because a heavier fluid2 will keep z2 small and the open tube can be shorter. One can, of course, apply thebasic hydrostatic formula (2.20). Or, more simply, one can begin at A, apply Eq. (2.32)down to z1, jump across fluid 2 (see Fig. 2.9) to the same pressure p1, and then useEq. (2.32) up to level z2:pA1zAz12z1z2p2patm(2.33)The physical reason that we can jump across at section 1 in that a continuous lengthof the same fluid connects these two equal elevations. The hydrostatic relation (2.20)requires this equality as a form of Pascals law:Any two points at the same elevation in a continuous mass of the same static fluidwill be at the same pressure.This idea of jumping across to equal pressures facilitates multiple-fluid problems.EXAMPLE 2.3|vvThe classic use of a manometer is when two U-tube legs are of equal length, as in Fig. E2.3,and the measurement involves a pressure difference across two horizontal points. The typical ap-|e-Text Main Menu|Textbook Table of Contents|Study Guide72Chapter 2 Pressure Distribution in a FluidFlow device(a)(b)Lh12E2.3plication is to measure pressure change across a flow device, as shown. Derive a formula for thepressure difference pa pb in terms of the system parameters in Fig. E2.3.SolutionUsing our up-down concept as in Eq. (2.32), start at (a), evaluate pressure changes around theU-tube, and end up at (b):pa1gLorpa1ghpb2gh(21gLpbAns.1)ghThe measurement only includes h, the manometer reading. Terms involving L drop out. Note theappearance of the difference in densities between manometer fluid and working fluid. It is a common student error to fail to subtract out the working fluid density 1 a serious error if bothfluids are liquids and less disastrous numerically if fluid 1 is a gas. Academically, of course,such an error is always considered serious by fluid mechanics instructors.Although Ex. 2.3, because of its popularity in engineering experiments, is sometimes considered to be the manometer formula, it is best not to memorize it butrather to adapt Eq. (2.20) or (2.32) to each new multiple-fluid hydrostatics problem.For example, Fig. 2.10 illustrates a multiple-fluid manometer problem for finding the3z 2, p2zA, pAJump acrossz 2, p21ABz1, p1|vvFig. 2.10 A complicated multiplefluid manometer to relate pA to pB.This system is not especially practical but makes a good homeworkor examination problem.Jump acrossz1, p1z 3, p3Jump acrossz 3, p324|e-Text Main Menu|Textbook Table of Contents|Study GuidezB, pB2.4 Application to Manometry73difference in pressure between two chambers A and B. We repeatedly apply Eq. (2.20),jumping across at equal pressures when we come to a continuous mass of the samefluid. Thus, in Fig. 2.10, we compute four pressure differences while making three jumps:pApB(pAp1)1(zA(p1p2)z1)2(z1(p2p3)z2)(p33(z2pB)z3)4(z3zB)(2.34)The intermediate pressures p1,2,3 cancel. It looks complicated, but really it is merelysequential. One starts at A, goes down to 1, jumps across, goes up to 2, jumps across,goes down to 3, jumps across, and finally goes up to B.EXAMPLE 2.4Pressure gage B is to measure the pressure at point A in a water flow. If the pressure at B is 87kPa, estimate the pressure at A, in kPa. Assume all fluids are at 20C. See Fig. E2.4.SAE 30 oilGage B6 cmMercuryA5 cmWaterflow11 cm4 cmE2.4SolutionFirst list the specific weights from Table 2.1 or Table A.3:water9790 N/m3mercury133,100 N/m3oil8720 N/m3Now proceed from A to B, calculating the pressure change in each fluid and adding:pAorW(pAz)WM(3(9790 N/m )(pA489.5 Paz)MO(0.05 m)9317 Paz)OpB(133,100 N/m3)(0.07 m)523.2 PapB(8720 N/m3)(0.06 m)87,000 Pa2where we replace N/m by its short name, Pa. The value zM 0.07 m is the net elevationchange in the mercury (11 cm 4 cm). Solving for the pressure at point A, we obtainpA96,351 Pa96.4 kPa|vvThe intermediate six-figure result of 96,351 Pa is utterly fatuous, since the measurementscannot be made that accurately.|e-Text Main Menu|Textbook Table of Contents|Study GuideAns.74Chapter 2 Pressure Distribution in a FluidIn making these manometer calculations we have neglected the capillary-heightchanges due to surface tension, which were discussed in Example 1.9. These effectscancel if there is a fluid interface, or meniscus, on both sides of the U-tube, as in Fig.2.9. Otherwise, as in the right-hand U-tube of Fig. 2.10, a capillary correction can bemade or the effect can be made negligible by using large-bore ( 1 cm) tubes.2.5 Hydrostatic Forces onPlane SurfacesA common problem in the design of structures which interact with fluids is the computation of the hydrostatic force on a plane surface. If we neglect density changes inthe fluid, Eq. (2.20) applies and the pressure on any submerged surface varies linearlywith depth. For a plane surface, the linear stress distribution is exactly analogous tocombined bending and compression of a beam in strength-of-materials theory. The hydrostatic problem thus reduces to simple formulas involving the centroid and momentsof inertia of the plate cross-sectional area.Figure 2.11 shows a plane panel of arbitrary shape completely submerged in a liquid. The panel plane makes an arbitrary angle with the horizontal free surface, sothat the depth varies over the panel surface. If h is the depth to any element area dAof the plate, from Eq. (2.20) the pressure there is p pah.To derive formulas involving the plate shape, establish an xy coordinate system inthe plane of the plate with the origin at its centroid, plus a dummy coordinate downfrom the surface in the plane of the plate. Then the total hydrostatic force on one sideof the plate is given byFp dA(pah) dApaAh dAThe remaining integral is evaluated by noticing from Fig. 2.11 that hp = paFree surfaceh (x, y)hCGResultantforce:F = pCG A=hsin ySide viewCGx|vvFig. 2.11 Hydrostatic force andcenter of pressure on an arbitraryplane surface of area A inclined atan angle below the free surface.dA = dx dyCPPlan view of arbitrary plane surface|e-Text Main Menu|Textbook Table of Contents|Study Guide(2.35)sinand,2.5 Hydrostatic Forces on Plane Surfaces75by definition, the centroidal slant distance from the surface to the plate is1ACGTherefore, sincedA(2.36)is constant along the plate, Eq. (2.35) becomesFpaAsinFinally, unravel this by noticing thatthe surface to the plate centroid. ThusFpa AdACGhCG ApaAsinCGA(2.37)sinhCG, the depth straight down from( pahCG)ApCG A(2.38)The force on one side of any plane submerged surface in a uniform fluid equals thepressure at the plate centroid times the plate area, independent of the shape of the plateor the angle at which it is slanted.Equation (2.38) can be visualized physically in Fig. 2.12 as the resultant of a linear stress distribution over the plate area. This simulates combined compression andbending of a beam of the same cross section. It follows that the bending portion ofthe stress causes no force if its neutral axis passes through the plate centroid of area.Thus the remaining compression part must equal the centroid stress times the platearea. This is the result of Eq. (2.38).However, to balance the bending-moment portion of the stress, the resultant forceF does not act through the centroid but below it toward the high-pressure side. Its lineof action passes through the center of pressure CP of the plate, as sketched in Fig. 2.11.To find the coordinates (xCP, yCP), we sum moments of the elemental force p dA aboutthe centroid and equate to the moment of the resultant F. To compute yCP, we equateFyCPThe termyp dAy( pasin ) dAsiny dApay dA vanishes by definition of centroidal axes. IntroducingPressure distributionpav = pCGp (x, y)|vvFig. 2.12 The hydrostatic-pressureforce on a plane surface is equal,regardless of its shape, to the resultant of the three-dimensional linearpressure distribution on that surfaceF pCGA.|e-Text Main MenuCentroid of the plane surface|Textbook Table of Contents|Arbitraryplane surfaceof area AStudy Guide(2.39)CGy,76Chapter 2 Pressure Distribution in a Fluidwe obtainsinFyCPCGy dAy2 dAsinIxx(2.40)where again y dA 0 and Ixx is the area moment of inertia of the plate area about itscentroidal x axis, computed in the plane of the plate. Substituting for F gives the resultIxxyCPsin(2.41)pCGAThe negative sign in Eq. (2.41) shows that yCP is below the centroid at a deeper leveland, unlike F, depends upon angle . If we move the plate deeper, yCP approaches thecentroid because every term in Eq. (2.41) remains constant except pCG, which increases.The determination of xCP is exactly similar:xp dAFxCPsinx[paxy dA(CGy) sin ] dAsinIxy(2.42)where Ixy is the product of inertia of the plate, again computed in the plane of theplate. Substituting for F givesIxyxCPsin(2.43)pCGAFor positive Ixy, xCP is negative because the dominant pressure force acts in the third,or lower left, quadrant of the panel. If Ixy 0, usually implying symmetry, xCP 0and the center of pressure lies directly below the centroid on the y axis.L2yA = bLxIxx =L2A = R2ybL3x12RIx y = 0RIxx = R44Ix y = 0b2b2(a)(b)syIxx =xL3|vvFig. 2.13 Centroidal moments ofinertia for various cross sections:(a) rectangle, (b) circle, (c) triangle, and (d) semicircle.b2b2e-Text Main MenubL336Ixx = 0.10976R 4yIx y = 0b (b 2 s) L 2Ix y =72xR(c )|2A = R2A = bL22L3R4R3(d )|Textbook Table of Contents|Study Guide2.5 Hydrostatic Forces on Plane Surfaces77In most cases the ambient pressure pa is neglected because it acts on both sides of theplate; e.g., the other side of the plate is inside a ship or on the dry side of a gate or dam.In this case pCGhCG, and the center of pressure becomes independent of specific weightGage-Pressure FormulasFhCGAIxx sinhCGAyCPxCPIxy sinhCGA(2.44)Figure 2.13 gives the area and moments of inertia of several common cross sectionsfor use with these formulas.EXAMPLE 2.5The gate in Fig. E2.5a is 5 ft wide, is hinged at point B, and rests against a smooth wall at pointA. Compute (a) the force on the gate due to seawater pressure, (b) the horizontal force P exertedby the wall at point A, and (c) the reactions at the hinge B.WallpaSeawater:64 lbf/ft 315 ftApaGate6 ftBE2.5a8 ftHingeSolutionPart (a)By geometry the gate is 10 ft long from A to B, and its centroid is halfway between, or at elevation 3 ft above point B. The depth hCG is thus 15 3 12 ft. The gate area is 5(10) 50 ft2. Neglect pa as acting on both sides of the gate. From Eq. (2.38) the hydrostatic force on the gate isFPart (b)pCGAhCGA(64 lbf/ft3)(12 ft)(50 ft2)38,400 lbfFirst we must find the center of pressure of F. A free-body diagram of the gate is shown in Fig.E2.5b. The gate is a rectangle, henceIxy0andIxxbL312(5 ft)(10 ft)312417 ft4The distance l from the CG to the CP is given by Eq. (2.44) since pa is neglected.vvl||Ans. (a)e-Text Main Menu|yCPIxx sinhCGA6(417 ft4)( 10 )(12 ft)(50 ft2)Textbook Table of Contents|0.417 ftStudy GuideChapter 2 Pressure Distribution in a FluidAPF5 ftCGlBBxCPL = 10 ftBzE2.5bThe distance from point B to force F is thus 10terclockwise about B givesPL sinF(5Part (c)ll)P(6 ft)Por54.583 ft. Summing moments coun-(38,400 lbf)(4.583 ft)029,300 lbfAns. (b)With F and P known, the reactions Bx and Bz are found by summing forces on the gateFx0BxF sinorBxFz0BzF cosorBzPBx38,400(0.6)29,3006300 lbfBz38,400(0.8)30,700 lbfAns. (c)This example should have reviewed your knowledge of statics.EXAMPLE 2.6A tank of oil has a right-triangular panel near the bottom, as in Fig. E2.6. Omitting pa, find the(a) hydrostatic force and (b) CP on the panel.paOil: = 800 kg/m 35m3011 m4m6mpaCGCP4m8mv|2m4mE2.6v78|e-Text Main Menu|Textbook Table of Contents|Study Guide2.6 Hydrostatic Forces on Curved Surfaces79SolutionPart (a)The triangle has properties given in Fig. 2.13c. The centroid is one-third up (4 m) and one-thirdover (2 m) from the lower left corner, as shown. The area is12(6 m)(12 m)36 m2The moments of inertia arebL336Ixxandb(bIxy(6 m)(12 m)3362s)L272The depth to the centroid is hCGF2.54Part (b)(6 m)[6 m5ghCGA288 m442(6 m)](12 m)27272 m49 m; thus the hydrostatic force from Eq. (2.44) is(800 kg/m3)(9.807 m/s2)(9 m)(36 m2)106 (kg m)/s22.54106 N2.54 MNAns. (a)The CP position is given by Eqs. (2.44):yCPIxx sinhCGA(288 m4)(sin 30)(9 m)(36 m2)0.444 mxCPIxy sinhCGA( 72 m4)(sin 30)(9 m)(36 m2)0.111 mAns. (b)The resultant force F 2.54 MN acts through this point, which is down and to the right of thecentroid, as shown in Fig. E220.127.116.11 Hydrostatic Forces onCurved SurfacesThe resultant pressure force on a curved surface is most easily computed by separating it into horizontal and vertical components. Consider the arbitrary curved surfacesketched in Fig. 2.14a. The incremental pressure forces, being normal to the local areaelement, vary in direction along the surface and thus cannot be added numerically. WeWairdCurved surfaceprojection ontovertical planeFV|vvFig. 2.14 Computation of hydrostatic force on a curved surface:(a) submerged curved surface; (b)free-body diagram of fluid abovethe curved surface.|e-Text Main MenuF1bW2FHaFV(a)|Textbook Table of Contents(b)|F1W1cFHFHeStudy GuideFHChapter 2 Pressure Distribution in a Fluidcould sum the separate three components of these elemental pressure forces, but it turnsout that we need not perform a laborious three-way integration.Figure 2.14b shows a free-body diagram of the column of fluid contained in the vertical projection above the curved surface. The desired forces FH and FV are exerted bythe surface on the fluid column. Other forces are shown due to fluid weight and horizontal pressure on the vertical sides of this column. The column of fluid must be instatic equilibrium. On the upper part of the column bcde, the horizontal componentsF1 exactly balance and are not relevant to the discussion. On the lower, irregular portion of fluid abc adjoining the surface, summation of horizontal forces shows that thedesired force FH due to the curved surface is exactly equal to the force FH on the vertical left side of the fluid column. This left-side force can be computed by the planesurface formula, Eq. (2.38), based on a vertical projection of the area of the curvedsurface. This is a general rule and simplifies the analysis:The horizontal component of force on a curved surface equals the force on the planearea formed by the projection of the curved surface onto a vertical plane normal tothe component.If there are two horizontal components, both can be computed by this scheme.Summation of vertical forces on the fluid free body then shows thatFVW1W2Wair(2.45)We can state this in words as our second general rule:The vertical component of pressure force on a curved surface equals in magnitudeand direction the weight of the entire column of fluid, both liquid and atmosphere,above the curved surface.Thus the calculation of FV involves little more than finding centers of mass of a column of fluid perhaps a little integration if the lower portion abc has a particularlyvexing shape.;;;;;;EXAMPLE 2.7A dam has a parabolic shape z/z0 (x/x0)2 as shown in Fig. E2.7a, with x0 10 ft and z0 24ft. The fluid is water,62.4 lbf/ft3, and atmospheric pressure may be omitted. Compute thepa = 0 lbf/ft2 gageFVzFHz0CPxx0E2.7a|vv80|e-Text Main Menu((xz = z0 x0|2Textbook Table of Contents|Study Guide2.6 Hydrostatic Forces on Curved Surfaces81forces FH and FV on the dam and the position CP where they act. The width of the dam is50 ft.SolutionThe vertical projection of this curved surface is a rectangle 24 ft high and 50 ft wide, with itscentroid halfway down, or hCG 12 ft. The force FH is thusFH(62.4 lbf/ft3)(12 ft)(24 ft)(50 ft)hCGAproj899,000 lbf103 lbf899Ans.The line of action of FH is below the centroid by an amount112Ixx sinhCGAprojyCP(50 ft)(24 ft)3(sin 90)(12 ft)(24 ft)(50 ft)4 ftThus FH is 12 4 16 ft, or two-thirds, down from the free surface or 8 ft from the bottom,as might have been evident by inspection of the triangular pressure distribution.The vertical component FV equals the weight of the parabolic portion of fluid above thecurved surface. The geometric properties of a parabola are shown in Fig. E2.7b. The weight ofthis amount of water is(62.4 lbf/ft3)( 2 )(10 ft)(24 ft)(50 ft)3( 2 x0z0b)3FV499,000 lbf499103 lbfAns.z0Area =3z052 x0z 03FVParabola0E2.7bx0 = 10 ft3x 08This acts downward on the surface at a distance 3x0 /8 3.75 ft over from the origin of coordinates. Note that the vertical distance 3z0 /5 in Fig. E2.7b is irrelevant.The total resultant force acting on the dam isF2(FH2FV)1/2[(499)2(899)2]1/21028103 lbfAs seen in Fig. E2.7c, this force acts down and to the right at an angle of 29 tan 1 499 . The899force F passes through the point (x, z) (3.75 ft, 8 ft). If we move down along the 29 line until we strike the dam, we find an equivalent center of pressure on the dam atxCP5.43 ftzCP7.07 ftAns.|vvThis definition of CP is rather artificial, but this is an unavoidable complication of dealing witha curved surface.|e-Text Main Menu|Textbook Table of Contents|Study Guide82Chapter 2 Pressure Distribution in a FluidzResultant = 1028 103 1bf acts along z = 10.083 0.5555 x3.75 ft499899Parabola z = 0.24x229CGCP7.07 ft8 ftE2.7c02.7 Hydrostatic Forces inLayered Fluidsx5.43 ftThe formulas for plane and curved surfaces in Secs. 2.5 and 2.6 are valid only for afluid of uniform density. If the fluid is layered with different densities, as in Fig. 2.15,a single formula cannot solve the problem because the slope of the linear pressure distribution changes between layers. However, the formulas apply separately to each layer,and thus the appropriate remedy is to compute and sum the separate layer forces andmoments.Consider the slanted plane surface immersed in a two-layer fluid in Fig. 2.15. Theslope of the pressure distribution becomes steeper as we move down into the denserzF 1= pCG1PlanesurfaceA1z=0pa1 < 2Fluid 1p = p 1gzaz 1, p1F2 = pp1 = p 1gz1aACG 2 22Fluid 2z 2 , p2p = p1 2 g(z z 1)|vvFig. 2.15 Hydrostatic forces on asurface immersed in a layered fluidmust be summed in separate pieces.p2 = p1 2 g(z 2 z 1)|e-Text Main Menu|Textbook Table of Contents|Study Guide2.7 Hydrostatic Forces in Layered Fluids83second layer. The total force on the plate does not equal the pressure at the centroidtimes the plate area, but the plate portion in each layer does satisfy the formula, so thatwe can sum forces to find the total:FFipCGi Ai(2.46)Similarly, the centroid of the plate portion in each layer can be used to locate the center of pressure on that portionyCPiigsin i IxxipCGi AiigxCPisin i IxyipCGi Ai(2.47)These formulas locate the center of pressure of that particular Fi with respect to thecentroid of that particular portion of plate in the layer, not with respect to the centroidof the entire plate. The center of pressure of the total force FFi can then be foundby summing moments about some convenient point such as the surface. The following example will illustrate.EXAMPLE 2.8A tank 20 ft deep and 7 ft wide is layered with 8 ft of oil, 6 ft of water, and 4 ft of mercury.Compute (a) the total hydrostatic force and (b) the resultant center of pressure of the fluid onthe right-hand side of the tank.SolutionPart (a)Divide the end panel into three parts as sketched in Fig. E2.8, and find the hydrostatic pressureat the centroid of each part, using the relation (2.38) in steps as in Fig. E2.8:pa = 0Oiz=07 ft5.04 ft(1)l: 511 ftlbf/ft 38 ftWater(62.4)Mercury6 ft(846)16 ft(2)4 ft (3)E2.8v|e-Text Main Menu|(55.0)(8)62.4(3)627 lbf/ft2pCG3|(55.0 lbf/ft3)(4 ft)pCG2vPCG1(55.0)(8)62.4(6)846(2)Textbook Table of Contents220 lbf/ft2|2506 lbf/ft2Study Guide84Chapter 2 Pressure Distribution in a FluidThese pressures are then multiplied by the respective panel areas to find the force on each portion:F1(220 lbf/ft2)(8 ft)(7 ft)pCG1A1F2F3pCG2 A2627(6)(7)2506(4)(7)pCG3A326,300 lbf70,200 lbfFPart (b)12,300 lbfFi108,800 lbfAns. (a)Equations (2.47) can be used to locate the CP of each force Fi, noting that90 and sin1 for all parts. The moments of inertia are Ixx1 (7 ft)(8 ft)3/12 298.7 ft4, Ixx2 7(6)3/12126.0 ft4, and Ixx3 7(4)3/12 37.3 ft4. The centers of pressure are thus at1gIxx1yCP1F162.4(126.0)26,300yCP2(55.0 lbf/ft3)(298.7 ft4)12,300 lbf0.30 ft1.33 ft846(37.3)70,200yCP30.45 ftThis locates zCP14 1.335.33 ft, zCP211 0.3011.30 ft, and zCP316 0.4516.45 ft. Summing moments about the surface then givesFizCPior12,300( 5.33)or26,300( 11.30)zCP1,518,000108,800FzCP70,200( 16.45)108,800zCP13.95 ftAns. (b)The center of pressure of the total resultant force on the right side of the tank lies 13.95 ft below the surface.2.8 Buoyancy and StabilityThe same principles used to compute hydrostatic forces on surfaces can be applied tothe net pressure force on a completely submerged or floating body. The results are thetwo laws of buoyancy discovered by Archimedes in the third century B.C.:1. A body immersed in a fluid experiences a vertical buoyant force equal to theweight of the fluid it displaces.2. A floating body displaces its own weight in the fluid in which it floats.These two laws are easily derived by referring to Fig. 2.16. In Fig. 2.16a, the bodylies between an upper curved surface 1 and a lower curved surface 2. From Eq. (2.45)for vertical force, the body experiences a net upward forceFBFV (2)FV (1)(fluid weight above 2)(fluid weight above 1)weight of fluid equivalent to body volume(2.48)Alternatively, from Fig. 2.16b, we can sum the vertical forces on elemental verticalslices through the immersed body:|vvFB|body(p2e-Text Main Menu|p1) dAH(z2z1) dAHTextbook Table of Contents( )(body volume) (2.49)|Study Guide2.8 Buoyancy and StabilityFV (1)Horizontalelementalarea d AHp1Surface185z1 z 2Fig. 2.16 Two different approachesto the buoyant force on an arbitraryimmersed body: (a) forces on upper and lower curved surfaces; (b)summation of elemental verticalpressure forces.Surface2p2FV (2)(a)(b)These are identical results and equivalent to law 1 above.Equation (2.49) assumes that the fluid has uniform specific weight. The line of action of the buoyant force passes through the center of volume of the displaced body;i.e., its center of mass is computed as if it had uniform density. This point throughwhich FB acts is called the center of buoyancy, commonly labeled B or CB on a drawing. Of course, the point B may or may not correspond to the actual center of mass ofthe bodys own material, which may have variable density.Equation (2.49) can be generalized to a layered fluid (LF) by summing the weightsof each layer of density i displaced by the immersed body:(FB)LFig(displacedvolume)i(2.50)Each displaced layer would have its own center of volume, and one would have to summoments of the incremental buoyant forces to find the center of buoyancy of the immersed body.Since liquids are relatively heavy, we are conscious of their buoyant forces, but gasesalso exert buoyancy on any body immersed in them. For example, human beings havean average specific weight of about 60 lbf/ft3. We may record the weight of a personas 180 lbf and thus estimate the persons total volume as 3.0 ft3. However, in so doingwe are neglecting the buoyant force of the air surrounding the person. At standard conditions, the specific weight of air is 0.0763 lbf/ft3; hence the buoyant force is approximately 0.23 lbf. If measured in vacuo, the person would weigh about 0.23 lbf more.For balloons and blimps the buoyant force of air, instead of being negligible, is thecontrolling factor in the design. Also, many flow phenomena, e.g., natural convectionof heat and vertical mixing in the ocean, are strongly dependent upon seemingly smallbuoyant forces.Floating bodies are a special case; only a portion of the body is submerged, withthe remainder poking up out of the free surface. This is illustrated in Fig. 2.17, wherethe shaded portion is the displaced volume. Equation (2.49) is modified to apply to thissmaller volume|vvFB|e-Text Main Menu|( )(displaced volume)Textbook Table of Contentsfloating-body weight|Study Guide(2.51)86Chapter 2 Pressure Distribution in a FluidNeglect the displaced air up here.CGWFBBFig. 2.17 Static equilibrium of afloating body.(Displaced volume) ( of fluid) = body weightNot only does the buoyant force equal the body weight, but also they are collinearsince there can be no net moments for static equilibrium. Equation (2.51) is the mathematical equivalent of Archimedes law 2, previously stated.EXAMPLE 2.9A block of concrete weighs 100 lbf in air and weighs only 60 lbf when immersed in fresh water (62.4 lbf/ft3). What is the average specific weight of the block?SolutionA free-body diagram of the submerged block (see Fig. E2.9) shows a balance between the apparent weight, the buoyant force, and the actual weight60 lbfFzorFBFB40 lbf060FB1003(62.4 lbf/ft )(block volume, ft3)Solving gives the volume of the block as 40/62.4the block is0.641 ft3. Therefore the specific weight ofW = 100 lbfE2.9block100 lbf0.641 ft3156 lbf/ft3Ans.Occasionally, a body will have exactly the right weight and volume for its ratio toequal the specific weight of the fluid. If so, the body will be neutrally buoyant and willremain at rest at any point where it is immersed in the fluid. Small neutrally buoyantparticles are sometimes used in flow visualization, and a neutrally buoyant body calleda Swallow float  is used to track oceanographic currents. A submarine can achievepositive, neutral, or negative buoyancy by pumping water in or out of its ballast tanks.Stability|vvA floating body as in Fig. 2.17 may not approve of the position in which it is floating.If so, it will overturn at the first opportunity and is said to be statically unstable, likea pencil balanced upon its point. The least disturbance will cause it to seek anotherequilibrium position which is stable. Engineers must design to avoid floating instabil-|e-Text Main Menu|Textbook Table of Contents|Study Guide2.8 Buoyancy and StabilitySmall disturbanceangleLine ofsymmetry87SmalldisturbanceangleMGGGWFBWFig. 2.18 Calculation of the metacenter M of the floating bodyshown in (a). Tilt the body a smallangle . Either (b) B moves farout (point M above G denotes stability); or (c) B moves slightly(point M below G denotes instability).WFBB'B'BEither(a)MFBRestoring moment(b)orOverturning moment(c)ity. The only way to tell for sure whether a floating position is stable is to disturbthe body a slight amount mathematically and see whether it develops a restoring moment which will return it to its original position. If so, it is stable; if not, unstable. Suchcalculations for arbitrary floating bodies have been honed to a fine art by naval architects , but we can at least outline the basic principle of the static-stability calculation. Figure 2.18 illustrates the computation for the usual case of a symmetric floatingbody. The steps are as follows:1. The basic floating position is calculated from Eq. (2.51). The bodys center ofmass G and center of buoyancy B are computed.2. The body is tilted a small angle , and a new waterline is established for thebody to float at this angle. The new position B of the center of buoyancy is calculated. A vertical line drawn upward from B intersects the line of symmetry ata point M, called the metacenter, which is independent offor small angles.3. If point M is above G, that is, if the metacentric height MG is positive, a restoring moment is present and the original position is stable. If M is below G (negative MG, the body is unstable and will overturn if disturbed. Stability increaseswith increasing MG.Thus the metacentric height is a property of the cross section for the given weight, andits value gives an indication of the stability of the body. For a body of varying crosssection and draft, such as a ship, the computation of the metacenter can be very involved.|vvStability Related to WaterlineArea|Naval architects  have developed the general stability concepts from Fig. 2.18 intoa simple computation involving the area moment of inertia of the waterline area aboutthe axis of tilt. The derivation assumes that the body has a smooth shape variation (nodiscontinuities) near the waterline and is derived from Fig. 2.19.The y-axis of the body is assumed to be a line of symmetry. Tilting the body a smallangle then submerges small wedge Obd and uncovers an equal wedge cOa, as shown.e-Text Main Menu|Textbook Table of Contents|Study Guide88Chapter 2 Pressure Distribution in a FluidyOriginalwaterlineareaGVariable-widthL(x) into paperdA = x tanMdxcabOBGFig. 2.19 A floating body tiltedthrough a small angle . The movement x of the center of buoyancy Bis related to the waterline area moment of inertia.GxdBxeTilted floating bodyThe new position B of the center of buoyancy is calculated as the centroid of the submerged portion aObde of the body:x abOdex dx dcOdeaObd0x d0cOax L (x tandx)Obdx (L dA)ObdxL ( x tanx (L dA)cOadx)x2 dAwaterlinetancOaIO tanwaterlinewhere IO is the area moment of inertia of the waterline footprint of the body about itstilt axis O. The first integral vanishes because of the symmetry of the original submerged portion cOdea. The remaining two wedge integrals combine into IO whenwe notice that L dx equals an element of waterline area. Thus we determine the desired distance from M to B:xMBIOsubmergedMGGBorMGIOsubGB(2.52)The engineer would determine the distance from G to B from the basic shape anddesign of the floating body and then make the calculation of IO and the submergedvolume sub. If the metacentric height MG is positive, the body is stable for smalldisturbances. Note that if GB is negative, that is, B is above G, the body is alwaysstable.EXAMPLE 2.10|vvA barge has a uniform rectangular cross section of width 2L and vertical draft of height H, asin Fig. E2.10. Determine (a) the metacentric height for a small tilt angle and (b) the range ofratio L/H for which the barge is statically stable if G is exactly at the waterline as shown.|e-Text Main Menu|Textbook Table of Contents|Study Guide2.9 Pressure Distribution in Rigid-Body Motion89GOGLE2.10HBLSolutionIf the barge has length b into the paper, the waterline area, relative to tilt axis O, has a base band a height 2L; therefore, IO b(2L)3/12. Meanwhile, sub 2LbH. Equation (2.52) predicts8bL3/12HHIOL2MGGBAns. (a)2LbH22sub3HThe barge can thus be stable only ifL23H2/2or2L2.45HAns. (b)The wider the barge relative to its draft, the more stable it is. Lowering G would help also.Even an expert will have difficulty determining the floating stability of a buoyantbody of irregular shape. Such bodies may have two or more stable positions. For example, a ship may float the way we like it, so that we can sit upon the deck, or it mayfloat upside down (capsized). An interesting mathematical approach to floating stability is given in Ref. 11. The author of this reference points out that even simple shapes,e.g., a cube of uniform density, may have a great many stable floating orientations, notnecessarily symmetric. Homogeneous circular cylinders can float with the axis of symmetry tilted from the vertical.Floating instability occurs in nature. Living fish generally swim with their planeof symmetry vertical. After death, this position is unstable and they float with theirflat sides up. Giant icebergs may overturn after becoming unstable when their shapeschange due to underwater melting. Iceberg overturning is a dramatic, rarely seenevent.Figure 2.20 shows a typical North Atlantic iceberg formed by calving from a Greenland glacier which protruded into the ocean. The exposed surface is rough, indicatingthat it has undergone further calving. Icebergs are frozen fresh, bubbly, glacial waterof average density 900 kg/m3. Thus, when an iceberg is floating in seawater, whoseaverage density is 1025 kg/m3, approximately 900/1025, or seven-eighths, of its volume lies below the water.|vv2.9 Pressure Distribution inRigid-Body Motion|In rigid-body motion, all particles are in combined translation and rotation, and thereis no relative motion between particles. With no relative motion, there are no strainse-Text Main Menu|Textbook Table of Contents|Study Guide90Chapter 2 Pressure Distribution in a FluidFig. 2.20 A North Atlantic icebergformed by calving from a Greenland glacier. These, and their evenlarger Antarctic sisters, are thelargest floating bodies in the world.Note the evidence of further calving fractures on the front surface.(Courtesy of Soren Thalund, Green/land tourism a/s Iiulissat, Greenland.)or strain rates, so that the viscous term 2V in Eq. (2.13) vanishes, leaving a balancebetween pressure, gravity, and particle accelerationp(ga)(2.53)The pressure gradient acts in the direction g a, and lines of constant pressure (including the free surface, if any) are perpendicular to this direction. The general caseof combined translation and rotation of a rigid body is discussed in Chap. 3, Fig. 3.12.If the center of rotation is at point O and the translational velocity is V0 at this point,the velocity of an arbitrary point P on the body is given by2VV0r0whereis the angular-velocity vector and r0 is the position of point P. Differentiating, we obtain the most general form of the acceleration of a rigid body:adV0dt(r0)ddtr0(2.54)Looking at the right-hand side, we see that the first term is the translational acceleration; the second term is the centripetal acceleration, whose direction is from point|vv2|For a more detailed derivation of rigid-body motion, see Ref. 4, Sec. 2.7.e-Text Main Menu|Textbook Table of Contents|Study Guide2.9 Pressure Distribution in Rigid-Body Motion91P perpendicular toward the axis of rotation; and the third term is the linear acceleration due to changes in the angular velocity. It is rare for all three of these terms to apply to any one fluid flow. In fact, fluids can rarely move in rigid-body motion unlessrestrained by confining walls for a long time. For example, suppose a tank of wateris in a car which starts a constant acceleration. The water in the tank would begin toslosh about, and that sloshing would damp out very slowly until finally the particlesof water would be in approximately rigid-body acceleration. This would take so longthat the car would have reached hypersonic speeds. Nevertheless, we can at least discuss the pressure distribution in a tank of rigidly accelerating water. The following isan example where the water in the tank will reach uniform acceleration rapidly.EXAMPLE 2.11A tank of water 1 m deep is in free fall under gravity with negligible drag. Compute the pressure at the bottom of the tank if pa 101 kPa.SolutionBeing unsupported in this condition, the water particles tend to fall downward as a rigid hunkof fluid. In free fall with no drag, the downward acceleration is a g. Thus Eq. (2.53) for thissituation gives p(g g) 0. The pressure in the water is thus constant everywhere andequal to the atmospheric pressure 101 kPa. In other words, the walls are doing no service in sustaining the pressure distribution which would normally exist.Uniform Linear AccelerationIn this general case of uniform rigid-body acceleration, Eq. (2.53) applies, a havingthe same magnitude and direction for all particles. With reference to Fig. 2.21, the parallelogram sum of g and a gives the direction of the pressure gradient or greatest rateof increase of p. The surfaces of constant pressure must be perpendicular to this andare thus tilted at a downward angle such thatax1tang(2.55)azzaxaazx = tan 1aaxg + azFluidat restgp g a|vvFig. 2.21 Tilting of constantpressure surfaces in a tank ofliquid in rigid-body acceleration.|e-Text Main Menu|azaxSp2p = p1p3Textbook Table of Contents|Study GuideChapter 2 Pressure Distribution in a FluidOne of these tilted lines is the free surface, which is found by the requirement that thefluid retain its volume unless it spills out. The rate of increase of pressure in the direction g a is greater than in ordinary hydrostatics and is given bydpdsGwhere G[a2x(gaz)2]1/2(2.56)These results are independent of the size or shape of the container as long as thefluid is continuously connected throughout the container.EXAMPLE 2.12A drag racer rests her coffee mug on a horizontal tray while she accelerates at 7 m/s2. The mugis 10 cm deep and 6 cm in diameter and contains coffee 7 cm deep at rest. (a) Assuming rigidbody acceleration of the coffee, determine whether it will spill out of the mug. (b) Calculate thegage pressure in the corner at point A if the density of coffee is 1010 kg/m3.SolutionPart (a)The free surface tilts at the angleaz 0 and standard gravity,given by Eq. (2.55) regardless of the shape of the mug. With7.0axtan 135.59.81gIf the mug is symmetric about its central axis, the volume of coffee is conserved if the tilted surface intersects the original rest surface exactly at the centerline, as shown in Fig. E2.12.tan13 cmz7 cmax = 7 m/s2A3 cmE2.12Thus the deflection at the left side of the mug isz(3 cm)(tan )2.14 cmAns. (a)This is less than the 3-cm clearance available, so the coffee will not spill unless it was sloshedduring the start-up of acceleration.Part (b)When at rest, the gage pressure at point A is given by Eq. (2.20):pA|vv92|g(zsurfe-Text Main MenuzA)|(1010 kg/m3)(9.81 m/s2)(0.07 m)Textbook Table of Contents694 N/m2|694 PaStudy Guide2.9 Pressure Distribution in Rigid-Body MotionDuring acceleration, Eq. (2.56) applies, with G [(7.0)2 (9.81)2]1/2tance s down the normal from the tilted surface to point A iss(7.02.14)(cos )9312.05 m/s2. The dis-7.44 cmThus the pressure at point A becomespAGs1010(12.05)(0.0744)906 PaAns. (b)which is an increase of 31 percent over the pressure when at rest.Rigid-Body RotationAs a second special case, consider rotation of the fluid about the z axis without anytranslation, as sketched in Fig. 2.22. We assume that the container has been rotatinglong enough at constant for the fluid to have attained rigid-body rotation. The fluidacceleration will then be the centripetal term in Eq. (2.54). In the coordinates of Fig.2.22, the angular-velocity and position vectors are given bykr0irr(2.57)Then the acceleration is given by(r0)r2ir(2.58)as marked in the figure, and Eq. (2.53) for the force balance becomespprirkpz(ga)( gkr2ir)(2.59)Equating like components, we find the pressure field by solving two first-order partialdifferential equationsprrpz2(2.60)This is our first specific example of the generalized three-dimensional problem described by Eqs. (2.14) for more than one independent variable. The right-hand sides ofz, kr, irp = paa = r 2 iraStill-waterlevel|vvFig. 2.22 Development of paraboloid constant-pressure surfaces in afluid in rigid-body rotation. Thedashed line along the direction ofmaximum pressure increase is anexponential curve.|e-Text Main Menup = p1Axis ofrotationggap2p3|Textbook Table of Contents|Study Guide94Chapter 2 Pressure Distribution in a Fluid(2.60) are known functions of r and z. One can proceed as follows: Integrate the firstequation partially, i.e., holding z constant, with respect to r. The result is12pr22f(z)(2.61)where the constant of integration is actually a function f(z). Now differentiate thiswith respect to z and compare with the second relation of (2.60):pzor0f (z)f(z)zC(2.62a)where C is a constant. Thus Eq. (2.61) now becomespconst12zr22(2.62b)This is the pressure distribution in the fluid. The value of C is found by specifying thepressure at one point. If p p0 at (r, z) (0, 0), then C p0. The final desired distribution ispp0z12r22(2.63)The pressure is linear in z and parabolic in r. If we wish to plot a constant-pressuresurface, say, p p1, Eq. (2.63) becomeszp0p1r2 22gabr2(2.64)Thus the surfaces are paraboloids of revolution, concave upward, with their minimumpoint on the axis of rotation. Some examples are sketched in Fig. 2.22.As in the previous example of linear acceleration, the position of the free surface isfound by conserving the volume of fluid. For a noncircular container with the axis ofrotation off-center, as in Fig. 2.22, a lot of laborious mensuration is required, and asingle problem will take you all weekend. However, the calculation is easy for a cylinder rotating about its central axis, as in Fig. 2.23. Since the volume of a paraboloid isStill waterlevelh2Volume =2R 2hh222h= R2gFig. 2.23 Determining the freesurface position for rotation of acylinder of fluid about its centralaxis.RR|vvThis is because f(z) vanishes when differentiated with respect to r. If you dont see this, you shouldreview your calculus.|e-Text Main Menu|Textbook Table of Contents|Study Guide2.9 Pressure Distribution in Rigid-Body Motion95one-half the base area times its height, the still-water level is exactly halfway betweenthe high and low points of the free surface. The center of the fluid drops an amount22h/2R /(4g), and the edges rise an equal amount.EXAMPLE 2.13The coffee cup in Example 2.12 is removed from the drag racer, placed on a turntable, and rotated about its central axis until a rigid-body mode occurs. Find (a) the angular velocity whichwill cause the coffee to just reach the lip of the cup and (b) the gage pressure at point A for thiscondition.SolutionPart (a)The cup contains 7 cm of coffee. The remaining distance of 3 cm up to the lip must equal thedistance h/2 in Fig. 2.23. Thus222(0.03 m)2Rh0.03 m4(9.81 m/s2)24gSolving, we obtain2Part (b)z1308or36.2 rad/s345 r/minTo compute the pressure, it is convenient to put the origin of coordinates r and z at the bottomof the free-surface depression, as shown in Fig. E2.13. The gage pressure here is p0 0, andpoint A is at (r, z) (3 cm, 4 cm). Equation (2.63) can then be evaluatedpA0(1010 kg/m3)(9.81 m/s2)( 0.04 m)12 (10103 cm396 N/m2kg/m3)(0.03 m)2(1308 rad2/s2)594 N/m2990 PaThis is about 43 percent greater than the still-water pressure pA0Ans. (a)Ans. (b)694 Pa.r7 cmHere, as in the linear-acceleration case, it should be emphasized that the paraboloidpressure distribution (2.63) sets up in any fluid under rigid-body rotation, regardlessof the shape or size of the container. The container may even be closed and filled withfluid. It is only necessary that the fluid be continuously interconnected throughout thecontainer. The following example will illustrate a peculiar case in which one can visualize an imaginary free surface extending outside the walls of the container.A3 cm3 cmE2.13EXAMPLE 2.14|vvA U-tube with a radius of 10 in and containing mercury to a height of 30 in is rotated about itscenter at 180 r/min until a rigid-body mode is achieved. The diameter of the tubing is negligible. Atmospheric pressure is 2116 lbf/ft2. Find the pressure at point A in the rotating condition.See Fig. E2.14.|e-Text Main Menu|Textbook Table of Contents|Study Guide96Chapter 2 Pressure Distribution in a FluidzSolution10 inBConvert the angular velocity to radians per second:r0(180 r/min)30 in2 rad/r60 s/min18.85 rad/sFrom Table 2.1 we find for mercury that846 lbf/ft3 and hence846/32.2 26.3 slugs/ft3.At this high rotation rate, the free surface will slant upward at a fierce angle [about 84; checkthis from Eq. (2.64)], but the tubing is so thin that the free surface will remain at approximatelythe same 30-in height, point B. Placing our origin of coordinates at this height, we can calculate the constant C in Eq. (2.62b) from the condition pB 2116 lbf/ft2 at (r, z) (10 in, 0):ApB2116 lbf/ft2orImaginaryfree surfaceCC1202116(26.3 slugs/ft3)( 10 ft)2(18.85 rad/s)2121129 lbf/ft23245We then obtain pA by evaluating Eq. (2.63) at (r, z)(0,30 in):E2.14pA1129(846 lbf/ft3)(3012ft)11292115986 lbf/ft2Ans.This is less than atmospheric pressure, and we can see why if we follow the free-surface paraboloid down from point B along the dashed line in the figure. It will cross the horizontal portion of the U-tube (where p will be atmospheric) and fall below point A. From Fig. 2.23 the actual drop from point B will be22(18.85)2( 10 )2R12h3.83 ft 46 in2(32.2)2gThus pA is about 16 inHg below atmospheric pressure, or about 16 (846) 1128 lbf/ft2 below12pa 2116 lbf/ft2, which checks with the answer above. When the tube is at rest,pA2116846(30124231 lbf/ft2)Hence rotation has reduced the pressure at point A by 77 percent. Further rotation can reducepA to near-zero pressure, and cavitation can occur.An interesting by-product of this analysis for rigid-body rotation is that the lineseverywhere parallel to the pressure gradient form a family of curved surfaces, assketched in Fig. 2.22. They are everywhere orthogonal to the constant-pressure surfaces, and hence their slope is the negative inverse of the slope computed from Eq.(2.64):dzdr GL1(dz/dr)p1constr2/gwhere GL stands for gradient linedzdr|vvor|e-Text Main Menu|gr(2.65)2Textbook Table of Contents|Study Guide2.10 Pressure Measurement97Separating the variables and integrating, we find the equation of the pressure-gradientsurfaces2rzC1 expg(2.66)Notice that this result and Eq. (2.64) are independent of the density of the fluid. In theabsence of friction and Coriolis effects, Eq. (2.66) defines the lines along which the apparent net gravitational field would act on a particle. Depending upon its density, a smallparticle or bubble would tend to rise or fall in the fluid along these exponential lines,as demonstrated experimentally in Ref. 5. Also, buoyant streamers would align themselves with these exponential lines, thus avoiding any stress other than pure tension. Figure 2.24 shows the configuration of such streamers before and during rotation.2.10 Pressure MeasurementPressure is a derived property. It is the force per unit area as related to fluid molecular bombardment of a surface. Thus most pressure instruments only infer the pressureby calibration with a primary device such as a deadweight piston tester. There are manysuch instruments, both for a static fluid and a moving stream. The instrumentation textsin Refs. 7 to 10, 12, and 13 list over 20 designs for pressure measurement instruments.These instruments may be grouped into four categories:1. Gravity-based: barometer, manometer, deadweight piston.2. Elastic deformation: bourdon tube (metal and quartz), diaphragm, bellows,strain-gage, optical beam displacement.3. Gas behavior: gas compression (McLeod gage), thermal conductance (Pirani gage),molecular impact (Knudsen gage), ionization, thermal conductivity, air piston.4. Electric output: resistance (Bridgman wire gage), diffused strain gage, capacitative, piezoelectric, magnetic inductance, magnetic reluctance, linear variable differential transformer (LVDT), resonant frequency.|vvThe gas-behavior gages are mostly special-purpose instruments used for certain scientific experiments. The deadweight tester is the instrument used most often for calibrations; for example, it is used by the U.S. National Institute for Standards and Technology (NIST). The barometer is described in Fig. 2.6.The manometer, analyzed in Sec. 2.4, is a simple and inexpensive hydrostaticprinciple device with no moving parts except the liquid column itself. Manometer measurements must not disturb the flow. The best way to do this is to take the measurement through a static hole in the wall of the flow, as illustrated for the two instrumentsin Fig. 2.25. The hole should be normal to the wall, and burrs should be avoided. Ifthe hole is small enough (typically 1-mm diameter), there will be no flow into the measuring tube once the pressure has adjusted to a steady value. Thus the flow is almostundisturbed. An oscillating flow pressure, however, can cause a large error due to possible dynamic response of the tubing. Other devices of smaller dimensions are used fordynamic-pressure measurements. Note that the manometers in Fig. 2.25 are arrangedto measure the absolute pressures p1 and p2. If the pressure difference p1 p2 is de-|e-Text Main Menu|Textbook Table of Contents|Study Guide98Chapter 2 Pressure Distribution in a Fluid|vvFig. 2.24 Experimental demonstration with buoyant streamers of thefluid force field in rigid-body rotation: (top) fluid at rest (streamershang vertically upward); (bottom)rigid-body rotation (streamers arealigned with the direction of maximum pressure gradient). (From Ref.5, courtesy of R. Ian Fletcher.)|e-Text Main Menu|Textbook Table of Contents|Study Guide2.10 Pressure Measurement99FlowFlowp1p2Fig. 2.25 Two types of accuratemanometers for precise measurements: (a) tilted tube with eyepiece; (b) micrometer pointer withammeter detector.(a)(b)sired, a significant error is incurred by subtracting two independent measurements, andit would be far better to connect both ends of one instrument to the two static holes p1and p2 so that one manometer reads the difference directly. In category 2, elasticdeformation instruments, a popular, inexpensive, and reliable device is the bourdontube, sketched in Fig. 2.26. When pressurized internally, a curved tube with flattenedcross section will deflect outward. The deflection can be measured by a linkage attached to a calibrated dial pointer, as shown. Or the deflection can be used to driveelectric-output sensors, such as a variable transformer. Similarly, a membrane or diaphragm will deflect under pressure and can either be sensed directly or used to driveanother sensor.ASection AABourdontubeAPointer fordial gageFlattened tube deflectsoutward under pressureLinkage|vvFig. 2.26 Schematic of a bourdontube device for mechanical measurement of high pressures.|e-Text Main MenuHigh pressure|Textbook Table of Contents|Study Guide100Chapter 2 Pressure Distribution in a FluidFig. 2.27 The fused-quartz, forcebalanced bourdon tube is the mostaccurate pressure sensor used incommercial applications today.(Courtesy of Ruska InstrumentCorporation, Houston, TX.)An interesting variation of Fig. 2.26 is the fused-quartz, forced-balanced bourdontube, shown in Fig. 2.27, whose deflection is sensed optically and returned to a zeroreference state by a magnetic element whose output is proportional to the fluid pressure. The fused-quartz, forced-balanced bourdon tube is reported to be one of the mostaccurate pressure sensors ever devised, with uncertainty of the order of 0.003 percent.The last category, electric-output sensors, is extremely important in engineeringbecause the data can be stored on computers and freely manipulated, plotted, and analyzed. Three examples are shown in Fig. 2.28, the first being the capacitive sensorin Fig. 2.28a. The differential pressure deflects the silicon diaphragm and changesthe capacitancce of the liquid in the cavity. Note that the cavity has spherical endcaps to prevent overpressure damage. In the second type, Fig. 2.28b, strain gages andother sensors are diffused or etched onto a chip which is stressed by the applied pressure. Finally, in Fig. 2.28c, a micromachined silicon sensor is arranged to deformunder pressure such that its natural vibration frequency is proportional to the pressure. An oscillator excites the elements resonant frequency and converts it into appropriate pressure units. For further information on pressure sensors, see Refs. 7 to10, 12, and 13.Summary|vvThis chapter has been devoted entirely to the computation of pressure distributions andthe resulting forces and moments in a static fluid or a fluid with a known velocity field.All hydrostatic (Secs. 2.3 to 2.8) and rigid-body (Sec. 2.9) problems are solved in thismanner and are classic cases which every student should understand. In arbitrary viscous flows, both pressure and velocity are unknowns and are solved together as a system of equations in the chapters which follow.|e-Text Main Menu|Textbook Table of Contents|Study GuideCover flangeSeal diaphragmHigh-pressure sideLow-pressure sideFilling liquidSensing diaphragm(a)Wire bondingStitch bondedconnections fromchip to body plugStrain gagesDiffused into integratedsilicon chipFig. 2.28 Pressure sensors withelectric output: (a) a silicon diaphragm whose deflection changesthe cavity capacitance (Courtesy ofJohnson-Yokogawa Inc.); (b) a silicon strain gage which is stressedby applied pressure; (c) a micromachined silicon element which resonates at a frequency proportionalto applied pressure. [(b) and (c)are courtesy of Druck, Inc., Fairfield, CT.]Etched cavityMicromachinedsilicon sensor|vv(Druck, Inc., Fairfield, Connecticut)(b)|Temperature sensorOn-chip diode foroptimum temperatureperformancee-Text Main Menu|Textbook Table of Contents(c)|Study Guide101102Chapter 2 Pressure Distribution in a FluidP2.2 For the two-dimensional stress field shown in Fig. P2.1suppose thatProblemsMost of the problems herein are fairly straightforward. Moredifficult or open-ended assignments are indicated with an asterisk, as in Prob. 2.8. Problems labeled with an EES icon (forexample, Prob. 2.62), will benefit from the use of the Engineering Equation Solver (EES), while problems labeled with adisk icon may require the use of a computer. The standard endof-chapter problems 2.1 to 2.158 (categorized in the problemlist below) are followed by word problems W2.1 to W2.8, fundamentals of engineering exam problems FE2.1 to FE2.10, comprehensive problems C2.1 to C2.4, and design projects D2.1 andD2.2.xxP2.3P2.4Problem DistributionSectionTopicProblems2.1, 18.104.22.168.22.214.171.124.126.96.36.199.10Stresses; pressure gradient; gage pressureHydrostatic pressure; barometersThe atmosphereManometers; multiple fluidsForces on plane surfacesForces on curved surfacesForces in layered fluidsBuoyancy; Archimedes principlesStability of floating bodiesUniform accelerationRigid-body rotationPressure measurements2.1 2.62.7 2.232.24 2.292.30 2.472.48 2.812.82 2.1002.101 2.1022.103 2.1262.127 2.1362.137 2.1512.152 2.158NoneP2.6P2.1 For the two-dimensional stress field shown in Fig. P2.1 itis found thatP2.8xx3000 lbf/ft22000 lbf/ft2yyxyP2.5P2.7500 lbf/ft2Find the shear and normal stresses (in lbf/ft2) acting onplane AA cutting through the element at a 30 angle asshown.yyyx=xy*P2.9AxxA30xxP2.10xy=yxyy|vvP2.1|e-Text Main Menu|2000 lbf/ft2yy3000 lbf/ft2n(AA)2500 lbf/ft2Compute (a) the shear stress xy and (b) the shear stresson plane AA.Derive Eq. (2.18) by using the differential element in Fig.2.2 with z up, no fluid motion, and pressure varying onlyin the z direction.In a certain two-dimensional fluid flow pattern the linesof constant pressure, or isobars, are defined by the expression P0 Bz Cx2 constant, where B and C areconstants and p0 is the (constant) pressure at the origin,(x, z) (0, 0). Find an expression x f (z) for the familyof lines which are everywhere parallel to the local pressure gradient Vp.Atlanta, Georgia, has an average altitude of 1100 ft. On astandard day (Table A.6), pressure gage A in a laboratoryexperiment reads 93 kPa and gage B reads 105 kPa. Express these readings in gage pressure or vacuum pressure(Pa), whichever is appropriate.Any pressure reading can be expressed as a length or head,h p/ g. What is standard sea-level pressure expressed in(a) ft of ethylene glycol, (b) in Hg, (c) m of water, and (d)mm of methanol? Assume all fluids are at 20C.The deepest known point in the ocean is 11,034 m in theMariana Trench in the Pacific. At this depth the specificweight of seawater is approximately 10,520 N/m3. At thesurface,10,050 N/m3. Estimate the absolute pressureat this depth, in atm.Dry adiabatic lapse rate (DALR) is defined as the negative value of atmospheric temperature gradient, dT/dz,when temperature and pressure vary in an isentropic fashion. Assuming air is an ideal gas, DALRdT/dz whenT T0( p/p0)a, where exponent a (k 1)/k, k cp /cv isthe ratio of specific heats, and T0 and p0 are the temperature and pressure at sea level, respectively. (a) Assumingthat hydrostatic conditions exist in the atmosphere, showthat the dry adiabatic lapse rate is constant and is given byDALR g(k 1)/(kR), where R is the ideal gas constantfor air. (b) Calculate the numerical value of DALR for airin units of C/km.For a liquid, integrate the hydrostatic relation, Eq. (2.18),by assuming that the isentropic bulk modulus, B( p/ )s, is constantsee Eq. (9.18). Find an expressionfor p(z) and apply the Mariana Trench data as in Prob. 2.7,using Bseawater from Table A.3.A closed tank contains 1.5 m of SAE 30 oil, 1 m of water, 20 cm of mercury, and an air space on top, all at 20C.The absolute pressure at the bottom of the tank is 60 kPa.What is the pressure in the air space?Textbook Table of Contents|Study GuideProblems 103P2.11 In Fig. P2.11, pressure gage A reads 1.5 kPa (gage). Thefluids are at 20C. Determine the elevations z, in meters,of the liquid levels in the open piezometer tubes B and C.Gasoline1.5 mWaterABCh1m2mAir1.5 mGasoline1mGlycerinLiquid, SG = 1.60P2.13ABAirP2.11z= 04mP2.12 In Fig. P2.12 the tank contains water and immiscible oilat 20C. What is h in cm if the density of the oil is 898kg/m3?2mAir4mWater2mP2.1415 lbf/in2 absh6 cmAAir12 cm8 cm2 ftOil1 ftWaterBOilP2.121 ft|vvP2.13 In Fig. P2.13 the 20C water and gasoline surfaces areopen to the atmosphere and at the same elevation. What isthe height h of the third liquid in the right leg?P2.14 The closed tank in Fig. P2.14 is at 20C. If the pressureat point A is 95 kPa absolute, what is the absolute pressure at point B in kPa? What percent error do you makeby neglecting the specific weight of the air?P2.15 The air-oil-water system in Fig. P2.15 is at 20C. Knowing that gage A reads 15 lbf/in2 absolute and gage B reads1.25 lbf/in2 less than gage C, compute (a) the specificweight of the oil in lbf/ft3 and (b) the actual reading ofgage C in lbf/in2 absolute.|e-Text Main Menu|Water2 ftCP2.15P2.16 A closed inverted cone, 100 cm high with diameter 60 cmat the top, is filled with air at 20C and 1 atm. Water at20C is introduced at the bottom (the vertex) to compressthe air isothermally until a gage at the top of the cone reads30 kPa (gage). Estimate (a) the amount of water needed(cm3) and (b) the resulting absolute pressure at the bottomof the cone (kPa).Textbook Table of Contents|Study Guide104Chapter 2 Pressure Distribution in a FluidP2.17 The system in Fig. P2.17 is at 20C. If the pressure at pointA is 1900 lbf/ft2, determine the pressures at points B, C,and D in lbf/ft2.MercuryAirAir3 ftBAC2 ft10 cm10 cmAir4 ft10 cmP2.195 ftWaterD2 ft2000lbfP2.173-in diameterP2.18 The system in Fig. P2.18 is at 20C. If atmospheric pressure is 101.33 kPa and the pressure at the bottom of thetank is 242 kPa, what is the specific gravity of fluid X?1 in15 inF1-in diameterOil1mSAE 30 oilP2.20Water2mAir: 180 kPa absFluid XMercury0.5 m80 cmP2.18P2.21vvP2.19 The U-tube in Fig. P2.19 has a 1-cm ID and contains mercury as shown. If 20 cm3 of water is poured into the righthand leg, what will the free-surface height in each leg beafter the sloshing has died down?P2.20 The hydraulic jack in Fig. P2.20 is filled with oil at 56lbf/ft3. Neglecting the weight of the two pistons, what forceF on the handle is required to support the 2000-lbf weightfor this design?P2.21 At 20C gage A reads 350 kPa absolute. What is the heighth of the water in cm? What should gage B read in kPa absolute? See Fig. P2.21.||Waterh?3me-Text Main Menu|MercuryABP2.22 The fuel gage for a gasoline tank in a car reads proportional to the bottom gage pressure as in Fig. P2.22. If thetank is 30 cm deep and accidentally contains 2 cm of water plus gasoline, how many centimeters of air remain atthe top when the gage erroneously reads full?P2.23 In Fig. P2.23 both fluids are at 20C. If surface tension effects are negligible, what is the density of the oil, in kg/m3?P2.24 In Prob. 1.2 we made a crude integration of the densitydistribution (z) in Table A.6 and estimated the mass ofthe earths atmosphere to be m 6 E18 kg. Can this re-Textbook Table of Contents|Study GuideProblems 105VentAirh?GasolineSG = 0.6830 cmWatermental observations. (b) Find an expression for the pressure at points 1 and 2 in Fig. P2.27b. Note that the glassis now inverted, so the original top rim of the glass is atthe bottom of the picture, and the original bottom of theglass is at the top of the picture. The weight of the cardcan be neglected.2 cmCardpgageP2.22Top of glassOil8 cm6 cmP2.27aWaterBottom of glassOriginal bottom of glass10 cm1GP2.23|vv2Gsult be used to estimate sea-level pressure on the earth?Conversely, can the actual sea-level pressure of 101.35 kPabe used to make a more accurate estimate of the atmosP2.27bCardOriginal top of glasspheric mass?P2.25 Venus has a mass of 4.90 E24 kg and a radius of 6050 km.Its atmosphere is 96 percent CO2, but let us assume it to(c) Estimate the theoretical maximum glass height suchbe 100 percent. Its surface temperature averages 730 K,that this experiment could still work, i.e., such that the wadecreasing to 250 K at an altitude of 70 km. The averageter would not fall out of the glass.surface pressure is 9.1 MPa. Estimate the atmospheric P2.28 Earths atmospheric conditions vary somewhat. On a cerpressure of Venus at an altitude of 5 km.tain day the sea-level temperature is 45F and the sea-levelP2.26 Investigate the effect of doubling the lapse rate on atmospressure is 28.9 inHg. An airplane overhead registers anpheric pressure. Compare the standard atmosphere (Tableair temperature of 23F and a pressure of 12 lbf/in2. EstiA.6) with a lapse rate twice as high, B2 0.0130 K/m.mate the planes altitude, in feet.Find the altitude at which the pressure deviation is (a) 1 *P2.29 Under some conditions the atmosphere is adiabatic, ppercent and (b) 5 percent. What do you conclude?(const)( k), where k is the specific heat ratio. Show that,P2.27 Conduct an experiment to illustrate atmospheric pressure.for an adiabatic atmosphere, the pressure variation isNote: Do this over a sink or you may get wet! Find a drinkgiven bying glass with a very smooth, uniform rim at the top. Fillthe glass nearly full with water. Place a smooth, light, flat(k 1)gz k/(k 1)p p0 1plate on top of the glass such that the entire rim of thekRT0glass is covered. A glossy postcard works best. A small inCompare this formula for air at z 5000 m with the standex card or one flap of a greeting card will also work. Seedard atmosphere in Table A.6.Fig. P2.27a.(a) Hold the card against the rim of the glass and turn the P2.30 In Fig. P2.30 fluid 1 is oil (SG 0.87) and fluid 2 is glycerin at 20C. If pa 98 kPa, determine the absolute presglass upside down. Slowly release pressure on the card.sure at point A.Does the water fall out of the glass? Record your experi-|e-Text Main Menu|Textbook Table of Contents|Study Guide106Chapter 2 Pressure Distribution in a FluidpaAir B1SAE 30 oil32 cmALiquid, SG = 1.453 cm5 cm10 cm24 cmAWaterP2.306 cm8 cm3 cmP2.31 In Fig. P2.31 all fluids are at 20C. Determine the pressure difference (Pa) between points A and B.P2.33*P2.34 Sometimes manometer dimensions have a significant ef-KerosineAirBenzeneB40 cmA9 cm20 cmfect. In Fig. P2.34 containers (a) and (b) are cylindrical andconditions are such that pa pb. Derive a formula for thepressure difference pa pb when the oil-water interface onthe right rises a distance h h, for (a) d D and (b) d0.15D. What is the percent change in the value of p?14 cm8 cmMercuryWaterDDP2.31(b)( a)P2.32 For the inverted manometer of Fig. P2.32, all fluids are at20C. If pB pA 97 kPa, what must the height H bein cm?Meriamred oil,SG = 0.827LSAE 30 oilHWaterh18 cmWaterdHMercuryAP2.3435 cmBP2.32|vvP2.33 In Fig. P2.33 the pressure at point A is 25 lbf/in2. All fluids are at 20C. What is the air pressure in the closed chamber B, in Pa?|e-Text Main Menu|P2.35 Water flows upward in a pipe slanted at 30, as in Fig.P2.35. The mercury manometer reads h 12 cm. Both fluids are at 20C. What is the pressure difference p1 p2 inthe pipe?P2.36 In Fig. P2.36 both the tank and the tube are open to theatmosphere. If L 2.13 m, what is the angle of tilt ofthe tube?P2.37 The inclined manometer in Fig. P2.37 contains Meriamred manometer oil, SG 0.827. Assume that the reservoirTextbook Table of Contents|Study GuideProblems 107with manometer fluid m. One side of the manometer is opento the air, while the other is connected to new tubing whichextends to pressure measurement location 1, some height Hhigher in elevation than the surface of the manometer liquid.For consistency, let a be the density of the air in the room,t be the density of the gas inside the tube, m be the density of the manometer liquid, and h be the height differencebetween the two sides of the manometer. See Fig. P2.38.(a) Find an expression for the gage pressure at the measurement point. Note: When calculating gage pressure, usethe local atmospheric pressure at the elevation of the measurement point. You may assume that h H; i.e., assumethe gas in the entire left side of the manometer is of density t. (b) Write an expression for the error caused by assuming that the gas inside the tubing has the same densityas that of the surrounding air. (c) How much error (in Pa)is caused by ignoring this density difference for the following conditions: m 860 kg/m3, a 1.20 kg/m3,1.50 kg/m3, H 1.32 m, and h 0.58 cm? (d) Cantyou think of a simple way to avoid this error?(2)30(1)h2mP2.35OilSG = 0.850 cmLWaterSG = 1.050 cmP2.361is very large. If the inclined arm is fitted with graduations1 in apart, what should the angle be if each graduationcorresponds to 1 lbf/ft2 gage pressure for pA?(tubing gas)p1pa at location 1ta(air)H1 inpAD=516U-tubemanometerinhmP2.38ReservoirP2.37|vvP2.38 An interesting article appeared in the AIAA Journal (vol. 30,no. 1, January 1992, pp. 279 280). The authors explain thatthe air inside fresh plastic tubing can be up to 25 percentmore dense than that of the surroundings, due to outgassingor other contaminants introduced at the time of manufacture.Most researchers, however, assume that the tubing is filledwith room air at standard air density, which can lead to significant errors when using this kind of tubing to measurepressures. To illustrate this, consider a U-tube manometer|e-Text Main Menu|P2.39 An 8-cm-diameter piston compresses manometer oil intoan inclined 7-mm-diameter tube, as shown in Fig. P2.39.When a weight W is added to the top of the piston, the oilrises an additional distance of 10 cm up the tube, as shown.How large is the weight, in N?P2.40 A pump slowly introduces mercury into the bottom of theclosed tank in Fig. P2.40. At the instant shown, the airpressure pB 80 kPa. The pump stops when the air pressure rises to 110 kPa. All fluids remain at 20C. What willbe the manometer reading h at that time, in cm, if it is connected to standard sea-level ambient air patm?Textbook Table of Contents|Study Guide108Chapter 2 Pressure Distribution in a FluidpAD = 8 cmPistonpB1W10 cm1h1h1d = 7 mmMeriam redoil, SG = 0.827P2.39h152P2.42patm8 cmAir: pB9 cmWaterhP2.44 Water flows downward in a pipe at 45, as shown in Fig.P2.44. The pressure drop p1 p2 is partly due to gravityand partly due to friction. The mercury manometer readsa 6-in height difference. What is the total pressure dropp1 p2 in lbf/in2? What is the pressure drop due to friction only between 1 and 2 in lbf/in2? Does the manometer reading correspond only to friction drop? Why?PumpMercury10 cmHg2 cmP2.40P2.41 The system in Fig. P2.41 is at 20C. Compute the pressure at point A in lbf/ft2 absolute.4515 ftWaterFlow2Oil, SG = 0.855 inApa = 14.7Waterlbf/in210 in6 in6 inWaterMercuryP2.44P2.41Mercury|vvP2.42 Very small pressure differences pA pB can be measuredaccurately by the two-fluid differential manometer in Fig.P2.42. Density 2 is only slightly larger than that of theupper fluid 1. Derive an expression for the proportionality between h and pA pB if the reservoirs are very large.*P2.43 A mercury manometer, similar to Fig. P2.35, records h1.2, 4.9, and 11.0 mm when the water velocities in the pipeare V 1.0, 2.0, and 3.0 m/s, respectively. Determine ifthese data can be correlated in the form p1 p2 Cf V2,where Cf is dimensionless.|e-Text Main Menu|P2.45 In Fig. P2.45, determine the gage pressure at point A inPa. Is it higher or lower than atmospheric?P2.46 In Fig. P2.46 both ends of the manometer are open to theatmosphere. Estimate the specific gravity of fluid X.P2.47 The cylindrical tank in Fig. P2.47 is being filled with water at 20C by a pump developing an exit pressure of 175EESkPa. At the instant shown, the air pressure is 110 kPa andH 35 cm. The pump stops when it can no longer raisethe water pressure. For isothermal air compression, estimate H at that time.P2.48 Conduct the following experiment to illustrate air pressure. Find a thin wooden ruler (approximately 1 ft inTextbook Table of Contents|Study GuideProblems 109patm50 cmAirAir20 COil,SG = 0.8575 cm30 cm45 cm40 cmHWaterPumpP2.4715 cmANewspaperWaterP2.45MercuryRulerSAE 30 oilDesk9 cm10 cmP2.48Water5 cm7 cm6 cmFluid XP2.494 cmP2.5012 cmP2.46|vvlength) or a thin wooden paint stirrer. Place it on the edgeof a desk or table with a little less than half of it hang- P2.51ing over the edge lengthwise. Get two full-size sheets ofnewspaper; open them up and place them on top of theruler, covering only the portion of the ruler resting on the *P2.52desk as illustrated in Fig. P2.48. (a) Estimate the totalforce on top of the newspaper due to air pressure in theroom. (b) Careful! To avoid potential injury, make surenobody is standing directly in front of the desk. Perform|e-Text Main Menu|a karate chop on the portion of the ruler sticking out overthe edge of the desk. Record your results. (c) Explainyour results.A water tank has a circular panel in its vertical wall. Thepanel has a radius of 50 cm, and its center is 2 m belowthe surface. Neglecting atmospheric pressure, determinethe water force on the panel and its line of action.A vat filled with oil (SG 0.85) is 7 m long and 3 m deepand has a trapezoidal cross section 2 m wide at the bottom and 4 m wide at the top. Compute (a) the weight ofoil in the vat, (b) the force on the vat bottom, and (c) theforce on the trapezoidal end panel.Gate AB in Fig. P2.51 is 1.2 m long and 0.8 m into thepaper. Neglecting atmospheric pressure, compute the forceF on the gate and its center-of-pressure position X.Suppose that the tank in Fig. P2.51 is filled with liquid X,not oil. Gate AB is 0.8 m wide into the paper. Suppose thatliquid X causes a force F on gate AB and that the momentof this force about point B is 26,500 N m. What is thespecific gravity of liquid X?Textbook Table of Contents|Study Guide110Chapter 2 Pressure Distribution in a Fluidpa6mOil,SG = 0.82Waterpa4mh8mA1mX1.2 mAB4 ftF40P2.51BP2.55P2.53 Panel ABC in the slanted side of a water tank is an isosceles triangle with the vertex at A and the base BC 2 m,as in Fig. P2.53. Find the water force on the panel and itsline of action.200 kghmBA30 cmAWaterWater3mP2.584mP2.53B, C*P2.59 Gate AB has length L, width b into the paper, is hinged atB, and has negligible weight. The liquid level h remainsat the top of the gate for any angle . Find an analytic expression for the force P, perpendicular to AB, required tokeep the gate in equilibrium in Fig. P2.59.3m|vvP2.54 If, instead of water, the tank in Fig. P2.53 is filled with liqPuid X, the liquid force on panel ABC is found to be 115 kN.What is the density of liquid X? The line of action is foundAto be the same as in Prob. 2.53. Why?P2.55 Gate AB in Fig. P2.55 is 5 ft wide into the paper, hingedat A, and restrained by a stop at B. The water is at 20C.Compute (a) the force on stop B and (b) the reactions athLA if the water depth h 9.5 ft.P2.56 In Fig. P2.55, gate AB is 5 ft wide into the paper, and stopB will break if the water force on it equals 9200 lbf. ForHingewhat water depth h is this condition reached?P2.57 In Fig. P2.55, gate AB is 5 ft wide into the paper. SupposeBP2.59that the fluid is liquid X, not water. Hinge A breaks whenits reaction is 7800 lbf, and the liquid depth is h 13 ft.*P2.60 Find the net hydrostatic force per unit width on the recWhat is the specific gravity of liquid X?tangular gate AB in Fig. P2.60 and its line of action.P2.58 In Fig. P2.58, the cover gate AB closes a circular opening80 cm in diameter. The gate is held closed by a 200-kg *P2.61 Gate AB in Fig. P2.61 is a homogeneous mass of 180 kg,1.2 m wide into the paper, hinged at A, and resting on amass as shown. Assume standard gravity at 20C. At whatsmooth bottom at B. All fluids are at 20C. For what wawater level h will the gate be dislodged? Neglect the weightter depth h will the force at point B be zero?of the gate.|e-Text Main Menu|Textbook Table of Contents|Study GuideProblems 111P2.63 The tank in Fig. P2.63 has a 4-cm-diameter plug at thebottom on the right. All fluids are at 20C. The plug willpop out if the hydrostatic force on it is 25 N. For this condition, what will be the reading h on the mercury manometer on the left side?1.8 m1.2 mAWater2mWaterGlycerin50B2mHP2.60h2 cmPlug,D = 4 cmMercuryWaterP2.63Glycerin*P2.64 Gate ABC in Fig. P2.64 has a fixed hinge line at B and is2 m wide into the paper. The gate will open at A to releasewater if the water depth is high enough. Compute the depthh for which the gate will begin to open.h2mA1mC60BP2.61P2.62 Gate AB in Fig. P2.62 is 15 ft long and 8 ft wide into thepaper and is hinged at B with a stop at A. The water is atEES20C. The gate is 1-in-thick steel, SG 7.85. Computethe water level h for which the gate will start to fall.PulleyAhBvvP2.62||Bh1mWater at 20C*P2.65 Gate AB in Fig. P2.65 is semicircular, hinged at B, andheld by a horizontal force P at A. What force P is requiredfor equilibrium?P2.66 Dam ABC in Fig. P2.66 is 30 m wide into the paper andmade of concrete (SG 2.4). Find the hydrostatic forceon surface AB and its moment about C. Assuming no seepage of water under the dam, could this force tip the damover? How does your argument change if there is seepageunder the dam?Water6020 cmP2.6410,000 lb15 ftAe-Text Main Menu|Textbook Table of Contents|Study Guide112Chapter 2 Pressure Distribution in a Fluid5m3mOil, SG = 0.831mWaterAAPGate:Side view3mGate2mBP2.65;;;;50 BP2.68AWater 20C80 m80 cm1mDamB5mWater,20CCHg, 20C60 mP2.66A2mP2.69|vv*P2.67 Generalize Prob. 2.66 as follows. Denote length AB as H,length BC as L, and angle ABC as . Let the dam material have specific gravity SG. The width of the dam is b.Assume no seepage of water under the dam. Find an analytic relation between SG and the critical angle c forwhich the dam will just tip over to the right. Use your relation to compute c for the special case SG 2.4 (concrete).P2.68 Isosceles triangle gate AB in Fig. P2.68 is hinged at A andweighs 1500 N. What horizontal force P is required at pointB for equilibrium?*P2.69 The water tank in Fig. P2.69 is pressurized, as shown bythe mercury-manometer reading. Determine the hydrostatic force per unit depth on gate AB.P2.70 Calculate the force and center of pressure on one side ofthe vertical triangular panel ABC in Fig. P2.70. Neglectpatm.*P2.71 In Fig. P2.71 gate AB is 3 m wide into the paper and isconnected by a rod and pulley to a concrete sphere (SG|e-Text Main Menu|B2 ftAWater6 ftCB4 ftP2.70Textbook Table of Contents|Study GuidePProblems 1132.40). What diameter of the sphere is just sufficient to keep *P2.74 In soft liquids (low bulk modulus ), it may be necesthe gate closed?sary to account for liquid compressibility in hydrostaticcalculations. An approximate density relation would be;;Concretesphere, SG = 2.4dpda2 dorpp0a2(0)6mwhere a is the speed of sound and (p0, 0) are the conditions at the liquid surface z 0. Use this approximationto show that the density variation with depth in a soft liqgz/a2uid iswhere g is the acceleration of gravity0e8mAand z is positive upward. Then consider a vertical wall ofwidth b, extending from the surface (z 0) down to depthzh. Find an analytic expression for the hydrostatic4mWaterforce F on this wall, and compare it with the incompress2Bible result F0gh b/2. Would the center of pressure bebelow the incompressible position z2h/3?P2.71*P2.75 Gate AB in Fig. P2.75 is hinged at A, has width b into thepaper, and makes smooth contact at B. The gate has density s and uniform thickness t. For what gate density s,P2.72 The V-shaped container in Fig. P2.72 is hinged at A andexpressed as a function of (h, t, , ), will the gate just beheld together by cable BC at the top. If cable spacing isgin to lift off the bottom? Why is your answer indepen1 m into the paper, what is the cable tension?dent of gate length L and width b?CableCBA1mWater3mLh110P2.72AtP2.73 Gate AB is 5 ft wide into the paper and opens to let freshwater out when the ocean tide is dropping. The hinge at Ais 2 ft above the freshwater level. At what ocean level hwill the gate first open? Neglect the gate weight.ATiderange10 fthSeawater, SG = 1.025StopB|vvP2.73|e-Text Main Menu|P2.75B*P2.76 Consider the angled gate ABC in Fig. P2.76, hinged at Cand of width b into the paper. Derive an analytic formulafor the horizontal force P required at the top for equilibrium, as a function of the angle .P2.77 The circular gate ABC in Fig. P2.77 has a 1-m radius andis hinged at B. Compute the force P just sufficient to keepthe gate from opening when h 8 m. Neglect atmosphericpressure.P2.78 Repeat Prob. 2.77 to derive an analytic expression for Pas a function of h. Is there anything unusual about yoursolution?P2.79 Gate ABC in Fig. P2.79 is 1 m square and is hinged at B.It will open automatically when the water level h becomeshigh enough. Determine the lowest height for which theTextbook Table of Contents|Study Guide114;;;;;;Chapter 2 Pressure Distribution in a FluidAPSpecific weight hAir1 atm2mSAE3B200oilh60WaterpaWaterpahcmMercury80CP2.76cmcmP2.80Panel, 30 cm high, 40 cm wideP2.81 Gate AB in Fig. P2.81 is 7 ft into the paper and weighs3000 lbf when submerged. It is hinged at B and restsagainst a smooth wall at A. Determine the water level h atthe left which will just cause the gate to open.A1mB1mChP4 ftAP2.77Water8 ftWaterhBWater6 ftA60 cmC40 cmP2.81B*P2.82 The dam in Fig. P2.82 is a quarter circle 50 m wide intoP2.79|vvgate will open. Neglect atmospheric pressure. Is this resultindependent of the liquid density?P2.80 For the closed tank in Fig. P2.80, all fluids are at 20C, andthe airspace is pressurized. It is found that the net outwardhydrostatic force on the 30-by 40-cm panel at the bottom ofthe water layer is 8450 N. Estimate (a) the pressure in theairspace and (b) the reading h on the mercury manometer.|e-Text Main Menu|the paper. Determine the horizontal and vertical components of the hydrostatic force against the dam and the pointCP where the resultant strikes the dam.*P2.83 Gate AB in Fig. P2.83 is a quarter circle 10 ft wide intothe paper and hinged at B. Find the force F just sufficientto keep the gate from opening. The gate is uniform andweighs 3000 lbf.P2.84 Determine (a) the total hydrostatic force on the curved surface AB in Fig. P2.84 and (b) its line of action. Neglect atmospheric pressure, and let the surface have unit width.Textbook Table of Contents|Study Guide;;;;;;;;;20 m20 mProblems 115pa = 0CPWaterP2.82Water10 ftP2.862 ftP2.87 The bottle of champagne (SG 0.96) in Fig. P2.87 is under pressure, as shown by the mercury-manometer reading. Compute the net force on the 2-in-radius hemispherical end cap at the bottom of the bottle.FAWaterr = 8 ftBP2.83BWater at 20 Cz1m4 inz = x32 in6 inxAP2.84P2.87r = 2 inMercuryP2.85 Compute the horizontal and vertical components of the hydrostatic force on the quarter-circle panel at the bottom of *P2.88 Gate ABC is a circular arc, sometimes called a Tainter gate,which can be raised and lowered by pivoting about pointthe water tank in Fig. P2.85.O. See Fig. P2.88. For the position shown, determine (a)the hydrostatic force of the water on the gate and (b) itsline of action. Does the force pass through point O?6mC5mWaterR=6mWater2mB6mO2mP2.856m|vvP2.86 Compute the horizontal and vertical components of the hydrostatic force on the hemispherical bulge at the bottomof the tank in Fig. P2.86.|e-Text Main Menu|AP2.88Textbook Table of Contents|Study Guide116Chapter 2 Pressure Distribution in a FluidP2.89 The tank in Fig. P2.89 contains benzene and is pressurized to 200 kPa (gage) in the air gap. Determine the vertical hydrostatic force on circular-arc section AB and itsline of action.3cm4m60 cmp = 200 kPa30 cmSixboltsB2mWaterBenzeneat 20 C60 cmP2.91P2.89A2mP2.90 A 1-ft-diameter hole in the bottom of the tank in Fig. P2.90is closed by a conical 45 plug. Neglecting the weight ofthe plug, compute the force F required to keep the plug inthe hole.Waterp = 3 lbf/in 2 gageBolt spacing 25 cm2mP2.921 ftAir :zWater3 ft1 ft, 45conehRFP2.90|vvP2.91 The hemispherical dome in Fig. P2.91 weighs 30 kN andis filled with water and attached to the floor by six equallyspaced bolts. What is the force in each bolt required tohold down the dome?P2.92 A 4-m-diameter water tank consists of two half cylinders,each weighing 4.5 kN/m, bolted together as shown in Fig.P2.92. If the support of the end caps is neglected, determine the force induced in each bolt.*P2.93 In Fig. P2.93, a one-quadrant spherical shell of radius Ris submerged in liquid of specific gravity and depthh R. Find an analytic expression for the resultant hydrostatic force, and its line of action, on the shell surface.|e-Text Main Menu|RzRxP2.93P2.94 The 4-ft-diameter log (SG 0.80) in Fig. P2.94 is 8 ftlong into the paper and dams water as shown. Computethe net vertical and horizontal reactions at point C.Textbook Table of Contents|Study GuideProblems 117wall at A. Compute the reaction forces at points Aand B.Log2ftWater2ftP2.94WaterCSeawater, 10,050 N/m3*P2.95 The uniform body A in Fig. P2.95 has width b into the paper and is in static equilibrium when pivoted about hingeO. What is the specific gravity of this body if (a) h 0and (b) h R?4mA2m45BAP2.97hP2.98 Gate ABC in Fig. P2.98 is a quarter circle 8 ft wide intothe paper. Compute the horizontal and vertical hydrostaticforces on the gate and the line of action of the resultantforce.RRWaterAOP2.95r = 4 ftP2.96 The tank in Fig. P2.96 is 3 m wide into the paper. Neglecting atmospheric pressure, compute the hydrostatic (a)horizontal force, (b) vertical force, and (c) resultant forceon quarter-circle panel BC.BCP2.98P2.99 A 2-ft-diameter sphere weighing 400 lbf closes a 1-ft-diameter hole in the bottom of the tank in Fig. P2.99. Compute the force F required to dislodge the sphere from thehole.A6mWaterWater45454mBWater4m3 ft1 ftP2.96C|vvP2.97 Gate AB in Fig. P2.97 is a three-eighths circle, 3 m wideinto the paper, hinged at B, and resting against a smooth|e-Text Main Menu|1 ftP2.99Textbook Table of ContentsF|Study Guide118Chapter 2 Pressure Distribution in a FluidP2.100 Pressurized water fills the tank in Fig. P2.100. Computethe net hydrostatic force on the conical surface ABC.2mAP2.106C4m7mBP2.107150 kPagageP2.108WaterP2.100P2.101 A fuel truck has a tank cross section which is approximately elliptical, with a 3-m horizontal major axis and a2-m vertical minor axis. The top is vented to the atmosphere. If the tank is filled half with water and half withgasoline, what is the hydrostatic force on the flat elliptical end panel?P2.102 In Fig. P2.80 suppose that the manometer reading is h25 cm. What will be the net hydrostatic force on the complete end wall, which is 160 cm high and 2 m wide?P2.103 The hydrogen bubbles in Fig. 1.13 are very small, lessthan a millimeter in diameter, and rise slowly. Their dragin still fluid is approximated by the first term of Stokesexpression in Prob. 1.10: F 3 VD, where V is the risevelocity. Neglecting bubble weight and setting bubblebuoyancy equal to drag, (a) derive a formula for the terminal (zero acceleration) rise velocity Vterm of the bubbleand (b) determine Vterm in m/s for water at 20C if D30 m.P2.104 The can in Fig. P2.104 floats in the position shown. Whatis its weight in N?P2.109whether his new crown was pure gold (SG 19.3).Archimedes measured the weight of the crown in air to be11.8 N and its weight in water to be 10.9 N. Was it puregold?It is found that a 10-cm cube of aluminum (SG 2.71)will remain neutral under water (neither rise nor fall) if itis tied by a string to a submerged 18-cm-diameter sphereof buoyant foam. What is the specific weight of the foam,in N/m3?Repeat Prob. 2.62, assuming that the 10,000-lbf weight isaluminum (SG 2.71) and is hanging submerged in thewater.A piece of yellow pine wood (SG 0.65) is 5 cm squareand 2.2 m long. How many newtons of lead (SG 11.4)should be attached to one end of the wood so that it willfloat vertically with 30 cm out of the water?A hydrometer floats at a level which is a measure of thespecific gravity of the liquid. The stem is of constant diameter D, and a weight in the bottom stabilizes the bodyto float vertically, as shown in Fig. P2.109. If the positionh 0 is pure water (SG 1.0), derive a formula for h asa function of total weight W, D, SG, and the specific weight0 of water.DSG = 1.0hFluid, SG > 1WP2.1093 cm8 cmWaterD = 9 cmP2.104|vvP2.105 It is said that Archimedes discovered the buoyancy lawswhen asked by King Hiero of Syracuse to determine|e-Text Main Menu|P2.110 An average table tennis ball has a diameter of 3.81 cm anda mass of 2.6 g. Estimate the (small) depth at which thisball will float in water at 20C and sea level standard airif air buoyancy is (a) neglected and (b) included.P2.111 A hot-air balloon must be designed to support basket, cords,and one person for a total weight of 1300 N. The balloonmaterial has a mass of 60 g/m2. Ambient air is at 25C and1 atm. The hot air inside the balloon is at 70C and 1 atm.What diameter spherical balloon will just support the totalweight? Neglect the size of the hot-air inlet vent.P2.112 The uniform 5-m-long round wooden rod in Fig. P2.112is tied to the bottom by a string. Determine (a) the tensionTextbook Table of Contents|Study GuideProblems 119in the string and (b) the specific gravity of the wood. Is itpossible for the given information to determine the inclination angle ? Explain.HingeD = 4 cmB= 301m8mD = 8 cm2 kg of leadWater at 20CP2.1144mBString8 ftWoodSG = 0.6P2.112P2.113 A spar buoy is a buoyant rod weighted to float and protrudevertically, as in Fig. P2.113. It can be used for measurementsor markers. Suppose that the buoy is maple wood (SG0.6), 2 in by 2 in by 12 ft, floating in seawater (SG 1.025).How many pounds of steel (SG 7.85) should be added tothe bottom end so that h 18 in?SeawaterARockP2.115P2.116 The homogeneous 12-cm cube in Fig. 2.116 is balancedby a 2-kg mass on the beam scale when the cube is immersed in 20C ethanol. What is the specific gravity of thecube?h2 kgWsteelP2.113|vvP2.114 The uniform rod in Fig. P2.114 is hinged at point B on thewaterline and is in static equilibrium as shown when 2 kgof lead (SG 11.4) are attached to its end. What is thespecific gravity of the rod material? What is peculiar aboutthe rest angle30?P2.115 The 2-in by 2-in by 12-ft spar buoy from Fig. P2.113 has 5lbm of steel attached and has gone aground on a rock, as inFig. P2.115. Compute the angle at which the buoy willlean, assuming that the rock exerts no moments on the spar.|e-Text Main Menu|12 cmP2.116P2.117 The balloon in Fig. P2.117 is filled with helium and pressurized to 135 kPa and 20C. The balloon material has aTextbook Table of Contents|Study Guide120Chapter 2 Pressure Distribution in a Fluidmass of 85 g/m2. Estimate (a) the tension in the mooringline and (b) the height in the standard atmosphere to whichthe balloon will rise if the mooring line is cut.P2.121 The uniform beam in Fig. P2.121, of size L by h by b andwith specific weight b, floats exactly on its diagonal whena heavy uniform sphere is tied to the left corner, as shown.Show that this can only happen (a) when b/3 and (b)when the sphere has sizeLhb(SG 1)D1/3D = 10 mWidth b << LAir:100 kPa at20CP2.117P2.118 A 14-in-diameter hollow sphere is made of steel (SG7.85) with 0.16-in wall thickness. How high will thisEESsphere float in 20C water? How much weight must beadded inside to make the sphere neutrally buoyant?P2.119 When a 5-lbf weight is placed on the end of the uniformfloating wooden beam in Fig. P2.119, the beam tilts at anangle with its upper right corner at the surface, as shown.Determine (a) the angle and (b) the specific gravity ofthe wood. (Hint: Both the vertical forces and the momentsabout the beam centroid must be balanced.);Lh << LbDiameter DSG > 1P2.121P2.122 A uniform block of steel (SG 7.85) will float at amercury-water interface as in Fig. P2.122. What is theratio of the distances a and b for this condition?5 lbfWater9 ftWater4 in 4 inaSteelblockP2.119bMercury: SG = 13.56P2.120 A uniform wooden beam (SG 0.65) is 10 cm by 10 cmby 3 m and is hinged at A, as in Fig. P2.120. At what angle will the beam float in the 20C water?P2.123 In an estuary where fresh water meets and mixes with seawater, there often occurs a stratified salinity condition withfresh water on top and salt water on the bottom, as in Fig.P2.123. The interface is called a halocline. An idealizationof this would be constant density on each side of the halocline as shown. A 35-cm-diameter sphere weighing 50 lbfwould float near such a halocline. Compute the sphereposition for the idealization in Fig. P2.123.P2.124 A balloon weighing 3.5 lbf is 6 ft in diameter. It is filledwith hydrogen at 18 lbf/in2 absolute and 60F and is released. At what altitude in the U.S. standard atmospherewill this balloon be neutrally buoyant?A1mWatervvP2.120||e-Text Main MenuP2.122|Textbook Table of Contents|Study GuideProblems 121SG = 1.0HaloclineSG = 1.02535/Salinity0IdealizationP2.123P2.125 Suppose that the balloon in Prob. 2.111 is constructed tohave a diameter of 14 m, is filled at sea level with hot airat 70C and 1 atm, and is released. If the air inside the balloon remains constant and the heater maintains it at 70C,at what altitude in the U.S. standard atmosphere will thisballoon be neutrally buoyant?*P2.126 A cylindrical can of weight W, radius R, and height H isopen at one end. With its open end down, and while filledwith atmospheric air (patm, Tatm), the can is eased downvertically into liquid, of density , which enters and compresses the air isothermally. Derive a formula for the heighth to which the liquid rises when the can is submerged withits top (closed) end a distance d from the surface.*P2.127 Consider the 2-in by 2-in by 10-ft spar buoy of Prob. 2.113.How many pounds of steel (SG 7.85) should be addedat the bottom to ensure vertical floating with a metacentric height MG of (a) zero (neutral stability) or (b) 1 ft(reasonably stable)?P2.128 An iceberg can be idealized as a cube of side length L, asin Fig. P2.128. If seawater is denoted by S 1.0, thenglacier ice (which forms icebergs) has S 0.88. Determine if this cubic iceberg is stable for the position shownin Fig. P2.128.Fig. P2.128 suppose that the height is L and the depth intothe paper is L, but the width in the plane of the paper isH L. Assuming S 0.88 for the iceberg, find the ratioH/L for which it becomes neutrally stable, i.e., about tooverturn.P2.130 Consider a wooden cylinder (SG 0.6) 1 m in diameterand 0.8 m long. Would this cylinder be stable if placed tofloat with its axis vertical in oil (SG 0.8)?P2.131 A barge is 15 ft wide and 40 ft long and floats with a draftof 4 ft. It is piled so high with gravel that its center of gravity is 2 ft above the waterline. Is it stable?P2.132 A solid right circular cone has SG 0.99 and floats vertically as in Fig. P2.132. Is this a stable position for thecone?Water :SG = 1.0SG = 0.99P2.132P2.133 Consider a uniform right circular cone of specific gravityS 1, floating with its vertex down in water (S 1). Thebase radius is R and the cone height is H. Calculate andplot the stability MG of this cone, in dimensionless form,versus H/R for a range of S 1.P2.134 When floating in water (SG 1.0), an equilateral triangular body (SG 0.9) might take one of the two positionsshown in Fig. P2.134. Which is the more stable position?Assume large width into the paper.Specific gravity=SM?GBhWaterS = 1.0P2.134P2.135 Consider a homogeneous right circular cylinder of lengthL, radius R, and specific gravity SG, floating in water(SG 1). Show that the body will be stable with its axisvertical ifLP2.128vvP2.129 The iceberg idealization in Prob. 2.128 may become unstable if its sides melt and its height exceeds its width. In||(b)(a)e-Text Main Menu|Textbook Table of ContentsRL[2SG(1|SG)]1/2Study Guide122Chapter 2 Pressure Distribution in a FluidP2.136 Consider a homogeneous right circular cylinder of lengthL, radius R, and specific gravity SG 0.5, floating in water (SG 1). Show that the body will be stable with itsaxis horizontal if L/R 2.0.P2.137 A tank of water 4 m deep receives a constant upward acceleration az. Determine (a) the gage pressure at the tankbottom if az 5 m2/s and (b) the value of az which causesthe gage pressure at the tank bottom to be 1 atm.P2.138 A 12-fl-oz glass, of 3-in diameter, partly full of water, isattached to the edge of an 8-ft-diameter merry-go-roundwhich is rotated at 12 r/min. How full can the glass be before water spills? (Hint: Assume that the glass is muchsmaller than the radius of the merry-go-round.)P2.139 The tank of liquid in Fig. P2.139 accelerates to the rightwith the fluid in rigid-body motion. (a) Compute ax inm/s2. (b) Why doesnt the solution to part (a) depend uponthe density of the fluid? (c) Determine the gage pressureat point A if the fluid is glycerin at 20C.Va?15 cm100 cm28 cmAz30P2.141xB9 cmWater at 20 CA24 cmaxP2.14228 cm15 cm100 cmAApa = 15 lbf/in2 absFig. P2.139ax2ft|vvP2.140 Suppose that the elliptical-end fuel tank in Prob. 2.101 is10 m long and filled completely with fuel oil (890kg/m3). Let the tank be pulled along a horizontal road. Forrigid-body motion, find the acceleration, and its direction,for which (a) a constant-pressure surface extends from thetop of the front end wall to the bottom of the back end and(b) the top of the back end is at a pressure 0.5 atm lowerthan the top of the front end.P2.141 The same tank from Prob. 2.139 is now moving with constant acceleration up a 30 inclined plane, as in Fig.P2.141. Assuming rigid-body motion, compute (a) thevalue of the acceleration a, (b) whether the acceleration isup or down, and (c) the gage pressure at point A if the fluidis mercury at 20C.P2.142 The tank of water in Fig. P2.142 is 12 cm wide into thepaper. If the tank is accelerated to the right in rigid-bodymotion at 6.0 m/s2, compute (a) the water depth on sideAB and (b) the water-pressure force on panel AB. Assumeno spilling.P2.143 The tank of water in Fig. P2.143 is full and open to the atmosphere at point A. For what acceleration ax in ft/s2 will thepressure at point B be (a) atmospheric and (b) zero absolute?|e-Text Main Menu|WaterB1ftP2.1431ft2ftP2.144 Consider a hollow cube of side length 22 cm, filled completely with water at 20C. The top surface of the cube ishorizontal. One top corner, point A, is open through a smallhole to a pressure of 1 atm. Diagonally opposite to pointA is top corner B. Determine and discuss the various rigidbody accelerations for which the water at point B beginsto cavitate, for (a) horizontal motion and (b) vertical motion.P2.145 A fish tank 14 in deep by 16 by 27 in is to be carriedin a car which may experience accelerations as high as6 m/s2. What is the maximum water depth which will avoidTextbook Table of Contents|Study GuideProblems 123spilling in rigid-body motion? What is the proper alignment of the tank with respect to the car motion?P2.146 The tank in Fig. P2.146 is filled with water and has a venthole at point A. The tank is 1 m wide into the paper. Inside the tank, a 10-cm balloon, filled with helium at 130kPa, is tethered centrally by a string. If the tank accelerates to the right at 5 m/s2 in rigid-body motion, at whatangle will the balloon lean? Will it lean to the right or tothe left?60 cmA1 atmWater at 20CD = 10 cmwith the child, which way will the balloon tilt, forward orbackward? Explain. (b) The child is now sitting in a carwhich is stopped at a red light. The helium-filled balloonis not in contact with any part of the car (seats, ceiling,etc.) but is held in place by the string, which is in turn heldby the child. All the windows in the car are closed. Whenthe traffic light turns green, the car accelerates forward. Ina frame of reference moving with the car and child, whichway will the balloon tilt, forward or backward? Explain.(c) Purchase or borrow a helium-filled balloon. Conduct ascientific experiment to see if your predictions in parts (a)and (b) above are correct. If not, explain.P2.149 The 6-ft-radius waterwheel in Fig. P2.149 is being used tolift water with its 1-ft-diameter half-cylinder blades. If thewheel rotates at 10 r/min and rigid-body motion is assumed, what is the water surface angle at position A?He40 cm20 cmString10 r/minP2.146A6 ftP2.147 The tank of water in Fig. P2.147 accelerates uniformly byfreely rolling down a 30 incline. If the wheels are frictionless, what is the angle ? Can you explain this interesting result?1 ftP2.149P2.150 A cheap accelerometer, probably worth the price, can bemade from a U-tube as in Fig. P2.150. If L 18 cm andD 5 mm, what will h be if ax 6 m/s2? Can the scalemarkings on the tube be linear multiples of ax?DhRest levelax301212LLP2.147LP2.150|vvP2.148 A child is holding a string onto which is attached a helium-filled balloon. (a) The child is standing still and suddenly accelerates forward. In a frame of reference moving|e-Text Main Menu|P2.151 The U-tube in Fig. P2.151 is open at A and closed at D.If accelerated to the right at uniform ax, what accelerationTextbook Table of Contents|Study Guide124Chapter 2 Pressure Distribution in a Fluidwill cause the pressure at point C to be atmospheric? Thefluid is water (SG 1.0).AD1 ft1 ftBP2.156 Suppose that the U-tube of Fig. P2.151 is rotated aboutaxis DC. If the fluid is water at 122F and atmosphericpressure is 2116 lbf/ft2 absolute, at what rotation rate willthe fluid within the tube begin to vaporize? At what pointwill this occur?P2.157 The 45 V-tube in Fig. P2.157 contains water and is openat A and closed at C. What uniform rotation rate in r/minabout axis AB will cause the pressure to be equal at pointsB and C? For this condition, at what point in leg BC willthe pressure be a minimum?CAC1 ftP2.151P2.152 A 16-cm-diameter open cylinder 27 cm high is full of water. Compute the rigid-body rotation rate about its centralaxis, in r/min, (a) for which one-third of the water willspill out and (b) for which the bottom will be barely exposed.P2.153 Suppose the U-tube in Fig. P2.150 is not translated butrather rotated about its right leg at 95 r/min. What will bethe level h in the left leg if L 18 cm and D 5 mm?P2.154 A very deep 18-cm-diameter can contains 12 cm of wateroverlaid with 10 cm of SAE 30 oil. If the can isrotated in rigid-body motion about its central axis at150 r/min, what will be the shapes of the air-oil and *P2.158oil-water interfaces? What will be the maximum fluid pressure in the can in Pa (gage)?P2.155 For what uniform rotation rate in r/min about axis C willthe U-tube in Fig. P2.155 take the configuration shown?EESThe fluid is mercury at 20C.A30 cm45BP2.157It is desired to make a 3-m-diameter parabolic telescopemirror by rotating molten glass in rigid-body motion until the desired shape is achieved and then cooling the glassto a solid. The focus of the mirror is to be 4 m from themirror, measured along the centerline. What is the propermirror rotation rate, in r/min, for this task?CB20 cm12 cm10 cm|vvP2.1555 cm|e-Text Main Menu|Textbook Table of Contents|Study GuideFundamentals of Engineering Exam Problems125Word ProblemsW2.1Consider a hollow cone with a vent hole in the vertex atthe top, along with a hollow cylinder, open at the top, withthe same base area as the cone. Fill both with water to thetop. The hydrostatic paradox is that both containers havethe same force on the bottom due to the water pressure, although the cone contains 67 percent less water. Can youexplain the paradox?W2.2 Can the temperature ever rise with altitude in the real atmosphere? Wouldnt this cause the air pressure to increaseupward? Explain the physics of this situation.W2.3 Consider a submerged curved surface which consists of atwo-dimensional circular arc of arbitrary angle, arbitrarydepth, and arbitrary orientation. Show that the resultant hydrostatic pressure force on this surface must pass throughthe center of curvature of the arc.W2.4 Fill a glass approximately 80 percent with water, and add alarge ice cube. Mark the water level. The ice cube, havingSG 0.9, sticks up out of the water. Let the ice cube meltwith negligible evaporation from the water surface. Will thewater level be higher than, lower than, or the same as before?W2.5A ship, carrying a load of steel, is trapped while floatingin a small closed lock. Members of the crew want to getout, but they cant quite reach the top wall of the lock. Acrew member suggests throwing the steel overboard in thelock, claiming the ship will then rise and they can climbout. Will this plan work?W2.6 Consider a balloon of mass m floating neutrally in the atmosphere, carrying a person/basket of mass M m. Discuss the stability of this system to disturbances.W2.7 Consider a helium balloon on a string tied to the seat ofyour stationary car. The windows are closed, so there is noair motion within the car. The car begins to accelerate forward. Which way will the balloon lean, forward or backward? (Hint: The acceleration sets up a horizontal pressuregradient in the air within the car.)W2.8 Repeat your analysis of Prob. W2.7 to let the car move atconstant velocity and go around a curve. Will the balloonlean in, toward the center of curvature, or out?Fundamentals of Engineering Exam ProblemsFE2.1 A gage attached to a pressurized nitrogen tank reads agage pressure of 28 in of mercury. If atmospheric pressure is 14.4 psia, what is the absolute pressure in the tank?(a) 95 kPa, (b) 99 kPa, (c) 101 kPa, (d) 194 kPa,(e) 203 kPaFE2.2 On a sea-level standard day, a pressure gage, moored below the surface of the ocean (SG 1.025), reads an absolute pressure of 1.4 MPa. How deep is the instrument?(a) 4 m, (b) 129 m, (c) 133 m, (d) 140 m, (e) 2080 mFE2.3 In Fig. FE2.3, if the oil in region B has SG 0.8 and theabsolute pressure at point A is 1 atm, what is the absolutepressure at point B?(a) 5.6 kPa, (b) 10.9 kPa, (c) 106.9 kPa, (d) 112.2 kPa,(e) 157.0 kPaAOilWaterSG = 15 cmB3 cm8 cmMercurySG = 13.564 cm|vvFE2.3|e-Text Main Menu|FE2.4 In Fig. FE2.3, if the oil in region B has SG 0.8 and theabsolute pressure at point B is 14 psia, what is the absolute pressure at point B?(a) 11 kPa, (b) 41 kPa, (c) 86 kPa, (d) 91 kPa, (e) 101 kPaFE2.5 A tank of water (SG 1,.0) has a gate in its vertical wall5 m high and 3 m wide. The top edge of the gate is 2 mbelow the surface. What is the hydrostatic force on the gate?(a) 147 kN, (b) 367 kN, (c) 490 kN, (d) 661 kN,(e) 1028 kNFE2.6 In Prob. FE2.5 above, how far below the surface is thecenter of pressure of the hydrostatic force?(a) 4.50 m, (b) 5.46 m, (c) 6.35 m, (d) 5.33 m, (e) 4.96 mFE2.7 A solid 1-m-diameter sphere floats at the interface betweenwater (SG 1.0) and mercury (SG 13.56) such that 40 percent is in the water. What is the specific gravity of the sphere?(a) 6.02, (b) 7.28, (c) 7.78, (d) 8.54, (e) 12.56FE2.8 A 5-m-diameter balloon contains helium at 125 kPa absoluteand 15C, moored in sea-level standard air. If the gas constant of helium is 2077 m2/(s2 K) and balloon material weightis neglected, what is the net lifting force of the balloon?(a) 67 N, (b) 134 N, (c) 522 N, (d) 653 N, (e) 787 NFE2.9 A square wooden (SG 0.6) rod, 5 cm by 5 cm by 10 mlong, floats vertically in water at 20C when 6 kg of steel(SG 7.84) are attached to one end. How high above thewater surface does the wooden end of the rod protrude?(a) 0.6 m, (b) 1.6 m, (c) 1.9 m, (d) 2.4 m, (e) 4.0 mTextbook Table of Contents|Study Guide126Chapter 2 Pressure Distribution in a FluidFE2.10 A floating body will be stable when its(a) center of gravity is above its center of buoyancy,(b) center of buoyancy is below the waterline, (c) centerof buoyancy is above its metacenter, (d) metacenter isabove its center of buoyancy, (e) metacenter is above itscenter of gravityComprehensive ProblemsC2.1 Some manometers are constructed as in Fig. C2.1, whereone side is a large reservoir (diameter D) and the other sideis a small tube of diameter d, open to the atmosphere. Insuch a case, the height of manometer liquid on the reservoirside does not change appreciably. This has the advantagethat only one height needs to be measured rather than two.The manometer liquid has density m while the air has density a. Ignore the effects of surface tension. When there isno pressure difference across the manometer, the elevationson both sides are the same, as indicated by the dashed line.Height h is measured from the zero pressure level as shown.(a) When a high pressure is applied to the left side, themanometer liquid in the large reservoir goes down, whilethat in the tube at the right goes up to conserve mass. Writean exact expression for p1gage, taking into account the movement of the surface of the reservoir. Your equation shouldgive p1gage as a function of h, m, and the physical parameters in the problem, h, d, D, and gravity constant g.(b) Write an approximate expression for p1gage, neglectingthe change in elevation of the surface of the reservoir liquid. (c) Suppose h 0.26 m in a certain application. If pa101,000 Pa and the manometer liquid has a density of 820kg/m3, estimate the ratio D/d required to keep the errorof the approximation of part (b) within 1 percent of the exact measurement of part (a). Repeat for an error within 0.1percent.To pressure measurement locationpaa(air)Dhp1Zero pressure levelmdC2.1|vvC2.2 A prankster has added oil, of specific gravity SG0, to theleft leg of the manometer in Fig. C2.2. Nevertheless, the|e-Text Main Menu|U-tube is still useful as a pressure-measuring device. It isattached to a pressurized tank as shown in the figure. (a)Find an expression for h as a function of H and other parameters in the problem. (b) Find the special case of yourresult in (a) when ptank pa. (c) Suppose H 5.0 cm, pais 101.2kPa, ptank is 1.82 kPa higher than pa, and SG00.85. Calculate h in cm, ignoring surface tension effects andneglecting air density effects.paPressurized air tank,with pressure ptankOilHhWaterC2.2C2.3 Professor F. Dynamics, riding the merry-go-round with hisson, has brought along his U-tube manometer. (You neverknow when a manometer might come in handy.) As shownin Fig. C2.3, the merry-go-round spins at constant angularvelocity and the manometer legs are 7 cm apart. Themanometer center is 5.8 m from the axis of rotation. Determine the height difference h in two ways: (a) approximately, by assuming rigid body translation with a equal tothe average manometer acceleration; and (b) exactly, usingrigid-body rotation theory. How good is the approximation?C2.4 A student sneaks a glass of cola onto a roller coaster ride.The glass is cylindrical, twice as tall as it is wide, and filledto the brim. He wants to know what percent of the cola heshould drink before the ride begins, so that none of it spillsduring the big drop, in which the roller coaster achieves0.55-g acceleration at a 45 angle below the horizontal.Make the calculation for him, neglecting sloshing and assuming that the glass is vertical at all times.Textbook Table of Contents|Study Guide7.00 cm6.00 rpmWaterhR5.80 m (to center of manometer)Center ofrotationC2.3Design Projectsh, mD2.1 It is desired to have a bottom-moored, floating systemwhich creates a nonlinear force in the mooring line as thewater level rises. The design force F need only be accuratein the range of seawater depths h between 6 and 8 m, asshown in the accompanying table. Design a buoyant system which will provide this force distribution. The systemshould be practical, i.e., of inexpensive materials and simple construction.D2.2 A laboratory apparatus used in some universities is shownin Fig. D2.2. The purpose is to measure the hydrostaticforce on the flat face of the circular-arc block and compare it with the theoretical value for given depth h. Thecounterweight is arranged so that the pivot arm is horizontal when the block is not submerged, whence the weightW can be correlated with the hydrostatic force when thesubmerged arm is again brought to horizontal. First showthat the apparatus concept is valid in principle; then derivea formula for W as a function of h in terms of the systemparameters. Finally, suggest some appropriate values of Y,L, etc., for a suitable appartus and plot theoretical W versus h for these values.F, Nh, mF, N6.006.256.506.757.004004374715025307.257.507.758.00554573589600LCounterweightWPivotPivot armRSide viewof block faceFluid:hYCircular arc blockbD2.2References188.8.131.52.6.7.U.S. Standard Atmosphere, 1976, Government Printing Office, Washington, DC, 1976.G. Neumann and W. J. Pierson, Jr., Principles of PhysicalOceanography, Prentice-Hall, Englewood Cliffs, NJ, 1966.T. C. Gillmer and B. Johnson, Introduction to Naval Architecture, Naval Institute Press, Annapolis, MD, 1982.D. T. Greenwood, Principles of Dynamics, 2d ed., PrenticeHall, Englewood Cliffs, NJ, 1988.R. I. Fletcher, The Apparent Field of Gravity in a RotatingFluid System, Am. J. Phys., vol. 40, pp. 959 965, July 1972.National Committee for Fluid Mechanics Films, IllustratedExperiments in Fluid Mechanics, M.I.T. Press, Cambridge,MA, 1972.J. P. Holman, Experimental Methods for Engineers, 6th ed.,McGraw-Hill, New York, 1993.|vv1.|e-Text Main Menu|184.108.40.206.12.13.R. P. Benedict, Fundamentals of Temperature, Pressure, andFlow Measurement, 3d ed., Wiley, New York, 1984.T. G. Beckwith and R. G. Marangoni, Mechanical Measurements, 4th ed., Addison-Wesley, Reading, MA, 1990.J. W. Dally, W. F. Riley, and K. G. McConnell, Instrumentation for Engineering Measurements, Wiley, New York, 1984.E. N. Gilbert, How Things Float, Am. Math. Monthly, vol.98, no. 3, pp. 201 216, 1991.R. J. Figliola and D. E. Beasley, Theory and Design for Mechanical Measurements, 2d ed., Wiley, New York, 1994.R. W. Miller, Flow Measurement Engineering Handbook, 3ded., McGraw-Hill, New York, 1996.Textbook Table of Contents127|Study Guide...
View Full Document
- Spring '08