Chapt02 - Roosevelt Dam in Arizona Hydrostatic pressure due...

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Unformatted text preview: Roosevelt Dam in Arizona. Hydrostatic pressure, due to the weight of a standing fluid, can cause enormous 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.) | v v 58 | e-Text Main Menu | Textbook Table of Contents | Study Guide Chapter 2 Pressure Distribution in a Fluid Motivation. Many fluid problems do not involve motion. They concern the pressure distribution 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 pressure variation is due only to the weight of the fluid. Assuming a known fluid in a given gravity 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 curved surfaces, (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 been spinning for a long time, the pressure also can be easily calculated, because the fluid is free of shear stress. We apply this idea here to simple rigid-body accelerations in Sec. 2.9. Pressure measurement instruments are discussed in Sec. 2.10. As a matter of fact, 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 Pressure Gradient In Fig. 1.1 we saw that a fluid at rest cannot support shear stress and thus Mohr’s circle reduces to a point. In other words, the normal stress on any plane through a fluid element at rest is equal to a unique value called the fluid pressure p, taken positive for compression by common convention. This is such an important concept that we shall review it with another approach. Figure 2.1 shows a small wedge of fluid at rest of size x by z by s and depth b into 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. Fx Fz 0 0 pzb pxb z x pnb s sin pnb s cos 1 2 (2.1) b x z | v v 59 | e-Text Main Menu | Textbook Table of Contents | Study Guide 60 Chapter 2 Pressure Distribution in a Fluid z (up) pn ∆s θ Element weight: d W = ρ g ( 1 b ∆ x ∆ z) 2 ∆z px ∆x θ x O Fig. 2.1 Equilibrium of a small wedge of fluid at rest. Width b into paper pz but the geometry of the wedge is such that s sin z s cos x (2.2) Substitution into Eq. (2.1) and rearrangement give px pn pz 1 2 pn z (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 a vertical change in pressure proportional to the density, gravity, and depth change. We shall 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) become px pz pn p (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 a moving 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 normal stresses ii on the element 1 3 p ( xx yy zz) (2.5) The minus sign occurs because a compression stress is considered to be negative whereas 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 Fluid Element Pressure (or any other stress, for that matter) causes no net force on a fluid element unless it varies spatially.1 To see this, consider the pressure acting on the two x faces in Fig. 2.2. Let the pressure vary arbitrarily p p(x, y, z, t) (2.6) 1 | v v An interesting application for a large element is in Fig. 3.7. | e-Text Main Menu | Textbook Table of Contents | Study Guide 2.2 Equilibrium of a Fluid Element 61 y dz p dy dz (p+ dy ∂p d x) dy dz ∂x x Fig. 2.2 Net x force on an element due to pressure variation. dx z The net force in the x direction on the element in Fig. 2.2 is given by dFx p dy dz p dx dy dz x p p dx dy dz x (2.7) In like manner the net force dFy involves p/ y, and the net force dFz concerns p/ z. The total net-force vector on the element due to pressure is dFpress p x i p y j k p dx dy dz z (2.8) We recognize the term in parentheses as the negative vector gradient of p. Denoting f as the net force per unit element volume, we rewrite Eq. (2.8) as ∇p fpress (2.9) Thus it is not the pressure but the pressure gradient causing a net force which must be balanced by gravity or acceleration or some other effect in the fluid. 2.2 Equilibrium of a Fluid Element The pressure gradient is a surface force which acts on the sides of the element. There may also be a body force, due to electromagnetic or gravitational potentials, acting on the entire mass of the element. Here we consider only the gravity force, or weight of the element dFgrav or g dx dy dz fgrav (2.10) g In 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 is 2 fVS V x2 2 2 V y2 V z2 ∇2V (2.11) | v v where 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 Guide Chapter 2 Pressure Distribution in a Fluid ing toward the center of the earth. On earth the average magnitude of g is 32.174 ft/s2 9.807 m/s2. The total vector resultant of these three forces — pressure, gravity, and viscous stress — must either keep the element in equilibrium or cause it to move with acceleration a. From Newton’s law, Eq. (1.2), we have a f fpress fgrav ∇p fvisc ∇2V g (2.12) This is one form of the differential momentum equation for a fluid element, and it is studied further in Chap. 4. Vector addition is implied by Eq. (2.12): The acceleration reflects the local balance of forces and is not necessarily parallel to the local-velocity vector, 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 up the more general problem where pressure, velocity, and acceleration are all unknown. Rewrite Eq. (2.12) as ∇p (g a) ∇2V B(x, y, z, t) (2.13) where B is a short notation for the vector sum on the right-hand side. If V and a dV/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 equations p x Bx(x, y, z, t) p y By(x, y, z, t) p z Bz(x, y, z, t) (2.14) Since the right-hand sides are known functions, they can be integrated systematically to obtain the distribution p(x, y, z, t) except for an unknown function of time, which remains 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 knowledge of a single known pressure p0 at a point (x0, y0, z0). If this sounds complicated, it is not; we shall illustrate with many examples. Finding the pressure distribution from a known velocity distribution is one of the easiest problems in fluid mechanics, which is 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 vanish identically, and p depends only upon gravity and density. This is the hydrostatic condition. 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, and an exact integral called Bernoulli’s equation can be found for the pressure distribution. See Sec. 4.9. 4. Arbitrary viscous motion: Nothing helpful happens, no general rules apply, but still the integration is quite straightforward. See Sec. 6.4. Let us consider cases 1 and 2 here. | v v 62 | e-Text Main Menu | Textbook Table of Contents | Study Guide 2.3 Hydrostatic Pressure Distributions 63 p (Pascals) High pressure: p = 120,000 Pa abs = 30,000 Pa gage 120,000 30,000 Local atmosphere: p = 90,000 Pa abs = 0 Pa gage = 0 Pa vacuum 90,000 40,000 Vacuum pressure: p = 50,000 Pa abs = 40,000 Pa vacuum 50,000 50,000 Fig. 2.3 Illustration of absolute, gage, and vacuum pressure readings. Gage Pressure and Vacuum Pressure: Relative Terms Absolute zero reference: p = 0 Pa abs = 90,000 Pa vacuum 0 (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 of differential type and record, not an absolute magnitude, but the difference between the fluid pressure and the atmosphere. The measured pressure may be either higher or lower than the local atmosphere, and each case is given a name: 1. p 2. p pa pa Gage pressure: Vacuum pressure: p(gage) p(vacuum) p pa pa p This is a convenient shorthand, and one later adds (or subtracts) atmospheric pressure to 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 at an altitude of 1000 m. Thus, on this day, pa 90,000 Pa absolute 0 Pa gage 0 Pa vacuum. Suppose gage 1 in a laboratory reads p1 120,000 Pa absolute. This value may be reported as a gage pressure, p1 120,000 90,000 30,000 Pa gage. (One must 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 you alert to this common engineering practice. 2.3 Hydrostatic Pressure Distributions 0 and ∇2V If the fluid is at rest or at constant velocity, a the pressure distribution reduces to ∇p 0. Equation (2.13) for g (2.15) | v v This is a hydrostatic distribution and is correct for all fluids at rest, regardless of their viscosity, 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 Guide 64 Chapter 2 Pressure Distribution in a Fluid is perpendicular everywhere to surfaces of constant p. Thus Eq. (2.15) states that a fluid in hydrostatic equilibrium will align its constant-pressure surfaces everywhere normal to the local-gravity vector. The maximum pressure increase will be in the direction of gravity, 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 is g gk (2.16) where g is the magnitude of local gravity, for example, 9.807 m/s2. For these coordinates Eq. (2.15) has the components p x p y 0 p z 0 g (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 to dp dz 2 or p2 p1 dz (2.18) 1 Equation (2.18) is the solution to the hydrostatic problem. The integration requires an assumption about the density and gravity distribution. Gases and liquids are usually treated differently. We state the following conclusions about a hydrostatic condition: Pressure in a continuously distributed uniform static fluid varies only with vertical distance and is independent of the shape of the container. The pressure is the same at all points on a given horizontal plane in the fluid. The pressure increases with depth 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 | v v Fig. 2.4 Hydrostatic-pressure distribution. Points a, b, c, and d are at equal depths in water and therefore have identical pressures. Points A, B, and C are also at equal depths in water and have identical pressures higher than a, b, c, and d. Point D has a different pressure from A, B, and C because it is not connected to them by a water path. Water Depth 1 a b c d Mercury Depth 2 | A e-Text Main Menu B | C Textbook Table of Contents D | Study Guide 2.3 Hydrostatic Pressure Distributions 65 tal plane and are interconnected by the same fluid, water; therefore all points have the same pressure. The same is true of points A, B, and C on the bottom, which all have the same higher pressure than at a, b, c, and d. However, point D, although at the same depth as A, B, and C, has a different pressure because it lies beneath a different fluid, mercury. Effect of Variable Gravity For a spherical planet of uniform density, the acceleration of gravity varies inversely as the square of the radius from its center g g0 r0 r 2 (2.19) where r0 is the planet radius and g0 is the surface value of g. For earth, r0 3960 statute mi 6400 km. In typical engineering problems the deviation from r0 extends from the deepest ocean, about 11 km, to the atmospheric height of supersonic transport operation, about 20 km. This gives a maximum variation in g of (6400/6420)2, or 0.6 percent. We therefore neglect the variation of g in most problems. Hydrostatic Pressure in Liquids Liquids 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 the deepest part of the ocean. Its effect on hydrostatics would be about half of this, or 2.3 percent. Thus we assume constant density in liquid hydrostatic calculations, for which Eq. (2.18) integrates to Liquids: p2 or p1 z1 z2 (z2 p2 z1) (2.20) p1 We use the first form in most problems. The quantity is called the specific weight of the fluid, with dimensions of weight per unit volume; some values are tabulated in Table 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, with z 0 at the free surface, where p equals the surface atmospheric pressure pa. When Table 2.1 Specific Weight of Some Common Fluids Specific weight at 68°F 20°C v | | lbf/ft3 N/m3 Air (at 1 atm) Ethyl alcohol SAE 30 oil Water Seawater Glycerin Carbon tetrachloride Mercury v Fluid 000.0752 049.2 055.5 062.4 064.0 078.7 099.1 846 000,011.8 007,733 008,720 009,790 010,050 012,360 015,570 133,100 e-Text Main Menu | Textbook Table of Contents | Study Guide 66 Chapter 2 Pressure Distribution in a Fluid Z p ≈ pa – b γ air +b Air Free surface: Z = 0, p = pa 0 Water g Fig. 2.5 Hydrostatic-pressure distribution in oceans and atmospheres. p ≈ pa + hγ water –h we introduce the reference value ( p1, z1) (negative) depth z, Lakes and oceans: p ( pa, 0), Eq. (2.20) becomes, for p at any pa z (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 2.1 Newfound Lake, a freshwater lake near Bristol, New Hampshire, has a maximum depth of 60 m, and the mean atmospheric pressure is 91 kPa. Estimate the absolute pressure in kPa at this maximum depth. Solution From Table 2.1, take 9790 N/m3. With pa the pressure at this depth will be 91 kN/m2 p 91 kPa (9790 N/m3)( 60 m) 587 kN/m2 By omitting pa we could state the result as p 60 m, Eq. (2.21) predicts that 1 kN 1000 N 678 kPa Ans. 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 upper end because mercury has an extremely small vapor pressure at room temperatures (0.16 Pa at 20°C). Since atmospheric pressure forces a mercury column to rise a distance h into the tube, the upper mercury surface is at zero pressure. | v v The Mercury Barometer 91 kPa and z | e-Text Main Menu | Textbook Table of Contents | Study Guide 2.3 Hydrostatic Pressure Distributions 67 p1 ≈ 0 ( Mercury has a very low vapor pressure.) z1 = h p2 ≈ pa ( The mercury is in contact with the atmosphere.) p h= γa M z pa z2 = 0 ρ M Mercury (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 of Fig. 2.28c. (Courtesy of Paul Lupke, Druck Inc.) From Fig. 2.6, Eq. (2.20) applies with p1 pa or 0 0 at z1 M(0 h and p2 0: h) pa h pa at z2 (2.22) M At 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 United States 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 Gases Gases are compressible, with density nearly proportional to pressure. Thus density must be considered as a variable in Eq. (2.18) if the integration carries over large pressure changes. It is sufficiently accurate to introduce the perfect-gas law p RT in Eq. (2.18) | v v dp dz | e-Text Main Menu | p g RT g Textbook Table of Contents | Study Guide Chapter 2 Pressure Distribution in a Fluid Separate the variables and integrate between points 1 and 2: 2 dp p 1 ln p2 p1 g R 2 1 dz T (2.23) The integral over z requires an assumption about the temperature variation T(z). One common approximation is the isothermal atmosphere, where T T0: p2 g(z2 z1) RT0 p1 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 earth’s mean atmospheric temperature drops off nearly linearly with z up to an altitude of about 36,000 ft (11,000 m): T T0 Bz (2.25) Here T0 is sea-level temperature (absolute) and B is the lapse rate, both of which vary somewhat from day to day. By international agreement [1] the following standard values are assumed to apply from 0 to 36,000 ft: T0 518.69°R B 288.16 K 0.003566°R/ft 15°C 0.00650 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 relation p pa 1 Bz T0 g/(RB) where g RB 5.26 (air) (2.27) in the troposphere, with z 0 at sea level. The exponent g/(RB) is dimensionless (again it must be) and has the standard value of 5.26 for air, with R 287 m2/(s2 K). The U.S. standard atmosphere [1] is sketched in Fig. 2.7. The pressure is seen to be nearly zero at z 30 km. For tabulated properties see Table A.6. EXAMPLE 2.2 If 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 temperature of 15°C. Is the isothermal approximation adequate? Solution Part (a) Use absolute temperature in the exact formula, Eq. (2.27): p pa 1 (0.00650 K/m)(5000 m) 288.16 K 101,350(0.52388) v | | e-Text Main Menu | 5.26 (101,350 Pa)(0.8872)5.26 54,000 Pa This is the standard-pressure result given at z v 68 Ans. (a) 5000 m in Table A.6. Textbook Table of Contents | Study Guide 2.3 Hydrostatic Pressure Distributions 50 40 40 Altitude z, km 60 50 30 20 20.1 km –56.5°C Altitude z, km 60 Part (b) 30 Eq. (2.24) Eq. (2.27) 10 Eq. (2.26) Troposphere 0 1.20 kPa 20 11.0 km 10 Fig. 2.7 Temperature and pressure distribution in the U.S. standard atmosphere. (From Ref. 1.) 69 – 60 – 40 101.33 kPa 15°C – 20 Temperature, °C 0 +20 40 80 Pressure, kPa 0 120 If the atmosphere were isothermal at 288.16 K, Eq. (2.24) would apply: pa exp p gz RT (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 Pa Ans. (b) This is 11 percent higher than the exact result. The isothermal formula is inaccurate in the troposphere. Is the Linear Formula Adequate for Gases? The linear approximation from Eq. (2.20) or (2.21), p z, is satisfactory for liquids, which are nearly incompressible. It may be used even over great depths in the ocean. For gases, which are highly compressible, it is valid only over moderate changes in altitude. The error involved in using the linear approximation (2.21) can be evaluated by expanding the exact formula (2.27) into a series 1 Bz T0 n 1 n Bz T0 n(n 1) Bz 2! T0 2 (2.28) where n g/(RB). Introducing these first three terms of the series into Eq. (2.27) and rearranging, we obtain | v v p | e-Text Main Menu | pa az 1 1 Bz T0 n Textbook Table of Contents 2 | Study Guide (2.29) 70 Chapter 2 Pressure Distribution in a Fluid Thus 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 if z (n 2T0 1)B 20,800 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 Manometry From 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 liquids or 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 in the formula as we move from one fluid to another. Figure 2.8 illustrates the use of the formula with a column of multiple fluids. The pressure change through each fluid is calculated 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 p cancel, and we have, for the example of Fig. 2.8, p5 p1 0(z2 z1) w(z3 z2) G(z4 z3) M(z5 z4) (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 lightest on top to the heaviest at bottom. This is the only stable configuration. If we attempt to layer them in any other manner, the fluids will overturn and seek the stable arrangement. A Memory Device: Up Versus Down The 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 by simply having the pressure increase downward and decrease upward. Thus they use the following mnemonic, or memory, device, first suggested to the writer by Professor John z = z1 z2 z z3 z4 | v v Fig. 2.8 Evaluating pressure changes through a column of multiple fluids. z5 | Known pressure p1 Oil, ρo p2 – p1 = – ρog(z 2 – z1) Water, ρw p3 – p2 = – ρw g(z 3 – z 2) Glycerin, ρG Mercury, ρM e-Text Main Menu | p4 – p3 = – ρG g(z 4 – z 3) p5 – p4 = – ρM g(z 5 – z 4) Sum = p5 – p1 Textbook Table of Contents | Study Guide 2.4 Application to Manometry 71 Open, pa zA, pA A Jump across z1, p1 Fig. 2.9 Simple open manometer for measuring pA relative to atmospheric pressure. z 2 , p2 ≈ pa ρ1 p = p1 at z = z1 in fluid 2 ρ2 Foss of Michigan State University: pdown pup 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 moving down or up. For example, Eq. (2.31) could be rewritten in the following “multiple increase” mode: p5 p1 0z1 z2 wz2 z3 Gz3 z4 Mz4 z5 That is, keep adding on pressure increments as you move down through the layered fluid. A different application is a manometer, which involves both “up” and “down” calculations. Figure 2.9 shows a simple open manometer for measuring pA in a closed chamber relative to atmospheric pressure pa, in other words, measuring the gage pressure. The chamber fluid 1 is combined with a second fluid 2, perhaps for two reasons: (1) to protect the environment from a corrosive chamber fluid or (2) because a heavier fluid 2 will keep z2 small and the open tube can be shorter. One can, of course, apply the basic 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 use Eq. (2.32) “up” to level z2: pA 1zA z1 2z1 z2 p2 patm (2.33) The physical reason that we can “jump across” at section 1 in that a continuous length of the same fluid connects these two equal elevations. The hydrostatic relation (2.20) requires this equality as a form of Pascal’s law: Any two points at the same elevation in a continuous mass of the same static fluid will be at the same pressure. This idea of jumping across to equal pressures facilitates multiple-fluid problems. EXAMPLE 2.3 | v v The 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 Guide 72 Chapter 2 Pressure Distribution in a Fluid Flow device (a) (b) L h 1 2 E2.3 plication is to measure pressure change across a flow device, as shown. Derive a formula for the pressure difference pa pb in terms of the system parameters in Fig. E2.3. Solution Using our “up-down” concept as in Eq. (2.32), start at (a), evaluate pressure changes around the U-tube, and end up at (b): pa 1gL or pa 1gh pb 2gh ( 2 1gL pb Ans. 1)gh The measurement only includes h, the manometer reading. Terms involving L drop out. Note the appearance 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 both fluids 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 but rather 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 the ρ3 z 2, p2 zA, pA Jump across z 2, p2 ρ1 A B z1, p1 | v v Fig. 2.10 A complicated multiplefluid manometer to relate pA to pB. This system is not especially practical but makes a good homework or examination problem. Jump across z1, p1 z 3, p3 Jump across z 3, p3 ρ2 ρ4 | e-Text Main Menu | Textbook Table of Contents | Study Guide zB, pB 2.4 Application to Manometry 73 difference 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 same fluid. Thus, in Fig. 2.10, we compute four pressure differences while making three jumps: pA pB (pA p1) 1(zA (p1 p2) z1) 2(z1 (p2 p3) z2) (p3 3(z2 pB) z3) 4(z3 zB) (2.34) The intermediate pressures p1,2,3 cancel. It looks complicated, but really it is merely sequential. 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 2.4 Pressure gage B is to measure the pressure at point A in a water flow. If the pressure at B is 87 kPa, estimate the pressure at A, in kPa. Assume all fluids are at 20°C. See Fig. E2.4. SAE 30 oil Gage B 6 cm Mercury A 5 cm Water flow 11 cm 4 cm E2.4 Solution First list the specific weights from Table 2.1 or Table A.3: water 9790 N/m3 mercury 133,100 N/m3 oil 8720 N/m3 Now proceed from A to B, calculating the pressure change in each fluid and adding: pA or W( pA z)W M( 3 (9790 N/m )( pA 489.5 Pa z)M O( 0.05 m) 9317 Pa z)O pB (133,100 N/m3)(0.07 m) 523.2 Pa pB (8720 N/m3)(0.06 m) 87,000 Pa 2 where we replace N/m by its short name, Pa. The value zM 0.07 m is the net elevation change in the mercury (11 cm 4 cm). Solving for the pressure at point A, we obtain pA 96,351 Pa 96.4 kPa | v v The intermediate six-figure result of 96,351 Pa is utterly fatuous, since the measurements cannot be made that accurately. | e-Text Main Menu | Textbook Table of Contents | Study Guide Ans. 74 Chapter 2 Pressure Distribution in a Fluid In making these manometer calculations we have neglected the capillary-height changes due to surface tension, which were discussed in Example 1.9. These effects cancel 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 be made or the effect can be made negligible by using large-bore ( 1 cm) tubes. 2.5 Hydrostatic Forces on Plane Surfaces A 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 in the fluid, Eq. (2.20) applies and the pressure on any submerged surface varies linearly with depth. For a plane surface, the linear stress distribution is exactly analogous to combined bending and compression of a beam in strength-of-materials theory. The hydrostatic problem thus reduces to simple formulas involving the centroid and moments of 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, so that the depth varies over the panel surface. If h is the depth to any element area dA of the plate, from Eq. (2.20) the pressure there is p pa h. To derive formulas involving the plate shape, establish an xy coordinate system in the plane of the plate with the origin at its centroid, plus a dummy coordinate down from the surface in the plane of the plate. Then the total hydrostatic force on one side of the plate is given by F p dA (pa h) dA paA h dA The remaining integral is evaluated by noticing from Fig. 2.11 that h p = pa Free surface θ h (x, y) hCG Resultant force: F = pCG A ξ= h sin θ y Side view CG x | v v Fig. 2.11 Hydrostatic force and center of pressure on an arbitrary plane surface of area A inclined at an angle below the free surface. dA = dx dy CP Plan view of arbitrary plane surface | e-Text Main Menu | Textbook Table of Contents | Study Guide (2.35) sin and, 2.5 Hydrostatic Forces on Plane Surfaces 75 by definition, the centroidal slant distance from the surface to the plate is 1 A CG Therefore, since dA (2.36) is constant along the plate, Eq. (2.35) becomes F paA sin Finally, unravel this by noticing that the surface to the plate centroid. Thus F pa A dA CG hCG A paA sin CGA (2.37) sin hCG, the depth straight down from ( pa hCG)A pCG A (2.38) The force on one side of any plane submerged surface in a uniform fluid equals the pressure at the plate centroid times the plate area, independent of the shape of the plate or 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 and bending of a beam of the same cross section. It follows that the “bending’’ portion of the 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 plate area. This is the result of Eq. (2.38). However, to balance the bending-moment portion of the stress, the resultant force F does not act through the centroid but below it toward the high-pressure side. Its line of 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 about the centroid and equate to the moment of the resultant F. To compute yCP, we equate FyCP The term yp dA y( pa sin ) dA sin y dA pay dA vanishes by definition of centroidal axes. Introducing Pressure distribution pav = pCG p (x, y) | v v Fig. 2.12 The hydrostatic-pressure force on a plane surface is equal, regardless of its shape, to the resultant of the three-dimensional linear pressure distribution on that surface F pCGA. | e-Text Main Menu Centroid of the plane surface | Textbook Table of Contents | Arbitrary plane surface of area A Study Guide (2.39) CG y, 76 Chapter 2 Pressure Distribution in a Fluid we obtain sin FyCP CG y dA y2 dA sin Ixx (2.40) where again y dA 0 and Ixx is the area moment of inertia of the plate area about its centroidal x axis, computed in the plane of the plate. Substituting for F gives the result Ixx yCP sin (2.41) pCGA The negative sign in Eq. (2.41) shows that yCP is below the centroid at a deeper level and, unlike F, depends upon angle . If we move the plate deeper, yCP approaches the centroid because every term in Eq. (2.41) remains constant except pCG, which increases. The determination of xCP is exactly similar: xp dA FxCP sin x[pa xy dA ( CG y) sin ] dA sin Ixy (2.42) where Ixy is the product of inertia of the plate, again computed in the plane of the plate. Substituting for F gives Ixy xCP sin (2.43) pCGA For 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 0 and the center of pressure lies directly below the centroid on the y axis. L 2 y A = bL x Ixx = L 2 A = π R2 y bL3 x 12 R Ix y = 0 R Ixx = π R4 4 Ix y = 0 b 2 b 2 (a) (b) s y Ixx = x L 3 | v v Fig. 2.13 Centroidal moments of inertia for various cross sections: (a) rectangle, (b) circle, (c) triangle, and (d) semicircle. b 2 b 2 e-Text Main Menu bL3 36 Ixx = 0.10976R 4 y Ix y = 0 b (b – 2 s) L 2 Ix y = 72 x R (c ) | 2 A = πR 2 A = bL 2 2L 3 R 4R 3π (d ) | Textbook Table of Contents | Study Guide 2.5 Hydrostatic Forces on Plane Surfaces 77 In most cases the ambient pressure pa is neglected because it acts on both sides of the plate; 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 pCG hCG, and the center of pressure becomes independent of specific weight Gage-Pressure Formulas F hCGA Ixx sin hCGA yCP xCP Ixy sin hCGA (2.44) Figure 2.13 gives the area and moments of inertia of several common cross sections for use with these formulas. EXAMPLE 2.5 The gate in Fig. E2.5a is 5 ft wide, is hinged at point B, and rests against a smooth wall at point A. Compute (a) the force on the gate due to seawater pressure, (b) the horizontal force P exerted by the wall at point A, and (c) the reactions at the hinge B. Wall pa Seawater: 64 lbf/ft 3 15 ft A pa Gate 6 ft θ B E2.5a 8 ft Hinge Solution Part (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 is F Part (b) pCGA hCGA (64 lbf/ft3)(12 ft)(50 ft2) 38,400 lbf First 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, hence Ixy 0 and Ixx bL3 12 (5 ft)(10 ft)3 12 417 ft4 The distance l from the CG to the CP is given by Eq. (2.44) since pa is neglected. v v l | | Ans. (a) e-Text Main Menu | yCP Ixx sin hCGA 6 (417 ft4)( 10 ) (12 ft)(50 ft2) Textbook Table of Contents | 0.417 ft Study Guide Chapter 2 Pressure Distribution in a Fluid A P F 5 ft CG l θ B Bx CP L = 10 ft Bz E2.5b The distance from point B to force F is thus 10 terclockwise about B gives PL sin F(5 Part (c) l l) P(6 ft) P or 5 4.583 ft. Summing moments coun- (38,400 lbf)(4.583 ft) 0 29,300 lbf Ans. (b) With F and P known, the reactions Bx and Bz are found by summing forces on the gate Fx 0 Bx F sin or Bx Fz 0 Bz F cos or Bz P Bx 38,400(0.6) 29,300 6300 lbf Bz 38,400(0.8) 30,700 lbf Ans. (c) This example should have reviewed your knowledge of statics. EXAMPLE 2.6 A 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. pa Oil: ρ = 800 kg/m 3 5m 30° 11 m 4m 6m pa CG CP 4m 8m v | 2m 4m E2.6 v 78 | e-Text Main Menu | Textbook Table of Contents | Study Guide 2.6 Hydrostatic Forces on Curved Surfaces 79 Solution Part (a) The triangle has properties given in Fig. 2.13c. The centroid is one-third up (4 m) and one-third over (2 m) from the lower left corner, as shown. The area is 1 2 (6 m)(12 m) 36 m2 The moments of inertia are bL3 36 Ixx and b(b Ixy (6 m)(12 m)3 36 2s)L2 72 The depth to the centroid is hCG F 2.54 Part (b) (6 m)[6 m 5 ghCGA 288 m4 4 2(6 m)](12 m)2 72 72 m4 9 m; thus the hydrostatic force from Eq. (2.44) is (800 kg/m3)(9.807 m/s2)(9 m)(36 m2) 106 (kg m)/s2 2.54 106 N 2.54 MN Ans. (a) The CP position is given by Eqs. (2.44): yCP Ixx sin hCGA (288 m4)(sin 30°) (9 m)(36 m2) 0.444 m xCP Ixy sin hCGA ( 72 m4)(sin 30°) (9 m)(36 m2) 0.111 m Ans. (b) The resultant force F 2.54 MN acts through this point, which is down and to the right of the centroid, as shown in Fig. E2.6. 2.6 Hydrostatic Forces on Curved Surfaces The resultant pressure force on a curved surface is most easily computed by separating it into horizontal and vertical components. Consider the arbitrary curved surface sketched in Fig. 2.14a. The incremental pressure forces, being normal to the local area element, vary in direction along the surface and thus cannot be added numerically. We Wair d Curved surface projection onto vertical plane FV | v v Fig. 2.14 Computation of hydrostatic force on a curved surface: (a) submerged curved surface; (b) free-body diagram of fluid above the curved surface. | e-Text Main Menu F1 b W2 FH a FV (a) | Textbook Table of Contents (b) | F1 W1 c FH FH e Study Guide FH Chapter 2 Pressure Distribution in a Fluid could sum the separate three components of these elemental pressure forces, but it turns out 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 by the 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 in static equilibrium. On the upper part of the column bcde, the horizontal components F1 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 the desired 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 curved surface. This is a general rule and simplifies the analysis: The horizontal component of force on a curved surface equals the force on the plane area formed by the projection of the curved surface onto a vertical plane normal to the 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 that FV W1 W2 Wair (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 magnitude and 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 particularly vexing shape. ;; ;; ;; EXAMPLE 2.7 A dam has a parabolic shape z/z0 (x/x0)2 as shown in Fig. E2.7a, with x0 10 ft and z0 24 ft. The fluid is water, 62.4 lbf/ft3, and atmospheric pressure may be omitted. Compute the pa = 0 lbf/ft2 gage FV z FH z0 CP x x0 E2.7a | v v 80 | e-Text Main Menu (( x z = z0 x 0 | 2 Textbook Table of Contents | Study Guide 2.6 Hydrostatic Forces on Curved Surfaces 81 forces FH and FV on the dam and the position CP where they act. The width of the dam is 50 ft. Solution The vertical projection of this curved surface is a rectangle 24 ft high and 50 ft wide, with its centroid halfway down, or hCG 12 ft. The force FH is thus FH (62.4 lbf/ft3)(12 ft)(24 ft)(50 ft) hCGAproj 899,000 lbf 103 lbf 899 Ans. The line of action of FH is below the centroid by an amount 1 12 Ixx sin hCGAproj yCP (50 ft)(24 ft)3(sin 90°) (12 ft)(24 ft)(50 ft) 4 ft Thus 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 the curved surface. The geometric properties of a parabola are shown in Fig. E2.7b. The weight of this amount of water is (62.4 lbf/ft3)( 2 )(10 ft)(24 ft)(50 ft) 3 ( 2 x0z0b) 3 FV 499,000 lbf 499 103 lbf Ans. z0 Area = 3z0 5 2 x0z 0 3 FV Parabola 0 E2.7b x0 = 10 ft 3x 0 8 This 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 is F 2 (FH 2 FV)1/2 [(499)2 (899)2]1/2 1028 103 lbf As seen in Fig. E2.7c, this force acts down and to the right at an angle of 29° tan 1 499 . The 899 force 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 at xCP 5.43 ft zCP 7.07 ft Ans. | v v This definition of CP is rather artificial, but this is an unavoidable complication of dealing with a curved surface. | e-Text Main Menu | Textbook Table of Contents | Study Guide 82 Chapter 2 Pressure Distribution in a Fluid z Resultant = 1028 × 103 1bf acts along z = 10.083 – 0.5555 x 3.75 ft 499 899 Parabola z = 0.24x2 29° CG CP 7.07 ft 8 ft E2.7c 0 2.7 Hydrostatic Forces in Layered Fluids x 5.43 ft The formulas for plane and curved surfaces in Secs. 2.5 and 2.6 are valid only for a fluid 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 and moments. Consider the slanted plane surface immersed in a two-layer fluid in Fig. 2.15. The slope of the pressure distribution becomes steeper as we move down into the denser z F 1= p CG1 Plane surface A1 z=0 p a ρ1 < ρ2 Fluid 1 p = p – ρ1gz a z 1, p1 F2 = p p1 = p – ρ1gz1 a A CG 2 2 ρ2 Fluid 2 z 2 , p2 p = p1 – ρ2 g(z – z 1) | v v Fig. 2.15 Hydrostatic forces on a surface immersed in a layered fluid must be summed in separate pieces. p2 = p1 – ρ 2 g(z 2 – z 1) | e-Text Main Menu | Textbook Table of Contents | Study Guide 2.7 Hydrostatic Forces in Layered Fluids 83 second layer. The total force on the plate does not equal the pressure at the centroid times the plate area, but the plate portion in each layer does satisfy the formula, so that we can sum forces to find the total: F Fi pCGi Ai (2.46) Similarly, the centroid of the plate portion in each layer can be used to locate the center of pressure on that portion yCPi ig sin i Ixxi pCGi Ai ig xCPi sin i Ixyi pCGi Ai (2.47) These formulas locate the center of pressure of that particular Fi with respect to the centroid of that particular portion of plate in the layer, not with respect to the centroid of the entire plate. The center of pressure of the total force F Fi can then be found by summing moments about some convenient point such as the surface. The following example will illustrate. EXAMPLE 2.8 A 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 on the right-hand side of the tank. Solution Part (a) Divide the end panel into three parts as sketched in Fig. E2.8, and find the hydrostatic pressure at the centroid of each part, using the relation (2.38) in steps as in Fig. E2.8: pa = 0 Oi z=0 7 ft 5.0 4 ft (1) l: 5 11 ft lbf /ft 3 8 ft Wa ter (62 .4) Me rcu ry 6 ft (84 6) 16 ft (2) 4 ft (3) E2.8 v | e-Text Main Menu | (55.0)(8) 62.4(3) 627 lbf/ft2 pCG3 | (55.0 lbf/ft3)(4 ft) pCG2 v PCG1 (55.0)(8) 62.4(6) 846(2) Textbook Table of Contents 220 lbf/ft2 | 2506 lbf/ft2 Study Guide 84 Chapter 2 Pressure Distribution in a Fluid These pressures are then multiplied by the respective panel areas to find the force on each portion: F1 (220 lbf/ft2)(8 ft)(7 ft) pCG1A1 F2 F3 pCG2 A2 627(6)(7) 2506(4)(7) pCG3A3 26,300 lbf 70,200 lbf F Part (b) 12,300 lbf Fi 108,800 lbf Ans. (a) Equations (2.47) can be used to locate the CP of each force Fi, noting that 90° and sin 1 for all parts. The moments of inertia are Ixx1 (7 ft)(8 ft)3/12 298.7 ft4, Ixx2 7(6)3/12 126.0 ft4, and Ixx3 7(4)3/12 37.3 ft4. The centers of pressure are thus at 1gIxx 1 yCP1 F1 62.4(126.0) 26,300 yCP2 (55.0 lbf/ft3)(298.7 ft4) 12,300 lbf 0.30 ft 1.33 ft 846(37.3) 70,200 yCP3 0.45 ft This locates zCP1 4 1.33 5.33 ft, zCP2 11 0.30 11.30 ft, and zCP3 16 0.45 16.45 ft. Summing moments about the surface then gives FizCPi or 12,300( 5.33) or 26,300( 11.30) zCP 1,518,000 108,800 FzCP 70,200( 16.45) 108,800zCP 13.95 ft Ans. (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 Stability The same principles used to compute hydrostatic forces on surfaces can be applied to the net pressure force on a completely submerged or floating body. The results are the two 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 the weight 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 body lies 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 force FB FV (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 vertical slices through the immersed body: | v v FB | body (p2 e-Text Main Menu | p1) dAH (z2 z1) dAH Textbook Table of Contents ( )(body volume) (2.49) | Study Guide 2.8 Buoyancy and Stability FV (1) Horizontal elemental area d AH p1 Surface 1 85 z1 – z 2 Fig. 2.16 Two different approaches to the buoyant force on an arbitrary immersed body: (a) forces on upper and lower curved surfaces; (b) summation of elemental verticalpressure forces. Surface 2 p2 FV (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 through which 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 of the body’s own material, which may have variable density. Equation (2.49) can be generalized to a layered fluid (LF) by summing the weights of each layer of density i displaced by the immersed body: (FB)LF ig(displaced volume)i (2.50) Each displaced layer would have its own center of volume, and one would have to sum moments 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 gases also exert buoyancy on any body immersed in them. For example, human beings have an average specific weight of about 60 lbf/ft3. We may record the weight of a person as 180 lbf and thus estimate the person’s total volume as 3.0 ft3. However, in so doing we 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 the controlling factor in the design. Also, many flow phenomena, e.g., natural convection of heat and vertical mixing in the ocean, are strongly dependent upon seemingly small buoyant forces. Floating bodies are a special case; only a portion of the body is submerged, with the remainder poking up out of the free surface. This is illustrated in Fig. 2.17, where the shaded portion is the displaced volume. Equation (2.49) is modified to apply to this smaller volume | v v FB | e-Text Main Menu | ( )(displaced volume) Textbook Table of Contents floating-body weight | Study Guide (2.51) 86 Chapter 2 Pressure Distribution in a Fluid Neglect the displaced air up here. CG W FB B Fig. 2.17 Static equilibrium of a floating body. (Displaced volume) × ( γ of fluid) = body weight Not only does the buoyant force equal the body weight, but also they are collinear since there can be no net moments for static equilibrium. Equation (2.51) is the mathematical equivalent of Archimedes’ law 2, previously stated. EXAMPLE 2.9 A 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? Solution A free-body diagram of the submerged block (see Fig. E2.9) shows a balance between the apparent weight, the buoyant force, and the actual weight 60 lbf Fz or FB FB 40 lbf 0 60 FB 100 3 (62.4 lbf/ft )(block volume, ft3) Solving gives the volume of the block as 40/62.4 the block is 0.641 ft3. Therefore the specific weight of W = 100 lbf E2.9 block 100 lbf 0.641 ft3 156 lbf/ft3 Ans. Occasionally, a body will have exactly the right weight and volume for its ratio to equal the specific weight of the fluid. If so, the body will be neutrally buoyant and will remain at rest at any point where it is immersed in the fluid. Small neutrally buoyant particles are sometimes used in flow visualization, and a neutrally buoyant body called a Swallow float [2] is used to track oceanographic currents. A submarine can achieve positive, neutral, or negative buoyancy by pumping water in or out of its ballast tanks. Stability | v v A 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, like a pencil balanced upon its point. The least disturbance will cause it to seek another equilibrium position which is stable. Engineers must design to avoid floating instabil- | e-Text Main Menu | Textbook Table of Contents | Study Guide 2.8 Buoyancy and Stability Small ∆θ disturbance angle Line of symmetry 87 Small disturbance angle ∆θ M G G G W FB W Fig. 2.18 Calculation of the metacenter M of the floating body shown in (a). Tilt the body a small angle . Either (b) B moves far out (point M above G denotes stability); or (c) B moves slightly (point M below G denotes instability). W FB B' B' B Either (a) M FB Restoring moment (b) or Overturning moment (c) ity. The only way to tell for sure whether a floating position is stable is to “disturb’’ the 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. Such calculations for arbitrary floating bodies have been honed to a fine art by naval architects [3], 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 floating body. The steps are as follows: 1. The basic floating position is calculated from Eq. (2.51). The body’s center of mass G and center of buoyancy B are computed. 2. The body is tilted a small angle , and a new waterline is established for the body 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 at a point M, called the metacenter, which is independent of for 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 increases with increasing MG. Thus the metacentric height is a property of the cross section for the given weight, and its value gives an indication of the stability of the body. For a body of varying cross section and draft, such as a ship, the computation of the metacenter can be very involved. | v v Stability Related to Waterline Area | Naval architects [3] have developed the general stability concepts from Fig. 2.18 into a simple computation involving the area moment of inertia of the waterline area about the axis of tilt. The derivation assumes that the body has a smooth shape variation (no discontinuities) 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 small angle then submerges small wedge Obd and uncovers an equal wedge cOa, as shown. e-Text Main Menu | Textbook Table of Contents | Study Guide 88 Chapter 2 Pressure Distribution in a Fluid y Original waterline area G Variable-width L(x) into paper dA = x tan M dx c a b O BG Fig. 2.19 A floating body tilted through a small angle . The movement x of the center of buoyancy B is related to the waterline area moment of inertia. G x d B x e Tilted floating body The new position B of the center of buoyancy is calculated as the centroid of the submerged portion aObde of the body: x υabOde x dυ x dυ cOdea Obd 0 x dυ 0 cOa x L (x tan dx) Obd x (L dA) Obd xL ( x tan x (L dA) cOa dx) x2 dAwaterline tan cOa IO tan waterline where IO is the area moment of inertia of the waterline footprint of the body about its tilt axis O. The first integral vanishes because of the symmetry of the original submerged portion cOdea. The remaining two “wedge” integrals combine into IO when we notice that L dx equals an element of waterline area. Thus we determine the desired distance from M to B: x MB IO υsubmerged MG GB or MG IO υsub GB (2.52) The engineer would determine the distance from G to B from the basic shape and design of the floating body and then make the calculation of IO and the submerged volume υsub. If the metacentric height MG is positive, the body is stable for small disturbances. Note that if GB is negative, that is, B is above G, the body is always stable. EXAMPLE 2.10 | v v A barge has a uniform rectangular cross section of width 2L and vertical draft of height H, as in Fig. E2.10. Determine (a) the metacentric height for a small tilt angle and (b) the range of ratio 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 Guide 2.9 Pressure Distribution in Rigid-Body Motion 89 G O G L E2.10 H B L Solution If the barge has length b into the paper, the waterline area, relative to tilt axis O, has a base b and a height 2L; therefore, IO b(2L)3/12. Meanwhile, υsub 2LbH. Equation (2.52) predicts 8bL3/12 H H IO L2 MG GB Ans. (a) 2LbH 2 2 υsub 3H The barge can thus be stable only if L2 3H2/2 or 2L 2.45H Ans. (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 buoyant body 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 may float 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, not necessarily 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 plane of symmetry vertical. After death, this position is unstable and they float with their flat sides up. Giant icebergs may overturn after becoming unstable when their shapes change due to underwater melting. Iceberg overturning is a dramatic, rarely seen event. 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, indicating that it has undergone further calving. Icebergs are frozen fresh, bubbly, glacial water of average density 900 kg/m3. Thus, when an iceberg is floating in seawater, whose average density is 1025 kg/m3, approximately 900/1025, or seven-eighths, of its volume lies below the water. | v v 2.9 Pressure Distribution in Rigid-Body Motion | In rigid-body motion, all particles are in combined translation and rotation, and there is no relative motion between particles. With no relative motion, there are no strains e-Text Main Menu | Textbook Table of Contents | Study Guide 90 Chapter 2 Pressure Distribution in a Fluid Fig. 2.20 A North Atlantic iceberg formed by calving from a Greenland glacier. These, and their even larger Antarctic sisters, are the largest 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 balance between pressure, gravity, and particle acceleration ∇p (g a) (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 case of 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 by2 V V0 r0 where is 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: a dV0 dt ( r0) d dt r0 (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 | v v 2 | For a more detailed derivation of rigid-body motion, see Ref. 4, Sec. 2.7. e-Text Main Menu | Textbook Table of Contents | Study Guide 2.9 Pressure Distribution in Rigid-Body Motion 91 P 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 unless restrained by confining walls for a long time. For example, suppose a tank of water is in a car which starts a constant acceleration. The water in the tank would begin to slosh about, and that sloshing would damp out very slowly until finally the particles of water would be in approximately rigid-body acceleration. This would take so long that 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 is an example where the water in the tank will reach uniform acceleration rapidly. EXAMPLE 2.11 A 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. Solution Being unsupported in this condition, the water particles tend to fall downward as a rigid hunk of fluid. In free fall with no drag, the downward acceleration is a g. Thus Eq. (2.53) for this situation gives ∇p (g g) 0. The pressure in the water is thus constant everywhere and equal 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 Acceleration In this general case of uniform rigid-body acceleration, Eq. (2.53) applies, a having the 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 rate of increase of p. The surfaces of constant pressure must be perpendicular to this and are thus tilted at a downward angle such that ax 1 tan g (2.55) az z ax a az x θ = tan –1 θ –a ax g + az Fluid at rest g p µg – a ∆ | v v Fig. 2.21 Tilting of constantpressure surfaces in a tank of liquid in rigid-body acceleration. | e-Text Main Menu | az ax S p2 p = p1 p3 Textbook Table of Contents | Study Guide Chapter 2 Pressure Distribution in a Fluid One of these tilted lines is the free surface, which is found by the requirement that the fluid 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 by dp ds G where G [a2 x (g az)2]1/2 (2.56) These results are independent of the size or shape of the container as long as the fluid is continuously connected throughout the container. EXAMPLE 2.12 A drag racer rests her coffee mug on a horizontal tray while she accelerates at 7 m/s2. The mug is 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 the gage pressure in the corner at point A if the density of coffee is 1010 kg/m3. Solution Part (a) The free surface tilts at the angle az 0 and standard gravity, given by Eq. (2.55) regardless of the shape of the mug. With 7.0 ax tan 1 35.5° 9.81 g If 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. tan 1 3 cm ∆z θ 7 cm ax = 7 m/s2 A 3 cm E2.12 Thus the deflection at the left side of the mug is z (3 cm)(tan ) 2.14 cm Ans. (a) This is less than the 3-cm clearance available, so the coffee will not spill unless it was sloshed during the start-up of acceleration. Part (b) When at rest, the gage pressure at point A is given by Eq. (2.20): pA | v v 92 | g(zsurf e-Text Main Menu zA) | (1010 kg/m3)(9.81 m/s2)(0.07 m) Textbook Table of Contents 694 N/m2 | 694 Pa Study Guide 2.9 Pressure Distribution in Rigid-Body Motion During acceleration, Eq. (2.56) applies, with G [(7.0)2 (9.81)2]1/2 tance ∆s down the normal from the tilted surface to point A is s (7.0 2.14)(cos ) 93 12.05 m/s2. The dis- 7.44 cm Thus the pressure at point A becomes pA Gs 1010(12.05)(0.0744) 906 Pa Ans. (b) which is an increase of 31 percent over the pressure when at rest. Rigid-Body Rotation As a second special case, consider rotation of the fluid about the z axis without any translation, as sketched in Fig. 2.22. We assume that the container has been rotating long enough at constant for the fluid to have attained rigid-body rotation. The fluid acceleration 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 by k r0 irr (2.57) Then the acceleration is given by ( r0) r 2 ir (2.58) as marked in the figure, and Eq. (2.53) for the force balance becomes ∇p p r ir k p z (g a) ( gk r 2 ir) (2.59) Equating like components, we find the pressure field by solving two first-order partial differential equations p r r p z 2 (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 of z, k r, ir p = pa Ω a = –r Ω 2 ir –a Still-water level | v v Fig. 2.22 Development of paraboloid constant-pressure surfaces in a fluid in rigid-body rotation. The dashed line along the direction of maximum pressure increase is an exponential curve. | e-Text Main Menu p = p1 Axis of rotation g g–a p2 p3 | Textbook Table of Contents | Study Guide 94 Chapter 2 Pressure Distribution in a Fluid (2.60) are known functions of r and z. One can proceed as follows: Integrate the first equation “partially,’’ i.e., holding z constant, with respect to r. The result is 1 2 p r2 2 f(z) (2.61) † where the “constant’’ of integration is actually a function f(z). Now differentiate this with respect to z and compare with the second relation of (2.60): p z or 0 f (z) f(z) z C (2.62a) where C is a constant. Thus Eq. (2.61) now becomes p const 1 2 z r2 2 (2.62b) This is the pressure distribution in the fluid. The value of C is found by specifying the pressure at one point. If p p0 at (r, z) (0, 0), then C p0. The final desired distribution is p p0 z 1 2 r2 2 (2.63) The pressure is linear in z and parabolic in r. If we wish to plot a constant-pressure surface, say, p p1, Eq. (2.63) becomes z p0 p1 r2 2 2g a br2 (2.64) Thus the surfaces are paraboloids of revolution, concave upward, with their minimum point 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 is found by conserving the volume of fluid. For a noncircular container with the axis of rotation off-center, as in Fig. 2.22, a lot of laborious mensuration is required, and a single 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 is Still water level h 2 Volume = π 2 R 2h h 2 22 h= Ω R 2g Ω Fig. 2.23 Determining the freesurface position for rotation of a cylinder of fluid about its central axis. R R | v v † This is because f(z) vanishes when differentiated with respect to r. If you don’t see this, you should review your calculus. | e-Text Main Menu | Textbook Table of Contents | Study Guide 2.9 Pressure Distribution in Rigid-Body Motion 95 one-half the base area times its height, the still-water level is exactly halfway between the high and low points of the free surface. The center of the fluid drops an amount 22 h/2 R /(4g), and the edges rise an equal amount. EXAMPLE 2.13 The 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 which will cause the coffee to just reach the lip of the cup and (b) the gage pressure at point A for this condition. Solution Part (a) The cup contains 7 cm of coffee. The remaining distance of 3 cm up to the lip must equal the distance h/2 in Fig. 2.23. Thus 2 22 (0.03 m)2 R h 0.03 m 4(9.81 m/s2) 2 4g Solving, we obtain 2 Part (b) z 1308 or 36.2 rad/s 345 r/min To compute the pressure, it is convenient to put the origin of coordinates r and z at the bottom of the free-surface depression, as shown in Fig. E2.13. The gage pressure here is p0 0, and point A is at (r, z) (3 cm, 4 cm). Equation (2.63) can then be evaluated pA 0 (1010 kg/m3)(9.81 m/s2)( 0.04 m) 1 2 (1010 3 cm 396 N/m2 kg/m3)(0.03 m)2(1308 rad2/s2) 594 N/m2 990 Pa This is about 43 percent greater than the still-water pressure pA 0 Ans. (a) Ans. (b) 694 Pa. r 7 cm Here, as in the linear-acceleration case, it should be emphasized that the paraboloid pressure distribution (2.63) sets up in any fluid under rigid-body rotation, regardless of the shape or size of the container. The container may even be closed and filled with fluid. It is only necessary that the fluid be continuously interconnected throughout the container. The following example will illustrate a peculiar case in which one can visualize an imaginary free surface extending outside the walls of the container. Ω A 3 cm 3 cm E2.13 EXAMPLE 2.14 | v v A U-tube with a radius of 10 in and containing mercury to a height of 30 in is rotated about its center 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 Guide 96 Chapter 2 Pressure Distribution in a Fluid z Solution 10 in B Convert the angular velocity to radians per second: r 0 (180 r/min) 30 in 2 rad/r 60 s/min 18.85 rad/s From Table 2.1 we find for mercury that 846 lbf/ft3 and hence 846/32.2 26.3 slugs/ft3. At this high rotation rate, the free surface will slant upward at a fierce angle [about 84°; check this from Eq. (2.64)], but the tubing is so thin that the free surface will remain at approximately the 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): Ω A pB 2116 lbf/ft2 or Imaginary free surface C C 1 2 0 2116 (26.3 slugs/ft3)( 10 ft)2(18.85 rad/s)2 12 1129 lbf/ft2 3245 We then obtain pA by evaluating Eq. (2.63) at (r, z) (0, 30 in): E2.14 pA 1129 (846 lbf/ft3)( 30 12 ft) 1129 2115 986 lbf/ft2 Ans. 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 be 22 (18.85)2( 10 )2 R 12 h 3.83 ft 46 in 2(32.2) 2g Thus pA is about 16 inHg below atmospheric pressure, or about 16 (846) 1128 lbf/ft2 below 12 pa 2116 lbf/ft2, which checks with the answer above. When the tube is at rest, pA 2116 846( 30 12 4231 lbf/ft2 ) Hence rotation has reduced the pressure at point A by 77 percent. Further rotation can reduce pA to near-zero pressure, and cavitation can occur. An interesting by-product of this analysis for rigid-body rotation is that the lines everywhere parallel to the pressure gradient form a family of curved surfaces, as sketched 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): dz dr GL 1 (dz/dr)p 1 const r 2 /g where GL stands for gradient line dz dr | v v or | e-Text Main Menu | g r (2.65) 2 Textbook Table of Contents | Study Guide 2.10 Pressure Measurement 97 Separating the variables and integrating, we find the equation of the pressure-gradient surfaces 2 r z C1 exp g (2.66) Notice that this result and Eq. (2.64) are independent of the density of the fluid. In the absence 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 small particle 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 Measurement Pressure 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 pressure by calibration with a primary device such as a deadweight piston tester. There are many such instruments, both for a static fluid and a moving stream. The instrumentation texts in 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. | v v The 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 instruments in Fig. 2.25. The hole should be normal to the wall, and burrs should be avoided. If the 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 almost undisturbed. 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 for dynamic-pressure measurements. Note that the manometers in Fig. 2.25 are arranged to measure the absolute pressures p1 and p2. If the pressure difference p1 p2 is de- | e-Text Main Menu | Textbook Table of Contents | Study Guide 98 Chapter 2 Pressure Distribution in a Fluid | v v Fig. 2.24 Experimental demonstration with buoyant streamers of the fluid force field in rigid-body rotation: (top) fluid at rest (streamers hang vertically upward); (bottom) rigid-body rotation (streamers are aligned with the direction of maximum pressure gradient). (From Ref. 5, courtesy of R. Ian Fletcher.) | e-Text Main Menu | Textbook Table of Contents | Study Guide 2.10 Pressure Measurement 99 Flow Flow p1 p2 Fig. 2.25 Two types of accurate manometers for precise measurements: (a) tilted tube with eyepiece; (b) micrometer pointer with ammeter detector. (a) (b) sired, a significant error is incurred by subtracting two independent measurements, and it would be far better to connect both ends of one instrument to the two static holes p1 and p2 so that one manometer reads the difference directly. In category 2, elasticdeformation instruments, a popular, inexpensive, and reliable device is the bourdon tube, sketched in Fig. 2.26. When pressurized internally, a curved tube with flattened cross 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 drive electric-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 drive another sensor. A Section AA Bourdon tube A Pointer for dial gage Flattened tube deflects outward under pressure Linkage | v v Fig. 2.26 Schematic of a bourdontube device for mechanical measurement of high pressures. | e-Text Main Menu High pressure | Textbook Table of Contents | Study Guide 100 Chapter 2 Pressure Distribution in a Fluid Fig. 2.27 The fused-quartz, forcebalanced bourdon tube is the most accurate pressure sensor used in commercial applications today. (Courtesy of Ruska Instrument Corporation, Houston, TX.) An interesting variation of Fig. 2.26 is the fused-quartz, forced-balanced bourdon tube, shown in Fig. 2.27, whose deflection is sensed optically and returned to a zero reference 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 most accurate pressure sensors ever devised, with uncertainty of the order of 0.003 percent. The last category, electric-output sensors, is extremely important in engineering because 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 sensor in Fig. 2.28a. The differential pressure deflects the silicon diaphragm and changes the capacitancce of the liquid in the cavity. Note that the cavity has spherical end caps to prevent overpressure damage. In the second type, Fig. 2.28b, strain gages and other 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 deform under pressure such that its natural vibration frequency is proportional to the pressure. An oscillator excites the element’s resonant frequency and converts it into appropriate pressure units. For further information on pressure sensors, see Refs. 7 to 10, 12, and 13. Summary | v v This chapter has been devoted entirely to the computation of pressure distributions and the 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 this manner 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 Guide Cover flange Seal diaphragm High-pressure side Low-pressure side Filling liquid Sensing diaphragm (a) Wire bonding Stitch bonded connections from chip to body plug Strain gages Diffused into integrated silicon chip Fig. 2.28 Pressure sensors with electric output: (a) a silicon diaphragm whose deflection changes the cavity capacitance (Courtesy of Johnson-Yokogawa Inc.); (b) a silicon strain gage which is stressed by applied pressure; (c) a micromachined silicon element which resonates at a frequency proportional to applied pressure. [(b) and (c) are courtesy of Druck, Inc., Fairfield, CT.] Etched cavity Micromachined silicon sensor | v v (Druck, Inc., Fairfield, Connecticut) (b) | Temperature sensor On-chip diode for optimum temperature performance e-Text Main Menu | Textbook Table of Contents (c) | Study Guide 101 102 Chapter 2 Pressure Distribution in a Fluid P2.2 For the two-dimensional stress field shown in Fig. P2.1 suppose that Problems Most of the problems herein are fairly straightforward. More difficult or open-ended assignments are indicated with an asterisk, as in Prob. 2.8. Problems labeled with an EES icon (for example, Prob. 2.62), will benefit from the use of the Engineering Equation Solver (EES), while problems labeled with a disk icon may require the use of a computer. The standard endof-chapter problems 2.1 to 2.158 (categorized in the problem list 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 and D2.2. xx P2.3 P2.4 Problem Distribution Section Topic Problems 2.1, 2.2 2.3 2.3 2.4 2.5 2.6 2.7 2.8 2.8 2.9 2.9 2.10 Stresses; pressure gradient; gage pressure Hydrostatic pressure; barometers The atmosphere Manometers; multiple fluids Forces on plane surfaces Forces on curved surfaces Forces in layered fluids Buoyancy; Archimedes’ principles Stability of floating bodies Uniform acceleration Rigid-body rotation Pressure measurements 2.1 – 2.6 2.7 – 2.23 2.24 – 2.29 2.30 – 2.47 2.48 – 2.81 2.82 – 2.100 2.101 – 2.102 2.103 – 2.126 2.127 – 2.136 2.137 – 2.151 2.152 – 2.158 None P2.6 P2.1 For the two-dimensional stress field shown in Fig. P2.1 it is found that P2.8 xx 3000 lbf/ft2 2000 lbf/ft2 yy xy P2.5 P2.7 500 lbf/ft2 Find the shear and normal stresses (in lbf/ft2) acting on plane AA cutting through the element at a 30° angle as shown. σyy σyx = σxy *P2.9 A σxx A 30° σxx P2.10 σxy = σyx σyy | v v P2.1 | e-Text Main Menu | 2000 lbf/ft2 yy 3000 lbf/ft2 n(AA) 2500 lbf/ft2 Compute (a) the shear stress xy and (b) the shear stress on plane AA. Derive Eq. (2.18) by using the differential element in Fig. 2.2 with z “up,’’ no fluid motion, and pressure varying only in the z direction. In a certain two-dimensional fluid flow pattern the lines of constant pressure, or isobars, are defined by the expression P0 Bz Cx2 constant, where B and C are constants and p0 is the (constant) pressure at the origin, (x, z) (0, 0). Find an expression x f (z) for the family of lines which are everywhere parallel to the local pressure gradient Vp. Atlanta, Georgia, has an average altitude of 1100 ft. On a standard day (Table A.6), pressure gage A in a laboratory experiment 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 20°C. The deepest known point in the ocean is 11,034 m in the Mariana Trench in the Pacific. At this depth the specific weight of seawater is approximately 10,520 N/m3. At the surface, 10,050 N/m3. Estimate the absolute pressure at 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, DALR dT/dz when T T0( p/p0)a, where exponent a (k 1)/k, k cp /cv is the ratio of specific heats, and T0 and p0 are the temperature and pressure at sea level, respectively. (a) Assuming that hydrostatic conditions exist in the atmosphere, show that the dry adiabatic lapse rate is constant and is given by DALR g(k 1)/(kR), where R is the ideal gas constant for air. (b) Calculate the numerical value of DALR for air in 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 constant—see Eq. (9.18). Find an expression for 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 20°C. 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 Guide Problems 103 P2.11 In Fig. P2.11, pressure gage A reads 1.5 kPa (gage). The fluids are at 20°C. Determine the elevations z, in meters, of the liquid levels in the open piezometer tubes B and C. Gasoline 1.5 m Water A B C h 1m 2m Air 1.5 m Gasoline 1m Glycerin Liquid, SG = 1.60 P2.13 A B Air P2.11 z= 0 4m P2.12 In Fig. P2.12 the tank contains water and immiscible oil at 20°C. What is h in cm if the density of the oil is 898 kg/m3? 2m Air 4m Water 2m P2.14 15 lbf/in2 abs h 6 cm A Air 12 cm 8 cm 2 ft Oil 1 ft Water B Oil P2.12 1 ft | v v P2.13 In Fig. P2.13 the 20°C water and gasoline surfaces are open to the atmosphere and at the same elevation. What is the height h of the third liquid in the right leg? P2.14 The closed tank in Fig. P2.14 is at 20°C. If the pressure at point A is 95 kPa absolute, what is the absolute pressure at point B in kPa? What percent error do you make by neglecting the specific weight of the air? P2.15 The air-oil-water system in Fig. P2.15 is at 20°C. Knowing that gage A reads 15 lbf/in2 absolute and gage B reads 1.25 lbf/in2 less than gage C, compute (a) the specific weight of the oil in lbf/ft3 and (b) the actual reading of gage C in lbf/in2 absolute. | e-Text Main Menu | Water 2 ft C P2.15 P2.16 A closed inverted cone, 100 cm high with diameter 60 cm at the top, is filled with air at 20°C and 1 atm. Water at 20°C is introduced at the bottom (the vertex) to compress the air isothermally until a gage at the top of the cone reads 30 kPa (gage). Estimate (a) the amount of water needed (cm3) and (b) the resulting absolute pressure at the bottom of the cone (kPa). Textbook Table of Contents | Study Guide 104 Chapter 2 Pressure Distribution in a Fluid P2.17 The system in Fig. P2.17 is at 20°C. If the pressure at point A is 1900 lbf/ft2, determine the pressures at points B, C, and D in lbf/ft2. Mercury Air Air 3 ft B A C 2 ft 10 cm 10 cm Air 4 ft 10 cm P2.19 5 ft Water D 2 ft 2000 lbf P2.17 3-in diameter P2.18 The system in Fig. P2.18 is at 20°C. If atmospheric pressure is 101.33 kPa and the pressure at the bottom of the tank is 242 kPa, what is the specific gravity of fluid X? 1 in 15 in F 1-in diameter Oil 1m SAE 30 oil P2.20 Water 2m Air: 180 kPa abs Fluid X Mercury 0.5 m 80 cm P2.18 P2.21 v v P2.19 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 be after the sloshing has died down? P2.20 The hydraulic jack in Fig. P2.20 is filled with oil at 56 lbf/ft3. Neglecting the weight of the two pistons, what force F on the handle is required to support the 2000-lbf weight for this design? P2.21 At 20°C gage A reads 350 kPa absolute. What is the height h of the water in cm? What should gage B read in kPa absolute? See Fig. P2.21. | | Water h? 3m e-Text Main Menu | Mercury A B P2.22 The fuel gage for a gasoline tank in a car reads proportional to the bottom gage pressure as in Fig. P2.22. If the tank is 30 cm deep and accidentally contains 2 cm of water plus gasoline, how many centimeters of air remain at the top when the gage erroneously reads “full’’? P2.23 In Fig. P2.23 both fluids are at 20°C. 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 density distribution (z) in Table A.6 and estimated the mass of the earth’s atmosphere to be m 6 E18 kg. Can this re- Textbook Table of Contents | Study Guide Problems 105 Vent Air h? Gasoline SG = 0.68 30 cm Water mental observations. (b) Find an expression for the pressure at points 1 and 2 in Fig. P2.27b. Note that the glass is now inverted, so the original top rim of the glass is at the bottom of the picture, and the original bottom of the glass is at the top of the picture. The weight of the card can be neglected. 2 cm Card pgage P2.22 Top of glass Oil 8 cm 6 cm P2.27a Water Bottom of glass Original bottom of glass 10 cm 1G P2.23 | v v 2G sult be used to estimate sea-level pressure on the earth? Conversely, can the actual sea-level pressure of 101.35 kPa be used to make a more accurate estimate of the atmosP2.27b Card Original top of glass pheric 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 such be 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 average ter would not fall out of the glass. surface pressure is 9.1 MPa. Estimate the atmospheric P2.28 Earth’s atmospheric conditions vary somewhat. On a cerpressure of Venus at an altitude of 5 km. tain day the sea-level temperature is 45°F and the sea-level P2.26 Investigate the effect of doubling the lapse rate on atmospressure is 28.9 inHg. An airplane overhead registers an pheric pressure. Compare the standard atmosphere (Table air temperature of 23°F and a pressure of 12 lbf/in2. EstiA.6) with a lapse rate twice as high, B2 0.0130 K/m. mate the plane’s altitude, in feet. Find the altitude at which the pressure deviation is (a) 1 *P2.29 Under some conditions the atmosphere is adiabatic, p percent 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 is Note: Do this over a sink or you may get wet! Find a drinkgiven by ing glass with a very smooth, uniform rim at the top. Fill the glass nearly full with water. Place a smooth, light, flat (k 1)gz k/(k 1) p p0 1 plate on top of the glass such that the entire rim of the kRT0 glass 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. See dard 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 20°C. 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 Guide 106 Chapter 2 Pressure Distribution in a Fluid pa Air B ρ1 SAE 30 oil 32 cm A Liquid, SG = 1.45 3 cm 5 cm 10 cm ρ2 4 cm A Water P2.30 6 cm 8 cm 3 cm P2.31 In Fig. P2.31 all fluids are at 20°C. Determine the pressure difference (Pa) between points A and B. P2.33 *P2.34 Sometimes manometer dimensions have a significant ef- Kerosine Air Benzene B 40 cm A 9 cm 20 cm fect. In Fig. P2.34 containers (a) and (b) are cylindrical and conditions are such that pa pb. Derive a formula for the pressure difference pa pb when the oil-water interface on the right rises a distance h h, for (a) d D and (b) d 0.15D. What is the percent change in the value of p? 14 cm 8 cm Mercury Water D D P2.31 (b) ( a) P2.32 For the inverted manometer of Fig. P2.32, all fluids are at 20°C. If pB pA 97 kPa, what must the height H be in cm? Meriam red oil, SG = 0.827 L SAE 30 oil H Water h 18 cm Water d H Mercury A P2.34 35 cm B P2.32 | v v P2.33 In Fig. P2.33 the pressure at point A is 25 lbf/in2. All fluids are at 20°C. 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 20°C. What is the pressure difference p1 p2 in the pipe? P2.36 In Fig. P2.36 both the tank and the tube are open to the atmosphere. If L 2.13 m, what is the angle of tilt of the tube? P2.37 The inclined manometer in Fig. P2.37 contains Meriam red manometer oil, SG 0.827. Assume that the reservoir Textbook Table of Contents | Study Guide Problems 107 with manometer fluid m. One side of the manometer is open to the air, while the other is connected to new tubing which extends to pressure measurement location 1, some height H higher 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 difference between 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, use the local atmospheric pressure at the elevation of the measurement point. You may assume that h H; i.e., assume the 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 density as 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) Can t you think of a simple way to avoid this error? (2) 30 (1) h 2m P2.35 Oil SG = 0.8 50 cm L Water SG = 1.0 50 cm P2.36 1 is very large. If the inclined arm is fitted with graduations 1 in apart, what should the angle be if each graduation corresponds to 1 lbf/ft2 gage pressure for pA? (tubing gas) p1 pa at location 1 t a (air) H 1 in pA θ D= 5 16 U-tube manometer in h m P2.38 Reservoir P2.37 | v v P2.38 An interesting article appeared in the AIAA Journal (vol. 30, no. 1, January 1992, pp. 279 – 280). The authors explain that the air inside fresh plastic tubing can be up to 25 percent more dense than that of the surroundings, due to outgassing or other contaminants introduced at the time of manufacture. Most researchers, however, assume that the tubing is filled with room air at standard air density, which can lead to significant errors when using this kind of tubing to measure pressures. To illustrate this, consider a U-tube manometer | e-Text Main Menu | P2.39 An 8-cm-diameter piston compresses manometer oil into an inclined 7-mm-diameter tube, as shown in Fig. P2.39. When a weight W is added to the top of the piston, the oil rises 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 the closed tank in Fig. P2.40. At the instant shown, the air pressure pB 80 kPa. The pump stops when the air pressure rises to 110 kPa. All fluids remain at 20°C. What will be 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 Guide 108 Chapter 2 Pressure Distribution in a Fluid pA D = 8 cm Piston pB ρ1 W 10 cm ρ1 h1 h1 d = 7 mm Meriam red oil, SG = 0.827 P2.39 h 15˚ ρ 2 P2.42 patm 8 cm Air: pB 9 cm Water h P2.44 Water flows downward in a pipe at 45°, as shown in Fig. P2.44. The pressure drop p1 p2 is partly due to gravity and partly due to friction. The mercury manometer reads a 6-in height difference. What is the total pressure drop p1 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? Pump Mercury 10 cm Hg 2 cm P2.40 P2.41 The system in Fig. P2.41 is at 20°C. Compute the pressure at point A in lbf/ft2 absolute. 45˚ 1 5 ft Water Flow 2 Oil, SG = 0.85 5 in A pa = 14.7 Water lbf/in2 10 in 6 in 6 in Water Mercury P2.44 P2.41 Mercury | v v P2.42 Very small pressure differences pA pB can be measured accurately by the two-fluid differential manometer in Fig. P2.42. Density 2 is only slightly larger than that of the upper 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 h 1.2, 4.9, and 11.0 mm when the water velocities in the pipe are V 1.0, 2.0, and 3.0 m/s, respectively. Determine if these 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 in Pa. Is it higher or lower than atmospheric? P2.46 In Fig. P2.46 both ends of the manometer are open to the atmosphere. Estimate the specific gravity of fluid X. P2.47 The cylindrical tank in Fig. P2.47 is being filled with water at 20°C by a pump developing an exit pressure of 175 EES kPa. At the instant shown, the air pressure is 110 kPa and H 35 cm. The pump stops when it can no longer raise the 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 in Textbook Table of Contents | Study Guide Problems 109 patm 50 cm Air Air 20˚ C Oil, SG = 0.85 75 cm 30 cm 45 cm 40 cm H Water Pump P2.47 15 cm A Newspaper Water P2.45 Mercury Ruler SAE 30 oil Desk 9 cm 10 cm P2.48 Water 5 cm 7 cm 6 cm Fluid X P2.49 4 cm P2.50 12 cm P2.46 | v v length) or a thin wooden paint stirrer. Place it on the edge of a desk or table with a little less than half of it hang- P2.51 ing over the edge lengthwise. Get two full-size sheets of newspaper; open them up and place them on top of the ruler, covering only the portion of the ruler resting on the *P2.52 desk as illustrated in Fig. P2.48. (a) Estimate the total force on top of the newspaper due to air pressure in the room. (b) Careful! To avoid potential injury, make sure nobody is standing directly in front of the desk. Perform | e-Text Main Menu | a karate chop on the portion of the ruler sticking out over the edge of the desk. Record your results. (c) Explain your results. A water tank has a circular panel in its vertical wall. The panel has a radius of 50 cm, and its center is 2 m below the surface. Neglecting atmospheric pressure, determine the 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 deep and has a trapezoidal cross section 2 m wide at the bottom and 4 m wide at the top. Compute (a) the weight of oil in the vat, (b) the force on the vat bottom, and (c) the force on the trapezoidal end panel. Gate AB in Fig. P2.51 is 1.2 m long and 0.8 m into the paper. Neglecting atmospheric pressure, compute the force F 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 that liquid X causes a force F on gate AB and that the moment of this force about point B is 26,500 N m. What is the specific gravity of liquid X? Textbook Table of Contents | Study Guide 110 Chapter 2 Pressure Distribution in a Fluid pa 6m Oil, SG = 0.82 Water pa 4m h 8m A 1m X 1.2 m A B 4 ft F 40° P2.51 B P2.55 P2.53 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 its line of action. 200 kg h m B A 30 cm A Water Water 3m P2.58 4m P2.53 B, C *P2.59 Gate AB has length L, width b into the paper, is hinged at B, and has negligible weight. The liquid level h remains at the top of the gate for any angle . Find an analytic expression for the force P, perpendicular to AB, required to keep the gate in equilibrium in Fig. P2.59. 3m | v v P2.54 If, instead of water, the tank in Fig. P2.53 is filled with liqP uid 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 found A to be the same as in Prob. 2.53. Why? P2.55 Gate AB in Fig. P2.55 is 5 ft wide into the paper, hinged at A, and restrained by a stop at B. The water is at 20°C. Compute (a) the force on stop B and (b) the reactions at h L A if the water depth h 9.5 ft. P2.56 In Fig. P2.55, gate AB is 5 ft wide into the paper, and stop B will break if the water force on it equals 9200 lbf. For Hinge what water depth h is this condition reached? P2.57 In Fig. P2.55, gate AB is 5 ft wide into the paper. Suppose B P2.59 that the fluid is liquid X, not water. Hinge A breaks when its 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 opening 80 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 a mass as shown. Assume standard gravity at 20°C. At what smooth bottom at B. All fluids are at 20°C. For what wawater level h will the gate be dislodged? Neglect the weight ter depth h will the force at point B be zero? of the gate. | e-Text Main Menu | Textbook Table of Contents | Study Guide Problems 111 P2.63 The tank in Fig. P2.63 has a 4-cm-diameter plug at the bottom on the right. All fluids are at 20°C. The plug will pop 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 m 1.2 m A Water 2m Water Glycerin 50° B 2m H P2.60 h 2 cm Plug, D = 4 cm Mercury Water P2.63 Glycerin *P2.64 Gate ABC in Fig. P2.64 has a fixed hinge line at B and is 2 m wide into the paper. The gate will open at A to release water if the water depth is high enough. Compute the depth h for which the gate will begin to open. h 2m A 1m C 60° B P2.61 P2.62 Gate AB in Fig. P2.62 is 15 ft long and 8 ft wide into the paper and is hinged at B with a stop at A. The water is at EES 20°C. The gate is 1-in-thick steel, SG 7.85. Compute the water level h for which the gate will start to fall. Pulley A h B v v P2.62 | | B h 1m Water at 20°C *P2.65 Gate AB in Fig. P2.65 is semicircular, hinged at B, and held by a horizontal force P at A. What force P is required for equilibrium? P2.66 Dam ABC in Fig. P2.66 is 30 m wide into the paper and made of concrete (SG 2.4). Find the hydrostatic force on surface AB and its moment about C. Assuming no seepage of water under the dam, could this force tip the dam over? How does your argument change if there is seepage under the dam? Water 60˚ 20 cm P2.64 10,000 lb 15 ft A e-Text Main Menu | Textbook Table of Contents | Study Guide 112 Chapter 2 Pressure Distribution in a Fluid 5m 3m Oil, SG = 0.83 1m Water A A P Gate: Side view 3m Gate 2m B P2.65 ;; ;; 50˚ B P2.68 A Water 20˚C 80 m 80 cm 1m Dam B 5m Water, 20˚C C Hg, 20˚C 60 m P2.66 A 2m P2.69 | v v *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 for which 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 and weighs 1500 N. What horizontal force P is required at point B for equilibrium? *P2.69 The water tank in Fig. P2.69 is pressurized, as shown by the 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 of the vertical triangular panel ABC in Fig. P2.70. Neglect patm. *P2.71 In Fig. P2.71 gate AB is 3 m wide into the paper and is connected by a rod and pulley to a concrete sphere (SG | e-Text Main Menu | B 2 ft A Water 6 ft C B 4 ft P2.70 Textbook Table of Contents | Study Guide P Problems 113 2.40). 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 hydrostatic calculations. An approximate density relation would be ;; Concrete sphere, SG = 2.4 dp d a2 d or p p0 a2( 0) 6m where a is the speed of sound and (p0, 0) are the conditions at the liquid surface z 0. Use this approximation to show that the density variation with depth in a soft liqgz/a2 uid is where g is the acceleration of gravity 0e 8m A and z is positive upward. Then consider a vertical wall of width b, extending from the surface (z 0) down to depth z h. Find an analytic expression for the hydrostatic 4m Water force F on this wall, and compare it with the incompress2 B ible result F 0gh b/2. Would the center of pressure be below the incompressible position z 2h/3? P2.71 *P2.75 Gate AB in Fig. P2.75 is hinged at A, has width b into the paper, 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 and expressed as a function of (h, t, , ), will the gate just beheld together by cable BC at the top. If cable spacing is gin 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? Cable C B A 1m Water 3m L h 110˚ P2.72 A t P2.73 Gate AB is 5 ft wide into the paper and opens to let fresh water out when the ocean tide is dropping. The hinge at A is 2 ft above the freshwater level. At what ocean level h will the gate first open? Neglect the gate weight. A Tide range 10 ft h Seawater, SG = 1.025 Stop B | v v P2.73 | e-Text Main Menu | P2.75 B *P2.76 Consider the angled gate ABC in Fig. P2.76, hinged at C and of width b into the paper. Derive an analytic formula for 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 and is hinged at B. Compute the force P just sufficient to keep the gate from opening when h 8 m. Neglect atmospheric pressure. P2.78 Repeat Prob. 2.77 to derive an analytic expression for P as a function of h. Is there anything unusual about your solution? 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 becomes high enough. Determine the lowest height for which the Textbook Table of Contents | Study Guide 114 ;; ;; ;; Chapter 2 Pressure Distribution in a Fluid A P θ θ Specific weight γ h Air 1 atm 2m SA E3 B 20 0o il h 60 Wa ter pa Water pa h cm Mercury 80 C P2.76 cm cm P2.80 Panel, 30 cm high, 40 cm wide P2.81 Gate AB in Fig. P2.81 is 7 ft into the paper and weighs 3000 lbf when submerged. It is hinged at B and rests against a smooth wall at A. Determine the water level h at the left which will just cause the gate to open. A 1m B 1m C h P 4 ft A P2.77 Water 8 ft Water h B Water 6 ft A 60 cm C 40 cm P2.81 B *P2.82 The dam in Fig. P2.82 is a quarter circle 50 m wide into P2.79 | v v gate will open. Neglect atmospheric pressure. Is this result independent of the liquid density? P2.80 For the closed tank in Fig. P2.80, all fluids are at 20°C, and the airspace is pressurized. It is found that the net outward hydrostatic force on the 30-by 40-cm panel at the bottom of the water layer is 8450 N. Estimate (a) the pressure in the airspace 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 point CP where the resultant strikes the dam. *P2.83 Gate AB in Fig. P2.83 is a quarter circle 10 ft wide into the paper and hinged at B. Find the force F just sufficient to keep the gate from opening. The gate is uniform and weighs 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 m 20 m Problems 115 pa = 0 CP Water P2.82 Water 10 ft P2.86 2 ft P2.87 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. F A Water r = 8 ft B P2.83 B Water at 20° C z 1m 4 in z = x3 2 in 6 in x A P2.84 P2.87 r = 2 in Mercury P2.85 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 point the 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) its line of action. Does the force pass through point O? 6m C 5m Water R=6m Water 2m B 6m O 2m P2.85 6m | v v P2.86 Compute the horizontal and vertical components of the hydrostatic force on the hemispherical bulge at the bottom of the tank in Fig. P2.86. | e-Text Main Menu | A P2.88 Textbook Table of Contents | Study Guide 116 Chapter 2 Pressure Distribution in a Fluid P2.89 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 its line of action. 3cm 4m 60 cm p = 200 kPa 30 cm Six bolts B 2m Water Benzene at 20 C 60 cm P2.91 P2.89 A 2m P2.90 A 1-ft-diameter hole in the bottom of the tank in Fig. P2.90 is closed by a conical 45° plug. Neglecting the weight of the plug, compute the force F required to keep the plug in the hole. Water p = 3 lbf/in 2 gage Bolt spacing 25 cm 2m P2.92 1 ft Air : z Water 3 ft 1 ft ρ, γ 45˚ cone h R F P2.90 | v v P2.91 The hemispherical dome in Fig. P2.91 weighs 30 kN and is filled with water and attached to the floor by six equally spaced bolts. What is the force in each bolt required to hold 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 R is submerged in liquid of specific gravity and depth h R. Find an analytic expression for the resultant hydrostatic force, and its line of action, on the shell surface. | e-Text Main Menu | R z R x P2.93 P2.94 The 4-ft-diameter log (SG 0.80) in Fig. P2.94 is 8 ft long into the paper and dams water as shown. Compute the net vertical and horizontal reactions at point C. Textbook Table of Contents | Study Guide Problems 117 wall at A. Compute the reaction forces at points A and B. Log 2ft Water 2ft P2.94 Water C Seawater, 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 hinge O. What is the specific gravity of this body if (a) h 0 and (b) h R? 4m A 2m 45° B A P2.97 h P2.98 Gate ABC in Fig. P2.98 is a quarter circle 8 ft wide into the paper. Compute the horizontal and vertical hydrostatic forces on the gate and the line of action of the resultant force. R R Water A O P2.95 r = 4 ft P2.96 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 force on quarter-circle panel BC. B C P2.98 P2.99 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 the hole. A 6m Water Water 45° 45° 4m B Water 4m 3 ft 1 ft P2.96 C | v v P2.97 Gate AB in Fig. P2.97 is a three-eighths circle, 3 m wide into the paper, hinged at B, and resting against a smooth | e-Text Main Menu | 1 ft P2.99 Textbook Table of Contents F | Study Guide 118 Chapter 2 Pressure Distribution in a Fluid P2.100 Pressurized water fills the tank in Fig. P2.100. Compute the net hydrostatic force on the conical surface ABC. 2m A P2.106 C 4m 7m B P2.107 150 kPa gage P2.108 Water P2.100 P2.101 A fuel truck has a tank cross section which is approximately elliptical, with a 3-m horizontal major axis and a 2-m vertical minor axis. The top is vented to the atmosphere. If the tank is filled half with water and half with gasoline, what is the hydrostatic force on the flat elliptical end panel? P2.102 In Fig. P2.80 suppose that the manometer reading is h 25 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, less than a millimeter in diameter, and rise slowly. Their drag in still fluid is approximated by the first term of Stokes’ expression in Prob. 1.10: F 3 VD, where V is the rise velocity. Neglecting bubble weight and setting bubble buoyancy equal to drag, (a) derive a formula for the terminal (zero acceleration) rise velocity Vterm of the bubble and (b) determine Vterm in m/s for water at 20°C if D 30 m. P2.104 The can in Fig. P2.104 floats in the position shown. What is its weight in N? P2.109 whether his new crown was pure gold (SG 19.3). Archimedes measured the weight of the crown in air to be 11.8 N and its weight in water to be 10.9 N. Was it pure gold? It is found that a 10-cm cube of aluminum (SG 2.71) will remain neutral under water (neither rise nor fall) if it is tied by a string to a submerged 18-cm-diameter sphere of buoyant foam. What is the specific weight of the foam, in N/m3? Repeat Prob. 2.62, assuming that the 10,000-lbf weight is aluminum (SG 2.71) and is hanging submerged in the water. A piece of yellow pine wood (SG 0.65) is 5 cm square and 2.2 m long. How many newtons of lead (SG 11.4) should be attached to one end of the wood so that it will float vertically with 30 cm out of the water? A hydrometer floats at a level which is a measure of the specific gravity of the liquid. The stem is of constant diameter D, and a weight in the bottom stabilizes the body to float vertically, as shown in Fig. P2.109. If the position h 0 is pure water (SG 1.0), derive a formula for h as a function of total weight W, D, SG, and the specific weight 0 of water. D SG = 1.0 h Fluid, SG > 1 W P2.109 3 cm 8 cm Water D = 9 cm P2.104 | v v P2.105 It is said that Archimedes discovered the buoyancy laws when 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 and a mass of 2.6 g. Estimate the (small) depth at which this ball will float in water at 20°C and sea level standard air if 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 balloon material has a mass of 60 g/m2. Ambient air is at 25°C and 1 atm. The hot air inside the balloon is at 70°C and 1 atm. What diameter spherical balloon will just support the total weight? Neglect the size of the hot-air inlet vent. P2.112 The uniform 5-m-long round wooden rod in Fig. P2.112 is tied to the bottom by a string. Determine (a) the tension Textbook Table of Contents | Study Guide Problems 119 in the string and (b) the specific gravity of the wood. Is it possible for the given information to determine the inclination angle ? Explain. Hinge D = 4 cm B = 30 1m 8m D = 8 cm 2 kg of lead θ Water at 20°C P2.114 4m B String θ 8 ft Wood SG = 0.6 P2.112 P2.113 A spar buoy is a buoyant rod weighted to float and protrude vertically, as in Fig. P2.113. It can be used for measurements or markers. Suppose that the buoy is maple wood (SG 0.6), 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 to the bottom end so that h 18 in? Seawater A Rock P2.115 P2.116 The homogeneous 12-cm cube in Fig. 2.116 is balanced by a 2-kg mass on the beam scale when the cube is immersed in 20°C ethanol. What is the specific gravity of the cube? h 2 kg Wsteel P2.113 | v v P2.114 The uniform rod in Fig. P2.114 is hinged at point B on the waterline and is in static equilibrium as shown when 2 kg of lead (SG 11.4) are attached to its end. What is the specific gravity of the rod material? What is peculiar about the rest angle 30? P2.115 The 2-in by 2-in by 12-ft spar buoy from Fig. P2.113 has 5 lbm of steel attached and has gone aground on a rock, as in Fig. P2.115. Compute the angle at which the buoy will lean, assuming that the rock exerts no moments on the spar. | e-Text Main Menu | 12 cm P2.116 P2.117 The balloon in Fig. P2.117 is filled with helium and pressurized to 135 kPa and 20°C. The balloon material has a Textbook Table of Contents | Study Guide 120 Chapter 2 Pressure Distribution in a Fluid mass of 85 g/m2. Estimate (a) the tension in the mooring line and (b) the height in the standard atmosphere to which the 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 and with specific weight b, floats exactly on its diagonal when a 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 size Lhb (SG 1) D 1/3 D = 10 m Width b << L Air: 100 kPa at 20°C P2.117 P2.118 A 14-in-diameter hollow sphere is made of steel (SG 7.85) with 0.16-in wall thickness. How high will this EES sphere float in 20°C water? How much weight must be added inside to make the sphere neutrally buoyant? P2.119 When a 5-lbf weight is placed on the end of the uniform floating wooden beam in Fig. P2.119, the beam tilts at an angle with its upper right corner at the surface, as shown. Determine (a) the angle and (b) the specific gravity of the wood. (Hint: Both the vertical forces and the moments about the beam centroid must be balanced.) ; L h << L γb γ Diameter D SG > 1 P2.121 P2.122 A uniform block of steel (SG 7.85) will “float’’ at a mercury-water interface as in Fig. P2.122. What is the ratio of the distances a and b for this condition? 5 lbf θ Water 9 ft Water 4 in × 4 in a Steel block P2.119 b Mercury: SG = 13.56 P2.120 A uniform wooden beam (SG 0.65) is 10 cm by 10 cm by 3 m and is hinged at A, as in Fig. P2.120. At what angle will the beam float in the 20°C water? P2.123 In an estuary where fresh water meets and mixes with seawater, there often occurs a stratified salinity condition with fresh water on top and salt water on the bottom, as in Fig. P2.123. The interface is called a halocline. An idealization of this would be constant density on each side of the halocline as shown. A 35-cm-diameter sphere weighing 50 lbf would “float’’ near such a halocline. Compute the sphere position for the idealization in Fig. P2.123. P2.124 A balloon weighing 3.5 lbf is 6 ft in diameter. It is filled with hydrogen at 18 lbf/in2 absolute and 60°F and is released. At what altitude in the U.S. standard atmosphere will this balloon be neutrally buoyant? A 1m θ Water v v P2.120 | | e-Text Main Menu P2.122 | Textbook Table of Contents | Study Guide Problems 121 SG = 1.0 Halocline SG = 1.025 35°/°° Salinity 0 Idealization P2.123 P2.125 Suppose that the balloon in Prob. 2.111 is constructed to have a diameter of 14 m, is filled at sea level with hot air at 70°C and 1 atm, and is released. If the air inside the balloon remains constant and the heater maintains it at 70°C, at what altitude in the U.S. standard atmosphere will this balloon be neutrally buoyant? *P2.126 A cylindrical can of weight W, radius R, and height H is open at one end. With its open end down, and while filled with atmospheric air (patm, Tatm), the can is eased down vertically into liquid, of density , which enters and compresses the air isothermally. Derive a formula for the height h to which the liquid rises when the can is submerged with its 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 added at 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, as in Fig. P2.128. If seawater is denoted by S 1.0, then glacier ice (which forms icebergs) has S 0.88. Determine if this “cubic’’ iceberg is stable for the position shown in Fig. P2.128. Fig. P2.128 suppose that the height is L and the depth into the paper is L, but the width in the plane of the paper is H L. Assuming S 0.88 for the iceberg, find the ratio H/L for which it becomes neutrally stable, i.e., about to overturn. P2.130 Consider a wooden cylinder (SG 0.6) 1 m in diameter and 0.8 m long. Would this cylinder be stable if placed to float 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 draft of 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 the cone? Water : SG = 1.0 SG = 0.99 P2.132 P2.133 Consider a uniform right circular cone of specific gravity S 1, floating with its vertex down in water (S 1). The base radius is R and the cone height is H. Calculate and plot 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 positions shown in Fig. P2.134. Which is the more stable position? Assume large width into the paper. Specific gravity =S M? G B h Water S = 1.0 P2.134 P2.135 Consider a homogeneous right circular cylinder of length L, radius R, and specific gravity SG, floating in water (SG 1). Show that the body will be stable with its axis vertical if L P2.128 v v P2.129 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 Contents R L [2SG(1 | SG)]1/2 Study Guide 122 Chapter 2 Pressure Distribution in a Fluid P2.136 Consider a homogeneous right circular cylinder of length L, radius R, and specific gravity SG 0.5, floating in water (SG 1). Show that the body will be stable with its axis 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 tank bottom if az 5 m2/s and (b) the value of az which causes the 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, is attached to the edge of an 8-ft-diameter merry-go-round which is rotated at 12 r/min. How full can the glass be before water spills? (Hint: Assume that the glass is much smaller than the radius of the merry-go-round.) P2.139 The tank of liquid in Fig. P2.139 accelerates to the right with the fluid in rigid-body motion. (a) Compute ax in m/s2. (b) Why doesn’t the solution to part (a) depend upon the density of the fluid? (c) Determine the gage pressure at point A if the fluid is glycerin at 20°C. V a? 15 cm 100 cm 28 cm A z 30° P2.141 x B 9 cm Water at 20° C A 24 cm ax P2.142 28 cm 15 cm 100 cm A A pa = 15 lbf/in2 abs Fig. P2.139 ax 2ft | v v P2.140 Suppose that the elliptical-end fuel tank in Prob. 2.101 is 10 m long and filled completely with fuel oil ( 890 kg/m3). Let the tank be pulled along a horizontal road. For rigid-body motion, find the acceleration, and its direction, for which (a) a constant-pressure surface extends from the top 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 lower than 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) the value of the acceleration a, (b) whether the acceleration is up or down, and (c) the gage pressure at point A if the fluid is mercury at 20°C. P2.142 The tank of water in Fig. P2.142 is 12 cm wide into the paper. If the tank is accelerated to the right in rigid-body motion at 6.0 m/s2, compute (a) the water depth on side AB and (b) the water-pressure force on panel AB. Assume no 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 the pressure at point B be (a) atmospheric and (b) zero absolute? | e-Text Main Menu | Water B 1ft P2.143 1ft 2ft P2.144 Consider a hollow cube of side length 22 cm, filled completely with water at 20°C. The top surface of the cube is horizontal. One top corner, point A, is open through a small hole to a pressure of 1 atm. Diagonally opposite to point A is top corner B. Determine and discuss the various rigidbody accelerations for which the water at point B begins to 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 carried in a car which may experience accelerations as high as 6 m/s2. What is the maximum water depth which will avoid Textbook Table of Contents | Study Guide Problems 123 spilling 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 vent hole at point A. The tank is 1 m wide into the paper. Inside the tank, a 10-cm balloon, filled with helium at 130 kPa, is tethered centrally by a string. If the tank accelerates to the right at 5 m/s2 in rigid-body motion, at what angle will the balloon lean? Will it lean to the right or to the left? 60 cm A 1 atm Water at 20°C D = 10 cm with the child, which way will the balloon tilt, forward or backward? Explain. (b) The child is now sitting in a car which is stopped at a red light. The helium-filled balloon is not in contact with any part of the car (seats, ceiling, etc.) but is held in place by the string, which is in turn held by the child. All the windows in the car are closed. When the traffic light turns green, the car accelerates forward. In a frame of reference moving with the car and child, which way will the balloon tilt, forward or backward? Explain. (c) Purchase or borrow a helium-filled balloon. Conduct a scientific 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 to lift water with its 1-ft-diameter half-cylinder blades. If the wheel rotates at 10 r/min and rigid-body motion is assumed, what is the water surface angle at position A? He 40 cm 20 cm String 10 r/min θ P2.146 A 6 ft P2.147 The tank of water in Fig. P2.147 accelerates uniformly by freely rolling down a 30° incline. If the wheels are frictionless, what is the angle ? Can you explain this interesting result? 1 ft P2.149 P2.150 A cheap accelerometer, probably worth the price, can be made from a U-tube as in Fig. P2.150. If L 18 cm and D 5 mm, what will h be if ax 6 m/s2? Can the scale markings on the tube be linear multiples of ax? θ D h Rest level ax 30° 1 2 1 2 L L P2.147 L P2.150 | v v P2.148 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 acceleration Textbook Table of Contents | Study Guide 124 Chapter 2 Pressure Distribution in a Fluid will cause the pressure at point C to be atmospheric? The fluid is water (SG 1.0). A D 1 ft 1 ft B P2.156 Suppose that the U-tube of Fig. P2.151 is rotated about axis DC. If the fluid is water at 122°F and atmospheric pressure is 2116 lbf/ft2 absolute, at what rotation rate will the fluid within the tube begin to vaporize? At what point will this occur? P2.157 The 45° V-tube in Fig. P2.157 contains water and is open at A and closed at C. What uniform rotation rate in r/min about axis AB will cause the pressure to be equal at points B and C? For this condition, at what point in leg BC will the pressure be a minimum? C A C 1 ft P2.151 P2.152 A 16-cm-diameter open cylinder 27 cm high is full of water. Compute the rigid-body rotation rate about its central axis, in r/min, (a) for which one-third of the water will spill out and (b) for which the bottom will be barely exposed. P2.153 Suppose the U-tube in Fig. P2.150 is not translated but rather rotated about its right leg at 95 r/min. What will be the 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 water overlaid with 10 cm of SAE 30 oil. If the can is rotated in rigid-body motion about its central axis at 150 r/min, what will be the shapes of the air-oil and *P2.158 oil-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 will the U-tube in Fig. P2.155 take the configuration shown? EES The fluid is mercury at 20°C. A 30 cm 45˚ B P2.157 It is desired to make a 3-m-diameter parabolic telescope mirror by rotating molten glass in rigid-body motion until the desired shape is achieved and then cooling the glass to a solid. The focus of the mirror is to be 4 m from the mirror, measured along the centerline. What is the proper mirror rotation rate, in r/min, for this task? C B Ω 20 cm 12 cm 10 cm | v v P2.155 5 cm | e-Text Main Menu | Textbook Table of Contents | Study Guide Fundamentals of Engineering Exam Problems 125 Word Problems W2.1 Consider a hollow cone with a vent hole in the vertex at the top, along with a hollow cylinder, open at the top, with the same base area as the cone. Fill both with water to the top. The hydrostatic paradox is that both containers have the same force on the bottom due to the water pressure, although the cone contains 67 percent less water. Can you explain the paradox? W2.2 Can the temperature ever rise with altitude in the real atmosphere? Wouldn’t this cause the air pressure to increase upward? Explain the physics of this situation. W2.3 Consider a submerged curved surface which consists of a two-dimensional circular arc of arbitrary angle, arbitrary depth, and arbitrary orientation. Show that the resultant hydrostatic pressure force on this surface must pass through the center of curvature of the arc. W2.4 Fill a glass approximately 80 percent with water, and add a large ice cube. Mark the water level. The ice cube, having SG 0.9, sticks up out of the water. Let the ice cube melt with negligible evaporation from the water surface. Will the water level be higher than, lower than, or the same as before? W2.5 A ship, carrying a load of steel, is trapped while floating in a small closed lock. Members of the crew want to get out, but they can’t quite reach the top wall of the lock. A crew member suggests throwing the steel overboard in the lock, claiming the ship will then rise and they can climb out. 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 of your stationary car. The windows are closed, so there is no air 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 pressure gradient in the air within the car.) W2.8 Repeat your analysis of Prob. W2.7 to let the car move at constant velocity and go around a curve. Will the balloon lean in, toward the center of curvature, or out? Fundamentals of Engineering Exam Problems FE2.1 A gage attached to a pressurized nitrogen tank reads a gage 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 kPa FE2.2 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 m FE2.3 In Fig. FE2.3, if the oil in region B has SG 0.8 and the absolute pressure at point A is 1 atm, what is the absolute pressure at point B? (a) 5.6 kPa, (b) 10.9 kPa, (c) 106.9 kPa, (d) 112.2 kPa, (e) 157.0 kPa A Oil Water SG = 1 5 cm B 3 cm 8 cm Mercury SG = 13.56 4 cm | v v FE2.3 | e-Text Main Menu | FE2.4 In Fig. FE2.3, if the oil in region B has SG 0.8 and the absolute 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 kPa FE2.5 A tank of water (SG 1,.0) has a gate in its vertical wall 5 m high and 3 m wide. The top edge of the gate is 2 m below the surface. What is the hydrostatic force on the gate? (a) 147 kN, (b) 367 kN, (c) 490 kN, (d) 661 kN, (e) 1028 kN FE2.6 In Prob. FE2.5 above, how far below the surface is the center 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 m FE2.7 A solid 1-m-diameter sphere floats at the interface between water (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) 12.56 FE2.8 A 5-m-diameter balloon contains helium at 125 kPa absolute and 15°C, moored in sea-level standard air. If the gas constant of helium is 2077 m2/(s2 K) and balloon material weight is neglected, what is the net lifting force of the balloon? (a) 67 N, (b) 134 N, (c) 522 N, (d) 653 N, (e) 787 N FE2.9 A square wooden (SG 0.6) rod, 5 cm by 5 cm by 10 m long, floats vertically in water at 20°C when 6 kg of steel (SG 7.84) are attached to one end. How high above the water 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 m Textbook Table of Contents | Study Guide 126 Chapter 2 Pressure Distribution in a Fluid FE2.10 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) center of buoyancy is above its metacenter, (d) metacenter is above its center of buoyancy, (e) metacenter is above its center of gravity Comprehensive Problems C2.1 Some manometers are constructed as in Fig. C2.1, where one side is a large reservoir (diameter D) and the other side is a small tube of diameter d, open to the atmosphere. In such a case, the height of manometer liquid on the reservoir side does not change appreciably. This has the advantage that 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 is no pressure difference across the manometer, the elevations on 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, the manometer liquid in the large reservoir goes down, while that in the tube at the right goes up to conserve mass. Write an exact expression for p1gage, taking into account the movement of the surface of the reservoir. Your equation should give 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, neglecting the change in elevation of the surface of the reservoir liquid. (c) Suppose h 0.26 m in a certain application. If pa 101,000 Pa and the manometer liquid has a density of 820 kg/m3, estimate the ratio D/d required to keep the error of the approximation of part (b) within 1 percent of the exact measurement of part (a). Repeat for an error within 0.1 percent. To pressure measurement location pa a (air) D h p1 Zero pressure level m d C2.1 | v v C2.2 A prankster has added oil, of specific gravity SG0, to the left 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 is attached 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 your result in (a) when ptank pa. (c) Suppose H 5.0 cm, pa is 101.2kPa, ptank is 1.82 kPa higher than pa, and SG0 0.85. Calculate h in cm, ignoring surface tension effects and neglecting air density effects. pa Pressurized air tank, with pressure ptank Oil H h Water C2.2 C2.3 Professor F. Dynamics, riding the merry-go-round with his son, has brought along his U-tube manometer. (You never know when a manometer might come in handy.) As shown in Fig. C2.3, the merry-go-round spins at constant angular velocity and the manometer legs are 7 cm apart. The manometer 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 to the average manometer acceleration; and (b) exactly, using rigid-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 filled to the brim. He wants to know what percent of the cola he should drink before the ride begins, so that none of it spills during the big drop, in which the roller coaster achieves 0.55-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 Guide 7.00 cm 6.00 rpm Water h R 5.80 m (to center of manometer) Center of rotation C2.3 Design Projects h, m D2.1 It is desired to have a bottom-moored, floating system which creates a nonlinear force in the mooring line as the water level rises. The design force F need only be accurate in the range of seawater depths h between 6 and 8 m, as shown in the accompanying table. Design a buoyant system which will provide this force distribution. The system should be practical, i.e., of inexpensive materials and simple construction. D2.2 A laboratory apparatus used in some universities is shown in Fig. D2.2. The purpose is to measure the hydrostatic force on the flat face of the circular-arc block and compare it with the theoretical value for given depth h. The counterweight is arranged so that the pivot arm is horizontal when the block is not submerged, whence the weight W can be correlated with the hydrostatic force when the submerged arm is again brought to horizontal. First show that the apparatus concept is valid in principle; then derive a formula for W as a function of h in terms of the system parameters. Finally, suggest some appropriate values of Y, L, etc., for a suitable appartus and plot theoretical W versus h for these values. F, N h, m F, N 6.00 6.25 6.50 6.75 7.00 400 437 471 502 530 7.25 7.50 7.75 8.00 554 573 589 600 L Counterweight W Pivot Pivot arm R Side view of block face Fluid: h Y Circular arc block b D2.2 References 2. 3. 4. 5. 6. 7. U.S. Standard Atmosphere, 1976, Government Printing Office, Washington, DC, 1976. G. Neumann and W. J. Pierson, Jr., Principles of Physical Oceanography, 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 Rotating Fluid System,’’ Am. J. Phys., vol. 40, pp. 959 – 965, July 1972. National Committee for Fluid Mechanics Films, Illustrated Experiments in Fluid Mechanics, M.I.T. Press, Cambridge, MA, 1972. J. P. Holman, Experimental Methods for Engineers, 6th ed., McGraw-Hill, New York, 1993. | v v 1. | e-Text Main Menu | 8. 9. 10. 11. 12. 13. R. P. Benedict, Fundamentals of Temperature, Pressure, and Flow 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, 3d ed., McGraw-Hill, New York, 1996. Textbook Table of Contents 127 | Study Guide ...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.

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