Water Pressure and Pressure Forces

Transcription

1 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page 14 2 Water Pressure and Pressure Forces 2.1 Te Free Surface of Water Wen water fills a containing vessel, it automaticall seeks a orizontal surface on wic te pressure is constant everwere. In practice, a free water surface is one tat is not in contact wit an overling vessel cover. free water surface ma be subjected to atmosperic pressure (open vessel) or an oter pressure tat is exerted witin te vessel (closed vessel). 2.2 bsolute and Gauge Pressures water surface in contact wit te eart s atmospere is subjected to atmosperic pressure, wic is approximatel equal to a m-ig column of water at sea level. In still water, an object located below te water surface is subjected to a pressure greater tan atmosperic pressure. Tis additional pressure is often referred to as drostatic pressure. More precisel, it is te force per unit area acting in a normal direction on te surface of a bod immersed in te fluid (in tis case water). To determine te variation of drostatic pressure between an two points in water (wit a specific weigt of ), we ma consider two arbitrar points and along an arbitrar x-axis, as sown in Figure 2.1. Consider tat tese points lie in te ends of a small prism of water aving a cross-sectional area d and a lengt L. P and P are te pressures at eac 14

2 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page 15 Sec. 2.2 bsolute and Gauge Pressures 15 x P L = L sinθ θ P Figure 2.1 Hdrostatic pressure on a prism end, were te cross-sectional areas are normal to te x-axis. ecause te prism is at rest, all forces acting on it must be in equilibrium in all directions. For te force components in te x-direction, we ma write F x = P d - P d + Ld sin u = 0 Note tat L # sin u = is te vertical elevation difference between te two points. Te above equation reduces to P - P = (2.1) Terefore, te difference in pressure between an two points in still water is alwas equal to te product of te specific weigt of water and te difference in elevation between te two points. If te two points are on te same elevation, = 0 and P = P. In oter words, for water at rest, te pressure at all points in a orizontal plane is te same. If te water bod as a free surface tat is exposed to atmosperic pressure, P atm, we ma position point on te free surface and write 1P 2 abs = + P = + P atm (2.2) Tis pressure, 1P 2 abs, is commonl referred to as te absolute pressure. Pressure gauges are usuall designed to measure pressures above or below te atmosperic pressure. Pressure so measured, using atmosperic pressure as a base, is called gauge pressure, P. bsolute pressure is alwas equal to gauge pressure plus atmosperic pressure: P = P abs - P atm (2.3) Figure 2.2 diagrammaticall sows te relationsip between te absolute and gauge pressure and two tpical pressure-gauge dials. Comparing Equations 2.2 and 2.3, we ave P = (2.4)

3 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page Water Pressure and Pressure Forces Cap. 2 bsolute pressure (water column, m) P 1 +(P 1 ) abs P 2 tmosperic pressure +(P 2 ) abs bsolute pressure Gauge pressure Meters H 2 O bsolute pressure Gauge pressure Meters H 2 O bsolute vacuum 0 Figure 2.2 bsolute and gauge pressure or = P g (2.5) Here te pressure is expressed in terms of te eigt of a water column. In draulics it is known as te pressure ead. Equation 2.1 ma tus be rewritten in a more general form as P g - P g = (2.6) meaning tat te difference in pressure eads at two points in water at rest is alwas equal to te difference in elevation between te two points. From tis relationsip we can also see tat an cange in pressure at point would cause an equal cange at point, because te difference in pressure ead between te two points must remain te same value. In oter words, a pressure applied at an point in a liquid at rest is transmitted equall and undiminised in all directions to ever oter point in te liquid. Tis principle, also know as Pascal s law, as been made use of in te draulic jacks tat lift eav weigts b appling relativel small forces.

4 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page 17 Sec. 2.3 Surfaces of Equal Pressure 17 Example 2.1 Te diameters of clindrical pistons and are 3 cm and 20 cm, respectivel. Te faces of te pistons are at te same elevation, and te intervening passages are filled wit an incompressible draulic oil. force P of 100 N is applied at te end of te lever, as sown in Figure 2.3. Wat weigt W can te draulic jack support? P 80 cm 20 cm F 3 cm W 20 cm Measurements in cm Figure 2.3 Hdraulic jack Solution alancing te moments produced b P and F around te pin connection ields Tus, From Pascal s law, te pressure P applied at is te same as tat of P applied at. Terefore, P = 1100 N21100 cm2 = F120 cm2 F 31p # 3 2 2>44 cm 2 P = 500 N 7.07 cm 2 = W 314 cm 2 F = 500 N 314 cm2 W = 500 N 7.07 cm 2 = 2.22 * 104 N W 31p # >44 cm Surfaces of Equal Pressure Te drostatic pressure in a bod of water varies wit te vertical distance measured from te free water surface. In general, all points on a orizontal surface in a static bod of water are subjected to te same drostatic pressure, according to Equation 2.4. For example, in Figure 2.4 (a), points 1, 2, 3, and 4 ave equal pressure, and te orizontal surface tat contains tese four points is a surface of equal pressure. However, in Figure 2.4 (b), points 5 and 6 are on te same orizontal plane but te pressures are not equal. Tis is because te water in te two tanks is not connected and te overling depts to te free surfaces are different. ppling Equation 2.4 would produce different pressures. Figure 2.4 (c) displas tanks filled wit two immiscible liquids of different densities. (Note: Immiscible liquids do not readil mix under normal conditions.) Te orizontal surface (7, 8) tat passes troug te interface of te two liquids is an

5 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page Water Pressure and Pressure Forces Cap. 2 ES = equal pressure surface NES = nonequal pressure surface ES 5 6 NES 9 7 H 2 O Oil 10 8 NES ES Valve closed (a) (b) (c) Figure 2.4 Hdraulic pressure in vessels equal pressure surface. ppling Equation 2.4 at bot points leads to te same pressure; we ave te same fluid (water) at bot locations (just below te interface at point 8), and bot points are te same distance beneat te free water surface. However, points 9 and 10 are not on an equal pressure surface because te reside in different liquids. Verification would come from te application of Equation 2.4 using te different depts from te free surface to points 9 and 10 and te different specific weigts of te fluids. In summar, a surface of equal pressure requires tat (1) te points on te surface be in te same liquid, (2) te points be at te same elevation (i.e., reside on a orizontal surface), and (3) te liquid containing te points be connected. Te concept of equal pressure surface is a useful metod in analzing te strengt or intensit of te drostatic pressure at various points in a container, as demonstrated in te following section. 2.4 Manometers manometer is a pressure-measurement device. It usuall is a tube bent in te form of a U tat contains a fluid of known specific gravit. Te difference in elevations of te liquid surfaces under pressure indicates te difference in pressure at te two ends. asicall, tere are two tpes of manometers: 1. n open manometer as one end open to atmosperic pressure and is capable of measuring te gauge pressure in a vessel. 2. differential manometer as eac end connected to a different pressure tap and is capable of measuring te pressure difference between te two taps. Te liquid used in a manometer is usuall eavier tan te fluids to be measured. It must form a distinct interface tat is, it must not mix wit te adjacent liquids (i.e., immiscible liquids). Te most frequentl used manometer liquids are mercur 1sp. gr. = 13.62, water 1sp. gr. = 1.002, alcool 1sp. gr. = 0.92, and oter commercial manometer oils of various specific gravities (e.g., from Meriam* Red Oil, sp. gr. = to Meriam No. 3 Fluid, sp. gr. = 2.95). * Meriam Process Tecnologies, Cleveland, Oio 44102

6 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page 19 Sec. 2.4 Manometers 19 tmospere Water Water Water 1 2 C 1 D 2 Liquid M Liquid M (a) (b) Figure 2.5 Tpes of manometers: (a) open manometer and (b) differential manometer Figure 2.5 (a) sows a scematic of a tpical open manometer; Figure 2.5 (b) sows a scematic of a tpical differential manometer. It is obvious tat te iger te pressure in vessel, te larger te difference,, in te surface elevations in te two legs of te manometer. matematical calculation of pressure in, owever, involves te densities of te fluids and te geometr involved in te entire measuring sstem. simple step-b-step procedure is suggested for pressure computation. Step 1. Make a sketc of te manometer sstem, similar to tat in Figure 2.5, and approximatel to scale. Step 2. Draw a orizontal line troug te lower surface of te manometer liquid (point 1). Te pressure at points 1 and 2 must be te same since te sstem is in static equilibrium. Step 3. (a) For open manometers, te pressure on 2 is exerted b te weigt of te liquid M column above 2; and te pressure on 1 is exerted b te weigt of te column of water above 1 plus te pressure in vessel. Te pressures must be equal in value. Tis relation ma be written as follows: (b) For differential manometers, te pressure on 2 is exerted b te weigt of te liquid M column above 2, te weigt of te water column above D, and te pressure in vessel, wereas te pressure on 1 is exerted b te weigt of te water column above 1 plus te pressure in vessel. Tis relationsip ma be expressed as: or M = + P or P = M - 1g2 M P = + P P = P - P = 1 M - 2 Eiter one of tese equations can be used to solve for P. Of course, in te case of te differential manometer, P must be known. Te same procedure can be applied to an complex geometr, as demonstrated in te following example. Example 2.2 mercur manometer (sp. gr. 13.6) is used to measure te pressure difference in vessels and, as sown in Figure 2.6. Determine te pressure difference in pascals 1N/m 2 2.

7 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page Water Pressure and Pressure Forces Cap. 2 + = 123 cm cm 3 4 = 15 cm Figure 2.6 Solution Te sketc of te manometer sstem (step 1) is sown in Figure 2.6. Points 3 and 4 1P 3, P 4 2 are on a surface of equal pressure (step 2) and so are te vessel and points 1 and 2 1P 1, P 2 2: P 3 = P 4 Te pressures at points 3 and 4 are, respectivel (step 3), Now and noting tat M (sp. gr.) P = P 1 = P 2 P 3 = P cm2 = P cm2 P 4 = P cm2 + M 115 cm2 P 3 = P cm2 = P 4 = P cm2 + M 115 cm2 P = P - P = 1135 cm - 27 cm2 + M 115 cm2 P = cm = N>m m2 P = 30,500 N>m 2 1pascals2 or 30.6 kilo-pascals Te open manometer, or U-tube, requires readings of liquid levels at two points. In oter words, an cange in pressure in te vessel causes a drop of liquid surface at one end and a rise in te oter. single-reading manometer can be made b introducing a reservoir wit a larger cross-sectional area tan tat of te tube into one leg of te manometer. tpical single-reading manometer is sown in Figure 2.7. ecause of te large area ratio between te reservoir and te tube, a small drop of surface elevation in te reservoir will cause an appreciable rise in te liquid column of te oter leg. If tere is an increase in pressure, P will cause te liquid surface in te reservoir to drop b a small amount. Ten = a (2.7)

8 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page 21 Sec. 2.4 Manometers 21 Δ 1 2 Figure 2.7 Single-reading manometer were and a are cross-sectional areas of te reservoir and te tube, respectivel. ppling step 2 to points 1 and 2, we ma generall write P = Simultaneous solution of Equations 2.7 and 2.8 give te value of P, te pressure in te vessel, in terms of. ll oter quantities in Equations 2.7 and 2.8, a,,, and are quantities predetermined in te manometer design. single reading of will tus determine te pressure. ecause can be made negligible b introducing a ver large /a ratio, te above relationsip ma be furter simplified to + P = Tus, te eigt reading is a measure of te pressure in te vessel. Te solution of practical draulic problems frequentl requires te difference in pressure between two points in a pipe or a pipe sstem. For tis purpose, differential manometers are frequentl used. tpical differential manometer is sown in Figure 2.8. (2.8) (2.9) c d Liquid a b 2 1 Liquid Figure 2.8 differential manometer installed in a flow-measurement sstem

9 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page Water Pressure and Pressure Forces Cap. 2 Te same computation steps (steps 1, 2, and 3) suggested previousl can be readil applied ere, too. Wen te sstem is in static equilibrium, te pressure at te same elevation points, 1 and 2, must be equal. We ma tus write g P c = g + g + P d Hence, te pressure difference, P, is expressed as P = P c - P d = 1g - g 2 (2.10) 2.5 Hdrostatic Forces on Flat Surfaces Determining te total (or resultant) drostatic force on structures produced b drostatic pressure is often critical in engineering design and analsis. To determine te magnitude of tis force, let s examine an arbitrar area (Figure 2.9) on te back face of a dam tat inclines at an angle. Next, place te x-axis on te line were te surface of te water intersects wit te dam surface (i.e., into te page) wit te -axis running downward along te surface or face of te dam. Figure 2.9 (a) sows a plan (front) view of te area and Figure 2.9 (b) sows te projection of on te dam surface. We ma assume tat te plane surface is made up of an infinite number of orizontal strips, eac aving a widt of d and an area of d. Te drostatic pressure on eac strip ma be considered constant because te widt of eac strip is ver small. For a strip at dept below te free surface, te pressure is P = g = g sin u θ p C.G. C.P. d (a) (b) Figure 2.9 Hdrostatic pressure on a plane surface

10 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page 23 Sec. 2.5 Hdrostatic Forces on Flat Surfaces 23 Te total pressure force on te strip is te pressure times te area df = g sin u d Te total pressure force (resultant force) over te entire plane surface is te sum of pressure on all te strips F = L df = L g sin u d = g sin u L d = g sin u (2.11) were = d> is te distance measured from te x-axis to te centroid (or te center of L gravit, C.G.) of te plane (Figure 2.9). Substituting, te vertical distance of te centroid below te water surface, for sin u, we ave F = g (2.12) Tis equation states tat te total drostatic pressure force on an submerged plane surface is equal to te product of te surface area and te pressure acting at te centroid of te plane surface. Pressure forces acting on a plane surface are distributed over ever part of te surface. Te are parallel and act in a direction normal to te surface. Tese parallel forces can be analticall replaced b a single resultant force F of te magnitude sown in Equation Te resultant force also acts normal to te surface. Te point on te plane surface at wic tis resultant force acts is known as te center of pressure (C.P., Figure 2.9). Considering te plane surface as a free bod, we see tat te distributed forces can be replaced b te single resultant force at te pressure center witout altering an reactions or moments in te sstem. Designating p as te distance measured from te x-axis to te center of pressure, we ma tus write Hence, F p = L df df p = L F (2.13) Substituting te relationsips df = g sin u d and F = g sin u, we ma write Equation 2.13 as p = L 2 d (2.14) in wic 2 d = I x and = M x are, respectivel, te moment of inertia and te static L moment of te plane surface wit respect to te x-axis. Terefore, p = I x M x (2.15)

11 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page Water Pressure and Pressure Forces Cap. 2 Wit respect to te centroid of te plane, tis ma be written as p = I = I 0 + (2.16) were I 0 is te moment of inertia of te plane wit respect to its own centroid, is te plane surface area, and is te distance between te centroid and te x-axis. Te center of pressure of an submerged plane surface is alwas below te centroid of te surface area (i.e., p 7 ). Tis must be true because all tree variables in te first term on te rigt-and side of Equation 2.16 are positive, making te term positive. Tat term is added to te centroidal distance ( ). Te centroid, area, and moment of inertia wit respect to te centroid of certain common geometrical plane surfaces are given in Table 2.1. TLE 2.1 Surface rea, Centroid, and Moment of Inertia of Certain Simple Geometrical Plates Sape rea Centroid Moment of Inertia bout te Neutral x-xis Rectangle b x = 1 2 b I 0 = 1 12 b3 x C.G. x = 1 2 b Triangle 1 2 b x = b + c 3 c = x 3 C.G. b x I 0 = 1 36 b3 Circle 1 4 pd2 x = 1 2 d I 0 = 1 64 pd4 d = 2r = 1 2 d r C.G. x Trapezoid a x 1a + b2 2 = 12a + b2 31a + b2 I 0 = 3 1a 2 + 4ab + b a + b2 C.G. b x Ellipse p b x = b = I 0 = p 4 b3 C.G. x b b

12 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page 25 Sec. 2.5 Hdrostatic Forces on Flat Surfaces 25 Sape rea Centroid Semi-ellipse C.G. b b x p 2 b x = b = 4 3p Moment of Inertia bout te Neutral x-xis I 0 = 19p2-642 b 3 72p Parabolic section 2 3 b = 2 5 I 0 = b3 x = ( 1 2 b 2 ) x = 3 x 8 b 0 C.G. b x Semicircle 1 2 pr2 = 4r 3p I 0 = 19p2-642r 4 72 p r C.G. x Example 2.3 vertical trapezoidal gate wit its upper edge located 5 m below te free surface of water is sown in Figure Determine te total pressure force and te center of pressure on te gate. 5 m 3 m 2 m 1 m Figure 2.10

13 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page Water Pressure and Pressure Forces Cap. 2 Solution Te total pressure force is determined using Equation 2.12 and Table 2.1. Te location of te center of pressure is F = g = 9, R R = 2.28 * 10 5 N = 228 kn p = I 0 + were (from Table 2.1) Tus, below te water surface. I 0 = = 5.83 m = 4.00 m 2 p = = 1.22 m = 5.88 m Example 2.4 n inverted semicircular gate (Figure 2.11) is installed at 45 wit respect to te free water surface. Te top of te gate is 5 ft below te water surface in te vertical direction. Determine te drostatic force and te center of pressure on te gate ft 4 ft _ p 5 ft _ C.G. F p C.P. Figure 2.11

14 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page 27 Sec. 2.6 Hdrostatic Forces on Curved Surfaces 27 Solution Te total pressure force is were F = g sin u and Terefore, Tis is te total drostatic force acting on te gate. Te location of te center of pressure is were (from Table 2.1) Terefore, p = = 1 2 3p = 25.1 ft 2 = 5 sec = 8.77 ft 3p F = 62.31sin = 9,700 lbs p = I 0 + I 0 = 19p2-642 r 4 = 28.1 ft 4 72p = 8.90 ft Tis is te inclined distance measured from te water surface to te center of pressure. 2.6 Hdrostatic Forces on Curved Surfaces Te drostatic force on a curved surface can be analzed best b resolving te total pressure force on te surface into its orizontal and vertical components. (Remember tat drostatic pressure acts normal to a submerged surface.) Figure 2.12 sows te curved wall of a container gate tat as a unit widt normal to te plane of te page. ecause te water bod in te container is stationar, ever part of te water bod must be in equilibrium or eac of te force components must satisf te equilibrium conditions tat is, F x = 0 and F = 0. In te free bod diagram of te water contained in ', equilibrium requires te orizontal pressure exerted on plane surface ' (te vertical projection of ) to be equal and opposite te orizontal pressure component F H (te force tat te gate wall exerts on te fluid). Likewise, te vertical component, F V, must equal te total weigt of te water bod above gate. Hence, te orizontal and vertical pressure force on te gate ma be expressed as F x = F - F H = 0 F H = F F = F V - 1W + W 2 = 0 F V = W + W

15 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:03 PM Page Water Pressure and Pressure Forces Cap. 2 Free bod W W W W F F F H F V Figure 2.12 Hdrostatic pressure on a curved surface Terefore, we ma make te following statements. 1. Te orizontal component of te total drostatic pressure force on an surface is alwas equal to te total pressure on te vertical projection of te surface. Te resultant force of te orizontal component can be located troug te center of pressure of tis projection. 2. Te vertical component of te total drostatic pressure force on an surface is alwas equal to te weigt of te entire water column above te surface extending verticall to te free surface. Te resultant force of te vertical component can be located troug te centroid of tis column. Example 2.5 Determine te total drostatic pressure and te center of pressure on te 5-m-long, 2-m-ig quadrant gate in Figure m F V 1.33 m 0.85 F H Figure 2.13

16 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page 29 Sec. 2.6 Hdrostatic Forces on Curved Surfaces 29 Solution Te orizontal component is equal to te drostatic pressure force on te projection plane '. F H = g = 19,790 N>m 3 2a 1 12 m2b312 m215m24 = 97,900 N 2 Te location of te orizontal component is, were = 10 m 2 p = I 0 / + (projected area) and I 0 = 315 m212 m2 3 4>12 = 3.33 m 4, p = m 4 2>3110 m m = 1.33 m below te free surface. Te vertical component is equal to te weigt of te water in te volume O. Te direction of tis pressure component is downward. F V = g (Vol) = 19,790 N>m 3 2a 1 4 p 12 m22 b15 m2 = 154,000 N Te pressure center is located at 4122>3p = 0.85 m (Table 2.1), and te resultant force is F = 2197, , = 182,000 N u = tan -1 a F V F H b = tan -1154, ,000 = 57.6 Example 2.6 Determine te total drostatic pressure and te center of pressure on te semiclindrical gate sown in Figure Solution Te orizontal component of te drostatic pressure force on te projection plane '' per unit widt can be expressed as F H = g = ga H 2 b H = 1 2 gh2 γ H ' F V1 F H C O γh ' F V2 Figure 2.14

17 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap. 2 Te pressure center of tis component is located at a distance of H/3 from te bottom. Te vertical component can be determined as follows. Te volume 'C over te upper alf of te gate, C, produces a downward vertical pressure force component: Te vertical pressure force component exerted b te water on te lower alf of te gate, C, is upward and equivalent to te weigt of water replaced b te volume 'C: combining tese two components, one can see tat te direction of te resultant vertical force is upward and equal to te weigt of te water replaced b te volume C. Te resultant force is ten F V1 F V2 = -g H2 4 - ph2 16 = g H2 4 + ph2 16 F V = F V1 + F V2 = g - H2 4 - ph H2 4 + ph2 16 R = g p 8 H2 F = gh p2 64 u = tan -1 F V F H = tan -1 a p 4 b = 38.1 ecause all pressure forces are concurrent at te center of te gate, point O, te resultant force must also act troug point O. 2.7 uoanc rcimedes discovered (~250.C.) tat te weigt of a submerged bod is reduced b an amount equal to te weigt of te liquid displaced b te bod. rcimedes principle, as we now call it, can be easil proven b using Equation ssume tat a solid bod of arbitrar sape,, is submerged in water as sown in Figure vertical plane MN ma ten be drawn troug te bod in te direction normal to te page. One observes tat te orizontal pressure force components in te direction of te paper, F H and F H, must be equal because te bot are calculated using te same vertical projection area MN. Te orizontal pressure force components in te direction normal to te page must also be equal for te same reason; te sare te same projection in te plane of te page. Te vertical pressure-force component can be analzed b taking a small vertical prism ab wit a cross-sectional area d. Te vertical pressure force on top of te prism 1g 1 d2 acts downward. Te vertical force on te bottom of te prism 1g 2 d2 acts upward. Te difference gives te resultant vertical force component on te prism (buoanc force) F V = g 2 d - g 1 d = g d c wic is exactl equal to te weigt of te water column ab replaced b te prism. In oter words, te weigt of te submerged prism is reduced b an amount equal to te weigt of te liquid replaced b te prism. summation of te vertical forces on all te prisms tat make up te entire submerged bod gives te proof of rcimedes principle.

18 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page 31 Sec. 2.8 Flotation Stabilit 31 S 1 S 2 1 F Va M 2 a F H b F H N F Vb Figure 2.15 uoanc of a submerged bod rcimedes principle ma also be viewed as te difference of vertical pressure forces on te two surfaces N and M. Te vertical pressure force on surface N is equal to te weigt of te potetical water column (volume of S 1 NS 2 ) acting upward; and te vertical pressure force on surface M is equal to te weigt of te water column S 1 MS 2 acting downward. ecause te volume S 1 NS 2 is larger tan te volume S 1 MS 2 b an amount exactl equal to te volume of te submerged bod MN, te net difference is a force equal to te weigt of te water tat would be contained in te volume MN acting upward. Tis is te buoanc force acting on te bod. floating bod is a bod partiall submerged resulting from a balance of te bod weigt and buoanc force. 2.8 Flotation Stabilit Te stabilit of a floating bod is determined b te relative positions of te center of gravit of bod G and te center of buoanc, wic is te center of gravit of te liquid volume replaced b te bod, as sow in Figure Te bod is in equilibrium if its center of gravit and its center of buoanc lie on te same vertical line, as in Figure 2.16 (a). Tis equilibrium ma be disturbed b a variet of causes (e.g., wind or wave action), and te floating bod is made to eel or list troug an angle as sown in Figure 2.16 (b). Wen te floating bod is in te eeled position, te center of gravit of te bod remains uncanged, but te center of buoanc, wic is now te center of gravit of area a'cb', as been canged from to '. Te buoant force g # Vol, acting upward troug ', and te weigt of te bod W, acting downward troug G, constitute a couple, W # X, wic resists furter overturning and tends to restore te bod to its original equilibrium position. extending te line of action of te buoant force troug te center of buoanc ', we see tat te vertical line intersects te original axis of smmetr c-t at a point M. Te point M is known as te metacenter of te floating bod, and te distance between te center of gravit and te metacenter is known as te metacentric eigt. Te metacentric eigt is a measure of te flotation stabilit of te bod. Wen te angle of inclination is small, te position of M does not

19 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap. 2 L a t W O G. Vol c (a) Figure 2.16 a a b G θ b c W O (b) F 2. Vol Center of buoanc and metacenter of a floating bod F 1 S X M t b cange materiall wit te tilting position. Te metacentric eigt and te rigting moment can be determined in te following wa. ecause tilting a floating bod does not cange te total bod weigt, te total displacement volume is not canged. Te roll troug an angle onl canges te sape of te displaced volume b adding te immersion wedge bob' and subtracting te emersion wedge aoa'. In tis new position, te total buoanc force 1g # Vol2 is sifted troug a orizontal distance S to '. Tis sift creates a couple F 1 and F 2 because of te new immersion and emersion wedges. Te moment of te resultant force 1g # Vol) about point must equal te sum of te moments of te component forces: or 1gVol2 1S2 = 1gVol2 1zero2 + moment of te force couple = 0 + gvol wedge L 1gVol2 S = gvol wedge L S = Vol wedge L 1a2 Vol were Vol is te total volume submerged, Vol wedge is te volume of wedge bob' (or aoa'), and L is te orizontal distance between te centers of gravit of te two wedges. ut, according to te geometric relation, we ave S = M sin u or M = S sin u (b) Combining Equations (a) and (b), we get M = Vol wedgel Vol sin u

20 M02_HOUG6380_04_SE_C02.qxd 7/8/09 5:31 PM Page 33 Sec. 2.8 Flotation Stabilit 33 O θ x d b b Figure 2.17 For a small angle, sin u L u, te previous relationsip ma be simplified to M = Vol wedgel Vol u Te buoanc force produced b wedge bob', as depicted in Figure 2.17, can be estimated b considering a small prism of te wedge. ssume tat te prism as a orizontal area, d, and is located at a distance x from axis of rotation O. Te eigt of te prism is x(tan u). For a small angle, it ma be approximated b xu. Tus, te buoanc force produced b tis small prism is gxud. Te moment of tis force about te axis of rotation O is gx 2 ud. Te sum of te moments produced b eac of te prisms in te wedge gives te moment of te immersed wedge. Te moment produced b te force couple is, terefore, gvol wedge L = FL = L gx 2 ud = gu L x 2 d ut L x 2 d is te moment of inertia of te waterline cross-sectional area of te floating bod about te axis of rotation O. Hence, we ave I 0 = L x 2 d Vol wedge L = I 0 u For small angles of tilt, te moment of inertia for uprigt cross section aob, and te inclined cross section a'ob' ma be approximated b a constant value. Terefore, M = I 0 Vol (2.17) Te metacentric eigt, defined as te distance between te metacenter M and te center of gravit G, can be estimated: GM = M ; G = I 0 Vol ; G (2.18) Te distance between te center of gravit and te center of buoanc G in te uprigt position, sown in Figure 2.16, can be determined b te sectional geometr or te design data of te vessel.

21 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap. 2 Te ; sign indicates te relative position of te center of gravit wit respect to te center of buoanc. For greater flotation stabilit, it is advantageous to make te center of gravit as low as possible. If G is lower tan, ten G would be added to te distance to M and produce a larger value of GM. Te rigting moment, wen tilted as depicted in Figure 2.16 (b), is M = W GM sin u (2.19) Te stabilit of buoant bodies under various conditions ma be summarized as follows. 1. floating bod is stable if te center of gravit is below te metacenter. Oterwise, it is unstable. 2. submerged bod is stable if te center of gravit is below te center of buoanc. Example m * 4 m rectangular box caisson is 2 m deep (Figure 2.18). It as a draft of 1.2 m wen it floats in an uprigt position. Compute (a) te metacentric eigt and (b) te rigting moment in seawater 1sp. gr. = wen te angle of eel (list) is 8. Solution From Equation 2.18 were GM = M - G and I 0 is te waterline area moment of inertia of te box about its longitudinal axis troug O. Terefore, (Note: L = 4 m, w = 3 m, = 2 m.) GM = 1 12 Lw3 Lw = m M = I 0 Vol 8 M 4 m G 2 m O x 3 m Figure 2.18

22 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page 35 Sec. 2.8 Flotation Stabilit 35 Te specific gravit of seawater is 1.03; from Equation 2.19, te rigting moment is M = WGM sin u = 319,790 N>m m213 m211.2 m m21sin 8 2 = 4,550 N# m PROLEMS (SECTION 2.2) Collapse dept (or crus dept) is te submerged dept tat a submarine cannot exceed witout collapsing because of te surrounding water pressure. Te collapse dept of modern submarines is not quite a kilometer 1730 m2. ssuming seawater to be incompressible 1sp. gr. = 1.032, wat is te crus dept pressure in N/m 2 and psi 1lb/in. 2 2? Is te pressure ou computed absolute or gauge pressure? clindrical water tank (Figure P2.2.2) is suspended verticall b its sides. Te tank as a 10-ft diameter and is filled wit 20 C water to a 3-ft dept. Determine te force exerted on te tank bottom using two separate calculations (a) based on te weigt of te water, and (b) based on te drostatic pressure on te bottom of te tank. Figure P Te simple barometer in Figure P2.2.3 uses water at 30 C as te liquid indicator. Te liquid column rises to a eigt of 9.8 m from an original eigt of 8.7 m in te vertical tube. Compute te new atmosperic pressure, neglecting surface tension effects. Wat is te percentage error if te direct reading is used and vapor pressure is ignored? 8.7 m Figure P Mercur is often used in barometers as depicted in Figure P Tis is because te vapor pressure of mercur is low enoug to be ignored, and because it is so dense 1sp. gr. = te tube can be sortened considerabl. Wit te atmosperic pressure found in Problem kn/m 2 at 30 C2, determine te column eigt in meters (and feet) if mercur is used storage tank 16 m * 6 m * 6 m2 is filled wit water. Determine te force on te bottom and on eac side.

23 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap ft-ig, 1-ft-diameter pipe is welded to te top of a cubic container 13 ft * 3 ft * 3 ft2. Te container and pipe are filled wit water at 20 C. Determine te weigt of te water and te pressure forces on te bottom and sides of te container closed tank contains a liquid 1sp. gr. = under pressure. Te pressure gauge depicted in Figure P2.2.7 registers a pressure of 4.50 * 10 4 N/m 2 (pascals). Determine te pressure at te bottom of te tank and te eigt of te liquid column tat will rise in te vertical tube. 1 m 1.4 m Figure P n underwater storage tank was constructed to store natural gas offsore. Determine te gas pressure in te tank (in pascals and psi; lb/in 2 ) wen te water elevation in te tank is 6 m below sea level (Figure P2.2.8). Te specific gravit of seawater is m Figure P closed tank contains oil wit a specific gravit If te gauge pressure at a point 10 feet below te oil surface is 23.7 psi 1lb/in. 2 2, determine te absolute pressure and gauge pressure (in psi) in te air space at te top of te oil surface multiple-piston draulic jack as two output pistons, eac wit an area of 250 cm 2. Te input piston, wose area is 25 cm 2, is connected to a lever tat as a mecanical advantage of 9:1. If a 50-Nforce is exerted on te lever, ow muc pressure 1kN/m 2 2 is developed in te sstem? How muc force (kn) will be exerted b eac output piston? (SECTION 2.4) Referring to Figure 2.4 (c), if te eigt of water (at 4 C) above point 7 is 52.3 cm, wat is te eigt of te oil 1sp. gr. = above point 8? (Note: Point 9 is 42.5 cm above point 7.) significant amount of mercur is poured into a U-tube wit bot ends open to te atmospere. If water is poured into one leg of te U-tube until te water column is 3 feet above te mercur water meniscus, wat is te elevation difference between te mercur surfaces in te two legs? n open tank in a petroleum compan lab contains a laer of oil on top of a laer of water. Te water eigt is 4 times te oil eigt. Te oil as a specific gravit of If te gauge pressure at te bottom of te tank indicates 26.3 cm of mercur, wat is te oil eigt?

24 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page 37 Sec. 2.8 Flotation Stabilit mercur 1sp. gr. = manometer is used to measure water pressure in a pipe. Referring to Figure 2.5 (a), te value of is 3.40 cm and te value of is 2.60 cm. Determine te pressure in te pipe manometer is mounted on a cit water-suppl pipe to monitor water pressure, as sown in Figure P However, te manometer reading of 3 ft 1Hg2 ma be incorrect. If te pressure in te pipe is measured independentl and found to be 16.8 lb/in. 2 1psi2, determine te correct value of te reading. 2 ft H 2 O Hg Figure P n open manometer, sown in Figure P2.4.6, is installed to measure pressure in a pipe carring an oil 1sp. gr. = If te monometer liquid is carbon tetracloride 1sp. gr. = 1.602, determine te pipe pressure (in meters of water-column eigt). ir 66 cm Oil 58 cm Manometer liquid Figure P In Figure P2.4.7, a single-reading mercur manometer is used to measure water pressure in te pipe. Wat is te pressure (in psi) if 1 = 6.9 in. and 2 = 24.0 in.?

25 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap Figure P In Figure P2.4.7, determine te water pressure [in kilo-pascals (kpa)] in te pipe if 1 = 20.0 cm and 2 = 67.0 cm. lso determine te cange in liquid eigt 1 for a 10 cm cange in 2 if te diameter of te manometer tube is 0.5 cm and te diameter of te manometer fluid reservoir is 5 cm In Figure P2.4.9, water is flowing in pipe and oil 1sp. gr. = is flowing in pipe. If mercur is used as te manometer liquid, determine te pressure difference between and in psi. 32 in. 12 in. 20 in. Manometer liquid Figure P micromanometer consists of two reservoirs and a U-tube as sown in Figure P Given tat te densities of te two liquids are r 1 and r 2, determine an expression for te pressure difference 1P 1 - P 2 2 in terms of r 1, r 2,, d 1, and d 2.

26 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page 39 Sec. 2.8 Flotation Stabilit 39 P 1 P 2 Δ ρ 1 ρ 1 Diameter d 1 Diameter d 2 ρ 2 Figure P For te sstem of manometers sown in Figure P2.4.11, determine te differential reading. Two different manometr fluids are being used wit different specific gravities. Water Water 46 cm 60 cm 23 cm 20 cm 16 cm 20 cm sp. gr. = 13.6 sp. gr. = 0.8 sp. gr. = 13.6 sp. gr. = 0.8 Figure P Determine te air pressure (kpa and cm of Hg) in te sealed left tank of Figure P if E = 32.5 m.

27 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap m 35 m ir Oil sp. gr. = 0.8 ir Water 37 m 20 kn/m 2 sp. gr. = 1.6 E Figure P (SECTION 2.5) vertical gate keeps water from flowing in a triangular irrigation cannel. Te cannel as a 4-m top widt and a 3-m dept. If te cannel is full, wat is te magnitude of te drostatic force on te triangular gate and its location? concrete dam wit a triangular cross section (Figure P2.5.2) is built to old 30 ft of water. Determine te drostatic force on a unit lengt of te dam and its location. lso, if te specific gravit of concrete is 2.67, wat is te moment generated wit respect to te toe of te dam,? Is te dam safe? 30 ft 10 ft Figure P m-diameter circular (plane) gate is mounted into an inclined wall (45 ). Te center of te gate is located 1 m (verticall) below te water surface. Determine te magnitude of te drostatic force and its location wit respect to te surface of te water along te incline vertical plate, composed of a square and a triangle, is submerged so tat its upper edge coincides wit te water surface (Figure P2.5.4). Wat is te eigt to lengt ratio suc tat te pressure force on te square is equal to te pressure force on te triangle?

28 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page 41 Sec. 2.8 Flotation Stabilit 41 L L H Figure P2.5.4 Front view of verticall submerged plate Te rectangular gate in Figure P2.5.5 is inged at and separates water in te reservoir from te tail water tunnel. If te uniforml tick gate as a dimension of 2 m * 3 m and weigs 20 kn, wat is te maximum eigt for wic te gate will sta closed? (Hint: ssume te water level does not rise above te inge.) Hinged 45 1 m 2 m Figure P circular gate is installed on a vertical wall as sown in Figure P Determine te orizontal force P necessar to old te gate closed if te gate diameter is 6 ft and = 7 ft. Neglect friction at te pivot. Pivot D P Figure P2.5.6

29 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap Figure P2.5.7 sows a 10-ft ig (H), vertical, rectangular gate. Te gate opens automaticall wen increases to 4 ft. Determine te location of te orizontal axis of rotation 0 0'. H 0 0 Figure P Calculate te magnitude and te location of te resultant pressure force on te annular gate sown in Figure P m Hub d = 3 m Hub 1 m Figure P Calculate te magnitude and te location of te resultant pressure force on te annular gate sown in Figure P2.5.8 if te round central ub is replaced b a square ub ( 1 m b 1 m) In Figure P2.5.10, te wicket dam is 5 m ig and 3 m wide and is pivoted at its center. Determine te reaction force in te supporting member. Hinge 5 m Figure P2.5.10

30 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page 43 Sec. 2.8 Flotation Stabilit Determine te dept of te water (d) in Figure P tat will cause te gate to open (la down). Te gate is rectangular and is 8 ft wide. Neglect te weigt of te gate in our computations. t wat dept will it close? 5,000 lbs d Hinge ft Figure P Neglecting te weigt of te inged gate, determine te dept at wic te gate will open in Figure P Hinge 1 m Figure P Te circular gate sown in Figure P is inged at te orizontal diameter. If it is in equilibrium, wat is te relationsip between and as a function of,, and d? γ γ d Figure P2.5.13

31 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap sliding gate 10 ft wide and 6 ft ig is installed in a vertical plane and as a coefficient of friction against te guides of 0.2. Te gate weigs 3 tons, and its upper edge is 27 ft below te water surface. Calculate te vertical force required to lift te gate. (SECTION 2.6) m-long curved gate depicted in Figure P2.6.1 is retaining a 6-m dept of water in a storage tank. Determine te magnitude and direction of te total drostatic force on te gate. Does te force pass troug point? Explain. 6 m 2 m Gate Figure P emisperical viewing port in a marine museum (Figure P2.6.2) as a 1-mradius, and te top of te port is 3 meters below te surface of te water (). Determine te magnitude, direction, and location of te total drostatic force on te viewing port. (ssume te saltwater as a sp. gr. = 1.03.) Figure P n inverted emisperical sell of diameter d as sown in Figure P2.6.3 is used to cover a tank filled wit water at 20 C. Determine te minimum weigt te sell needs to be to old itself in place if te diameter is 6 feet.

32 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page 45 Sec. 2.8 Flotation Stabilit 45 d H 2 O Figure P Te corner plate of a barge ull is curved, wit a radius of 1.75 m. Te barge is leaking, and te dept of submergence (draft) is 4.75 m as depicted in Figure P Te water on te inside is up to level, producing drostatic pressure on te inside as well as te outside. Determine te resultant orizontal drostatic pressure force and te resultant vertical drostatic force on plate per unit lengt of ull. 3 m 1.75 m Figure P Te tainter gate section sown in Figure P2.6.5 as a clindrical surface wit a 12-m radius; it is supported b a structural frame inged at O. Te gate is 10 m long (in te direction perpendicular to te page). Determine te magnitude, direction, and location of te total drostatic force on te gate (Hint: Determine te orizontal and vertical force components.) 45 0 Figure P2.6.5

33 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap Calculate te magnitude, direction, and location of te total drostatic pressure force (per unit lengt) on te gate sown in Figure P ft r = 10 ft 8 ft Figure P ft-diameter clindrical tank las on its side wit its central axis orizontal. 1.5-ft diameter pipe extends verticall upward from te middle of te tank. Oil 1sp. gr. = 0.92 fills te tank and pipe to a level of 8 ft above te top of te tank. Wat is te drostatic force on one end of te tank? Wat is te total drostatic pressure force on one side (semicircle) of te tank if it is 10 ft long? Calculate te orizontal and vertical forces acting on te curved surface C in Figure P R R C Figure P Calculate te magnitude and location of te vertical and orizontal components of te drostatic force on te surface sown in Figure P2.6.9 (quadrant on top of te triangle, bot wit a unit widt). Te liquid is water, and te radius R = 4 ft. R 6 ft 8 ft Figure P2.6.9

34 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page 47 Sec. 2.8 Flotation Stabilit In Figure P a omogeneous cone plugs te 0.1-m-diameter orifice between reservoir tat contains water and reservoir tat contains oil 1sp. gr. = Determine te specific weigt of te cone if it unplugs wen 0 reaces 1.5 m. 0.3 m 0.5 m m 0.1 m Figure P Wat would be te specific weigt of te cone if reservoir in Figure P contains air at a pressure of 8,500 N/m 2 instead of oil? Te omogeneous clinder 1sp. gr. = 2.02 in Figure P is 1 m long and 12 m in diameter and blocks a 1-m 2 opening between reservoirs and 1sp. gr. = 0.8, sp. gr. = Determine te magnitude of te orizontal and vertical components of te drostatic force on te clinder. 6 m 5 m 2 m u 45 l 1 m Figure P (SECTION 2.8) piece of irregularl saped metal weigs 301 N. Wen te metal is completel submerged in water, it weigs 253 N. Determine te specific weigt and te specific gravit of te metal Te solid floating prism sow in Figure P2.8.2 as two components. Determine g and g in terms of if g = 1.5 # g.

35 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap. 2 H γ 1.5 H 2 H Figure P solid brass spere of 30-cm diameter is used to old a clindrical buo in place (Figure P2.8.3) in seawater 1sp. gr. = Te buo 1sp. gr. = as a eigt of 2 m and is tied to te spere at one end. Wat rise in tide,, will be required to lift te spere off te bottom? 50 cm 30 cm sp. gr Figure P Tree people are in a boat wit an ancor. If te ancor is trown overboard, will te lake level rise, fall, or sta te same teoreticall? Explain freswater clindrical ancor 1 = 1.2 ft and D = 1.5 ft2 is made of concrete 1sp. gr. = Wat is te maximum tension in te ancor line before te ancor is lifted from te lake bottom if te ancor line is at an angle of 60 wit respect to te bottom? In Figure P2.8.6, te sperical buo of radius R opens te square gate wen water rises to te alf-buo eigt. Determine R if te weigt of te buo and te gate are negligible. R 1m Hinge 1m 1m Figure P2.8.6

36 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page 49 Sec. 2.8 Flotation Stabilit Te floating rod sown in Figure P2.8.7 weigs 150 lbs, and te water s surface is 7 ft above te inge. Calculate te angle assuming a uniform weigt and buoanc distribution. 6'' 6'' 12' θ 7' Hinge Figure P rectangular barge is 14 m long, 6 m wide, and 2 m deep. Te center of gravit is 1.0 m from te bottom, and te barge draws 1.5 m of seawater 1sp. gr. = Find te metacentric eigt and te rigting moments for te following angles of eel (or list): 4, 8, and Figure P2.8.9 sows a buo tat consists of a wooden pole 25 cm in diameter and 2 m long wit a sperical weigt at te bottom. Te specific gravit of te wood is 0.62, and te specific gravit of te bottom weigt is Determine (a) ow muc of te wooden pole is submerged in te water, (b) te distance to te center of buoanc from te water level, (c) te distance to te center of gravit from te water level, and (d) te metacentric eigt. 2 m 0.25 m 0.5 m Figure P wooden block is 2 m long, 1 m wide, and 1 m deep. Is te floating block stable if te metacenter is at te same point as te center of gravit? Explain subwa tunnel is being constructed across te bottom of a arbor. Te process involves tugboats tat pull floating clindrical sections (or tubes as te are often called) across te arbor and sink tem in place, were te are welded to te adjacent section alread on te arbor bottom. Te clindrical tubes are 50 feet long wit a diameter of 36 feet. Wen in place for te tugboats, te tubes are submerged verticall to a dept of 42 feet, and 8 feet of te tube is above te water 1sp. gr. = To accomplis tis, te tubes are flooded wit 34 feet of water on te inside. Determine te metacentric eigt and estimate te rigting moment wen te tubes are tipped

37 M02_HOUG6380_04_SE_C02.qxd 7/3/09 7:04 PM Page Water Pressure and Pressure Forces Cap. 2 troug a eel (list) angle of 4 b te tugboats. (Hint: ssume te location of te center of gravit can be determined based on te water contained inside te tubes and te container weigt is not tat significant.) m-long, 4.8-m-wide, and 4.2-m-deep rectangular pontoon as a draft of 2.8 m in seawater 1sp. gr. = ssuming te load is uniforml distributed on te bottom of te pontoon to a dept of 3.4 m, and te maximum design angle of list is 15, determine te distance tat te center of gravit can be moved from te center line toward te edge of te pontoon.