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Chapter 2 FLUID STATICS

????? :?????? ??? ??? ??? ?? ????? ??? ??? ??? ??? ???? ?? ???? ???? ??? ??? ???? ??? ????? ??? ????? ???? ??? ?????. ???? ?? ???? ??? ??? ????? ?? ????? ????? ???? ?? ????.???? ??????? ??? ?? ????? ? ??? ???? ???? ??? ?? ????? ???.. 2.1 PRESSURE AT A POINT. ????? ??? ?? ???? ? ???? ??? ????. ?

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Chapter 2 FLUID STATICS

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    1. Chapter 2 FLUID STATICS

    2. ????? : ?????? ??? ??? ??? ?? ?? ??? ??? ??? ??? ??? ? ??? ?? ???? ???? ??? ??? ???? ??? ????? ??? ????? ???? ??? ?????. ???? ?? ???? ??? ??? ????? ?? ????? ????? ???? ?? ????. ???? ??????? ??? ?? ????? ? ??? ???? ???? ??? ?? ????? ???.

    3. 2.1 PRESSURE AT A POINT ????? ??? ?? ???? ? ???? ??? ????. ? ???? ?? ?? ? ?? ???? ????? 0?? ??? ? ? ?? ???? ???? ?? ? ?? ?? ?????. ?????? ?? ?? ???? ??? ?? ??? ? ?? ??? ?? ???. ??? ??? ?????? ?? ?? ?? ?? ??? ? ??? ????? ???? ??? ?. ? ??? ???? ??? ??? ?? ?? ??? ????? ?? ????. ??? ??? ???? ??? ?????? ? ? (x,y)?? ???? ?? ????? ????? ??? ??? ?? ?? ??? ???? ??(?? 2.1).

    5. ? ?? ???? ???? ??, ?? ?????? ???? ?????. x? y ??? ?? ???? ?? ??? ?? ????. ??? px, py, ps? ?? ? ?? ???? ??????, r? ??? ???, ?? ??, ax, ay? ???? x, y ???? ? ? ??? ????? ???? ? (x, y)? ?? ??? ???? ????? ??? 0?? ???? ??? ??? ???? ??? ? ???? ???? ??? ?? ??? ?? ??? ?? ??? ?? ??? ?????? ??? ? ??.

    6. dy?dx? ?? ?? ? ???? ???? ?? ??? ???. ? ? ??? ??? ?????? ?? ?? ??????? ? ?? ???? ??? ?? ???? ??? ??? ?? ??. ? ??? 2??? ??? ??? ????, 3?? ????? 1?? ?? ???? ?? ?? ???? ????? ????? ? ???? 3??? ???? ??? ??? ? ??. ??? ???? ? ?? ???? ??? ????? ??? ????? ???? ?? ????? ? ?? ???? ????? ?? ???? ???? ?? ??. ? ?? ? ?? ???? ??? ? ?? ???? ?? ?? ?? ??????? ???? ????. ????? 0? ????, ? ??? ?? ????? ??? ?? ??? ???? ????? ???? ???? ? ???? ??? ?? ???? ??.

    7. 2.2 BASIC EQUATION OF FLUID STATICS ???????? ???? ?? Force balance: ?????? ??? ???? ? (Fig. 2.2) : ???(surface forces ), ???( body forces) ?????? ???? ????? ??? ?, ? y??? ??? ???? y???? -? dx dy dz ? ????. ?? (x, y, z)??? ??? p? ? ? y?? ???? ??? ?? ??? ?? ???? ?? ???? ???? ?? ??? dy/2? ?????? y?? ???? ???? ????.

    9. y???? ???? ?? ??? ??? ??. x? z??? ????, ? ??? ?????? ???? ???? ??2.2? ??? ???? ? ?? dF? ??? ??. dx dy dz = dV ? ??? ???, ??? ??? 0?? ????? ? ?? ? ??? ????? ???? ???? ????, ???? ???? ? ??? 0? ??? ?? ?? ? ?

    10. p? ???? -?p? ??? ?? ????? ???? ???? ??? f??. ????? ?? ?????? ??? ?? ?? ???? ?? ??????, ?? ??? ?? ??? ????? 0? ??? ??? ??? Newton? ?? ? 2??? ??? ?? ??? ????. ??? a? ????? ?????. (f - jr)? ??? ??????? ??? ? ??? ???? ?????.

    11. ?(2.2.4)? ???? ??? ???? ??? ?? ??. ????? ????? ???? ????? ? ??? Pascal? ??? ??? ???. ?, ???? ?? ??????? ?? ??? ?? ? ???? ??? ??? ?? ????. p? y?? ????? ?????? ???, ?????? ??? ???? ???, ???? ????? ???? ??? ??? ???? ???? ??. ??, ????????? r? ????? ? (2.2.7)? ???? ????? c? ?? ??? ?? ?? ??. ????? ?? ????? ??? ?? ?? ??? ?? ??. ??? h ? ???????? ???? ?? (h = -y )?? p? ??????? ???? ??? ??? ??? ????

    12. [?? 2.1] ????? ?? 5m? ?????? ???? ??. ???? ??? ?? ? 100m?? ?? ????? ?? ? ??? ??. ??? ??? 1.020?? ?? ????? ??? ????? ????? ???. At the top, h = 100 m, and ??? ?????? ??? ? ??? ? ?? ?????

    13. ????????? ???? ? ??? ?? ??? ????? ?? ????? ? ? ? (1.6.2)??? ?? ?? ?? ? ??. ? (2.2.7)?? ? ? ?? ?g? ????, ? (2.2.7)? ? (2.2.9)??? ? ? ???? ?? ?? ?? ? ??. ??P = P0 ?? ? = ?0? ?? ? ????? ??? ???? ?? ??? ?? ?? ? ??? ??? ?? ????? ???? ?????. ??? ?? ??? ?? ???????? ???? ????. ??????? ???? ????? (-0.00651 K/m)??. ????? ??????, ??? ??? ??? ??? ??? ? ??.

    14. [?? 2.2] ??? ???? ????? ????, 2,000m ????? ??? ?? ? ????. ????? ??? ??? ?? Pa abs, 1.24kg/ ??. ? (2.2.12)? ?? ? (2.2.9)? ?? #

    15. 2.3 UNITS AND SCALES OF PRESSURE MEASUREMENT ??? ?? ??? ???? ??? ? ??. ?? ???? ?? ????? ?? 0? ??(absolute zero) ?? ???(local atmospheric pressure) ????(Absolute pressure): ?????? ??? ??? ? ?? ????(Gage pressure):??????? ??? ?? ? ? ??

    17. The bourdon ??? (Fig. 2.3): ????? ???? ???? ??? ??????? ??? ??? ???? ???? ??? ?????. ?? ?? ?? ??, ??? ?? ????? ?? ??? ?????? ?? ??. ????? ??? ???? ?? ?? ???? ?? ??? ????? ?? ?? ???? ???? ???? ??. ??? ??? ???? ?? ?? ?? ??? ?? ? ?? ??? 0?? ????. ?? ???? ? ??? ????? ??? ??? ??? ??? ????. ??????? ?? ?????? ??? ??.

    19. ?? 2.4: ????? ?? ??? ??? ??? ?? ???? ?????. ?????? ????? ?????? 29.92inHg. ??? ?? ? ??? ??? ?? ???? ????? ???? ?? ????. ????? ??? ?? ????? ??? p = ?h? ??? ?? ??? ?? h? ?? p??? ??? ????. ?: ?? 9806 N/m3.

    20. ?? ???(Local atmospheric pressure) ?????(mercury barometer) Aneroid ???(Aneroid barometer): ???? ???? ??, ?? bourdon ???? ???? ?? ?? ?? ??? ??? ???? ?????? ????. ?? ???(Mercury barometer): ?? ?? ?? ?? ???? ??? ?? ??? ??? ??? ?? ?? ?? ?? ?? ???? ? ???. ???? ???? R? ?? ?? ? ??? ??? ??? ??. ?? ??? ???? ????? ???? ??, ????? ?? hv? mmHg? ????, R? mm ??? ?? A?? ??? ??? ?? ??? ? ??.

    21. ?? 2.4?? ??? ????? ????? ????, ?? 0? ??? ??? ?????? ???? ?? ??? ???? ???. ????? ??????? ?? ???????? ???? ?? ???? ??(??????? ?? ??)? ?? ?? ??(vacuum)?? ??. ???, ???? ??? 720mmHg ? ? ? 1??? ?? ??? 460mmHg ?? ?? ????? -260mmHg, 11inHg ?? ??? ?? ? ? ??.???? ????? ?????? ??? ?? ??? ????. . Note: Pabs = pbar + pgage ???? : P, ????? : p.

    22. [??2.3]???? ??? ?? ?????? ?????? ??. ??? ?? ??? ??? ? ?? ???? ?? ?? ????. ??? ? ??? ??? ? ??? ???? ??? ???? ??? ????????(dry adiabatic laps late)?? ?? ??? ????. ?????? 30ft ???? ????? ? ??? ??? ????? ???? 20oF ?? ??? ????. ?? ????? ??? ?? ??? ??? ???? (a) (a) ????? = -0.00651oC/m, t0 =20oC. (b) ????? = 0.00365oC/m??.

    23. ?(2.2.7)? (2.2.14)? ???? ??? ?? (???????, 7.1?)???? ??? ??? ?? ??? ????? ??? T1? ??? ?? ???? P0? ?? ?? ????, K? ?????. ??? 2???? ???? K? 1.4??. . ? ??? P/P0 ? ???? ??? ? ????? ??? ? ?? ?? ? ???. ??? ? ??? ?? y? ??? = -0.00651oC per metre, R = 287 m?N/(kg?K), a = 2.002, ?? y = 3201 m. ???? ??? = -0.00365oC per metre??, a = -0.2721, ?? y = 809.2 m. #

    24. 2.4 MANOMETERS ???? ?? ??? ???? ???? ???? ????. ?? 2.6a: ???? ??? ? ???? ????? 0?? ? ????? ???? ? ????. ???? ??? ????? ?? ??? ???? ???? ???? ?? ??? ??? ??? ??? ???? ???? ?? ??. ??? ??? ??(????)???? ???? ?????? ???? h? ????. ?? h ? ??? ??? ??????? ????. Piezometer?? ?? ???? ???? ???? ?? ??. ??? ??? ??(????)???? ???? ?????? ???? h ? ????. ???? ?? ??? ????? ??? ???? ? ??? ????. ?? A?? ??? ?? ??? ??? ????? ???? ?? ??? ??? ???? ??. ??? ??? S? ?? A?? ??? ??????? hS???.

    26. ?? 2.6b: ?? ? ?? ?? ????? ???? ???? ???? ?? ???? ????(???)? ? ??? ? ?? ??. ??????? ?????? 0? ?? ? h? ??? ??? A?? ??? ????? units of length H2O ??? ? ?? ?? ????? ???? ???? ?? 2.6(c)? ?? ??? ?? ? ?2? ??? ????. ?2? ??? ????? ?? ?1? ??? ???? ?? ???? ?? ?1? ??? ??? ??? ??. A?? ?? ?1??? ??? S1 ???? ?? ?2??? ??? S2 ? ?? A? ?? ??? ????? ???? ???? ?? ????? A?? ??? ?? ???? ??? ??? ?? ??? ? ??. hA ????? ?? A ?? ??? ??? ??? ??? ?, h1, h2 - ????

    28. ?????(??2.7)? A, B ?? ???? ????? ??? ? ?? ? ? ? ??? ???? ??? ? ?? ??. ??? ??? ?? ??? ??2.7(a)? ????, For Fig. 2.7b: ?? A, B??? ?? ??? ??? ??? ? ??? Fig. 2.7a, For Fig 2.7b:

    29. [?? 2.4] ?? 2.7(a)?? A? B? ?? ??? ???, ????? ?? S=0.8? ??? ?? ??. h1=300mm, h2=200mm, h3=600mm ??. (a) pA-pB? ? Pa??? (b) pB=50kPa ?? ??? ??? 730mmhg ??? A??? ?? ??? ????? ? m? ?????? (a) (b) (a)?

    30. Micromanometers ?? ?? ???? ? ???? ???? ???? ??? ???? ? ?? ??? ???? ???? ??. ?? ?? ???? ? ???? ??? ???? ???? ??? ? ??? ??? ?? ??. ?, ? ?? ????? ? ??? ?? ???? ?? ? ?? ?? ????? ???(vernier)? ???? ?? ? ??? ?? ??. ???? ???(pinion)? ???? ??? ??? ??? ? ?? ?(rack)? ???? ??, ????? ??? ???? ??? ? ??? ?? ??.

    32. Fig. 2.8: ?? ???? ?? ? ?? ??? ?? ?? ?? ???? ??? ??? ? ??R? ???? ? ??. ??? ???? U??? 0-0 ??? ?? ??, ?? ??? ???? ?? ?? ???? ? ???(reservoir)? 1-1??? ???. ????? ?? ??? C? D? ???? ????? ?? ?? ??? 1-1? ??? ??? ???. C??? ??? D?? ?? ??? ???? ????? ?? 2.8? ?? ?? ???. ? ????? ??? ??? ??? ??? Manometer ? ?1, ?2 and ?3 are the unit gravity force

    33. Example 2.5 ?? 2.8? ???? ? ???? ???? ????? ??. S2 = 1.0, S3 = 1.10, a/A = 0.01, R = 5 mm, t = 20oC, and ??? 760 mm Hg??. ??? ???? ? Pa?? ?1(a/A)? ?? ???? ????. ?. (2.4.1)? ???? #

    34. The inclined manometer: frequently used for measuring small differences in gas pressures. Adjusted to read zero, by moving the inclined scale, when A and B are open. Since the inclined tube requires a greater displacement of the meniscus for given pressure difference than a vertical tube, it affords greater accuracy in reading the scale. Surface tension causes a capillary rise in small tubes. If a U tube is used with a meniscus in each leg, the surface-tension effects cancel.

    35. 2.5 FORCES ON PLANE AREAS ??? ?????? ???? ??? ?????(inclined manometer)? ?? ????(?? 2.9). A? B? ????? ? ????? ??? ??? 0? ??? ????. ?? ?????? ???? ??? ???? ???? ?? ??? ?? ???? ??? ?? ? ??. ?? ???? ?????? ?? ?????? ????. U??? ??? ? ????? ????? ???? ? ?? ??????? ?? ????. ?? 0.5in??? ???? ?????? ??? ? ??.

    36. Horizontal Surfaces ?? ?? ?? ???? ?? ?? ???? ???? ??? ??? ??? ????. ??? ? ?? ???? ?? ??? ? ?? ??? A? ???? ??? ? pdA? ?? ???? ??? ???? ??? ?? ??? ? pdA? ????(scalar summation)? ??? ??? ????. ? ??? ??? ??? p? ??? ??? ??? ?? ???. . ??? ??? ?? ???(??? ?? ?? ???? 0? ?)? ?? ???? ?? 2.10? ?? ??? xy?? ???. ???? ??? ???, ??? ?? ?? ??? ???? ???? ???? ?? ??? ??? y?? ? ??? ???? ??? ??. ??? x'? y????? ??? ?????? ????. p? ????? ??? ??. ? ??? ??(?? A??)??? ????. ??? ??????? ???? ??? ??? ??? ??? ????. x y????? ??? ?????? ??

    38. Momentum (1) First moment The moment of an area A about the y axis

    40. (2) Second moment The second moment of an area A (the moment of inertia of the area)

    41. The product of inertia Ixy of an area

    42. Inclined Surfaces Fig. 2.11: a plane surface is indicated by its trace A'B;it is inclined ?o from the horizontal. x axis: intersection of the plane of the area and the free surface. y axis: taken in the plane of the area, with origin O in the free surface. The xy plane portrays the arbitrary inclined area. The magnitude, direction, and line of action of the resultant force due to the liquid, acting on one side of the area, are sought. For dA: Since all such elemental forces are parallel, the integral over the area yields the magnitude of force F, acting on one side of the area, Magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid The presence of a free surface is unnecessary

    44. Center of Pressure Fig. 2.11: the line of action of the resultant force has its piercing point in the surface at a point called the pressure center, with coordinates (xp, yp). Center of pressure of an inclined surface is not at the centroid. To find the pressure center, the moments of the resultant xpF, ypF are equated to the moment of the distributed forces about the y axis and x axis, respectively ? - may be evaluated conveniently through graphical integration, for simple areas they may be transformed into general formulas:

    45. When either of the centroidal axes is an axis of symmetry for the surface, vanishes and the pressure center lies on x = x- . Since may be either positive or negative, the pressure center may lie on either side of the line x = x-. To determine yp by formula, with Eqs. (2.5.2) and (2.5.6) In the parallel-axis theorem for moments of inertia in which IG is the second moment or the area about its horizontal centroidal axis. If IG is eliminated from Eq. (2.5.9)

    47. (a) Fig. 2.12? ???? ????, x, y? ?? ????? ????, y = 4??, x = 0, y = 6.5??, x = 3, ??? a, b? ??? ??, a, b? ?? ?1?? ???? x? ?? y? ??? ??? ? ?? ?? ???? , y = 6.5, x = 3; y = 9, x = 0; x = 6/5(9 - y). ???, ? ?? ???? ?sin?? ?? ????, ??? ??? ???. ?. (2.5.2)????

    48. (b) ??? ???? ??? ?. (2.5.8)?? ???? x?? ??? ???? ??? ????? I-xy= 0 ??? In Eq. (2.5.11), ? ????? ???? ???? ?? ???? 0.16m ??? ????. (c) ??? ?? ??? ???? ????, CD? ?? ?????? ??? ,

    49. The Pressure Prism ??? ???? ??? ???? ??? ? ??? ????? ?????? ??? ???? ??? ??. ??????? ??? ???? ???? ??, ??? ?????? ??? ?? p = ?h? ?? ???? ??? ???. ???, h? ??? ??????? ??????? ??????(??2.13).(??? ????? ???? ?? ?? h? ???? ??? ??????? ??? ? ??.) ???? ?h? ??? ??? ??? ?? ? ???? OM? ??? ??. ?? ?? ? ???? dA ?? (2.5.12) ? ?? ?????? ? ???? ?? ??. ??? ???? F = ? ? ??. ??,? ?????? ??? ??? ?? ?? ???? ??? ??? ?? ????. ?. (2.5.5)? (2.5.6)???, (2.5.13) ??? xp, yp? ?????? ??[centroid, ? (A.5)]??? ??? ????. ??? ??? ???? ?????? ??? ????. ? ?? ??? ??? ???? ????? ??? ????. ??? ???? ??? ?? ????. ?? ??, ? ???? ????? ???? ???? ????? ????(wedge-shaped)? ???? ????, ? ??? ?????? 1/3 ??? ??. ??? ????? ?????? 1/3 ??? ????

    51. Effects of Atmospheric Pressure on Forces on Plane Areas ??? ?? ?? ????? ??? ??? ???? ???. ???????? ?? ?? ??? ??? p = ?h ? ?? ???. ??? ? ?? ?? ??? ???? 0, ? ??????? ????. ??? ???? ?? ?? ???? ??? ???? 0? ?????? ?? ??? P0 ? ? ?? A?? ?, ? P0A ?? ? ?? ???? ? ???. ??, ??? ?? ?? ?? ??? ???? ?? ???? ???? ?? ? P0A ? ??? ??? ???? ?? ??? ??? ????? ??? ??? ??? ??? ?? ??. ????? ?? ?? ???? ??? ??? ????? ???? ??, ? ???? ??? 0? ?? ????? ???? ??? ???? ??? ???? ??? ???? ??? ? ??.

    52. [?? 2.8] ??? ???? ??? ?? ?? ? ????? ???(gravity dam)? ??? ??. ?? ???? ????? ? ????? ??????? ??????? ??? ? ??. ?? 2.15? ???? ?? ? ??? ??? ????. ????? ???? 2.5???, ?? ?? ?????. ? 1ft? ?? ????? ????. ????? ???? ??? ????, ?, ????(foundation pressure)??? (hydrostatic uplift)?? ?? ?? ? ? ??. ????? ??? ??? ???? ??? ? ?? ??? ?????, ???? ?? ???? (hydrostatic head)? 1/2? ????, ??? ?????? ????? ???? ????? 0? ??? ????. ??? ???, ?? ??? ????? ?? ???? ???? ??? ??? ?? ???? ?? ??? ???? ??. ?, Rx = 5000? ??. ????? ?? ?? ???? ???? ?? ???? ?????(?????)???(???)? ? ?? ??. ?, Ry = 6750? + 2625? - 1750? = 7625? N ??, ???? ????? ??? ??? ???. 0?? ?? ???? ???

    54. ?? ????? ? ??? ?? ????? ???? ??? ????. ?, ?? ???? Ry ? ?? ??? ?? ????? ??. ???, ??? smax ? smin ? ? ??? ??? ??, ????????. ?????? ??(??)? x = 44.8 m ??? ????. ??? ??? smax ? smin ? ??? ???? ??? ?0? ?? ???? ???. ??? ???? ? ???. ??? ???? ??? ???? ? ??? 3?? ?? ? ????? ??? ?? smin ? ?? ????? ??. ????? ??? ?? ????? ?? ??? ?? ???? ??? ??? 3?? ? ??? ????? ???? ?? ????.

    55. 2.6 FORCE COMPONENTS ON CURVED SURFACES ???? ?? ??? ? ???? ?? p dA ? ??? ???. ???? ??? ??? ? ?? ?? ? ????? ???? ???? ????? ??. ?,?? ???? 3?? ??? ?? ????? ?? ? ?? ? ??? ????? ??? ??. ?? ???? ? ?? ????? ??? ????(? ??? ??? ?? ??? ? ??.)? ?? ??? ?? ? ??. ? ??? ???? ?? ????.

    56. Horizontal Component of Force on a Curved Surface ??? ???? ???? ????? ??? ???? ??? ???? ???? ??. ?????? ????? ??? ?? ????. ?? 2.16? ??? ??? 3?? ??? ???? ??. dA? ?? ????? dA? ??? ???? ?? x?? ? ??? ???. dA? ?? ?? ???? ???? x?????

    58. Vertical Component of Force on a Curved Surface The vertical component of pressure force on a curved surface is equal to the weight surface and extending up to the free surface Can be determined by summing up the vertical components of pressure force on elemental areas dA of the surface In Fig.2.18 an area element is shown with the force p dA acting normal to it. Let ? be the angle the normal to the area element makes with the vertical. Then the vertical component of force acting on the area element is p cos ? dA, and the vertical component of force on the curved surface is given by (2.6.2) p replaced by its equivalent ?h; cos ? dA is the projection of dA on a horizontal plane ? Eq. (2.6.2): (2.5.3-4) in which d? is the volume of the prism of height h and base cos ? dA, or the volume of liquid vertically above the area element

    60. ?? ??? ??? ? ????? ????????? ????? ???? ??? ????. ?? ???? ???? ???? ????? ?? ??? ???? ??? ????? ??? ????. ??????? ??? ? ????? ??? ??? ?? ??? ??? ???? ?? ????? ??. ???? ??? ??? ???? ????? ???? ???? ?? ????. ????? ?? ?? ??? ?? ??? ??? ???? ?? ??? ???? ????? ??? ?? ??? ??? ? ???. ???? ?? ???? ??? ??? ??? ??? ???? ???? ????? ?? ??? ?? ?? ?? ??? ??? ??? ?? ??? ????(??)? ??. ???? ???? ????? ?? ??? ????? ???? ??? ??. ??? ???? ?? ??? ???? ? ?? ?? ???? ?? ??? ???. ?? ??? ???? ??? ?? ???? ????? ???? ?? ?? ??? ?? ?? ???? ????? ????? ??.

    61. ????? ???? ??? ??? ?? ?? ???? ???? ???? ???? ????? ???? ?? ?? ????? ?? ? ??. ? O? ??? ??? ?? ???? ??? (Fig.2.18), x? O? ?? ?????? ???? Fv = ?? ? x ????? ????. ???? ????? ???? ??? ????? ?? ?? ???? ??? ??? ??? ????.

    62. [?? 2.11] ?? ???? ??? ?? ?? ?? ??(??2.20). ?? ??? ??? ????. ? 1m? ??? ??? ??? ???. (a)??? ?? (b)??? ??? ?. (a) ??????? ??? ??? ???? ??? ???? ???? ????? ????.(CD??? ?? ?? ????? A? ?? ??? ??.) BCD? ???? ???? AB? ???? ???? ??? ?? ??? ??? ??? (b)??? ???? ?? ABC? ???? ???? ??????? CD??? ????? ? ???. BC? CD??? ????? ?? ????? ???? ???? BCD? ????? 0??. ???? ?? ???? ????? 2?? ??? ????? ??? 9806Pa ??.

    64. Tensile Stress in a Pipe and Spherical Shell ??? ?? ?????? ?????? ???? ???. ????? ????? ???? ???? ????, ? ?? ?? 2.21? ??? ?? ?? ????? ????? ??? ??. ????? ????, ?? ??? ???? ???? ????? ??? ?? ? ????? ??? ???? ???? ???? ?? ????? ???? ??. ???? ?? ????? ??? ???? ???? ????? ???? ??? ?? ?? ? ??. ??? ?? ??? ????? ????? ????? ???? ? ??? 2pr ??. ??? p? ?????? ???? r? ???? ?????. ?? ??? ??? ????? ??????? ?? ? ???? T1 = T2=T? ??. ??? T? ?????? ?????. ? ??? e? ?? ??? ????? ???? ??? ?? ????.

    66. [?? 2.12] ?? 100 mm-ID ? ??? ? ??? in??. ?? ????? 70 MPa ??? ? ????? ???? ?????

    67. 2.7 BUOYANT FORCE ?? ???? ?? ??? ? ?? ??? ??? ???? ???? ??? ????? ??. ??? ?? ?????? ????. ?? ?? ?? ?? ????? ?? ?? ??? ???? ??? ????? ?? ?? 0??? ????? ????? ???? ???. ?? ?? ??? ???? ??? ??? ??? ???? ???? ????? ??? ???? ???? ?????? ???. ?? 2.22?? ??? ???? ??? ?? ABC?? ???? ??? ?? ?? ?? ??? ??? ??.

    69. ?? ???, ABCEFA ?? ?? ?? ??? ????. ??? ???? ??? ABCDEFA ??? ?? ??? ??. ? ? ?? ?? ???? ???, ? ?? ?? ??? ?? ???? ?????. ??? ? ?? ??? ?? ??? ??? ??? ??? ??? ABCD? ??? ??. ??? ???? FB ??, V ???????, ? ??? ??? ???? ??? ? ??? ??? ???? ?? ?(2.7.1)? ??? ??? ? ??. ?? 2.22? ? ??? ??? ???? ??.

    70. ?? 2.23?? ??? dA? ???? ??? ?? ?? ???? ???? ???? ??? dV? ??? ????. ? ? ??? ??? ??? ? ? ??? ?? ???? ? : ????? ??? ???? ??? ??? ?? ? O? ?? ???? ?? ??? ? ?? ?? ???? ?? ???. ?, ??? x? ? 0??? ?????? ????. ? ?? ?? ????? ????? ??? ????. ???? ??? ???? ??? ?? ??? ??? ??? ????. ?? ?? ??? ? ?? ??? ??? ????. ??? ????? ??? ??(center of buoyance)?? ??.

    72. ?? ??? ? ??? ?? ?? ? ?? ???? ?? ?? ??? ?? ? ??? ??, ??, ??? ? ??? ??? ? ??. ?? 2.24? ?? ??? ? ?? ???? ?? ??? ?? ???????. F1,, F2 ? ?? ? ????? ? ??? ?? ?1 , ?2 ? ? ??? ????? ?? ??? ?? W ? ?? V? ??????

    74. ?????(hydrometer)? ??? ??? ???? ??? ??? ???? ????. ?? 2.25? ? ??? ?? ?? ???? ?? ?? ????. ??? ??? ???? a?? ????. ?? ??? ?? ??? ?? S=1.0? ???? ?? ???? ? ? ??? ???, ??? ? ?? ??. ??? V0? ?? ?? ??? ??, ?? ?? ?????. ?? ?? ????? ???? ?? ??? ???? S? ???? 1.0? ????. ??? ?? S? ?? ?? ?? ?? ?? ???, ???? ?? ?????? ?V = a?h? ???? ?. (2.7.2) ? (2.7.3)? ??

    76. [?? 2.13] ?? ??? 1.5N? ?????? ???? ??? ?, ??? 1.1N???. ?????? ??? ? ???? ? ??? ?????

    77. 2.8 STABILITY OF FLOATING AND SUBMERGED BODIES ????? ? ?? ??? ?????? ????. ?????? ?? ??? ???? ????? ????, ??? ???? ???? ??? ?? ??? ?????. ???? ?? ???? ?? ??? ???? ??? ??? ????, ??? ???? ??? ?? ?? ??? ????. ?? ?? ?? ???? ??? ?? ????? ??? ? ???? ????? ???? ??? ? ??? ?? ???? ????? ???. ????? ??? ?? ???? ??? ?? ???? ?? ? ? ??? ?? ??? ??? ????? ???.

    78. ?????? ???? ??? ??? ????? ??. ??? ????? ????, ????? ?? ?????? ? ?? ? ???? ? ?? ? ??. ??? ???????? ?? ??, ?? ?? ? ??? ???? ???? ??? ?? ???? ????. ??????? ?? ??? ?? ?? ? ??? ??? ???? ???? ???. ?? 2.26? 3?? ????? ??? ???? ???. (a)? ??? ?? ???? ???? ?? ?? ???? ??????? ????. ?? (b)? ???? ?? ???? ?? ?? ???? ??? ???????? ??. ???? ?? ? ??? ??? ??? (a)? ?? ????? ???? ??. ?? (c) ? ??? ? ?? ??? ?? ?? ? ?? ???? ?? ? ??? ??? ??? ???? ????????. ?, ? ??? ??? ???? ???? ???.

    80. ???? ??? ?? ?? ??? ?? 2.27(a)? ?? ??? ???? ??? ???? ??? ?? ??? ??????. ?? 2.27(b)? ?? ??? ??????? ???? ??? ??? ?????? ?????? ???? ?????.

    81. ?? ?? ???? ? ? ?? ??? ???? ?? ??? ?? ??? ????. ??? ????? ??? ???? ?????, ?? ??? ?? ???? ??? ????? ?? ?? ?? ??? ??? ???? ??? ????? ??? ????. ???, ? ? ?? ?? ??? ?? ??? ???? ?????. ???? ??? ?????? ??? ??? ???? ???? ????

    82. Determination of Rotational Stability of Floating Objects ?? 2.26(a)?? ?? ?? ?? ??? ??? ??(????? ??) ??? ???? ???? ??????? ? ?? ??. ??? ?? ???? ?? ??? ??? ???? ??? ??? ??? ???????? ? ?? ? ??. ?? ??? ???? ?? ????, ??? ???? ???? ?? ???? ? ?? ???? ?? ??? ????? ??. ?? 2.38(a)? ?? ?? ???? ?? ??? ?? ??? ?????. ??? ?? ????? ??? ????. ???? ? ?? ??? ???? ??? ?? ?? ???? ??? ????.

    84. ??? ??? ?? 2.30(b)? ?? ???? ? ??? ???? ABCD? ?? B'?? ??? ????, ??? ?? G?? ??? ????. B'? ??? ???? ??? ???? G ??? ? M?? ????, ?????? ???? ??? ?? ??? ???. ??? ???? ??? ????? ??? ???? ??, M?? ????. M? G?? ????, ??? ????, G?? ???? ?????. G? ?? ?? ??????. ?? MG? ????? ??. ? ?? ??? ???? ???? ??? ??? ??. ?????? ??? ??. ? : ???, W : ???? ??

    85. [?? 2.14] ?? 2.30??? ???? ?? 6m?? ?? 20m, ? ??? 200Mg??. ??? ?? ? 30cm? ?? ??. ?y=30 cm? ? ?? ??? ?????? ???. ? ?? ?? ??? ?? h? ??. ???? ????? ?? AB? BC? ?? ?????? ???

    86. G? ???? 7.0ft??? ???? ? ?? ?? ???? ??(???)? ???? ?????? ??? ??. #

    87. Nonprismatic Cross Sections ??? ??[??2.29(a)] ??? ??? ??? ???? ??, ?? ???? ? ?? ?? ??? ????? ??? ??? ??? ? ??. ??? ?? ???? r[??2.29(b)]? ?? ??? ?? ??? ???? ??? ??? ?????? ??? ? ???, ?? ?? ??? ?? ??? ?? ?? ?FB? ? ???? ????, ?? ??? ????? ?? ??? ?FB?? ??? ?????. ??? B? ???? ??? ??? ???? ???? ???? ?? ?FBs? ???? ?? ? B'? ???? ?? ??? ???? ???? ??? ??? ??. ??? B? ?? ???? ??? ??? ?? r? ??? ?? ? ??.

    89. ???? ??? ????? ???? ???, ? ???? ???? ?O? ?? ???? ??? ?? ? ?? ????? ? dA ? ?? ????? ??? x? dA??. ? ??? ?? ??? ?x2? dA??. ??? ?? ?? ???? ????? ??? ???. ? ???? ???? ?? ???? ??? ?? ???? ?? ? ??. ??? I? ???? ??? yy? ?? ?? ???????[??2.29(a). ? ??? ? (2.8.1)? ???? V : ???? ??? ? ???? T? ?? ????

    90. [?? 2.15] ????? 1Mkg ?? ????? ????(???)? ?? 2.32? ?? ???? ??. ??? ?? ?? 2.0m? ?? ??? ???? 0.5m ??? ??. ?? y-y? ??? ??? ??? ?(rolling)? x-x?? ?? ??? ??? ?(pitching) ??? ??? ???. GB = 2 0.5 = 1.5 m For rolling For pitching #

    92. 2.9 RELATIVE EQUILIBRIUM Fluid statics: no shear stresses ? the variation of pressure is simple to compute For fluid motion such that no layer moves relative to an adjacent layer, the shear stress is also zero throughout the fluid A fluid with a translation at uniform velocity still follows the laws of static variation of pressure. When a fluid is being accelerated so that no layer moves relative to an adjacent one (when the fluid moves as if it were a solid), no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body Two cases are of interest, a uniform linear acceleration and a uniform rotation about a vertical axis When moving thus, the fluid is said to be in relative equilibrium

    93. Uniform Linear Acceleration Fig. 2.31: a liquid in an open vessel is given a uniform linear acceleration a After some time the liquid adjusts to the acceleration so that it moves as a solid, i.e., the distance between any two fluid particles remains fixed ? no shear stresses occur By selecting a cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane (Fig. 2.31a), the z axis is normal to a and there is no acceleration component in that direction Fig. 2.31b: the pressure gradient ?p is then the vector sum of -?a and -j? Since ?p is in the direction of maximum change in p (the gradient), at right angles to ?p there is no change in p. Surfaces of constant pressure, including the free surface, must therefore be normal to ?p

    95. To obtain a convenient algebraic expression for variation of p with x, y, and z, that is, p = p(x, y, z), Eq. (2.2.5) is written in component form : Since p is a function of position (x, y, z), its total differential is Substituting for the partial differentials gives which can be integrated for an incompressible fluid,

    96. To evaluate the constant of integration c: let x = 0, y = 0, p = p0; then c = p0 and When the accelerated incompressible fluid has a free surface, its equation is given by setting p = 0 in the above eq. Solving it for y gives The lines of constant pressure, p = const, have the slope and are parallel to the free surface. The y intercept of the free surface is

    97. Example 2.15 The tank in Fig. 2.32 is filled with oil, relative density 0.8, and accelerated as shown. There is a small opening in the rank at A. Determine the pressure at B and C; and the acceleration ax required to make the pressure at B zero. By selecting point A as origin and by applying Eq. (2.9.2) for ay = 0 At B, x = 1.8 m, y = - 1.2 m, and p = 2.35 kPa. Ft C, x = -0.15 m, y = -1.35 m, and p = 11.18 kPa. For zero pressure at B, from Eq. (2.9.2) with origin at A,

    99. Example 2.16 A closed box with horizontal base 6 by 6 units and a height of 2 units is half-filled with liquid (Fig. 2.33). It is given a constant linear acceleration ax = g/2, ay = -g/4. Develop an equation for variation of pressure along its base. The free surface has the slope: hence, the free surface is located s shown in the figure. When the origin is taken at 0, Eq. (2.9.2) becomes Then, for y = 0, along the bottom,

    101. Uniform Rotation about a Vertical Axis Forced-vortex motion: rotation of a fluid, moving as a solid, about an axis Every particle of fluid has the same angular velocity This motion is to be distinguished from free-vortex motion, in which each particle moves in a circular path with a speed varying inversely as the distance from the center A liquid in a container, when rotated about a vertical axis at constant angular velocity, moves like a solid alter some time interval. No shear stresses exist in the liquid, and the only acceleration that occurs is directed radially inward toward the axis of rotation. By selecting a coordinate system (Fig. 2.34a) with the unit vector i in the r direction and j in the vertical upward direction with y the axis of rotation, the following equation may be applied to determine pressure variation throughout the fluid: (2.2.5)

    103. For constant angular velocity w, any particle of fluid P has an acceleration w2r directed radially inward (a = -iw2r) Vector addition of -j? and -?a (Fig. 2.34b) yields ?p, the pressure gradient. The pressure does not vary normal to this line at a point ? if P is taken at the surface, the free surface is normal to ?p Expanding Eq. (2.2.5) k is the unit vector along the z axis (or tangential direction). Then p is a function of y and r only: For a liquid (? const) integration yields c is the constant of integration

    104. If the value of pressure at the origin (r = 0, y = 0) is p0, then c = p0 and When the particular horizontal plane (y = 0) for which p0 = 0 is selected and the above eq. is divided by ?, : the head, or vertical depth, varies as the square of the radius. The surfaces of equal pressure are paraboloids of revolution.

    105. When a free surface occurs in a container that is being rotated, the fluid volume underneath the paraboloid of revolution is the original fluid volume The shape of the paraboloid depends only upon the angular velocity with respect to the axis (Fig. 2.35). The rise of liquid from its vertex to the wall of the cylinder is w2r02/rg (Eq. (2.9.6)), for a circular cylinder rotating about its axis. Since a paraboloid of revolution has a volume equal to one-half its circumscribing cylinder, the volume of the liquid above the horizontal plane through the vertex is When the liquid is at rest, this liquid is also above the plane through the vertex to a uniform depth of Hence, the liquid rises along the walls the same amount as the center drops, thereby permitting the vertex to be located when w, r0, and depth before rotation are given

    107. Example 2.17 A liquid, relative density 1.2, is rotated at 200 rpm about a vertical axis. At one point A in the fluid 1 m from the axis, the pressure is 70 kPa. What is the pressure at a point B which is 2 m higher than A and 1.5 m from the axis? When Eq. (2.9.5) is written for the two points, Then w = 200 x 2p/60 = 20.95 rad/s, ? = 1.2 x 9806 = 11.767 N/m3, rA = 1 m, and rB = 1.5 m. When the second equation is subtracted from the first and the values are substituted, Hence

    108. Example 2.18 A straight tube 2 m long, closed at the bottom and filled with water, is inclined 30o with the vertical and rotated about a vortical axis through its midpoint 6.73 rad/s. Draw the paraboloid of zero pressure, and determine the pressure at the bottom and midpoint of the tube. In Fig. 2.36, the zero-pressure paraboloid passes through point A. If the origin is taken at the vertex, that is, p0 = 0, Eq. (2.9.6) becomes which locates the vertex at O, 0.577 m below A. The pressure at the bottom of the tube is or At the midpoint, =.289 m and

    110. Fluid Pressure Forces in Relative Equilibrium The magnitude of the force acting on a plane area in contact with a liquid accelerating as a rigid body can be obtained by integrating over the surface The nature of the acceleration and orientation of the surface governs the particular variation of p over the surface When the pressure varies linearly over the plane surface (linear acceleration), the magnitude of force is given by the product of pressure at the centroid and area, since the volume of the pressure prism is given by pGA For nonlinear distributions the magnitude and line of action can be found by integration.

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