FluidMechWhite5e Ch02b - Chapter 2 Pressure Distribution in...

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Chapter 2 Pressure Distribution in a Fluid 105 champagne 6 inches above the bottom: AA atmosphere 24 p (0.96 62.4) ft (13.56 62.4) ft p 0 (gage), 12 12 ±² × = = ³´ µ¶ 2 AA or: P 272 lbf/ft (gage) = Then the force on the bottom end cap is vertical only (due to symmetry) and equals the force at section AA plus the weight of the champagne below AA: V AA AA 6-in cylinder 2-in hemisphere 22 3 FF p( A r e a ) W W (272) (4/12) (0.96 62.4) (2/12) (6/12) (0.96 62.4)(2 /3)(2/12) 4 23.74 2.61 0.58 . Ans π ππ == + =+ × × 25.8 lbf 2.88 Circular-arc Tainter gate ABC pivots about point O. For the position shown, determine (a) the hydrostatic force on the gate (per meter of width into the paper); and (b) its line of action. Does the force pass through point O? Solution: The horizontal hydrostatic force is based on vertical projection: Fig. P2.88 HC G v e r t F h A (9790)(3)(6 1) 176220 N at 4 m below C γ × = The vertical force is upward and equal to the weight of the missing water in the segment ABC shown shaded below. Reference to a good handbook will give you the geometric properties of a circular segment, and you may compute that the segment area is 3.261 m 2 and its centroid is 5.5196 m from point O, or 0.3235 m from vertical line AC, as shown in the figure. The vertical (upward) hydrostatic force on gate ABC is thus VA B C F A (unit width) (9790)(3.2611) 31926 N at 0.4804 m from B =
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106 Solutions Manual Fluid Mechanics, Fifth Edition The net force is thus 22 1 / 2 HV F[ F F ] =+ = 179100 N per meter of width, acting upward to the right at an angle of 10.27 ° and passing through a point 1.0 m below and 0.4804 m to the right of point B. This force passes, as expected, right through point O . 2.89 The tank in the figure 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. Solution: Assume unit depth into the paper. The vertical force is the weight of benzene plus the force due to the air pressure: Fig. P2.89 2 (0.6) (1.0)(881)(9.81) (200,000)(0.6)(1.0) . 4 V FA n s π = N 122400 m Most of this (120,000 N/m) is due to the air pressure, whose line of action is in the middle of the horizontal line through B. The vertical benzene force is 2400 N/m and has a line of action (see Fig. 2.13 of the text) at 4R/(3 ) = 25.5 cm to the right or A. The moment of these two forces about A must equal to moment of the combined (122,400 N/m) force times a distance X to the right of A: += X = 29.9 cm (120000)(30 cm) (2400)(25.5 cm) 122400( ), . X solve for Ans The vertical force is 122400 N/m (down), acting at 29.9 cm to the right of A. 2.90 A 1-ft-diameter hole in the bottom of the tank in Fig. P2.90 is closed by a 45 ° conical plug. Neglecting plug weight, compute the force F required to keep the plug in the hole. Solution: The part of the cone that is inside the water is 0.5 ft in radius and h = 0.5/tan(22.5 ° ) = 1.207 ft high. The force F equals the air gage pressure times the hole Fig. P2.90
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Chapter 2 Pressure Distribution in a Fluid 107 area plus the weight of the water above the plug: gage hole 3-ft-cylinder 1.207-ft-cone 22 2 Fp A W W 1 (3 144) (1 ft) (62.4) (1) (3) (62.4) (1) (1.207) 44 3 4 339.3 147.0 19.7 . Ans ππ π =+ ± ² ³´ + µ¶ · ¸ ¹º » ¼ =+− = 467 lbf 2.91 The hemispherical dome in Fig. P2.91
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This homework help was uploaded on 03/19/2008 for the course ASE 320 taught by Professor Raman during the Spring '08 term at University of Texas at Austin.

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FluidMechWhite5e Ch02b - Chapter 2 Pressure Distribution in...

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