By Jack B. Evett, Cheng Liu
This robust problem-solver delivers 2,500 difficulties in fluid mechanics and hydraulics, totally solved step by step! From Schaum’s, the originator of the solved-problem consultant, and students’ favourite with over 30 million learn publications sold—this timesaver is helping you grasp all sorts of fluid mechanics and hydraulics challenge that you'll face on your homework and in your assessments, from houses of fluids to pull and raise. paintings the issues your self, then money the solutions, or move on to the solutions you wish utilizing the entire index. suitable with any lecture room textual content, Schaum’s 2500 Solved difficulties in Fluid Mechanics and Hydraulics is so whole it’s the suitable software for graduate or specialist examination evaluation!
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Extra info for 2500 Solved Problems in Fluid Mechanics and Hydraulics (Schaum's Solved Problems Series)
Prob. 9396 ft2 (App. 711 m2 : Assume std. atmosphere above water. 31 (See Prob. 8 mm A(m2) y (m) Ay (m3 ) Ii(m4) h(m) Ah2 1 Rect. 4 × 10−7 2 Semicirc. 54 in 2 ) Lp − Lc = Sum moments about hinge at top of gate. 524 m ) × See Prob. 19 for data. See Prob. 20 for data. 209 ft ⎝4⎠ ⎝4⎠ ha = pa = See Prob. 26. 46 3 1728 in 3 ft See Prob. 28. 97 in [See Prob. 389 m2 FH = 0 because horiz. 50) FV = 17328 lb = FR FH = 0 because horiz. 55 (See Prob. 99) Eq. 00 FH (See Prob. 65 kN/ m 2 Eq. 11, net vertical force equals the weight of the displaced fluid acting upward and the weight of the cylinder acting downward.
57. 59 See Prob. 57. 28 lb up But this indicates that the cylinder would float, as expected. Then, the force exerted by the cylinder on the bottom of the tank is zero. 60 The specific weight of the cylinder must be less than or equal to that of the fluid if no force is to be exerted on the tank bottom. 61 (See Prob. 1 lb down. This is true as long as the fluid depth is greater than or equal to the diameter of the cylinder. 62 (See Prob. 3 lb Force (downward) on upper part of cylinder = wt. of volume of cross-hatched volume.
524 m ) × See Prob. 19 for data. See Prob. 20 for data. 209 ft ⎝4⎠ ⎝4⎠ ha = pa = See Prob. 26. 46 3 1728 in 3 ft See Prob. 28. 97 in [See Prob. 389 m2 FH = 0 because horiz. 50) FV = 17328 lb = FR FH = 0 because horiz. 55 (See Prob. 99) Eq. 00 FH (See Prob. 65 kN/ m 2 Eq. 11, net vertical force equals the weight of the displaced fluid acting upward and the weight of the cylinder acting downward. 58 See Prob. 57. 59 See Prob. 57. 28 lb up But this indicates that the cylinder would float, as expected.
2500 Solved Problems in Fluid Mechanics and Hydraulics (Schaum's Solved Problems Series) by Jack B. Evett, Cheng Liu