Applied Fluid Mechanics. Solutions Manual by Robert L. Mott

By Robert L. Mott

For undergraduate-level classes in Fluid Mechanics or Hydraulics in Mechanical, Chemical, and Civil Engineering know-how and Engineering courses. the preferred applications-oriented method of engineering expertise fluid mechanics, this article covers all the uncomplicated rules of fluid mechanics-both statics and dynamics-in a transparent, functional presentation that ties idea on to genuine units and platforms utilized in chemical approach industries, production, plant engineering, waste water dealing with and product layout. Readable and obviously written, the hot sixth version brings a way more beautiful visual appeal to the e-book and comprises many updates and extra gains.

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Sample text

52 54 From Prob. 22, Fig.

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. 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.

30 kN FU > FD It will float. 2740 kN = 274 N; because w = 277 N > Fb—It will sink. 14 From Prob. 9 lb/ft > γw - cube would tend to float. 81 kN/m —it would float. 8836 m3 Entire hemisphere is submerged. 082 m) ( ) Wt. 30 From Prob. 6875 ft3 ends Vol. 196 ft3 sub. 34 4 Drums Weigh 4(30 lb) = 120 lb Wt. of platform and load = 1801 − 120 = 1681 lb See Prob. 63 for method of computing AS. wdrums + wwood + wload − FbD − Fbw = 0 wdrums = 4(30 lb) = 120 lb (Prob. 5 lb (Prob. 32) FbD = 1801 lb (Prob. 6 lb (Prob.

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