By Robert L. Mott

ISBN-10: 0131146807

ISBN-13: 9780131146808

For undergraduate-level classes in Fluid Mechanics or Hydraulics in Mechanical, Chemical, and Civil Engineering expertise and Engineering courses. the most well-liked applications-oriented method of engineering expertise fluid mechanics, this article covers the entire uncomplicated ideas of fluid mechanics-both statics and dynamics-in a transparent, useful presentation that ties idea on to genuine units and structures utilized in chemical technique industries, production, plant engineering, waste water dealing with and product layout. Readable and obviously written, the recent sixth version brings a way more appealing visual appeal to the ebook and comprises many updates and extra positive factors.

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For undergraduate-level classes in Fluid Mechanics or Hydraulics in Mechanical, Chemical, and Civil Engineering know-how and Engineering courses. the most well-liked applications-oriented method of engineering expertise fluid mechanics, this article covers the entire easy rules of fluid mechanics-both statics and dynamics-in a transparent, useful presentation that ties thought on to genuine units and structures utilized in chemical technique industries, production, plant engineering, waste water dealing with and product layout.

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**Additional info for Applied Fluid Mechanics. Solutions Manual**

**Sample text**

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. 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.

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. 00 ft3) = 256 lb wC = 600 lb; VC = wC = Fnet = 600 − 256 − 300 = 44 lb down—OK—block sits on bottom. 00 in would tend to submerge entire float. But additional buoyant force on float is sufficient to lift concrete block off sea floor. 33 in above surface With concrete block suspended, float is unrestrained and it would drift with the currents.

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. 58 See Prob. 57. 59 See Prob. 57. 28 lb up But this indicates that the cylinder would float, as expected.

### Applied Fluid Mechanics. Solutions Manual by Robert L. Mott

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