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

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 know-how and Engineering courses. the most well-liked applications-oriented method of engineering know-how 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 platforms utilized in chemical method industries, production, plant engineering, waste water dealing with and product layout. Readable and obviously written, the hot sixth variation brings a way more beautiful visual appeal to the ebook and contains many updates and extra good points.

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Extra resources for Applied Fluid Mechanics. Solutions Manual

Sample text

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.

A gage pressure can be no lower than one atmosphere below the prevailing atmospheric pressure. On earth, the atmospheric pressure would never be as high as 150 kPa. 7 psia; from App. E by interpolation. 84 psia; from App. E by interpolation. 13 Zero gage pressure. 77 kPa(abs) By interpolation - App. 72 a. 4 kPa(gage) b. 73 A barometer measures atmospheric pressure. 74 See Fig. 7. 75 The height of the mercury column is convenient. 79 The vapor pressure above the mercury column and the specific weight of the mercury change.

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. 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.