FTFS Chap11 P038

FTFS Chap11 P038 - Chapter 11 Fluid Statics Fluids in Rigid...

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Chapter 11 Fluid Statics Fluids in Rigid Body Motion 11-38C A moving body of fluid can be treated as a rigid body when there are no shear stresses (i.e., no motion between fluid layers relative to each other) in the fluid body. 11-39C A glass of water is considered. The water pressure at the bottom surface will be the same since the acceleration for all four cases is zero. 11-40C The pressure at the bottom surface is constant when the glass is stationary. For a glass moving on a horizontal plane with constant acceleration, water will collect at the back but the water depth will remain constant at the center. Therefore, the pressure at the midpoint will be the same for both glasses. But the bottom pressure will be low at the front relative to the stationary glass, and high at the back (again relative to the stationary glass). Note that the pressure in all cases is the hydrostatic pressure, which is directly proportional to the fluid height. 11-41C When a vertical cylindrical container partially filled with water is rotated about its axis and rigid body motion is established, the fluid level will drop at the center and rise towards the edges. Noting that hydrostatic pressure is proportional to fluid depth, the pressure at the mid point will drop and the pressure at the edges of the bottom surface will rise due to rotation. 11-28

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Chapter 11 Fluid Statics 11-42 A water tank is being towed by a truck on a level road, and the angle the free surface makes with the horizontal is measured. The acceleration of the truck is to be determined. Assumptions 1 The road is horizontal so that acceleration has no vertical component ( a z = 0). 2 Effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be secondary, and are not considered. 3 The acceleration remains constant. Analysis We take the x- axis to be the direction of motion, the z- axis to be the upward vertical direction. The tangent of the angle the free surface makes with the horizontal is z x a g a tan Solving for a x and substituting, 2 m/s 2.63 15 tan ) 0 m/s 81 . 9 ( tan ) ( 2 z x a g a Discussion Note that the analysis is valid for any fluid with constant density since we used no information that pertains to fluid properties in the solution. 11-29 a x = 15 Water tank
Chapter 11 Fluid Statics 11-43 Two water tanks filled with water, one stationary and the other moving upwards at constant acceleration. The tank with the higher pressure at the bottom is to be determined. Assumptions 1 The acceleration remains constant. 2 Water is an incompressible substance. Properties We take the density of water to be 1000 kg/m 3 . Analysis The pressure difference between two points 1 and 2 in an incompressible fluid is given by ) )( ( ) ( 1 2 1 2 1 2 z z a g x x a P P z x or ) )( ( 1 2 2 1 z z a g P P z since a x = 0. Taking point 2 at the free surface and point 1 at the tank bottom, we have atm P P 2 and h z z 1 2 and thus h a g P P z ) ( bottom gage , 1 Tank A : We have a z

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