Fluid Mechanics (Fluidmekanik) Course Code: 1TV024 5 hp Fluid Mechanics Spring 2011 Instruct: Chris Hieronymus Office: Geocentrum Dk255 Phone: 471 2383 email: christoph.hieronymus@geo.uu.se Literature: Engineering Fluid Mechanics by Clayton T. Crowe, Donald F. Elger, and John A. Robertson 9th edition, 2010 (alternatively 8th ed., 2005) Wiley & Sons, Inc., Hoboken, NJ, USA Grading: Grades will be based on final examination. Attendance at Lab Exercises towards end of term is mandaty. Lab rept to be handed in is also mandaty. Further Infmation: Course Outline Additional infmation regarding schedule, assignments, etc. can be found at http://studentptalen.uu.se Past student evaluations of this course can be found at the web page listed above. The exam will be closed-book. In other wds, you cannot use your book your notes during the exam. However, you may (and should!) prepare a cheat sheet to bring to the exam. One page (A-4, double-sided) of equations and whatever you think might be useful. (A list of equations will also be provided with the exam. But I think you ll learn me if you make your own.) Fluid Statics (Chapter 3) Flowing Fluids and Pressure Variation (Chapter 4) Control Volume Approach and Continuum Principle (Chapter 5) Momentum Principle (Chapter 6) Energy Principle (Chapter 7) Surface Resistance (Chapter 9) Flow in Conduits (Chapter 10) Varied Flow in Open Channels (Chapter 15)
Chapter 3 FLUID STATICS Shear Fces Nmal Fces (pressure) Fluid Mechanics, Spring Term 2011 where F is a fce nmal to area A Pressure is a scalar quantity Flow of an unconfined viscous fluid down an incline. Figure 3.1 (p. 31) Flowing viscous fluid exert shear fces. Static fluids only exert nmal fces. Moving fluids (dynamics) will be covered later. Fce balance in the x-direction:
From last slide: Divide through by to get Fce balance in the z-direction: Now shrink the element to a point: Vertical fce on!a Vertical fce on lower boundary Total weight of wedge element = specific weight This can be done f any ientation ", so F your Culture (i.e., not required f this course ) = It is possible to have different nmal stresses. Consider a small cubic fluid element that is part of a larger fluid mass: Different nmal fces in one (codinate) ientation are equal to shear fces in another ientation. (contʼd:) So what is the pressure then? Pressure is the average of the nmal fces acting at a point. Differences between nmal fces are due to fluid motion. In this case, if the fce vects are equal in magnitude, then p = 0
Pressure Transmission Absolute Pressure, Gage Pressure, and Vacuum Figure 3.3 (p. 34) Example of pressure relations Hydraulic Lift Figure 3.2 (p. 32) In a closed system, pressure changes from one point are transmitted throughout the entire system (Pascalʼs Law). Pressure in a vacuum is p = 0. Absolute pressure is referenced to perfect vacuum. Gage pressure is referenced to another pressure, typically atmospheric pressure (most gages measure relative pressures). Pressure Variation with Elevation Static fluid: All fces must balance as there are no accelerations. Look at fce balance in direction of! l From figure, note that Shrink cylinder to zero length: Figure 3.4 (p. 35) (from previous slide)
Pressure Variation f a Unifm-Density Fluid The pressure-elevation relation derived on the previous slide, F an incompressible fluid, # is constant. Pressure and elevation at one point can thus be related to pressure and elevation at another point: is perfectly general (applies also to variable #). But if # is constant, the above equation is easy to integrate: f The quantity is known as the piezometric pressure and is called the piezometric head. Example 3.3: What is the water pressure at a depth of 35 ft? With the infmation given, all we can calculate is the pressure difference between points 1 and 2. Example 3.4: What is the gage pressure at point 3? Two step solution: 1) Calculate 2) Calculate (Do yourself a fav and wk in SI-units!) (relative to atmospheric pressure at point 1)
Pressure Measurements Find pressure at center of pipe: Can start either at open end inside pipe. Here we start at open end: Figure 3.6 (p. 41) Piezometer simple manometer Figure 3.7 (p. 42) U-tube manometer Better f higher pressures. Possible to measure pressure in gases. p at open end Change in p from 1 to 2 Change in p from 3 to 4 p in pipe The complete path from point 1 to point 2 may include several U-tubes. Differential Manometer In general: Figure 3.8 (p. 44) From example 3.9 (p. 44) Used f measuring pressure differences between points along a pipe.
Example 3.10: Find the change in piezometric pressure and in piezometric head between points 1 and 2. The to give cancel out (piezometric pressure) ( from ) (piezometric head) Hydrostatic Fces on Plane Surfaces The white area AB in the figure is a plane of irregular shape. Line A-B is an edge view of that area. What is the net fce due to pressure acting on the sloping plane AB? This figure is absolutely awful Line AB represents the true location of the surface. The white surface is not drawn in its actual location. Line 0-0 is hizontal; the white area has been rotated about axis A-B from its proper location. In other wds, the apparent depth of the white area within the fluid is not as it appears. First, note that hydrostatic pressure increases along y as (since y is not vertical)
From the definition of pressure: so that the total fce on a plane area A is But the first moment of the area is defined as so that the total fce can be written as, since # and sin " are constants where is the pressure at the centroid of the area. The boxed equation is known as the hydrostatic fce equation. We have thus replaced an integral involving a variable pressure by a constant resultant pressure: Vertical Location of Line of Action of Resultant Hydrostatic Fce So, f the moment about a point at y cp we have In English: We wish to represent the distributed pressure fce by an equivalent point fce. Where (in the vertical) does that fce act? But with and we get The integral on the right-hand side is the second moment of the area (about point y=0): 2 weights on a beam suppted at y cp
The book just refers to the parallel axis theem to write Mathematically, I think it is easy to see that using Identifying as the 2nd moment about y=0 and as the 2nd moment about We have thus proved the parallel-axis theem: Notice that the last term is zero because As an aside, you may recall that The moment of inertia of an object about an axis through its center of mass I cm is the minimum moment about any axis in that direction. The moment about any other parallel axis is equal to I cm plus the moment of inertia about distance d of the entire object treated as a point mass located at the center of mass. Our system of pressures has nothing to do with rotations, but the equations are of the same fm Back to the problem at hand: Recall from a few slides ago that so that Note that at great depth, the difference between the centroid and the center of mass gets very small.
Example 3.12: Now calculate the slant distance between and Find the nmal fce required to open the elliptical gate if it is hinged at the top. First find F total, the total hydrostatic fce acting on the plate: The slant distance to the hinge is 8m x 5m/4m = 10m, and the slant distance from the hinge to the centroid is 2.5m. Hence, With (Appendix p. A-5) we get The two moments about the hinge must add to zero: Hydrostatic Fces on Curved Surfaces We could integrate the vect fces along segment AB, but it is often easier to find equivalent fces on a free body as illustrated above. F AC acts at the center of pressure as from previous section, F CB acts at centroid of area CB, and W acts at the center of mass of the free body ABC.
Example 3.14: Find magnitude and line of action of equivalent fce F. Fce balance in x and y: The line of action of the hizontal fce is Where we just read directly off the figure. The line of action f the vertical fce can be found by summing the moments about C ( any other point ) (notice that we could add a constant to every x-codinate since ) From Appendix p. A-5 (Figure A.1): The complete result is summarized below: Distance from C to centroid is: So that x cp is found to be