1. FLUID STATICS: Hydrostatic Force on Plane Surfaces slide 18

3. Pressure at a Point: Pascal’s Law How does the pressure at a point vary with orientation of the plane passing through the point? Pressure is the normal force per unit area at a given point acting on a given plane within a fluid mass of interest. Blaise Pascal (1623-1662) p is average pressure in the x, y, and z direction. P s is the average pressure on the surface  is the plane inclination  is the length is each coordinate direction, x, y, z  s is the length of the plane  is the specific weight Wedged Shaped Fluid Mass F.B.D. Pressure Forces Gravity Force V = (1/2  y  z)*  x

4. Pressure at a Point: Pascal’s Law For simplicity in our Free Body Diagram, the x-pressure forces cancel and do not need to be shown. Thus to arrive at our solution we balance only the the y and z forces: Pressure Force in the y-direction on the y-face Pressure Force on the plane in the y-direction Rigid body motion in the y- direction Pressure Force in the z-direction on the z-face Pressure Force in the plane in the z-direction Weight of the Wedge Rigid body motion in the z- direction Now, we can simplify each equation in each direction, noting that  y and  z can be rewritten in terms of  s:

5. Pressure at a Point: Pascal’s Law Substituting and rewriting the equations of motion, we obtain: Math Now, noting that we are really interested at point only, we let  y and  z go to zero: Pascal’s Law: the pressure at a point in a fluid at rest, or in motion, is independent of the direction as long as there are no shearing stresses present.

6. Pressure at a Point: Pascal’s Law p1xsp1xs psxspsxs p2xsp2xs p s = p 1 = p 2 Note: In dynamic system subject to shear, the normal stress representing the pressure in the fluid is not necessarily the same in all directions. In such a case the pressure is taken as the average of the three directions.

7. Pressure Field Equations How does the pressure vary in a fluid or from point to point when no shear stresses are present? Consider a Small Fluid Element Surface Forces Body Forces Taylor Series V =  y  z  x For simplicity the x-direction surface forces are not shown p is pressure  is specific weight

8. Pressure Field Equations Looking at the resultant surface forces in the y-direction: Similarly, looking at the resultant surface forces in the x and z-direction, we obtain: Expressing these results in vector form:

20. Hydrostatic Condition: Incompressible Fluids We can immediately integrate since  is a constant: where the subscripts 1 and 2 refer two different vertical levels as in the schematicschematic.

22. Hydrostatic Condition: Incompressible Fluids As in the schematic, noting the definition of h = z 2 –z 1 : h is known as the pressure head. The type of pressure distribution is known as a hydrostatic distribution. The pressure must increase with depth to hold up the fluid above it, and h is the depth measured from the location of p 2. Linear Variation with Depth The equation for the pressure head is the following: Physically, it is the height of the column of fluid of a specific weight, needed to give the pressure difference p 1 – p 2.

23. Hydrostatic Condition: Incompressible Fluids If we are working exclusively with a liquid, then there is a free surface at the liquid-gas interface. For most applications, the pressure exerted at the surface is atmospheric pressure, p o. Then the equation is written as follows: The Pressure in a homogenous, incompressible fluid at rest depends on the depth of the fluid relative to some reference and is not influenced by the shape of the container. p = p o p = p 1 p = p 2 Lines of constant Pressure For p 2 = p =  h + p o h1h1 For p 1 = p =  h 1 + p o

26. Hydrostatic Condition: Compressible Fluids Gases such as air, oxygen and nitrogen are thought of as compressible, so we must consider the variation of density in the hydrostatic equation: Note:  =  g and not a constant, then By the Ideal gas law: Thus, R is the Gas Constant T is the temperature  is the density Then, For Isothermal Conditions, T is constant, T o :

30. Measurement of Pressure: Barometers Evangelista Torricelli (1608-1647) Animation of Experiment: Torricelli’s Sketch Schematic: Note, often p vapor is very small, 0.0000231 psia at 68° F, and p atm is 14.7 psi, thus:

34. HYDROSTATIC FORCES F=PA

35. WHAT IS A RESULTANT FORCE AND CENTRE OF PRESSURE ?. در اين بخش راجع به مقدار نيروي برآيند و خط اثر آن (مركز فشار) بحث مي كنيم..

36. HOW TO DETERMINE THE RESULTANT FORCE

39. Review of Centroid & Area Moment of Inertia

40. Hydrostatic Forces on Curved Surfaces Find the magnitude and line of action of the hydrostatic force acting on surface AB 1.What is the shape of the curve? 2. How deep is the curved surface? 3.Where does the curve intersect straight surfaces? 4.What is the radius of the curve? Important Questions to Ask

41. Hydrostatic Forces on Curved Surfaces

43. Hydrostatic forces on Curved surfaces. Find the magnitude and line of action of the hydrostatic force acting on surface AB 1.F V : Force on the fluid element due to the weight of water above CB 2.F H : Force on the fluid element due to horizontal hydrostatic forces on AC 3.W : Weight of the water in fluid element ABC 4.F : The force that counters all other forces - F has a horizontal component: Fx - F has a vertical component: Fy Forces acting on the fluid element

44. Hydrostatic forces on Curved surfaces Find the magnitude and line of action of the hydrostatic force acting on surface AB - Given: Surface AB with a width of 1 m 1.By inspection, curve is a ¼ circle. 2.The depth to the beginning of the curve (4 m depth to B) 3.The curve radius (2 m horizontal curve projection distance = curve radius) 4.Label relevant points: BCDE is water above fluid element defined by the curve ABC is the fluid element defined by the curve Problem Solving Preparation

45. Example 3.11: Find F v, F H, W, F x, F y, F, Line of action for F H & F v Given: Surface AB goes 1 m into the paper The hydrostatic force acting on AB is equal and opposite to the force F shown F x = F H = (5 x 9810) (2 x 1) = 98.1 kN Pres. at the cenroid AC side area F y = W + F v F v = 9810 x 4 x 2 x 1 = 78.5 kN W= γV ABC = 9810 (1/4 x  r 2 ) 1 = 30.8 kN Fy= 78.5 + 30.8 = 109.3 kN

46. The centroid of the quadrant Location of the resultant force جایزه

49. FLUID STATICS: Buoyancy

50. Buoyancy, Flotation & Stability Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. Archimedes’ Principle buoyant force. وزن حجمی از سیال که جابه جا شد Fv2-Fv1 وزن حجمی از سیال که جابه جا شده

51. Buoyancy: Floating Object where g is the specific weight of the fluid and V D is the volume of the body Depends on submerged portion of the volume V D is the submerged volume Buoyant force Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

52. Example 3.12: Bouyant force on a metal part Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved. - Wood block (1) has dimensions of 10 ‐ mm x 50 mm x 50 mm -Specific gravity of 0.3 - Metal object (2) has volume of 6600 mm 3 – Find the tension in the cable and mass of object 2. Steps Find the buoyant forces. Find the weight of the block. Perform force balances on both objects.

53. Solution:

55. Basic Equations in Integral Form for a Control Volume

56. روش های مطالعه سیال روش دیفرانسیلی روش لانتگرالی آنالیز ابعادی