Civil Engineering Hydraulics Mechanics of Fluids. Pressure and Fluid Statics. The fastest healing part of the body is the tongue.

Transcription

1 Civil Engineering Hydraulics Mechanics of Fluids and Fluid Statics The fastest healing part of the body is the tongue. Common Units 2 In order to be able to discuss and analyze fluid problems we need to be able to understand some fundamental terms commonly used 1

2 Common Units The most used term in hydraulics and fluid mechanics is probably pressure is defined as the normal force exerted by a fluid per unit of area 3 The important part of that definition is the normal (perpendicular) to the unit of area Common Units 4 The Pascal is a very small unit of pressure so it is most often encountered with a prefix to allow the numerical values to be easy to display Common prefixes are the Kilopascal (kpa=103pa), the Megapascal (MPa=106Pa), and sometimes the Gigapascal (GPa=109Pa) 2

3 Common Units A bar is defined as 105 Pa so a millibar (mbar) is defined as 10-3 bar so the millibar is 102 Pa The word bar finds its origin in the Greek word báros, meaning weight. 5 Common Units 6 Standard atmospheric pressure or "the standard atmosphere" (1 atm) is defined as kilopascals (kpa). 3

4 Common Units This "standard pressure" is a purely arbitrary representative value for pressure at sea level, and real atmospheric pressures vary from place to place and moment to moment everywhere in the world. 7 Common Units is usually given in reference to some datum 8 Absolute pressure is given in reference to a system with no pressure whatsoever (a vacuum) 4

5 Common Units is usually given in reference to some datum 9 Gage pressure, the more commonly used pressure, is the difference between local atmospheric pressure and the absolute pressure of the system being measured Common Units If the gage pressure of the system being measured is less than local atmospheric pressure, the pressure may be termed a vacuum pressure This does not imply that it has no pressure, just that the pressure is less then local atmospheric 10 5

6 Common Units 11 Common Units To be the most precise, when giving pressure, you should state if it is an absolute or a gage pressure There is a difference 12 6

7 at a point What can appear as non-intuitive is that the pressure at any point in a fluid is the same in all directions It would seem to make more sense if the pressure was greater on the top of the point than on the bottom and not at all on the sides but remember we are talking about a point. 13 at a point 14 The statics (yes I said statics) of the situation can be used to define just what is happening 7

8 at a point We can start with a fluid at rest and therefore at static equilibrium In that fluid, we can pick a segment with a triangular cross section and a unit depth, into the fluid or into the page, with a thickness of 1 unit 15 at a point The shape was a thickness of 1 in the ydirection. Remember the pressure is defined as the force exerted normal to the surface. Since we have the FBD of this wedge, we are showing forces action on the wedge. 16 8

9 at a point We need to include one more force on this FBD and that is the weight of this wedge of fluid. 17 W at a point We can start by writing our expressions for force balances along each axis. F F F F 18 z =0 z = W + P2 ( Δx )(1) P3 ( l ) (1)( cos θ ) x =0 x = P1 ( Δz )(1) P3 ( l ) (1)( sin θ ) W 9

10 at a point The weight of the wedge can be found by solving for the volume of the wedge and then using the mass density and gravity to find the weight. W = ρvg W = ρg {( W } 1 Δx )( Δz ) 1 2 ρ is the mass density of the fluid and g is the universal gravitational constant. 19 at a point Substituting for W in our initial expressions we have F =0 F = ρ g z z Fx = 0 F x {( W } 1 Δx )( Δz ) 1 + P2 ( Δx )(1) P3 ( l ) (1)( cos θ ) 2 = P1 ( Δz )(1) P3 ( l ) (1)( sin θ ) Since the left side of all the expressions are equal to 0, we can divide both sides by 1 and get rid of the 1 s in both expressions

11 at a point Now we have { } 1 ( Δx )( Δz ) + P2 ( Δx ) - P3 ( l ) ( cos θ ) 2 0 = P1 ( Δz ) P3 ( l ) ( sin θ ) 0 = -ρ g W In the second expression, the term l sin θ is equal to Δz so we have {( } 1 Δx )( Δz ) + P2 ( Δx ) - P3 ( l ) ( cos θ ) 2 0 = P1 ( Δz ) P3 ( Δz ) 0 = -ρ g P1 = P3 21 at a point {( } 1 Δx )( Δz ) + P2 ( Δx ) - P3 ( l ) ( cos θ ) 2 0 = P1 ( Δz ) P3 ( Δz ) 0 = -ρ g W P1 = P3 In the first expression, the term l cos θ is equal to Δx so we have "1 % 0 = -! g # (!x ) (!z )& + P2 (!x ) - P3 (!x ) $2 ' 22 11

12 at a point 0 = -ρ g {( } 1 Δx )( Δz ) + P2 ( Δx ) - P3 ( Δx ) 2 W P1 = P3 Dividing all the terms by Δx we have "1 % 0 = -! g # (!z )& + P2 - P3 $2 ' 23 at a point Remember that we are talking about pressure at a point in the fluid. W We can reduce our wedge to a point by allowing the Δx and Δz dimensions to approach 0. As Δz approaches 0, the weight term also approaches 0. "1 % 0 = -! g # (!z )& + P2 - P3 $2 ' 0 = P2 - P3 This reduces our expression to P1 = P3 ( P2 = P3 = P

13 at a point Notice that the value for θ was not critical for our derivation and the density of the fluid did not enter into our calculation at the end. W At any point in a fluid, the magnitude of the pressure is the same in all 0 = P2 - P3 directions. P1 = P3 P2 = P3 = P1 25 at a point Note that this statement is made about any point, it not made about any two points having the same pressure. That is a different problem and not covered by the assumptions that we just made. W 0 = P2 - P3 P1 = P3 P2 = P3 = P

14 Variation with If we consider a fluid with a constant density over a depth 27 We can consider most gasses as having a constant density with depth over reasonable depths and most liquids also Variation with 28 We can start by remembering that pressure is the force exerted per unit of area. 14

15 Variation with We can start with a cylinder of diameter d and find the pressure at some depth h in the cylinder. d The fluid has a mass density of ρ and the pressure at the top of the cylinder is patm (atmospheric pressure). h The pressure at the depth h is going to be the sum of the patm and the pressure exerted by the weight of the fluid about the depth h. 29 Variation with The weight of the fluid can be determined by taking the volume of the fluid and multiplying that by the product of the mass density ρ and the gravitational constant g. d h V= π d2 4 h F = Vρ g = 30 π d2 4 hρ g 15

16 Variation with The area that the force is acting normal to is the cross sectional area of the cylinder. d V= h π d2 4 h F = Vρ g = A= 31 π d2 4 hρ g π d2 4 Variation with So the added pressure of the fluid is the force exerted by the fluid divided by the area it is acting over. d F = V!g = h A= The development in the text uses z for the depth of the fluid rather than h. 32 " d2 h! g 4 " d2 4! d2 h! g F 4 p= = = h! g " d2 A 4 16

17 Variation with We can check for dimensional consistency. d p = h! g mass! length mass length time 2 = length!! 2 length length 3 time 2 h 33 Variation with The pressure is not dependent on the area. d If we assume that density remains constant with depth and that the gravitational constant also remains constant with depth the pressure becomes a linear function of depth. h As you go deeper in a fluid, the pressure increases linearly. p = h! g 34 17

18 Variation with p = h! g In your text, the symbol z is used for depth because h is often also used as a variable in thermodynamics. We may use both depending on the problem. The important thing is to take the variable as measuring the depth below the surface in the fluid. 35 Variation with p = h! g In the previous slide, we made two assumptions The first was that we were starting from 0 gage pressure at the top of the fluid which was the reference for our depth measurement The second was that the density of the fluid did not change with depth 36 18

19 Variation with A more formal expression for change of pressure with changing depth would be z2 h2 z2 h2 z1 h1 z1 h1 p2! p1 = "! g dz = "! g dh = "! dz = "! dh 37 Variation with This would allow for a fluid which might change density as a function of depth z2 h2 z2 h2 z1 h1 z1 h1 p2! p1 = "! g dz = "! g dh = "! dz = "! dh 38 19

20 Problem 39 pressure increases downward in a given fluid and decreases upward pressure increases downward in a given fluid and decreases upward Problem 40 20

21 Reading 41 Sections 2-2 and 2-3 Homework 42 21

22 Homework 43 Homework 44 22