Civil Engineering Hydraulics Pressure and Fluid Statics Leonard: It wouldn't kill us to meet new people. Sheldon: For the record, it could kill us to meet new people. Common Units ¢ In order to be able to discuss and analyze fluid problems we need to be able to understand some fundamental terms commonly used 2 Pressure 1 Common Units ¢ The most used term in hydraulics and fluid mechanics is probably pressure ¢ Pressure is defined as the normal force exerted by a fluid per unit of area l The important part of that definition is the normal (perpendicular) to the unit of area 3 Pressure Common Units ¢ 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) 4 Pressure 2 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 Pressure Common Units ¢ Standard atmospheric pressure or "the standard atmosphere" (1 atm) is defined as 101.325 kilopascals (kPa). 6 Pressure 3 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 Pressure Common Units ¢ Pressure is usually given in reference to some datum l Absolute pressure is given in reference to a system with no pressure whatsoever (a vacuum) 8 Pressure 4 Common Units ¢ Pressure is usually given in reference to some datum l Gage pressure, the more commonly used pressure, is the difference between local pressure and the absolute pressure of the system being measured 9 Pressure 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 Pressure 5 Common Units 11 Pressure 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 Pressure 6 Pressure 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 Pressure Pressure at a point ¢ The statics (yes I said statics) of the situation can be used to define just what is happening 14 Pressure 7 Pressure 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 Pressure Pressure at a point The shape was a thickness of 1 in the y- direction. 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 Pressure 8 Pressure at a point We need to include one more force on this FBD and that is the weight of W this wedge of fluid. 17 Pressure Pressure at a point We can start by writing our expressions for force balances along each axis. W ∑ Fz = 0 ∑ FWPxz =−+23(Δ)(11cos) − Pl( )( )( θ ) ∑ Fx = 0 ∑ FPzx =13(Δ)(11sin) − Pl( )( )( θ ) 18 Pressure 9 Pressure 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 W to find the weight. W= ρ Vg 1 Wg=ρ (ΔΔ xz)( ) 1 {2 } ρ is the mass density of the fluid and g is the universal gravitational constant. 19 Pressure Pressure at a point Substituting for W in our initial expressions we W have ∑ Fz = 0 1 FgxzPxPlz =−ρθ( ΔΔ)( ) 1+23(Δ)( 1) − ( )( 1)( cos ) ∑ {2 } ∑ Fx = 0 ∑ FPzx =13(Δ)(1) − Pl( )( 1)( sinθ ) 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. 20 Pressure 10 Pressure at a point Now we have W 1 0-=ρθgxzPxPl(ΔΔ)( ) +23(Δ) -( )( cos) {2 } 0=PzPl13(Δ−) ( )( sinθ ) In the second expression, the term l sin θ is equal to Δz so we have 1 0-=ρθgxzPxPl(ΔΔ)( ) +(Δ) -( )( cos) {2 } 23 0 =PzPz13(Δ−Δ) ( ) ∴PP13= 21 Pressure Pressure at a point 1 0-=ρθgxzPxPl(ΔΔ)( ) +23(Δ) -( )( cos) W {2 } 0 =PzPz13(Δ−Δ) ( ) ∴PP13= In the first expression, the term l cos θ is equal to Δx so we have ⎧1 ⎫ 0 = -ρg ⎨ (Δx)(Δz)⎬ + P2 (Δx) - P3 (Δx) ⎩2 ⎭ 22 Pressure 11 Pressure at a point 1 0-=ρgxzPxPx(ΔΔ)( ) +(ΔΔ) - ( ) W {2 } 23 PP13= Dividing all the terms by Δx we have ⎧1 ⎫ 0 = -ρg ⎨ (Δz)⎬ + P2 - P3 ⎩2 ⎭ 23 Pressure Pressure at a point Remember that we are talking W about pressure at a point in the fluid. 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 = -ρg ⎨ (Δz)⎬ + P2 - P3 This reduces our expression to ⎩2 ⎭ 0 = P2 - P3 P1 = P3 ∴P2 = P3 = P1 24 Pressure 12 Pressure at a point Notice that the value for θ was W not critical for our derivation and the density of the fluid did not enter into our calculation at the end. At any point in a fluid, the magnitude of the pressure is the same in all directions. 0-= PP23 PP13= ∴PPP231= = 25 Pressure Pressure at a point Note that this statement is W 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. 0-= PP23 PP13= ∴PPP231= = 26 Pressure 13 Pressure Variation with Depth ¢ Consider a fluid with a constant density over a depth ¢ We can start by remembering that pressure is the force exerted per unit of area. 27 Pressure Pressure Variation with Depth We can start with a cylinder of d diameter d and find the pressure at some depth h in the cylinder. The fluid has a mass density of ρ and the pressure at the top of the cylinder is p (atmospheric h atm pressure). 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 above the depth h. 28 Pressure 14 Pressure Variation with Depth The weight of the fluid can be d determined by taking the volume of the fluid and multiplying that by the product of the mass density ρ and the gravitational constant g. h π d2 Vh= 4 π d2 FVg==ρρ hg 4 29 Pressure Pressure Variation with Depth The area that the force is acting d normal to is the cross sectional area of the cylinder. π d2 Vh= h 4 π d2 FVg==ρρ hg 4 π d2 A = 4 30 Pressure 15 Pressure Variation with Depth So the added pressure of the d fluid is the force exerted by the fluid divided by the area it is acting over. πd2 F = Vρg = hρg h 4 πd2 A = 4 πd2 hρg F 4 p = = 2 = hρg The development in the text uses z for A πd the depth of the fluid rather than h. 4 31 Pressure Pressure Variation with Depth We can check for dimensional d consistency. p = hρg mass × length h 2 mass length time = length × × length2 length3 time2 32 Pressure 16 Pressure Variation with Depth The pressure is not dependent d on the area. If we assume that density remains constant with depth and that the gravitational constant h also remains constant with depth the pressure becomes a linear function of depth. As you go deeper in a fluid, the pressure increases linearly. p = hρg 33 Pressure Pressure Variation with Depth p h g = ρ 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. 34 Pressure 17 Pressure Variation with Depth p = hρg ¢ In the previous slide, we made two assumptions l 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 l The second was that the density of the fluid did not change with depth 35 Pressure Pressure Variation with Depth ¢ A more formal expression for change of pressure with changing depth would be z2 h2 z2 h2 p − p = ρg dz = ρg dh = γ dz = γ dh 2 1 ∫ ∫ ∫ ∫ z1 h1 z1 h1 36 Pressure 18 Pressure Variation with Depth ¢ This would allow for a fluid which might change density as a function of depth z2 h2 z2 h2 p − p = ρg dz = ρg dh = γ dz = γ dh 2 1 ∫ ∫ ∫ ∫ z1 h1 z1 h1 37 Pressure Problem pressure increases downward in a 38 Pressure given fluid and decreases upward 19 Problem pressure increases downward in a 39 Pressure given fluid and decreases upward Reading ¢ Sections 2-2 and 2-3 40 Pressure 20 Homework – Problem 3-1 41 Pressure Homework – Problem 3-2 42 Pressure 21 Homework – Problem 3-3 43 Pressure 22 .