9. Hydrostatik I (1.2 1.5)

9. Hydrostatik I (1.2 1.5) Vätsketryck, tryck-densitet-höjd Tryck mot plana ytor Övningstal: H10 och H12 HYDROSTATICS Hydrostatics: Study of fluids (water) at rest No motion no shear stress viscosity non-significant Only existing stress for a fluid at rest is normal (compression) stress, i.e. pressure 1

Characteristics of pressure 1. Measurement unit [Pa]=[N/m 2 ] 2. Pressure is transmitted normal to solid boundaries or arbitrary sections 3. Pressure is transmitted undiminished to all other points in a fluid at rest 4. Pressure has the same magnitude in all directions at a point in a fluid at rest (scalar quantity) arbitrary = godtycklig skalar => ingen vektor Relation between pressure and depth in an incompressible liquid Assuming constant density and no horizontal pressure variation, the liquid column in the fig below can be used to determine the pressure as a function of depth. Vertical forces acting on column (V, A, and y are volume, area, and height of column, respectively): A y Upward pressure force: P A (EQ 1.2) Weight (downward): wv = way Vertical force balance: P A = way P= wy= ρgy (EQ 1.8) Pressure often quoted as heads, h = P/w (in m H 2 O or mm Hg) heads = tryckhöjd 2

ABSOLUTE AND RELATIVE PRESSURE Pressures are measured and quoted in two different systems, one relative (gauge) and one absolute. The relation between them is: P abs = P atm + P gauge (EQ 1.9) Negative gauge pressures are often termed vacuum pressures Often only relative pressures are of interest 100 kpa gauge = tryckgivare EXAMPLE ON GAUGE AND ABSOLUTE PRESSURES A pressure gauge registers a vacuum of 310 mm of mercury when the atmospheric pressure is 100 kpa, absolute. Calculate the corresponding absolute pressure. Solution: P atmospheric = 100 kpa (P gauge /w Hg ) = -310 mm Hg P gauge = -0.31w Hg w Hg = 133.0 kn/m 2 (e.g., page 4; ρg) P absolute = P atmospheric + P gauge P absolute = 100-0.31 133.0 = 58.8 kpa 3

Fig. 1.5 Typical examples of situations where hydrostatic force may have to be calculated FORCE ON SUBMERGED PLANE SURFACES mini summary Example of applications: - Design of dams, ships, gates, and tanks. Characteristics of pressure in a fluid at rest: Constant pressure on plane horizontal surface Linear pressure variation with depth for constant density liquid Pressure acts perpendicular to the surface 4

Pressure prism: volume of pressure on the plane surface F = P A = ρgh A (EQ 1.11) Resultant force is equal to the volume of the pressure prism and acts through its centroid γ = w (tunghet) prism = VVR145 prisma Vatten centroid = geometriskt centrum ( yt-tyngdpunkt ) RELEVANT EQUATIONS FORCES ON PLANE SUBMERGED SURFACES Resultant force: F= wh G A= ρgh G A (EQ 1.11) Point of action of resultant force: L P = I G /(A L G ) + L G (EQ 1.13) L P A L G L L p Compare with Figure A1.1, page 554! P=ρgh A: area of plane surface; h G : vertical distance liquid surface - area center; L P : distance O - pressure center; L G : distance O area center; I G : second moment of area about area center axis; L = h/sinθ I G = second moment of the area: yttröghetsmoment Point of action = angreppspunkt 5

Page 10 Median line gives lateral position for center of pressure for regular plane areas 6

H10 A rectangular gate 1.8 m long and 1.2 m high lies in a vertical plane with its centre 2.1 m below a water surface. Calculate magnitude, direction and location of the total force on the gate. H12 This rectangular gate will open automatically when the depth of water, d, becomes large enough. What is the minimum depth that will cause the gate to open? 7

UPPGIFT 1 (6 poäng) Vilken kraft P behövs för att hålla kvar den 5 m breda (in i pappret) rektangulära luckan i sin position enligt figuren nedan? Luckans längd är L = 4 m och vattendjupet till vänster om luckan (till vänster om leden) är 2 m. Antag att leden är friktionsfri f i och att vi har luft på höger sida om luckan. Försumma luckans egentyngd. LÖSNING steg 1: Rita ut relevanta krafter som verkar på luckan. 8

10. Hydrostatik II (1.6 1.7) Tryck mot buktiga ytor Flytkraft / Archimedes princip Övningstal: H15 och H18 Fig. 1.27 Pressure on a sphere 1

FORCES ON CURVED SUBMERGED SURFACES (1) Resolve the force into two components, one vertical and one horizontal Pressure intensity on a curved surface. F passes through the center of curvature. curvature = krökning (2) The horizontal force is obtained by projecting the curved surface onto a vertical plane. The horizontal force is equal to the force on this projected area: F H = ρg h G,proj A proj G P Projection of the curved surface onto a vertical plane 2

(3) The vertical force is equal to the weight of the volume of liquid above the curved surface F V ==ρ = ρ g V Kom ihåg: V= volym ovanför The vertical force component, F V, caused by the weight of liquid above the surface (4) The resultant force is given by: F = F 2 + F 2 V H Eq. 1.15 and the direction of the resultant force by: tanφ = F V F H Eq. 1.16 The direction of the resultant force, F, which must also pass through C (5) Remember that there is an equal and opposite force acting on the other side of the surface. 3

ARCHIMEDES PRINCIPLE BUOYANCY FORCE Law of buoyancy (Archimedes principle): The upthrust t (buoyancy force) on a body immersed in a fluid is equal to the weight of the fluid displaced Law of flotation: A floating body displaces its own weight of the liquid in which it floats Proof of Archimedes principle Vertical forces acting cylinder surface: Downwards P 1 : p 1 A = ρ g ya = w ya F B Upwards P 2 : p 2 A = ρ g (y+l)a= w (y+l)a Net pressure force (upthrust), F B : F B = w(y+l)a - wya= wla = = wv = ρ g V Eq. 1.14 4

H15: The quarter cylinder AB is 3 m long. Calculate magnitude, direction, and location of the resultant force of the water on AB. Z X C H18: The weightless sphere of diameter d is in equilibrium in the position shown. Calculate d as a function of w 1, h 1, w 2, and h 2. w 1 w 2 Sfärs volym = πd 3 /6 Area = πd 2 /4 equilibrium = jämvikt 5

11. Hydrostatik III (1.8, 1.9, 2.1-2.7) Hydrostatiska jämviktsekvationen Tryckmätning, manometri Övningstal: H1, H3-4 och H8 Hydrostatiska Jämviktsekvationen (samband mellan tryck, densitet och vertikalt avstånd) The general relation for pressure in a static fluid is: dp = w γ = ρ g dz dp = -w dz => z Implication: pressure varies only with depth and is constant in a horizontal plane OBS: z pekar uppåt 1

For a fluid with constant density: p 1 p 2 = γ (z 2 z 1 ) = γ h or p p 1 2 h = γ (p 2, z 2 ) h (p 1, z 1 ) Implications: pressure varies linearly with depth pressure may be expressed as head of fluid of weight density w pressure are often quoted as head in mm Hg or m H 2 O z p p 1 2 + z = + z = Const, for all points in a fluid at rest γ 1 γ 2 H1: The weight density (w = ρ g) of water in the ocean may be calculated l from the empirical i relation w = w 0 + K (h) 1/2, in which h = the depth (m) below the ocean surface. Derive an expression for the pressure at any point h and calculate weight density and pressure at a depth of 3220 m assuming w = 3 0 10 kn/m,k= 7.08 N/m 7/2. 2

Sample problem: An open tank contains water 1.4 m deep covered dby a 2 m thick layer of oil (r.d.=0.855). What is the pressure head at the bottom of the tank, in term of a water column? (r.d. = relative density) Mini summary: 3

MEASUREMENT OF PRESSURE MANOMETRY Pressure is constant over horizontal planes within continuous columns of the same fluid Conversion of manometer readings to Pressure (γ = w) (a) p 1 = p 2 p 1 = p x + γ l p 2 = p atm + γ 1 h p x = p atm + γ 1 h - γ l (absolute) (b) p 4 = p 5 p 4 = p x + γ 1 l 1 p 5 = p y + γ 2 l 2 + γ 3 h p x -p y = γ 2 l 2 + γ 3 h - γ 1 l 1 4

H3: With the manometer reading as shown, calculate p x. L R (r.d. = relative density) H4: Calculate p x -p y for this inverted U-tube manometer. ρ = r.d. ρ water ). L R (r.d. = relative density) 5

H8* The sketch shows a sectional view through a submarine. Calculate the depth of submergence, y. Assume that the weight density of sea water is 10.0 kn/m 3. L R 6