MET 212 Module 2-hydrostatics


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MET 212 Module 2-hydrostatics

  1. 1. Hydrostatics1- Fluid pressure Pressure is the force per unit area Pressure =Unit is Newton/square meter bar = 105 N/m2 Pound/square foot (fps)Example:A mass m of 50 kg acts on a piston of area 100 cm2 what is the intensity ofpressure on the water in contact with the underside of the position if the pistonis in equilibriumSolution.Force = mg = 50 x 9.81 = 490.5 NArea of piston A = 100 cm2 = 0.01 m2Pressure = = N/m2 = 4.905 x 104 N/m2 For liquids or gases at rest the pressure gradient in the vertical direction at any point in a fluid depends only on the specific weight of the fluid at that point Dr. Adel Afify 1 MET 212
  2. 2. 2- Pascals Law for Pressure at a Point Triangular prismatic element of fluid acts perpendicular to surface ABCD, acts perpendicular to surface ABFE and acts perpendicular to surface FECD.And, as the fluid is at rest, in equilibrium, the sum of the forces in any directionis zero. Pressure at any point is the same in all directions. This is known as Pascals Law and applies to fluids at rest. Dr. Adel Afify 2 MET 212
  3. 3. 3. Variation of pressure vertically in a fluid under gravity Vertical elemental cylinder of fluidThe fluid is at rest and in equilibrium so all the forces in the vertical directionsum to zero. We haveTaking upward as positive, in equilibrium we have Thus in a fluid under gravity, pressure decreases with increase in height . Dr. Adel Afify 3 MET 212
  4. 4. 4- pressure head Free surface (pressure = p0) Z p2 z2 h = z2 – z1 p1 z1 y X p1 - p2 = ρghh = pressure head Dr. Adel Afify 4 MET 212
  5. 5. example:A diver is working at a depth of 18 m below the surface of the sea. How muchgreater is the pressure intensity at this depth than at the surface? Specificweight of sea water is 10000 N/m3Solution: p1 - p2 = ρgh Δ p = 10000 x 18 Δ p=180000 N/m2 Example:Find the head height (h) of water corresponding to an intensity of pressure (p)of 340000 N/m2. The specific weight of water is 9.81 x 103 N/m3 Solution: p = ρgh h= h= = 34.7 mThe hydraulic jack F1 = PA1 F2 = PA2 Dr. Adel Afify 5 MET 212
  6. 6. Example: If the force F1 is 850 N is applied to the smaller cylinder of ahydraulic jack. The area A1 of a small piston is 15 cm2 and the area A2 of thelarge piston is 150 cm2. What load can be lifted on the larger piston? Thespecific weight of the liquid in the jack is 9.81 x 103 N/m3Solution: F1 = PA1850 = P (15/10000) P = 5.667 X 105 N/m2F2 = PA2F2 = 5.667 X 105 N/m2 X (150/10000) F2 = 8500 NMass of lifted = Mass of lifted = = 866.5 kg Dr. Adel Afify 6 MET 212
  7. 7. Measurement of pressure Pressure 1 Gage pressure @ 1 Local atmospheric Pressure reference Absolute pressure @ 1 2 Gage pressure @ 2 Pressure is designated as either absolute pressure or gage pressure A barometer is used to measure atmospheric pressure Absolute pressure = gauge pressure + atmospheric pressureGauge pressure isAbsolute pressure isExample:A mountain lake has an average temperature of 10 0c and a maximum depth of40 m. for a barometric pressure of 598 mm Hg. Determine the absolutepressure (in Pascal’s) at the deepest part of the lake.Solution: Dr. Adel Afify 7 MET 212
  8. 8. P = Ϫ H2O h + p0p0 is the pressure at the surfacep0 = Ϫ Hg hp0 = (133 kN/m3) (0.598m) = 79.5 kN/m2P = Ϫ H2O h + p0P = (9.81x1000 N/m3)(40 m) + 79.5 5 kN/m2 = 392. 5 kN/m2 + 79.5 5 kN/m2 = 472 kN/m2 = 472 kpa (abs)Manometer 1- Piezometer tube Consists of a vertical tube, open at the top, and attached to the container in which the pressure is desired P = Ϫ h + p0 Pressure will increase as we move downward And will decrease as we move upward A simple piezometer tube manometer Dr. Adel Afify 8 MET 212
  9. 9. Example:A pressure tube is used to measure the pressure of oil (mass density ρ = 640kg/m3) in a pipeline. If the oil rises to a height of 1.2 m above the center of thepipe, what is the gage pressure in N/m2 at that point? Solution: P=Ϫ h P=ρgh P = (640 kg/m3) x (9.81 m/s2) x (1.2 m) P = 7.55 kN/m2 2- U- Tube manometerPressure in a continuous static fluid is the same at any horizontal level so, Dr. Adel Afify 9 MET 212
  10. 10. For the left hand armFor the right hand armAs we are measuring gauge pressure we can subtract givingExample:1What will be the (a) the gauge pressure and (b) the absolute pressure ofwater at depth 12m below the surface? ρwater = 1000 kg/m3, and p atmosphere =101kN/m2.a)b)Example:2At what depth below the surface of oil, relative density 0.8, will produce apressure of 120 kN/m2? What depth of water is this equivalent to? Dr. Adel Afify 10 MET 212
  11. 11. a)b)Example:3What would the pressure in kN/m2 be if the equivalent head is measuredas 400mm of (a) mercury relative density =13.6 (b) water ( c) oil specificweight 7.9 kN/m3?a) Mercuryb) Waterc) Oil Dr. Adel Afify 11 MET 212
  12. 12. Example 4A manometer connected to a pipe indicates a negative gauge pressure of50mm of mercury. What is the absolute pressure in the pipe in Newton’sper square meter is the atmospheric pressure is 1 bar?Example 5What height would a water barometer need to be to measure atmosphericpressure?Measurement of pressure difference by using a "U"-TubeManometer. Dr. Adel Afify 12 MET 212
  13. 13. Pressure difference measurement by the "U"-Tube manometerExampleIn the figure below two pipes containing the same fluid of density ρ= 990 kg/m are 3connected using a u-tube manometer. What is the pressure between the two pipes ifthe manometer contains fluid of relative density13.6h1 = 1.5 m , h2 = 0.5 m , h3 = 0.76 m & ρ2 = SG * ρwater Dr. Adel Afify 13 MET 212
  14. 14. ρ2 = 13.6 * 1000 kg/m3Ϫ 1 = ρ1g = 990 * 9.81 = 9711.9 N/m3Ϫ 2 = ρ2 g = 13.6 * 1000 * 9.81 = 133416 N/m3pC = pDpC = pA + g hApD = pB + g (hB - h) + man g hpA - pB = g (hB - hA) + hg(man - )= 990 x9.81x(0.75-1.5) + 0.5x9.81 x(13.6-0.99) x 103= -7284 + 61852= 54 568 N/m2 (or Pa or 0.55 bar)Forces on Submerged Surfaces in Static Fluids1. Fluid pressure on a surfacePressure is defined as force per unit area. If a pressure p acts on a small area then theforce exerted on that area will beSince the fluid is at rest the force will act at right-angles to the surface.General submerged planeConsider the plane surface shown in the figure below. The total area is made up of manyelemental areas. The force on each elemental area is always normal to the surface but, ingeneral, each force is of different magnitude as the pressure usually varies. Dr. Adel Afify 14 MET 212
  15. 15. We can find the total or resultant force, R, on the plane by summing up all of the forces onthe small elements i.e.This resultant force will act through the centre of pressure, hence we can sayIf the surface is a plane the force can be represented by one single resultant force,acting at right-angles to the plane through the centre of pressure.Horizontal submerged planeFor a horizontal plane submerged in a liquid (or a plane experiencing uniform pressure overits surface), the pressure, p, will be equal at all points of the surface. Thus the resultant forcewill be given byCurved submerged surfaceIf the surface is curved, each elemental force will be a different magnitude and in differentdirection but still normal to the surface of that element. The resultant force can be found byresolving all forces into orthogonal co-ordinate directions to obtain its magnitude anddirection. This will always be less than the sum of the individual forces, .2. Resultant Force and Centre of Pressure on a submerged plane surface in aliquid.This plane surface is totally submerged in a liquid of density and inclined at an angle ofto the horizontal. Taking pressure as zero at the surface and measuring down from thesurface, the pressure on an element , submerged a distance z, is given by Dr. Adel Afify 15 MET 212
  16. 16. and therefore the force on the element isThe resultant force can be found by summing all of these forces i.e.(assuming and g as constant).The term is known as the 1st Moment of Area of the plane PQ about the free surface.It is equal to i.e.where A is the area of the plane and is the depth (distance from the free surface) to thecentroid, G. This can also be written in terms of distance from point O ( as )The resultant force on a planeThis resultant force acts at right angles to the plane through the centre of pressure, C, at adepth D. The moment of R about any point will be equal to the sum of the moments of theforces on all the elements of the plane about the same point. We use this to find theposition of the centre of pressure.It is convenient to take moments about the point where a projection of the plane passesthrough the surface, point O in the figure.We can calculate the force on each elemental area:And the moment of this force is: Dr. Adel Afify 16 MET 212
  17. 17. are the same for each element, so the total moment isWe know the resultant force from above , which acts through the centre ofpressure at C, soEquating gives,Thus the position of the centre of pressure along the plane measure from the point O is:It look a rather difficult formula to calculate - particularly the summation term. Fortunately thisterm is known as the 2nd Moment of Area , , of the plane about the axis through O and itcan be easily calculated for many common shapes. So, we know:And as we have also seen that 1st Moment of area about a line through O,Thus the position of the centre of pressure along the plane measure from the point O is:and depth to the centre of pressure isHow do you calculate the 2nd moment of area?To calculate the 2nd moment of area of a plane about an axis through O, we use the parallelaxis theorem together with values of the 2nd moment of area about an axis though thecentroid of the shape obtained from tables of geometric properties.The parallel axis theorem can be written Dr. Adel Afify 17 MET 212
  18. 18. where is the 2nd moment of area about an axis though the centroid G of the plane.Using this we get the following expressions for the position of the centre of pressureThe second moment of area of some common shapes.The table blow given some examples of the 2 nd moment of area about a line through thecentroid of some common shapes. 2nd moment of area, , about Shape Area A an axis through the centroid Rectangle Triangle Circle Semicircle Dr. Adel Afify 18 MET 212
  19. 19. Example: 1The 4- m diameter circular gate is located in the inclined wall of a large reservoir containingwater (Ϫ = 9.8 kN/m3). The gate is mounted on a shaft along its horizontal diameter. For awater depth of hc = 10 m above the shaft determine: (a) the magnitude and location of theresultant force exerted on the gate by the water. (b) the moment that would have to beapplied to the shaft to open the gate.Solution F = Ϫ hc A F = 9.8 X 103 X 10 X (4π) F = 1230 X 103 N mA = π R2 = π (22) = 12.57 m2IGG = R4 = 12.567Sc = +11.55 = 11.637 mD = Sc * sin 60D = 11.637 * sin60 = 10.0775 m(b) the momentM = F * (Sc – X)M = 1230 X 103 * (11.637 – 11.55)M = 1.07 X 105 N . mExample: 2Determine the resultant force due to the water acting on the 1m by 2m rectangular area ABshown in the diagram below. Dr. Adel Afify 19 MET 212
  20. 20. The magnitude of the resultant force on a submerged plane is:R = pressure at centroid area of surfaceThis acts at right angle to the surface through the centre of pressure.By the parallel axis theorem (which will be given in an exam), , where IGG isthe 2nd moment of area about a line through the centroid and can be found in tables. For a rectangleAs the wall is vertical, , Dr. Adel Afify 20 MET 212
  21. 21. Example: 3Determine the resultant force due to the water acting on the 1.25m by 2.0m triangular areaCD shown in the figure above. The apex of the triangle is at C. For a triangleDepth to centre of gravity is .Distance from P isDistance from P to centre of pressure isForces on submerged surfacesExample: 4Obtain an expression for the depth of the centre of pressure of a plane surface whollysubmerged in a fluid and inclined at an angle to the free surface of the liquid.A horizontal circular pipe, 1.25m diameter, is closed by a butterfly disk which rotates about ahorizontal axis through its centre. Determine the torque which would have to be applied to Dr. Adel Afify 21 MET 212
  22. 22. the disk spindle to keep the disk closed in a vertical position when there is a 3m head of freshwater above the axis.Answer:The question asks what is the moment you have to apply to the spindle to keep the discvertical i.e. to keep the valve shut?So you need to know the resultant force exerted on the disc by the water and the distance xof this force from the spindle.We know that the water in the pipe is under a pressure of 3m head of water (to the spindle)Diagram of the forces on the disc valve, based on an imaginary water surface. , the depth to the centroid of the disch = depth to the centre of pressure (or line of action of the force)Calculate the force:Calculate the line of action of the force, h. Dr. Adel Afify 22 MET 212
  23. 23. By the parallel axis theorem 2nd moment of area about O (in the surface)where IGG is the 2nd moment of area about a line through the centroid of the disc and IGG = 4r /4.So the distance from the spindle to the line of action of the force isAnd the moment required to keep the gate shut is Dr. Adel Afify 23 MET 212