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This 58 page Class Notes was uploaded by Gerhard Leuschke II on Saturday October 3, 2015. The Class Notes belongs to ME311 at California State Polytechnic University taught by MaryamShafahi in Fall. Since its upload, it has received 20 views. For similar materials see /class/218326/me311-california-state-polytechnic-university in Mechanical Engineering at California State Polytechnic University.
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Date Created: 10/03/15
ME 311 Fluid Mechanics Fluid statics Maryam Shafahi Winter 2011 r9 CAL POLY POMONA Fluid Statics Study of fluids in which there is no relative motion between particles When fluid particles are in static equilibrium what forces are acting on them Pressure dimension Unit in SI In the next slide let us take a look at the fluid static forces acting on a triangular element in 2D Forces acting on a Static Element Take a look at the element below only weight and compressive forces are acting on this element The depth is unit PAs As PXAV Av PyAx Sum of Forces If we sum of all forces for this element we will get the following equation AxAy EFxmax prypAssm0p ax AxAy 39 AxAy 2Fymayz pyAx pg 2 pAscosQp 2 ay And if we make some substitutions into the above equation Assin0Ay AseosbAx it will take the form of p 2961A x P 2 i A Prp piay 2539 y Pressure at a Point Now let us shrink the element to a point which is to say that we let Ax and Ay approach zero then the equation below PAS 32 A 39 As P p z 12 39 My ML in mm will become PXPyP This important result implies that the magnitude of pressure is the same at a given point regardless from which direction the pressure is applied As a result the pressure is a scalar function not a vector Pressure Variation So we now know that pressure is a scalar quantity We also know that pressure is defined as FA The question is then what kind of force caused pressure To find out we want to look at a small fluid element surrounded by fluid ofthe same kind Forces around the Element Let s now take a look at the forces acting on this element 7 51 6 5 V av6v l 7 pigcc vl av5zgt i 7439 255 A 6quot 75X 5 5 J3 gt Since pressure is a scalar quantity we assume that pressure forces act all the way around this element in x y and z direction In the zdirection body force weight of the element acts on it as well Pgdrdydz Forces on the element To further understand the forces let us take a look at the force acting on the right face of the element The area for the right face is dxdz If we assume the pressure at center of the element is Pxyz as a function of position then the pressure at a distance O5dy away is p 615 V 6P 5 d Px7y7yazPx7y7ZE pl624d 6zgt p 5 5 yb39xxir 43v i 2 4 J Forces on the Element cont So the force acting on the right face of the element is 2 3de F Px dxd Ly H M 6w 2 gi M z i The Sign of the force IS negative 5 r 5 l7 t 5x5 quot I quot 7 6 5 because the arrow of Pforce 39 ii 6 ya quot5 X OJ Z is pointing to the negative y direction 5quot A vJ5x539 kL z i Forces on the Element cont If we follow the same procedure for other faces we will have 3de F P dd xayaz ax 6P dx F P dd xayaz ax 6P dy F PxyZ57dde 6P dy F P 77 d d xayaz x Z 2 Ed my 2 Egdxdy T 32 2 E Px y z ai dey 32 2 QE My Ii 2 5 t apay r 5 W 39 53b litgj j Z39lexb quot ID dv 2 V n A 54 y x o 5 6y 916 I A p7 9f ans kL 1 J 9 Forces on the Element cont If the uid element is at static equilibrium then all the accelerations are zero therefore equation above becomes We can now apply Newton s 2ncl law ZFma and will arrive at the follwing equations Pressure at a Point Pressure varies as a function of depth ifthe density is constant Ppghyh Where g is the gravitational constant Characteristics of Pressure So pressure is a scalar quantity Pressure is dependent on fluid density and depth Pressure is independent of crosssectional area Pressure always acts toward the body Let s take a look at the three containers below which one has the highest pressure at point indicated assume all of them contain the same liquid Characteristics Of Pressure Let s take a look at the three containers below which one has the highest pressure at point indicated assume all of them contain the same liquid Manometer Manometers use columns of liquids to measure pressures Let s take a look at the following different kinds of manometers U tube same fluid with specific gravity v 9 For the Utube on the right pressure at point B is open to atmosphere so PB Patm0gage 9 Since pressure increases with depth and A is deeper than B so PA is higher than PB o PAyh PB Manometer with Different Fluids There can be different fluids in the manometer too Utu be two different fluids with specific gravities v1 and v2 w B J For the Utube on the right pressure at 39 point B is open to atmosphere so CZPathOgage 4 Let s now try to set up an equation from point A to point B 1313 Yzhz39 thlsz Move down to fluid interface since point of interface is lowerthan B so pressure is larger y2 Manometer Example Start from either point right or left watch out for units make sure you convert the heights to m before proceeding Use v for water in SI unit 98 kNm3 6 for Hg is 136 and 6 for oil is 092 Reminder ng S and same for oil We are going to move from the left point at 40 kPa to the point on the right at 16 kPa Every time we move downward we add pressure to equation If we move upward we subtract vwater 226 For the setup shown in Fig P2226 calculate the manometer reading H FIGURE P226 Fluid Mech by Potter and Wiggert 3rd ed Manometer Example 226 For the setup shown if Fig P2126 calculate the The equatlon 1s below manometer reading H 4 Note the unit is in kPa 4 You can solve for H 40 kPa FIGURE P226 40 71120 SGHgVHZOH SGOz HZO 4O 98X2 136gtlt98gtlt H O92gtlt98gtlt3 16kPa Manometer Example O Move from water to oil 229 Determine the pressure difference between the When WC move downward we add Water pipe and the oil pipe shown in Fig P229 to the pressure equation When we move upward we subtract from the pressure equation Use the following equation you can solve for the pressure differential between water and oil pipes FIGURE P229 98X015 l36gtlt98gtltOl O68gtlt98gtlt2O86gtlt98gtlt15IQUcPa P water summary characteristics of pressure pressure varies with depth only how to do manometer problems Fluid Statics ME311 Maryam Shafahi Winter 2011 Pressure Force on Submerged Plane Surface Let us now consider application of pressure force in a different case a surface area is submerged under fluid what will be magnitude and direction of the pressure force Where will the pressure force be acting at Application 439 Consider the car below is partially submerged under water in a flooded street If you are stranded inside the car what force do you need to apply to the car door in order for you to open it Application II A plug is at the bottom of a sink In order to drain the sink what force is required to unplug the plug Application lll You are building a dam What does the dimension of the dam has to be so that it will not topple over What is the maximum depth that is allowed for the reservoir Application IV You are building roadways on a hill side When it is raining hard the soil becomes saturated with water and starts to exhibit characteristics of fluids and the pressure force starts to push on the concrete slab that is protecting the slope How big is the pressure force and will it destroy the slab Road surface I Soil Concrete slab Calculate Pressure Force on Inclined Plane In order to calculate the pressure forces in the above 4 scenarios we will need to figure out how to calculate pressure forces on an inclined plane Let s take a look at the following slide for a figure of a plane that is submerged Centroid c Location of resultant force center of pressure CP Calculate Pressure Force on Inclined Plane The force acting on the differential area dA is a pressure force Thus we can use an integral to define the resultant force FR IpdA 4 Pressure is dependent on depth so we can define P as u Substitute into force equation Calculate Pressure Force on Inclined Plane Define distance from fluids surface to the centroid along the direction of the inclined plane in u I ll 0 I r y Iquot 4 Substitute to force quation Which simply states that the pressure force acting on an inclined plane is directly depended on specific weight of the fluid height from centroid of plane to free surface and the plane area Graphical Representation Let us take a look at the following example WW 4 The picture on the left shows the original pressure distribution on the inclined surface The picture on the right shows that the trapezoidal distribution has been replaced by a single force In order to make the two systems equivalent the single force needs to be applied at a particular location Center of Pressure From Statics we know that in order for two systems to be equivalent the sum of the forces must be equivalent and the sum of the moments needs to be equivalent as well WWW So for the above two systems to be equivalent the singleforce replacement is at a specific point where the moments of the two systems are equivalent as well The point is called center of pressure Center of Pressure If we equate moment about the x axis if A The integral in the numerator is the second moment of the area moment of inertia xc yR yc up Check Fig 218 to find the moments for some geometrical shapes Center of Pressure xc f 39gtyca The point through which the resultant force acts is called the center of pressure Example The inclined surface shown hinged along edge A is 5 m wide Determine the resultant force FR of the water on the inclined surface Net hydrostatic pressure distribution on gate Given Rectangular gate hinged along A w5m Find Resultant force FR We solve this problem by using idirect integration ii the algebraic equations FR 2 IpdA hDysini9 dAle A Integral equations Net hydrostatic aressure dwstr rbution on gate L FRyR wayD ysin 6dy 0 2 3 L sz Dy y sin6 FR 2 3 0 Algebraic Equations FE prA pghCA pgD sin6Lw Maryam Shafahi Winter 2011 ME 113 Fluid Mechanics ME 311 Textbook Fundamentals of Fluid Mechanics Munson Young OkiishiHuebsch Class TuThu 10451200 Instructor Maryam Shafahi Email mshafahicsupomonaedu Office 9 222 Office hours TuThu 9301030 230330 2 midterms 25 each Quiz 10 Final 40 Class activity 5 HW submit them in Bb ME 113 Fluid Mechanicsl Dimensions and units Basic dimensions M L TG Or F L T 9 System of units SI BG EE SI International system kg m s K N m s K BGBritish Gravitational system slug ft 5 R lb ft 5 R EEEnglish Engineering system Ibm ft 5 R lb ft 5 R Newton s second law Fma 1 slug 32174 Ibm ME 113 IFluici Mechanics I Example 12 A tank of liquid having a total mass of 36 kg rests on a support in the equipment bay ofthe space shuttle Find the force in newtons tank exerts on the support shortly after lift off when shuttle is accelerating upward at 15 ftsz 2F ma Ff WzmagtFfmga i l Be careful about the unit system ft03048 m Ff 518 kgms2 518 N The direction is downward since the force shown on the free body diagram in the force of the support on the tank ME 113 Fluid Mechanics l Physical properties Density p mass per unit volume kgm3 slugsft3 Sped cvohnnevyvohuneperunnrnass v1p n kg Specific weighty weight per unit volume V pg N m3 Specific gravity 56 p pHZO4C IdealGasLaun pWRT Viscosity u describes the fluidity of the fluid ifquot du TZIle y t shear stress dimension FtA Newtonian fluid Example This sled slides along on a thin horizontal layer of water between the ice and the runners The horizontal force that the water puts on the runner is equal to 12 lb when the sled s speed is 50 fts the surface area in contact with the water is 008 ft2 and water viscosity is 3510395 lbsft2 Determine the thickness of the water layer under the runners Assume a linear velocity distribution in the water layer Given F A uu FindD FTA 1 39udy 39uD u Fu A D D11710394 ft ME 113 IFluici Mechanics I Viscosity NonNewtonian fluids Bingham plastic Shear thinning fluid the harder sheared the less viscous polymer solution latex paint Shear thinning HNewtonian Shearing stress 7 Shear thickening fluid the harder sheared the more viscous watercorn starch watersand starch Shear thickening Bingham plastic yield stress without motion du Rate of shearing strain dy toothpaste mayonnaise Viscosity is a function of temperature CT32 S and C are empirical constants Sutherland equation T S D and B are constants Andrade s equation a DeBT ME 113 lFluid Mechanics l Compressibility of the fluid Bulk modulus EV dpdvv ME 113 IFlu ci Mechanics I Surface tension Interface of a liquid and a gas or between two immiscible liquids surface behaves like a skin stretched over the fluid mass The intensity of the molecular attaraction per unit length along any line in the surface is called the surface tension 0 ME 113 Fluid Mechanics I Surface tension Pressure inside a drop of liquid 27rR0Ap7rR2 gt Ap 19 pe ZYO Capillary WrRZh 27rR0 cos 6 20 cos 6 yR h wetting Nonwetting VIE 113 Fluid Mechanics I Example Capillary raise in a tube What diameter of clean glass tubing is required so that the rise of water at 20 C in a tube due to capillary action as opposed to pressure in the tube is less than hl0 mm 11 Zacos oSe3 R00149m nk 3911 Fluiillrlerlialwsl Example 39 A layer of water flows down an inclined fixed surface with the velocity profile shown in the figure Determine the magnitude and direction of the shearing stress that the water exerts on the fixed surface for U2ms and h01 m Tdu dy ziy dy h h2 du U Tw 2 dyyZO h tw ME 113 Fluid Mechanicsl Summary Systems and dimensions Specific weight Specific gravity Newtonian fluid shear stress Bulk modulus Capillary rise in a tube Shear stress ME 113 Fluid Mechanics I ME 311 Fluid Statics Forces on Plane Surfaces Maryam Shafahi Mechanical Enginaaring Dapartmant Example 27 of the book Example For a vertical gate shown below if the top of the gate is 4 m below water level determine magnitude and location of the pressure force if the gate is of the following shapes gate ll 5m 5m Find Centroids First You can find information on centroid and moment of inertia from Fig 218 Find Moment of Inertia Next Note semicircle and quartercircle are special cases which we will discuss on a separate slide gate 5m 4 D T AV Em C d 3 3 3 3 3 3 KW jzs 5 In M 25 5 5621 55 Ixc 20054881134 4 E36 5 36 312 Plug in the formulas to get your answers 439 Note In this case since the gates are vertical hC and yC are the same J Case a FzyhcA984130552 54 I E yR yCA yc10 ye 43552 4 Try to calculate for Cases b and c on your own Quarter and Semicircles So for part d in the prior case 45 F7hCA9845 3557Z394 7r 1 00554 gtXlt yRycxj445 i5 45 ye 45 3 557r4 7239 Determine the force P needed to hold thew wide gate in the position shown in Fig
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