Pressure in Stationary and Moving Fluid
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Pressure in stationary and moving fluid Lab-On-Chip: Lecture 2 Fluid Statics • No shearing stress •.no relative movement between adjacent fluid particles, i.e. static or moving as a single block Pressure at a point Question: How pressure depends on the orientation of the plane ? Newton’s second law: δδδxyz Fpxzpxs=−δδ δδsin θ = ρ a ∑ yy s2 y δδδx y zx δδδy z Fpx=−δδy pxs δδcos θ − γ = ρ a ∑ zz s22 z δ ys==δθδδθcos zs sin δ y pppa−= ys y2 δ z pp−=() pag +ρ zs z 2 if δδδx,,y zpppp→== 0,yszs , • Pascal’s law: pressure doesn’t depend on the orientation of plate (i.e. a scalar number) as long as there are no shearing stresses Basic equation for pressure field Question: What is the pressure distribution in liquid in absence shearing stress variation from point to point • Forces acting on a fluid element:: – Surface forces (due to pressure) – Body forces (due to weight) Surface forces: ∂∂pyδδ py δ Fp=−()()δδ xzp −+ δδ xz y ∂∂yy22 ∂p δδδδFyxz=− y ∂y ∂p δδδδFyxz=− x ∂x ∂p δδδδFyxz=− z ∂z Basic equation for pressure field G ∂pppG ∂∂G G Resulting surface force in vector form: δ F =−()ijkyxz + + δδδ ∂∂xy ∂ z G ∂∂∂G G G δ F If we define a gradient as: ∇ =++ijk =−∇ p ∂∂∂xyzδδδ yxz G G The weight of element is: −=−δWkρδδδ g y x z k G G G Newton’s second law: δδFWkma−= δ G G −∇pδ yxzδδ − ρ g δδδ yxzk = − ρ g δδδ yxza General equation of G G motion for a fluid w/o −∇pgkga −ρ = −ρ shearing stresses Pressure variation in a fluid at rest G •At rest a=0 −∇−pgkρ =0 ∂ppp∂∂ ===−00 ρg ∂∂∂xyz • Incompressible fluid p12= pgh+ ρ Fluid statics Same pressure – much higher force! Fluid equilibrium Transmission of fluid pressure, e.g. in hydraulic lifts • Pressure depends on the depth in the solution not on the lateral coordinate Compressible fluid • Example: let’s check pressure variation in the air (in atmosphere) due to compressibility: – Much lighter than water, 1.225 kg/m3 against 1000kg/m3 for water – Pressure variation for small height variation are negligible – For large height variation compressibility should be taken into account: nRT pRT==ρ V dp gp =−ρg =− dz RT ~8 km ⎡⎤gz()12− z −hH/ assumingT=⇒= const p21 p exp⎢⎥ ; p ( h ) = p 0 e ⎣⎦RT0 Measurement of pressure • Pressures can be designated as absolute or gage (gauge) pressures phpatm= γ + vapor very small! Hydrostatic force on a plane surface • For fluid in rest, there are no shearing stresses present and the force must be perpendicular to the surface. • Air pressure acts on both sides of the wall and will cancel. h Force acting on a side wall in rectangular container: F ==pA ρg bh Rav 2 Example: Pressure force and moment acting on aquarium walls • Force acting on the wall H H 2 FghdAgHybdygb==−⋅=ρρ ρ R ∫∫() A 0 2 Fg==ρθsin ydAgyA ρθ sin • Generally: Rc∫ A Centroid (first moment of the area) • Momentum of force acting on the wall H H 3 F y==−⋅=ρρ gh ydA g H y y bdy ρ g b RR ∫∫() A 0 6 yHR = 3 ∫ ydA2 • Generally, A yR = yAc Pressure force on a curved surface Buoyant force: Archimedes principle • when a body is totally or partially submerged a fluid force acting on a body is called buoyant force FFFWB =−−21 FF21−=ρ ghhA() 21 − FB =−−−−ρρgh()21 hA g[] () h 21 hA V FB = ρgV Buoyant force is acting on the centroid of displaced volume Stability of immersed bodies • Totally immersed body Stability of immersed bodies • Floating body Elementary fluid dynamics: Bernoulli equation Bernuolli equation – ”the most used and most abused equation in fluid mechanics” Assumptions: • steady flow: each fluid particle that passes through a given point will follow a the same path • inviscid liquid (no viscosity, therefore no thermal conductivity FF== mmaa Net pressure force + Net gravity force Streamlines Streamlines: the lines that are tangent to velocity vector through the flow field dv ∂v ∂s ∂v Acceleration along the a = = = v streamline: s dt ∂s ∂t ∂s v2 Centrifugal acceleration: a = n R Along the streamline ∂v ∂p δF = ma = ρδV v δF = δW +δF = −ρgδV sin(θ ) − δV ∑ s s ∂s ∑ s s ps ∂s ∂p ∂v − ρg sin(θ ) − = ρ v ∂s ∂s Balancing ball Pressure variation along the streamline • Consider inviscid, incompressible, steady flow along the horizontal streamline A-B in front of a sphere of radius a. Determine pressure variation along the streamline from point A to point B. Assume: ⎛⎞a3 VV=+0 ⎜⎟1 3 ⎝⎠x Equation of motion: ∂∂p v =−ρv ∂∂s s 33 ∂vaa2 ⎛⎞ vv=−310 ⎜⎟ + 34 ∂sxx⎝⎠ 32 3 ∂pa3ρav0 ⎛⎞ =+43⎜⎟1 ∂xx⎝⎠ x −a 32 33⎛⎞6 3ρav0 ⎛⎞aaa2 1 ⎛⎞ ∆=pdxv43⎜⎟1 + =−ρ 0 ⎜⎟ 3 + ⎜⎟ ∫ xx⎜⎟ xx2 −∞ ⎝⎠ ⎝⎠⎝⎠ Raindrop shape The actual shape of a raindrop is a result of balance between the surface tension and the air pressure Bernoulli equation ∂p ∂v Integrating − ρg sin(θ ) − = ρ v ∂s ∂s dz dp , n=const along streamline ds ds dz dp1 dv2 −−=ρρg ds ds2 ds 1 2 We find dp + ρd(v ) + ρgdz = 0 Along a streamline 2 Assuming 1 p + ρv2 + ρgz = const Along a streamline incompressible 2 flow: Bernoulli equation Example: Bicycle • Let’s consider coordinate system fixed to the bike. Now Bernoulli equation can be applied to 1 p − p = ρv 2 2 1 2 0 Pressure variation normal to streamline v2 ∂p F W F g V cos( ) V δFn = man = ρδV ∑δ n = δ n +δ pn = −ρ δ θ − δ ∑ R ∂n dz ∂p v2 − ρg − = ρ dn ∂n R v2 p + ρ dn + ρgz = const Across streamlines ∫ R compare 1 2 p + ρv + ρgz = const Along a streamline 2 Free vortex Example: pressure variation normal to streamline • Let’s consider 2 types of vortices with the velocity distribution as below: solid body free vortex rotation dz ∂p v2 − ρg − = ρ dn ∂n R ∂∂∂pVρ 2 ∂pVρ 2 C 2 as =− , ==ρCr2 ==ρ 1 ∂∂nr∂rr 1 ∂rr r3 1 22 2 1112 ⎛⎞ p = ρCr10()−+ r p 0 p = ρCp20⎜⎟22−+ 2 2 ⎝⎠rr0 1 p + ρv2 + ρgz = const 2 Static, Stagnation, Dynamic and TotalPressure • each term in Bernoulli equation has dimensions of pressure and can be interpreted as some sort of pressure 1 p + ρv2 + ρgz = const 2 hydrostatic pressure, dynamic pressure, static pressure, point (3) 1 p = p + ρv2 Velocity can be determined from stagnation pressure: 2 1 2 Stagnation pressure On any body in a flowing fluid there is a stagnation point. Some of the fluid flows "over" and some "under" the body. The dividing line (the stagnation streamline) terminates at the stagnation point on the body. As indicated by the dye filaments in the water flowing past a streamlined object, the velocity decreases as the fluid approaches the stagnation point. The pressure at the stagnation point (the stagnation pressure) is that pressure obtained when a flowing fluid is decelerated to zero speed by a frictionless process Pitot-static tube Steady flow into and out of a tank. ρ1 A1v1 = ρ2 A2v2 Determine the flow rate to keep the height constant 1 1 p + ρv 2 + ρgz = p + ρv 2 + ρgz 1 2 1 1 2 2 2 2 Q = A1v1 = A2v2 Venturi channel Measuring flow rate in pipes 1 1 p + ρv 2 = p + ρv 2 1 2 1 2 2 2 Q = A1v1 = A2v2 Probelms • 2.43 Pipe A contains gasoline (SG=0.7), pipe B contains oil (SG=0.9). Determine new differential reading of pressure in A decreased by 25 kPa. • 2.61 An open tank contains gasoline r=700kg/cm at a depth of 4m. The gate is 4m high and 2m wide. Water is slowly added to the empty side of the tank. At what depth h the gate will open. Problems • 3.71 calculate h. Assume water inviscid and incompressible. .