Pressure in stationary and moving fluid. LabOnChip: Lecture 2


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1 Pressure in stationary and moving fluid LabOnChip: Lecture
2 Fluid Statics No shearing stress.no relative movement between adjacent fluid particles, i.e. static or moving as a single block
3 Pressure at a point Question: How pressure depends on the orientation of the plane? Newton s second law: δxδyδz Fy = pyδxδz psδxδssinθ = ρ ay δxδyδz δxδyδz F = p δxδy pδxδscosθ γ = ρ a z z s z δ y = δscosθ δz = δssinθ δ y py ps = pay δ z pz ps = ( paz + ρ g) if δx, δy, δz 0, p = p, p = p y s z s 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
4 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: pδy pδy δ Fy = ( p ) δxδz ( p+ ) δxδz y y p δfy = δyδxδz y p δfx = δyδxδz x p δfz = δyδxδz z
5 Basic equation for pressure field Resulting surface force in vector form: If we define a gradient as: p p p δ F = ( i + j + k) δyδxδz x y z δ F = i + j + k = p x y z δyδxδz The weight of element is: δwk = ρg δ yδ xδ z k Newton s second law: δf δwk = δma pδ yδxδz ρgδyδxδzk = ρgδyδxδza General equation of motion for a fluid w/o shearing stresses p ρgk = ρga
6 Pressure variation in a fluid at rest At rest a=0 p ρgk = 0 Incompressible fluid p1 = p + ρgh p p p = 0 = 0 = ρ g x y z
7 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
8 Compressible fluid Example: let s check pressure variation in the air (in atmosphere) due to compressibility: Much lighter than water, 1.5 kg/m 3 against 1000kg/m 3 for water Pressure variation for small height variation are negligible For large height variation compressibility should be taken into account: nrt p= = ρrt V dp gp = ρg = dz RT gz ( 1 z) assuming T = const p = p1exp ; p( h) = p0e RT0 h/ H ~8 km
9 Measurement of pressure Pressures can be designated as absolute or gage (gauge) pressures p = γ h+ p atm vapor very small!
10 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. Force acting on a side wall in rectangular container: h FR = pava= ρg bh
11 Example: Pressure force and moment acting on aquarium walls Force acting on the wall R A H H F = ρghda= ρg H y bdy = ρg b ( ) 0 Generally: R = sin = sin F ρg θ yda ρg θ y A Centroid (first moment of the area) A c Momentum of force acting on the wall R R R A = H 3 H 3 H F y = ρgh yda = ρg H y y bdy = ρg b y Generally, ( ) 0 6 y R = A yda ya c
12 Pressure force on a curved surface
13 Buoyant force: Archimedes principle when a body is totally or partially submerged a fluid force acting on a body is called buoyant force F = F F W B 1 F F = ρg( h h) A [ ] F = ρg( h h) A ρg ( h h) A V B FB = ρgv Buoyant force is acting on the centroid of displaced volume
14 Stability of immersed bodies Totally immersed body
15 Stability of immersed bodies Floating body
16 Elementary fluid dynamics: Bernoulli equation
17 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 F= ma Net pressure force + Net gravity force
18 Streamlines Streamlines: the lines that are tangent to velocity vector through the flow field Acceleration along the streamline: a s = dv dt = v s s t = v s v Centrifugal acceleration: v a n = R
19 v s v V ma F s s = = ρδ δ V s p V g F W F ps s s δ θ δ ρ δ δ δ = + = ) sin( v s v s p g = ρ θ ρ ) sin( Along the streamline
20 Balancing ball
21 Pressure variation along the streamline Consider inviscid, incompressible, steady flow along the horizontal streamline AB in front of a sphere of radius a. Determine pressure variation along the streamline from point A to point B. Assume: 3 Equation of motion: p s v = ρv s V = V 1+ a x 0 3 v a a v = 3v 1 + s x x p 3ρav 0 a = x x x a ρav 0 a a 1 a p= 1 dx ρv = x x x x
22 Raindrop shape The actual shape of a raindrop is a result of balance between the surface tension and the air pressure
23 Bernoulli equation Integrating ρ g sin( θ ) p s = v ρ v s dz ds dp ds 1 ρ g dz dp = ρ dv ds ds ds, n=const along streamline 1 We find dp + ρd( v ) + ρgdz = 0 Along a streamline Assuming incompressible flow: 1 p + ρv + ρgz = const Along a streamline Bernoulli equation
24 Example: Bicycle Let s consider coordinate system fixed to the bike. Now Bernoulli equation can be applied to p p 1 = 1 ρ v 0
25 Pressure variation normal to streamline δ F = ma = ρδv n n v R δfn = δwn + δfpn = ρgδv cos( θ ) p δv n dz p v ρ g = ρ dn n R v p + ρ dn + ρgz = R const Across streamlines compare 1 p + ρv + ρgz = const Along a streamline
26 Free vortex
27 Example: pressure variation normal to streamline Let s consider types of vortices with the velocity distribution as below: solid body rotation free vortex dz p v ρ g = ρ dn n R as n = r, p ρv = = ρ r r C r 1 p ρv = = ρ C r r r p = ρc ( r r ) + p p = ρc + p 0 r0 r
28 1 p + ρv + ρgz = const
29 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 + ρv + ρgz = static pressure, point (3) const hydrostatic pressure, dynamic pressure, Velocity can be determined from stagnation pressure: 1 p = p v 1 + ρ Stagnation pressure
30 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
31 Pitotstatic tube
32 Steady flow into and out of a tank. ρ = 1A1 v1 ρ Av
33 Determine the flow rate to keep the height constant p ρ v1 + ρgz1 = p + ρv + ρgz Q = A v = 1 1 Av
34 Venturi channel
35 Measuring flow rate in pipes p ρ v = p + Q = A v = Av ρv
36 Probelms.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 5 kpa..61 An open tank contains gasoline r=700kg/cm at a depth of 4m. The gate is 4m high and m wide. Water is slowly added to the empty side of the tank. At what depth h the gate will open.
37 Problems 3.71 calculate h. Assume water inviscid and incompressible.
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