# Pressure in stationary and moving fluid Lab- Lab On- On Chip: Lecture 2

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1 Pressure in stationary and moving fluid Lab-On-Chip: Lecture

2 Lecture plan what is pressure e and how it s distributed in static fluid water pressure in engineering problems buoyancy y and archimedes law; stability of floating bodies fluid kinematics. nd Newton law for fluid particles. Bernoulli equation and its application

3 Fluid Statics No shearing stress No relative movement between adjacent fluid particles, i.e. static or moving as a single block Main question: How pressure is distributed through the fluid

4 Pressure at a point Question: How pressure depends on the orientation of a plane in fluid? Newton s second law: δxδyδz Fy = pyδxδz psδxδssinθ = ρ ay δ xδy δz δxδ yδz Fz = pz δx δy ps δx δs cos θ γ = ρ az δy = δscos θ δz = δssin θ δ y py ps = pay δ z pz ps = ( paz + ρ g) if δ x, δ y, δz 0, p = p, p = p y s z s Pascal s s law:pressure doesn t depend on the orientation of plate (i.e. a scalar number) as long as there are no shearing stresses

5 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: p δy p δy δ Fy ( p = ) δxδz ( p+ ) δxδz y y p δfy = δyδxδz y p δ Fx = δ y δ x δ z x p δfz = δyδxδz z

6 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: δ F δwk = δma pδ yδxδz ρgδyδxδzk = ρgδyδxδza General equation of motion for a fluid without shearing stresses p ρgk = ρga

7 Pressure variation in a fluid at rest At rest a=0 p ρgk = 0 Incompressible fluid p1 = p + ρgh p p p = 0 = 0 = ρg x y z

8 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

9 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= = ρ RT V dp gp = ρg = ~8 km dz RT gz ( 1 z) h/ H assuming T = const p = p1exp ; p( h) = p0e RT 0

10 Measurement of pressure Pressures can be designated as absolute or gage (gauge) pressures p = γ h + p atm vapor very small!

11 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

12 Example: Pressure force and moment acting on aquarium walls Force acting on the wall H H F = ρghda= ρg H y bdy = ρg b R A ( ) 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 H 3 H F y = ρ gh yda = ρ g H y y bdy = ρ g b y R R R = H ( ) A 3 Generally, 0 6 y R = A y da ya c

13 Pressure force on a curved surface

14 Buoyant force: Archimedes principle when a body is totally t 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 B t f i ti th t id Buoyant force is acting on the centroid of displaced volume

15 Stability of immersed bodies Totally immersed body

16 Stability of immersed bodies Floating body

17 Elementary fluid dynamics: Bernoulli equation

18 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 the same path inviscid liquid (no viscosity, therefore no thermal conductivity F= ma Net pressure force + Net gravity force

19 Streamlines Streamlines: the lines that are tangent to velocity vector through the flow field Acceleration e along the streamline: a s = dv dt = v s s t = v s v Centrifugal acceleration: a n = v R

20 Along the streamline 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( s s v s v s p g = ρ θ ρ ) sin(

21 Balancing ball

22 Pressure variation along the streamline Consider inviscid, i id incompressible, ibl steady flow along the horizontal streamline A-B in front of a sphere of radius a. Determine pressure variation along the streamline from point At to point tb. Assume: 3 Equation of motion: p s v = ρv s V = V 1+ a x v a a v = 3 v s x x 3 3 p 3ρav 0 a = x x x a ρav a a 1 a 0 Δ p= dx v 3 = ρ + 3 x x x x

23 Raindrop shape The actual shape of a raindrop is a result of balance between the surface tension and the air pressure

24 Bernoulli equation Integrating ρ g sin( θ ) p s = v ρ s v 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

25 Example: Bicycle Let s consider coordinate system fixed to the bike. Now Bernoulli equation can be applied to p p 1 = 1 ρv 0

26 Pressure variation normal to streamline δ F = ma = ρδ V 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 + ρ gz = R const Across streamlines compare 1 p + ρv + ρgz = const Along a streamline

27 Free vortex

28 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 p ρv C = = 1 ρc r = = 1 ρ 3 r r r r r p = ρ C ( ) p = ρ 0 C + r p 1 r0 r + p0 0 r

29 1 p + ρv + ρgz = const

30 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 + ρgρ gz = static pressure, point (3) const hydrostatic pressure, dynamic pressure, Velocity can be determined from stagnation pressure: 1 p = p + 1 ρv Stagnation pressure

31 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

32 Pitot-static tube

33 Steady flow into and out of a tank. ρ 1 A 1v 1 = ρ A v

34 Determine the flow rate to keep the height constant p ρ v1 + ρ gz1 = p + ρ v + ρ gz Q = A1 v1 = A v

35 Venturi channel

36 Measuring flow rate in pipes p ρ v = p + Q = A v = Av ρv

37 Restriction on use of Bernoulli equation Incompressible e flow Steady flow Application of Bernoulli equation across the stream line is possible only in irrotational flow Energy should be conserved along the streamline (inviscid flow + no active devices).

38 Probelms.4 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. The initial differential reading is 30cm as shown..39 An open tank contains gasoline ρ=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.

39 Problems 3.9 The circular stream of water from a faucet is observed to taper from a diameter 0 mm (at the faucet) down to 10 mm in a distance of 50 cm. Determine the flow rate. Water flows through a pp pipe contraction as shown below. Calculate flowrate as a function of smaller pipe diameter for both manometer configuration

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