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School
Department
Physics
Course
Physics 1029A/B
Professor
Martin Zinke- Allmang
Semester
Winter

Description
Chapter 11: Liquid water and aqueous solutions, static fluids Chapter 12: Cardiovascular System: Fluid flow - p. 368 – - Fluid = deformable so evolve toward mechanical eq’m in given - ﬂow no longer stationary ﬂuid model because dynamic process .: not in mechanical eq’m space. Intermolecular forces dominate = liquid ﬂuid. Thermal - ideal stationary ﬂuid AND dynamic ﬂuid = incompressible/deformable energy of system dominates = gas ﬂuid. Ideal ﬂuid model: ﬂuid is - ideal dynamic ﬂuid: (1) no turbulences (.: no air through trachea), (2) no sound waves (b/c density ﬂuctuations, automatic if incompressible, but incompressible. If in mechanical eq’m, ideal stationary ﬂuid. If in physiologically bun dis), (3) no friction with walls or objects that move @ diff speeds .: interaction of ﬂuid particles = elastic collisions only condensed liquid state, Pascal’s law: pressure increase linearly - laminar ﬂow is ﬂuid ﬂow satisfying (1)(2); established when (i) ﬂow lines in the ﬂuid never cross each other, and (ii) ﬂow tubes never penetrate with depth below surface (Pressure at liquid surface = pressure in each other adjacent gas space) - conservation of ﬂuid mass: ∆m /∆t =1∆m /∆t; 2 D * ∆V 1∆t = D * ∆V /∆2 ; mass ﬂow rate vs volume ﬂow rate ; rewrite volume --> D * - Surface of condensed ideal stationary ﬂuid has properties A 1 ∆l 1∆t = D * A *2∆l /∆2 ; using velocity --> A *1v | 1 A *|2 | .2 A *|v| = const (= ∆V/∆t) = equation of continuity distinct from those of bulk material due to excess energy required to - Equation of continuity = expression of conservation of mass or volume of an incompressible ﬂuid. It states that the volume ﬂuid rate is form surface (surface tension/energy), which causes pressure diff constant along a tube. The ﬂuid ﬂows faster when it passes through a section of tube with smaller cross-section. Apples to laminar ﬂow whether ideal across bubble / droplet surface inversely proportional to radius dynamic ﬂuid or not (b/c type interactions irrelevant). - Model system: ideal stationary ﬂuid = systems that yield to any force that attempts to alter their shape, causing system to ﬂow until - venturi meter measures changes of ﬂuid speed based on pressure (pressure ~ 1/speed) reaches mechanical eq’m in which ﬂuid then conforms to shape of - Bernoulli’s law: expression of conservation of (total) energy for closed system. States that increase in speed of ideal dynamic ﬂuid in tube is container accompanied by drop in pressure. Correct pressure locally due to tube diameter variations. Only applies to ideal dynamic ﬂuid (conservation of E)  - Liquids and gases, but not solids (don’t change shape for const = p + ½ D * v 2 ; p = pressure in ﬂuid ; Work = – (p – p )∆V 2 1 container) - Liquids: molecules in condensed state, maintain - Newtonian ﬂuid model: describes ﬂuid movement by removing restrictive assumption, since must have friction interactions constant intermolecular distance, forms surface in container of larger - Viscosity: interaction between neighbouring layers of moving ﬂuid in direction perpendicular to ﬂow lines (F ext= Resistance volume than liquid occupying volume - F ext= n * A |∆v| / ∆y ; n = viscosity coefficient (N * s/m ) ; ∆y = distance between the 2 layers ; v = velocity ; A = area ; F = force required ; n for - Gases: adjust intermolecular distance and ﬁll any gases INCREASE with temp (more interaction with layers); n for liquids DECREASE with temp (molecules have to squeeze past) provided space uniformly, have no natural surfaces - viscosity is dynamic effect requires velocity gradient perpendicular to direction of ﬂow lines in ﬂuid ∆v/∆y .: no role in stationary (v = 0) or ideal - Fluids in mechanical eq’m = stationary ﬂuid (all “essential dynamic (∆v/∆y = 0) ﬂuids .: all ﬂuids even ideal gas, behave as Newtonian ﬂuid once velocity gradient introduced. In Newtonian ﬂuid, inelastic parameters” time-independent) interaction with ﬂuid-conﬁning walls causes velocity gradients. Viscosity replaces assumption of elastic collisions required between ideal dynamic ﬂuid - Liquid: no obvious changes occur while observe system and its conﬁning walls. In Newtonian ﬂuids, the ﬂow is laminar and viscous. - Gas: microscopic level =lot of motion of particles -.: macroscopic eq’m excludes microscopic - Newtonian assumes steady-state .: only 2 forces = pressure diff force pushing forward (1), and viscosity force pushing back (2). (1) dominates in - Properties of ideal stationary ﬂuid: middle, (2) dominates close to walls 2 2  = incompressible: V (volume) and p (density) = m/V are constant - velocity proﬁle: v = ( r tube– r / 4 * n ) * ( ∆p / l ) ;tube= radius of tube, r = distance to container, ∆p = pressure diff along segment of tube length l, (pressure n = coefficient (always > 0), ∆p/l = constant pressure gradient along tube independent) -----Good for liquids, doesn’t apply to gases (retain 4 ideal gas model) - Poiseuille’s law: ∆V / ∆t = ( π / 8 * n ) * r tube* ( ∆p / l ), pi/8 = factor for cylindrical tube = states that volume ﬂow rate of Newtonian ﬂuid through th  = deformable under forces and seeks mechanical eq’m, and only cylindrical tube is proportional to 4 power of radius of tube. Actual ﬂow through the tube. .: narrow tube reduces ﬂow severely i.e. reduce diameter 2 then = ﬂow reduced by 16 does ﬂuid become stationary----- Good for both liquids and gases - Ohm’s law: ∆p = R * ( ∆V / ∆t ) ; R = ﬂow resistance (Pa * s / m ) ; Ohm’s law states that volume ﬂow rate of Newtonian ﬂuid ~ pressure diff along 4 - No condition for interactions with ﬂuid molecules or container tube, and that proportionality constant = ﬂow resistance . .: R = n * (8 * l / π * r tube) walls, no elastic collision limitation - Pascal’s law: p 2 p =1–density * g (y – y2) 1 - Q. water ﬂow in sink: volume ﬂow rate constant, gravity accelerates it .: jet narrows as it falls - difference between pressures at 2 diff positions in ﬂuid is - Q. In a decorative fountain in a garden, water is shot vertically from a pipe. The uprising jet of water A) broadens as it rises 4 proportional to vertical distance between 2 positions. Proportionality - Q. diver: pressure @ depth = D g h ; too wide breathing tube = too much dead space; too narrow = too much resistance (1/r ) = exhausting factor is product of density of ﬂuid and gravitational acceleration - Q. cardboard/pin = Bernoulli preventing cardboard from falling - in systems with identiﬁable surface of ﬂuid, y = 1, p = 1 - Q. vascular ﬂutter: constriction (lower area)  lower pressure  collapse  build up of pressure  forced open p atm, make downward direction positive .: p = p atm+ density * g * - Q. prairie dog mound / hurricane and ROOF: mound causes increase speed = lower pressure = air sucked in to tunnels depth = p atm+ p gauge - Q. WRONG: poiseuille’s law applies as derived only to ideal dynamic ﬂuids; RIGHT: Poseuille’s law applies as derived only to laminar ﬂow, - doesn’t apply to gases, since gases are compressible, density incompressible ﬂuids, to Newtonian ﬂuids; Poseuille’s law and the equation of continuity can be used together for the same system depends on pressure - Q. water tank exposed to air, hole __ m below, leaks at rate ___, speed?(a) Diameter?(b) (a) around hole: p + ½ D * v = p + ½ D * v 2 ; inside, v = 2 2 - doesn’t contain any info on shape of container .: regardless 0, p = p.atm + Dgh; outside p = p.atm .:  p.atm + ½ D*g*h = p.atm + ½ D *v  gh = ½ v (b) A * v = V/t of shape, pressure increases below surface at ﬁxed values - Q. hypodermic needle, velocity out ? p = F / A + p  Bernouille  v = 0, p = 1 atm ; v = = 1 ext 1 atm 1 2 2 - gauge pressure: p gauge= p absolutep air can be positive or - Q. hypodermic needle, length = , radius = , ﬂow =  pressure? Poseuille negative, density * g * d = gauge pressure - Q. blood vessel splits into 2  take new area, because splits into 2 multiply area by 2 before calculate velocity o 5 - p atm= 1 atm @ 20 C = 1.01325 x 10 Pa = 760 torr = 760 - Q. volume ﬂow rate in = volume ﬂow rate out, if nothing else open and affecting mmHg - Q. suction, range of y it’ll work @? Y max = p atm(D*g) - 1.0 g / cm = 1.0 kg/L = 1 x 10 kg/m 3 3 - Q. pipe underground. Faucet. Change in pressure using Bernouille + Dgh 2 - Pressure = force/area = N/m - Q. aorta volume = total cap volume .: given # cap, A cap, cap, Aaorta solve for v aortahen multiply by #cap - 1 Pa = 1 N / m = 1 kg / m*s = J / m 3 - Q. venture-meter / tapering: v = ; = = when R >> r (in thick part, with Area A) - Q. Streamlined race cars as the one at right usually have “spoilers”, which are basically upside–down wings. Due to the Equation of Continuity and - Diastolic blood pressure: 10.7 kPa, 80 mmHg  veins, Bernoulli’s law in what direction will an additional force act on the spoiler at high speed? DOWN pulmonary circulation - Q. A viscous ﬂuid is forced through a pipe to obtain a certain volume ﬂow rate (experiment 1). If the same ﬂuid is then forced through - Systolic blood pressure: 16.0 kPa, 120 mmHg a pipe of same length with double the cross–sectional area (experiment 2), by how much has the pressure difference )p2 along the - Supine position: species that person is lying down, and blood pipe dropped from the previous value )p1 to obtain the same volume ﬂow rate? /\p = ¼ /\p 2 1 pressure has maximum variation of 15% atmospheric pressure - Q. Bird doesn’t beneﬁt from Bernoulli when tailwind pushes forward - Q. ﬁsh can ﬂoat in water, .: their density = density of water - For box in ﬂuid… - Q. Laplace’s law describes pressure in hollow bubble (e.g. alveoli in lungs) in form p inside poutside 4o/r, in which o is surface tension and r is radius of - Fnet= F up– Fdown – W B W Fluid– WBlock curvature of bubble. In healthy alveoli, surfactant is used to reduce surface tension by coating parts of surface. WRONG = surfactant must coat - Fnet> 0. Weight of block < weight of displaced liquid, block particularly areas with large radius of curvature. RIGHT: surfactant must coat areas with SMALL radius curvature, does not change pressure in bubble B rises to surface - Fnet= 0 .lock ﬂoats at current depth below ﬂuid surface, - Q. We introduced Pascal’s law in 2 forms, one is , can NOT be used when ﬂuid is compressible weight of block = weight of displaced ﬂuid - Q. Turtle of common shape, if swims fast horizontally under water, in what direction will force act on body due to Equation of Continuity and - Fnet< 0 .eight of block > weight of displaced ﬂuid, block Bernoulli’s law? Note: turtle has a ﬂat bottom and domed top. UP sinks to bottom - Q. Lady ﬂoating effortless on surface of water, what is true is water is extremely salty, increasing density to above that of lady’s body - Archimedes principle: when object is immersed in a ﬂuid, - Q. how does mass ﬂow rate m/t of an incompressible, viscous ﬂuid ﬂowing through a cylindrical pipe depend on pipe’s diameter d? m/t ~ d 4 ﬂuid exerts an upward force on object equal to weight of ﬂuid - Q. A viscous ﬂuid is forced through a pipe to obtain a certain volume ﬂow rate (experiment 1). If the same ﬂuid is then forced through a pipe of same displaced by object cross–sectional area but triple the length (experiment 2), how has the pressure difference along the pipe changed from the value in experiment 1?.: Δp2 - Buoyant force = magnitude of weight of displaced ﬂuid: = 3 Δp1 F bouyant= density ﬂuid* Vobject g - Q. The viscosity coefficient is a material’s constant Always against gravity .: up ; Always non-zero, CAN USE - Q. The sixth ﬁgure shows a neat experiment you can do at home. Stick a pin through the middle of a small postcard. Hold it under a thread spool so FORMULA IN AIR that the pin projects up into the hole. If you blow down the hole and let go of the card, it won’t fall down. How many of the following laws do you need to explain this experiment? (I) Bernoulli’s law, (II) Pascal’s law, (III) Equation of continuity (IV) Poiseuille’s law, (V) Laplace’s law. 1 or 2 - Surface tension = ﬂuid property associated with presence of - Q. ideal dynamical ﬂuid is ﬂowing in a circular pipe which maintains a ﬁxed diameter and remains level while it is routed around a large external surface toward air Chapter 16: Elastic Tissue: Elasticity and Vibrations SPRINGS Chemical Bonds - Interfacial tension used to describe analogous phenomena for p = W / A (pressure = weight / area) ; SPHERE: - Hooke’s law applies in elastic regime of material & - HCl  Cl is a lot bigger .: acts like a ﬂuids having interfaces with solids or other liquids SA = 4πr2 ; V = (4/3)πr3 states that stress and strain of material = linear wall, H is like an object on a spring - spring makes obect move to eq’m in response to attached to it (model dies at high - surface/interface don’t have symmetry of interactions of molecules with immediate neighbours broken ELASTICITY displacement temps and high separation) + - - sufficiently far below surface, water molecule attractive interactions - stress = force exerted per unit area of surface on - HOOKES LAW for object attached to horizontal - Na Cl  when too close, repulsion; in all directions, .: eq’m, but within 1 nm of surface sphere of extended object, leads to strain = relative change in spring: (note F.elast = opposite F.external) -F elastic otherwise obviously attract size of object (tensile stress  stretching for length, Fext= k(x – x eqm) ; k = spring constant (N/m) - Eattract - (1 / 4 π ε0) * (e / r) ; neighbours not complete .: net force acts pulling molecule down away from surface shearing stress  twisting for angle, hydraulic stress - .: elastic -k(x – x eqm) = m · a Erepulsive b · e -a · (a and b are - force pulls molecules away from surface, molecules can’t leave  compression/expansion for volume) - an object on an ideal springs that undergoes a experimental constants, ε 0 surface as surface area can’t shrink .: brought to surface AGAINST small displacement from its eqm position = Hooke’s permittivity, e = elementary charge) net force down - stretching strain = length change caused by tensile law - .: energy associated with formation of surface .: surface stress = F/A (Pa) ; A = area of object on which force - W = area under F*displacement  W = ½ F extx ﬁnal 2 energy σ = ∆E/∆A ; Energy needed to increase surface of ﬂuid by acts W = ½ k * (x – x eq) = elastic potential energy - Q. Cable and shit  Steel sphere, 2 2 - elastic response = strain/stress are proportional diameter = 1.5 m, thickness 5 cm. area A .: = energy required to form 1 m of new surface (J/m ) - W = ∆E = σ *∆A ; W = work required to increase surface by A; (linear)  F/A = Y · Δl/l ; Y = Young’s modulus (Pa), - amplitude (A) = maximum displacement of Weight = 4900 N. STEEL Rope = 3 cm 3 W = F * d = F *∆l y ∆A = l x∆ y materials constant (~ temp) vibrating object .: kinetic 0 = E total E spring E total diameter, density = 7.5 g/cm . σ = 2 2 - σ = F / lx ; units = N/m (.: = surface tension) = J/m .: is - σ (stress) = F/A ; ε (strain) = Δl/l .: σ = Y · ½ k A 6.87E8 Pa. SA sphere * thickness = energy to increase surface or force to stretch ﬁlm ε  strong materials = HIGH Y (large force = little - At minimum displacement, all E = kinetic .: E total ½ volume .: 4 π r d * ρ g = weight - wire example for soap with high cohesiveness; water change e.g. metal); soft = LOW Y m v 2 ; v = MAXIMUM speed  .: v = sqrt (k / m) * A sphere 25980. Buoyant = 4/3 π r ρ 3 measured by measuring energy to lift solid off ﬂuid surface - elastic limit = threshold above which permanent g (volume obj*densityﬂuid*g) = deformation 17320. .: sum force down – Fb = - sphere has least surface : volume, .: raindrops release E when 17320 N needed to be supported by reshaping to sphere, or need E to change from sphere Plastic Deformations - position of object on spring: x(t) = A · cos( sqrt(k/ rope. Check max weight  F = σ A = 2 - when droplet on ﬂat surface, shape complex --> ﬂat frame = - equation doesn’t always hold  irreversible because m) · t ) σ r π = 4.86E5. .: that – 17 = ﬂat layer ﬁlm, drops on surface = partial sphere (alveoli) strain doesn’t return to 0 when stress removed (e.g.  x(t) = A · cos(ω · t) (simple harmonic motion); 4.72E5 of length. Translates to W = π playdough) ω = angular frequency (1/s, rad/s) r l (ρsteel ρﬂuidg (to integrate Fb into - bubbles have increased pressure inside them. Internal p > - stress increases beyond linearity  microscopic - ω = sqrt(k/m) (1/s, rad/s) - period (T) = full it)  10480 m = l structural alterations  signiﬁcant strain increase  cycle - Q. Muscle has no elastic regime @ external p - Laplace’s Law: pressure difference between inside and outside of permanent plastic deformation  material no longer  ω = 2 · π / T  f = 1/T .: ω = 2πf resting length ﬂuid with curved surface is inversely proportional to the radius of the withstand stress, pass through ultimate stress point - ***note (lectures): may have to include phase curvature of the curved surface. This means that smaller bubble,  beyond which material approaches rupture point angle in cos( ) if initial displacement*** quickly, without need for further increase in stress droplet, or cylinder has a larger pressure difference ∆p. *negative if curve is U  material = elastic when responds to external forces - bubbles: ∆p = p insidep outside 4*σ / r = (stress) with linear deformation (strain). Elastic transmural pressure deformations are reversible. A material is called - droplet / hollow cylinder: ∆p = p inside poutside 2*σ / r plastic when it responds to stress non-linearly - homogenous cylinde
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