Week 3 Notes
Week 3 Notes PHYS 5B
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This 3 page Class Notes was uploaded by Shanee Dinay on Monday February 8, 2016. The Class Notes belongs to PHYS 5B at University of California - Santa Cruz taught by A.Steinacker in Fall 2015. Since its upload, it has received 12 views. For similar materials see Intro to Physics II in Physics 2 at University of California - Santa Cruz.
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Date Created: 02/08/16
Day 7 1/20/2016 Phys 5b Homework Submission Date: Homework Due Monday 1/25 Fluid Flow Lagrange Formalism r = F(ro t) v = dt a = ddt Each point within the volume of fluid, we can associate a density, density can be changing in time ρ(t) measure the temperature in the volume, make a temperature map of class/air Velocity field: v = f(r, t) this is the Euler Formalism density distributed in this volume, we integrate to calculate the mass If we want to assess what is the mass in this volume m = ρ∫V V we take an area element on the surface dA, area vector → points outward perpendicular to the surface dm dt = ∫ρv • dA A if we have the mass m in the volume, how can the mass change Looking at a situation where we have: no sinks, no sources of fluid dm d dt= dt∫ρdv = −∫ρv • dA V A if we combine the two integrals together Integral form of continuity equation: d ∫ρdv + ∫ ρv • dA = 0 ← continuity equation that accounts for when there is not dt V A sources available we have an area integral and a volume integral then we get a differential equation but we do not need to know this because this level of math is not required to know Momentum Equation take controlled volume integral with closed circle: over closed surface Fp= − ∮ PdA force and area vector have opposite direction A dP Fnet dt momentum, pushed around due to all the force Momentum Equation in Differential form dv = (v •▽)v − ▽P + g + ... dt ρ Pipe A1= A2 C.E.: vA = vA → v = v 11ρv122 ρv2 1 2 B.E.: P1+ 2 = P2+ 2 → P1= P2 really P2 P1hy? Wp1+ Wp2 Wvisc >0 <0 <0 P1dx P2dx = |Wvisc It is necessary to consider viscosity in any fluid Viscosity when is it more viscous than another fluid. Test the viscosity: take two plates and sandwich the two plates and the bottom plate is fixed while we pull the top plate. Bottom v = 0. Force pulls the plate, v = velocity plate. Fluid some P everywhere. Distance between two slabs 0 is d, what does the flow look like between the two plates. Question: what is my velocity? v = (v(y), 0 ,0) x Boundary conditions: v yy=d) = v0v xy=0) = 0 → + CE + ME We get a differentiation equation: d x dy2= 0 o If we work through this we will find that v xy) = dy Strength of the force depends on the area A F ~ v 0d A F = ηv 0d the greek letter is a constant eta coefficient of dynamic viscosity What are the units of eta? [η] = m/s •m m2s = Pa•s Laminar Flow through a cylindrical pipe of radius R r, φ, z CE + ME dP = m ρ(r ) dt r ρr ρr P2−P1 2 2 v(r) = 4Lη (r R) ← velocity profile, we know it is parabolic R2π Volume Flow Rate = ∫∫ vdA 0 0 we know how v varies with r in v(r) π(P1−P2) 4 Q = 8Lη R HagenPoiseuille
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