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by: Shanee Dinay

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3

# Week 3 Notes PHYS 5B

Shanee Dinay
UCSC
GPA 3.94

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Week 3 notes of Intro to Physics 2. Topics include Fluid Flow, Momentum, and Laminar Flow
COURSE
Intro to Physics II
PROF.
A.Steinacker
TYPE
Class Notes
PAGES
3
WORDS
CONCEPTS
physics 2, Fluids, momentum, Laminar Flow
KARMA
25 ?

## Popular in Physics 2

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(r​o​ 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.: v​A​ = v​A​ → v​  = v​  1​1​ρv12​2 ρv2 1​ 2 B.E.: P1​+  2  = P2​+  2  → P​1​= P2  really P2 ​P​1​hy?  W​p1​+ W​p2​ W​visc​ >0        <0     <0  P1​dx ­ P​2​dx = |W​visc​ 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​ y​y=d) = v​0v​ x​y=0) = 0  → + CE + ME  We get a differentiation equation:  d x dy2= 0  o If we work through this we will find that v​ x​y) =  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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