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41 Cards in this Set
- Front
- Back
Fluid Dynamics
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aka hydrodynamics
the study of fluids in motion |
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Factors that complicate hydrodynamics
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friction of particles on vessel walls
viscosity flow turbulence |
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4 Basic Characteristics of Fluid Flow
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Steady vs non-steady
Compressible vs Incompressible Irrotational vs Rotational Viscous vs Nonviscous |
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Steady flow
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Iff every particle passes a point at the same velocity at every point in the flow
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non-steady flow
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particles moving at different velocities and / or in different directions
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Incompressible
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Iff ρ of fluid is constant throughout
particles can't squish together |
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Compressible
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Iff ρ changes at different points in the flow
Generally, smaller diameter pipe yields greater ρ |
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Irrotational
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Iff no particles exhibit rotational flow
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Rotational flow
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some particles exhibit rotational motion
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streamline
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defined path of a family of fluid particles in a fluid flow
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laminar flow
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particles assume specific smooth paths called streamlines
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turbulence
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condition of flow in which some particles exhibit rotational motion
sets up eddy currents |
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Nonviscous
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flow in which frictional (drag) forces between fluid particles & with pipeline walls are negligible
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Viscous
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flow in which particles interact with each other and / or vessel walls
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η
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(greek eta)
describes flow viscosity η = 0 = perfectly nonviscous η_water < η_molasses |
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Ideal Fluid Flow Model
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Steady, Incompressible, Irrotational, Nonviscous
applies to low η fluids over short distances |
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measures used in an ideal fluid flow model
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pressure, velocity
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Continuity Equation
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measures ∆V in an ideal fluid flow
A₁V₁ = A₂V₂ The volume of water in a given area is constant |
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Deriving Continuity
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Given a pipe in which the diameter changes, measure a cylinder of fluid in area 1, then watch how it moves over ∆t through areas of changing diameter
The cylinder has a particular density, mass, and volume which are constant in an ideal fluid flow m₁ = m₂ and V₁ = V₂ and ρ₁ = ρ₂, so ρ₁m₁∆V₁ = ρ₂m∆V₂, canceling density & mass yields ∆V₁ = ∆V₂ V can be expressed in terms of the cylinder so V₁ = πr₂∆x = A∆x this movement occurs over time, so A∆x₁ / ∆t = A∆x₂ / ∆t, factoring yields A(∆x₁ / ∆t) = A(∆x₂ / ∆t), and ∆x/∆t=V, so A₁V₁ = A₂V₂ |
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relationship between area & pressure
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inverse
as area decreases, pressure increases |
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Q
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Flow Rate
time rate of change of volume flow |
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Q equation
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Q = ∆V/∆t
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Derive Q
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using Continuity we know
∆V₁ / ∆t = ∆V₂ / ∆t so Q₁ = Q₂ and so Q = ∆V/∆t in m³/s or ft³/s or CFS (cubic feet per second) |
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Important Q equation variants
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Q = ∆V₁/∆t = A₁V₁ = A₂V₂ = Q₂
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m (with a dot over it)
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mass rate
describes how mass moves in a particular region over time |
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m equation
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m = ρQ in kg / s
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m derived
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m = ∆m/∆t = ρ∆v/∆t = ρQ
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Find the flow rate for blood moving through an artery with a mass rate of 1.5x10⁻⁵ kg/s with a density of 1057 kg/m³.
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Q=flow rate
m=1.5x10⁻⁵ kg/s ρ=1057 kg/m³ Given m=ρQ, then Q = m/ρ Q = 1.5x10⁻⁵ kg/s / 1057 kg/m³ = 1.42x10⁻⁸ m/w *note that kg's cancel |
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Bernoulli's
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Apply the Work-Energy theorem to understand how an ideal fluid moves through a pipe of changing diameters and heights
Allows measurement of ∆p |
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Bernoulli's Equation
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p₁ + 1/2ρV₁² + ρgy¹ = p₂ + 1/2ρV₂² + ρgy₂
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Bernoulli Equation variants
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p₁ = ρ₂ + 1/2ρ∆V + ρg∆y where ∆ is 2-1
p₂ = p₁ + 1/2ρ∆V + ρg∆y where ∆ is 1-2 |
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Continuity Equation
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A₁V₁=A₂V₂
shows area & velocity are inversely related |
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Flow rate
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Q
Q=∆v/∆t=AV=constant |
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mass rate
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m with a dot over it
mass rate = m/∆t = ρQ = ρAV |
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Bernoulli Equation
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p₁ + 1/2ρV₁² + ρgy₁ = p₂ + 1/2ρV₂² + ρgy₂
or p₁-p₂ = 1/2ρ(V₂²-V₁²) + ρg(y₂-y₁) |
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If an adult aorta has a radius of .9 cm that feeds arterioles with a combined cross sectional area of 20 cm², find the average velocity of blood flow through an arteriole if the flow rate in the aorta is 5 L/min.
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5 L/min * 1m³/1000L * 1min/60 s = 8.3x10⁻⁵ m³/s = Q of aorta
20 cm² = .002m² = cross sectional area of all arterioles Quantity of arterioles unknown, but if it is assumed all have the same cross sectional area, then the velocity in each is equal. V_arteriole = Q_aorta / A_arterioles V_arteriole = 8.3x10⁻⁵ m³/s / .002 m² = 0.041665 m/s |
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Consider the flow of blood where ρ=1057 kg/m³ through a vessel occluded by plaque to half its width. The diameter of the clean area is .2 cm and the speed of flow in that area is .4 m/s. Find the change in pressure from the clear area to the occluded area.
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Since r₁>r₂, then V₁<V₂, ∴ a≠0
r₁=.2 cm / 2 * 1m/100cm = .001 cm r₂=.2 cm / 2 / 2 * 1m/100cm = .0005m From here, use Bernoulli. ∆y exists because the plaque makes it like a pipe going downhill. V₂=A₁V₁/A₂ → π.1cm² * 4 m/s² / π.05cm² = 1.6 m/s pi's cancel, UoM for radius cancel, so no need to convert p₁-p₂ = 1/2ρ(V₂²-V₁²) + ρg(y₂-y₁) ∆p = 1/2ρ(4 m/s² - 1.6 m/s²) + ρg(.0005m - .001m) =1263.28 N/m² |
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UoM for pressure in Bernoulli
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Pressure is measured in N/m²
The 1/2ρV² term yields kg / m*s². The ρgy term yields kg / m*s² if g is measured in m/s² Given that 1 N = kg * m / s², a Newton per m² is kg (m/s²) / m² which reduces to kg / m*s², the units for the KE and UE terms in Bernoulli |
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A building has a flat roof of A=200m². A wind of 45 m/s blows. The air pressure in the building is 1 atm, with a density of 1.29 kg/m³. All windows and doors are shut. Find the net force exerted on the roof by the air in the building and the wind blowing over the roof.
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use Bernoulli, with UE terms = 0
p₁ + 1/2ρV₁² = p₂ + 1/2ρV₂² Since the windows & doors are shut, V₁=0, simplify p₁ = p₂ + 1/2ρV₂² or p₁ - p₂ = 1/2ρV₂² Recall that F_net = F₁-F₂ and that F=pA so F_net=p₁A-p₂A = A(p₁-p₂) Substituting from Bernoulli F_net = A(1/2ρV₂²) =200m² * 1/2 * 1.29kg/m³ * (45m/s)² =261,225 N =58725 lbs |
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Poiseuille's Law
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calculates the resistance offered to flow due to viscosity
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Poiseuille's Law equations
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Q = vA = (πR⁴∆p) / (8ηL)
where vA = velocity * area R=radius η=coefficient of viscosity L=length of vessel Also R = (8ηL) / (πR⁴) ∆p = (8ηLQ) / (πR⁴) |