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41 Cards in this Set

  • Front
  • Back
Fluid Dynamics
aka hydrodynamics

the study of fluids in motion
Factors that complicate hydrodynamics
friction of particles on vessel walls
viscosity
flow turbulence
4 Basic Characteristics of Fluid Flow
Steady vs non-steady

Compressible vs Incompressible

Irrotational vs Rotational

Viscous vs Nonviscous
Steady flow
Iff every particle passes a point at the same velocity at every point in the flow
non-steady flow
particles moving at different velocities and / or in different directions
Incompressible
Iff ρ of fluid is constant throughout

particles can't squish together
Compressible
Iff ρ changes at different points in the flow

Generally, smaller diameter pipe yields greater ρ
Irrotational
Iff no particles exhibit rotational flow
Rotational flow
some particles exhibit rotational motion
streamline
defined path of a family of fluid particles in a fluid flow
laminar flow
particles assume specific smooth paths called streamlines
turbulence
condition of flow in which some particles exhibit rotational motion

sets up eddy currents
Nonviscous
flow in which frictional (drag) forces between fluid particles & with pipeline walls are negligible
Viscous
flow in which particles interact with each other and / or vessel walls
η
(greek eta)

describes flow viscosity

η = 0 = perfectly nonviscous

η_water < η_molasses
Ideal Fluid Flow Model
Steady, Incompressible, Irrotational, Nonviscous

applies to low η fluids over short distances
measures used in an ideal fluid flow model
pressure, velocity
Continuity Equation
measures ∆V in an ideal fluid flow

A₁V₁ = A₂V₂

The volume of water in a given area is constant
Deriving Continuity
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
relationship between area & pressure
inverse

as area decreases, pressure increases
Q
Flow Rate

time rate of change of volume flow
Q equation
Q = ∆V/∆t
Derive Q
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)
Important Q equation variants
Q = ∆V₁/∆t = A₁V₁ = A₂V₂ = Q₂
m (with a dot over it)
mass rate

describes how mass moves in a particular region over time
m equation
m = ρQ in kg / s
m derived
m = ∆m/∆t = ρ∆v/∆t = ρQ
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.
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
Bernoulli's
Apply the Work-Energy theorem to understand how an ideal fluid moves through a pipe of changing diameters and heights

Allows measurement of ∆p
Bernoulli's Equation
p₁ + 1/2ρV₁² + ρgy¹ = p₂ + 1/2ρV₂² + ρgy₂
Bernoulli Equation variants
p₁ = ρ₂ + 1/2ρ∆V + ρg∆y where ∆ is 2-1

p₂ = p₁ + 1/2ρ∆V + ρg∆y where ∆ is 1-2
Continuity Equation
A₁V₁=A₂V₂

shows area & velocity are inversely related
Flow rate
Q

Q=∆v/∆t=AV=constant
mass rate
m with a dot over it

mass rate = m/∆t = ρQ = ρAV
Bernoulli Equation
p₁ + 1/2ρV₁² + ρgy₁ = p₂ + 1/2ρV₂² + ρgy₂

or

p₁-p₂ = 1/2ρ(V₂²-V₁²) + ρg(y₂-y₁)
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.
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
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.
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
UoM for pressure in Bernoulli
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
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.
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
Poiseuille's Law
calculates the resistance offered to flow due to viscosity
Poiseuille's Law equations
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)