Dimensional Analysis and Similitude April 1, 2008

Dimensionless Groups and Outline

Similitude • Review momentum and energy
balances
Larry Caretto • Dimensionless analysis basics
Mechanical Engineering 390 • Dimensionless groups in fluid
Fluid Mechanics mechanics
– Reynolds number
– Drag coefficients
April 1, 2008
– Others
• Experimental data and similitude
2

Review Momentum Review Momentum II

• Have balance equation for each • Note differences in computation of mass
component of momentum (k = x, y, z) flow and momentum components y
V = 10 m/s
– Use Vx = u, Vy =v, Vz = w, V = ui + vj + wk • Example at right
θ = 30o
– Speed = V = (u2 + v2 + w2)1/2 – m & = ρVA
x

∂ (mVk )cv N outlets N inlets – Find x and y momentum A = 0.01 m2

∂t
+ ∑ ρoVo AoVk ,o − ∑ ρiVi AiVk ,i = ∑ Fk flows for ρ = 1000 kg/m3
⎞ − 866 kg ⋅ m
o =1 i =1
m& u =
1000 kg 10 m
(
0.01 m 2 )⎛⎜ − 10sm cos 30 o
⎟= = −866 N
∂ (mVk )cv N outlets N inlets m3 s ⎝ ⎠ s2

∂t
+ ∑ m& oVk ,o − ∑ m& iVk ,i = ∑ Fk m& v =
1000 kg 10 m
( ⎛ 10 m
0.01 m 2 ⎜ ) ⎞ 500 kg ⋅ m
sin 30 o ⎟ = = 500 N
o =1 i =1 m3 s ⎝ s ⎠ s2
3 4

Review Energy Equation Review Energy and Bernoulli

2 2
mV ( E V ( • Steady energy equation with ρ = ρo = ρi
E= + mgz + mu ⇒ e = = + gz + u
2 m 2 is Bernoulli equation with added terms
N outlets N inlets
∂Ecv
∂t
+ ∑ ρoVo Aoeo − ∑ ρiVi Ai ei = Q&innet + W&innet Vo2 Po Vi2 Pi
W& shaft ⎛
1⎜( (
Q& net ⎞
⎟
o =1 i =1 + + zo = + + zo +
net in
− ⎜ u o − ui −
in
⎟
2g γ 2g γ g⎜
⎟⎟
m& g m&
∂Ecv
N outlets N inlets ⎜

∂t
+ ∑ m& oeo − ∑ m& i ei = Q&innet + W&innet Vo2 Po V2 P
⎝ ⎠
o =1 i =1 + + zo = i + i + zo + hs − hL
( 2g γ 2g γ
• Introduce shaft work and enthalpy, h = u( + P ρ
• Head loss, hL is always positive
∂Ecv
N outlets
⎛V 2 ( ⎞ Ninlets ⎛ Vi2 (⎞
∂t
+ ∑ m& o ⎜ o + gzo + ho ⎟ −
⎜ 2 ⎟ ∑
m& i ⎜
⎜ 2
+ gzi + hi ⎟ = Q& net + W& shaft
⎟ – See example from last class at end of
o =1 ⎝ ⎠ i =1 ⎝ ⎠ in net in
these notes
5 6

ME 390 – Fluid Mechanics 1

Dimensional Analysis and Similitude April 1, 2008

Dimensionless Analysis Another Approach

• Buckingham Pi Theorem • Write governing differential equations
– Can resolve a problem with N variables involving velocity components, u, v, w,
and D dimensions in to a set of N – D pressure, and coordinates, x, y, z
dimensionless variables
• Define dimensionless variables, u/V∞,
– Part of chapter seven not assigned v/V∞, w/V∞, x/Lref, y/Lref, z/Lref, and p/ρV∞2
discusses general approach for doing this
for general problems • Substitute dimensionless variables into
– Results for problems considered here are differential equations and obtain
known and we will not cover this general quantities like Reynolds number (Re =
approach VLrefρ/μ) by rearrangement
7 8

Example of Problem Example of Result

• Expect • Two
pressure drop, dimensionless
Δploss/L in variables:
smooth pipes – Reynolds
to depend on number =
variables D, ρ, ρVD/m
μ, and V – Loss
– 5 variables parameter
DΔP/(LρV2) =
– 3 dimensions
DΔPl /(ρV2)
Figure 7.1, Fundamentals of Fluid Mechanics, 5/E by Bruce
Munson, Donald Young, and Theodore Okiishi, Copyright9 © Figure 7.2, Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, 10
2005 by John Wiley & Sons, Inc. All rights reserved. and Theodore Okiishi, Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

More Generally Reynolds Number

• We would like to find the head loss in a • Re = ρVLc/m, where Lc is characteristic
pipe flow as a pressure head, Δp/γ length for the geometry
• We expect that this will include the – Lc = D for pipe flow and external flows
following variables: V, L, D, m, and pipe about cylinders and spheres
roughness, ε • Measures ratio of inertia (momentum)
• In next chapter we will use the following forces to viscous forces
relationship among dimensionless momentum (ρVA)V ρV 2 ρVLc
= ≈ =
variables (ΔP ) D ⎛ ρVD ε ⎞ viscous dV V μ
loss
= f ⎜⎜ , ⎟⎟ μA μ
ρV 2 L ⎝ μ D⎠ dy Lc
2 11 12

ME 390 – Fluid Mechanics 2

Dimensional Analysis and Similitude April 1, 2008

Loss Coefficients Example: Sphere Drag

• Typically written as ΔP/(ρV2) or
D
ΔP/(ρV2/2)
Figure 7.7, Fundamentals of Fluid
CD = Mechanics, 5/E by Bruce Munson,
ρV 2 Donald Young, and Theodore Okiishi,
A Copyright © 2005 by John Wiley &

• Measure of head loss due to viscous 2 Sons, Inc. All rights reserved.

forces or form drag D = Drag force A = Projected

πd 2
• Drag coefficient CD in external flows Area =
4
– D is drag force in N or lbf
– A is frontal area (projected area on plane
perpendicular to the flow)
– CD = (D/A) / (ρV2/2)
13 14

Problem More Complicated Problem

• Estimate the drag force on a 1-cm- • Use the data on the drag ceofficient
diameeter sphere in a 10-m/s air flow chart to find the terminal velocity of a
– Use air properties of air at atmospheric 0.08 kg sphere with d = 0.01 m in air
pressure and 300 K: ρ = 1,164 kg/m3 and μ – Terminal velocity occurs when drag force
= 184.6x10-7 N·s/m2. equals force of gravity: D = CDAρV2/2 = mg
1.164 kg 10 m
(1cm) .01 m – Use air properties for ρ and μ in trial and
ρVD m 3 s cm error solution
Re = = = 6,309 CD = 0.5
μ 184.5 x10 −7 N ⋅ s 1 kg ⋅ m • Guess V
m2 N ⋅ s2 • Compute Re = ρVd/μ, find CD from chart, and
ρV 2 πd 2 ρV 2 π(0.01 m )2 1 1.165 kg ⎛ 10 m ⎞ 1N ⋅ s 2
2 compute D = (πd2/4)(ρV2/2)
D = CD A = CD = 0.5 ⎜ ⎟ = 0.0023 N • Continue until D is close enough to mg
2 4 2 4 2 m3 ⎝ s ⎠ kg ⋅ m
15 16

Other Dimensionless Variables More Dimensionless Variables

• Froude number, Fr = V/(gLc)1/2 is ratio of • Mach number, Ma = V/c (c is sound
inertia force to gravitational force used speed) is square root of Cauchy number
in flows with a free surface when Es is used [c = (Es/ρ)1/2]
• Euler number, Eu = p/ρV2 is ratio of • Stroudahl number = ωLc/V, is ratio of
pressure force to inertia force, used in two inertia forces: local oscillation with
calculating pressure differences radian frequency ω and main flow
• Cauchy number, Ca = ρV2/Ev is ratio of • Weber number, We = ρV2Lc/σ, is ratio of
inertia force to compressibility force inertia force to surface tension force
used where compressibility important
1 ⎛ ∂p ⎞ 1 ⎛ ∂p ⎞
ET = ⎜ ⎟ Es = ⎜ ⎟
ρ ⎜⎝ ∂ρ ⎟⎠T ρ ⎜⎝ ∂ρ ⎟⎠ s
17 18

ME 390 – Fluid Mechanics 3

Dimensional Analysis and Similitude April 1, 2008

Experimental Data Measurements and Calculations

14

• Use dimensionless parameters to 12

correlate experimental data Measured
10
• Example MS student obtained data on 8
Calculated

pressure drop for 4 tube diameters and Friction

factor 6
Calculation line
extension
five different fluids (different ρ and μ)
4
• Initial data analysis showed lack of
agreement between theory and data with
2

no pattern in disagreement 0
0 20 40 60 80 100 120 140

• Replot as f = (DΔp/L)/(ρV2) vs Re = ρVd/μ Reynolds Number

19 20

Models Similitude
• Experiments on scale models in wind • Assume that we want to measure drag
tunnels or towing tanks used to predict force in an incompressible flow by
fluid flows and forces testing a model
• How are models and test procedures • Drag forces depend on variables like
designed? velocity, density, viscosity, and size
• We saw results for CD vs. Re for a
• Basic idea: keep all important
sphere; expect similar results for model
dimensionless parameters the same testing
between model and prototype (pre-
• Want Re for model to be same as Re
production physical system to be built) for prototype to get same CD
21 22

Typical Similarity Requirements Model Scales

• For flow in full ducts and flows over • Given idea is that model size will be
external objects Reynolds number smaller than prototype
similarity is required • Have length scale that applies to all
– Forces found from by assuming model dimensions
drag coefficient is same as prototype – E.g. in a 1/50 scale model, all dimensions
• Open channel flows require Froude would 1/50th of the dimensions in the
number similarity prototype
– Driving force is slope of channel captured – Look at effect that this has on parameters
by model scale such are Re = ρVLc/μ
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ME 390 – Fluid Mechanics 4

Dimensional Analysis and Similitude April 1, 2008

Model Scales II Model Scales III

• If we want Lc,model = Lc,prototype/50 with • What are implications of this equation?
Remodel = Reprototype, we must have Vm Lc , p
=
ρ mVm Lc ,m ρ pV p Lc , p ρ mVm μ p Lc , p V p Lc ,m
= ⇒ =
μm μp ρ pV p μ m Lc ,m • If we have a small scale model, the test
• If model test uses the same fluid as the velocity for the model must be much
prototype will use, we must have greater than that for the model
Vm Lc , p • For flows in air the higher model velocity
= may have compressibility effects that
V p Lc ,m
are not present in the prototype
25 26

Distorted Models Free Surface Models

• Because of experimental difficulties in • Froude number similarity important
matching all dimensionless groups Vm Vp Vm Lc ,m
Frm = = = Frp ⇒ =
model testing is often done without only g m Lc ,m g p Lc , p Vp Lc , p
partial matching • For viscous effects want similar Reynolds
• This necessary practice is an art and number
there is an extensive literature on such ρ mVm Lc ,m ρ pV p Lc , p ρ mVm μ p Lc , p
= ⇒ =
tests μm μp ρ pV p μ m Lc ,m
• Open channel flows and building flows Lc , p ρ mVm μ p ρ m μ p Lc ,m ρm μ p Lc , p
= = ⇒ =3
use distorted models Lc ,m ρ pV p μ m ρ p μ m Lc , p ρ p μm Lc ,m
27 28

Free Surface Models II Free Surface Models III

• Last relation requires specific fluid • For most free-surface models, the
property ratio that cannot easily be met Froude number is the important similarity
• For surface tension effects want similar parameter
Weber number • Matching Reynolds number is less
– Same Weber and Froude numbers require
important
– Exception is ship models
ρ mVm2 Lc ,m ρ pV p2 Lc , p ρ mVm2σ p Lc , p
= ⇒ = • Surface tension effects are usually not
σm σp ρ pV p2σ m Lc ,m significant
2
Lc , p ρ mσ p Vm2 ρ mσ p Lc ,m ρ mσ p ⎛ Lc , p ⎞ • Open channel flow models often use
= = ⇒ = ⎜ ⎟
Lc ,m ρ pσ m V p2 ρ pσ m Lc , p ρ pσ m ⎜⎝ Lc ,m ⎟⎠ distorted scales with vertical scale
smaller than the horizontal scale
29 30

ME 390 – Fluid Mechanics 5

Dimensional Analysis and Similitude April 1, 2008

Problem Problem Solution

• It is desired to test a model of a car with • For same air properties in model and
a maximum dimension of 20 ft in a wind prototype Vm Lc , p 20 ft
= = =5
tunnel that can accommodate a Vp Lc ,m 4 ft
maximum length of 4 ft. What air • Testing prototype speed of 20 mph
velocity is required to test a prototype requires model speed of 100 mph
speed of (a) 20 mph, (b) 90 mph
• 90 mph for prototype requires 450 mph
• Reynolds number similarity is required for model which is not valid because of
ρ mVm Lc ,m ρ pV p Lc , p ρ mVm μ p Lc , p compressibility effects
= ⇒ =
μm μp ρ pV p μ m Lc ,m
31 32

Energy Example and Solution Example Solution II

• From last class • Energy equation
Vo2 Po
• Given: flow as + + zo =
2g γ
shown in diagram
Vi 2 Pi
• Find: flow direction + + zi + hs − hL
2g γ
and head loss
• Solve energy equation for head loss • Subscript “o” is outlet and “i” is inlet
• Head loss must be positive • Assume flow is downhill (left to right)
– Assume a flow direction and find head loss – No shaft work so hs = 0
• If hL > 0 assumption is correct and we know hL – For incompressible fluid ViAi = VoAo and with
• If hL < 0 assumption is wrong, change hL sign Ai = Ao, Vi = Vo, so V2 terms cancel
33 34

Example Solution II Example Solution III

• Solve for head loss • Assumed downhill
hL =
pi
+ zi flow direction gives
γ negative head loss
⎛p ⎞ – Original
− ⎜⎜ o + zo ⎟⎟
⎝ γ ⎠ assumption wrong!
• Piezometer tubes measure z + p/γ • Conclude that flow is uphill (from right to
po pi
+ zo = 3 m + z i = 2.5 m left) and correct hL is 0.5 m (same
γ γ magnitude as computed previously, but
pi ⎛ po ⎞ opposite sign).
hL = + z i − ⎜⎜ + z o ⎟⎟ = 2.5 m − 3 m = −0.5 m
γ ⎝ γ ⎠
35 36

ME 390 – Fluid Mechanics 6