The Role of Modelling in Engineering Design...The Design Challenge Doubling the size has the...

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The Role of Modelling in Engineering Design John Fitzpatrick Professor of Mechanical Engineering Trinity College Dublin

Transcript of The Role of Modelling in Engineering Design...The Design Challenge Doubling the size has the...

Page 1: The Role of Modelling in Engineering Design...The Design Challenge Doubling the size has the following consequences : a. The mass increases by a factor of 8. b. The lift increases

The Role of Modelling in Engineering Design

John FitzpatrickProfessor of Mechanical Engineering

Trinity College Dublin

Page 2: The Role of Modelling in Engineering Design...The Design Challenge Doubling the size has the following consequences : a. The mass increases by a factor of 8. b. The lift increases

Equation of Fluid Motion(Navier-Stokes Equations)

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

21zv

yv

xv

yp

zvw

yvv

xvu

tv ν

ρ

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

21zu

yu

xu

xp

zuw

yuv

xuu

tu ν

ρ

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

21zw

yw

xw

zp

zww

ywv

xwu

tw ν

ρ

0=∂∂

+∂∂

+∂∂

zw

yv

xu

Page 3: The Role of Modelling in Engineering Design...The Design Challenge Doubling the size has the following consequences : a. The mass increases by a factor of 8. b. The lift increases

Navier Stokes EquationsNo general solution existsComputational methods – (CFD)Experiments requiredHow do we conduct the experiments?

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CFD Study of a Jet

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ModellingMost experiments conducted for design purposes involve modelling.The model is often either smaller (or bigger) than the real construction. How do we conduct model tests and relate the results to the “real thing” (i.e. the prototype)

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The A380 ParadoxAerospatiale proposal 1994 –“How do we build an aircraft twice the size of A340 & 747?”

(Boeing 747 & Airbus 340 aircraft were considered to be at the limit

of aircraft size.)

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The Design ChallengeDoubling the size has the following consequences :

a. The mass increases by a factor of 8.b. The lift increases by a factor of 4.The Challenge

a. Increase the lift effectivenessb. Reduce the weightc. Do bothThe Answer

The first A380 was delivered last year to Singapore Airlines (18 months late!)

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Fundamentals

force = mass x acceleration (ma) work = force x distance (Fs)energy== workpower = rate of work (FU)

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Basic Dimensions

Space (m)Time (s)Mass (kg)Temperature (K)

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Derived Units

Force (1N=1kgm/s2) Work (1J=1Nm=1kgm2/s2) Energy (J)Power (1W=1J/s)

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Functional RelationshipsThese are expressions in which a dependent variable is writen as function of the other independent variables which affect the dependent variable.

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A functional relationshipThe force (F) required to move a body through a fluid can be expressed as :

F=f(d,L,U,ρ,µ)Where d = diameter

L = lengthU = velocityρ = densityµ = viscosity (friction)

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Important ParametersGeometric

Length, Area, VolumeKinematic/Dynamic Variables

Velocity, Time, TemperatureForces, Pressure, Stress, Strain

PropertiesDensity (ρ)Viscosity (µ)Elastic modulusSpecific heat, conductivity etc.

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How to design a Model Test

Geometric ScalingForce Scaling

We are going to consider applications in Fluid Flows

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A QuestionA series of tests are conducted on a model of a windmill which is 1/5th the size of the proposed system (prototype) to determine the power output. What do you think the power output for the full size (prototype) windmill will be?

5 times, 10 times, 100 times!(We will return to this at the end)

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Geometric ScalingRelatively easy to achieve as all geometric attributes can readily be scaled and a physical model constructed.

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Force ScalingThis is more conceptually more difficult, but is relatively straightforward.Unfortunately, there are a number of applications where we cannot scale all the important forces in the same model test.

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Forces in Fluid SystemsInertial Forces :these are due to the “mass” (density) of the moving fluid. Friction Forces : these are due to the viscosity of the fluid which resists motion.Gravity Forces : these are due to gravitational action and arise in water type flows (e.g. waves).Elastic Forces : these are due to compressibility effects and are important at very high speeds.

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Aerospace Systems

Key parametersGeometric

length & width – L & dKinematic/ Flow Variable

speed – UFluid Properties

density – ρ viscosity – µ

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A Functional Relationship

F=f(d,L,U,ρ,µ)F is dependent on 5 variables.To determine the nature of each dependency, it would be necessary to conduct tests in which all the parameters but one are kept constant so that the dependence of F on this variable is established. This would then be repeated for the remaining four.

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Dimensonal AnalysisIt is possible to regroup the 5 dimensional variables into 3 “dimensionless” groups as follows.e.g.

, ,22dU

Fρ d

ρUd

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Dimensionless GroupsThese groups represent the following :

Required Force/Inertia ForceLength/Diameter

Inertia Force/Viscous(Friction) ForceThe Inertia Force (~ρU2) is the most important in fluid flows and groups are usually formed as the ratio of this to other forces acting.

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Dimensionless RelationshipWe can now write a dimensionless functional form as :

So that we can now investigate the forces required to move bodies with different slenderness ratios (L/d) through a fluid

),(22 µρφ

ρUd

dL

dUF

=

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ModellingUsing dimensionless relationships enables us to perform model tests. The key is that in addition to having a geometric model, the force ratios must be scaled properly.

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ExampleA series of tests are to be conducted on a model car of 1/5 scale to investigate the drag force resulting from posible design modifications. If the prototype speed is 160 km/hr, at what speed should the test be conducted? It the results show that the drag force is 1.75kN, what is the drag on the protype?

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A geometric scale model is constructed to 1/5 scale.

i.e. (L/d)prot=(L/d)model

Assume the density and viscosity of the air are the same for the test and full scale conditions.

Uprot=45m/s

then(ρUd/µ)prot= (ρUd/µ)model

so thatUmodel=225m/s

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The force on the prototype (full scale) is then determined from

(F/ρd2U2)prot= (F/ρd2U2)model

so thatFprot=1.75kN

and the power can be obtained asPower = Force x Velocity

so that Power = 78.75kW

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Some Limitations Some dimensionless groups will not scale simultaneously. For boats, viscous, gravity and inertia forces are important.

inertia/viscous = (ρUd/µ)

inertia/gravity = (ρU2d2/ρgd3) = (U2/gd)

(Consider a 1/10th scale model!)

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ConsequencesSeparate tests are conducted to determine a. the power required to overcome the

gravity (wave) forcesb. the power required to overcome the

viscous forces

The results are then combined to determine the overall power required for the vessel.

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Further Problems!Scaling aircraft at cruise conditions with U=250m/s (900km/hr, 560mph).Using a wind tunnel atmospheric conditions gives 1/10th scale model testing velocities at ~2000m/s. This is equivalent to 6 times the speed of sound (M=6) and means the “elastic” or compressible forces become important.

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Wright Brothers

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Wright Flyer

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German/Dutch Low Speed Wind TunnelCapacity for Testing both full size & scale

modelsTest Section Dimensions

9.5m x 9.5mFlow Velocity

80 m/s

Aircraft models are tested for take off and landing conditions.

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The Propeller/WindmillParameters

Power (dependent) Diameter (d) & Chord width (c)Rotational Speed (N)Wind Speed (U)Density (ρ)

i.e. Power, W = f(d,c,N,U,ρ)

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Scaling for Props/Windmills

The groups for scaling are:c/d

U/Nd&

W/ρN3d5

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Scaling for WindmillsThe last grouping shows that the power scales with the 5th power of diameter!

Thus, a full size windmill which rotates at 1/5th of the model can generate 3125 times that of a

1/5th scale model!

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Thank you for your attention

Any Questions?