CompanyLOGO Modeling and

Similitude

Department of Mechanical Engineering

Muhammadiyah Universityof Pontianak

Gunarto, S.T.,M.Eng.

of Pontianak

Topics Reviewed

Dimensional Analysis Similitude Common Dimensionless Groups

1.DIMENSION ANALYSIS In this section, a method called

dimensional analysis along with the Buckingham Pi theorem will be introduced to identify the important dimensionless parameters governing a particular problem.

Since analytical solutions are not Since analytical solutions are not available for the majority of real fluids problems, experimental work plays a vital role in the study of fluid mechanics.

Before setting up an experiment, it is important to identify all dimensionless parameters that govern a system or problem, and ultimately obtain a correlation to relate those parameters.

Dimensionless Parameters

for Pipe Flow

Cont.... Dimensional analysis is one such method that can be

used to minimize the time and expenses spent on experiments and yet obtain valuable information from the fewest number of experiments possible.

To better understand the need for dimensional analysis, consider flow through a simple pipe. As one might imagine, pressure is critical to pipe flowimagine, pressure is critical to pipe flow

But how is the change in pressure related to the pipe and liquid? What pipe parameters effects pressure changes in a pipe? Some obvious parameters are velocity, density, pipe diameter, and viscosity.

To isolate and test for each of these parameters independently with an experiment may be impossible. Besides, the number of experiments would be vast.

Cont...To minimize the experimental effort, the

dependent variables are group together to form dimensionless parameters that can be used for other similar pipes.

While simple, this is a important tool in developing models for fluid dynamics. developing models for fluid dynamics.

An example of five factors that influence pipe flow is shown in the diagram on the left.

Two possible dimensionless parameters are plotted, using experimental data points.

Basic Dimensions of Common Parameters

QuantityBasic Dimensions

FLTSystem(US)

MLTSystem

(SI)Acceleration LT -2 LT -2

Angular Velocity T -1 T -1

Area L2 L2

Mass Density FL-4 T 2 ML-3

Weight Density FL-3 ML-2T -2

Force (weight) F MLT -2Force (weight) F MLT -2Kinematic Viscosity L2T -1 L2T -1

Length L L Mass FL-1T 2 MPower FLT -1 ML2T -3

Pressure FL-2 ML-1T -2

Surface Tension FL-1 MT -2

Velocity LT -1 LT -1

Viscosity FL-2T ML-1T -1

Volume L3 L3

Volume Flowrate L3T -1 L3T -1

Work, Energy FL ML2T -2

Dimensional Analysis Procedure Steps(using Buckingham Pi Theorem)

Step 1:The first step of dimensional analysis is to identify all independent parameters for the system or study. These parameters generally include fluid properties (e.g., density, viscosity and surface tension), system geometry (e.g., length, area and volume) or flow conditions (e.g., velocity, pressure change and applied force).velocity, pressure change and applied force).

Step 2: The second step is to determine the number of basic dimensions involved. Two sets of basic dimensions can be used:

(i) Force (F), Length (L) and Time (T)(ii) Mass (M), Length (L) and Time (T)

Cont...For example, the basic dimensions for

density can be expressed in terms of FL-4T 2 or ML-3.

Some of the basic dimensions for common parameters encountered in fluid mechanics parameters encountered in fluid mechanics problems are summarized in the table.

Temperature (T) and charge (q) dimensions can be added when considering heat transfer and electrical problems, respectively.

Cont... Step 3: The next step is to determine the number of

dimensionless parameters (pi terms, denoted by ) through the Buckingham pi theorem.

According to the Buckingham pi theorem, the number of pi terms is equal to (n-k) where n is the number of pi terms is equal to (n-k) where n is the number of independent parameters involved (determined in step 1) and k is the number of basic dimensions involved (determined in step 2).

Hence, for a given system, one can write 1 = function (2, 3, ..., n-k)

Cont...Step 4: From the list of parameters determined in step 1,

select k number of repeating parameters. These repeating parameters must include all the

basic dimensions, but they cannot be dimensionless or have the same basic dimensions (i.e., do not include both L (length) and L3 (volume) as repeating parameters).

In other words, the repeating parameters cannot form dimensionless parameters by themselves.

Cont... The pi terms are then formed by

multiplying the remaining parameters with the repeating parameters raised to a certain power.

The exponents of the repeating parameters are determined such

Quantity Symbol MLTPressure Drop p ML-1T-2

Pipe Length l L

Pipe Diameter D Lparameters are determined such that the pi terms are dimensionless.

Once the pi terms are determined, this concludes the dimensional analysis.

Experiments can then be conducted to find a correlation among the pi terms.

Fluid Velocity V LT -1

Fluid Density ML-3

Fluid Viscosity ML-1T-1

Pipe Surface Roughness

L

An Example: Flow in a Circular Pipe

Cont...For a better illustration of the use of dimensional analysis, take fluid flow in a circular pipe.Step 1:

The parameters that are involved in the The parameters that are involved in the pipe flow problem are the pressure drop (p), pipe length (l), pipe diameter (D), fluid velocity (V), fluid density (), fluid viscosity () and pipe surface roughness (). A total of 7 parameters (n = 7) is involved in this problem.

Cont..Step 2:

The basic dimensions involved are summarized in the table on the left. All three of the basic dimensions, M, L, and T (k = 3) are involved in this problem.(k = 3) are involved in this problem.

Step 3:According to the Buckingham pi theorem, the number of pi terms is 4 (n - k = 7 - 3 = 4).

Cont...Step 4:The next task is to determine

the form of the pi terms. Select the pipe diameter (D),

fluid velocity (V) and density () as the repeating parameters.

The pi terms are then given by:1 = pDa1Vb1c12 = lDa2Vb2c23 = Da3Vb3c34 = Da4Vb4c4

Fluid Flow in a Circular Pipe

Cont.. The exponents of the first pi terms are determined as

follows: 1 = pDa1Vb1c1 = (ML-1T-2)(L)a1(LT-1)b1(ML-3)c1

= M(1 + c1) L(-1 + a1 + b1 - 3c1) T(-2 - b1) In order for 1 to be dimensionless:

M: 1 + c1 = 0M: 1 + c1 = 0c1 = -1

T: -2 - b1 = 0b1 = -2

L: -1 + a1 + b1 -3c1 = 0a1 = 3(-1) - (-2) + 1 = 0

Hence, 1 is determined to be p/V2.

Cont.. Since the basic dimension for the pipe length l is L, by

inspection, the second pi term is given by (a2 = -1, b2 = 0 and c2 = 0):2 = l/D

Similarly, the last pi term is given by (a4 = -1, b4 = 0 and c4 = 0):c4 = 0):4 = /D

The exponents of the third pi terms are determined as follows:3 = Da3Vb3c3= (ML-1T-1)(L)a3(LT-1)b3(ML-3)c3

= M(1 + c3) L(-1 + a3 + b3 - 3c3) T(-1 - b3)

Cont...In order for 3 to be dimensionless:

M: 1 + c3 = 0c3 = -1

T: -1 - b3 = 0b3 = -1b3 = -1

L: -1 + a3 + b3 -3c3 = 0a3 = 3(-1) - (-1) + 1 = -1

Cont...Hence, 3 is determined to be /DV.

Recognizing that the inverse of the pi term is also dimensionless, the third pi term can also be written as DV/, which is the Reynolds number (Re).Reynolds number (Re).

For flow in a circular pipe, the pressure drop is then given byp/V2 = function(l/D, /D, Re)

Cont..As is discussed in the Viscous FLow in

Pipe section, flow in a circular pipe can be solved with the aid of the Moody chart (obtained from experiments) where the pressure drop term is expressed in terms pressure drop term is expressed in terms of the friction factor (f).

As shown in the Moody chart, the friction factor is a function of the entrance length (l/D), relative roughness of the pipe (/D) and the Reynolds number (Re).

CASE STUDY 1 To illustrate that dimensional

analysis introduced in fluid mechanics can also be used in other engineering problems (e.g., statics, dynamics and solid mechanics), students are challenged to find the important challenged to find the important dimensionless parameters governing the deflection of a circular cantilever beam due to an applied force at its tip, as shown in the figure.

Deflection of a Cantilever Beam

Cont...Questions Determine the dimensionless parameters involved using

dimensional analysis. Approach First, list all the parameters of the problem. Hint: One of

the parameters is the modulus of elasticity (E) , which is the parameters is the modulus of elasticity (E) , which is a material property of the beam.

Determine the number of basic dimensions involved. Find the number of pi terms using Buckingham Pi

theorem. Determine the pi terms.

CASE STUDY SOLUTION 1 Step 1:

In the study of the deflection of a cantilever beam, the parameters involved are the applied force (P), deflection (), modulus of elasticity (E), beam radius (r) and beam length (l). A total of 5 parameters (n = 5) is involved total of 5 parameters (n = 5) is involved in this problem.

Step 2:The basic dimensions involved are summarized in the following table

Deflection of a Cantilever Beam

Cont..

Quantity Symbol MLT

Applied Force P F

Deflection L

Hence, 2 basic

dimensions (k = 2) are

involved in this problem.

Step 3:Modulus of Elasticity E FL-2

Beam Radius r L

Beam Length I L

Step 3:According to the Buckingham pi theorem, the number of pi terms is 3 (n - k = 5 - 2 = 3).

Cont... Step 4: The next task is to determine the form of the pi

terms. Select the modulus of elasticity (E) and beam radius (r) as the repeating parameters.

The pi terms are then given by:1 = PEa1rb12 = Ea2rb22 = E r3 = lEa3rb3

The exponents of the first pi terms are determined as follows: 1 = PEa1rb1 = (F)(FL-2)a1(L)b1

= F(1 + a1) L(-2a1 + b1)

Cont... In order for 1 to be dimensionless:

F: 1 + a1 = 0a1 = -1

L: -2a1 + b1 = 0b1 = 2(-1) = -2

Hence, 1 is determined to be P/Er2. Hence, 1 is determined to be P/Er . By inspection, the second and third pi terms are given by

(a2 = 0, b2 = -1, a3 = 0 and b3 = -1):2 = /r3 = l/r

Hence according to the dimensional analysis, the important parameters are/r = function(P/Er2, l/r)

Cont... Note: According to the beam bending theory, the

deflection of a circular beam is given by

Recast the above equation in terms of the Recast the above equation in terms of the dimensionless parameters to yield

which is in agreement with the results obtained from dimensional analysis.

2.SIMILITUDEIn the Dimensional Analysis section,

methods to identify dimensionless parameters governing a particular model were presented.

The discussion in this section is focused on The discussion in this section is focused on how to relate a model to an actual prototype (i.e., the real full scale structure).

In other words, it is desired to find out under what conditions will testing or experiments done on a model accurately represent or predict the actual phenomena.

Cont.. In the study of fluid

mechanics, models are frequently used for testing and development purposes in laboratories before a full scale prototype is built.

The model can be either The model can be either smaller than the prototype (e.g., design of dam, airplane and automobiles) or larger than the prototype (e.g., study of interaction between red blood cells and the vessel wall).

Wind tunnel testing

Cont...

Take the aircraft industry for example. It is a common practice to design and

develop an aircraft by testing the model in a wind tunnel.

This way, the expenses of product development will be reduced considerably compared to conducting the design and development process on a full scale aircraft.

Cont...The next question would be under what

conditions will the model accurately portray the actual full scale prototype.

In order to yield useful information from the model being tested, the model and the model being tested, the model and prototype should fulfill the requirement of geometric, kinematic and dynamic similitude, as discussed in the subsequent paragraphs.

Geometric Similitude

The first requirement is for the model and prototype to be geometrically similar.

The model and prototype should have the same shape, and their dimensions should be scaled proportionally. be scaled proportionally.

In addition, their orientation and surface roughness should also have the same scale.

Kinematic SimilitudeTo obtain useful information from the model

testing, the second requirement is for the model and prototype to be kinematically similar.

In order to satisfy this condition, the flow conditions should be the same.

In other words, the velocities and accelerations at corresponding points should have the same directions and scaled magnitudes.

Model and prototype that are kinematic similar are also geometrically similar.

Dynamic SimilitudeIn order to achieve dynamic similitude, all

forces (e.g., pressure, shear, viscous, and surface tension forces) at corresponding points of the model and prototype should have the same directions and scaled have the same directions and scaled magnitudes.

Model and prototype that are dynamically similar are both geometrically and kinematically similar.

An Example: Flow Past a Sphere Take flow past a sphere for

example. A prototype and model satisfying

geometric similitude are shown in the figure.

The diameters of the spheres are proportional. Based on dimensional analysis, it can be

Flow Past a Sphere: Geometric and dimensional analysis, it can be determined that the drag force (D) can be modeled as,

D/(d2V2) = function (Re) where Re is the Reynolds number, and is given by

Re = Vd/

Flow Past a Sphere: Geometric and Kinematic Similitude

Flow Past a Sphere: Dinamic Similitude

Cont...Both the model and prototype should

follow the same relationship.In order to satisfy the dynamic similitude

requirement, the Reynolds number and dimensionless drag for the model and dimensionless drag for the model and prototype should be the same:

Rem = Rep andDm/(dm2 mVm2) = Dp/(dp2 pVp2)

where the subscripts m and p refer to the model and prototype, respectively.

CASE STUDY 2 A 1:10 scale model of an attack

submarine is being tested in a water tunnel.

The drag force ratio the submarine in the deep sea and the submarine model in the tunnel need to be estimated. tunnel need to be estimated.

The submarine cruises at 4 m/s in seawater (viscosity, = 1.210-3 N-s/m2 and density, = 1,030 kg/m3).

The fresh water in the tunnel is maintained at a temperature of 40 oC.

An Attack Submarine in the Deep Sea

Cont...QuestionsWhat should the velocity of the model be in

the tunnel? Determine the drag force ratio of the submarine prototype and model.

ApproachApproachAssume the submarine is far from the water

surface.Assume the compressibility of water is

negligible.

CASE STUDY SOLUTION 2 The submarine model is built with a

scale of 1:10. That is, Lp/Lm = 10 Since submarines usually cruise well

below the water surface, the Froude number is not important in this study.

In addition, it is assumed that the In addition, it is assumed that the compressibility effects are negligible, hence the Mach number has no role in this study as well.

In order for dynamic similitude, both the Reynolds number and dimensionless drag should be identical for the model and prototype.

An Attack Submarine

Cont...The Reynolds number for the prototype is

calculated asRep = pVpLp/p

= (1,030)(4)(10Lm)/(1.2)(10-3)= 34.3106 L= 34.3106 Lm

From the properties table of water, the viscosity and density of fresh water at a temperature of 40oC are 0.710-3 N-s/m2and 992 kg/m3, respectively.

CONT..Equate the Reynolds number for the

prototype and model and rearrange terms to yield an expression for the model's velocity

Rem = RepRem = RepmVmLm/m = 34.3106 Lm992 Vm / 0.710-3= 34.3106Vm = 24.2 m/s

Now, equating the dimensionless drag force for the prototype and model yields

Cont...

In this problem, it was assumed that both the Re and dimensionless drag force relationship where known.

They could have been derived using the Buckingham Pi Theorem

3. DIMENSIONLESS GROUPS Some of the dimensionless

groups encountered frequently in the study of fluid mechanics are introduced in this section.

As discussed in another section, dimensional analysis section, dimensional analysis can be used to identify dimensionless groups (pi terms) governing a system.

Some common imensionless groups in fluid mechanics are introduced here.

Laminar Flow

Turbulent Flow

Cont... Reynolds Number (Re): The Reynolds number perhaps is the

most common dimensionless parameter used in fluid mechanics. It is defined as

Re = VL/ where is the density, V is the velocity, L Hydraulic Radius where is the density, V is the velocity, L is the characteristic length, and is the viscosity.

The L term is different for each flow type. For example, for a pipe, L is the diameter

of the pipe. For open channel flow, the hydraulic radius, Rh (see diagram) is commonly used.

Hydraulic Radius

(Used with Open Channel

Flow)

Hydraulic Depth

Cont... Physically, Re represents the ratio of the inertial force to

the viscous force. A small Reynolds number implies that the viscous

effects are important, while the inertial effects are dominant when the Reynolds number is large.

The Reynolds number is commonly used to characterize The Reynolds number is commonly used to characterize if a flow in a pipe is laminar or turbulent .

The flow is generally assumed to be laminar when Re < 2,100, and turbulent when Re > 4,000.

The flow is referred to as transition flow when the Reynolds number is between 2,100 and 4,000.

Cont...

Froude Number (Fr): The Froude number is an important

dimensionless parameter in the study of open-channel flow, and it is given by

Fr = V / (gL)0.5where V is the average velocity, L is the where V is the average velocity, L is the characteristic length associated with the depth (hydraulic depth for open channel flow), and g is the gravitational acceleration.

For rectangular cross sections, the hydraulic depth is the water depth.

Physically, the Froude number represents the ratio of inertial forces to gravitational forces.

Open-Channel Flow

Cont...As discussed in the open-channel

sections, open-channel flow can be classified according to the Froude number in the following manner:(a) Fr < 1: subcritical (tranquil) flow(a) Fr < 1: subcritical (tranquil) flow(b) Fr = 1: critical flow(c) Fr > 1: supercritical (rapid) flow

Cont...It is also common to write Fr as V/c, where c is

the wave celerity, c (speed of a wave in the fluid).

This form is similar to the Mach Number in air. For subcritical flow (V < c), the waves created by any surface disturbances (e.g., throwing a stone any surface disturbances (e.g., throwing a stone in the water) at the downstream can travel upstream.

On the other hand, for supercritical flow (V > c), all surface disturbances will be swept downstream.

The wave will remain stationary for critical flow (V = c).

Cont.. Mach Number (Ma): For high speed flows in some fluids,

density is highly dependent on the pressure, and the compressibility effects become important.

The Mach number is used to indicate if a flow is incompressible or compressible, and it is given by

B-2 Spirit Bomber: High compressible, and it is given by

Ma = V/c where c is the speed of sound (343 m/s at 20oC) and V is the fluid velocity.

The Mach number represents the ratio of inertia forces to compressibility forces.

B-2 Spirit Bomber: High

Subsonic Speed

Concorde from British Airways:

Supersonic Speed (Mach 2)

Cont.. Flow can be characterized using the Mach number as

folllows:(a) Ma 0.3: incompressible(b) 0.3 < Ma < 1.0: compressible subsonic flow(c) Ma 1.0: compresible supersonic flow

The Mach number is often used to classify the top speed of a fighter or passenger jet. a fighter or passenger jet.

For example, the B-2 bomber shown in the picture is capable of reaching high subsonic speed.

The Concorde of British Airways is a supersonic passenger jet, which cruises at Mach 2, and it takes only approximately 4 hours from Los Angeles to Tokyo.

Compressible flow is beyond the scope of this introductory fluid mechanics eBook, and is discussed in an advanced level fluid mechanics course.

Cont...Weber Number (We): The dimensionless parameter associated with

surface tension effects is the Weber number, and it is defined as

We = V2L/ where is the surface tension.

The Weber number denotes the ratio of the inertial forces to surface tension forces.

The Weber number becomes an important parameter when dealing with applications involve two fluid interfaces such as the flow of thin films of liquid and bubble formation.

Cont...Parameter Definition Qualitative Use

Reynoldsnumber Almost all flow

Mach Compressible flowMachnumber

Compressible flow

Froudenumber

Free-surface flow

Webernumber

Free-surface flow

CASE STUDY 3 Bioengineer Rebecca is modeling

the flow of blood plasma in capillary blood vessels.

One of the important dimensionless parameters in modeling blood flow is the Reynolds number. Bioengineer Rebecca Studies

Blood Flow in Capillary Vessels She wonders what Reynolds

number is associated with the blood flow, and if the flow is laminar or turbulent.

The capillary vessel has a diameter of 10-5 m, and the plasma flows at a speed of 2.25 mm/s.

Blood Flow in Capillary Vessels

Blood Flow in Vessel

Cont...Questions

Determine the Reynolds number for the blood flow in a capillary vessel.

ApproachAssume the blood vessel is not collapsible (i.e., Assume the blood vessel is not collapsible (i.e.,

the vessel wall is not deformable, and it maintains the circular cross section).

The density and viscosity of the blood plasma is 980 kg/m3 and 0.0014 N-s/m2, respectively.

Determine if the flow is laminar or turbulent based on flow in a circular tube model.

CASE STUDY SOLUTION 3 The Reynolds number is given by

Re = VL/where L is the characteristic length.

It is assumed that the blood vessel is not collapsible, which means the vessel wall is not deformable, and vessel wall is not deformable, and the stenosis scenario (i.e., constriction or narrowing of coronary vessels due to the buildup of fat and cholesterol) is not considered.

Since the blood vessel maintains the circular cross section, the diameter (D) is chosen as the characteristic length, which is 10-5 m.

Blood Flow in Vessel

Cont..At body temperature, the density () and viscosity

() of the plasma is given as 980 kg/m3 and 0.0014 N-s/m2, respectively.

Since the plasma consists 90% of water, the density and viscosity of plasma are close to thase of water ( = 999 kg/m3 and = 0.0011 N-s/m2). of water ( = 999 kg/m3 and = 0.0011 N-s/m2).

The Reynolds number is calculated to beRe = (980)(2.25)(10-3)(10-5)/(0.0014)

= 0.0158Based on the pipe flow model, the calculated

Reynolds number is less than 2,100, hence the flow is laminar.

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