Dimensional Analysis and Similitude

A

n important consideration in all equations written thus far has been dimensional homogeneity. At times it has been necessary to use proper conversion factors in order that an answer be correct numerically and have the proper units. The idea of dimensional consistency can be used in another way, by a procedure known as dimensional analysis, to group the variables in a given situation into dimensionless parameters that are less numerous than the original variables. Such a procedure is very helpful in experimental work, in which the very number of significant variables presents an imposing task of correlation. By combining the variables into a smaller number of dimensionless parameters, the work of experimental data reduction is considerably reduced.

This chapter will include means of evaluating dimensionless parameters both in situations in which the governing equation is known, and in those in which no equation is available. Certain dimensionless groups emerging from this analysis will be familiar, and some others will be encountered for thefirst time. Finally, certain aspects of similarity will be used to predict theflow behavior of equipment on the basis of experiments with scale models.

▶ 11.1

DIMENSIONS

In dimensional analysis, certain dimensions must be established as fundamental, with all others expressible in terms of these. One of these fundamental dimensions is length, symbolized L. Thus, area and volume may dimensionally be expressed as L² and L³, respectively. A second fundamental dimension is time, symbolized t. The kinematic quantities, velocity and acceleration, may now be expressed asL/tand L/t², respectively.

Another fundamental dimension is mass, symbolizedM. An example of a quantity whose dimensional expression involves mass is the density that would be expressed as M/L³. Newton’s second law of motion gives a relation between force and mass and allows force to be expressed dimensionally asFˆMaˆML/t². Some texts reverse this procedure and consider force fundamental, with mass expressed in terms ofF,L, andtaccording to Newton’s second law of motion. Here, mass will be considered a fundamental unit.

The significant quantities in momentum transfer can all be expressed dimensionally in terms ofM,L, andt; thus these comprise the fundamental dimensions we shall be concerned with presently. The dimensional analysis of energy problems in Chapter 19 will require the addition of two more fundamental dimensions, heat and temperature.

141

Some of the more important variables in momentum transfer and their dimensional representations in terms ofM,L, andtare given in Table 11.1.

▶ 11.2

DIMENSIONAL ANALYSIS OF GOVERNING DIFFERENTIAL EQUATIONS The differential equations that describe fluid behavior as developed in Chapter 9 are powerful tools for analyzing and predictingfluid phenomena and their effects. The Navier– Stokes equations have been solved analytically for a few simple situations. For more complex applications, these relationships provide the basis for a number of sophisticated and powerful numerical codes.

In this section, we will use the differential forms of the continuity and momentum (Navier–Stokes) equations to develop some useful dimensionless parameters that will be valuable tools for subsequent analysis. This process will now be illustrated as we examine two-dimensional incompressibleflow.

The governing differential equations are the following.

Continuity:

@vx

@x ‡@vy

@y ˆ0 (9-3)

Momentum:

r @v

@t‡vx@v

@x‡vy@v

@y

ˆrg Ñr‡m @²v

@x²‡@²v

@y²

(9-19) We now stipulate the reference values for length and velocity:

• reference length L

• reference velocity u_∞

and, accordingly, specify nondimensional quantities for the variables in equations (9-3) and (9-19) as

xˆx=L vx ˆvx=v_∞ yˆy=L vy ˆvy=v_∞ tˆtv_∞

L vˆv=v_∞ Ñ ˆLÑ Table 11.1 Important variables in momentum transfer

Variable Symbol Dimension

Mass M M

Length L L

Time t t

Velocity u L/t

Gravitational acceleration g L/t²

Force F ML/t²

Pressure P ML/t²

Density r M/L³

Viscosity m M/Lt

Surface tension s M/t²

Sonic velocity a L/t

The last quantity in this list,Ñ^*, is the dimensionless gradient operator. AsÑis composed of first derivatives with respect to space coordinates, the product LÑ is seen to be dimensionless.

The next step is to nondimensionalize our governing equations by introducing the specified dimensionless variables. This process involves the chain rule for differentiation;

for example, the two terms in equation (9-3) are transformed as follows:

@vx

@x ˆ@v_x

@x

@vx

@vx

@x

@x ˆ@v_x

@x…v_∞†…1=L† ˆv_∞ L

@v_x

@x

@vy

@y ˆ@vy

@y

@vy

@v_y

@y

@y ˆv_∞ L

vx

@x Substitution into equation (9-3) gives

@v_x

@x‡@vy

@y ˆ0 (11-1)

and we see that the continuity equation has the same form in terms of dimensionless variables as it had originally.

Utilizing the chain rule in the same manner as just discussed, the equation of motion becomes

rv²_∞ L

@v

@t ‡vx

@v

@x‡vy

@v

@y

ˆrg‡1

LÑP‡mv_∞ L²

@²v

@x²‡@²v

@y²

(11-2) In equation (11-2), we note that each term has the units M/L²t²or F/L³. Also, it should be observed that each term represents a certain kind of force—that is,

• rv²_∞

L is an inertial force

• mv²_∞

L is a viscous force

• rgis a gravitational force

• P/Lis a pressure force

If we next divide through by the quantity,rv²_∞=L, our dimensionless equation becomes

@v

@t ‡vx

@v

@x‡vy

@v

@yˆg L v²_∞

ÑP rv²_∞ ‡ m

Lv_∞r

@²v

@x²‡@²v

@y²

(11-3) This resulting dimensionless equation has the same general characteristics as its original except that, as a result of its transformation into dimensionless form, each of the original force terms (those on the right-hand side) has a coefficient composed of a combination of variables. An example of these coefficients reveals that each is dimensionless. Addition- ally, because of the manner in which they were formed, the parameters can be interpreted as a ratio of forces.

Consideration of thefirst term,gL=v²_∞, reveals that it is, indeed, dimensionless. The choice ofgL=v²_∞ orv²_∞=gLis arbitrary; clearly both forms are dimensionless.

The conventional choice is the latter form. TheFroude number is defined as

Frv²_∞=gL (11-4)

This parameter can be interpreted as a measure of the ratio of inertial to gravitational forces.

The Froude number arises in analyzing flows involving a free liquid surface. It is an important parameter when dealing with open-channelflows.

11.2 Dimensional Analysis of Governing Differential Equations ◀ 143

The next parameter,P=rv²_∞;is observed to be the ratio of pressure forces to inertial forces. In this form it is designated the Euler number,

EuP=rv²_∞ (11-5)

A modified form of equation (11-5), also clearly dimensionless, is the coefficient of drag, CDˆ F=A

rv²_∞=2 (11-6)

which, we will see directly, has application to both internal and externalflows.

The third dimensionless ratio that has been generated is theReynolds number, which is conventionally expressed as

ReLv_∞r=m (11-7)

In this form the Reynolds number is observed to represent the ratio of inertial forces to viscous forces. The Reynolds number is generally considered the most important dimen- sionless parameter in the field offluid mechanics. It is ubiquitous in all of the transport processes. We will encounter it frequently throughout the remainder of this text.

If equation (11-3) can be solved, the results will provide the functional relationships between applicable dimensionless parameters. If direct solution is not possible, then one must resort to numerical modeling or experimental determination of these functional relationships.

▶ 11.3

THE BUCKINGHAM METHOD

The procedure introduced in the previous section is, obviously, quite powerful when one knows the differential equation that pertains to a specificfluid-flow process. There are, however, many situations of interest in which the governing equation is not known. In these cases, we need an alternative method for dimensional analysis. In this section, we discuss a more general approach for generating dimensionless groups of variables. This procedure was proposed by Buckingham¹in the early part of the twentieth century. It is generally referred to as theBuckingham method.

The initial step in applying the Buckingham method requires the listing of the variables significant to a given problem. It is then necessary to determine the number of dimensionless parameters into which the variables may be combined. This number may be determined using the Buckingham pitheorem, which states

The number of dimensionless groups used to describe a situation involvingn variables is equal ton r, whereris the rank of the dimensional matrix of the variables.

Thus,

iˆn r (11-8)

where

i ˆthe number of independent dimensionless groups nˆthe number of variables involved

1E. Buckingham,Phys. Rev.2, 345 (1914).

and

r ˆ the rank of the dimensional matrix

The dimensional matrix is simply the matrix formed by tabulating the exponents of the fundamental dimensionsM,L, andt, which appear in each of the variables involved.

An example of the evaluation ofrandi, as well as the application of the Buckingham method, follows.

Example 1

Determine the dimensionless groups formed from the variables involved in theflow offluid external to a solid body. The force exerted on the body is a function ofu,r,m,andL(a significant dimension of the body).

A usualfirst step is to construct a table of the variables and their dimensions.

Variable Symbol Dimensions

Force F ML/t²

Velocity u L/t²

Density r M/L³

Viscosity m M/Lt

Length L L

Before determining the number of dimensionless parameters to be formed, we must knowr.The dimensional matrix that applies is formed from the following tabulation:

F v r m L

M 1 0 1 1 0

L 1 1 3 1 1

t 2 1 0 1 0

The numbers in the table represent the exponents ofM,L, andtin the dimensional expression for each variable involved. For example, the dimensional expression ofF is ML/t², hence the exponents 1, 1, and 2 are tabulated versusM, L, and t, respectively, the quantities with which they are associated. The matrix is then the array of numbers shown below

1 0 1 1 0

1 1 3 1 1

2 1 0 1 0

0

@

1 A

The rank,r, of a matrix is the number of rows (columns) in the largest nonzero determinant that can be formed from it. The rank is 3 in this case. Thus, the number of dimensionless parameters to be formed may be found by applying equation (11-4). In this example,iˆ5 3ˆ2.

The two dimensionless parameters will be symbolizedp1andp2and may be formed in several ways. Initially, acore group ofrvariables must be chosen, which will consist of those variables that will appear in each pi group and, among them, contain all of the fundamental dimensions. One way to choose a core is to exclude from it those variables whose effect one desires to isolate.

In the present problem it would be desirable to have the drag force in only one dimensionless group, hence it will not be in the core. Let us arbitrarily let the viscosity be the other exclusion from the core. Our core group now consists of the remaining variablesv, r, andL, which, we observe, includeM,L, andtamong them.

We now know thatp1andp2both includer,L, andu; that one of them includesFand the otherm; and that they are both dimensionless. In order that each be dimensionless, the variables must be raised to certain exponents. Writing

p1ˆv^ar^bL^cF and p2ˆv^dr^eL^fm

we shall evaluate the exponents as follows. Considering eachpgroup independently, we write p1ˆv^ar^bL^cF

11.3 The Buckingham Method ◀ 145

and dimensionally

M⁰L⁰t⁰ˆ1ˆ L t

a M

L³

b

…L†^cML t²

Equating exponents ofM,L, andton both sides of this expression, we have, forM, 0ˆb‡1

forL,

0ˆa 3b‡c‡1 and fort,

0ˆ a 2

From these wefind thataˆ 2,bˆ 1, andcˆ 2, giving

p1ˆ F

L²rv²ˆF=L² rv² ˆEu Similarly forp2we have, in dimensional form,

1ˆ L t

d M

L³

e

…L†^fM Lt

and for exponents ofM,

0ˆe‡1 forL,

0ˆd 3e‡f 1 and fort,

0ˆ d 1

givingdˆ 1, eˆ 1 andfˆ 1. Thus, for our second dimensionless group we have p2ˆm=rvLˆ1=Re

Dimensional analysis has enabled us to relate the originalfive variables in terms of only two dimensionless parameters in the form

Euˆf…Re† (11-9)

CDˆf…Re† (11-10)

The two parameters, Eu and C_D, were also generated in the previous section by an alternate method. The functions f(Re) andf(Re) must be determined by experiment.

Table 11.2 lists several dimensionless groups that pertain tofluidflow. Similar tables will be include in later chapters that list dimensionless parameters common to heat transfer and to mass transfer.

▶ 11.4

GEOMETRIC, KINEMATIC, AND DYNAMIC SIMILARITY

An important application and use of the dimensionless parameters listed in Table 11.2 is in using experimental results obtained using models to predict the performance of full-size prototypical systems. The validity of suchscalingrequires that the models and prototypes possesssimilarity.Three types of similarity are important in this regard; they are geometric, kinematic, and dynamic similarity.

Geometric similarityexists between two systems if the ratio of all significant dimen- sions is the same for each system. For example, if the ratio a/bfor the diamond-shaped section in Figure 11.1 is equal in magnitude to the ratioa/bfor the larger section, they are Table 11.2 Common dimensionless parameters in momentum transfer

Name/symbol

Dimensionless

group Physical meaning Area of application Reynolds number, Re Lvr

m

Inertial force Viscous force

Widely applicable in a host offluid flow situations

Euler number, Eu P

rv²

Pressure force

Inertial force Flows involving pressure differences due to frictional effects

Coefficient of skin friction,Cf

F=A rv²=2

Drag force

Dynamic force Flows in aerodynamics and hydrodynamics Froude number, Fr v²

gL

Inertial force

Gravitational force Flows involving free liquid surfaces

Weber number, We rv²L s

Inertial force

Surface tension force Flows with significant surface tension effects

Mach number,M v

C

Inertial force

Compressibility force Flows with significant compressibility effects

b a

1 2

Figure 11.1 Two geometrically similar objects.

11.4 Geometric, Kinematic, and Dynamic Similarity ◀ 147

geometrically similar. In this example, there are only two significant dimensions. For more complex geometries, geometric similarity would be achieved when all geometric ratios between model and prototype are equal.

Kinematic similaritysimilarly exists when, in geometrically similar systems➀and➁, the velocities at the same locations are related according to

vx

vy 1

ˆ vx

vy 2

vx

vz 1

ˆ vx

vz 2

The third type of similarity, dynamic similarity, exists when, in geometrically and kinematically similar systems, the ratios of significant forces are equal between model and prototype. These force ratios that are important in fluid-flow applications include the dimensionless parameters listed in Table 11.2.

The process of scaling using these similarity requirements will be presented in Section 11.5.

▶ 11.5

MODEL THEORY

In the design and testing of large equipment involvingfluidflow, it is customary to build small models geometrically similar to the larger prototypes. Experimental data achieved for the models are then scaled to predict the performance of full-sized prototypes according to the requirements of geometric, kinematic, and dynamic similarity. The following examples will illustrate the manner of utilizing model data to evaluate the conditions for a full-scale device.

Example 2

A cylindrical mixing tank is to be scaled up to a larger size such that the volume of the larger tank isfive times that of the smaller one. What will be the ratios of diameter and height between the two?

Geometric similarity between tanks A and B in Figure 11.2 requires that Da

ha

ˆDb

hb

or

hb

ha

ˆDb

Da

The volumes of the two tanks are

Vaˆp

4D²_aha and Vbˆp 4D²_bhb

The scaling ratio between the two is stipulated asVb

Va

ˆ5;thus, Vb

Vaˆ…p=4†D²_bhb

…p=4†D²_aha

ˆ5

D_a

D_b

h_a

h_b

A

B

Figure 11.2 Cylindrical mixing tanks for Example 2.

and we get

Db

Da 2hb

haˆ5 We now substitute the geometric similarity requirement that gives

Db

Da 3

ˆ Lb

La 3

ˆ5 and the two ratios of interest become

Db

DaˆLb

Laˆ5¹⁼³ˆ1:71

Example 3

Dynamic similarity may be obtained by using a cryogenic wind tunnel in which nitrogen at low temperature and high pressure is employed as the workingfluid. If nitrogen at 5 atm and 183 K is used to test the low-speed aerodynamics of a prototype that has a 24.38 m wing span and is tofly at standard sea-level conditions at a speed of 60 m/s, determine

1. The scale of the model to be tested

2. The ratio of forces between the model and the full-scale aircraft

Conditions of dynamic similarity should prevail. The speed of sound in nitrogen at 183 K is 275 m/s.

For dynamic similarity to exist, we know that both model and prototype must be geometrically similar and that the Reynolds number and the Mach number must be the same. A table such as the following is helpful.

Model Prototype

Characteristic length L 24.38 m

Velocity v 60 m/s

Viscosity m 1.78910 ⁵Pa?s

Density r 1.225 kg/m³

Speed of sound 275 m/s 340 m/s

The conditions listed for the prototype have been obtained from Appendix I. Equating Mach numbers, we obtain MmˆMp

vˆ275

34060ˆ48:5 m=s Equating the Reynolds numbers of the model and the prototype, we obtain

RemˆRep

r48:5L

m ˆ1:225_?60_?24:38

1:789?10 ⁵ ˆ1:00210⁸

11.5 Model Theory ◀ 149

Using equation (7-10), we may evaluatemfor nitrogen. From Appendix K,e/kˆ91.5 K andsˆ3:681 A for nitrogen so that kT/eˆ2 andWmˆ1.175 (Appendix K). Thus,

mˆ2:6693?10 ⁶

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 28?183 p

…3:681†²…1:175†ˆ1:200?10 ⁵Pa?s The density may be approximated from the perfect gas law

rˆ P P1

M M1

T1

T r1

so that

rˆ5 28 28:96

288 183

1:225ˆ9:32 kg=m³ Solving for the wing span of the model, we obtain

Lˆ3:26 m…10:7 ft†

The ratio of the forces on the model to the forces experienced by the prototype may be determined equating values of Eu between the model and the prototype. Hence

F rV²AR

model

ˆ F

rV²AR

prototype

whereARis a suitable reference area. For an aircraft, this reference area is the projected wing area. The ratio of model force to prototype force is then given by

Fm

Fp ˆrm

rp

V²_m V²_p

AR;m

AR;pˆ…rV²†m

…rV²†p

lm

lp 2

where the ratio of reference areas can be expressed in terms of the scale ratio. Substituting numbers, Fm

Fp ˆ 9:32 1:225

48:5 60:0

2 3:26 24:38

2

ˆ0:089

The forces on the model are seen to be 8.9% the prototype forces.

▶ 11.6

CLOSURE

The dimensional analysis of a momentum-transfer problem is simply an application of the requirement of dimensional homogeneity to a given situation. By dimensional analysis, the work and time required to reduce and correlate experimental data are decreased substantially by the combination of individual variables into dimensionlesspgroups, which are fewer in number than the original variables. The indicated relations between dimensionless parameters are then useful in expressing the performance of the systems to which they apply.

It should be kept in mind that dimensional analysiscannotpredict which variables are important in a given situation, nor does it give any insight into the physical transfer

mechanism involved. Even with these limitations, dimensional analysis techniques are a valuable aid to the engineer.

If the equation describing a given process is known, the number of dimensionless groups is automatically determined by taking ratios of the various terms in the expression to one another. This method also gives physical meaning to the groups thus obtained.

If, on the contrary, no equation applies, an empirical method, theBuckinghammethod, may be used. This is a very general approach, but gives no physical meaning to the dimensionless parameters obtained from such an analysis.

The requirements of geometric, kinematic, and dynamic similarity enable one to use model date to predict the behavior of a prototype or full-size piece of equipment.Model theory is thus an important application of the parameters obtained in a dimensional analysis.

▶

PROBLEMS

11.1 The power output of a hydraulic turbine depends on the diameterDof the turbine, the densityrof water, the heightHof water surface above the turbine, the gravitational accelerationg, the angular velocitywof the turbine wheel, the dischargeQof water through the turbine, and the efficiencyhof the turbine. By dimensional analysis, generate a set of appropriate dimension- less groups.

11.2 Through a series of tests on pipeflow,H.Darcy derived an equation for the friction loss in pipeflow as

hLˆfL D

v² 2g;

in whichfis a dimensionless coefficient that depends on (a) the average velocity uof the pipeflow; (b) the pipe diameter D;

(c) the fluid density r; (d) the fluid viscosity m; and (e) the average pipe wall unevenesse(length). Using the Buckinghamp theorem,find a dimensionless function for the coefficientf.

11.3 The pressure rise across a pumpP(this term is propor- tional to the head developed by the pump) may be considered to be affected by thefluid densityr, the angular velocity w, the impeller diameterD, the volumetric rate offlowQ, and thefluid viscositym. Find the pertinent dimensionless groups, choosing them so thatP,Q, andmeach appear in one group only. Find similar expressions, replacing the pressure risefirst by the power input to the pump, then by the efficiency of the pump.

11.4 The maximum pitching moment that is developed by the water on aflying boat as it lands is noted ascmaxThe following are the variables involved in this action:

aˆangle made by flight path of plane with horizontal b ˆangle defining attitude of plane

Mˆmass of plane L ˆlength of hull r ˆdensity of water g ˆacceleration of gravity

R ˆradius of gyration of plane about axis of pitching

a. According to the Buckinghamptheorem, how many inde- pendent dimensionless groups should there be characteriz- ing this problem?

b. What is the dimensional matrix of this problem? What is its rank?

c. Evaluate the appropriate dimensionless parameters for this problem.

11.5 The rate at which metallic ions are electroplated from a dilute electrolytic solution onto a rotating disk electrode is usually governed by the mass diffusion rate of ions to the disk. This process is believed to be controlled by the following variables:

Dimensions kˆmass-transfer coefficient L/t

Dˆdiffusion coefficient L²/t

dˆdisk diameter L

aˆangular velocity 1/t

rˆdensity M/L³

mˆviscosity M/Lt

Obtain the set of dimensionless groups for these variables where k, m, and D are kept in separate groups. How would you accumulate and present the experimental data for this system?

11.6 The performance of a journal bearing around a rotating shaft is a function of the following variables:Q, the rate offlow lubricating oil to the bearing in volume per unit time;D, the bearing diameter;N, the shaft speed in revolutions per minute;m, the lubricant viscosity;r, the lubricant density; ands, the surface tension of the lubricating oil. Suggest appropriate parameters to be used in correlating experimental data for such a system.

11.7 The massMof drops formed by liquid discharging by gravity from a vertical tube is a function of the tube diameterD, liquid density, surface tension, and the acceleration of gravity.

Problems ◀ 151