LECTURE NOTES - VIII. Prof. Dr. Atıl BULU

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1 LECTURE NOTES - VIII «LUID MECHNICS» Istanbul Technical Universit College of Civil Engineering Civil Engineering Department Hdraulics Division

2 CHPTER 8 DIMENSIONL NLYSIS 8. INTRODUCTION Dimensional analsis is one of the most important mathematical tools in the stud of fluid mechanics. It is a mathematical technique, which makes use of the stud of dimensions as an aid to the solution of man engineering problems. The main advantage of a dimensional analsis of a problem is that it reduces the number of variables in the problem b combining dimensional variables to form non-dimensional parameters. B far the simplest and most desirable method in the analsis of an fluid problem is that of direct mathematical solution. But, most problems in fluid mechanics such complex phenomena that direct mathematical solution is limited to a few special cases. Especiall for turbulent flow, there are so man variables involved in the differential equation of fluid motion that a direct mathematical solution is simpl out of question. In these problems dimensional analsis can be used in obtaining a functional relationship among the various variables involved in terms of non-dimensional parameters. Dimensional analsis has been found useful in both analtical and experimental work in the stud of fluid mechanics. Some of the uses are listed: ) Checking the dimensional homogeneit of an equation of fluid motion. ) Deriving fluid mechanics equations expressed in terms of non-dimensional parameters to show the relative significance of each parameter. ) Planning tests and presenting experimental results in a sstematic manner. ) nalzing complex flow phenomena b use of scale models (model similitude). 8. DIMENSIONS ND DIMENSIONL HOMOGENEITY Scientific reasoning in fluid mechanics is based quantitativel on concepts of such phsical phenomena as length, time, velocit, acceleration, force, mass, momentum, energ, viscosit, and man other arbitraril chosen entities, to each of which a unit of measurement has been assigned. or the purpose of obtaining a numerical solution, we adopt for computation the quantities in SI or MKS units. In a more general sense, however, it is desirable to adopt a consistent dimensional sstem composed of the smallest number of dimensions in terms of which all the phsical entities ma be expressed. The fundamental dimensions of mechanics are length [L], time [T], mass [M], and force [], related b Newton s second law of motion, ma.

3 Dimensionall, the law ma also be written as, ML T [ ] or T ML (8.) Which indicates that when three of the dimensions are known, the fourth ma be expressed in the terms of the other three. Hence three independent dimensions are sufficient for an phsical phenomenon encountered in Newtonian mechanics. The are usuall chosen as either [MLT] (mass, length, time) or [LT] (force, length, time). or example, the specific mass (ρ) ma be expressed either as [M/L ]or as [T /L ], and a fluid pressure (p), which is commonl expressed as force per unit area [/L ] ma also be expressed as [ML/T ] using the (mass, length, time) sstem. summar of some of the entities frequentl used in fluid mechanics together with their dimensions in both sstems is given in Table 8.. TBLE 8. ENTITIES COMMONLY USED IN LUID MECHNICS ND THEIR DIMENSIONS Entit MLT Sstem LT Sstem Length (L) L L rea () L L Volume (V) L L Time (t) T T Velocit (v) LT - LT - cceleration (a) LT - LT - orce () and weight (W) MLT - Specific weight (γ) ML - T - L - Mass (m) M L - T - Specific mass (ρ) ML - L - T Pressure (p) and stress (τ) ML - T - L - Energ (E) and work ML T - L Momentum (mv) MLT - T Power (P) ML T - LT - Dnamic viscosit (μ) ML - T - L - T Kinematic viscosit (υ) L T - L T - With the selection of three independent dimensions either [MLT] or [LT]- it is possible to express all phsical entities of fluid mechanics. n equation which expresses the phsical phenomena of fluid motion must be both algebraicall correct and dimensionall homogenous. dimensionall homogenous equation has the unique characteristic of being independent of units chosen for measurement. Equ. (8.) demonstrates that a dimensionall homogenous equation ma be transformed to a non-dimensional form because of the mutual dependence of fundamental dimensions. lthough it is alwas possible to reduce dimensionall homogenous equation to a non-dimensional form, the main difficult in a complicated flow problem is in setting up the correct equation of motion. Therefore, a special mathematical method called dimensional analsis is required to determine the functional relationship among all the variables involved in an complex phenomenon, in terms of non-dimensional parameters.

4 8. DIMENSIONL NLYSIS The fact that a complete phsical equation must be dimensionall homogenous and is, therefore, reducible to a functional equation among non-dimensional parameters forms the basis for the theor of dimensional analsis. 8.. Statement of ssumptions The procedure of dimensional analsis makes use of the following assumptions: ) It is possible to select m independent fundamental units (in mechanics, m, i.e., length, time, mass or force). ) There exist n quantities involved in a phenomenon whose dimensional formulae ma be expressed in terms of m fundamental units. ) The dimensional quantit can be related to the independent dimensional quantities,,..., n- b, (,,..., ) K... (8.) n n n Where K is a non-dimensional constant, and,,..., n- are integer components. ) Equ. (8.) is independent of the tpe of units chosen and is dimensionall homogenous, i.e., the quantities occurring on both sides of the equation must have the same dimension. EXMPLE 8.: Consider the problem of a freel falling bod near the surface of the earth. If x, w, g, and t represent the distance measured from the initial height, the weight of the bod, the gravitational acceleration, and time, respectivel, find a relation for x as a function of w, g, and t. SOLUTION: Using the fundamental units of force, length L, and time T, we note that the four phsical quantities, x, w, g, and t, involve three fundamental units; hence, m and n in assumptions () and (). B assumption () we assume a relation of the form: ( w, g, t) Kw g t x (a) Where K is an arbitrar non-dimensional constant. Let [ ] denote dimensions of a quantit. Then the relation above can be written (using assumption ()) as, or [] [ ] [] [ ] x w g t L T ( ) ( LT ) ( T ) L T

5 Equating like exponents, we obtain : : L : T or Therefore, Equ. (a) becomes t g Kw x or Kgt x ccording to the elementar mechanics we have xgt /. The constant K in this case is ½, which cannot be obtained from dimensional analsis. EXMPLE 8.: Consider the problem of computing the drag force on a bod moving through a fluid. Let D, ρ, μ, l, and V be drag force, specific mass of the fluid, dnamic viscosit of the fluid, bod reference length, and bod velocit, respectivel. SOLUTION: or this problem m, n5, D, ρ, μ, l, and V. Thus, according to Equ (8.), we have ( ),,, V l K V l D μ ρ μ ρ (a) or [ ] [ ] [ ] [ ] [ ] ( ) ( ) ( ) ( ) T L T L LT L T L T L T L V l D μ ρ Equating like exponents, we obtain : : : T L In this case we have three equations and four unknowns. Hence, we can onl solve for three of the unknowns in terms of the fourth unknown (a one-parameter famil of solutions exists). or example, solving for, and in terms of, one obtains

6 The required solution is or D Kρ μ D l V ρvl μ ρv ( K ) l If the Renolds number is denoted b ReρVl/μ, dnamic pressure b qρv /, and area b l, we have where D C D K q C ( Re) K ( Re) D q Theoretical considerations show that for laminar flow K.8 and 8.. Buckingham-π (Pi) Theorem It is seen from the preceding examples that m fundamental units and n phsical quantities lead to a sstem of m linear algebraic equations with n unknowns of the form a a a m a a a m... a... a n n... a mn n n n b b... b m (8.) or, in matrix form, where b (8.) aa... a n aa... an... amam... amn. n and b b b. b n

7 is referred to as the coefficient matrix of order m n, and and b are of order n and m respectivel. The matrix in Equ. (8.) is rectangular and the largest determinant that can be formed will have the order n or m, whichever is smaller. If an matrix C has at least one determinant of order r, which is different from zero, and nonzero determinant of order greater than r, then the matrix is said to be of rank r, i.e., R ( C) r (8.5) In order to determine the condition for the solution of the linear sstem of Equ. (8.) it is convenient to define the rank of the augmented matrix B. The matrix B is defined as B b aa... a nb aa... anb... amam... amnb m (8.6) or the solution of the linear sstem in Equ. (8.), three possible cases arise: ) R ()<R (B): No solution exists, ) R()R(B) r n: unique solution exits, ) R ()R (B)r<n: n infinite number of solutions with (n-r) arbitrar unknowns exist. Example 8. falls in case () where R ( ) R( B) < n and ( n r) ( ) an arbitrar unknown exits. The mathematical reasoning above leads to the following Pi theorem due to Buckingham. Let n quantities,,.., n be involved in a phenomenon, and their dimensional formulae be described b (m<n) fundamental units. Let the rank of the augmented matrix B be R (B) r m. Then the relation (,,..., n ) (8.7) is equivalent to the relation ( π ) π,,..., π nr (8.8) Where π, π,..,π n-r are dimensionless power products of,,.., n taken r at a time. Thus, the Pi theorem allows one to take n quantities and find the minimum number of non-dimensionless parameter, π, π,.., π n-r associated with these n quantities. 5

8 8... Determination of Minimum Number of π Terms In order to appl the Buckingham π Theorem to a given phsical problem the following procedure should be used: Step. Given n quantities involving m fundamental units, set up the augmented matrix B b constructing a table with the quantities on the horizontal axis and the fundamental units on the vertical axis. Under each quantit list a column of numbers, which represent the powers of each fundamental units that makes up its dimensions. or example, ρ p d Q L - - T - Where ρ, p, d, and Q are the specific mass, pressure, diameter, and discharge, respectivel. The resulting arra of numbers represents the augmented matrix B in Buckingham s π Theorem, i.e., B The matrix B is sometimes referred to as dimensional matrix. Step. Having constructed matrix B, find its rank. rom step, since and no larger nonzero determinant exists, then ( B) r R Step. Having determined the number of π dimensionless (n-r) terms, following rules are used to combine the variables to form π terms. a) rom the independent variables select certain variables to use as repeating variables, which will appear in more than π term. The repeating variables should contain all the dimensions used in the problem and be quantities, which are likel to have substantial effect on the dependent variable. b) Combine the repeating variables with remaining variables to form the required number of independent dimensionless π terms. c) The dependent variable should appear in one group onl. d) variable that is expected to have a minor influence should appear in one group onl. 6

9 Define π as a power product of r of the n quantities raised to arbitrar integer exponents and an one of the excluded (n-r) quantities, i.e., π r... r r Step. Define π, π,..., π n-r as power products of the same r quantities used in step raised to arbitrar integer exponents but a different excluded quantit for each π term, i.e., π π nr nr, nr,... r r r r... π nr, r r r r n Step 5. Carr out dimensional analsis on each π term to evaluate the exponents. EXMPLE 8.: Rework Example 8. using the π theorem. SOLUTION: Step. With, L, and T as the fundamental units, the dimensional matrix of the quantities w, g, t and x is, W g t x L T - Where B Step. Since and no larger nonzero determinant exists, then ( B) r R Step. rbitraril select x, w, and g as the r m base quantities. The number n-r of independent dimensional products that can be formed b the four quantities is therefore, i.e., π x w g t 7

10 Step. Dimensional analsis gives, or [ π ] [ x] [ w] [ g] [ t] L T ( L) ( ) ( LT ) ( T ) Which results in,, and Hence, π gt x EXMPLE 8.: Rework Example 8. using the π theorem. SOLUTION: Step. With, L and T as the fundamental units, the dimensional matrix of the quantities D, ρ, μ, l, and V is. Where D ρ μ l V L - - T - B Step. Since and no larger nonzero determinant exists, then ( B) r R 8

11 Step. Select l, V, and ρ as the r base quantities. B the π-theorem (n-r),(5-) π terms exist. Hence, π l π l V V ρ ρ D μ Step. Dimensional analsis gives -, -, - -, -, - π π D ρv l μ ρvl 8. THE USE O DIMENSIONLESS π-terms IN EXPERIMENTL INVESTIGTIONS Dimensional analsis can be of assistance in experimental investigation b reducing the number of variables in the problem. The result of the analsis is to replace an unknown relation between n variables b a relationship between a smaller numbers, n-r, of dimensionless π-terms. n reduction in the number of variables greatl reduces the labor of experimental investigation. or instance, a function of one variable can be plotted as a single curve constructed from a relativel small number of experimental observations, or the results can be represented as a single table, which might require just one page. function of two variables will require a chart consisting of a famil of curves, one for each value of the second variable, or, alternativel the information can be presented in the form of a book of tables. function of three variables will require a set of charts or a shelffull of books of tables. s the number of variables increases, the number of observations to be taken grows so rapidl that the situation soon becomes impossible. n reduction in the number of variables is extremel important. Considering, as an example, the resistance to flow through pipes, the shear stress or resistance R per unit area at the pipe wall when fluid of specific mass ρ and dnamic viscosit μ flows in a smooth pipe can be assumed to depend on the velocit of flow V and the pipe diameter D. Selecting a number of different fluids, we could obtain a set of curves relating frictional resistance (measured as R/ρV ) to velocit, as shown in ig

12 .5 ll measurements at 5.5 C. Water, d 5.8 mm Water, d 5. mm Water, d 6.7 mm ir, d 5. mm ir, d.7 mm ir, d 6.5 mm R/pv.5.. ig v ( ms ) Such a set of curves would be of limited value both for use and for obtaining a proper understanding of the problem. However, it can be shown b dimensional analsis that the relationship can be reduced to the form R ρv ρvd μ φ φ ( Re) or, using the Darc resistance coefficient f R ρv, f ρvd φ φ μ ( Re) 7 - riction coefficient, f Renolds number, Re ig

13 If the experimental points in ig. 8. are used to construct a new graph of Log (f) against Log (Re) the separate sets of experimental data combine to give a single curve as shown in ig. 8.. or low values of Renolds number, when flow is laminar, the slope of this graph is (-) and f 6/Re, while for turbulent flow at higher values of Renolds number, f.8(re) -/.