流体力学Chapter-7-Dimensional-Analysis

1、Chapter 7 Dimensional AnalysisModeling, and Similitude,2,MAIN TOPICS,7.1 Dimensional Analysis 7.2 Buckingham Pi Theorem 7.3 Determination of Pi Terms 7.4 Comments about Dimensional Analysis 7.5 Determination of Pi Terms by Inspection 7.6 Common Dimensionless Groups in Fluid Mechanics 7.7 Correlation

2、 of Experimental Data 7.8 Modeling and Similitude 7.9 Typical Model Studies 7.10 Similitude Based on Governing Differential Equation,3,7.1 Dimensional Analysis_1,A typical fluid mechanics problem in which experimentation is required, consider the steady flow of an incompressible Newtonian fluid thro

3、ugh a long, smooth-walled, horizontal, circular pipe. An important characteristic of this system, which would be interest to an engineer designing a pipeline, is the pressure drop per unit length that develops along the pipe as a result of friction.,4,Dimensional Analysis_2,The first step in the pla

4、nning of an experiment to study this problem would be to decide on the factors, or variables, that will have an effect on the pressure drop. Pressure drop per unit length,Pressure drop per unit length depends on FOUR variables:piping diameter (D); velocity (V); fluid density (); fluid viscosity (m),

5、5,Dimensional Analysis_3,To perform the experiments in a meaningful and systematic manner, it would be necessary to change on of the variable, such as the velocity, which holding all other constant, and measure the corresponding pressure drop. Difficulty to determine the functional relationship betw

6、een the pressure drop and the various factors that influence it.,6,Series of Tests,7,Dimensional Analysis_4,Fortunately, there is a much simpler approach to the problem that will eliminate the difficulties described above. Collecting these variables into two nondimensional combinations of the variab

7、les (called dimensionless product or dimensionless groups),Only one dependent and one independent variable Easy to set up experiments to determine dependency Easy to present results (one graph),8,Plot of PressureDrop Data Using ,dimensionless product or dimensionless groups,Pumping Power,The h-s dia

8、gram is usually used to calculate the net work and efficiency of RC. This process can be described as follows.,10,7.2 Buckingham Pi Theorem_1,A fundamental question we must answer is how many dimensionless products are required to replace the original list of variables ? The answer to this question

9、is supplied by the basic theorem of dimensional analysis that states If an equation involving k variables is dimensionally homogeneous, it can be reduced to a relationship among k-r independent dimensionless products, where r is the minimum number of reference dimensions required to describe the var

10、iables.,Buckingham Pi Theorem,Pi terms,11,Buckingham Pi Theorem_2,Given a physical problem in which the dependent variable is a function of k-1 independent variables. Mathematically, we can express the functional relationship in the equivalent form,Where g is an unspecified function, different from

11、f.,12,Buckingham Pi Theorem_3,The Buckingham Pi theorem states that: Given a relation among k variables of the form The k variables may be grouped into k-r independent dimensionless products, or terms, expressible in functional form by,r ? ?,13,Buckingham Pi Theorem_4,The number r is usually, but no

12、t always, equal to the minimum number of independent dimensions required to specify the dimensions of all the parameters. Usually the reference dimensions required to describe the variables will be the basic dimensions M, L, and T or F, L, and T. The theorem does not predict the functional form of o

13、r . The functional relation among the independent dimensionless products must be determined experimentally. The k-r dimensionless products terms obtained from the procedure are independent.,14,Buckingham Pi Theorem_5,A term is not independent if it can be obtained from a product or quotient of the o

14、ther dimensionless products of the problem. For example, if then neither 5 nor 6 is independent of the other dimensionless products or dimensionless groups.,15,7.3 Determination of Pi Terms_1,Several methods can be used to form the dimensionless products, or pi term, that arise in a dimensional anal

15、ysis. The method we will describe in detail is called the METHOD of repeating variables. Regardless of the method to be used to determine the dimensionless products, one begins by listing all dimensional variables that are known (or believed) to affect the given flow phenomenon. Eight steps listed b

16、elow outline a recommended procedure for determining the terms.,16,Determination of Pi Terms_2,Step 1 List all the variables. 1 List all the dimensional variables involved. Keep the number of variables to a minimum, so that we can minimize the amount of laboratory work. All variables must be indepen

17、dent. For example, if the cross-sectional area of a pipe is an important variable, either the area or the pipe diameter could be used, but not both, since they are obviously not independent.,=g, that is, , and g are not independent.,17,Determination of Pi Terms_3,Step 1 List all the variables. 2 Let

18、 k be the number of variables. Example: For pressure drop per unit length, k=5. (All variables are p, D, and V ),18,Determination of Pi Terms_4,Step 2 Express each of the variables in terms of basic dimensions. Find the number of reference dimensions. Select a set of fundamental (primary) dimensions

19、. For example: MLT, or FLT. Example: For pressure drop per unit length , we choose FLT.,r=3,19,Determination of Pi Terms_5,Step 3 Determine the required number of pi terms. Let k be the number of variables in the problem. Let r be the number of reference dimensions (primary dimensions) required to d

20、escribe these variables. The number of pi terms is k-r Example: For pressure drop per unit length k=5, r = 3, the number of pi terms is k-r=5-3=2.,20,Determination of Pi Terms_6,Step 4 Select a number of repeating variables, where the number required is equal to the number of reference dimensions. S

21、elect a set of r dimensional variables that includes all the primary dimensions ( repeating variables). These repeating variables will all be combined with each of the remaining parameters. No repeating variables should have dimensions that are power of the dimensions of another repeating variable.

22、Example: For pressure drop per unit length ( r = 3) select , V, D.,21,Determination of Pi Terms_7,Step 5 Form a pi term by multiplying one of the nonrepeating variables by the product of the repeating variables, each raised to an exponent that will make the combination dimensionless. 1 Set up dimens

23、ional equations, combining the variables selected in Step 4 with each of the other variables (nonrepeating variables) in turn, to form dimensionless groups or dimensionless product. There will be k r equations. Example: For pressure drop per unit length,22,Determination of Pi Terms_8,Step 5 (Continu

24、ed) 2,Determination of Pi Terms_9,Step 6 Repeat Step 5 for each of the remaining nonrepeating variables.,24,Determination of Pi Terms_10,Step 7 Check all the resulting pi terms to make sure they are dimensionless. Check to see that each group obtained is dimensionless. Example: For pressure drop per

25、 unit length .,24,25,Determination of Pi Terms_11,Step 8 Express the final form as a relationship among the pi terms, and think about what is means. Express the result of the dimensional analysis. Example: For pressure drop per unit length .,Dimensional analysis will not provide the form of the func

26、tion. The function can only be obtained from a suitable set of experiments.,26,Determination of Pi Terms_12,The pi terms can be rearranged. For example, 2, could be expressed as,27,Ex 7.1 Method of Repeating Variables,A thin rectangular plate having a width w and a height h is located so that it is

27、normal to a moving stream of fluid. Assume that the drag, D, that the fluid exerts on the plate is a function of w and h, the fluid viscosity, ,and , respectively, and the velocity, V, of the fluid approaching the plate. Determine a suitable set of pi terms to study this problem experimentally.,28,E

28、xample 7.1 Solution1/5,Drag force on a PLATE Step 1:List all the dimensional variables involved. D,w,h, ,V k=6 dimensional parameters. Step 2:Select primary dimensions M,L, and T. Express each of the variables in terms of basic dimensions,29,Example 7.1 Solution2/5,Step 3: Determine the required num

29、ber of pi terms. k-r=6-3=3 Step 4:Select repeating variables w,V,. Step 56:combining the repeating variables with each of the other variables in turn, to form dimensionless groups or dimensionless products.,30,Example 7.1 Solution3/5,31,Example 7.1 Solution4/5,32,Example 7.1 Solution5/5,Step 7: Chec

30、k all the resulting pi terms to make sure they are dimensionless. Step 8: Express the final form as a relationship among the pi terms.,The functional relationship is,33,7.4 Selection of Variables_1,One of the most important, and difficult, steps in applying dimensional analysis to any given problem

31、is the selection of the variables that are involved. There is no simple procedure whereby the variable can be easily identified. Generally, one must rely on a good understanding of the phenomenon involved and the governing physical laws. If extraneous variables are included, then too many pi terms a

32、ppear in the final solution, and it may be difficult, time consuming, and expensive to eliminate these experimentally.,34,Selection of Variables_2,If important variables are omitted, then an incorrect result will be obtained; and again, this may prove to be costly and difficult to ascertain. Most en

33、gineering problems involve certain simplifying assumptions that have an influence on the variables to be considered. Usually we wish to keep the problems as simple as possible, perhaps even if some accuracy is sacrificed,35,Selection of Variables_3,A suitable balance between simplicity and accuracy

34、is an desirable goal. Variables can be classified into three general group: Geometry: lengths and angles. Material Properties: relate the external effects and the responses. External Effects: produce, or tend to produce, a change in the system. Such as force, pressure, velocity, or gravity.,36,Selec

35、tion of Variables_4,Points should be considered in the selection of variables: Clearly define the problem. Whats the main variable of interest? Consider the basic laws that govern the phenomenon. Start the variable selection process by grouping the variables into three broad classes. Consider other

36、variables that may not fall into one the three categories. For example, time and time dependent variables. Be sure to include all quantities that may be held constant (e.g., g). Make sure that all variables are independent. Look for relationships among subsets of the variables.,37,Determination of R

37、eference Dimension 1/3,When to determine the number of pi terms, it is important to know how many reference dimensions are required to describe the variables. In fluid mechanics, the required number of reference dimensions is three, but in some problems only one or two are required. In some problems

38、, we occasionally find the number of reference dimensions needed to describe all variables is smaller than the number of basic dimensions. Illustrated in Example 7.2.,38,Ex 7.2 Determination of Pi Terms,An open, cylindrical tank having a diameter D is supported around its bottom circumference and is

39、 filled to a depth h with a liquid having a specific weight . The vertical deflection, , of the center of the bottom is a function of D, h, d, , and E, where d is the thickness of the bottom and E is the modulus of elasticity of the bottom material. Perform a dimensional analysis of this problem.,39

40、,Example 7.2 Solution1/3,The vertical deflection,For F,L,T. Pi terms=6-2=4 For M,L,T Pi terms=6-3=3,40,Example 7.2 Solution2/3,For F,L,T system, Pi terms=6-2=4,D and are selected as repeating variables,41,Example 7.2 Solution3/3,For M,L,T system, Pi terms=6-3=3 ?,A closer look at the dimensions of t

41、he variables listed reveal that only two reference dimensions, L and MT-2 are required.,42,Uniqueness of Pi Terms_1,The Pi terms obtained depend on the somewhat arbitrary selection of repeating variables. For example, in the problem of studying the pressure drop in a pipe.,Selecting D,V, and as repe

42、ating variables:,Selecting D,V, and as repeating variables:,43,Uniqueness of Pi Terms_2,Both are correct, and both would lead to the same final equation for the pressure drop. There is not a unique set of pi terms which arises from a dimensional analysis. The functions 1 and 2 are will be different

43、because the dependent pi terms are different for the two relationships.,44,Uniqueness of Pi Terms_3,EXAMPLE,Form a new pi term,All are correct,45,Uniqueness of Pi Terms_4,Selecting D,V, and as repeating variables:,46,7.5 Determination of Pi Terms by Inspection_1,This method provides a step-by-step p

44、rocedure that if executed properly will provide a correct and complete set of pi terms. Since the only restrictions placed on the pi terms are that they be (1) correct in number, (2) dimensionless,and (3) independent, it is possible to simply form the pi terms by inspection,without resorting to the

45、more formal procedure.,47,One possibility is to first divide by so that To eliminate the dependence on T, we can divide by V2 so Pi terms can be formed by inspection by simply making use of the fact that each pi term must be dimensionless.,Determination of Pi Terms by Inspection_2,48,Finally, to mak

46、e the combination dimensionless we by D so that Thus And, therefore,Determination of Pi Terms by Inspection_3,49,If 2 can be formed by a combination of say 3 , 4, and 5 such as then 2 is not an independent pi term.,Determination of Pi Terms by Inspection_4,50,7.6 Common Dimensionless Groups,A list o

47、f variables that commonly arise in fluid mechanical problems. Possible to provide a physical interpretation to the dimensionless groups which can be helpful in assessing their influence in a particular application.,51,Froude Number 1/2,In honor of William Froude (18101879), a British civil engineer,

48、 mathematician, and naval architect who pioneered the use of towing tanks for the study of ship design. Froude number is the ratio of the forces due to the acceleration of a fluid particles (inertial force) to the force due to gravity (gravity forces). Froude number is significant for flows with fre

49、e surface effects. Froude number less than unity indicate subcritical flow and values greater than unity indicate supercritical flow.,52,Froude Number 2/2,53,Reynolds Number 1/2,In honor of Osborne Reynolds (18421912), the British engineer who first demonstrated that this combination of variables co

50、uld be used as a criterion to distinguish between laminar and turbulent flow. The Reynolds number is a measure of the ration of the inertia forces to viscous forces. If the Reynolds number is small (Re1), this is an indication that the viscous forces are dominant in the problem, and it may be possib

51、le to neglect the inertial effects; that is, the density of the fluid will no be an important variable.,54,Reynolds Number 2/2,Flows with very small Reynolds numbers are commonly referred to as “creeping flows”. For large Reynolds number flow, the viscous effects are small relative to inertial effec

52、ts and for these cases it may be possible to neglect the effect of viscosity and consider the problem as one involving a “nonviscous” fluid. Flows with “large” Reynolds number generally are turbulent. Flows in which the inertia forces are “small” compared with the viscous forces are characteristical

53、ly laminar flows.,55,Euler number,In honor of Leonhard Euler (17071783), a famous Swiss mathematician who pioneered work on the relationship between pressure and flow. Eulers number is the ratio of pressure force to inertia forces. It is often called the pressure coefficient, Cp.,56,Cavitation Numbe

54、r,For problems in which cavitation is of concern, the dimensionless group is commonly used, where pv is the vapor pressure and pr is some reference pressure. The cavitation number is used in the study of cavitation phenomena. The smaller the cavitation number, the more likely cavitation is to occur.

55、,57,Weber Number 1/2,Named after Moritz Weber (18711951), a German professor of naval mechanics who was instrumental in formalizing the general use of common dimensionless groups as a basis for similitude studies. Weber number is important in problem in which there is an interface between two fluids

56、. In this situation the surface tension may play an important role in the phenomenon of interest.,58,Weber Number 2/2,Weber number is the ratio of inertia forces to surface tension forces. Common examples of problems in which Weber number may be important include the flow of thin film of liquid, or

57、the formation of droplets or bubbles. The flow of water in a river is not affected significantly by surface tension, since inertial and gravitational effects are dominant (We1).,59,7.7 Correlation of Experimental Data,Dimensional analysis only provides the dimensionless groups describing the phenome

58、non, and not the specific relationship between the groups. To determine this relationship, suitable experimental data must be obtained. The degree of difficulty depends on the number of pi terms.,60,Problems with One Pi Term,The functional relationship for one Pi term.,where C is a constant. The val

59、ue of the constant must still be determined by experiment.,61,Ex 7.3 Flow with One Pi Term,Assume that the drag, D, acting on a spherical particle that falls very slowly through a viscous fluid is a function of the particle diameter, d, the particle velocity, V, and the fluid viscosity, . Determine, with the aid the dimensional analysis, how the drag depends on the particle velocity.,62,Example 7.3 Solution,The drag,For a given particle and fluids, the drag varies directly with the velocity,63,Problems with Two or More Pi Term_1,Problems with two