流體力學chapter-7-dimensional-analysis

1、Chapter 7 Dimensional AnalysisModeling, and Similitude,,2,MAIN TOPICS,7.1 Dimensional Analysis7.2 Buckingham Pi Theorem7.3 Determination of Pi Terms7.4 Comments about Dimensional Analysis7.5 Determination of Pi Ter

2、ms by Inspection7.6 Common Dimensionless Groups in Fluid Mechanics7.7 Correlation of Experimental Data7.8 Modeling and Similitude7.9 Typical Model Studies7.10 Similitude Based on Governing Differential Equation,3,7.

3、1 Dimensional Analysis_1,A typical fluid mechanics problem in which experimentation is required, consider the steady flow of an incompressible Newtonian fluid through a long, smooth-walled, horizontal, circular pipe. An

4、 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,T

5、he first step in the planning 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 le

6、ngth depends on FOUR variables:piping diameter (D); velocity (V); fluid density (ρ); fluid viscosity (m),5,Dimensional Analysis_3,To perform the experiments in a meaningful and systematic manner, it would be necessary t

7、o 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 between the pressure drop and the vari

8、ous 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

9、 two nondimensional combinations of the variables (called dimensionless product or dimensionless groups),Only one dependent and one independent variableEasy to set up experiments to determine dependencyEasy to present

10、results (one graph),8,Plot of PressureDrop Data Using …,dimensionless product or dimensionless groups,,Pumping Power,The h-s diagram is usually used to calculate the net work and efficiency of RC. This process can be d

11、escribed 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 is suppli

12、ed by the basic theorem of dimensional analysis that statesIf an equation involving k variables is dimensionally homogeneous, it can be reduced to a relationship among k-r independent dimensionless products, where r is

13、the minimum number of reference dimensions required to describe the variables.,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 ind

14、ependent variables.Mathematically, we can express the functional relationship in the equivalent form,Where g is an unspecified function, different from f.,12,Buckingham Pi Theorem_3,The Buckingham Pi theorem states th

15、at: Given a relation among k variables of the form The k variables may be grouped into k-r independent dimensionless products, or Π terms, expressible in functional form by,r ?? Π??,13,Buc

16、kingham Pi Theorem_4,The number r is usually, but not always, equal to the minimum number of independent dimensions required to specify the dimensions of all the parameters. Usually the reference dimensions required to

17、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 ? or ? . The functional relation among the independent dimensionless products Π must be

18、determined experimentally.The k-r dimensionless products Π terms obtained from the procedure are independent.,14,Buckingham Pi Theorem_5,A Π term is not independent if it can be obtained from a product or quotient of th

19、e other dimensionless products of the problem. For example, if then neither Π5 nor Π6 is independent of the other dimensionless products or dimensionless groups.,15,7.3 Determination of Pi Terms_1,Several method

20、s can be used to form the dimensionless products, or pi term, that arise in a dimensional analysis.The method we will describe in detail is called the METHOD of repeating variables.Regardless of the method to be used t

21、o 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 below outline a recommended procedure for deter

22、mining the Π terms.,16,Determination of Pi Terms_2,Step 1 List all the variables. 1List all the dimensional variables involved.Keep the number of variables to a minimum, so that we can minimize the amount of laborator

23、y work.All variables must be independent. 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 indep

24、endent.,γ=ρ×g, that is, γ,ρ, and g are not independent.,17,Determination of Pi Terms_3,Step 1 List all the variables. 2Let k be the number of variables.Example: For pressure drop per unit length, k=5. (All varia

25、bles 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.

26、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.

27、Let r be the number of reference dimensions (primary dimensions) required to describe these variables.The number of pi terms is k-rExample: For pressure drop per unit length k=5, r = 3, the number of pi terms is k-r=

28、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.Select a set of r dimensional variables that includes all th

29、e 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

30、 repeating variable.Example: For pressure drop per unit length ( r = 3) select ρ , 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 repea

31、ting variables, each raised to an exponent that will make the combination dimensionless. 1Set up dimensional equations, combining the variables selected in Step 4 with each of the other variables (nonrepeating variables

32、) 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 (Continued) 2,,Determination of Pi Terms

33、_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 obtain

34、ed is dimensionless.Example: For pressure drop per 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

35、of the dimensional analysis.Example: For pressure drop per unit length .,Dimensional analysis will not provide the form of the function. The function can only be obtained from a suitable set of experiments.,26,Determi

36、nation of Pi Terms_12,The pi terms can be rearranged. For example, Π2, 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 no

37、rmal 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, µ ,and ρ, respectively, and the velocity, V, of the fluid approaching the

38、plate. Determine a suitable set of pi terms to study this problem experimentally.,28,Example 7.1 Solution1/5,Drag force on a PLATEStep 1:List all the dimensional variables involved. D,w,h, ρ,μ,V k=6 dimensional para

39、meters.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 number of pi terms. k-r=6-3=3Step 4:Select

40、repeating variables w,V,?.Step 5~6: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 Solution

41、4/5,,32,Example 7.1 Solution5/5,Step 7: Check 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 Se

42、lection of Variables_1,One of the most important, and difficult, steps in applying dimensional analysis to any given problem is the selection of the variables that are involved.There is no simple procedure whereby the v

43、ariable 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 appear in the f

44、inal 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 agai

45、n, this may prove to be costly and difficult to ascertain. Most engineering problems involve certain simplifying assumptions that have an influence on the variables to be considered.Usually we wish to keep the problems

46、 as simple as possible, perhaps even if some accuracy is sacrificed,35,Selection of Variables_3,A suitable balance between simplicity and accuracy is an desirable goal.~~~~~Variables can be classified into three general

47、 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 gravi

48、ty.,36,Selection of Variables_4,Points should be considered in the selection of variables:Clearly define the problem. What’s the main variable of interest?Consider the basic laws that govern the phenomenon.Start the v

49、ariable selection process by grouping the variables into three broad classes.Consider other variables that may not fall into one the three categories. For example, time and time dependent variables.Be sure to include a

50、ll 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 Reference Dimension 1/3,When to determine the n

51、umber 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

52、two are required.In some problems, 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 Determin

53、ation of Pi Terms,An open, cylindrical tank having a diameter D is supported around its bottom circumference and is filled to a depth h with a liquid having a specific weight ?. The vertical deflection, ? , of the cente

54、r 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,Example 7.2 Solutio

55、n1/3,The vertical deflection,For F,L,T. Pi terms=6-2=4For 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 γ are selected as repeating variables,,41,Example 7.2 Solution3/3,For M,L

56、,T system, Pi terms=6-3=3 ?,A closer look at the dimensions of the 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 som

57、ewhat arbitrary selection of repeating variables. For example, in the problem of studying the pressure drop in a pipe.,Selecting D,V, and ? as repeating variables:,Selecting D,V, and ? as repeating variables:,,,43,Unique

58、ness 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 Φ1 and Φ2 are will

59、be different 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 repeati

60、ng variables:,46,7.5 Determination of Pi Terms by Inspection_1,This method provides a step-by-step procedure that if executed properly will provide a correct and complete set of pi terms. Since the only restrictions

61、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 more formal procedure.,47,One poss

62、ibility is to first divide by ρ so thatTo 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 dimensionles

63、s.,Determination of Pi Terms by Inspection_2,48,Finally, to make the combination dimensionless we by D so thatThusAnd, therefore,Determination of Pi Terms by Inspection_3,49,If П 2 can be formed by a combinatio

64、n of say П3 , П4, and П5 such as then П2 is not an independent pi term.,Determination of Pi Terms by Inspection_4,50,7.6 Common Dimensionless Groups,A list of variables that commonly arise in fluid mechanical problems.

65、 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 (1810~1879), a B

66、ritish civil engineer, 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 (inerti

67、al force) to the force due to gravity (gravity forces).Froude number is significant for flows with free surface effects.Froude number less than unity indicate subcritical flow and values greater than unity indicate sup

68、ercritical flow.,52,Froude Number 2/2,,,,53,Reynolds Number 1/2,In honor of Osborne Reynolds (1842~1912), the British engineer who first demonstrated that this combination of variables could be used as a criterion to dis

69、tinguish 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 (Re<<1), this is an indication that the viscous f