1. DIMENSIONAL ANALYSIS

2. Dimensional Analysis
Dimensions and Units
P Theorem

3. What is Dimensional Analysis?
Have you ever used a map?
Since the map is a small-scale representation of a large area, there is a scale that you can use to convert from small-scale units to large-scale units—for example, going from inches to miles or from cm to km.

4. What is Dimensional Analysis?
Ex: 3 cm = 50 km

5. What is Dimensional Analysis?
Have you ever been to a foreign country?
One of the most important things to do when visiting another country is to exchange currency.
For example, one United States dollar equals 0.39 Omani Ryals.

6. What is Dimensional Analysis?
Whenever you use a map or exchange currency, you are utilizing the scientific method of dimensional analysis.

7. What is Dimensional Analysis?
Dimensional analysis is a problem-solving method that uses the idea that any number or expression can be multiplied by another one without changing its value.
It is used to go from one unit to another.

8. How Does Dimensional Analysis Work?
A conversion factor, or a fraction that is equal to one, is used, along with what you’re given, to determine what the new unit will be.

9. How Does Dimensional Analysis Work?
In our previous discussions, you could say that 3 cm equals 50 km on the map
Or:
that $1 equals 0.39 Omani Ryal (OR).

10. How Does Dimensional Analysis Work?
If we write these expressions mathematically, they would look like
3 cm = 50 km
$1 = 0.39 OR

11. Examples of Conversions
60 s = 1 min
60 min = 1 h
24 h = 1 day

12. Examples of Conversions
You can write any conversion as a fraction.
Be careful how you write that fraction.
For example, you can write
60 s = 1 min
as:

13. Primary purposes of dimensional analysis
To generate nondimensional parameters that help in the design of experiments (physical and/or numerical) and in reporting of results
To obtain scaling laws so that prototype performance can be predicted from model performance.
To predict trends in the relationship between parameters.

14. Dimensional Analysis Example: Pressure drop per unit length
Pressure drop per unit length depends on FOUR parameters: Diameter (D); speed (V); fluid density (ρ); fluid viscosity (µ)
Difficult to know how to set up experiments to determine dependencies
Difficult to know how to present results (four graphs?)

15. Plot of Pressure Drop Data Using …

16. Dimensional Analysis Example: Pressure drop per unit length
Only one dependent and one independent variable
Easy to set up experiments to determine dependency
Easy to present results (one graph)

17. What do we gain by using Dimensional Analysis?
Any consistent set of units will work
We don’t have to conduct an experiment on every single size and type of pipe at every velocity
Our results will even work for different fluids

18. Buckingham Pi Theorem 1/5
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 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 variables.
Buckingham Pi Theorem
Pi terms

19. Buckingham Pi Theorem 2/5
Given a physical problem in which the dependent parameter is a function of k-1 independent parameters.
Mathematically, we can express the functional relationship in the equivalent form
Where g is an unspecified function, different from f.

20. Buckingham Pi Theorem 3/5
The Buckingham Pi theorem states that: Given a relation among k parameters of the form
The k parameters may be grouped into k-r independent dimensionless ratios, or Π parameters, expressible in functional form by

21. Buckingham Pi Theorem 4/5
Where 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 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 Π parameters must be determined experimentally.
The k-r dimensionless Π parameters obtained from the procedure are independent.

22. Buckingham Pi Theorem 5/5
A Π parameter is not independent if it can be obtained from a product or quotient of the other parameters of the problem. For example, if
then neither Π5 nor Π6 is independent of the other dimensionless parameters.

23. Determination of Pi Terms1/11
Regardless of the method to be used to determine the dimensionless parameters, one begins by listing all dimensional parameters that are known (or believed) to affect the given flow phenomenon.
Do not be afraid to include all the parameters that you feel are important. If the parameter is extraneous, an extra Π parameter may result, but experiments will later show that it may be eliminated from consideration.
Six steps listed below outline a recommended procedure for determining the Π parameter.

24. Determination of Pi Terms 2/11
Step 1
List all the dimensional parameters involved.
Let “k” be the number of parameters.
Example: For pressure drop per unit length pl, D,,,V and k=5

25. Determination of Pi Terms 3/11
Step 2
Select a set of fundamental (primary) dimensions.
For example: MLT, or FLT.
Example: For pressure drop per unit length , we choose FLT.

26. Determination of Pi Terms 4/11
Step 3
List the dimensions of all parameters in terms of primary dimensions.
Let “r” be the number of primary dimensions.
Example: For pressure drop per unit length r = 3.

27. Determination of Pi Terms 5/11
Step 4
Select a number of repeating variables, where the number required is equal to the number of reference dimensions.
These repeating parameters will all be combined with each of the remaining parameters. No repeating parameters should have dimensions that are power of the dimensions of another repeating parameter.
Example: For pressure drop per unit length ( r = 3) select ρ , V, D.

28. Determination of Pi Terms 6/11
Step 5
Set up dimensional equations, combining the parameters selected in Step 4 with each of the other parameters in turn, to form dimensionless groups.
There will be k – r equations.
Example: For pressure drop per unit length .

29. Determination of Pi Terms 7/11
Step 5 (Continued)
Example: For pressure drop per unit length .

30. Determination of Pi Terms 8/11
Step 5 (Continued)
Set up dimensional equations, combining the parameters selected in Step 4 with each of the other parameters in turn, to form dimensionless groups.
There will be k – r equations.
Example: For pressure drop per unit length .

31. Determination of Pi Terms 9/11
Step 5 (Continued)
Example: For pressure drop per unit length .

32. Determination of Pi Terms 10/11
Step 6
Check to see that each group obtained is dimensionless.
Example: For pressure drop per unit length .

33. Determination of Pi Terms 11/11
Step 7
Express the result of the dimensional analysis.
Example: For pressure drop per unit length .

34. EXAMPLE-I 1/4 Drag force on a PLATE
Step 1:List all the dimensional parameters involved. D,w,h, ρ,μ,V k=6 dimensional parameters.
Step 2:Select primary dimensions M,L, and T.
Step 3:List the dimensions of all parameters in terms of primary dimensions. r=3 primary dimensions

35. EXAMPLE-I 2/4
Step 4:Select repeating parameters w,V,. r=3 repeating parameters.
Step 5:combining the repeating parameters with each of the other parameters in turn, to form dimensionless groups.

36. EXAMPLE-I 3/4

37. EXAMPLE-I 4/4
Step 6: Check to see that each group obtained is dimensionless
The functional relationship is

38. EXAMPLE-II 1/3 Drag force on a smooth sphere
Step 1:List all the dimensional parameters involved. Df,V,D, ρ,μ k=5 dimensional parameters.
Step 2:Select primary dimensions M,L, and T.
Step 3:List the dimensions of all parameters in terms of primary dimensions. r=3 primary dimensions

39. EXAMPLE-II 2/3
Step 4:Select repeating parameters ρ,V,D. r=3 repeating parameters.
Step 5:combining the repeating parameters with each of the other parameters in turn, to form dimensionless groups.

40. EXAMPLE-II 3/3
Step 6: Check to see that each group obtained is dimensionless
The functional relationship is

41. Selection of Variables 1/3
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 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 appear in the final solution, and it may be difficult, time consuming, and expensive to eliminate these experimentally. If important variables are omitted, then an incorrect result will be obtained; and again, this may prove to be costly and difficult to ascertain.

42. Selection of Variables 2/3
Most engineering 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.
A suitable balance between simplicity and accuracy 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.

43. Selection of Variables 3/3
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 variable selection process by grouping the variables into three broad classes (geometry, material, external force)
Consider other 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.

44. Determination of Reference Dimension 1/3
Typically, in fluid mechanics, the required number of reference dimensions is three, but in some problems only one or two are required.
Example 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 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.
For F,L,T. Pi terms=6-2=4
For M,L,T Pi terms=6-3=3

45. Determination of Reference Dimension 2/3
For F,L,T system, Pi terms=6-2=4
For M,L,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.

46. Determination of Reference Dimension 3/3
EXAMPLE
MLT system
FLT system
Pi term=4-2=2
Pi term=4-3=1
Dimensional matrix
MLT system
FLT system
Rank=2
Pi term=4-2=2

47. Uniqueness of Pi Terms1/2
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 repeating variables:
Selecting D,V, and  as repeating variables:
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.

48. Uniqueness of Pi Terms2/2
EXAMPLE
Form a new pi term or

49. Common Dimensionless Groups

50. Reynolds Number
Discovered by Osborne Reynolds, the British engineer, in the 1880s.
This parameter is a criterion by which the tube flow may be categorized as laminar or turbulent regime.
The Reynolds number is the ratio of inertia forces to viscous forces.
Flows with “large” Reynolds number generally are turbulent. Flows in which the inertia forces are “small” compared with the viscous forces are characteristically laminar flows.

51. Euler number
This ratio is named after Leonhard Euler, the Swiss mathematician. Euler was the first to recognize the role of pressure in fluid mechanics.
Euler’s number is the ratio of pressure force to inertia forces. It is often called the pressure coefficient, Cp.

52. Cavitation Number
Where p is the pressure in the liquid stream, and pv is the liquid vapor pressure at the test temperature.
The cavitation number is used in the study of cavitation phenomena.
The smaller the cavitation number, the more likely cavitation is to occur.

53. Froude Number Discovered by Froude and this son, Robert Edmund Froude.
Froude number is the ratio of inertia forces to gravity forces.
Froude number is significant for flows with free surface effects since gravity principally affects this type of flow.
Typical problems would include the study of the flow of water around ships or flow through rivers or open conduits
Froude number less than unity indicate subcritical flow and values greater than unity indicate supercritical flow.

54. Weber Number
Weber number is the ratio of inertia forces to surface tension forces.
Weber number is indicative of the existence of, and frequency of, capillary waves at a free surface.
Common examples of problems in which this parameter may be important include the flow of thin films of liquid, or in the formation of droplets or bubbles.

55. Mach Number
Introduced by Ernst Mach, the Austrian physicist, in the 1870s.
Mach number is the ratio of inertia forces to forces due to compressibility, where V is the flow speed and c is the local sonic speed.
Mach number is a key parameter that characterizes compressibility effects in a flow.
For truly incompressible flow, c=∞ so that M=0.

56. Correlation of Experimental Data
Problems with one Pi term (c is constant)
Problems with two or more terms.

57. Example: Valve Coefficient
The pressure coefficient, , for a 600-mm-diameter valve is to be determined for 5 ºC water at a maximum velocity of 2.5 m/s. The model is a 60-mm-diameter valve operating with water at 5 ºC. What water velocity is needed?

58. Example: Valve Coefficient
Note: roughness height should scale!
Reynolds similarity
ν = 1.52 x 10-6 m2/s
Vm = 25 m/s

59. Example: Valve Coefficient (Reduce Vm?)
What could we do to reduce the velocity in the model and still get the same high Reynolds number?
Decrease kinematic viscosity:
Use a different fluid
Use water at a higher temperature

60. Example: Valve Coefficient
Change model fluid to water at 80 ºC
νm = ______________
0.367 x 10-6 m2/s
νp = ______________
1.52 x 10-6 m2/s
Vm = 6 m/s