Fluid Mechanics: Dimensional Analysis: Buckingham Pi Theorem — Transcript

Full transcript

1. 0:04dimensional analysis Buckingham PI
2. 0:06theorem
3. 0:13the Buckingham PI theorem was named
4. 0:15after the American Edgar Buckingham
5. 0:23you
6. 0:27it states that if there are n
7. 0:29dimensional variables in a dimensionally
8. 0:31homogeneous equation described by J
9. 0:33fundamental dimensions they may be
10. 0:35grouped in K equals n minus J
11. 0:37dimensionless groups
12. 0:43each dimensionless group is called pi
13. 0:52and the relation among the dimensionless
14. 0:55groups will be written as follows
15. 0:59function of pi 1 pi 2 pi 3 until pi k
16. 1:03equals 0
17. 1:06or PI one is a function of pi 2 pi 3
18. 1:09until PI K
19. 1:12or pi/2 is a function of PI 1 PI 3 until
20. 1:16PI K
21. 1:18and so on and so forth all these
22. 1:22presentations are similar
23. 1:27this method is also known as method of
24. 1:29repeating variables
25. 1:35as an illustration of the Buckingham PI
26. 1:37theorem let's take the case of the drag
27. 1:39force on a smooth sphere
28. 1:46step1 consists of listing the parameters
29. 1:49in the problem
30. 1:52the drag force on a smooth sphere is
31. 1:55mainly influenced by the sphere diameter
32. 1:57D the fluid velocity V the fluid density
33. 2:00Rho in the fluid viscosity mu
34. 2:05our five dimensional variables in this
35. 2:07case thus N equals five
36. 2:16step2 consists of listing the primary
37. 2:19dimensions of each parameter
38. 2:23to obtain the dimension of a parameter
39. 2:25sometimes it is easier to start by
40. 2:27obtaining its unit
41. 2:30from the unit the translation to the
42. 2:33dimension can be easily done
43. 2:37the unit for drag forces Newton or
44. 2:40kilograms meter per second square
45. 2:45the FM represents the dimension for mass
46. 2:46L for length and T for time then the
47. 2:51dimension for the drag force is ml T
48. 2:53power negative 2
49. 2:57the unit for diameter is meter then it's
50. 3:00dimension is L
51. 3:03for fluid velocity it is meter per
52. 3:05second the dimension is Lt power
53. 3:08negative one
54. 3:11the unit for fluid density is kilograms
55. 3:13per meter cube then it's dimension is ml
56. 3:16power negative 3
57. 3:19and the unit for fluid viscosity mu is
58. 3:22kilograms per meter second hence its
59. 3:25dimension is ml power negative one T
60. 3:27power negative one
61. 3:32here we observe that there are only
62. 3:33three primary dimensions involved in all
63. 3:36parameters which are M L and T thus J
64. 3:39equals three
65. 3:46step3 consists of calculating the number
66. 3:48of dimensionless groups
67. 3:51having N equals 5 and J equals 3 then
68. 3:54the number of dimensionless groups will
69. 3:56be 5 minus 3 or K equals 2
70. 4:01the two dimensionless group are pi-1 and
71. 4:03pi2
72. 4:09step four consists of choosing the
73. 4:11repeating variables
74. 4:14as a hint it is generally advantageous
75. 4:17to choose primary variables that relate
76. 4:19to mass geometry and kinematics
77. 4:23among these parameters the one that is
78. 4:26related to mass is the fluid density Rho
79. 4:32the one which is related to geometry is
80. 4:34the sphere diameter D
81. 4:39and the one which is related to
82. 4:41kinematics is the fluid velocity v
83. 4:48hence the three repeating variables are
84. 4:50row D and V
85. 4:56Step five consists of constructing the
86. 4:58dimensionless groups
87. 5:01each of pi must contains the three
88. 5:04repeating variables
89. 5:07and we add to each pie each of the
90. 5:09remaining variables
91. 5:15pi-1 and pi2 are both dimensionless
92. 5:17groups for this we put powers for each
93. 5:20of the repeating variables
94. 5:23let's put a 1 and a 2 as the power for
95. 5:26row B 1 and B 2 as the power for D and C
96. 5:321 and C 2 is the power for V
97. 5:41so our pies are PI 1 equals Rho power a
98. 5:451 times D power B 1 times V power C 1
99. 5:49times f
100. 5:52pi 2 equals Rho power a 2 times D power
101. 5:55B 2 times V power C 2 times mu
102. 6:00let's solve for pi one
103. 6:05pi-1 is dimensionless group we can say
104. 6:08that it's dimension is empower 0l power
105. 6:100 and t power 0
106. 6:16we replace row with its dimension
107. 6:23deal with its dimension
108. 6:29v with its dimension
109. 6:34and F with its dimension
110. 6:39now let's rearrange this by dimensions
111. 6:48some up the power for each dimension
112. 6:53from here we obtain three equations with
113. 6:55three unknowns the three equations our
114. 6:59equation from the power of M
115. 7:04equation from the power of L
116. 7:08and equation from the power of tea
117. 7:12and the three unknowns are a1 b1 and c1
118. 7:19these equations can be easily solved
119. 7:26giving a1 equals minus one c1 equals -2
120. 7:30and b1 equals -2
121. 7:34let's replace the power coefficients in
122. 7:36the PI one with these values
123. 7:44the final expression for PI 1 is death
124. 7:47over Rho D square v square
125. 7:52now let's work on pi/2
126. 7:56just like what we did with Pi 1 here we
127. 7:59analyze the dimension of this relation
128. 8:00we determinate to B 2 and C 2 in order
129. 8:04to have dimensionless group PI 2
130. 8:10we replace row with its dimension
131. 8:17deal with its dimension
132. 8:21v with its dimension
133. 8:25and new with its dimension
134. 8:29rearranging the power and solving it
135. 8:38you
136. 8:45we obtained a to equals negative 1 c2
137. 8:52equals negative 1
138. 8:55and b2 equals negative 1
139. 9:00hence the expression of pi/2 is mu over
140. 9:03Rho VD
141. 9:09we know that Rho VD over mu is Reynolds
142. 9:12number
143. 9:15hence pi 2 equals 1 over Reynolds number
144. 9:18or Reynolds number power negative 1
145. 9:25finally step six is to write the final
146. 9:28functional relationship
147. 9:31we have both the expression for pi-1 and
148. 9:34pi2
149. 9:36pi-1 equals F over Rho V Square D square
150. 9:40PI 2 equals Reynolds number power
151. 9:43negative 1
152. 9:45by taking the relationship in the form
153. 9:47of PI one is a function of PI to the
154. 9:50final relation is f over Rho V Square D
155. 9:53square is a function of Reynolds number
156. 10:03you