# Ch7 Similitude, Dimensional Analysis, and Modeling by nak14542

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```									Ch7 Similitude, Dimensional Analysis, and Modeling
Experimentally obtained Data

A. Dimensional Analysis

Pipeline

∆Pl     Pressure force against friction
D       pipe diameter

ρ       fluid density

µ       fluid viscosity

V       mean velocity

1
∆Pl = f (D, ρ , µ , V)

(a) Measured      fixed     varied

Q1: hard to carry out: vary ρ        keeping µ = constant
Q2: validation (for any pipe system)?

Simpler approach: dimensionless products or dimensionless groups.

ρVD      D‧ Pl ~ ρV 2
Re =
µ
∆Pl ⋅ D    ⎛ ρVD ⎞
∴           = φ⎜
⎜ µ ⎟ ⎟
ρV  2
⎝     ⎠
5 variables → two variables (non-dimension)

2
∆P1 ⋅ D
Experiments to be performed, vary Re, measure
ρV 2
a single, universal curve.
if the flow is self similar

-2        -2 -1             -4       -2     -1
F = M L T ; ∆P = FL L ; D = L; ρ =FL T; µ= FL T; V=L T
∆Pl ⋅ D               ρVD
= F 0 L0T 0 ;     = F 0 L0T 0
ρV  2
µ

B. Buckingham PI Theorem
例               數                           參數                       理論

數                              參數
If an equation involving K variables is dimensionally homogeneous, it
can be reduced to a relationship among K-             independent dimensionless

products, where            is the minimum number of reference dimension
required to describe the variables.

“Pi term”      “Buckingham Pi theorem”.u1 = f (u2, u3, …… ,uk )

p1 = φ (p2, p3, …… ,pk-r)               了r

3
r is determined by the minimum number of reference dimensions required
to describe the original list of variables.

M, L, T           3
F, L, T           3        Proof. Hard ref 1 - 15 P.404
L, T              2        A little complicated
M/T 2 , L         2        to determine π terms.
L                 1

C. Determination of PI terms

Method of repeating variables.          (Appendix D)
An alternate method.

Step1         List all variables which are independent.             (K)
“Most difficult one”.                                   5
“Geometry”;          “Fluid properties”; “external effects”

(pipe flow) D, L                                            Pl

It is important that all variables be independent.
(e.g. Area or Diameter not independent use only one.)

(e.g.       or   =   g not independent use only one.)

Step 2      Basic dimensions for each variable.           (r)
M, L, T or F, L, T                              3
Step 3      # of Pi terms (K–r)                            2 (for pipe)
Step 4      Repeating variables       D, V, ρ       u1, u2, u3)

Don’t use the dependent variables as repeating variable. Choose
simplest, dimensionally.

4
Step 5   π i = ui u1ai u 2 u 3ci
bi
( pipe flow )

term = nonrepeating variable ´ the product of repeating

∆Pi D
variables. π 1 = ∆Pi D aV b ρ c =
ρV 2

∆Pl      Da       Vb               ρc               0 0 0
=FLT
FL−3 La (LT −1 )               (FL      T2)
b        −4     c

F : 1+ C = 0                     C = 1 L : –3 + a + b – 4c = 0      a=1

T : –b+2c=0                      b = –2

Step 6   Repeat step 5. for other nonrepeating variables. Step 7
∆Pl ⋅ D
Dimensionless check π 1 =                      = M 0 L0 T 0 = F 0 L0 T 0   (pipe flow)
ρV  2

Step 8   Find form and what it means.
π 1 = φ (π 2 ,π 3 ,.....π k −r ) 1

2 dimensionless variables; rather than 5 variables.

Example 7.1.

Step 1    D = f (w, h, µ , ρ ,V )

Step 2   M, L, T           r=3
Step 3   K–r = 3             # of Pi terms
Step 4   Repeating Variables

Step 5   π 1 = Dw aV b ρ c → a = −2, b = −2, c = −1
5
h
Step 6   π2 =
w
Step 7 Dimensionless check
D      ~⎛ h µ ⎞       ⎛ w wVρ ⎞
Step 8            =φ ⎜ ,     ⎟ = φ⎜ ,
⎜ w wVρ ⎟    ⎜h µ ⎟  ⎟
w 2V 2 ρ    ⎝       ⎠    ⎝       ⎠

6

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