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

steady, incompressible, Newtonian fluid

∆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    ⎝       ⎠    ⎝       ⎠




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