Chapter06

Mechanics of Fluids

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103 CHAPTER 6 Dimensional Analysis and Similitude 6.1 g V V g p g z g V V g p g z 1 2 1 2 1 1 1 2 2 2 2 2 2 2 + + = + + r r . 1 2 1 2 1 1 2 1 1 2 2 2 1 2 2 1 2 2 1 2 + + = + + p V gz V V V p V gz V r r or 1 2 1 2 1 1 2 1 1 2 2 2 2 2 2 2 2 2 1 2 + + = + + p V gz V p V gz V V V r r 6.2 a) [ ] & . . m FT L = = = kg s N s m s N s m 2 b) [ ] p F L = N m 2 . 2 c) [ ] r = = = kg m N s m m N s m 3 2 3 2 4 . . FT L 2 4 d) [ ] m = N s m 2 . FT L 2 e) [ ] W FL = N m. f) [ ] & . W FL T = N m s g) [ ] s = N/m . F L 6.3 ( A ) The dimensions on the variables are as follows: 22 2 3 / [] [ ] , [ ] , [ ] ,[] LL M L M L T ML W F V M d L pV TT T T L LT = × = = = == & First, eliminate T by dividing W & by p . That leaves T in the denominator so divide by V leaving L 2 in the numerator. Then divide by d 2 . That provides 2 W pVd p = & 6.4 1 2 52 , , . T er f RRR RR m r w rw  ∴=   l
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104 6.5 ( A ) ( , , , , ) . The units on the variables on the rhs are as follows: V fdlg wm = 1 2 [ ] , [ ] [ ] [ ] [] L ML d LlL gT T T - == === Because mass M occurs in only one term, it cannot enter the relationship. 6.6 [ ] [ ] [ ] [ ] V f V L T L M L M LT = = = = = ( , , ). , , , . l l r m r m 3 There is one p - term: p r m 1 = V l . = = = = p p r m 1 1 2 0 f V C ( ) , Re Const. or Const. l 6.7 [ ] [ ] [ ] [ ] V f d V L T M T M L d L = = = = = ( , , ). , , , . s r s r 2 3 = = = = p s r p p s r 1 2 1 1 2 0 2 V d f C V d C . ( ) , onst. or We = Const. 6.8 [ ] [ ] [ ] [ ] V f H g m V L T g L T m M H L = = = = = ( , , , , , . 2 = = = p p 1 0 2 1 gHm V C V gH C . . / . 6.9 [ ] [ ] [ ] [ ] [ ] [ ] V f H g m V L T H L g L T m M M L M LT = = = = = = = ( , , , , ). , , , , , . r m r m 2 3 Choose repeating variables H g , , r (select ones with simple dimensions-we couldn’t select V, H, and g since M is not contained in any of those terms): p r p r p m r 1 2 3 1 1 1 2 2 2 3 3 3 = = = VH g mH g H g a b c a b c a b c , , . = = = = = p r p r p m r m r 1 0 2 3 3 3 2 3 V g H V gH m H gH gH . . . / = V gH f m H gH 1 3 3 r m r , Note: The above dimensionless groups are formed by observation: simply combine the dimensions so that the p - term is dimensionless. We could have set up equations similar to those of Eq. 6.2.11 and solved for 11 1 22 2 333 , , and , , c and , , . ab c a b abc But the method of observation is usually successful. 6.10 [ ] [ ] [ ] [ ] [ ] F f d V F ML T d L V L T M LT M L D D = = = = = = ( , , , , ). , , , , . l m r m r 2 3
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105 p r p r p m r 1 2 3 1 1 1 2 2 2 3 3 3 = = = F V dV V D a b c b c a a b c l l l , , . = = = p r p p m r 1 2 2 2 3 F V d V D l l l , , . = F V f d V D r m r l l l 2 2 1 , . We could write p p p p p 1 2 2 2 2 3 2 1 = f , or F d V f d dV D r m r 2 2 2 = l , This is equivalent to the above. Either functional form must be determined by experimentation. 6.11 [ ] [ ] [ ] [ ] [ ] F f d V F ML T d L V L T M LT M L D D = = = = = = ( , , , , ). , , , , . l m r m r 2 3 p m p m p r m 1 2 3 1 1 1 2 2 2 3 3 3 = = = F d V d V d V D a b c a b c a b c , , . l By observation we have 1 23 , . D F Vd V dd r p pp mm = == l = F Vd f d Vd D m r m 1 l , . Rather than p p p 1 1 2 3 = f ( , ), we could write p p p p 1 3 2 2 3 1 = f , , an acceptable form: F V d f d Vd D r m r 2 2 2 = l , 6.12 [ ] [ ] [ ] [ ] [ ] [ ] h f d g h L M T d L M L T g L T = = = = = = = ( , , , , ). , , , , , . s g b s g b 2 2 2 2 1 Select d g , , g as repeating variables.
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Chapter06 - CHAPTER 6 Dimensional Analysis and Similitude...

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