Dimensional Analysis

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(1)409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. A Quick Look  Dimensional Analysis Dimensions and units of measurement The principle of dimensional homogeneity The  theorem Dimensionless form of conservation laws.  Modeling Similitude Applications of modeling aerodynamic drag of vehicles resistance of ships propellers and wind turbines pumps and hydroturbines complex piping system vessel propulsion system Dimensional Analysis. (2) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. Dimensional Analysis  Dimensional Analysis  Analytical procedure to organize empirical information of fluid flow  Relies on dimensional homogeneity  Does not produce new information, but reorders to provide a clearer picture of the phenomenon at hand.  Example  Show that each term has the same dimension V 1  (V   ) V   p  g   2 V t .   .   V  V LT 1   LT 2 , V   V  [ V 2 ]   (LT 1 ) 2 L1  LT 2   T   t  t . 1  [][p] (L1 ) (M L1 T 2 )  LT 2 , [ g ]  LT 2   p     3 ML  . [2 V]  [][ 2 ][V]  (L2 T 1 ) (L2 ) (LT 1 )  LT 2. Dimensional Analysis. (3) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. Principle of Dimensional Homogeneity  Pitot Tube. V. V  f ( h , ,  m , g ). (1). V f ( h, ,  m , g )  gh gh    V  F   gh  m . S.  (2). (3). g.  Eq.3 is a considerable advance over eq.1  Need only run a set of experiments in which  / m is varied while V is fixed at a known value.  We would then find    F     m . 2(m  ) . Dimensional Analysis. h. m. (4) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. The Theorem - Principle  The Equational Form v1  f ( v 2 , v 3 , L , v n ) dependent variable. independent variables.  The principle of dimensional homogeneity → this relationship can be expressed in dimensionless form with fewer (dimensionless) independent variables  But how many fewer dimensionless variables will there be? → the  theorem determines the number. 1  F   2 , 3 , L,  n r  → usually r = d (# of fundamental dimensions) Dimensional Analysis. (5) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. The Theorem - Procedure  List the dimensions of the n physical variables, and determine # of fundamental dimensions of ( M, L, T,  ).  Assuming r = d , select r independent variables containing all the d fundamental dimensions that cannot form a dimensionless  by themselves.. vs , L , vs  r 1  For each of the remaining n - r unselected variables, form a .  i  vi ( vs ) L ( vs  r  1 ) as. where as. [ vi ( vs ) L ( vs  r  1 )  The final formula then takes the form. . i  F  2 ( 3 )2 ,.  3 , L,  n  r. as. . Dimensional Analysis. as.  r 1.  r 1. ] 1. (6) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. The Theorem - Example  Experimental determination of p p  f ( D , ε , V , μ , ρ ). [ D]  L [ ]  L   [  ]  M L1 T 1 [  ]  M L3 .  [p]  M L2 T 2. [V ]  L T 1. n6 M , L ,T  d 3r.  Good idea to pick each variable from a different category, i.e. geometry, fluid and flow  pick: D,  , V among which they have M, L, T.  a b c. 1  p D  V. [ p Daμ b V c ]  (M L2 T 2 ) (L) a (M L1 T 1 ) b (L T 1 ) c  M1 b L2  a  b  c T 2  b  c  M o Lo T o 1  b  0  b  1   2  a  b  c  0  c  1 2bc  0   a2. p D 2  1  V. Dimensional Analysis. (7) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. The Theorem - Example.  D  3   Da  b V c. , . 2 . [  Da b Vc ]  (M L3 ) (L)a (M L1 T 1 ) b (L T 1 )c. Now 1  F (  2 , 3 ) or.  3 . DV  Re .   VD  p D 2   F  , V D   . p D 2 1  is not a familiar dimensionless variable, however, so divide by  3 V 1 p D 2 p D f    2 3  V  V D  V 2. Darcy friction factor.  1  F (  2 ,  3 ) or f  F   V D ,     3 D   /D 64 1 2.51  f  (Re  2300 )   2 . 0 log  Recall that  3.7 Re f Re f  Dimensional Analysis.   (Re  2300)  . (8) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. Dimensional Analysis - Example A solid sphere of radius “a” and density “ s ” is dropped into a container of liquid with  ,  . The sphere descends into the liquid at a final velocity of Vf. . l. l.  [ V f ]  L T 1 [ s ]  M L3 [ l ]  M L3. [  l ]  M L1 T 1. [a ]  L. [g ]  L T  2.  M, L,T    d 3r  n r  63  3     1 ,  2 ,  3. , ,  1 . 2 . S. Vf ga. s l. choose l , a , g. inverse Re #   3 . We thus have. l l  1/ 2 3 / 2  l g a a l g a. l  l. Vf.  s l g1/ 2 a 3 / 2    F  , l ga  l . Vf. Alternatively, had we chosen l , a , l :.  s l g1/ 2 a 3 / 2  1 l Vf a    F '  , 3 l   l  l . Dimensional Analysis. a. (9) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. Dimensionless Form of Conservation Equations Continuity and Momentum Equations V  0 ;. V 1  (V  )V   p   2 V t . t'   t , R' . R V p , '  L , V'  , p'  L V  V2.  '  V'  0 ; St.  V' 1 VL L  (V'  ' )V'  ' p' (' ) 2 V' ; Re  , St   t' Re  V.  Very convenient for numerical computation since the differential terms are of order unity.. V'  FV ( R ' , t ' ; Re, St ) , p'  Fp ( R ' , t ' ; Re, St )  Can also calculate nondimensional thrust and torque. Dimensional Analysis. (10) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. Dimensionless Form of Boundary Conditions Boundary conditions define additional dimensionless parameters.  If the incompressible fluid has a free surface,  p  p B  gz B  r1  r2. surface tension. * B. (p*B ) ' . pB gL '  1  z   B  V2 V2  V 2 L r1 '  r2 '. air. p *B. water.  For flow around a ship’s propeller or in a pump, the fluid may vaporize, then the cavitation number is used to characterize the appearance of vapor bubbles. Ca . p  p v  V2.  cp V  For compressible flow, Ma  , Pr  a k.  Other dimensionless groups may include. w q"S L F  CF  , CT  , Cf  , Nu  2 2 3 2 ½  V AF  V RT ½ V k T Dimensional Analysis. (11) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. Modeling - Dimensionless Parameters Reynolds number Strouhal number Weber number Froude number Cavitation number Mach number Prandtl number Force coefficient Torque coefficient Friction coefficient Nusselt number. Re St We Fr Ca Ma Pr CF Cτ Cf Nu. VL/. L/ V  V2 L / . V/ gL (p   p v ) /  V 2 V/a  cp / k. F / (  V2 AF ) / 2  /  V 2 R 3T w /  V 2 / 2 q w L / k T. Dimensional Analysis. (12) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. Modeling - Similitude  Geometric similitude scaling ratio: SR = Lm / Lp. m: model, p: prototype.  Kinematic similitude (e.g. wind turbine of radius L in Vw of wind)  2 L   2 L   2    2   V / L m  V / L p.  Vw   Vw         L m   L p.  . (Vw / ) m L m   SR (Vw / ) p L p. Vw.  Dynamic similitude (for the entire flow field) ( 1 ) m  ( 1 ) p. L. (  2 )m  (  2 )p  (  n  r )m  (  n  r )p Dimensional Analysis. (13) 409.319A. Aero Thermo Hydro Engineers Nexus Application Department of Nuclear Engineering, Seoul National University. Modeling - Example 1. Construct a lab scale model of a complex piping system to investigate p ~ m D , ) f  F ( V  D 2. A model of the vessel of Lp=100m is built to a scale of 1:50,. Lm  2m , m  5102m3 , (A )  0.9m3 Tested in a towing tank of  103kg / m3 at Vm=1.1m/s , total drag force is 2.66N. (a) Vp (b) Dp in.  1.03103 kg / m3. (c) Pp=DpVp: propulsive power Dimensional Analysis. (14)