KUNDU Fluid Mechanics 2 Episode 13 ppt
... eddies irrotational fluid irrotational fluid turbulent fluid turbulent fluid 4 Figure 13. 14 htrainmcnt of a nonturbulent fluid and its assimilation into turhuleni fluid by viscous ... reached after n - 1 steps. Using rule (13. 84) successively, we get R; = R;-~ + L~ = R' ,, -2 + 2L2 - = R: + (n - l)Lz = nL2. The rms distance...
Ngày tải lên: 13/08/2014, 16:21
KUNDU Fluid Mechanics 2 Episode 1 ppt
... 2 . Analogy between Heat and Vorticity Diffusion 3 . Pressure Change Due to Dynamic Effects 20 9 21 3 21 6 21 8 22 1 22 5 22 7 23 0 23 2 23 4 23 8 24 0 24 2 24 5 24 a 25 0 25 4 25 5 ... 22 5 22 7 23 0 23 2 23 4 23 8 24 0 24 2 24 5 24 a 25 0 25 4 25 5 25 6 25 7 26 1 26 2 26 4 26 6 26 8 27 0 27 0 27 0 27 1 27 3 27 3 theory is also given...
Ngày tải lên: 13/08/2014, 16:21
KUNDU Fluid Mechanics 2 Episode 3 ppt
... vectors il, i2, and i3 changc with time. To this observer the time derivative of P is d ( $)F = Z(plil+ P2i2 + - . dPi . dP2 . dP3 dir di2 di3 - 11 - + 12- + 1.3 ... dy, (4 .26 ) because the x-directional velocity at these surfaces is nearly U,. Combining Eqs. (4 .22 H4 .26 ) gives a net outflow of x-momentum of: The momentum balance (4 .23 )...
Ngày tải lên: 13/08/2014, 16:21
KUNDU Fluid Mechanics 2 Episode 11 ppt
... sinhq*z, (0‘ - K’)2W = A(qi + K’)’ COSqoZ + B(q2 - K2 )2~ ~~hq~ + C(q*? - K2j2 cosh 4*z. The boundary conditions ( 12. 27) then require 4* cosh - 2 4 cash - 2 4 q sinh - ... substitution K Equations (1 .2. 16) and ( 12. 17) bccome ( 12. 16) ( 12. 17) (1 2. 18) ( 12. I 9) where gord4 RaG-, KV The basic slate satisfies d2T dZ2 0 = K ( 12....
Ngày tải lên: 13/08/2014, 16:21
KUNDU Fluid Mechanics 2 Episode 2 potx
... whcrc the two are related as (2. 26) (2. 27) As a check, Eq. (2. 27) givcs R11 = 0 and R 12 = -~ 123 ~3 = -w, which is in agree- ment with Eq. (2. 26). (In Chapter 3 we shall call ... (Figure 2. 7). Using Eq. (2. 12) , the components of he stress tensor in the rotated frame are 431 I& -43 .;I = CllC21t 12 + C2ICllt21 = TZU + TlU - TU, ti...
Ngày tải lên: 13/08/2014, 16:21
KUNDU Fluid Mechanics 2 Episode 4 pdf
... 'Itucsdcll, C. A. (19 52) . Stokes' principle of viscosity. JoumZ oJRationol Mechanics ud Analysi.s 1: physical Journul131: 4 424 47. Univemity Press. 22 8 -23 1. Supplernim?al Reading ... (5 .22 ) can be written as -V&nqi&ijkWk,jq = -V(&jSqk - &~haqj)wk.jq = -vWk,nk + vOn,jj = v%,jj. (5 .25 ) If we use Eqs. (5 .23 H5 .25 ), vor...
Ngày tải lên: 13/08/2014, 16:21
KUNDU Fluid Mechanics 2 Episode 5 potx
... 22 1 Logcr of ( hsiant Depth 22 3 Laycr of Vnriahle Depth H (.r) 22 4 Yotilir!fnr Stwpeniirg it1 CI .Voriili'pmii .v Mi~iiinti 22 5 H!rlruulir Jiunp 22 7 ... Water 21 2 It~hiem~ of.Su!$iri. TrLsirm 2 1 3 Slctrrrling Ilitrw 21 6 Cmup 1 hloixly rurd I.,'rictg- Fhx 2 1 8 CNJI.~) Irlocic! . cmd 1Iiri.v Dispmiirri...
Ngày tải lên: 13/08/2014, 16:21
KUNDU Fluid Mechanics 2 Episode 6 ppsx
... approximation tanh(2aHlA) 2 I valid for H > 0 .28 A, we obtain (7.67) For an &-water interface at 20 "Cy the surface tension is u = 0.074 N/in, giving = 23 .2 cm/s at A,,, ... the flow. To see this, consider the mechanical energy of a fluid particle at the surrace. E = u2 /2 + gH = Q2/2H2 + gH. Eliminatjng Q by Eq. (7.88) we obtain,...
Ngày tải lên: 13/08/2014, 16:21
KUNDU Fluid Mechanics 2 Episode 7 pdf
... (7.168) whcrc we have usccl p” = 6’ N4p ,2/ 2w2g2, found from Eq. (7.166) after taking its real part. Use of thc dispersion rclation w2 = k2N2/(k2 + m’) shows that Ek = Ep. (7.169) ... 26 8 Intcrd Fmiidt: Number 26 8 RiclmnLqm Aimher. 26 9 Mwh Niuihs 27 0 I’ra~idtl R’i~nhrx. 27 0 ficrc&es 27 0 Litemlure Cited 27 0 Siipplementul Rea...
Ngày tải lên: 13/08/2014, 16:21
KUNDU Fluid Mechanics 2 Episode 8 pot
... derivatives from Eq. (10 .22 ): rlll (10 .26 ) (10 .27 ) (10 .28 ) (10 .29 ) (1 0.30) dS 3 Uqf" dS - - - u- -[f - fq] = a2+ a+ - = Uf', ay i 12+ Uf" a$ 6 ' ... = p(a2+/ay2)oY where the subscript zero stands for y = 0. Ushg a2$/ay2 = Uf"/S given in Fq. (20 .29 ). we obtain to = pVf "(0)/6, and finally 0.33...
Ngày tải lên: 13/08/2014, 16:21