KUNDU Fluid Mechanics 2E Episode 16 ppt

kundu fluid mechanics 2e Fluid Mechanics, Second Edition doc

kundu fluid mechanics 2e Fluid Mechanics, Second Edition doc

... 586 588 591 595 598 (108 613 615 623 626 627 L SECOND EDITION PIJUSH K. KUNDU 0 IRA M. COHEN xviii PrcJacw fluid mechanics in the clearest way I have ever seen, and Professor ... provided. 1 have becn a studcnt 01 fluid mechanics since 1954 when I entered college to study aeronautical engineering. I have been teaching fluid mechanics sincc...

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kundu fluid mechanics 2e pdf

kundu fluid mechanics 2e pdf

... Asmmechanics, Ncw York Dover Publications. (A clear and simplc discussion of potential and adiibalic temperature gradients is given.) In memory of Pijush Kundu Pijush Kanti Kundu ... (p+dp)dxdy-pgdxdydz =o: which simplifies to - I I This shows that the pressure in a static fluid decreases with height. For a fluid of uniform density, Eq. (1.8) can be integrat...

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KUNDU Fluid Mechanics 2 Episode 1 ppt

KUNDU Fluid Mechanics 2 Episode 1 ppt

... Mapping 14 8 15 0 1 52 15 4 15 6 15 7 15 7 15 9 16 0 16 3 16 6 17 0 17 1 17 3 17 5 17 6 18 1 18 4 18 5 18 7 18 8 .I 89 19 0 1 92 1 92 chi!pter 7 Gravity Waves 1 . Introduction 19 4 2 . ... 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...

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KUNDU Fluid Mechanics 2 Episode 2 potx

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...

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KUNDU Fluid Mechanics 2 Episode 3 ppt

KUNDU Fluid Mechanics 2 Episode 3 ppt

... il, i2, and i3 changc with time. To this observer the time derivative of P is d ( $)F = Z(plil+ P2i2 + - . dPi . dP2 . dP3 dir di2 di3 - 11 - + 12- + 1 .3 - + ... that a@ aY u -, (3. 34) a$ ax ' v E then Eq. (3. 33) is automatically satisfied. Therefore, a streamfunction JI can be defined whenever Eq. (3. 33) is valid. (A...

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KUNDU Fluid Mechanics 2 Episode 4 pdf

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 42 4 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 ),...

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KUNDU Fluid Mechanics 2 Episode 5 potx

KUNDU Fluid Mechanics 2 Episode 5 potx

... ha2, 25 r 2n 2n m m 25 r 2? r =- 1n(x~-y*-u~+i2xy) 11na~. (6.48) Wc know that the logarithm of any complex quantity C = I< I exp (iQ) can be written as In 5 = In 15 ... 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 2...

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KUNDU Fluid Mechanics 2 Episode 6 ppsx

KUNDU Fluid Mechanics 2 Episode 6 ppsx

... dc/h = 0 in Eq. (7 .66 ), and assuming the deep-water approximation tanh(2aHlA) 2 I valid for H > 0 .28 A, we obtain (7 .67 ) For an &-water interface at 20 "Cy the surface ... 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,...

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KUNDU Fluid Mechanics 2 Episode 7 pdf

KUNDU Fluid Mechanics 2 Episode 7 pdf

... 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 Reading 27 0 ... 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) w...

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KUNDU Fluid Mechanics 2 Episode 11 ppt

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....

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KUNDU Fluid Mechanics 2 Episode 13 ppt

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...

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KUNDU Fluid Mechanics 2E Episode 15 doc

KUNDU Fluid Mechanics 2E Episode 15 doc

... = y/sin0, (1 5.6) where Eq. (15. 4) has been used. We now eliminate 0 between Eqs. (15. 5) and (15. 6). First note from Eq. (15. 6) that cos2e = (2btan fi - y)/2btan p, ... obtain (1 5.4) x2 sinze - y2 cosze = 4b2 sin% cos2e. (15. 5) To understand the shape of the curve represented by Eq. (15. 5) we must express 0 in terms of x, y,...

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KUNDU Fluid Mechanics 2E Episode 16 ppt

KUNDU Fluid Mechanics 2E Episode 16 ppt

... 13.715 I 13.9549 14.1 984 14.4456 14.6965 14.9513 15.2099 15.4724 15.7388 16. 0092 16. 2837 16. 5622 16. WY 17.1317 17.4228 17.71 81 18.01 78 18.3218 18.6303 18.9433 19.2608 19.5828 ... involving thc hc Lwo powers of (M: - 1) do not appear in Bq. (16. 36). U.siug the pressure ratio (16. 31), Eq. (16. 36) can be written 85 3 &-Si y2- 1...

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KUNDU Fluid Mechanics 2E Episode 17 potx

KUNDU Fluid Mechanics 2E Episode 17 potx

... mapping, 171 -173 douhlet/dipole. 157-159 forces on two-dimensional body, imp, mclhod or, 143, 170 -171 numcrical zolulion orplanc, 176 1 81 ovcr elliplic cylinder, 173 -174 past ... - . 1.33E - 7 1.38E - 7 1. 42E - 7 1.468-7 1. 52E- 7 1.58E - 7 CP R J@-'K-' V/K 4 217 13.4 4192 9.5 4182 7.1 4178 5.5 4178 4.3 4180 3.5 Latent heat...

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