Environmental Fluid Mechanics - Chapter 4 doc

... introducing Eq. (4. 4.22) into Eq. (4. 4.21) we obtain Q F D i  2      U C i 2z   z 2 CÐÐÐ     2 dz 4. 4.23 The residue of the complex function subject to integration in Eq. (4. 4.23) is ... quantity, along with Eqs. (4. 4.6) and (4. 4.7) into Eq. (4. 4.1), the pressure distribution is obtained along the surface as shown in Fig. 4. 16. By integrating the pressure Figure 4. 16...

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... are very small. Therefore the Navier–Stokes equations are approx- imated by  D 0 ∂ 4  ∂x 4 C 2 ∂ 2  ∂x 2 ∂ 2  ∂y 2 C ∂ 4  ∂y 4 D 0 3.5.1 Copyright 2001 by Marcel Dekker, Inc. All Rights ... Eq. (3 .4. 8). Using the definition (3 .4. 9), it is easy to find ∂u ∂y D du dÁ Á y ∂ 2 u ∂y 2 D d 2 u dÁ 2  Á y  2 ∂u ∂t D du dÁ   Á 2t  3 .4. 11 Introducing Eq. (3 .4. 11) into Eq....

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...  2  q 2  D  N 1 C  N 2  D  V10 .4. 4 Note that this result indicates that diffusion does not contribute to the total momentum (both Eqs. (10 .4. 4) and (10 .4. 1) can be satisfied together only when  F 1 D  F 2 D  0 ... in the x-direction are shown in Fig. 10.6, but similar fluxes may be defined for the y-andz-directions. Using a truncated Figure 10.6 Fluid element, showing fluxes o...

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... 100 110 45 267 840 00 105788800 31. 245 21.5151 20 .48 763 31. 245 15 Aug. 25 35 30 267 840 00 92396800 13.7396 13 .49 53 6.992327 13.73998 Sept. 40 40 45 0 92396800 13.951 13.9 141 20.90 643 13. 949 38 Oct. ... 12 .46 99 5.06051 9.93 948 9 12 .46 9 74 Feb. 180 178 10 48 3 840 0 109115200 8.1382 3.60256 6.33693 8.13821 Mar. 350 345 25 13392000 115811200 18.652 24. 63 048 30.96 741...

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... Reserved. Contents Preface Part 1 Principles of Environmental Fluid Mechanics 1. Preliminary Concepts 1.1 Introduction 1.2 Properties of Fluids 1.3 Mathematical Preliminaries 1 .4 Dimensional Reasoning Problems Supplemental ... students with appro- priate fluid mechanics backgrounds. Although the book addresses principles of fluid mechanics relevant to the entire environment, the emp...

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... with Eq. (2 .4. 7) indicates that the property Á of Eq. (2 .4. 7) was replaced by the density  in Eq. (2 .4. 8). We may, therefore, apply the transport theorem of Reynolds, namely Eq. (2 .4. 8), to obtain ∂ ∂t  U dUC  S  E V ...   ∂ 2 u i ∂x 2 j C 1 3 ∂ 2 u i ∂x i ∂x j  2.7 .4 For incompressible flow, Eq. (2.7 .4) reduces to  D E V Dt Drp C gZ C r 2 E V2.7.5a or  Du i Dt D...

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... t p D 2 ω 8 .4. 3 By combining Eqs. (8 .4. 2) and (8 .4. 3), it is seen that the wave propagates a distance of a single wavelength during one period, namely,  D ct p 8 .4. 4 The function F of Eq. (8 .4. 1) ... (8 .4. 15) and (8 .4. 16): ∂ ∂z D u D k 0 e kz sinωt kx 8 .4. 22 ∂ ∂x Dw Dk 0 e kz cosωt kx 8 .4. 23 Direct integration of these expressions yields  D  0 e kz...

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... differentiating Eqs. (9 .4. 9) and (9 .4. 10) with respect to z, using the shear boundary conditions, we obtain du dz     zD0 DC 2  p 2 sin C 4 C cos C 4  D 0 ) C 4 D  4 9 .4. 12 and d v dz     zD0 DC 2  p 2  cos  4 C ... D 2 24 ð 3600 D 7.27 ð10 5 s 1 By applying this value, we obtain the following table: Â 10 ° 20 ° 30 ° 40 ° 50 ° 60 ° 70 ° 80 ° 90 ° f ð...

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... D y ∂ 2 C ∂y 2 11 .4. 13 For steady-state conditions, we apply the following approximations (refer to Eqs. 11.3. 14 and 11.3.27): D y D a T V (11 .4. 14a) C D C 0 1 Á n Á D y υ (11 .4. 14b) where υ is ... kg/m 3  f D 1000 kg/m 3 11 .4. 10 where  f and  s are the density of fresh and saltwater, respectively. Introducing Eq. (11 .4. 10) into Eq. (11 .4. 3), B D 40 h f 11 .4. 11 Th...

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... K L (cm/h) K H SO 2 1,600 34, 420 0.038 N 2 O 1,900 20 1.6 CO 2 ,40 0 20 50 CH 4 3,180 20 42 CCl 4 1,030 10.7 1.08 CCl 3 F 1,085 11.3 5 Source: Adapted from Liss and Slater (19 74) . Because oxygen is ... Whitman, W. G., 19 24. Principles of gas absorption. Industrial Engrg. Chem. 16(12):1215. Liss, P. W. and Slater, P. G., 19 74. Flux of gases across the air-sea interface. Nature 247...

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