0.2. CONSERVATION LAW FOR FLUID 9

Transcription

1 0.2. CONSERVATION LAW FOR FLUID 9 Consider x-component of Eq. (26), we have D(ρu) + ρu( v) dv t = ρg x dv t S pi ds, (27) where ρg x is the x-component of the bodily force, and the surface integral is the x-component force exerted by the pressure of the neighboring fluid. Applying Green s theorem, we have D(ρu) + ρu( v) dv t = For y and z components, we have D(ρv) + ρv( v) dv t = D(ρw) + ρw( v) dv t = Combining the three equations, we have Since from mass conservation we have D(ρv) In the differential form or more explicitly ( ρg x p ) dv t. (28) x ( ρg y p ) dv t, (29) y ( ρg z p ) dv t. (30) z D(ρv) + ρv( v) dv t = (ρg p) dv t. (31) + ρv( v) = ρ Dv + v ( Dρ + ρ( v) ) = ρ Dv, ρ ( ρ Dv ) + p dv t = ρgdv t. (32) ( v ρ Dv + (v )v + p = ρg, (33) ) + p = ρg (34)

2 10 CONTENTS Conservation of Energy In a fluid, the energy consists of two parts, kinetic energy and internal energy (thermal energy). Assuming that e is the specific internal energy (internal energy per unit mass), the total energy density E (total energy per unit volume) is E = 1 2 ρ v 2 +ρe. The energy flux is, therefore, ( 1 2 ρ v 2 +ρe)v. For a fixed volume, the conservation of energy requires that the rate of the energy change equals the net influx of energy plus the work done at the boundary by neighboring fluid which equals -pv ds per unit area and the work done by bodily force. Thus the energy conservation gives us ( 1 ( 1 W 2 ρ v 2 + ρe) dv = S 2 ρ v 2 + ρe + p) v ds+ ρg vdv. (35) W From Green s theorem, we have W ( E ) + ((E + p)v) dv = W ρg vdv. (36) Since W is aritary, we have the differential form of the energy conservation as E + ((E + p)v) = ρg v. (37) Incompressible Fluid For an incompressible fluid, if one follows the fluid particle in the moving volume, this volume should not be changed dv t = JdV = W W J dv = 0. (38) This requires J = 0. (39) Another way to look at the incompressible fluid is that if one follows the particle motion, the fluid density should not be changed, that is Dρ/ = 0. From conservation of mass (Eq. (25)), this is equivalent to v = 0, (40)

3 0.2. CONSERVATION LAW FOR FLUID 11 therefore the complete set of equations for incompressible fluid is if density is uniform, it is simply ρ + (v )ρ = 0 ( ) v ρ + (v )v = p + ρg (41) v = 0 ρ ( v Isentropic Flow + (v )v ) v = 0 = p + ρg (42) In thermodynamics, the enthalpy of the fluid can be written as dw = T ds + V dp, (43) where T is the absolute temperature, S is entropy, and V is the specific volume V = 1. For isentropic flow, the entropy S is a constant, therefore we ρ have dw = V dp = dp p. It follows that = w, and the momentum equation ρ ρ becomes v + (v )v = w + g. (44) Since normally g can be expressed as the gradient of a potential function as g = φ, and (v )v = 1 2 v 2 v v, we have v v 2 v v = (w + φ). (45) Taking the curl of both sides, we have or v ω = (v v), (46) = (v ω), (47)

4 12 CONTENTS where ω is the vorticity of the fluid. If the fluid is initially irrotational, it will remain irrotational because of Eq. (47). In such case, we simply have For steady flow, / = 0, we have Integrating over space, we have Eq. (50) is called Bernoulli s law. v v 2 = (w + φ). (48) 1 2 v 2 + w + φ = 0. (49) 1 2 v 2 + w + φ = Constant, (50) 0.3 Conservation Law and Wave Equations Linear Wave Equation The general form of first order scalar wave equation has the following form u t + c(u)u x = 0, (51) Here the function c(u) serve as the propagation speed of hyperbolic wave. If c is independent of u, the equation is linear, otherwise it is nonlinear. When compared with conservation law Eq. (8), we have c(u) = F (u). For linear equation F (u) = au and c(u) = a, where a is a constant. We start from the linear wave equation: Du = u t + au x = 0 or u t + (au) x = 0 (52) and discuss the following aspects of the equation (1). The traveling wave solution in (, ): u(x, 0) = φ(x), u(x, t) = φ(x at), (53) where a is the speed of the traveling wave.

5 0.3. CONSERVATION LAW AND WAVE EQUATIONS 13 (2). The characteristics of the equation: dx dt = a, along which du dt = u t + au x = 0. (54) This tells us that along each characteristics, u is a constant. (3). Boundary condition: if a > 0, boundary condition can only be applied from the left side, if a < 0 boundary condition can only be applied from the right side. (4). Discontinuity: with smooth initial condition, linear equation will not produce discontinuity, but the traveling wave solution does allow discontinuous initial condition. The discontinuous solution does NOT satisfy the original equation at the discontinuity. Most of the physical waves, including sound wave, electromagnetic wave, etc, are described by the second order wave equation u tt a 2 u xx = 0 (55) We can factorize the differential operator 2 a 2 2 into ( a )( +a 2 x 2 x In the one dimensional infinite domain, the d Alembert solution gives: x ). u(x, t) = φ(x + at) + ψ(x at) (56) corresponding to the left-traveling and the right-traveling waves with characteristics dx = ±a (57) dt A typical initial value problem in the infinite domain is The d Alembert solution must satisfy u tt a 2 u xx = 0, (58) u(x, 0) = g(x), u t (x, 0) = h(x). (59) φ(x) + ψ(x) = g(x), aφ (x) aψ (x) = h(x). (60) After integration, we have u(x, t) = 1 1 x+at [g(x + at) + g(x at)] + h(ξ)dξ. (61) 2 2a x at

6 14 CONTENTS For solutions in finite domain, boundary condition must be considered. Two of the most frequently used boundary conditions are: (1) fixed boundary condition u(x B, t) = 0, and (2) free boundary condition u x (x B, t) = 0. Consider solution in a semi-infinite domain with a boundary at x B = 0, for the fix boundary condition (1), the solution should be an odd function about the point x = 0, while for free boundary condition (2), the solution should be an even function about the point x = Sound Wave Consider one dimensional Euler equations with no external force ρ t + (ρu) x = 0 (62) ρu t + ρuu x + p x = 0, (63) for small oscillation, we may assume p = p 0 + δp, ρ = ρ 0 + δρ and u = δu. Substitute this into Eqs. (62-63) and neglect all second order terms, we have δρ t + ρ 0 δu x = 0 (64) ρ 0 δu t + δp x = 0. (65) Now assume that p = p(ρ), we have δp x = p (ρ)ρ x. Substituting this into Eqs. (64-65), we have δρ t + ρ 0 δu x = 0 (66) ρ 0 δu t + δp x = 0. (67) Differentiating Eq. (66) with respect to t and Eq. (67) with respect to x and substituting the second one into the first one, we have δρ tt p (ρ)δρ xx = 0. (68) This gives the sound speed square a 2 = p (ρ). During the small oscillation of sound, the pressure responds to the change of density adiabatically, therefore, we have γp0 a = p (ρ) =, (69) ρ 0 where γ is the gas constant, p 0 is the ambient air pressure, and ρ 0 is the ambient air density.

7 0.3. CONSERVATION LAW AND WAVE EQUATIONS Numerical Diffusion Linear diffusion Consider viscous linear equations u t + au x = νu xx (70) with discontinuous initial condition u(x, 0) = { ul if x < 0 u r if x > 0 (71) A substitution of X = x at will transform Eq. (70) into a diffusion equation u t = νu XX (72) Diffusion equation For the parabolic diffusion equation u t = νu XX, and initial condition u(x, 0) = u 0 (X), (73) using Fourier transform here û(k, t) = 1 2π u(x, t)e ikx dx (74) û t = νk 2 û, û(k, t) = û 0 (k)e νk2t, (75) û 0 (k) = 1 2π is the Fourier transform of initial condition. u(x, t) = 1 2π u(x, 0)e ikx dx (76) û(k, t)e ikx dk = 1 u(y, 0)e νk2 t+ik(x Y ) dkdy. 2π (77) Integrating over k (see Appendix), we obtain u(x, t) = 1 4πνt u 0 (Y )e (X Y )2 /4νt dy (78) Solution to singular initial condition u 0 (X) = δ(x), u(x, t) = 1 4πνt e X2 /4νt (79)

8 46 CONTENTS Birkoff and Carter used conformal mapping to predict a 2D bubble has C 1 = Garabedian estimated C Numerical experiments have been in support to such terminal bubble velocities. Youngs and the α Paradox Read [12] and Youngs [15, 16, 17] first presented their expreiment of Rayleigh- Taylor instability with randomly perturbed interface and claimed that the perturbed interface will grow under constant acceleration. h h 0 = αagt 2 (192) His experimental value of α was in the range of , but numerical simulation failed to support this result. Front tracking simulation has a much closer agreement with the experiment. Sharp-Wheeler Merger Model Researchers have tried to explain why the random surface RT instability has constant acceleration while single bubble has constant velocity. The first premitive model was given by Sharp and Wheeler [8] as the so-called bubble merger model. With the merger criterion h 2 h 1 m(r 1 ) = C 2 r 1 (193) with appropriate choice (some how arbitary) of C 2, numerical simulation has suggested Youngs s constant acceleration. More sophisticated models include the renormalization model, buoyancydrag mode, average equation model etc. It is still an open area for imagination and creative thinking Richtmyer-Meshkov Instability The Impulsive Model The impulsive model was proposed by Richtmyer in In this model, the growth rate of the instability ȧ 0 (t) is given by a simple formula ȧ 0 (t) = kū ρ 1 ρ 2 ρ 1 + ρ 2 a 0 (0), (194)