Turbulence Solutions

Transcription

1 School of Mechanical, Aerospace & Civil Engineering 3rd Year/MSc Fluids Turbulence Solutions Question 1. Decomposing into mean and fluctuating parts, we write M = M + m and Ũ i = U i + u i a. The transport equation for M can then be written t M +m+u j +u j M +m = Rearranging gives: M t + m t + U M j = Averaging the equation leads to λ M + m +M +m1 M m M m m + u j + U j + u j λ M + λ m + M1 M 2mM + m m 2 1 M t + U M j m = u j + = λ M λ M u j m + M1 M m 2 + M1 M m 2 2 b. Subtracting equation 2 from the instantaneous one equation 1 gives the equation for the fluctuating m: m t +U m M j = u j + u j m u j m+ λ m 2mM +m+m x 2 m 2 j To obtain a transport equation for m 2, note that Dm 2 Dt = 2m Dm Dt So, multiplying the above equation by m and averaging: Dm 2 Dt = 2u j m M 2u j m x 2 + 2m λ m 4m j x 2 M + 2m 2 2m 3 j The terms involving λ can be written as 2m λ m = 2 λm m 2λ m m = λ m2 2λ m m 1

2 Hence we get the transport equation Dm 2 Dt = 2u j m M 2λ m m + λ m2 2u j m x 2 j + 2m 2 1 2M 2m 3 If M < 0.5, the term 2m 2 1 2M is positive, implying that fluctuations are enhanced. If M > 0.5, the term 2m 2 1 2M is negative, implying that fluctuations are damped by the chemical reaction. Question 2. P 5mm = 100 s 1 ρ = 1.2 kg/m 3 a. Using the MLH: µ t = ρlm 2 = ρκy2 = = kg/ms The turbulent shear stress is given by = = m/s 2 b. Assuming that the shear stress is constant across the near-wall layer, then the wall shear stress τ w /ρ = m/s 2. The non-dimensional distance y + is given by y + = yτ w /ρ 1/2 /ν = / /1.2 = 61.5 Since y + > 30, the point does lie in the fully turbulent region of the flow. We thus assume the log-law holds, and obtain U + = 1 κ logey+ so U = τ w/ρ 1/2 logey + κ = /2 log = 3.16 m/s c. At the point P, the turbulence energy is k = 0.2 m/s 2. Using the 1-equation model, with l = 2.55y: µ t = ρc µ k 1/2 l = / = kgms 2

3 The shear stress is then given by = = m/s 2 The generation and dissipation rates of k are: P k = uv = 5.13 m2 /s 3 ε = k 3/2 /l = 0.2 3/2 / = m 2 /s 3 Question 3. If the lengthscale is 0.1D and the turbulent kinetic energy k = 4 m/s 2, then the dissipation rate can be estimated as ε = k 3/2 /l = 4 3/2 / = 160 m 2 /s 3 If the velocity and length scales associated with the smallest eddies are v ε = εν 1/4 and η ε = ν 3 /ε 1/4, then v ε = /4 = m/s η ε = /160 1/4 = m Question 4. Using the MLH, the shear stress is modelled as uv = lm 2 = κ2 y 2 At the wall, / is O1, and since κ is a constant, we get uv y 2. If we introduce the Van Driest damping term: l m = κy[1 exp y + /A + ] Expanding the exponential term as a Taylor series to obtain its behaviour near the wall, we have l m = κy[1 1 y + /A + + = κyy + /A + + Since y + is proportional to y, hence l m y 2 near the wall, and uv y 4. Question 5. In decaying grid turbulence, we have Dk Dt = U dk = ε = ωk dx Dω Dt = U dω dx = c ω2ω 2 3

4 We are given that k should decay as k = cx n where the decay exponent n is 1.2. Subsituting this into the k equation gives so the decay of ω is given by ω = Unx 1. Substituting this into the ω equation gives U dk dx = Uncx n 1 = ωk = ωcx n U dω dx = U2 nx 2 = c ω2 U 2 n 2 x 2 Cancelling the factor U 2 nx 2 then gives c ω2 = 1/n = Question 6. a. In a zero pressure gradient boundary layer convection may be ignored as may be streamwise difusion. Thus the U momentum equation becomes ν uv = 0 The total shear stress ν/ uv is thus constant across the boundary layer. Since uv is zero at the wall, the total shear stress must be equal to the wall shear stress. Hence in the fully turbulent region, where viscous effects can be ignored, the turbulent shear stress uv must be equal to the wall shear stress. b. In the boundary layer the mixing length model will give 2 = l2 m If the velocity satisfies U + = 1/κ logey +, then = = 1 κy +τ w/ρ 1/2τ w/ρ 1/2 ν Hence the shear stress is represented by the model as uv = lm 2 τw /ρ κ 2 y 2 = τ w/ρ 1/2 κy If the Reynolds shear stress should be equal to the wall shear stress, we should thus have l 2 m/κ 2 y 2 = 1, or l m = κy. c. In the 1-equation model the turbulent viscosity is ν t = c µ k 1/2 l. For this to be equivalent to that in the MLH representation of the boundary layer above, we must have c µ k 1/2 l = lm 2 = κ2 y 2τ w/ρ 1/2 = κyτ w /ρ 1/2 κy But in a simple boundary layer we have uv /k = c 1/2 µ, so that k = uv /c 1/2 τ w /ρ/cµ 1/2, so that the above equation becomes or l = κ/c 3/4 µ y τ w /ρ 1/2 c 3/4 µ l = κyτ w/ρ 1/2 4 µ =

5 Question 7. a. With zero pressure gradient and neglecting convection and streamwise diffusion, the U momentum equation can be written as 0 = ν uv 1 i Integrating equation 1 results in ν uv = Constant and the total shear stress is thus constant across the layer. Since at the wall uv will be zero and ν/ = τ w /ρ, the constant in the above equation must be τ w /ρ. In the fully turbulent region we can neglect the viscous shear stress, so we get uv = τ w /ρ ii Using the mixing length we have ν t = lm / 2 and hence the turbulent shear stress is modelled as 2 = l2 m Since U satisfies the log-law, we have = = 1 κy τ w/ρ 1/2 τ w/ρ 1/2 + ν Since uv = τ w /ρ in the fully turbulent region we thus have τw /ρ 1/2 2 = τ w/ρ 1/2 κy τ w /ρ = l 2 m κy which gives l m = κy. iii Immediately adjacent to the wall we want uv y 3. Since uv = l 2 m /2 and the velocity gradient is O1 near the wall, we thus want to ensure that l 2 m y3 or l m y 3/2. Expanding the given expression as a Taylor series we get l m = κy1 1 y + /A n +... = κyy + /A n +... For the leading term of this to be proportional to y 3/2 we thus need n = 0.5. b. If we now have a non-zero pressure gradient, but still neglect convection and streamwise diffusion, the U momentum equation becomes 1 P ρ x = ν uv 5

6 i Integrating the above equation now gives ν uv = Constant + y P ρ x and if P/ x is negative, the total shear stress thus decreases as y increases. ii In such an accelerating flow, the viscous sublayer thickness would thus increase, and hence we would want to increase the constant A in the above model so that viscous effects would be influential at larger values of y +. Question 8. a. Defining the lengthscale as l = k 1/2 /ω, then a transport equation for l can be derived from Dl Dt = D 1 Dk Dt k1/2 /ω = 2k 1/2 ω Dt k1/2 Dω ω 2 Dt Ignoring diffusion terms, this gives ωp k c ω1 Dl Dt = 1 2k 1/2 ω P k kω k1/2 ω 2 = l 2k P k kω l c ω1 P k k c ω2ω = lp k k 1/2 c ω1 k 1/2 1/2 c ω2 k c ω2ω 2 b. In decaying grid turbulence, we have k = Cx n, and since the k generation rate is zero, the k, ω and l equations are Equation 1 results in U dk dx = kω 1 U dω dx = c ω2ω 2 2 U dl dx = k1/2 1/2 c ω2 3 UnCx n 1 = Cx n ω so that ω = nu/x. Substituting into equation 2 then gives nu 2 /x 2 = c ω2 n 2 U 2 /x 2 and hence c ω2 = 1/n. With c ω2 = 5/6, we thus have n = 6/5 = 1.2. The lengthscale variation is given by l = k 1/2 /ω = C 1/2 x n/2 nu/x 1 = C 1/2 /nux 1 n/2. With n = 1.2 the lengthscale thus increases as x 0.4. c. In local equilibrium the generation and dissipation rates of k are equal. Thus P k = kω, and the lengthscale equation reduces to U dl dx = lω1/2 c ω1 k 1/2 1/2 c ω2 = k 1/2 c ω2 c ω1 With the quoted coefficients, c ω2 c ω1 = 5/6 5/9 > 0. The lengthscale will thus increase with distance downstream. 6

7 Question 9. In a simple shear flow, we get and ν t = lm 2 a. In local equilibrium, the generation and dissipation rates of turbulence energy are equal. In the context of the above model in a simple shear flow, this leads to P k = uv 3 = ε or l2 m = ε b. If the total shear stress across the boundary layer is constant, then in the fully turbulent region where the viscous stress is negligible the turbulent shear stress must be equal to the wall shear stress. Hence uv = τ w /ρ. Using the above model thus results in 2 τ w /ρ = ν t = l2 m If the mixing length is taken as l m = κy, then we obtain Using the chain rule, = + + = τ w/ρ 1/2 κy + + = τ w/ρ + ν + Hence, and integrating gives + + = ν τ w /ρ 1/2 κy = 1 κy + U + = 1 κ logey+ c. Near the wall we should have uv y 3. i Since uv = l 2 m /2, and the mean velocity gradient / is O1 near the wall, we thus need l m y 3/2. Expanding the proposed expression for l m in a Taylor series: l m = κy[1 1 y + n /A + ] = κy[y + n /A + ] Since y + y, we thus need n = 0.5 to give the desired shear stress variation. ii If we want the mixing length to only be 5% away from the linear form, then we need 1 exp{ y + n /A} = 0.95 or y + n = A log0.05 With n = 0.5 and y + 30, this gives A =

8 Question 10. a. In decaying grid turbulence the k and ε transport equations become U dk dx dε = ε and U dx = c ε 2 ε2 k If k = Cx n, then dk/dx = ncx n+1, and so ε = ncux n+1. Substituting into the ε equation one then obtains and hence nn + 1CU 2 x n+2 = c ε2 n 2 C 2 U 2 x 2n+1 b. The lengthscale l is given by Cx n n + 1 = c ε2 n or n = 1/c ε2 1 l = k 3/2 /ε = C3/2 x 3n/2 C1/2 = ncux n+1 n x1 n/2 = c ε2 n 2 CU 2 x n+2 With c ε2 = 1.92, we get n = 1.09 and hence l increases proportional to x c. In the diffuser described, we have x = U o/l = V by continuity i The generation rate of k is thus given by P k = u 2 x = 4ν t x V v2 = 2ν t 2 = 4c µ k 2 /εu o /L V + 2ν t x = 4c µ k 1/2 lu o /L 2 = 4c µ k 1/2 l o 1 + x/lu o /L 2 ii Neglecting diffusion and dissipation, the k equation along the centreline then becomes U dk dx = U o1 + x/l dk dx = P k = 4c µ k 1/2 l o 1 + x/lu o /L 2 We thus get dk dx = 4c µk 1/2 l o U o /L 2 or 1 dk k 1/2 dx = 4c µl o U o /L 2 Integrating and applying the boundary condition k = k o at x = 0 gives k 1/2 = 2c µ xl o U o /L 2 + k 1/2 o or k = [k 1/2 o + 2c µ xl o U o /L 2 ] 2 8

9 Question 11. a. In the fully developed flow V = 0 and there are no gradients of streamwise velocity. Hence the U momentum equation becomes 0 = r P x + rµ r r ρruv i Integrating gives r µ r ρuv = 1 2 r2 P x + Constant and conditions at r = 0 imply that the constant is zero. Hence the total shear stress, µ/ r ρuv, is given by µ r ρuv = 1 2 r P x and hence increases linearly across the pipe. ii At the wall, we must have uv = 0 and µ/ r = τ w. Hence 1 2 R P x = τ w or P x = 2τ w/r Thus, in the fully turbulent region where viscous stress can be neglected, we have ρuv = r P 2 x = rτ w/r or uv = τ w /ρr/r b. The model gives r and ν t = lm 2 r i Since the velocity gradient is O1 near the wall, in order to get uv y 3 we need ν t y 3. Hence we require l 2 m y3 or l m y 3/2. Near the wall we will have l m = κyy + n /A and since y + is proportional to y, l m is thus proportional to y 1+n. We therefore need to take n = 0.5. ii In the model, viscous effects become negligible when y + n /A is equal to unity. At 5% of the pipe radius from the wall, we have y + = 0.05Rτ w /ρ 1/2 /ν. Since the Reynolds number Re τ = τ w /ρ 1/2 R/ν is 400, we thus have that at the edge of the viscous layer, y + = = 20 Hence we need to set the constant A = 20 = iii Since / r is the only non-zero velocity gradient in this fully developed flow, the model will return u 2 = v 2 = w 2 = 2/3k. Since the correct near wall variation is different for the various stress components, the model cannot return the correct behaviour for all components. 9

10 Question 12. a. In decaying grid turbulence, the k equation reduces to U dk dx = k3/2 /l If k = Cx n, then dk/dx = ncx n 1 and we get nucx n 1 = C 3/2 x 3n/2 /l or l = C1/2 nu If n = 1.1, then 1 n/2 = 0.45, and hence we require l x x 3n/2 C1/2 = x n 1 nu x n/2+1 b. In the flow shown we have / x = A, and so from continuity V/ = A. i The production term is now given by P k = u 2 x The EVM formulation for the stresses gives v2 V = u2 v 2 A u 2 = 2/3k 2ν t x = 2/3k 2c µk 1/2 la v 2 = 2/3k 2ν t V = 2/3k + 2c µk 1/2 la Hence the generation term can be written P k = 4c µ k 1/2 laa = 4c µ k 1/2 la 2 ii Neglecting dissipation and diffusion, on the centreline of the nozzle we will have U k x = P k = 4c µ k 1/2 la 2 Hence 1 k k 1/2 x = 4c µla 2 U o + Ax Integrating gives 2k 1/2 = 4c µ la logu o + Ax + C At x = 0 we have k = k o, so the constant C is given by C = 2k 1/2 o 4c µ la logu o The turbulence energy on the centreline is thus given by k 1/2 = 2c µ la logu o +Ax+ko 1/2 2c µ la logu o = 2c µ la log1+ax/u o +ko 1/2 and hence k = [ ] 2c µ la log1 + Ax/U o + ko 1/2 2 10