Solutions - Chapter 7 Solutions 2.1 Using the denitions for...

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Chapter 7 Solutions 2.1 Using the defnitions For Re t and Re λ , the relation between ε and λ ( ε =15 ν u 2 2 ), and the estimate ε = u 3 /L turb , we obtain: Re λ = u λ ν Re 2 λ = u 2 λ 2 ν 2 = u 2 15 ν u 2 ν 2 ε = 15 u 4 L turb ν u 3 Re t Re λ = 15 Re 1 / 2 t 2.2 Using u = 10 m/s and u / u =0 . 1, we obtain u = 1 m/s. With ν =1 . 5 × 10 5 m 2 /s and L turb . 01 m, we obtain: T turb = L turb / u . 01 s; T 0 turb = L turb / u . 001 s; Re t = u L turb = 667; Re λ = 15 Re 1 / 2 t = 100; ε = u 3 /L turb = 100 m 2 /s 3 ; k =3 / 2 u 2 . 5m 2 /s 3 ; λ 2 ν u 2 λ . 22 mm; η K = L turb Re 3 / 4 t =76 µ m; v K = u Re 1 / 4 t . 20 m/s; and τ K = T turb Re 1 / 2 t . 39 ms. The Fastest Frequency will be encountered when the smallest eddy passes past our probe at the mean velocity u , so it will be equal to u/η K = 131 . 6 kHz. To obtain 1000 statistically- independent samples, we need to sample For 1000 T 0 turb =1s . 2.3 The relation dX/dt v ( X ) should be understood as an approximation where X is some statistical measure oF the size oF the cloud (more on this in Chapter 4). Using the concept 56
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that dissipation is independent of scale, ε u 3 /L turb = v ( X ) 3 /X . Hence, dX dt = v ( X ) dX dt =( εX ) 1 / 3 3 2 [ X 2 / 3 ( t ) X 2 / 3 (0)] = ε 1 / 3 t If the initial size is very small, X (0) ± X ( t ), then X ( t )=(2 / 3) 3 / 2 ε 1 / 2 t 3 / 2 . 2.4 Using the deFnitions of the Kolmogorov quantities (Eqs. 2.15-2.17), it can be easily shown that v K η K = 1. The physical interpretation of this result is that the small scales are dominated by viscous forces. 3.1 Using ν turb = C u L turb ,with C 0 . 1, then ν turb =0 . 1 × 1 × 0 . 005 = 5 × 10 4 m 2 /s, i.e. two orders of magnitude larger than the laminar kinematic viscosity of water ( ν =1 × 10 6 m 2 /s). 3.2 The decay of k =3 / 2 u 2 for homogeneous isotropic turbulence is given by d (3 / 2 u 2 ) /dt = u 3 /L turb after using ε = u 3 /L turb . This can be very easily integrated, since we assume that L turb does not change with t . 3.3 Starting from the instantaneous equation and performing the Reynolds decomposition, a series of transport equations for the higher moments may be derived. Using φ = φ + φ 0 and u = u + u 0 in the instantaneous scalar equation we get (before averaging): ( φ + φ 0 ) ∂t + ( u j + u 0 j ) ( φ + φ 0 ) ∂x j = D 2 ( φ + φ 0 ) 2 j (7.1) Multiplying by φ 0
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Solutions - Chapter 7 Solutions 2.1 Using the denitions for...

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