Spectrum Models
Published:
This blog introduces the several turbulence (spectrum) models, including the derivations and coefficient determination for the modeling of the energy-containing range, the inertial range, and the dissipation range.
Miscellaneous
Gamma function $\Gamma(z)$ (definition):
\[\begin{align*} \Gamma(z) &\equiv \int_0^\infty { t^{z-1} e^{-t} } \mathrm{d}t \\ &= \int_0^\infty { 2 x^{2z-1} e^{-x^2} } \mathrm{d}x. \quad (t = x^2) \end{align*}\]Beta function $\mathrm{B}(m, n)$:
\[\begin{align*} \mathrm{B}(m,n) &= \int_0^1 { u^{m-1} (1-u)^{n-1} } \mathrm{d}u \\ &= \int_0^\infty { 2 \frac { x^{2n-1} } { (1+x^2)^{m+n} } } \mathrel{d}x \quad \left( u = \frac{1}{1+x^2} \right) \\ &= \int_0^\infty { 2 x^{-(2m+1)} \left( \frac{x^2}{1+x^2} \right)^{m+n} } \mathrm{d}x \\ &= \int_0^\infty { 2 x^{-(2m+1)} \left( \frac{x}{\sqrt{1+x^2}} \right)^{ (2m+1) + (2n-1) } } \mathrm{d}x. \end{align*}\]Now prove that
\[\begin{align*} \mathrm{B}(p, q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}. \end{align*}\]Let $u=\cos^2\theta$,
\[\begin{align*} \mathrm{B}(p, q) &= \int_{\frac{\pi}{2}}^0 { (\cos^{2p-2}\theta) (\sin^{2q-2}\theta) (-2\sin\theta\cos\theta) } \mathrm{d}\theta \\ &= 2 \int_{0}^{\frac\pi2} { (\cos^{2p-1}\theta) (\sin^{2q-1}\theta) } \mathrm{d}\theta. \end{align*}\]Consider $\Gamma(p)\Gamma(q)$,
\[\begin{align*} \Gamma(p) \Gamma(q) &= \int_0^\infty { 2 x^{2p-1} e^{-x^2} } \mathrm{d}x \int_0^\infty { 2 y^{2q-1} e^{-y^2} } \mathrm{d}y \\ &= 4 \int_0^\infty \int_0^\infty { x^{2p-1} y^{2p-1} e^{-(x^2+y^2)} } \mathrm{d}x \mathrm{d}y. \end{align*}\]Let $x = \rho \cos\theta$ and $y = \rho \sin\theta$,
\[\begin{align*} \Gamma(p) \Gamma(q) &= 4 \int_0^{\frac{\pi}2} \int_0^\infty { \rho^{2(p+q)-1} (\cos^{2p-1}\theta) (\sin^{2q-1}\theta) e^{-\rho^2} } \mathrm{d}\rho \mathrm{d}\theta \\ &= 2 \int_0^\infty { \rho^{2(p+q)-1} e^{-\rho^2} } \mathrm{d}\rho \cdot 2 \int_0^{\frac\pi2} { (\cos^{2p-1}\theta) (\sin^{2q-1}\theta) } \mathrm{d}\theta \\ & = \Gamma(p+q) \mathrm{B}(p+q). \end{align*}\]Spectrum Model
Turbulence Reynolds number:
\[\mathrm{Re}_t = \frac{ u_t L }{\nu} = \frac{ u_t^4 }{ \varepsilon \nu } = \left( \frac{L}{\eta} \right)^{4/3}.\]Dissipation Scale:
\[\begin{align*} \eta &\equiv \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}. \end{align*}\]Low $\mathrm{Re}_t$ Model
\[\begin{align*} E(\kappa) &= \alpha u_t^2 \frac{\kappa^4}{\kappa_0^5} \exp{\left( -2\frac{\kappa^2}{\kappa_0^2} \right)} \\ &= C \varepsilon^{2/3} \kappa^{-5/3} (\kappa L)^{7/3} (\kappa\eta)^{10/3} \exp{\left( -2\frac{\kappa^2}{\kappa_0^2} \right)}, \quad \textrm{(DO NOT USE)} \end{align*}\]where $\alpha = 16 \sqrt{ \frac{2}{\pi} }$ and the energy spectrum function peak at $\kappa = \kappa_0$.
High $\mathrm{Re}_t$ Model
General form:
\[\begin{align} E(\kappa) = C \varepsilon^{2/3}\kappa^{-5/3} f_L(\kappa L) f_\eta(\kappa \eta), \end{align}\]where
\[\begin{gather*} \lim_{\kappa \to \infty} f_L(\kappa L) = 1, \\ \lim_{\kappa \to 0} f_\eta(\kappa \eta) = 1. \end{gather*}\]Set $C = 1.5$.
Energy-containing Range Function
General form:
\[\begin{align} f_L(\kappa L) = \left\lbrace \frac { \kappa L } { [(\kappa L)^2 + c_L]^{1/2} } \right\rbrace^{5/3 + p_0}. \end{align}\]The integral of the model spectrum Eq.$(1)$ over all $\kappa$ yields
\[\begin{align*} k &= \int_0^\infty { C \varepsilon^{2/3}\kappa^{-5/3} f_L(\kappa L) f_\eta(\kappa \eta) } \mathrm{d}\kappa \\ &\approx C (\varepsilon L)^{2/3} \int_0^\infty { (\kappa L)^{-5/3} f_L(\kappa L) } \mathrm{d} (\kappa L) \\ &= \tfrac{2}{3} C k \cdot G_L, \end{align*}\]which requires $G_L = \frac{3}{2C} = 1$ provided that $C = 1.5$. Introduce Eq.$(2)$,
\[\begin{align*} G_L &= \int_0^\infty { t^{-5/3} f_L(t) } \;\mathrm{d}t \\ &= \int_0^\infty { \frac { t^{p_0} } { (c_L+t^2)^{\tfrac{1}{2} (5/3+p_0)} } } \;\mathrm{d}t \\ &= c_L^{-1/3} \int_0^\infty { \frac { x^{p_0} } { (1+x^2)^{\tfrac{1}{2} (5/3+p_0)} } } \;\mathrm{d}x \quad (x = c_L^{-1/2} t) \\ &= c_L^{-1/3}\cdot \tfrac{1}{2}\; \mathrm{B} \left( \tfrac13, \tfrac12 + \tfrac12p_0 \right) \\ &= c_L^{-1/3} \frac { \Gamma\left( \tfrac{1}{3} \right) \Gamma\left( \tfrac{1}{2} + \tfrac{1}{2}p_0 \right) } { 2\; \Gamma\left( \tfrac{5}{6} + \tfrac{1}{2}p_0 \right) } = \frac{3}{2C} = 1. \end{align*}\]For normal spectrum model, let $p_0 = 2$ then
\[\begin{align*} c_L = \left[ \frac { \Gamma\left( \tfrac{1}{3} \right) \Gamma\left( \tfrac{3}{2} \right) } { \Gamma\left( \tfrac{11}{6} \right) } \cdot \frac{C}{3} \right]^3 = 2.009744224711003. \end{align*}\]If $p_0 = 4$ (known as $\sf{von\, K\acute{a}rm\acute{a}n}$ spectrum),
\[\begin{align*} c_L = \left[ \frac { \Gamma\left( \tfrac{1}{3} \right) \Gamma\left( \tfrac{5}{2} \right) } { \Gamma\left( \frac{17}{6} \right) } \cdot \frac{C}{3} \right]^3 = 1.100753974315793. \end{align*}\]Dissipation Range Function
The expression for dissipation obtained from integration of the model spectrum Eq.$(1)$ is
\[\begin{align*} \varepsilon &= 2 \nu \int_0^\infty { \kappa^2 \cdot C \varepsilon^{2/3}\kappa^{-5/3} f_L(\kappa L) f_\eta(\kappa \eta) } \mathrm{d}\kappa \\ &\approx 2 \nu C \varepsilon^{2/3} \eta^{-4/3} \int_0^\infty { (\kappa \eta)^{1/3} f_\eta(\kappa \eta) } \mathrm{d} (\kappa \eta) \\ &= 2 C \varepsilon \int_0^\infty { (\kappa \eta)^{1/3} f_\eta(\kappa \eta) } \mathrm{d} (\kappa \eta) \quad \left( \eta = \varepsilon^{-1/4} \nu^{3/4} \right) \\ &= 2 C \varepsilon \cdot G_\eta, \end{align*}\]which requires $G_\eta = \frac{1}{2C} = \frac{1}{3}$ provided that $C = 1.5$.
Exponential decay model:
\[\begin{align} f_\eta(\kappa \eta) = \exp \left[ -\mathrm{B} (\kappa \eta)^{1/q_0} \right]. \\ \end{align}\]Using Eq.$(3)$, the integration becomes
\[\begin{align*} G_\eta &= \int_0^\infty { t^{1/3} \exp\left(-\mathrm{B} t^{1/q_0}\right) } \;\mathrm{d} t \\ &= \mathrm{B}^{-\frac{4}{3} q_0} q_0 \int_0^\infty { x^{\frac{4}{3}q_0 - 1} \exp(-x) } \;\mathrm{d} x \quad \left( x = \mathrm{B} t^{1/q_0} \right) \\ &= \mathrm{B}^{-\frac{4}{3} q_0} q_0\; \Gamma\left(\frac{4}{3}q_0\right) = \frac{1}{2C} = \frac{1}{3}. \end{align*}\]Thus,
\[\begin{align} \mathrm{B} = \left[ 2 C \cdot q_0\; \Gamma\left(\frac{4}{3}q_0\right) \right]^{3/(4q_0)}. \end{align}\]For simple spectrum model, $q_0 = 1$ and
\[\mathrm{B} = \left[ 2 C \cdot \Gamma\left(\frac{4}{3}\right) \right]^{3/4} = 2.093977746651470.\]If $q_0 = 3/4$ (the Pao spectrum),
\[\begin{align*} \mathrm{B} = \frac32 C \cdot \Gamma\left(1\right) = 2.25. \end{align*}\]If $q_0 = 1/2$ (FAKE Pao spectrum),
\[\begin{align*} \mathrm{B} = \left[ C \cdot \Gamma\left(\frac{2}{3}\right) \right]^{3/2} = 2.894820410941134. \end{align*}\]Set $\mathrm{B} = 2$ while keeping $q_0 = 1/2$, now
\[\begin{align*} C = \frac{\mathrm{B}^{2/3}}{\Gamma(\frac23)} = 1.172276805254214. \end{align*}\]Thus, $c_L$ of $\sf{von\, K\acute{a}rm\acute{a}n}$ spectrum ($p_0 = 4$) will be
\[c_L = 0.525420485368824.\]Unity-consistent exponential decay model:
\[\begin{align} f_\eta &= \exp \left\lbrace - \mathrm{B} \left\lbrace \left[ (\kappa \eta)^{4} + c_\eta^{4} \right]^{1/4} - c_\eta \right\rbrace \right\rbrace. \end{align}\]Using Eq.$(5)$, the integration becomes
\[\begin{align*} G_\eta &= \int_0^\infty { t^{1/3} \exp \left\lbrace - \mathrm{B} \left[ \left( t^{4} + c_\eta^{4} \right)^{1/4} - c_\eta \right] \right\rbrace } \;\mathrm{d} t \\ &= \int_0^\infty { t^{1/3} \exp \left\lbrace - \mathrm{B} \left[ \left( t^{4} + c_\eta^{4} \right)^{1/4} - c_\eta \right] \right\rbrace } \;\mathrm{d} t \\ &= e^{\mathrm{B} c_\eta} \int_0^\infty { t^{1/3} \exp \left\lbrace - \mathrm{B} c_\eta \left[ (t/c_\eta)^{4} + 1 \right]^{1/4} \right\rbrace } \;\mathrm{d} t \\ &= e^{\mathrm{B} c_\eta} c_\eta^{4/3} \int_0^\infty { x^{1/3} \exp \left[ - \mathrm{B} c_\eta \left( x^{4} + 1 \right)^{1/4} \right] } \;\mathrm{d} x \quad (x = t/c_\eta) \\ &= c_\eta^{4/3} \int_0^\infty { x^{1/3} C_\eta^{1-\left(x^{4}+1\right)^{1/4}} } \;\mathrm{d} x \quad \left(C_\eta = e^{\mathrm{B} c_\eta}\right) \\ &= \frac{1}{3}. \end{align*}\]Provided that $\mathrm{B} = 0.52$,
\[c_\eta = 0.401684789281759.\]Spectrum Peak
Using Eq.$(2)$,
\[\begin{align} f_L'(t) &= c_L \left( \frac{5}{3}+p_0 \right) \frac { t^{\frac23+p_0} } { \left( t^2 + c_L \right)^{\frac{11}{6} + \frac12 p_0} }. \end{align}\]Let $f_\eta(t) = \exp[g_\eta(t)]$, then
\[\begin{align} f_\eta'(t) &= \exp[g_\eta(t)] g_\eta'(t). \end{align}\]Introduce Eq.$(6)$ and Eq.$(7)$, then
\[\begin{alignat*} {2} &\frac{\mathrm{d} E(\kappa)}{\mathrm{d} \kappa} \bigg|_{\kappa_0} = 0 \\ \iff &-\frac53\kappa_0^{-1} f_L(\kappa_0 L) f_\eta(\kappa_0 \eta) + L f_L'(\kappa_0 L) f_\eta(\kappa_0 \eta) + \eta f_L(\kappa_0 L) f_\eta'(\kappa_0 \eta) = 0 \\ \iff &f_L(\kappa_0 L) \left[ -\frac53(\kappa_0 L)^{-1} + \left(\frac{L}\eta\right)^{-1} g_\eta'(\kappa_0 \eta) \right] + f_L'(\kappa_0 L) = 0 \\ \iff &\frac { (\kappa_0 L)^{\frac23 + p_0} } { [(\kappa_0 L)^2 + c_L]^{\frac{11}6 + \frac12p_0} } \times \\ &\left\lbrace c_L \left( \frac{5}{3}+p_0 \right) - \frac53 [(\kappa_0 L)^2 + c_L] + (\kappa_0 \eta) g_\eta'(\kappa_0 \eta) [(\kappa_0 L)^2 + c_L] \right\rbrace = 0. \end{alignat*}\]If $\kappa_0 \eta$ is samll enough, then
\[\begin{align} \kappa_0 L = \left( \frac35 p_0 c_L \right)^{1/2}. \end{align}\]Special Form Constant Conversion
Another form of spectrum was used in the literature:
\[E(\kappa) = \alpha \frac{u_t^2}{\kappa_e} \frac { \left(\kappa/\kappa_e\right)^4 } { \left[ 1 + \left(\kappa/\kappa_e\right)^2 \right]^{17/6} } \exp\left[-2\left(\frac{\kappa}{\kappa_\eta}\right)^2\right],\]where $\kappa_\eta = \varepsilon^{1/4} \nu^{-3/4}$. In this instance, it is easy to deduce that $p_0 = 4$, $q_0 = 0.5$, and $\mathrm{B} = 2$.
Notice that
\[L = \frac{c_L^{1/2}}{\kappa_e}\]must be fullfilled if special form is identical to general form essentially.
Through rearrangement,
\[\begin{align*} \alpha \frac{u_t^2}{\kappa_e} \frac { \left(\kappa/\kappa_e\right)^4 } { \left[ 1 + \left(\kappa/\kappa_e\right)^2 \right]^{17/6} } &= \alpha \varepsilon^{2/3} L^{5/3} c_L^{1/3} \frac { (\kappa L)^4 } { \left[ c_L + \left(\kappa L\right)^2 \right]^{17/6} } \\ &= \alpha c_L^{1/3} \varepsilon^{2/3} \kappa^{-5/3} \frac { (\kappa L)^{17/3} } { \left[ c_L + \left(\kappa L\right)^2 \right]^{17/6} }. \end{align*}\]Therefore, if $c_L = 0.525420485368824$,
\[\begin{gather*} \alpha = C \cdot c_L^{-1/3} = 1.452762112210974, \\ c_L^{1/2} = 0.724858941704401. \end{gather*}\]