Steady-state similarity velocity profiles for dense granular flow down inclined chutes

Abstract

In this paper we exploit the fact that dense granular flow down an inclined chute exhibits a simple scaling property. This implies that the fundamental governing partial differential equation admits similarity profiles which are applicable to any general friction law $\mu (I)$ of the material irrespective of the particular functional form under consideration, where $\mu (I)$ is a function of the dimensionless parameter, called the inertial number I. We adopt a widely accepted constitutive model describing the material response that draws its original inspiration from the realms of classical visco-plasticity. The governing steady-state similarity solutions are however applicable for any friction law describing dense granular media. These steady-state solutions embody the key aspects of dense granular flow down an inclined chute that find wide usage in various industrial applications and will further aid in modelling natural geophysical phenomenon. We provide steady-state analytical solutions for some special cases and we emphasize the importance of similarity solutions as a benchmark standard for various numerical studies concerning the complex material behaviour of dense granular flows.

Appendix A: Integral evaluation for analytical solution (\(\mu _{\infty } = \mu _{0}\))

In this appendix we evaluate the integral arising in Eq. (21) and given by,

$$\begin{aligned} J = \int \frac{\sin \zeta }{(\sigma + \sin \zeta )^2}d\zeta \end{aligned}$$

(24)

On making the successive substitutions,

$$\begin{aligned} t = \tan ({\zeta }/{2}), \quad x = \sigma ^{1/2}t + {1}/{\sigma ^{1/2}}, \end{aligned}$$

we might deduce

$$\begin{aligned} J = 4 \int \frac{t dt}{(\sigma t^2 +2t +\sigma )^2} = \frac{4}{\sigma }\int \frac{(x-{1}/{\sigma ^{1/2})}}{(x^2 - a^2)^2} dx \quad , \end{aligned}$$

where $a = {(1-\sigma ^2)^{1/2}}/{\sigma ^{1/2}}$. With the further substitution $x = a \cosh \varTheta$ we might deduce

$$\begin{aligned} J = - \frac{2}{\sigma (x^2 - a^2)} -\frac{4}{\sigma ^{3/2}a^3} \int {{\rm{csch}} ^3 \varTheta } d \varTheta \quad , \end{aligned}$$

so that

$$\begin{aligned} K = \int \frac{d\varTheta }{\sinh ^3 \varTheta } = - \frac{\coth \varTheta }{\sinh \varTheta } - \int \frac{\coth \varTheta }{\sinh ^2 \varTheta }\cosh \varTheta d\varTheta \quad , \end{aligned}$$

and therefore,

$$\begin{aligned} 2K = -\frac{\cosh \varTheta }{\sinh ^2 \varTheta } - \ln \tanh \left( \frac{\varTheta }{2} \right) \quad . \end{aligned}$$

The above expression may be simplified to give

$$\begin{aligned} \begin{aligned} J =&- \frac{2}{\sigma (\sigma t^2 + 2t + \sigma )} \\&+\frac{2}{(1-\sigma ^2)^{3/2}}\left[ \frac{(1-\sigma ^2)^{1/2}(1+\sigma t)}{\sigma (\sigma t^2 + 2t + \sigma )}\right. \\&\left. + \frac{1}{2}\ln \left( \frac{\sigma t +1 - (1-\sigma ^2)^{1/2}}{\sigma t +1 + (1-\sigma ^2)^{1/2}}\right) \right] , \end{aligned} \end{aligned}$$

where we have used,

$$\begin{aligned} \tanh ({\varTheta }/{2}) = {\sinh \varTheta }/{(1+\cosh \varTheta )} \quad , \end{aligned}$$

and ${x}/{a} = ({\sigma t +1 })/{(1-\sigma ^2)^{1/2}}$. Further simplification gives

$$\begin{aligned} J= & {} -\frac{2\cos ^2 ({\zeta }/{2})}{\sigma (\sigma + \sin \zeta )} + \frac{2}{(1-\sigma ^2)^{3/2}} \\&\left[ \frac{\left( 1-\sigma ^2\right) ^{1/2}\left( 1+\sigma \tan ({\zeta }/{2})\right) \cos ^2 ({{\zeta }/{2}})}{\sigma (\sigma + \sin \zeta )} \right. \\&\left. + \frac{1}{2}\ln \left( \frac{\sigma \tan ({\zeta }/{2}) +1 -(1-\sigma ^2)^{1/2}}{\sigma \tan ({\zeta }/{2}) +1 +(1-\sigma ^2)^{1/2}}\right) \right] + C \quad , \end{aligned}$$

where C denotes a constant of integration, and

$$\begin{aligned} J =&\frac{2 \cos ^2 ({\zeta }/{2})\left( \sigma + \tan ({\zeta }/{2})\right) }{(1-\sigma ^2)(\sigma + \sin \zeta )} \\&+\frac{1}{(1-\sigma ^2)^{3/2}} \ln \left( \frac{\sigma \tan ({\zeta }/{2}) +1 -(1-\sigma ^2)^{1/2}}{\sigma \tan ({\zeta }/{2}) +1 +(1-\sigma ^2)^{1/2}}\right) + C \quad . \end{aligned}$$

On using

$$\begin{aligned} 2\cos ^2({\zeta }/{2})\left( \sigma + \tan ({\zeta }/{2})\right) = (\sigma + \sin \zeta + \sigma \cos \zeta ) \end{aligned}$$

we finally obtain

$$\begin{aligned} J =&\frac{\sigma \cos \zeta }{(1-\sigma ^2)(\sigma + \sin \zeta )} + \frac{1}{(1-\sigma ^2)^{3/2}}\ln \\&\quad \left[ \frac{\sigma \tan ({\zeta }/{2})+1 - (1-\sigma ^2)^{1/2}}{\sigma \tan ({\zeta }/{2})+1 + (1-\sigma ^2)^{1/2}}\right] + C^{*} \quad , \end{aligned}$$

where $C^{*}$ denotes a different constant of integration.