Introduction

Shot blasting treatment integrates surface peening, rust removal and cleaning, and is considered to be a highly efficient mechanical equipment. Because of its high processing efficiency, wide processing range and flexibility, it is widely used for cleaning and maintenance of mechanical parts, road and bridge and ship hull shells, and shot blasting machine is the equipment to realize shot blasting treatment1. Mobile shot blasting machine broke the traditional fixed shot blasting machine can only be cleaned indoors traditional parts of the limitations of the shot blasting process to expand to the surface of the ship rust removal and strengthening2, airport runway debonding repair, highway and bridge paint removal maintenance and various other fields, compared with other maintenance equipment, shot blasting treatment in the economy, environmental protection and efficiency has huge advantages3. Mobile shot blasting machine recycling device can realize the recycling of steel shot and automatic recycling of processing waste, but in the actual production and processing, the recycling rate of waste generated after shot blasting process is low4, and the dust particles generated during the recycling process are easy to cause environmental pollution, construction personnel are prone to lung diseases caused by inhalation of large amounts of dust5. Studies have shown that a more efficient and environmentally friendly recycling efficiency is achieved when the waste recovery efficiency reaches 90%6. Therefore, the efficient recycling of waste materials has received a lot of attention from researchers7,8,9.

Solid particles flowing with the air stream inside the blast machine recovery unit causes the gas phase (air)10 and solid phase (solid particles)11 to form a gas–solid two-phase flow inside the pipe

$$\frac{{\partial \rho_{{\text{a}}} }}{\partial t} + \nabla \cdot \left( {\rho_{{\text{a}}} v_{{\text{a}}} } \right) = 0$$
(1)
$$\frac{{\partial \left( {\rho_{{\text{a}}} v_{{\text{a}}} } \right)}}{\partial t} + \nabla \cdot \left( {\rho_{{\text{a}}} v_{{\text{a}}}^{2} } \right) = - \nabla p_{{\text{a}}} + \nabla \tau_{{\text{a}}} + \rho_{{\text{a}}} g - f_{{\text{a}}}$$
(2)

where ρa is the air density, kg/m3; t is the time, s; va is the air speed, m/s; pa is the air pressure, Pa; g is the acceleration of gravity, m/s 2; τa is the air stress tensor, Pa; fa is the average resistance of air, N.

A standard k-ε turbulence model is used to describe the rotating airflow in the recovery bin of a horizontal mobile blast machine, which is represented in the model as follows:

$$\frac{\partial }{\partial t}(\rho_{g} k_{g} ) + \frac{\partial }{{\partial x_{i} }}\left( {\rho_{g} k_{g} u_{i} } \right) = \frac{\partial }{{\partial x_{j} }}\left[ {\left( {\mu + \frac{{\mu_{t} }}{{\sigma_{k} }}} \right)\frac{\partial k}{{\partial x_{j} }}} \right] + G_{k} + G_{b} - \rho_{g} \varepsilon - Y_{M}$$
(3)
$$\frac{\partial }{\partial t}\left( {\rho_{g} \varepsilon } \right) + \frac{\partial }{{\partial x_{i} }}\left( {\rho_{g} \varepsilon u_{i} } \right) = \frac{\partial }{{\partial x_{j} }}\left[ {\left( {\mu + \frac{{\mu_{t} }}{{\sigma_{\varepsilon } }}} \right)\frac{\partial \varepsilon }{{\partial x_{j} }}} \right] + C_{1\varepsilon } \frac{\varepsilon }{{k_{g} }}\left( {G_{k} + C_{3\varepsilon } G_{b} } \right) - C_{2\varepsilon } \rho_{g} \frac{{\varepsilon^{2} }}{{k_{g} }}$$
(4)

where, kg is the turbulent kinetic energy, ε is the turbulent dissipation rate, \(\mu\) is the dynamic viscosity of the gas, and the model constants C1ε、 C2ε、C3ε,、\(\sigma_{k}\) and \(\sigma_{\varepsilon }\) have the following default values: C1ε = 1.44, C2ε = 1.92, C3ε = 1.3, \(\sigma_{k}\) = 1.0, \(\sigma_{\varepsilon }\) = 1.3. \(\mu_{t}\) is the turbulent viscosity, which is expressed as follows

$$\mu_{t} = \rho_{g} C_{\mu } \frac{{k^{2} }}{\varepsilon },C_{\mu } = 0.09$$
(5)

Solid phase control equation

Given that the air in the blast machine recovery unit carries the solid particle motion, the solid particle population is considered as a discrete phase, so the flow of solid particles can be solved by DPM. Based on the Lagrange coordinate system37, the trajectory of the discrete phase is solved by the differential equation of particle forces to simulate the particle motion in turbulent flow. Then the expression of the differential equation of particle force in the discrete phase is:

$$\left\{ {\begin{array}{*{20}l} {\frac{{{\text{d}}v_{{\text{p}}} }}{{{\text{d}}t}} = F_{{\text{d}}} \left( {v_{{\text{a}}} - v_{{\text{p}}} } \right) + \frac{{g\left( {\rho_{{\text{p}}} - \rho_{{\text{a}}} } \right)}}{{\rho_{{\text{p}}} }} + F_{{\text{other }}} } \hfill \\ {F_{{\text{d}}} = \frac{18\mu }{{\rho_{{\text{p}}} d_{{\text{p}}}^{2} }}\frac{{C_{{\text{d}}} Re_{{\text{p}}} }}{24}} \hfill \\ {Re_{{\text{p}}} = \frac{{\rho_{{\text{a}}} d_{{\text{p}}} \left| {v_{{\text{p}}} - v_{{\text{a}}} } \right|}}{\mu }} \hfill \\ {C_{{\text{d}}} = a_{1} + \frac{{a_{2} }}{{Re_{{\text{p}}} }} + \frac{{a_{3} }}{{Re_{{\text{p}}}^{2} }}} \hfill \\ \end{array} } \right.$$
(6)

where vp is the particle velocity, m/s; Fd(va-vp)is the traction force per unit mass of particles, m/s2; ρp is the particle density, kg/m3; μis the aerodynamic viscosity, Pa·s; Cdis the traction coefficient; Rep is the relative Reynolds number of particles; dp is the particle size, m; Fotheris the other force per unit mass of particles, m/s2; a1、a2、a3 are constants.

Considering the actual working conditions, the two-way coupling method of CFD-DEM is used to simulate the transport process of gas and solid phases in the shot blasting machine recovery unit38. Based on the above controlling equations for the gas and discrete phases, the momentum value transmitted from the gas phase to the discrete phase is solved by calculating the momentum change of the particles when they pass through a defined spatial region in the flow field, thus realizing the two-way coupling calculation39,40.

The contact forces between particles a,b and the pipe wall on particle a, are shown in Fig. 1. The above rotational motion of particle a is described as:

$$I_{a} \frac{{d\omega_{a} }}{{d_{t} }} = \sum\limits_{b = 1}^{{k_{a} }} {(F_{c,ab} + F_{d,ab} )}$$
(7)

where, \(I_{a}\) and \(\omega_{a}\) denote the moment of inertia tensor and rotational velocity of particle a, respectively. \(F_{c,ab}\) is the tangential force acting on particle a by particle b. \(F_{c,ab}\) is defined as follows:

$$F_{c,ab} = R_{a,b} \left( {F_{n,ab} + F_{t,ab} } \right)$$
(8)
$$F_{n,ab} = \frac{4}{3}E^{ * } \sqrt {R^{ * } } \delta_{n,ab}^{1.5}$$
(9)
$$F_{t,ab} = - 2\sqrt{\frac{5}{6}} \frac{\ln e}{{\sqrt {\ln^{2} e + \pi^{2} } }}\sqrt {S_{n} m^{*} } v_{n,ab}$$
(10)
$$S_{n} = 2E^{ * } \sqrt {R^{ * } \delta_{n,ab} }$$
(11)

where,\(F_{n,ab}\) is the normal contact force, \(F_{t,ab}\) is the tangential contact force, \(R_{a,b}\) is the vector from the center of mass to the point of contact, \(S_{n}\) is the normal stiffness, \(\delta_{n,ab}^{{}}\) is the normal overlap, \(E^{ * }\),\(R^{ * }\), \(m^{*}\) \(v_{n,ab}\) and \(e\) is the equivalent Young's modulus, equivalent radius of the particle, equivalent mass, relative normal velocity and recovery factor.

Figure 1
figure 1

Schematic diagram of particle collision.

Full size image

\(F_{d,ab}\) is the rolling friction moment of particle b acting on particle a, \(F_{d,ab}\) defined as follows:

$$F_{d,ab} = - \mu_{r,ab} d_{a} \frac{{\omega_{ab} }}{{\left| {\omega_{ab} } \right|}}$$
(12)

where, \(\mu_{r,ab}\) and \(d_{a}\) are the rolling coefficient and particle diameter, respectively, \(\omega_{ab}\) and is the relative angular velocity of particle a to particle b. During the particle–particle collision, the forces and moments on the particle are similar to those in the particle–particle collision, that is, Eqs. (8)–(12).