Abstract In this article, an analytical simulation based on a new model incorporating surface interaction was conducted to study the slip phenomenon in Couette flow at different scales. The velocity profile was calculated by taking account micro-force between molecules and macro-force from the viscous shearing effect, as they contribute to the achievement of slip length. The calculated results were compared with those obtained from the MD simulation, showing an excellent agreement. Further, the effect of the shear rate on slip was investigated. The results can well predict the fluid flow behaviors on a solid substrate, but has to be proved by experiment.
PACS: 23.40.-s, 21.60.Cs, 23.40.Hc, 27.40.+z
The “no-slip” boundary condition, i.e., zero …show more content…
The inside fluid is trapped in two movable walls. The distance between two walls is set as H, and the inside fluid is divided into (n+1) layers in the thickness direction. Note that one wall is named as the 0th layer, the other wall is named as the (n+1)th layer, and distance between the ith and the (i+1)th layer is named as hi+1. The total number of molecules, N, in each layer is assumed as equal, and tends to be infinite, and as a result the fluid layer can be seen as an infinite plane for a molecule. All the molecules in a layer are assumed to have the same distance to the wall or the adjacent layer, and therefore, they own the same forcing state. Figure 1(b) shows the force schematic of the ith layer in the thickness direction. For the ith layer, it suffers forces from the wall and the adjacent fluid layers, and the force equilibrium equation can be expressed by Eq. (1). , (1) where Fw,j is the force acted on the ith layer from wall, j = 1 represents force from one wall and j = 2 indicates the force from the other wall, and Fl,k represents the force from the kth fluid layer.
For a certain layer, Fw,j and Fl,k are equal to the sum of the force acting on a molecule in this layer from the wall and fluid layers, respectively. According to the assumption that the forcing state of all the molecules in a layer remains the same, Eq. (1) can be transformed to Eq. …show more content…
As the total molecular quantity of each layer remains unchanged, σ is determined by Eq. (3) , (3) where σ0 and ρ0 represent the distance parameter of the potential function and the fluid density under the condition of ambient pressure, respectively, and ρ(p) represents the density under the conditions of working pressure.
For forces Fsw,j and Fsl,k acting on a single molecule, these two forces are exerted on a single molecule from an infinite wall surface and the kth layer, respectively. They can be deduced by the L–J potential function represented by Eqs. (4) and (5). (4)
PACS: 23.40.-s, 21.60.Cs, 23.40.Hc, 27.40.+z
The “no-slip” boundary condition, i.e., zero …show more content…
The inside fluid is trapped in two movable walls. The distance between two walls is set as H, and the inside fluid is divided into (n+1) layers in the thickness direction. Note that one wall is named as the 0th layer, the other wall is named as the (n+1)th layer, and distance between the ith and the (i+1)th layer is named as hi+1. The total number of molecules, N, in each layer is assumed as equal, and tends to be infinite, and as a result the fluid layer can be seen as an infinite plane for a molecule. All the molecules in a layer are assumed to have the same distance to the wall or the adjacent layer, and therefore, they own the same forcing state. Figure 1(b) shows the force schematic of the ith layer in the thickness direction. For the ith layer, it suffers forces from the wall and the adjacent fluid layers, and the force equilibrium equation can be expressed by Eq. (1). , (1) where Fw,j is the force acted on the ith layer from wall, j = 1 represents force from one wall and j = 2 indicates the force from the other wall, and Fl,k represents the force from the kth fluid layer.
For a certain layer, Fw,j and Fl,k are equal to the sum of the force acting on a molecule in this layer from the wall and fluid layers, respectively. According to the assumption that the forcing state of all the molecules in a layer remains the same, Eq. (1) can be transformed to Eq. …show more content…
As the total molecular quantity of each layer remains unchanged, σ is determined by Eq. (3) , (3) where σ0 and ρ0 represent the distance parameter of the potential function and the fluid density under the condition of ambient pressure, respectively, and ρ(p) represents the density under the conditions of working pressure.
For forces Fsw,j and Fsl,k acting on a single molecule, these two forces are exerted on a single molecule from an infinite wall surface and the kth layer, respectively. They can be deduced by the L–J potential function represented by Eqs. (4) and (5). (4)