# POLYFLOW - How can I set a proper value of the coefficient for the Navier's slipping law?

 Let us consider the boundary of a flow domain. Let Vs denote the relative (tangential) slipping velocity of a fluid point at a solid wall of the domain. A tangential force Fs also exist, which balances the wall shear stress, and can be written as:Fs = -k.Vswhere k is the friction coefficient, while the negative sign originates from the fact that the slipping force is opposed to the slipping velocity. Such a relationship has been suggested in the XIX:th century, and is known as the Navier's slipping law.How to select a proper value for this coefficient?An inverse approachIn most cases, the geometry is not that of a slit or a circular die, while the rheology significantly deviates from the Newtonian linear behaviour. Under some circumstances, it is still possible to derive an analytical solution along a similar procedure as above. In most cases however, the use of an evolution scheme applied on the slipping coefficient will offer a convenient alternative. From a sequence of solutions, it is then possible to estimate the best appropriate value for matching some requirements. A rigorous approach for axisymmetric flowLet us assume a cylindrical channel of radius R. We also assume that a fluid of constant viscosity M is flowing at an overall flow rate Q; the mean velocity is given by V = 4Q/(pi R^3). We presently try to evaluate the coefficient k of the Navier's slipping law, such that a relative wall velocity Vs is obtained and equals to a fraction b of the mean velocity V.By considering the symmetry, the balance of force and the volume flow rate, the velocity profile v(r) is given byv(r) = c - a.r2wherec = (2V - Vs) , a = 2(V - Vs)/R2At the wall, the balance of force requires2MaR = kVsThis leads tok = 4M(V - Vs)/(Vs.R) = 4(M/R)(1/b - 1) A quick and simplified approachLet us consider a point, and the flow cross-section passing through that point. Let V and L denote the typical fluid velocity passing through that cross-section whose typical dimension is L. A typical wall shear rate can be estimated by the ratio V/L. Let us also assume the shear viscosity h is constant and known.Let us assume that one wants a relative slipping velocity Vs whose magnitude is a fraction a of the mean velocity V. The possible relative slipping velocity results from a balance between the wall shear stress and the friction force. In other words:k.Vs = k.a.V = h.V/LIn such a relationship, all quantities are known, except the slipping coefficient k, which can be directly estimated. It is important to remind here that this is a rather simplified approach, and concerns mainly the order of magnitude. More sophisticated developments are possible, and suggested below. However, they involve some assumptions.A rigorous approach for planar flowLet us assume a flat and long channel, whose length and opening are given by L and 2H. We also assume that a fluid of constant viscosity M is flowing at an overall flow rate Q; the flow rate Q' per length unit in the transverse direction is given by Q' = Q/L. The mean velocity is given by V = Q/(2HL). We presently try to evaluate the coefficient k of the Navier's slipping law, such that a relative wall velocity Vs is obtained and equals to a fraction b of the mean velocity V.By considering the symmetry, the balance of force and the volume flow rate, the velocity profile v(x) is given byv(x) = c - a.x^2wherec = (3V - Vs)/2 , a = 3(V - Vs)/2H^2At the wall, the balance of force requires2MaH = kVsThis leads tok = 3M(V - Vs)/(Vs.H) = 3(M/H)(1/b - 1)

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