POLYFLOW - How can I set a proper value of the coefficient for the Navier's slipping law?
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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.Vs where 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 approach In 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 flow Let 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 by v(r) = c - a.r2 where c = (2V - Vs) , a = 2(V - Vs)/R2 At the wall, the balance of force requires 2MaR = kVs This leads to k = 4M(V - Vs)/(Vs.R) = 4(M/R)(1/b - 1) A quick and simplified approach Let 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/L In 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 flow Let 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 by v(x) = c - a.x^2 where c = (3V - Vs)/2 , a = 3(V - Vs)/2H^2 At the wall, the balance of force requires 2MaH = kVs This leads to k = 3M(V - Vs)/(Vs.H) = 3(M/H)(1/b - 1) |
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