Abstract
Bingham fluids constitute a very important class of non-Newtonian fluids. The modeling of Bingham materials is of crucial importance in industrial applications, since a large variety of materials (e.g. foams, pastes, slurries, oils, ceramics, etc.) exhibit the fundamental character of viscoplasticity, that is the capability of flowing only if the stress is above some critical value. The flow of these materials is difficult to predict, because of the presence of unknown interfaces separating the yielded and the unyielded regions which are difficult to track. This is particularly evident when the flow occurs in complex geometries and when major simplifications, such as lubrication approximation, can be applied. Indeed, in some cases the Bingham model may even lead to a paradox, known as the “lubrication paradox”. In this chapter we focus on some practical situations of Bingham flow which are the subject of a current mathematical research (lubrication flows, asymptotic expansions, etc.). Such issues and developments arise, for example, in the petroleum industry and in many natural contexts.
A chapter to appear in “New trends in non-newtonian fluid mechanics and complex flow”, Lecture Notes Centro Internazionale Matematico Estivo (C.I.M.E.) Series, Lecture Notes in Mathematics, Springer, 2017. The corresponding lectures were given at the CIME-CISM Course “New trends in non-newtonian fluid mechanics and complex flows” held in Levico Terme, Italia, August 28–September 2, 2016
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1 Introduction
Bingham fluids, or yield stress fluids, are encountered in a wide range of applications: toothpastes, cements, mortars, foams, muds, mayonnaise, etc. The fundamental character of these fluids is that they are able to deform indefinitely only if they are submitted to a stress above some critical value. Actually, toothpaste visually exhibits the fundamental character of such a fluid model: it flows when the applied stress exceeds a certain value, otherwise it does not flow, behaving as a solid body (actually below the critical stress toothpaste deforms in finite way). Despite the apparent simplicity in the constitutive modeling (especially within the implicit framework theory developed by Rajagopal and co-workers [22,23,24,25]), the flow characteristics of these materials are difficult to be predicted, since they involve unknown boundaries separating the liquid and the solid regions. Readers are referred to the books by Huilgol [14] and by Ionescu, Sofonea [15] where several issues concerning the Bingham model (constitutive equations, mathematical techniques, numerical methods and so on) are deeply analyzed.
A well known example of materials which are often modeled as Bingham Fluids are waxy crude oils (i.e. oils with an high paraffin content). These fluids are known to cause handling and pipelining difficulties. The flow properties depend strongly on the yield stress which, in turn, depends on the shear history [31]. This leads to a definable minimum operating point below which flow in a waxy crude oil pipeline would cease.
Familiar examples of non-Newtonian fluids described by the Bingham model include also mud (see [18] and the references therein cited), lubricated pipelining [13], and materials used in ceramic casting [16].
2 Constitutive Model
The simplest shear-stress experiment that characterizes the Bingham fluid is represented in Fig. 1. On the top surface of a layer of material a uniform shear force F is applied. If A is the area of the surface, the applied shear is F∕A. If the applied shear load (i.e. force per unit surface) does not exceed a certain threshold, τ o, the material does not move (the bottom of the layer is fixed on the “floor” ). When the applied shear load exceeds the τ o, the material flows as a linear viscous fluid.
A schematic representation of the 1D shear stress experiment. F is the shear force and A which is uniformly applied on a surface A so that the shear stress is F∕A. The threshold is τ o
Actually such a peculiar rheological behavior is well highlighted by the experimental tests performed in cylindrical viscometers. We just mention the recent review by Coussot [5] and the numerous experimental papers therein cited. Hence, considering a simple one-dimensional shear flow, if τ denote the modulus of the shear stress and \(\dot {\gamma }\) the modulus of the strain rate, the constitutive Bingham model writes as
and
Indeed (2) means, from the physical point of view, rigid behavior [11]. The threshold τ o is usually defined shear yield stress, or simply yield stress, and μ is referred to as viscosity. In Fig. 2 we have reported the shear stress-shear rate curve for a Bingham fluid.
The shear stress—shear rate curve for the Bingham model
The model (1), (2) was introduced for the first time by E.C. Bingham [1, 2]. We note that in the Bingham model is much more natural to express the modulus of the shear rate \(\dot {\gamma }\) in terms of the modulus of the shear stress τ. Indeed (1), (2) can be rewritten (see also Fig. 3 )
where \(\left ( \ \ \ \ \right ) _{+}\) denotes the positive part, namely
In the example of Fig. 1 it is easy to identify the shear stress and formulate the yield criterion. But how can we formulate the yield criterion in the general 3D case? What is in this case the shear stress? Let us consider a point P and a small facet of area ds surrounding. The normal to the facet is n. If we denote by F the force acting on ds we have
where \(\mathbf {T}\left ( P\right ) \) is the Cauchy stress. The force vector F can be splitted into its component along n and in a tangential vector τ (said tangential or shear force)
where \(\sigma =\boldsymbol {F\cdot n}=\mathbf {T}\left ( P\right ) \boldsymbol { n\cdot n}\) (see Fig. 4). Obviously \(\left \vert \boldsymbol {F} \right \vert ^2 = \sigma ^2 +\tau ^2 \), where \(\tau =\left \vert \boldsymbol {\tau }\right \vert \).
Shear stress—shear rate according to (3)
Decomposition of the force F
We then consider a reference frame in which \(\mathbf {T}\left ( P\right ) \) is diagonal, namely
and assume T 1 > T 2 > T 3. Hence
and
We thus get the system
which, once solved with respect to \(n_{1}^{2}\), \(n_{2}^{2}\), \(n_{3}^{2}\), gives
Thus, in the \(\left ( \sigma ,\tau \right ) \) plane a domain is defined
which corresponds to the area bordered by three circles, the so-called three Mohr circles (see Fig. 5).
Three points on the horizontal axis correspond to the three main stresses. The figure shows that the maximum shear is equal to the radius of the largest circle
Thus, since the maximum shear stress is τ MAX, a possible yield criterion is
where
Such a criterion is known as Tresca criterion [3]. The representation in the principal stress space of the surface (5), known as Tresca Surface, or
is a cylindrical unbounded surface with hexagonal section and axis (1,1,1). So, according to the Tresca criterion, at every point of the material we have to compute the eigenvalues of the stress tensor. If all the eigenvalues are within the Tresca surface then the material is in a rigid state. if at least one of the eigenvalues is outside the surface then the material is in a fluid state.
The Tresca criterion, unfortunately, is not very practical to use. For this reason it is preferred to use the Von Mises criterion [30], based on the second invariant of the stress tensor. The maximum shear stress criterion (4) is replaced by
where
Thus, splitting the Cauchy stress as T = −P I + S, where P = 1∕3trT and S is the extra-stress, we immediately realize that σ V M coincides with the second invariant of the extra-stress, namely
Therefore the generalization of the Bingham model (1), (2) to the 3D case is the following
and
where
and where v is the fluid velocity field. We remark that the Bingham constitutive equation can be written in the implicit form (see [22,23,24, 26])
The above constitutive equation allows to express S as a function of D only when II S > τ o, while D = 0, entails only II S ≤ τ o, the stress being constitutively undetermined.
3 Flow in a Channel
We consider the flow of an incompressible Bingham fluid in a symmetric channel of length L and width 2H. We consider a laminar flow so that the velocity field is
where, as shown in Fig. 6, x, y are the longitudinal and transversal coordinates respectively.
Sketch of the channel
Due to symmetry we consider just the upper part of the channel. The the inlet pressure ΔP is prescribed and we rescale the outlet pressure to 0. Hence the pressure gradient driving the flow is given by
In principle \(f_{o}=f_{o}\left ( t\right )\) since ΔP may depend on time. In such a simple setting
so that \(II_{\mathbf {S}}=\left \vert S_{12}\right \vert \) and \(II_{\mathbf {D} }=\left \vert v_{y}\right \vert \). Hence, (6), (7) give
Next, we assume that the viscous region, namely II S > τ o, and the rigid region, II S ≤ τ o, are separated by a sharp interface y = s(t) a priori unknown, i.e. a free boundary. In the fluid region, i.e. \(s\left ( t\right ) <y<H\), the motion equation reduces to
coupled with the no-slip condition on y = H,
and with the threshold condition on the free boundary \(s\left ( t\right ) \)
In particular, we assume no-slip also at the interface \(s\left ( t\right ) \), so that the velocity of the rigid core is
Figure 7 represents a sketch of the problem we have to solve.
The position s(t) of the “fluid – rigid” is unknown. Indeed we deal with a free boundary problem
We need an evolution for the interface \(s\left ( t\right ) \). According to the approach developed in [12], which is based on the pioneering works by Safronchik [29] and Rubinstein [28], the unyielded region is treated as an evolving non material volume, whose motion is determined by using the integral (or global) momentum balance. The dynamics of the unyielded domain
is thus given by (see, e.g., [3])
where w is the velocity of the boundaryFootnote 1 ∂Ω and n its outward normal.
Taking Fig. 8 into account relation (9 ) reduces toFootnote 2
The evolving non-material domain Ω
We thus end up with this free boundary problem
The domain of the problem is represented in Fig. 9.
Sketch of the domain of problem (10)
The problem (10) is a free boundary problem but not of Stefan type, because in the evolution equation for the free boundary \(\dot s\) does not appear. However if we consider as new dependent variable \( z\left (y.t\right ) =v_{t}\left (y,t\right ) \), problem (10 ) rewrites as
In [4] global well posedness of problem (11) has been proved.
4 Bingham Model with Deformable Core
The Bingham model predicts that the material behaves as a rigid body if the shear stress is less than the threshold τ o. It is however evident that the schematization of the rigid body is not plausible from a physical point of view (think, for example, to the mayonnaise). Oldroyd [20] and Yoshimura et al. [32] have proposed to treat the “solid phase” as deformable. In [7] Fusi et al. have studied an extension of the 1D problem (10) to the case of an elastic core. The model has been developed within the context of the theory of natural configurations [24, 25]. In [10] Fusi et al. have extended the problem studied in [7] to a 2D channel flow where the channel amplitude is not uniform.
Actually, other models have been proposed for Bingham-like fluids whose unyielded core is deformable. In [8, 9], for instance, the material in the non-fluid region has been modeled as a visco-elastic fluid.
4.1 Channel Flow of a Bingham-Like Fluid with Linear Elastic Core
Here we consider a channel flow of a Bingham-like fluid driven by a known pressure gradient, assuming that the continuum behaves as a linear viscous fluid if the stress is above the yield stress and as a linear elastic solid when the stress is below such a threshold. In particular we assume that the domain may be split in two sub-domains, see Fig. 10. The inner core, in which the material behaves as a linear elastic material, and the outer part (i.e. the one close to the channel walls) where a linear viscous behavior occurs. The two regions are separated by the unknown sharp interfaces y = ±σ(x, t). Moreover the channel width varies along x, so that the channel walls are given by the function y = H(x). The channel is finite and we denote its length by L. As in Sect. 3, we limit our analysis to the upper part of the channel because of symmetry.
A schematic representation of the channel flow with deformable core
The motion equations are obtained by imposing the mass and momentum balance. The evolution equation of the interface y = σ(x, t), as well as the boundary conditions, are derived imposing Rankine-Hugoniot conditions and Von Mises yield criterion. The general mathematical problem is therefore a two-dimensional free boundary problem. We develop the model assuming that the characteristic height of the upper part H is far less than L, i.e. the aspect ratio
is very small.
4.2 Kinematics and Constitutive Equation
Consider a two dimensional setting and assume that the motion is given by
where χ is a differentiable and invertible mapping from R 2 →R 2. The vectors x, X are the Eulerian and Lagrangian coordinates, respectively. The deformation tensor isFootnote 3
and mechanical incompressibility entails \(\det \mathbf {F}=1\). The Eulerian velocity and acceleration are defined as
respectively. The strain rate tensor is
and mechanical incompressibility gives
Splitting, as usual, the Cauchy stress as T = −P I + S, with P = 1∕3trT, we extend we extend the constitutive relation (6), (7) to the case in which the region II S < τ o behaves as a linear elastic materialFootnote 4 (see [7, 10]).
where
is the linearized strain tensor, f = χ −X is the displacement , η is the elastic modulus and Θ is the Heaviside function
From (14) it is clear that, whenever II S ≤ τ o, the continuum behaves as a linear elastic solid whereas the viscous behavior occurs when II S ≥ τ o.
4.3 Flow in a Channel
We rescale the longitudinal variable as
and we introduce the Reynolds and Bingham number
where U is the characteristic velocity and ρ is the material density. The we set
Concerning velocity and pressure, we introduce
where P c comes from the classical Poiseuille formula. Rescaling S as
the motion equations becomeFootnote 5
and mass conservation is
The displacement is rescaled as
so thatFootnote 6
The sharp interface y = σ(x, t) separates the elastic domain from the viscous domain, as shown in Fig. 2. We remark that the interface σ is unknown and it is not material. The dimensionless normal velocity of the interface is
The dimensionless strain rate tensor is
while the dimensionless invariant II D is
The yield criterion becomes
4.3.1 Boundary Conditions
Let 
denotes the “jump” across the surface y = σ. Assuming no-slip on σ we get
The continuity of the stress gives
and
Remark 1
If we neglect \(\mathcal {O}\left ( \varepsilon \right ) \) terms, then 
, provided that S
ij are bounded.
On the channel wall y = h the no-slip condition yields \(\boldsymbol {v}\left ( x,h ,t\right ) \equiv 0\), while the boundary conditions for the pressure are
Finally, we impose the symmetry conditions:
4.3.2 The Elastic Domain and the Viscous Domain
Let us first consider the elastic domain y ∈ [0, σ]. According to (14)
with
The momentum balance yields
In the viscous domain y ∈ [σ, 1]
The momentum equations are
The model is consistent if II S < B n for y ∈ [0, σ], i.e.
and if II S ≥B n for y ∈ [σ, 1], i.e. II D ≥ 0. Finally we notice that
4.3.3 Asymptotic Expansion
The assumption ε ≪ 1, allows to seek the unknown fields (i.e. \(v_{1}^{\,}\), v 2, P, etc.) in the following form
Substituting into the governing equations we get a hierarchy of problems that must be matched together. The matching procedure requires the specification of Re, Bi, and λ 2. We consider: \(\mathsf {Re}\leq \mathcal {O}\left ( 1\right ) ,\) i.e. laminar flow and \(\mathsf {Bn}=\mathcal {O}\left ( 1\right ) \). We introduce the dimensionless parameter
and consider two different cases which correspond to different behaviors of the inner core:
-
(1)
\(\varGamma =\mathcal {O}(1).\)
-
(2)
\(\varGamma =\mathcal {O}(\varepsilon ).\)
When \(\varGamma =\mathcal {O}\left ( 1\right ) \), namely \(\mathsf {Re}= \mathcal {O}\left ( \lambda ^{2}\right ) \), we have an almost rigid inner core, while the case \(\varGamma =\mathcal {O}\left ( \varepsilon \right ) \), i.e. \( \mathsf {Re}=\mathcal {O}\left ( \varepsilon \lambda ^{2}\right ) \), corresponds to a “ soft” inner core where deformations are non-negligible. Let us show that the first case leads to the classical Bingham model. Indeed, considering \(\mathsf {Re}=\mathcal {O}\left ( \lambda ^{2}\right ) \)
Evaluating the order of magnitude of the elastic stress S el and of the viscous stress S vis we have
where δ is the dimensionless order of magnitude of the longitudinal displacement. The continuity of the shear stress at the interface yields
Hence we find \(\delta =\mathcal {O}\left ( \varepsilon \right )\), meaning an almost uniform longitudinal displacement as in the classical Bingham model. In the second case
which entails \(\delta =\mathcal {O}\left ( 1\right ) \). As a consequence the longitudinal displacement is not uniform.
4.3.4 First Case: \(\varGamma =\mathcal {O}(1)\)
We start considering the elastic domain, and we focus on the zero order terms. From (27) we obtain \(P^{\left ( 0\right ) }=P^{\left ( 0\right ) }\left ( x,t\right ) \). Next, since 
on the interface, we have that in the whole elastic region \(P^{\left ( 0\right ) }=P^{\left ( 0\right ) }\left ( x,\sigma ^{+},t\right ) \). From (26) we have
Mass conservation in the elastic domain becomes
Recalling (24) it is easy to show that
Therefore setting
we have \(\kappa =\kappa \left ( t\right ) \). From (21) we find
From (35)
and the transversal velocity vanishes everywhere in the elastic region, so that
Let us now consider relation (31) at the zero order approximation. We get
The latter forces to analyze the first order term in (26), (27), namely
Hence (39) becomes
It is easy to show that
in the elastic region so that condition (30) is fulfilled when
We now focus on the viscous part σ < y < 1. Here (29) entails \(P^{\left ( 0\right ) }=P^{\left ( 0\right ) }\left ( x,t\right ) \), so that the pressure is uniform on any channel section. Next, we observe that
and (39) can be rewritten as
We conclude that
Focussing now on (28) we find
so that
Solving (43) we get
In particular, (37) yields
Remark 2
We remark that
Therefore the derivative w.r.t. x of the r.h.s. of (46) must vanish and
From (18)
since \(v_{2}^{\left ( 0\right ) }\left ( x,h,t\right ) =v_{2}^{\left ( 0\right ) }\left ( x,\sigma ^{+},t\right ) =0\). Hence, exploiting (44)
Thus (48) yields
In case h is uniform we get
and
from which we derive the classical Bingham flow condition we find in [4]
ensuring the flow within the channel.
Remark 3
In case h does depend on x (49) does not give rise to any solution consistent with (47) (a well known paradox of the Bingham model). Indeed, form (42)2
that inserted into (49) yields
The above implies
Inserting (50) into (47) we get h = −2σ, that is a contradiction (lubrication paradox, see [6, 17, 21, 27]).
4.3.5 Second Case \(\varGamma =\mathcal {O}( \varepsilon )\)
Set \(\varepsilon \hat {\varGamma }=\varGamma \) with \( \hat {\varGamma }=\mathcal {O}\left ( 1\right ) \). From (31)
Once again \(P^{\left ( 0\right ) }=P^{\left ( 0\right ) }\left ( x,t\right ) \) in the whole domain. In the elastic part
so that (51) implies
The term \(a\left ( x,t\right ) \) is unknown at this stage while \(v_{1}^{\left ( 0\right ) }\) is
where now
can be interpreted as the uniform part of the longitudinal velocity. Checking condition (30) we find again (41).
Remark 4
At the leading order the longitudinal displacement is a superposition of a uniform displacement \(a\left ( x,t\right )\) and a non uniform displacement modulated by the pressure gradient. The latter becomes negligible for large values of \(\hat {\varGamma }\) and the first case is recovered for \( \hat {\varGamma }\gg 1\).
In the fluid region
Moreover
The jump condition (21) yields
with ω given by (53). From mass balance and boundary conditions
which, after some of algebra, gives
Therefore the mathematical problem at the zero order approximation is the following
Example 1
Consider the stationary problem when h ≡ 1
We get
yielding σ = 1 (which we do not consider) and \(P^{\left ( 0\right ) }=P_{in}-\varDelta Px\), with ΔP known. Therefore
Requiring σ < 1 we get the usual Bingham flow condition.
4.3.6 Stationary Version of (54)
We show that the stationary version of system (54), namely
admits a unique solution when h is a smooth bounded function of x with h ∈ [h m, h M]. From (58)1
and from (58)2
Hence, system (58) can be rewritten as
In [10] it has been proved that there exists a unique pair of sufficiently regular functions \(\left ( \sigma \left ( x\right ) ,P^{\left ( 0\right ) }\left ( x\right ) \right ) \) such that:
-
(a)
\(0<\sigma \left ( x\right ) <h\left ( x\right ) \), for all \(x\in \left [ 0,1\right ] \).
-
(b)
\(P^{\left ( 0\right ) }\left ( 0\right ) =P_{in}\), and \(P^{\left ( 0\right ) }\left ( 1\right ) =P_{in}-\varDelta P\).
-
(c)
\(\sigma \left ( x\right ) \) and \(P^{\left ( 0\right ) }\left ( x\right ) \) fulfill the equations of (60).
Remark 5
Assuming that (60) is solvable according to the above definition, we find immediately a new “flow condition”. Indeed, integrating (60)1 between 0 and 1, we obtain
Since
if (60) admits a solution, the following inequality holds true
Going back to dimension variables we get
which generalizes the classical Bingham flow condition.
4.4 Numerical Simulations
We present some numerical simulations to investigate the stationary behavior of σ(x) when \(\varGamma =\mathcal {O}(1)\). Suppose following function
The channel profile and the free boundary separating the elastic and the viscous phase are shown in Figs. 11, 12, 13, 14 for different values of ΔP∕B n satisfying condition (62). We see that
and consequently we perform numerical simulation with ΔP∕B n > 2.2.
Plot of the free boundary (dashed line) y = σ(x) with ΔP∕B n = 2.5 and with condition (62) fulfilled
Plot of the free boundary (dashed line) y = σ(x) with ΔP∕B n = 3.5 and with condition (62) fulfilled
Plot of the free boundary (dashed line) y = σ(x) with ΔP∕B n = 5.5 and with condition (62) fulfilled
Plot of the free boundary (dashed line) y = σ(x) with ΔP∕B n = 9.5 and with condition (62) fulfilled
The amplitude of the inner core decreases as ΔP∕B n increases. When ΔP∕B n ≫ 1 (Fig. 14), the inner core approximately disappears and the system becomes almost purely viscous. In the same way, we observe that when ΔP∕B n is close to 2.2 then σ → h (Fig. 11 ) and the viscous part tends to disappear. Also in this case we speak of a limit, since condition (62) must be fulfilled.
5 Two Dimensional Channel Flow: A New Approach
To model the flow of a Bingham fluid one considers the balance of linear momentum written in the differential form
where ρ is density, v is velocity, P is pressure and S is the deviatoric part of the stress. Equation (63) is typically used in the whole domain, assuming that the velocity and the stress are continuous across the fluid/rigid interface. Within the liquid domain the fluid is assumed to behave as a viscous incompressible fluid, whereas in the rigid part the stress is indetermined. Indeed in the unyielded part we only know that the strain rate vanishes, i.e. D = 0. Assuming that Eq. (63) holds in every part of the domain may lead to paradoxes, as the one that occurs in lubrication regimes, [6, 17].
To avoid this occurrence we propose a novel approach which essentially consists in using a integral formulation for the balance of linear momentum within the unyielded part. We apply this approach to study the flow in a bidimensional channel of varying amplitude, with the driving force being an applied pressure gradient (Poiseuille flow). We assume that the aspect ratio of the channel is small, so that the lubrication approximation is suitable. In this case, Eq. (63) can be simplified introducing the ratio ε ≪ 1 between the length and the maximum amplitude and rescaling the problem in a nondimensional form. The solution can be sought as power series of ε, where the leading order is the one we are interested in. With this procedure one tacitly assumes that the nondimensional variables and their derivatives are \(\mathcal {O} (1)\) in the liquid and solid domain. In particular the stress components S ij are assumed to be everywhere \(\mathcal {O}(1)\), but this latter hypothesis can be checked “a posteriori” only in the liquid part, since in the rigid domain the stress is not determined.
This point is the central importance for our procedure. Indeed, the assumption \(S_{ij}= \mathcal {O}(1)\) and the use of (63) to derive the motion in the rigid part, leads to the well known “lubrication paradox” , which consists in a plug velocity that depends on the longitudinal coordinate. Note that the paradox disappears when one considers a deformable core, as shown in the previous sections. If one does not use Eq. (63) in the unyielded part and write the balance of linear momentum using an integral global approach similar to the one presented in [29] and in [28], the paradox is no longer present. Therefore, the unyielded part is treated as an evolving non material volume Ω t and its dynamics is modelled writing the balance of linear momentum as
where T = −P I + S, is the Cauchy stress tensor and w the velocity of the boundary ∂Ω t. The advantage of this approach lies in the fact that the knowledge of the stress inside the rigid part is no longer needed and no guess has to be made on the order of magnitude of the stress components. Only the stress acting on the boundary of Ω t is required.
Therefore we need to know: (1) the forces acting on the yield surface σ (see Fig. 15); (2) the forces acting on the inlet and outlet of the channel. On σ the viscous stress is given once the problem in the viscous domain is solved. On the channel inlet and outlet the applied pressure, assumed to be a given datum of the problem, is required.
Sketch of the domain of the problem
When dealing with the leading order approximation in the channel flow, Eq. (64) becomes an integro-differential equation for the pressure P, whose solution allows to determine explicit expressions for the velocity field v and the yield surface σ. We prove that the longitudinal velocity is spatially uniform, while the transversal velocity vanishes (no paradox). We also show that these results can be also extended to the case of fluids with constant density and pressure dependent viscosity.
5.1 The Physical Model
Let us consider the flow of an incompressible Bingham fluid in a channel of length L and amplitude \(2h\left ( x\right ) \). Because of symmetry, we may limit our analysis to the upper part of the layer [0, h(x)]. The velocity field is given by
The Cauchy stress is T = −P I + S, where the deviatoric part is the one of a Bingham fluid
In the above μ is the viscosity, τ o is the yield stress. If D ≠ 0 we get
which holds with II S ≥ τ o. Therefore, whenever D = 0, we have II S ≤ τ o and the stress is not determined. We assume that the viscous and the rigid regions are separated by a sharp interface y = ±σ(x, t). Assuming incompressibility we write
5.2 The Viscous Domain
We write the governing equations in the viscous region neglecting body forces. These are the incompressibility condition (66) and
5.3 The Rigid Domain
The rigid domain Ω t at some time t > 0 is given by
The integral momentum balance for the whole domain Ω t in the absence of body forces is given by (64). Focussing on the upper part of the domain (y > 0) we find that (64) can be rewritten as
The external forces acting on the boundary ∂Ω t are expressed by the surface integral on the r.h.s. Assuming that P in, P out are the (uniform) pressures acting on the inlet and outlet of the channel, we find
where σ in = σ(0, t), σ out = σ(L, t). Recalling that in the rigid plug velocity is
the dynamics of the whole rigid region is expressed only by the first component of (64), that is
Hence the prescribed pressure difference driving the flow is
The boundary condition on the channel wall is
i.e. the no-slip condition. On σ we impose
which express the continuity of the velocity and of the stress across the yield surface y = σ ( t and n are the tangent and normal unit vector to σ).
Remark 6
In Sect. 5.10 we will extend our model the case in which the viscosity depends on pressure, namely \(\mu =\mu \mu \left ( P\right )\).
5.4 Scaling
Set
and introduce
which is crucial for applying the classical thin film approach. We rescale the problem using the following non dimensional variables
where we select the reference pressure using the Poiseuille formula
After some algebra (and neglecting the tildas) we find
where
is the Bingham number. Moreover
where
is the Reynolds number. The inner core equation (70) becomes
The boundary conditions (72)–(74) become
In the rigid domain the non dimensional velocity field is
with k 1 = k 1∕U.
5.5 The Leading Order Approximation
We look for a solution in which the main variables of the problem can be expressed as power series of ε
We further assume that h(x) is such that
and we limit our analysis to the leading order, assuming that \(\mathsf {Bi=}\mathcal {O}\left ( 1\right ) \) and \(\mathsf { Re\leqslant }\mathcal {O}\left ( 1\right ) \). We begin by observing that
and, since we are looking for a solution with \(v_{1y}^{(0)}<0\) in the upper part of the channel, we write
The problem reduces to
with boundary conditions
Integrating we find
Exploiting the continuity equation we get
Evaluating the velocity components on the yield surface we get
which entails
Supposing \(k_{1}^{(0)}\neq 0\), we find
where C is unknown. Let us now consider the rigid core equation (80) at the zero order
which, after an integration by parts, reduces to
Substituting (90) into (91), we obtain
We thus have
or equivalently
where h in = h(0). Defining the viscous region width as
formula (92) entails
Hence
Now, differentiating(~96) with respect to x, we obtain
that is the integro-differential equation
The boundary conditions are \(\left . P^{\left ( 0\right ) }\right \vert _{x=0}=\varDelta P \), and \(\left . P^{\left ( 0\right ) }\right \vert _{x=1}=0\). The solution \(P^{\left ( 0\right ) }\) of (97) is then used to evaluate the \(v_{1}^{\left ( 0\right ) }\) via ( 87), \(v_{2}^{\left ( 0\right ) }\) via (88) and the yield surface σ (0) via (93).
Remark 7
From (92) we observe that \(\sigma _{x}^{(0)}=-2h_{x}\), meaning that the amplitude of the rigid core becomes larger as the channel narrows, whereas it shrinks as the channel becomes wider. This is in accordance with what found in [21].
5.6 Flow Condition
Let us investigate the conditions on ΔP that prevent the system from coming to a halt. Let \(h\left ( x\right ) \equiv h_{in}\). From (93) we get:
-
\(\varDelta P>{\displaystyle {{\displaystyle {\mathsf {Bn} \over h_{in}}}}}\), ⇒ σ (0) < h in (the fluid is flowing)
-
\(\varDelta P<{\displaystyle {{\displaystyle {\mathsf {Bn} \over h_{in}}}}}\), ⇒ σ (0) > h in (no flow)
When h(x) is not uniform we have to ensure that \(\sigma ^{\left ( 0\right ) }<h\left ( x\right ) \), in order to prevent the flow from stopping. Recalling (95) we must impose
where \(h_{\min }=\min _{x\in \lbrack 0,1]}h\). To estimate the integral in the r.h.s. we remark that \(P^{\left ( 0\right ) }\) fulfils Eq. (97), that is an equation of elliptic type. Maximum principle entails \(0\leq P^{\left ( 0\right ) }\leq \varDelta P\). So, writing
we have
where
In conclusion
Therefore, recalling (98), we have
If we assume
and require that
which implies
we are sure that the flow never comes to a stop.
Example 2
If we consider a linear wall profile
where h out > 0, there are two possibilities:
-
Δh > 0, ⇒ \(h_{\min }=h_{in}\), and \(\max \left \{ \overline {h}_{x};0\right \} =\varDelta h\). Condition (101) yields
$$\displaystyle \begin{aligned} {{{\displaystyle {\mathsf{Bn} \over \varDelta P}}}}<~\underset{2h_{out}-3h_{in}}{\underbrace{ h_{in}-2\varDelta h}}~,\ \ \ \ \ \Leftrightarrow \ \ \ \ \varDelta P>{{{\displaystyle { \mathsf{Bn} \over 2h_{out}-3h_{in}}},}} \end{aligned}$$where, of course, we assume 2h out − 3h in > 0
-
Δh < 0, ⇒ \(h_{\min }=h_{out}\), and \(\max \left \{ \overline {h}_{x};0\right \} =0\). Inequality (101) entails
$$\displaystyle \begin{aligned} {{{\displaystyle {\mathsf{Bn} \over \varDelta P}}}}<~\underset{3h_{out}-2h_{in}}{\underbrace{ 2\varDelta h+h_{out}}}~,\ \ \ \ \Leftrightarrow \ \ \ \ \varDelta P>{{{\displaystyle { \mathsf{Bn} \over 3h_{out}-2h_{in}}},}} \end{aligned}$$where now we require 3h out − 2h in > 0.
5.7 Inner Core Appearance or Disappearance
A non uniform channel profile may cause the appearance/disappearance of the rigid plug. These phenomena (highlighted also in [6] and [21]) are not possible when the channel profile is uniform, namely when \(h\left ( x\right ) \equiv h_{in}\). Recalling (93 ), we set
in order to avoid physical inconsistencies. Hence, \(\sigma ^{\left ( 0\right ) }\) vanishes when
The r.h.s. of (102) is a critical value, that we denote as h crt , such that, whenever h ≥ h crt the core disappears.
Example 3
Let us consider the channel profile
depicted with the dashed line in Fig. 20. We now estimate h crt exploiting (102), when ΔP = 10, and B i = 5,
The core-free region is thus obtained solving h ≥ h crt, which we approximate with h ≥ 0.67, whose solution is the interval 1 ≤ x ≤ 0.58. Looking at Fig. 20 the actual core-free region is 1 ≤ x ≤ 0.55, which substantially agrees with the above estimate.
5.8 Solution for an Almost Flat Channel
When \(h\left ( x\right ) \equiv h_{in}\) (i.e. uniform channel) equation (93) gives
Equation (97) yields
The velocity field becomesFootnote 7
and we also find
Let us now consider a non-uniform channel profile \(h\left ( x\right ) \), assuming that amplitude width variation is small. We thus set
where \(\left \langle h\right \rangle \) denotes the spatial average along the channel, i.e.
and assume that \(\max \left \vert \phi \left ( x\right ) \right \vert \) is small, that is we consider an almost flat channel. We look for \(P^{\left ( 0\right ) }\) in the form
where \(\left . \varPi \right \vert _{x=0}=\left . \varPi \right \vert _{x=1}=0\), and where we expect that both \(\max \left \vert \varPi \right \vert \), \(\max \left \vert \varPi _{x}\right \vert \) are small. Inserting (107) into (92) we obtain
Concerning \(\ell ^{\left ( 0\right ) }\) we have
Exploiting then (97) we compute the pressure field solving
Neglecting ϕ x Π x we end up with the following problem
so that
In conclusion
which yields
Example 4
Let us consider \(h\left ( x\right ) =1+mx\), with m small. We write
In this case
and
We see that \(\sigma _{x}^{(0)}=-2m\), i.e. the core amplitude widens for m < 0 and shrinks for m > 0.
Example 5
Let us consider a wavy channel
where δ > 0, and θ ≪ 1. We write
with \(\max \left \vert \phi \right \vert =\mathcal {O}\left ( \theta \right ) \ll 1\). Exploiting (108) we obtain
The behavior for θ = 0.1, and δ = 1∕5 is shown in Figs. 16 and 17. In particular in Fig. 17 a close-up showing the difference between the approximated solution (112) and the computed one (see next section) is displayed.
Plot of x-component of the velocity, h given by (111), δ = 0.1, θ = 0.02, and B n = 5 , ΔP = 10.5
5.9 Numerical Simulations
Setting \(F=P_{x}^{\left ( 0\right ) }\), the elliptic problem (97) can be transformed in the following integral equation
where, recalling (95),
Now, if the conditions ensuring that \(\ell ^{\left ( 0\right ) }\) is strictly positive (Sect. 5.6) are fulfilled, we can solve (113) through the following iterative procedure:
Step j = 0. Set F 0 = −ΔP, and \(\ell _{F,~0}=\min \left \{ h\left ( x\right ) ,~3h\left ( x\right ) -{{{\displaystyle { {\displaystyle {\mathsf {Bn} \over \varDelta P}}}}-}}2 \mathop {\displaystyle \int } _{0}^{1}h~dx\right \} .\)
Step j = 1. \(F_{1}=-\varDelta P{\displaystyle {\dfrac {\exp \left \{ - \operatorname *{\displaystyle \int }_{0}^{x}{\displaystyle {{\displaystyle { 6h_{x^{\prime }} \over \ell _{F,~0}}}}}dx^{\prime }\right \} }{ \operatorname *{ \displaystyle \int }_{0}^{1}\exp \left \{ - \operatorname *{\displaystyle \int }_{0}^{x}{ \displaystyle {{\displaystyle {6h_{x^{\prime }} \over \ell _{F,~0}}}}}dx^{\prime }\right \} dx} }}\).
……
Step j > 1. \(F_{j}=-\varDelta P{\displaystyle {\dfrac {\exp \left \{ - \operatorname *{\displaystyle \int }_{0}^{x}{\displaystyle {{\displaystyle { 6h_{x^{\prime }} \over \ell _{F,~j-1}}}}}dx^{\prime }\right \} }{ \operatorname *{\displaystyle \int }_{0}^{1}\exp \left \{ - \mathop {\displaystyle \int } _{0}^{x}{\displaystyle {{\displaystyle {6h_{x^{\prime }} \over \ell _{F,~j-1}}}}}dx^{\prime }\right \} dx}}}\), with
Iterating the procedure until the desired tolerance is reached, we determine the solution \(F=P^{(0)}_{x}\). Integration then provides the pressure field P (0). We can show that, under suitable hypotheses, the solution of (113) exists and is unique.
In Figs. 16, 17 we have plotted h(x) and σ (0)(x) for the wavy channel profile given by (111). In Figs. 18, 19 we have reported the contour plots of \(v_{1}^{(0)}\) , and \(v_{2}^{(0)}\), when h(x) is given by (111), with δ = 0.1, θ = 0.02, and B n = 5.
Plot of y-component of the velocity, h given by (111), δ = 0.1, θ = 0.02, and B n = 5 , ΔP = 10.5
Plot of σ (0) and h, when h is given by ( 103). B n = 5, ΔP = 10.5
The solid colored regions of Figs. 18 and 19 denote the core, with vanishing transversal velocity and uniform longitudinal velocity. Notice also the symmetry of the transversal velocity shown in Fig. 19. In Figs. 20, 21, 22 we have considered the profile (103). The yield surface σ (0) and the velocities \(v_{1}^{\left ( 0\right ) }\), \(v_{2}^{\left ( 0\right ) }\) are reported respectively.
Plot of x-component of the velocity, h given by (103). B n = 5, ΔP = 10.5
Plot of y-component of the velocity, h given by (103). B n = 5, ΔP = 10.5
5.10 Model with Pressure Dependent Viscosity
In this section we extend our model to the case of a pressure-dependent viscosity. Going back to dimensional variables equation (65) rewrites in this way
The viscosity is expanded considering
so that, around ε = 0, we get μ = μ (0) + εμ (1) + ε 2 μ (2) + …, where
Following the same procedure described in Sect. 5.5, the non-dimensional leading order problem becomes
whose boundary conditions are still given by (86). We get
and
The interface \(\sigma ^{\left ( 0\right ) }\) is still given by (92 ), while Eq. (97) modifies in this way
where \(\ell ^{\left ( 0\right ) }\) is given by (95).
Example 6
In case \(\mu \left ( P\right ) =e^{\gamma P}\), and h ≡ 1, we get
We now consider \(h^{\left ( 0\right ) }=1+mf\left ( x\right ) \), with m small perturbation. We look for a solution of (115) of the form
with Π = 0 on x = 0, 1. After inserting ( 117) into (115) and neglecting the m 2, we find
and
6 Planar Squeeze
We consider the flow of an incompressible Bingham fluid placed between parallel plates of length in a channel of length L. The gap between the plates occupied by the fluid has amplitude 2h(t), as depicted in Fig. 23 (see also [19]). Because of symmetry, we confine our analysis to the upper part of the layer, namely \([0,h\left ( t\right ) ]\). The velocity field is v = u(x, y, t)i + v(x, y, t)j, where x, y are the longitudinal and transversal coordinate respectively.
A schematic representation of the squeezing channel
We assume that the region where II S ≥ τ o (yielded) and the region where II S ≤ τ o (unyielded) are separated by a sharp interface y = ±Y (x, t) representing the yield surface. We also define the inner plug
We may have Y (x, t) = 0 for some \(x\in \left ( 0,L\right ) \) and for some t, so that Ω p becomes a segment of zero measure. The rigid plug Ω p moves uniformly with velocity
Neglecting inertia and body forces, the governing equations in the viscous phase are
and
The integral momentum balance for the domain Ω p is given by
where ρ is the material density. Neglecting the inertial terms, we get following equationFootnote 8
where \(P_{Y_{o}}\), \(P_{Y _{1}}\) represent the normal stresses on x = 0 and x = L. As usual we impose
where w ⋅n is the wall normal velocity and t is the wall tangent vector. On Y we write
while at x = 0
6.1 Squeezing Between Parallel Plates
We assume \(h=h\left ( t\right ) \) and we set
We define the aspect ratio ε = H∕L , assuming ε ≪ 1. Then we rescale the problem using the following non dimensional variables
where T is the characteristic time scale, i.e. the “squeezing time”. We define the characteristic transversal velocity as V = H∕T, and the longitudinal velocity as U = V∕ε, so that u = u∕U, v = v∕V = v ∗∕(εU). Pressure is again rescaled exploiting the Poiseuille formula P c = (μLU)∕H 2 and we set P = P∕P c, P out = P out∕P c, where P out is the (given) pressure field applied at the channel outlet. We suppose that P out is constant in time and space. Next we introduce
so that
where
is again the Bingham number. The mechanical incompressibility constraint and momentum balance become (neglect the tildas)
Equation (122) can be rewritten as
where \(P_{out}=P_{Y _{1}}\). Boundary conditions (123) become
since the squeezing velocity is
Jump conditions on Y become
while conditions (126) become
6.2 Problem at the Leading Order
As for the channel, we look for a solution expressed as power series of ε, assuming \(\mathsf {Bi=}\mathcal {O }\left ( 1\right ) \). We get
since we are looking for a solution with \(u_{y}^{(0)}<0\) in the upper part of the channel. Equations (127)–(129) reduces to
with boundary conditions
For the sake of simplicity we suppress the superscript (0). Since P = P(x, t) we get
Moreover exploiting mass conservation
Evaluating u, v on Y and recalling conditions (132), we obtain
The plug equation (130) becomes
Recalling (134) we have u y = 0 in x = 0 implying \(\left . P_{x}\right \vert _{x=0}=0\). The solid region must be detached from x = 0, since otherwise u p ≡ 0, i.e. no rigid domain motion. Accordingly there must be some \(s\left ( t\right ) \in \lbrack 0,1]\), not a priori known, such that \(Y \left ( x,t\right ) \equiv 0\), for \(0\leq x\leq s\left ( t\right ) \). Hence the spatial domain \(\left [ 0,1\right ] \) can be split in two sub-domains (see Fig. 23):
-
\(0\leq x\leq s\left ( t\right ) \), where Y ≡ 0;
-
\(s\left ( t\right ) <x\leq 1\), where Y does not vanish.
Assuming that the longitudinal velocity is continuous across \( s\left ( t\right ) \), we have
where \(P_{x}\left ( s,t\right ) \) is unknown at this stage. From (139) we get
that is
In order to avoid physical inconsistencies, we set
The local instantaneous discharge is given by
Mass conservation then requires \(Q\left ( x,t\right ) =-\overset {\cdot }{h}x\), so that \(Q\left ( s,t\right ) =-\overset {\cdot }{h}s\) so that
which is positive since \(\overset {\cdot }{h}<0\). As a consequence
Therefore the fluid squeezes out of the channel only if \(Y \left ( 1,t\right ) <h\left ( t\right ) \), namely when \(s\left ( t\right ) >2/3\). In \(x\in \left [ 0,s \right ]\) we have Y = 0 and the pressure fulfills Eq. (139) with the boundary condition \(P_{x}\left ( 0,t\right ) =0\)
Therefore
with \(A\left ( t\right ) \) still unknown at this stage. Recalling that Y is linear in x we integrate (138) between x and 1 getting
Then imposing the continuity of P across x = s we get
Finally rewriting (140) as
we get
Hence, solving (149) we find \(s\left ( t\right ) \) and we are able to determine the pressure field in the whole channel and the rigid domain as well. We observe that s(t) is not a material point so that, in principle, s(t) can also be still (i.e. \(\dot {s}(t)=0\)), while the rigid plug is moving with velocity u p(t). Figure 24 shows the behavior of the function f(s) in the l.h.s. of (149) with s ∈ (2∕3, 1). We easily realize that f(s) is monotonically decreasing for 2∕3 < s ≤ 1 and that its range is \(\left [ 0,+\infty \right ) \). So, given any \(-(2\mathsf {Bn} h^{2})/(3\dot {h})>0\), there exists one and only one s fulfilling (149).
Behavior of \(f \left ( s \right )\) for 2∕3 < s ≤ 1
The force acting on the unit surface of upper plate is
Exploiting (149) we get
Remark 8
When B n → 0, the solution of (149) is simply s = 1, i.e. the solid region does not exists at all (as physically expected for a Newtonian fluid). Furthermore formula (150) reduces to
corresponding to the Newtonian fluid planar squeezing, [17]. These results confirms the physical consistency of our model.
6.3 Numerical Simulation
We perform here some numerical simulations to investigate the behavior of our asymptotic solution at the leading order. To illustrate the dependence of the solution on the Bingham number we consider the cases: B n = 1 and B n = 25. We plot the yield surface Y , the pressure field P and the axial velocity u, assuming that the plates have constant velocity so that
We consider t ∈ [0, 0.6], which guarantees that the plates do not come in touch in the select time interval. We set h f = h(0.6), representing the half gap width at time t f = 0.6 and s f = s(0.6) representing the onset of the rigid plug at time t = 0.6. The yield surface Y and pressure field P are plotted for different times t belonging to the selected time interval and for x ∈ [0, 1]. The axial velocity u is plotted at time t = 0.6 (i.e. when h = h f) for a finite number of x ∈ [s f, 1] and for y ranging in [0, h f].
In Figs. 25, 26 we have plotted the yield surface Y (x, t) and the upper plate y = h(t) at different times in the time interval [0, t f]. We have plotted the upper plate only for x ∈ [s(t), 1] so that the evolution of the onset of the plug x = s(t) is visible. We notice that the slope of the unyielded plug becomes smaller as s(t) increases, as expected. In Figs. 27, 28 we have plotted the pressure field at different times in the time interval [0, t f] in the whole domain x ∈ [0, 1]. Also for this case the position x = s(t) has been put in evidence. We notice that the pressure within the gap increases as B n increases.
Y for B n = 1 and h given by (151)
Y for B n = 25 and h given by (151)
P for B n = 1 and h given by (151)
P for B n = 25 and h given by (151)
In Figs. 29, 30 we have plot the axial velocity profile at time t = 0.6 for some fixed x ∈ [s f, 1]. In particular velocity is plotted for x = 0.69, x = 0.73, x = 0.77, x = 0.81, x = 0.85. As one can easily observe the velocity of the plug is the same for each (x, y) belonging to the plug.
u for B n = 1 and h given by (151)
u for B n = 25 and h given by (151)
Finally in Figs. 31, 32 we have plotted the squeeze force given in (150) for different values of the Bingham number, B n. We have plotted (150) for the linear squeezing (151) and for the exponential squeezing
We observe that the linear squeezing requires a grater squeezing force than the exponential squeezing. This is physically consistent, since in the linear case the plates move faster than in the exponential case.
Squeeze force for linear h(t), (151)
Squeeze force for exponential h(t), (152)
6.4 Squeezing Between Surfaces
In this section we generalize the problem to the case in which the parallel plates are surfaces y = ±h(x, t) that are approaching the channel centerline, as shown in Fig. 33. In this case
and we again assume H∕L = ε ≪ 1. The theory develops exactly as in Sect. 6.1, so that (140) still holds. We split \(\left [ 0,1\right ] \) into \(\left [ 0,s\right ] \) and \(\left [ s ,1\right ] \), so that continuity of u across \(s\left ( t\right ) \) yields
A schematic representation of the squeezing channel
Recalling (139) we find
which generalizes (142). We thus get the following Cauchy problem
whose solution is
where s is still unknown. Following (143) we set
The local discharge is
while mass conservation h t + Q x = 0 implies
since Q(0, t) = 0. We find
implying
which is the generalization of (145). In conclusion, substituting (155) into (154), we find
In \(x\in \left [ 0,s\right ] \) the pressure fulfils
so that
with \(A\left ( t\right ) \) to be determined. In \(x\in \left [ s,1\right ] \) we have
so that
with Y given by (156). We observe that (157) and ( 159) yield \(\ \left . P_{x}\right \vert _{s^{-}}=\left . P_{x}\right \vert _{s^{+}}\). Let us now integrate (159) between x and 1 with the boundary condition \(P\left ( 1,t\right ) =P_{out}\). We find
Imposing \(\left . P\right \vert _{s^{-}}=\left . P\right \vert _{s^{+}}\), from ( 158), (160) we find A(t), so that the pressure can be written in terms of s throughout the whole domain. Substituting (156) and (159) into (140) we get
which provides an integral equation for the unknown s(t). Equation (161) can be solved once we know the explicit form of the function h(x, t).
When h(x, t) = f(x)g(t), with f ⋅ g > 0, then (161) can be rewritten as
Example 7
Let us consider
where α, and β both positive. Exploiting (162) we find
or equivalently
which is an implicit equation for s(t). Notice that taking the limit β → 0 of the l.h.s. of (164) we recover the l.h.s. of ( 149), as expected. We immediately realize that (164) admits a unique solution \(s(t)\in ( \hat {s}(\beta ),1]\), with
for each value of \(\left ( 2\mathsf {Bn}/3\alpha \right ) e^{-\alpha t}\). In particular it is easy to show that
so that s ∈ (2∕3, 1) for all t > 0. Recalling (156) we get
Clearly Y > 0 for every x > s, and Y = 0 at x = s. Actually we can show that Y and h never meets. Indeed, suppose that Y (x, t) < h(x, t), then
or analogously
Hence Y < h if and only if (166) holds true for each x ≥ s . Now recall that \(s\geq \hat {s}(\beta )>2/3\), for every finite time t > 0 and β > 0. Therefore
which proves that (166) holds true. As a consequence we get
We observe that Y → h, which in turn tends to 0, only in the limit t →∞. In Fig. 34 we have plot the advancing front x = s(t) for different values of B n ranging from B n = 0.1 to B n = 100. The parameters used are α = 2 and β = 0.4.
The advance of the front x = s(t) for B n ranging between 0.1 and 100
Notes
- 1.
In the 1D case, considering just the upper part of Ω, \(\boldsymbol {w} =\dot s\left ( t\right ) \boldsymbol {e}_{y}\).
- 2.
We remark that \(\dot \kappa \left ( t\right ) \neq v_{t}\left ( s\left ( t\right ) ,t\right ) \).
- 3.
Here grad denotes the gradient operator w.r.t. Lagrangian coordinates, while ∇ the gradient w.r.t. Eulerian coordinates.
- 4.
We are considering a linear elastic model even if \(\left \Vert \mathbf {F} \right \Vert \) may be, in general, not very “small” , as we shall see in Sect. 4.3.3.
- 5.
We omit “ \(\tilde {}\) ” to keep notation simple.
- 6.
Again, we have omitted “ ”.
- 7.
We set, for the sake of simplicity, h in = 1.
- 8.
The expression \(\left [ -Y _{x}T_{11}+T_{12} \right ] _{Y ^{+}}\) represents the force exerted by the viscous region on the lateral side of the inner rigid core.
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Farina, A., Fusi, L. (2018). Viscoplastic Fluids: Mathematical Modeling and Applications. In: Farina, A., Mikelić, A., Rosso, F. (eds) Non-Newtonian Fluid Mechanics and Complex Flows. Lecture Notes in Mathematics(), vol 2212. Springer, Cham. https://doi.org/10.1007/978-3-319-74796-5_5
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