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Unsteady shear flow of a viscoplastic material
J. Non-Newtonian Fluid Mech., 72 (1997) 87 – 100
Unsteady shear flow of a viscoplastic material S.L. Burgess *, S.D.R. Wilson Department of Mathematics, Uni6ersity of Manchester, Oxford Road, Manchester M13 9PL, UK Received 30 November 1996; accepted 11 February 1997
Abstract A viscoplastic, or yield-stress, liquid occupies the space between two infinite parallel plates. Initially the whole system is at rest. The lower plate is suddenly jerked into motion with given speed or shear stress, while the upper plate is kept fixed. The flow consists of two regions; (1) a lower sheared region bounded above by the yield surface, (2) an upper unyielded region bounded below by the yield surface. The yield surface propagates to the upper plate as time proceeds. We first consider the equivalent one plate problem of flow over a jerked plate, and find similarity solutions and small time asymptotic solutions for prescribed shear and speed cases respectively. These solutions are used as initial solutions for the two plate case. The motion of the yield surface and the time taken for the entire material to yield are investigated. The problems considered here are two dimensional representations of some control devices, for example the light duty clutch, which consists of two corotating, coaxial discs separated by a layer of electrorheological material. In this application it is useful to know the time taken for all the material to yield. © 1997 Elsevier Science B.V. Keywords: Electrorheological fluids; Shear stress; Viscoplastic material
1. Introduction Commercially, electrorheological (ER) fluids have been put to many uses, yet studies on the underlying fluid mechanics of these applications are scant. The problem investigated here is part of a more extensive study into the behaviour and possible uses of ER materials in control devices and the results should be of wider, more general interest. Our interest mainly lies with light duty clutches; these can be used as the control mechanisms in robotic arms for example. They consist of two corotating, coaxial discs, closely spaced and with ER fluid between them. Other applications include dampers, shock absorbers and alternators, more details of which can be found in Hartsock et al. [1]. An investigation into the use of ER materials in automotive engine mounts is presented by Williams et al. [2]. * Corresponding author. 0377-0257/97/$17.00 © 1997 Elsevier Science B.V. All rights reserved. PII S 0 3 7 - 0 2 5 7 ( 9 7 ) 0 0 0 1 8 - 9
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The current investment in ER fluids is due to the fact that the viscosity of such materials can be controlled by an externally applied electric field. Typically, ER materials consist of electrically polarisable particles suspended in an oil. The yield stress of the material is proportional to the applied electric field (see Stangroom [3]), so accounting for the ‘smart’ behaviour of ER materials. The manner in which the fluid yields and, in particular, the time taken for all the material to yield are of interest. In previous studies (e.g. Stangroom [3]) ER materials have been said to be represented well by a Bingham constitutive model. This model in its usual formulation asserts that the material is perfectly rigid until the yield stress is exceeded, thereafter behaving in a Newtonian-like manner. Barnes and Walters [4], however, give experimental evidence that materials which flow at high shear rates will also do so even at very low shear stresses. This means that some flow can occur below the yield point. Moreover, the use of the Bingham model can lead to difficulties arising from the fact that for an absolutely rigid material the stress distribution is indeterminate, in general. See, for example, Lipscomb and Denn [5] and Wilson [6] for a discussion of the so-called squeeze flow paradox. For these reasons we use a biviscosity model. This viscoplastic model assumes that the material behaves as a Newtonian fluid with very large viscosity until the critical ‘yield stress’ is exceeded. Above the yield stress, the material has a smaller, rapidly decreasing viscosity. The Bingham model can be recovered as a limiting case. We investigate the unsteady fluid motion in a layer of viscoplastic material contained between two closely spaced parallel plates. The upper plate is kept fixed whilst the lower is suddenly given constant speed or shear stress. We are particularly interested in the start-up process, that is, the fluid motion up to the time when all the fluid has yielded. First, we study the jerked plate problem for a single plate. This, a generalisation of the Rayleigh problem, is interesting in itself and provides the initial solutions needed for the two plate case. (The effect of the top plate is minimal in the first instants of motion). Particular attention is paid to the motion of the yield surface which separates the sheared and unyielded regions. For the case of prescribed shear stress, similarity solutions are found. It is not possible to find such self similar profiles for the prescribed speed case; a lengthscale is already available from the plate velocity and the characteristic yield stress of the material. However, we find asymptotic solutions valid at small times. Two phase flow, the regions separated by a moving interface, can be typified by the classic Stefan problem of ice melting into water, for example. The complicating factor is the unknown position of the moving solid/liquid interface (see Crank [7] for more details). In the two plate case, the solution must be obtained numerically and so we choose to fix the moving yield surface and the computational domains by a simple coordinate transformation. The velocity of the yield surface then appears in the governing equations. An estimate of the time taken for all the material to yield is made. The flow of a viscoplastic material over a plate having prescribed constant shear stress has been considered by Phan-Thien [8]. This work has been extended to the case of a time dependent plate shear stress by Phan-Thien and Gartling [9]. The viscoplastic model used in both papers, however, is inappropriate; it is proposed that the shear stress be simply proportional to the shear rate in both the yielded and unyielded regions (with different constants of proportionality). But this means that there must be a discontinuity in either the shear rate the shear stress at the yield surface. As a result, we do not consider these papers further.
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Sekimoto [10] has looked into the flow of a Bingham material over a single plate. Initially, the prescribed stress of the plate was reduced from a value greater than the yield stress to one less than it. Similarity solutions were found. Sekimoto also mentions that in the start-up problem for Bingham materials the yield surface propagates to infinity instantaneously. This work, however, is cited as being unpublished and appears to remain so. The present work shows that the basic idea is correct and gives an explicit estimate of the propagation speed in the ‘Bingham limit’.
2. Analysis and results Using Cartesian co-ordinates, the fluid velocity is given in dimensional form by (U(Y, T), 0), where Y is the co-ordinate perpendicular to the plate and T is the time. Neglecting body forces, the fluid velocity is governed by r
(U (t = , (T (Y
(1)
where r is the constant fluid density and t is the shear stress. The plates and fluid are quiescent until the lower plate is suddenly jerked into motion parallel to itself with constant velocity u0 or constant applied shear stress − tb(tb \ 0). Refer to Fig. 1 for a diagram of the problem. The boundary and initial conditions are U(d, T) =0
ÖT
(2)
and either U(0, T) =u0
T\0
(3)
or
Fig. 1. The two plate problem.
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Fig. 2. The biviscosity model; dependence of shear stress t on shear rate e= (U/(Y.
t Y = 0 = −tb
T\0
(4)
We now describe the constitutive model. Many fluids of industrial importance exhibit a yield stress, or apparent yield stress. This means that little or no motion of the material will occur until some critical yield stress has been exceeded. For reasons noted above, we choose the so-called biviscosity model. This supposes that
Á (U (U t0 t \t c Ãh 2 + sgn (Y (Y t=Í (U Ã h1 t B t c (Y Ä
(5)
Fig. 2 illustrates this model. Two viscosities are introduced by this law, and the ratio of the two, namely h2/h1 = e, is a key parameter. We have t0 =tc(1−e). The material is supposed to behave like a Newtonian fluid with very large viscosity h1 below the apparent yield stress tc, and has a rapidly decreasing viscosity above this critical stress. The usual Bingham model is obtained by letting e 0, and e = 1 gives the Newtonian model. For the jerked plate problem sgn(#U/#Y)= − 1. The biviscosity model indicates that there will be two regions of flow separated by a yield surface Yc. On Yc the shear stress equals the characteristic yield stress, tc, of the material. Region 1 is the layer of fluid bounded below by the yield surface Yc and above by the fixed plate. Here, little of the fluid occurs and the behaviour is like that of a Newtonian fluid
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with large viscosity h1. Region 2 is adjacent to the lower plate and bounded above by Yc, and is yielded. As time proceeds, we expect more fluid to yield and the yield surface to move towards the upper plate. An added complication is that the yield surface position Yc =Yc(T) is not known in advance and can only be determined by solving the system of governing equations and conditions. It is important to remember that both the velocity and the shear stress must be continuous on the yield surface, and that the stress also equals −tc on Yc. It will be seen that for the two plate problem, analytical solutions are not available; the governing systems are solved numerically. The fluid profiles and yield surface position must then be known initially in order to start the numerical procedure. However the sudden imposition of prescribed shear stress or speed at the lower boundary, as an idealisation, results in singularities at T=0 which are difficult to deal with numerically. This is one reason for searching for analytical solutions to the one-plate problem; they provide small-time starting profiles for the full problem. In addition, they are of some interest per se. The case of prescribed plate stress is dealt with first of all as this is easier mathematically. Then the prescribed speed problem is investigated; this is the most important industrially (most control devices are constructed as prescribed speed devices).
3. Prescribed shear
3.1. One plate case The plate is contained in a semi-infinite expanse of viscoplastic material and is given a constant shear stress −tb, where tb \ 0. Before starting the process of finding a solution, a comment must be made about the way in which the variables will be non-dimensionalised. In order to use the one plate solution as a starting solution for two plate case, the scalings used must be the same. We use the scalings appropriate for the two plate problem, namely Y =dy,
Ui =
dtb ui, h2
T=
d 2r t; h2
i =1, 2.
(6)
Of course, for the one plate problem the plate separation d must be irrelevant and it will be seen that the solution, when obtained, will not depend on it. The choice is made for later numerical convenience, as noted.We propose similarity solutions of the form u1 = t 1/2f1(n),
(7)
u2 = t 1/2f2(n),
(8)
where n=
y , t 1/2
The governing equations reduce to
(9)
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e { f1 − nf %1} =f ¦, 1 2
(10)
1 { f2 − nf %2} =f ¦, 2 2
(11)
f %2(0) =Ls(1 −e) −1,
(12)
f1( ) = 0,
(13)
f1(nc) = f2(nc),
(14)
f %1(nc) = f %2(nc)= − Lse,
(15)
where %= d/dn, nc = yc/t 1/2 is the position of the yield surface and Ls =tc/tb. We assume that the applied stress is greater than the yield stress, i.e. 0 BLs B1; otherwise the motion throughout is that of a Newtonian liquid of viscosity h1. Eqs. (10) and (11) can be integrated to give
! !
2
f1(n) =A1 e − en /4 − 2
f2(n) =A2 e − n /4 −
" "
(pe)1/2 e 1/2n n erfc 2 2
p 1/2 n n erfc 2 2
+B1n,
(16)
+ B2n.
(17)
The five conditions in Eqs. (12)–(15) determine the unknown constants A1, B1, A2 and B2 and the position of the yield surface nc. Eliminating the constants we find a single implicit equation for the yield surface nc, namely (1 −Ls) erfc
2 e 1/2nc nc = Lse 1/2 e(1 − e)nc /4 1−erfc 2 2
.
(18)
It is straightforward to evaluate nc numerically from this equation. The dependence of nc on the viscosity ratio e, for three fixed values of Ls is illustrated in Fig. 3. It is seen that the smaller e is in the range (0, 1) the larger nc is. In addition, the larger Ls is the closer the yield surface is to the plate. We expect this physically, as larger values of Ls correspond to smaller values of the applied shear stress. In the physical applications we are interested in, the viscosity ratio e is typically small, and so it is useful to solve Eq. (18) for nc in the limiting case e0; the position of the yield surface is then found to be given explicitly as
! !
yc = 2t 1/2 ln
1− Ls Lse 1/2
""
1/2
.
(19)
Note that as e 0, the ‘Bingham limit’, the yield surface goes to infinity at any t \0. To summarise, the velocity profiles at small times for the case of constant plate stress are u1 = 2Ls u2 = 2Ls
! ! et p
1/2
et p
1/2
2
e − en /4 − 2
" "
(pe)1/2 e 1/2n n erfc 2 2
eg2/4 e − n /4 −
(p)1/2 e n erfc 2 2
,
−eLsn.
(20) (21)
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As we now have initial solutions for the two plate case, this problem will be investigated.
3.2. Two plate case The lower plate is given prescribed shear stress whilst the top one is kept fixed. It is expected that the yield surface will move towards the upper plate as time progresses, and there will exist a critical time Tc at which it hits the upper plate. After this time, all the material will be yielded. The moving boundary Yc = Yc(T) must be tracked in some way. Various methods are available, here we choose to fix the yield surface by an appropriate coordinate transformation. It then turns out that the velocity of the yield surface appears in the governing equations, with the equations being valid on a finite fixed domain. The yields surface yc is fixed by introducing the transformation z=
y −1 yc(t) −1
(22)
in region 1, and z=
y yc(t)
(23)
in region 2. The upper region is mapped from [yc, 1] to [1, 0], and the lower region is mapped from [0, yc] to [0, 1]. The system is then governed by
Fig. 3. Self similar yield surface position nc for 10 − 3 5 e5 1; Ls = 10 − 3 ——, Ls = 10 − 2 — — —
, Ls = 0.5 - -.
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dyc (u1 ( 2u1 (u1 − e(yc − 1)2 =0, + e(yc − 1) z (z 2 dt (z (t
(24)
dyc (u2 (u2 ( 2u2 −y 2c = 0, 2 + ycz (z dt (z (t
)
(25)
1 (u2 =Ls(1− e)− 1, yc (z z = 0
(26)
u1(0, t) =0,
(27)
)
u1(1, t) =u2(1, t),
)
(u1 1 1 (u2 = = − Lse, yc (z z = 1 (yc − 1) (z z = 1
(28) (29)
where Ls = tc/tb. Notice that the velocity of the yield surface, dyc/dt, appears in the governing equations, and that the equations are defined on a fixed domain. The initial solution used is that found in Section 3.1 for the one plate problem. Suppose that the integration in the two plate case is started at a small time ts, say. Then the yield surface position yc(ts), velocity profiles u1(z, ts) and u2(z, ts) can be evaluated from Eqs. (19)–(21) respectively. Now consider the system of equations and conditions given in Eqs. (24)–(29). The parabolic equations are linear and include both the yield surface position yc and yield surface velocity dyc/dt. If yc at time t were known a priori then it would be straightforward to integrate the equations. This, however, is not the case, as yc is only known at time ts. The system is over determined and so the ‘extra’ condition is used to calculate yc at time t. The equations are solved by a standard numerical routine using finite differences and the method of lines. The boundary conditions incorporated in this routine are those given in Eqs. (26) – (28) and
)
)
(u1 1 1 (u2 = . yc (z z = 1 (yc − 1) (z z = 1
)
(30)
The method of bisection is used to calculate the yield surface position from the extra condition 1 (u2 = −Lse. yc (z z = 1
(31)
It is helpful to consider typical values that Ls might take. It is usual for the lower plate to be jerked with a shear stress several orders of magnitude greater than the yield stress of the material. It is pointless considering the case when the prescribed stress has magnitude less than the critical stress—in this circumstance none of the material will yield and little, if any, flow will occur. The parameter Ls is, therefore, thought to be at least O(10 − 3) and at most O(10 − 1). Fig. 4 illustrates how the critical time tc varies with e for the case of prescribed shear. We took Ls = 10 − 3, 10 − 2 and 0.5 and e= 10 − 3, 10 − 2, 5×10 − 2 and 10 − 1. For a particular Ls, the time taken for all the material to yield increases as e increases. This fits in with the result for a Bingham material (e = 0); in this case tc is infinitesimal. Even for a material with e=10 − 1, tc is only O(10 − 2). Also tc increases as Ls increases, for a given e. We expect the material to take
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Fig. 4. Viscosity ratio e against tc for prescribed shear; Ls = 10 − 3 ——, Ls = 10 − 2 — — — —, Ls = 0.5 - -.
longer to yield when the ratio of the yield stress to the plate stress is increased; this corresponds to reducing the plate stress. An idea of the real critical time Tc can be obtained by evaluating the timescale d 2r/h2. Typically1 d = O(10 − 3) m,
r = O(103) kg m − 3,
h2 =O(1) Pa s.
(32)
The timescale is calculated to be O(10 − 3); this means that the real time involved is merely O(10 − 5) s. For Bingham materials (e = 0), the yield surface has an infinite speed of propagation. The numerical routine was run for small e, and so we expect the yield surface to take a short time to reach the upper plate.
4. Prescribed plate speed
4.1. One plate case A plate is contained in a semi-infinite expanse of viscoplastic material. The system is at rest until the plate is suddenly given a constant speed u0 parallel to itself. In line with the analysis used for the prescribed shear problem, we now consider what scalings might be regarded as appropriate for the prescribed speed case. Because this problem includes the plate speed u0 and the characteristic yield stress tc, there is a natural lengthscale h2u0/tc, and so there will not be a similarity solution of the resulting system of equations and conditions. 1 Sources of data: Hartsock et [1], Barnes and Walters [4], Lord Corporation product information documentation for materials ER100 and ER200.
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However, we can find an asymptotic solution valid for small times. This solution will still then be a valid initial solution for the two plate case. The limit of interest is t0, e fixed. We choose to use the two plate scalings from the start for the sake of simplicity and so put Ui = u0ui,
Y = dy,
T=
d 2r t; h2
i =1, 2.
(33)
As a result we then expect the ratio of the two lengthscales to occur as a non-dimensional parameter of the flow, namely dtc/h2u0 = Ln. The flow is governed by e
(u1 ( 2u1 = 2, (t (y
(34)
(u2 ( 2u2 = 2, (t (y
(35)
u1( , t) = 0,
(36)
u2(0, t) =1,
(37)
)
)
u1(yc, t) = u2(yc, t),
(38)
(u2 (u1 = = − Lne. (y y c (y y c
(39)
Here yc = yc(t) denotes the yield surface. It is clear from the form of boundary conditions Eqs. (37) and (39) that this system will not admit an exact similarity solution. However it is possible to obtain an approximate solution valid at small times, that is, in the limit t 0, e fixed; this solution is of interest per se and, as noted, will serve to start off the numerical solution of the two plate problem. However the calculations become very elaborate and only the outline will be given here. We can begin by considering the case of a Newtonian fluid for which e=1. Here we do of course have an exact similarity solution u1 = u2 = erfc(n),
(40)
where n = y/2t 1/2; the ‘yield stress’ is purely notional and the condition Eq. (39) simply gives the position of the surface on which the shear stress takes a prescribed value. We find that the ‘yield surface’ is given by 2
e − nc =
Ln (pt)1/2 2
(41)
from which we conclude that nc as t0. When e B 1 we conjecture that nc is again large when t is small, and so will have u2 erfc(n)+ ···
(42)
as a first approximation. After some trial and error it turns out that the appropriate form for u1 is u1 t kl mf1(n ),
(43)
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where l= log (1/t), n = e 1/2n (motivated by the observation that the similarity variable contains the square root of the viscosity), k and m are constants to be determined, and the function f1(n ) satisfies f ¦1 + 2n f %1 −4kf1 = 0.
(44)
We require f1 0 as n . The equation can be solved in terms of the usual parabolic cylinder functions. Applying the conditions at the yield surface (Eqs. (38) and (39)) then gives 1 e m = k= − 2 2
(45)
and an estimate of the position of the yield surface nc: 2
e − nc =
L6 (ept)1/2. 2
(46)
We now turn to the problem of obtaining the correction terms to the estimates Eqs. (42) and (43). The main reason for undertaking this is that the possibility of continuing the expansion is evidence that the leading terms are in fact correct. It is found2 that Eq. (42) becomes u2(n) =erfc(n)+ t 1/2
An Bn log l f2(n) + 2 +··· , + l l2 l
(47)
where A and B are constants to be determined, and f2 satisfies the equation f ¦2 +2nf %2 −2f2 =4An with f2(0) = 0.
(49)
Meanwhile Eq. (43) becomes
u1(n ) = t kl m f1(n )+
log l 1 g1(n ) + h1(n )+··· , l l
(50)
where g ¦1 + 2n g %1 − 2(1− e)g1 = 0,
(51)
g1( ) = 0,
(52)
h ¦1 + 2n h %1 − 2(1− e)h1 = − 2(1− e),
(53)
h1( ) − 0.
(54)
Matching the solutions at the yield surface gives, after lengthy calculations,
2
A = 2(Ln −e),
(55)
B = − 2(Ln − e),
(56)
We are grateful to Mr E.J. Watson of Manchester University for his assistance in finding these solutions.
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and the correction to the form of nc, which was given to leading order in Eq. (46): nc =
(Ln − e) l − log(ep 1/2)− +··· 2 el
1/2
.
(57)
To summarise, the small time solutions for the velocities u1 and u2 are u1 =
tl 2
−
(1/2) − (e/2)
!
2 1 2k − 1 2k − 3 Lnp − e/2e 1 − (3e/2)e − n n − − k+ (k+1)n − 2
(1 − e) log l − 2 + e 1 − 2 + e (1−e) 2 pe 3 n (1−e) log n − 3− +log + n 2 2l 2 e l
"
+··· , (58)
u2 = erfc(n) +2t 1/2 +
!
2 2nLn 1 (Ln − e)n l− log l− g −log 2 + 2 1− 2 pe l e l
"
2(Ln − e) (− 1)nn 2n + 3 5 % (n+ 1) 3 2 n=0
+··· .
(59)
n
4.2. Two plate case We use the scalings given in Eq. (33) and the transformations given in Eqs. (22) and (23). The system of equations and boundary conditions governing the flow are those given in Eqs. (24) – (29), with the exceptions that Eq. (26) is replaced by u2(0, t) =1,
(60)
and Ls is replaced by Ln = dtc/h2u0 in Eq. (28). In the full two plate system there is, again, one ‘extra’ condition. The pdes are integrated using finite differences and the method of lines, and the method of bisection is used to determine the yield surface position from the condition
)
1 (u2 = −Lne. yc (z z = 1
(61)
The numerical solution is found in the same manner as described in Section 3.2, and the initial solution is the one plate solution given in Section 4.1. Before presenting any results, it is necessary to determine what values of the parameter Ln = dtc/h2u0 occur in practice. Typically3 d = O(10 − 3) m, h2 = O(1) Pa s, 3
tc = O(10) u0 = O(1)
Sources of data: as given previously.
or O(102) Pa,
(62)
or O(10) m s − 1
(63)
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The parameter Ln can be seen to be at least of O(10 − 3) and at most of O(10 − 1). We choose Ln = 10 − 3, 10 − 2 and 0.5. Results of the viscosity ratio e against the time taken for the yield surface to hit the top plate tc are presented in Fig. 5. (In practice, small e results are of interest so we take e =10 − 3, 10 − 2 and 10 − 1.) From Fig. 5 it can be seen that for a particular value of Ln the time taken for the yield surface to hit the upper plate increase as e gets larger. This is expected, since for a Bingham material (e= 0) the yield surface moves to the upper plate instantaneously. In addition we see that as the parameter Ln is increased for a particular e, the time taken for all the material to yield, tc, increases. Increasing Ln corresponds to increasing the plate separation or reducing the speed at which the plate is jerked. In both situations we would expect the critical time tc to increase; our results support this. Note that from the figure, tc is only O(10 − 2). The real time taken for the yield surface to hit the upper plate can be evaluated in the manner described in Section 3.2 for the prescribed shear stress case. Again it turns out that Tc is merely O(10 − 5) s. As only relatively small viscosity ratios have been considered, very small critical times are expected. The Bingham model provides the reference case for this. Bingham materials (e= 0) and Newtonian materials (e=1) are the two limiting cases of the biviscosity model. The time taken for all the material to yield increases as e is increased from 0– 1. It is found that the critical time is very small, regardless of whether the lower plate is given prescribed speed or shear stress. The time taken for the yield surface to hit the top plate is typically a fraction of a second for materials of small e. The analysis conducted here and the numerical methods used will provide a good basis for future work on the full problem of viscoplastic flow between rotating discs. The yielded material in this case will be contained inside a cylinder of critical radius, and this cylinder will move radially outwards as time proceeds and more material yields.
Fig. 5. Viscosity ratio e against tc for prescribed velocity; Ln = 10 − 3 ——, Ln = 10 − 2 — — — —, Ln -0.5 - -.
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4.3. Note on time taken to achie6e steady Couette flow For the two plate problem the velocity profile will reach the steady state Couette solution u= 1 − y in some timescale tf after the material has yielded. This happens in both the prescribed speed and prescribed shear stress cases. We can think of the jerked plate problem for a yield stress material as consisting of two steps; (1) the yield surface moves towards the upper plate and then hits it, (2) the material then behaves in a Newtonian fashion as the flow approaches the usual Couette profile. It is interesting to estimate the size of tf. For the Newtonian flow the timescale is rd 2/h, where h is the viscosity. This timescale can be evaluated to be O(10 − 3). So once the material has yielded, Couette flow will be achieved relatively quickly.
5. Conclusions The problem of unsteady shear flow has been investigated for a viscoplastic material. We modelled the material as obeying a biviscosity law and considered both the one plate and two plate problems. In both cases, the jerked plate was given constant prescribed speed or shear. Similarity solutions were found for the one plate prescribed shear case, and small time asymptotic solutions were found for the case of given plate speed. Numerical solutions were found for both the two plate cases. The yield surface was tracked and estimates of the time taken for the entire fluid to yield were made. For the two plate flow the yield surface takes a fraction of a second to reach the upper plate; the material then achieves the usual Couette profile within a short time. The larger the viscosity ratio e, the longer the material takes to yield completely. This result accords with that given by the standard Bingham model (e=0) for which the material yields instantaneously. In addition, as the non-dimensional parameters Ln or Ls increase so does the yield time. The motivation for this work was to gain a better understanding of the workings of control devices, the light duty clutch in particular. The work presented here will provide a sound foundation for further investigations.
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]
D.L. Hartsock, R.F. Novak, C.J. Chaundy, J. Rheol. 35 (1991) 1305. E.W. Williams, S.G. Rigby, J.L. Sproston, R. Stanwy, J. Non Newt. Fluid Mech. 14 (1993) 337. J.E. Stangroom, Proc. 2nd Int. Conf. on E.R. Fluids, Technomic, Lancaster, 1990, p. 41. H.A. Barnes, K. Walters, Rheol. Acta 24 (1985) 323. G.G. Lipscomb, M.M. Denn, J. Non Newt. Fluid Mech. 14 (1984) 337. S.D.R. Wilson, J. Non Newt. Fluid Mech. 47 (1993) 211. J. Crank, Free and Moving Boundary Problems, Clarendon Press, 1984, p. 187. N. Phan-Thien, ASME J. Appl. Mech. 50 (1983) 229. N. Phan-Thien, D.K. Gartling, 64 (1984) 196. K. Sekimoto, J. Non Newt. Fluid Mech. 39 (1991) 107.
.
Unsteady shear flow of a viscoplastic material S.L. Burgess *, S.D.R. Wilson Department of Mathematics, Uni6ersity of Manchester, Oxford Road, Manchester M13 9PL, UK Received 30 November 1996; accepted 11 February 1997
Abstract A viscoplastic, or yield-stress, liquid occupies the space between two infinite parallel plates. Initially the whole system is at rest. The lower plate is suddenly jerked into motion with given speed or shear stress, while the upper plate is kept fixed. The flow consists of two regions; (1) a lower sheared region bounded above by the yield surface, (2) an upper unyielded region bounded below by the yield surface. The yield surface propagates to the upper plate as time proceeds. We first consider the equivalent one plate problem of flow over a jerked plate, and find similarity solutions and small time asymptotic solutions for prescribed shear and speed cases respectively. These solutions are used as initial solutions for the two plate case. The motion of the yield surface and the time taken for the entire material to yield are investigated. The problems considered here are two dimensional representations of some control devices, for example the light duty clutch, which consists of two corotating, coaxial discs separated by a layer of electrorheological material. In this application it is useful to know the time taken for all the material to yield. © 1997 Elsevier Science B.V. Keywords: Electrorheological fluids; Shear stress; Viscoplastic material
1. Introduction Commercially, electrorheological (ER) fluids have been put to many uses, yet studies on the underlying fluid mechanics of these applications are scant. The problem investigated here is part of a more extensive study into the behaviour and possible uses of ER materials in control devices and the results should be of wider, more general interest. Our interest mainly lies with light duty clutches; these can be used as the control mechanisms in robotic arms for example. They consist of two corotating, coaxial discs, closely spaced and with ER fluid between them. Other applications include dampers, shock absorbers and alternators, more details of which can be found in Hartsock et al. [1]. An investigation into the use of ER materials in automotive engine mounts is presented by Williams et al. [2]. * Corresponding author. 0377-0257/97/$17.00 © 1997 Elsevier Science B.V. All rights reserved. PII S 0 3 7 - 0 2 5 7 ( 9 7 ) 0 0 0 1 8 - 9
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The current investment in ER fluids is due to the fact that the viscosity of such materials can be controlled by an externally applied electric field. Typically, ER materials consist of electrically polarisable particles suspended in an oil. The yield stress of the material is proportional to the applied electric field (see Stangroom [3]), so accounting for the ‘smart’ behaviour of ER materials. The manner in which the fluid yields and, in particular, the time taken for all the material to yield are of interest. In previous studies (e.g. Stangroom [3]) ER materials have been said to be represented well by a Bingham constitutive model. This model in its usual formulation asserts that the material is perfectly rigid until the yield stress is exceeded, thereafter behaving in a Newtonian-like manner. Barnes and Walters [4], however, give experimental evidence that materials which flow at high shear rates will also do so even at very low shear stresses. This means that some flow can occur below the yield point. Moreover, the use of the Bingham model can lead to difficulties arising from the fact that for an absolutely rigid material the stress distribution is indeterminate, in general. See, for example, Lipscomb and Denn [5] and Wilson [6] for a discussion of the so-called squeeze flow paradox. For these reasons we use a biviscosity model. This viscoplastic model assumes that the material behaves as a Newtonian fluid with very large viscosity until the critical ‘yield stress’ is exceeded. Above the yield stress, the material has a smaller, rapidly decreasing viscosity. The Bingham model can be recovered as a limiting case. We investigate the unsteady fluid motion in a layer of viscoplastic material contained between two closely spaced parallel plates. The upper plate is kept fixed whilst the lower is suddenly given constant speed or shear stress. We are particularly interested in the start-up process, that is, the fluid motion up to the time when all the fluid has yielded. First, we study the jerked plate problem for a single plate. This, a generalisation of the Rayleigh problem, is interesting in itself and provides the initial solutions needed for the two plate case. (The effect of the top plate is minimal in the first instants of motion). Particular attention is paid to the motion of the yield surface which separates the sheared and unyielded regions. For the case of prescribed shear stress, similarity solutions are found. It is not possible to find such self similar profiles for the prescribed speed case; a lengthscale is already available from the plate velocity and the characteristic yield stress of the material. However, we find asymptotic solutions valid at small times. Two phase flow, the regions separated by a moving interface, can be typified by the classic Stefan problem of ice melting into water, for example. The complicating factor is the unknown position of the moving solid/liquid interface (see Crank [7] for more details). In the two plate case, the solution must be obtained numerically and so we choose to fix the moving yield surface and the computational domains by a simple coordinate transformation. The velocity of the yield surface then appears in the governing equations. An estimate of the time taken for all the material to yield is made. The flow of a viscoplastic material over a plate having prescribed constant shear stress has been considered by Phan-Thien [8]. This work has been extended to the case of a time dependent plate shear stress by Phan-Thien and Gartling [9]. The viscoplastic model used in both papers, however, is inappropriate; it is proposed that the shear stress be simply proportional to the shear rate in both the yielded and unyielded regions (with different constants of proportionality). But this means that there must be a discontinuity in either the shear rate the shear stress at the yield surface. As a result, we do not consider these papers further.
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Sekimoto [10] has looked into the flow of a Bingham material over a single plate. Initially, the prescribed stress of the plate was reduced from a value greater than the yield stress to one less than it. Similarity solutions were found. Sekimoto also mentions that in the start-up problem for Bingham materials the yield surface propagates to infinity instantaneously. This work, however, is cited as being unpublished and appears to remain so. The present work shows that the basic idea is correct and gives an explicit estimate of the propagation speed in the ‘Bingham limit’.
2. Analysis and results Using Cartesian co-ordinates, the fluid velocity is given in dimensional form by (U(Y, T), 0), where Y is the co-ordinate perpendicular to the plate and T is the time. Neglecting body forces, the fluid velocity is governed by r
(U (t = , (T (Y
(1)
where r is the constant fluid density and t is the shear stress. The plates and fluid are quiescent until the lower plate is suddenly jerked into motion parallel to itself with constant velocity u0 or constant applied shear stress − tb(tb \ 0). Refer to Fig. 1 for a diagram of the problem. The boundary and initial conditions are U(d, T) =0
ÖT
(2)
and either U(0, T) =u0
T\0
(3)
or
Fig. 1. The two plate problem.
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Fig. 2. The biviscosity model; dependence of shear stress t on shear rate e= (U/(Y.
t Y = 0 = −tb
T\0
(4)
We now describe the constitutive model. Many fluids of industrial importance exhibit a yield stress, or apparent yield stress. This means that little or no motion of the material will occur until some critical yield stress has been exceeded. For reasons noted above, we choose the so-called biviscosity model. This supposes that
Á (U (U t0 t \t c Ãh 2 + sgn (Y (Y t=Í (U Ã h1 t B t c (Y Ä
(5)
Fig. 2 illustrates this model. Two viscosities are introduced by this law, and the ratio of the two, namely h2/h1 = e, is a key parameter. We have t0 =tc(1−e). The material is supposed to behave like a Newtonian fluid with very large viscosity h1 below the apparent yield stress tc, and has a rapidly decreasing viscosity above this critical stress. The usual Bingham model is obtained by letting e 0, and e = 1 gives the Newtonian model. For the jerked plate problem sgn(#U/#Y)= − 1. The biviscosity model indicates that there will be two regions of flow separated by a yield surface Yc. On Yc the shear stress equals the characteristic yield stress, tc, of the material. Region 1 is the layer of fluid bounded below by the yield surface Yc and above by the fixed plate. Here, little of the fluid occurs and the behaviour is like that of a Newtonian fluid
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with large viscosity h1. Region 2 is adjacent to the lower plate and bounded above by Yc, and is yielded. As time proceeds, we expect more fluid to yield and the yield surface to move towards the upper plate. An added complication is that the yield surface position Yc =Yc(T) is not known in advance and can only be determined by solving the system of governing equations and conditions. It is important to remember that both the velocity and the shear stress must be continuous on the yield surface, and that the stress also equals −tc on Yc. It will be seen that for the two plate problem, analytical solutions are not available; the governing systems are solved numerically. The fluid profiles and yield surface position must then be known initially in order to start the numerical procedure. However the sudden imposition of prescribed shear stress or speed at the lower boundary, as an idealisation, results in singularities at T=0 which are difficult to deal with numerically. This is one reason for searching for analytical solutions to the one-plate problem; they provide small-time starting profiles for the full problem. In addition, they are of some interest per se. The case of prescribed plate stress is dealt with first of all as this is easier mathematically. Then the prescribed speed problem is investigated; this is the most important industrially (most control devices are constructed as prescribed speed devices).
3. Prescribed shear
3.1. One plate case The plate is contained in a semi-infinite expanse of viscoplastic material and is given a constant shear stress −tb, where tb \ 0. Before starting the process of finding a solution, a comment must be made about the way in which the variables will be non-dimensionalised. In order to use the one plate solution as a starting solution for two plate case, the scalings used must be the same. We use the scalings appropriate for the two plate problem, namely Y =dy,
Ui =
dtb ui, h2
T=
d 2r t; h2
i =1, 2.
(6)
Of course, for the one plate problem the plate separation d must be irrelevant and it will be seen that the solution, when obtained, will not depend on it. The choice is made for later numerical convenience, as noted.We propose similarity solutions of the form u1 = t 1/2f1(n),
(7)
u2 = t 1/2f2(n),
(8)
where n=
y , t 1/2
The governing equations reduce to
(9)
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e { f1 − nf %1} =f ¦, 1 2
(10)
1 { f2 − nf %2} =f ¦, 2 2
(11)
f %2(0) =Ls(1 −e) −1,
(12)
f1( ) = 0,
(13)
f1(nc) = f2(nc),
(14)
f %1(nc) = f %2(nc)= − Lse,
(15)
where %= d/dn, nc = yc/t 1/2 is the position of the yield surface and Ls =tc/tb. We assume that the applied stress is greater than the yield stress, i.e. 0 BLs B1; otherwise the motion throughout is that of a Newtonian liquid of viscosity h1. Eqs. (10) and (11) can be integrated to give
! !
2
f1(n) =A1 e − en /4 − 2
f2(n) =A2 e − n /4 −
" "
(pe)1/2 e 1/2n n erfc 2 2
p 1/2 n n erfc 2 2
+B1n,
(16)
+ B2n.
(17)
The five conditions in Eqs. (12)–(15) determine the unknown constants A1, B1, A2 and B2 and the position of the yield surface nc. Eliminating the constants we find a single implicit equation for the yield surface nc, namely (1 −Ls) erfc
2 e 1/2nc nc = Lse 1/2 e(1 − e)nc /4 1−erfc 2 2
.
(18)
It is straightforward to evaluate nc numerically from this equation. The dependence of nc on the viscosity ratio e, for three fixed values of Ls is illustrated in Fig. 3. It is seen that the smaller e is in the range (0, 1) the larger nc is. In addition, the larger Ls is the closer the yield surface is to the plate. We expect this physically, as larger values of Ls correspond to smaller values of the applied shear stress. In the physical applications we are interested in, the viscosity ratio e is typically small, and so it is useful to solve Eq. (18) for nc in the limiting case e0; the position of the yield surface is then found to be given explicitly as
! !
yc = 2t 1/2 ln
1− Ls Lse 1/2
""
1/2
.
(19)
Note that as e 0, the ‘Bingham limit’, the yield surface goes to infinity at any t \0. To summarise, the velocity profiles at small times for the case of constant plate stress are u1 = 2Ls u2 = 2Ls
! ! et p
1/2
et p
1/2
2
e − en /4 − 2
" "
(pe)1/2 e 1/2n n erfc 2 2
eg2/4 e − n /4 −
(p)1/2 e n erfc 2 2
,
−eLsn.
(20) (21)
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As we now have initial solutions for the two plate case, this problem will be investigated.
3.2. Two plate case The lower plate is given prescribed shear stress whilst the top one is kept fixed. It is expected that the yield surface will move towards the upper plate as time progresses, and there will exist a critical time Tc at which it hits the upper plate. After this time, all the material will be yielded. The moving boundary Yc = Yc(T) must be tracked in some way. Various methods are available, here we choose to fix the yield surface by an appropriate coordinate transformation. It then turns out that the velocity of the yield surface appears in the governing equations, with the equations being valid on a finite fixed domain. The yields surface yc is fixed by introducing the transformation z=
y −1 yc(t) −1
(22)
in region 1, and z=
y yc(t)
(23)
in region 2. The upper region is mapped from [yc, 1] to [1, 0], and the lower region is mapped from [0, yc] to [0, 1]. The system is then governed by
Fig. 3. Self similar yield surface position nc for 10 − 3 5 e5 1; Ls = 10 − 3 ——, Ls = 10 − 2 — — —
, Ls = 0.5 - -.
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dyc (u1 ( 2u1 (u1 − e(yc − 1)2 =0, + e(yc − 1) z (z 2 dt (z (t
(24)
dyc (u2 (u2 ( 2u2 −y 2c = 0, 2 + ycz (z dt (z (t
)
(25)
1 (u2 =Ls(1− e)− 1, yc (z z = 0
(26)
u1(0, t) =0,
(27)
)
u1(1, t) =u2(1, t),
)
(u1 1 1 (u2 = = − Lse, yc (z z = 1 (yc − 1) (z z = 1
(28) (29)
where Ls = tc/tb. Notice that the velocity of the yield surface, dyc/dt, appears in the governing equations, and that the equations are defined on a fixed domain. The initial solution used is that found in Section 3.1 for the one plate problem. Suppose that the integration in the two plate case is started at a small time ts, say. Then the yield surface position yc(ts), velocity profiles u1(z, ts) and u2(z, ts) can be evaluated from Eqs. (19)–(21) respectively. Now consider the system of equations and conditions given in Eqs. (24)–(29). The parabolic equations are linear and include both the yield surface position yc and yield surface velocity dyc/dt. If yc at time t were known a priori then it would be straightforward to integrate the equations. This, however, is not the case, as yc is only known at time ts. The system is over determined and so the ‘extra’ condition is used to calculate yc at time t. The equations are solved by a standard numerical routine using finite differences and the method of lines. The boundary conditions incorporated in this routine are those given in Eqs. (26) – (28) and
)
)
(u1 1 1 (u2 = . yc (z z = 1 (yc − 1) (z z = 1
)
(30)
The method of bisection is used to calculate the yield surface position from the extra condition 1 (u2 = −Lse. yc (z z = 1
(31)
It is helpful to consider typical values that Ls might take. It is usual for the lower plate to be jerked with a shear stress several orders of magnitude greater than the yield stress of the material. It is pointless considering the case when the prescribed stress has magnitude less than the critical stress—in this circumstance none of the material will yield and little, if any, flow will occur. The parameter Ls is, therefore, thought to be at least O(10 − 3) and at most O(10 − 1). Fig. 4 illustrates how the critical time tc varies with e for the case of prescribed shear. We took Ls = 10 − 3, 10 − 2 and 0.5 and e= 10 − 3, 10 − 2, 5×10 − 2 and 10 − 1. For a particular Ls, the time taken for all the material to yield increases as e increases. This fits in with the result for a Bingham material (e = 0); in this case tc is infinitesimal. Even for a material with e=10 − 1, tc is only O(10 − 2). Also tc increases as Ls increases, for a given e. We expect the material to take
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Fig. 4. Viscosity ratio e against tc for prescribed shear; Ls = 10 − 3 ——, Ls = 10 − 2 — — — —, Ls = 0.5 - -.
longer to yield when the ratio of the yield stress to the plate stress is increased; this corresponds to reducing the plate stress. An idea of the real critical time Tc can be obtained by evaluating the timescale d 2r/h2. Typically1 d = O(10 − 3) m,
r = O(103) kg m − 3,
h2 =O(1) Pa s.
(32)
The timescale is calculated to be O(10 − 3); this means that the real time involved is merely O(10 − 5) s. For Bingham materials (e = 0), the yield surface has an infinite speed of propagation. The numerical routine was run for small e, and so we expect the yield surface to take a short time to reach the upper plate.
4. Prescribed plate speed
4.1. One plate case A plate is contained in a semi-infinite expanse of viscoplastic material. The system is at rest until the plate is suddenly given a constant speed u0 parallel to itself. In line with the analysis used for the prescribed shear problem, we now consider what scalings might be regarded as appropriate for the prescribed speed case. Because this problem includes the plate speed u0 and the characteristic yield stress tc, there is a natural lengthscale h2u0/tc, and so there will not be a similarity solution of the resulting system of equations and conditions. 1 Sources of data: Hartsock et [1], Barnes and Walters [4], Lord Corporation product information documentation for materials ER100 and ER200.
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However, we can find an asymptotic solution valid for small times. This solution will still then be a valid initial solution for the two plate case. The limit of interest is t0, e fixed. We choose to use the two plate scalings from the start for the sake of simplicity and so put Ui = u0ui,
Y = dy,
T=
d 2r t; h2
i =1, 2.
(33)
As a result we then expect the ratio of the two lengthscales to occur as a non-dimensional parameter of the flow, namely dtc/h2u0 = Ln. The flow is governed by e
(u1 ( 2u1 = 2, (t (y
(34)
(u2 ( 2u2 = 2, (t (y
(35)
u1( , t) = 0,
(36)
u2(0, t) =1,
(37)
)
)
u1(yc, t) = u2(yc, t),
(38)
(u2 (u1 = = − Lne. (y y c (y y c
(39)
Here yc = yc(t) denotes the yield surface. It is clear from the form of boundary conditions Eqs. (37) and (39) that this system will not admit an exact similarity solution. However it is possible to obtain an approximate solution valid at small times, that is, in the limit t 0, e fixed; this solution is of interest per se and, as noted, will serve to start off the numerical solution of the two plate problem. However the calculations become very elaborate and only the outline will be given here. We can begin by considering the case of a Newtonian fluid for which e=1. Here we do of course have an exact similarity solution u1 = u2 = erfc(n),
(40)
where n = y/2t 1/2; the ‘yield stress’ is purely notional and the condition Eq. (39) simply gives the position of the surface on which the shear stress takes a prescribed value. We find that the ‘yield surface’ is given by 2
e − nc =
Ln (pt)1/2 2
(41)
from which we conclude that nc as t0. When e B 1 we conjecture that nc is again large when t is small, and so will have u2 erfc(n)+ ···
(42)
as a first approximation. After some trial and error it turns out that the appropriate form for u1 is u1 t kl mf1(n ),
(43)
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where l= log (1/t), n = e 1/2n (motivated by the observation that the similarity variable contains the square root of the viscosity), k and m are constants to be determined, and the function f1(n ) satisfies f ¦1 + 2n f %1 −4kf1 = 0.
(44)
We require f1 0 as n . The equation can be solved in terms of the usual parabolic cylinder functions. Applying the conditions at the yield surface (Eqs. (38) and (39)) then gives 1 e m = k= − 2 2
(45)
and an estimate of the position of the yield surface nc: 2
e − nc =
L6 (ept)1/2. 2
(46)
We now turn to the problem of obtaining the correction terms to the estimates Eqs. (42) and (43). The main reason for undertaking this is that the possibility of continuing the expansion is evidence that the leading terms are in fact correct. It is found2 that Eq. (42) becomes u2(n) =erfc(n)+ t 1/2
An Bn log l f2(n) + 2 +··· , + l l2 l
(47)
where A and B are constants to be determined, and f2 satisfies the equation f ¦2 +2nf %2 −2f2 =4An with f2(0) = 0.
(49)
Meanwhile Eq. (43) becomes
u1(n ) = t kl m f1(n )+
log l 1 g1(n ) + h1(n )+··· , l l
(50)
where g ¦1 + 2n g %1 − 2(1− e)g1 = 0,
(51)
g1( ) = 0,
(52)
h ¦1 + 2n h %1 − 2(1− e)h1 = − 2(1− e),
(53)
h1( ) − 0.
(54)
Matching the solutions at the yield surface gives, after lengthy calculations,
2
A = 2(Ln −e),
(55)
B = − 2(Ln − e),
(56)
We are grateful to Mr E.J. Watson of Manchester University for his assistance in finding these solutions.
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and the correction to the form of nc, which was given to leading order in Eq. (46): nc =
(Ln − e) l − log(ep 1/2)− +··· 2 el
1/2
.
(57)
To summarise, the small time solutions for the velocities u1 and u2 are u1 =
tl 2
−
(1/2) − (e/2)
!
2 1 2k − 1 2k − 3 Lnp − e/2e 1 − (3e/2)e − n n − − k+ (k+1)n − 2
(1 − e) log l − 2 + e 1 − 2 + e (1−e) 2 pe 3 n (1−e) log n − 3− +log + n 2 2l 2 e l
"
+··· , (58)
u2 = erfc(n) +2t 1/2 +
!
2 2nLn 1 (Ln − e)n l− log l− g −log 2 + 2 1− 2 pe l e l
"
2(Ln − e) (− 1)nn 2n + 3 5 % (n+ 1) 3 2 n=0
+··· .
(59)
n
4.2. Two plate case We use the scalings given in Eq. (33) and the transformations given in Eqs. (22) and (23). The system of equations and boundary conditions governing the flow are those given in Eqs. (24) – (29), with the exceptions that Eq. (26) is replaced by u2(0, t) =1,
(60)
and Ls is replaced by Ln = dtc/h2u0 in Eq. (28). In the full two plate system there is, again, one ‘extra’ condition. The pdes are integrated using finite differences and the method of lines, and the method of bisection is used to determine the yield surface position from the condition
)
1 (u2 = −Lne. yc (z z = 1
(61)
The numerical solution is found in the same manner as described in Section 3.2, and the initial solution is the one plate solution given in Section 4.1. Before presenting any results, it is necessary to determine what values of the parameter Ln = dtc/h2u0 occur in practice. Typically3 d = O(10 − 3) m, h2 = O(1) Pa s, 3
tc = O(10) u0 = O(1)
Sources of data: as given previously.
or O(102) Pa,
(62)
or O(10) m s − 1
(63)
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The parameter Ln can be seen to be at least of O(10 − 3) and at most of O(10 − 1). We choose Ln = 10 − 3, 10 − 2 and 0.5. Results of the viscosity ratio e against the time taken for the yield surface to hit the top plate tc are presented in Fig. 5. (In practice, small e results are of interest so we take e =10 − 3, 10 − 2 and 10 − 1.) From Fig. 5 it can be seen that for a particular value of Ln the time taken for the yield surface to hit the upper plate increase as e gets larger. This is expected, since for a Bingham material (e= 0) the yield surface moves to the upper plate instantaneously. In addition we see that as the parameter Ln is increased for a particular e, the time taken for all the material to yield, tc, increases. Increasing Ln corresponds to increasing the plate separation or reducing the speed at which the plate is jerked. In both situations we would expect the critical time tc to increase; our results support this. Note that from the figure, tc is only O(10 − 2). The real time taken for the yield surface to hit the upper plate can be evaluated in the manner described in Section 3.2 for the prescribed shear stress case. Again it turns out that Tc is merely O(10 − 5) s. As only relatively small viscosity ratios have been considered, very small critical times are expected. The Bingham model provides the reference case for this. Bingham materials (e= 0) and Newtonian materials (e=1) are the two limiting cases of the biviscosity model. The time taken for all the material to yield increases as e is increased from 0– 1. It is found that the critical time is very small, regardless of whether the lower plate is given prescribed speed or shear stress. The time taken for the yield surface to hit the top plate is typically a fraction of a second for materials of small e. The analysis conducted here and the numerical methods used will provide a good basis for future work on the full problem of viscoplastic flow between rotating discs. The yielded material in this case will be contained inside a cylinder of critical radius, and this cylinder will move radially outwards as time proceeds and more material yields.
Fig. 5. Viscosity ratio e against tc for prescribed velocity; Ln = 10 − 3 ——, Ln = 10 − 2 — — — —, Ln -0.5 - -.
S.L. Burgess, S.D.R. Wilson / J. Non-Newtonian Fluid Mech. 72 (1997) 87–100
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4.3. Note on time taken to achie6e steady Couette flow For the two plate problem the velocity profile will reach the steady state Couette solution u= 1 − y in some timescale tf after the material has yielded. This happens in both the prescribed speed and prescribed shear stress cases. We can think of the jerked plate problem for a yield stress material as consisting of two steps; (1) the yield surface moves towards the upper plate and then hits it, (2) the material then behaves in a Newtonian fashion as the flow approaches the usual Couette profile. It is interesting to estimate the size of tf. For the Newtonian flow the timescale is rd 2/h, where h is the viscosity. This timescale can be evaluated to be O(10 − 3). So once the material has yielded, Couette flow will be achieved relatively quickly.
5. Conclusions The problem of unsteady shear flow has been investigated for a viscoplastic material. We modelled the material as obeying a biviscosity law and considered both the one plate and two plate problems. In both cases, the jerked plate was given constant prescribed speed or shear. Similarity solutions were found for the one plate prescribed shear case, and small time asymptotic solutions were found for the case of given plate speed. Numerical solutions were found for both the two plate cases. The yield surface was tracked and estimates of the time taken for the entire fluid to yield were made. For the two plate flow the yield surface takes a fraction of a second to reach the upper plate; the material then achieves the usual Couette profile within a short time. The larger the viscosity ratio e, the longer the material takes to yield completely. This result accords with that given by the standard Bingham model (e=0) for which the material yields instantaneously. In addition, as the non-dimensional parameters Ln or Ls increase so does the yield time. The motivation for this work was to gain a better understanding of the workings of control devices, the light duty clutch in particular. The work presented here will provide a sound foundation for further investigations.
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