The first problem of Stokes for an Oldroyd-B fluid

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International Journal of Non-Linear Mechanics 38 (2003) 1539 – 1544

The rst problem of Stokes for an Oldroyd-B *uid Constantin Fetecau∗ , Corina Fetecau Department of Mathematics, Technical University of Iasi, Iasi 6600, Romania Received 3 January 2002; accepted 28 June 2002

Abstract The velocity eld and the associated tangential tension corresponding to the *ow of an Oldroyd-B *uid over a suddenly moved *at plate are determined. The well-known solutions for a Navier–Stokes *uid, as well as those corresponding to a Maxwell *uid and a second-grade one, appear as limiting cases of our solutions. Finally, some comparative diagrams concerning the velocity and tension pro les are presented for di5erent values of the material constants. ? 2002 Elsevier Science Ltd. All rights reserved. Keywords: The rst problem of Stokes; Velocity elds; Tangential tensions

1. Introduction Due to its practical applications, the Stokes’ problem for the *at plate (1851) has received much attention. An interesting solution of this problem, within the context of Navier–Stokes theory, was obtained by Zierep [1]. Soundalgekar [2], Rajagopal and Na [3], Puri [4], Bandelli et al. [5], B@ohme [6], Tigoiu [7], Fetecau and Zierep [8] and Fetecau and Fetecau [9] have studied the problem for di5erent types of non-Newtonian *uids. In this paper, a new exact solution corresponding to the *ow of an Oldroyd-B *uid over a suddenly moved *at plate is determined for all values of relaxation and retardation times. The well-known solution for a Navier–Stokes *uid [1], as well as those corresponding to a Maxwell *uid and a second-grade one [8,9], appears as a limiting case of our solution. Finally, the associated tangential tension is determined and some ∗

Corresponding author. Tel.: +40-32-140513. E-mail address: [email protected] (C. Fetecau).

comparative diagrams are presented. The steady-state solutions appear as limiting cases for t → ∞, too. 2. Mathematical formulation The mechanical behavior of non-Newtonian *uids can be modeled by several constitutive equations. Here, we shall consider a rate type model due to Oldroyd. The Cauchy stress T in such a model is given by [6,10–13] 
S A ; (1) T = −pI + S; S +  =  A + r t t where  and r are relaxation and retardation times,  is the dynamic viscosity of the *uid, −pI denotes the indeterminate spherical stress, A = L + LT with L the velocity gradient and =t represents the upper-convected derivative de ned through B = B˙ − LB − BLT ; t

(2)

the over dot indicating the material time derivative.

0020-7462/03/$ - see front matter ? 2002 Elsevier Science Ltd. All rights reserved. PII: S 0 0 2 0 - 7 4 6 2 ( 0 2 ) 0 0 1 1 7 - 8

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C. Fetecau, C. Fetecau / International Journal of Non-Linear Mechanics 38 (2003) 1539 – 1544

The model characterized by the constitutive equations (1) and (2) is usually referred to as the Oldroyd-B *uid (cf. [10] or [11]). It includes as special cases the Maxwell model and the linearly viscous *uid model. For some special *ows the model of second-grade *uid is also included. Existence results for the *ow of an Oldroyd-B *uid have been established by GuillopNe and Saut [12] and the rst exact solutions seem to be obtained by Rajagopal and Bhatnagar [13]. In the following, we wish to determine the velocity eld and the associated tangential tension corresponding to the rst problem of Stokes for an Oldroyd-B *uid lying over an in nitely extended *at plate. Initially, the *uid is at rest and at time t = 0+ the plate is impulsively brought to the constant velocity V . By the in*uence of shear, the *uid above the plate is gradually moved. In a suitable Cartesian coordinate system Oxy z the velocity eld is of the form v = v(y; t)i;

(3)

where i is the unit vector along the x-coordinate direction. For this velocity eld, the constraint of incompressibility is automatically satis ed. Substituting (3) into (1)2 and having in mind the initial condition S(y; 0) = 0;

(4)

we get Sxz = Syy = Syz = Szz = 0. For the tangential tension Sxy it results in the di5erential equation Sxy (y; t) + 9t Sxy (y; t) = (1 + r 9t )9y v(y; t):

(5)

In the absence of a pressure gradient in the x-direction the balance of linear momentum leads to 9y Sxy (y; t) = 9t v(y; t);

(6)

where  is the constant density of the *uid. Eliminating Sxy between Eqs. (5) and (6), we attain to the linear partial di5erential equation 9t v(y; t) + 92t v(y; t) = (1 + r 9t )92y v(y; t); y; t ¿ 0;

v(0; t) = V;

t ¿ 0:

(9)

It is worth observing that the partial di5erential equation (7) is one order higher than the equation corresponding to Navier–Stokes *uids [1]. In order to obtain an exact unique solution, we need additional initial and boundary conditions. The appropriate boundary conditions (cf. [14,15]) v(y; t); 9y v(y; t) → 0

as y → ∞

(10)

are consequences of the fact that the *uid is at rest at in nity and there is no shear in the free stream. Moreover, we assume that (cf. [16]) 9t v(y; t) → 0

as t → 0:

(11)

3. Solution of the problem  Multiplying both sides of Eq. (7) by 2= sin(y), integrating with respect to y from 0 to ∞ and taking into account conditions (8)–(11), we nd that 92t vs (; t) + (1 + 2 )9t vs (; t) + 2 vs (; t)  = 2= V; t ¿ 0;

(12)

where  = r and the Fourier sine transform vs (; t) of v(y; t) has to satisfy vs (; 0) = 9t vs (; 0) = 0;  ¿ 0:

(13)

The solution of the ordinary di5erential equation (12) with the initial conditions (13) has one of the following three forms:    r1 exp(r2 t) − r2 exp(r1 t) 2 V 1− vs (; t) =   r1 − r2 if  ¡ r ;

(14)

(7)

where  = = is the kinematic viscosity of the *uid. Since the *uid has been at rest for all t 6 0 and the plate is maintained at the constant speed V for all t ¿ 0 we have v(y; 0) = 0;

and

y¿0

(8)

vs (; t) =    2 exp(−t=) − exp(−2 t) 2 V 1−   2 − 1 if  = r ;

(15)

C. Fetecau, C. Fetecau / International Journal of Non-Linear Mechanics 38 (2003) 1539 – 1544

or

and

vs (; t) =    r1 exp(r2 t) − r2 exp(r1 t) 2 V   1− ;      r1 − r 2        ∈ (0; a] ∪ [b; ∞)      
  1 + 2 2 V 1 − exp − t    2    
 
   2    × cos t + 1 +  sin t ;   2  2       ∈ (a; b);

v(y; t) = V − × ×

1 a= √ √ √ (  +  − r )

Inverting (14)–(16) by means of Fourier’s sine formula we nd that

2V ∞ v(y; t) = V −  0 r1 exp(r2 t) − r2 exp(r1 t) sin(y) × d r1 − r 2 

2V v(y; t) = V −  ×

a




r 2 exp − t 2


 t cos 2

∞

b

r1 exp(r2 t) − r2 exp(r1 t) sin(y) d r1 − r2 

if  ¿ r :

(19)

By making t → ∞ in anyone of the above expressions, we obtain the steady-state solution. In view of (4) and (5) we have for the tangential tension 


 t −t exp Sxy (y; t) =  0  (20)

By substituting (17), (18) and (19) into (20) we get

2V ∞ (1 + r1 r )(1 + r2 r ) Sxy (y; t) = −  0  − r ×

exp(r2 t) − exp(r1 t) cos(y) d r2 − r 1

if  ¡ r ;

(21)

 t  ∞ cos(y) 2V d exp −   0 2 − 1

2V ∞ cos(y) + exp(−2 t) d  0 2 − 1

Sxy (y; t) = − (17)

∞

if  = r

0

2 exp(−t=) − exp(−2 t) sin(y) d 2 − 1 

if  = r

0

r1 exp(r2 t) − r2 exp(r1 t) r1 − r2

×(1 + r 9 )9y v(y; ) d:

1 b= √ √ √ (  −  − r )

if  ¡ r ;

b

×

and

a


 1 + r 2 sin(y) t 2V sin d −  2  

+

if  ¿ r . In the above relations we have used the notations  −(1 + 2 ) ± (1 + 2 )2 − 42 ; r1; 2 = 2   = 42 − (1 + 2 )2 ;

 t  2V sin(y) d − exp −   2

(16)

2V 

1541

(18)

and Sxy (y; t) = −

2V 

0

a

(22)

(1 + r1 r )(1 + r2 r )  − r

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C. Fetecau, C. Fetecau / International Journal of Non-Linear Mechanics 38 (2003) 1539 – 1544

exp(r2 t) − exp(r1 t) cos(y) d r2 − r 1
  t  b 1 4V r 2 − exp − exp − t  2 a  2
 t cos(y) d ×sin 2

2V ∞ (1 + r1 r )(1 + r2 r ) −  b  − r

2 2

×

t = 10 s

v3 (y) −0.2

0 0 0

(a)

0.2

0.4

0.6

0.8

1 1

y

2 2

1

v2 (y)

t = 30 s

v3 (y)

(23)

−0.2

0 0 0

(b)

4. Limiting cases 1. Taking the limits of Eqs. (17) and (21) as  → 0, we get the velocity eld (see [8], Eq. (2.6)) 

2 ∞ sin(y) v(y; t) = V 1 −  0   
2 t d ; (24) ×exp − 1 + 2 and the tangential tension

2V ∞ cos(y) Sxy (y; t) = −  0 1 + 2
 2 ×exp − t d; 1 + 2

1

v2 (y)

v1 (y)

exp(r2 t) − exp(r1 t) × cos(y) d r2 − r 1 if  ¿ r :

v1 (y)

0.2

0.4

0.6

0.8

1 1

y

2 2 v1 (y) v2 (y) 1

t = 60 s

v3 (y) −0.2 (c)

0 0 0

0.2

0.4

0.6

0.8

y

1 1

Fig. 1. Velocity pro les v(y; t) corresponding to relations (17)— curves v1, (18)—curves v2 and (19)—curves v3 for V = 2; =0; 0011746 (glycerin), r =15 and =8, 15 and 30, respectively.

(25)

corresponding to the rst problem of Stokes for a second-grade *uid. 2. In the special case when r → 0, corresponding to a Maxwell *uid, our solutions (19) and (23) reduce to (see [9], Eq. (11)) √

2V 1=(2 ) r3 exp(r4 t) − r4 exp(r3 t) v(y; t) = V −  0 r3 − r 4  t  ∞ sin(y) 2V × d − exp −   2 1=(2√)     1  t  sin(y) t + sin × cos d;  2  2 (26)

respectively,

√ 1=(2 )

exp(r4 t) − exp(r3 t) r4 − r3 0  t  4V ×cos(y) d − exp −  2

∞ 1  t  cos(y) d; (27) × sin √ 2 1=(2 ) 

2V Sxy (y; t) = − 

where r3; 4 =

−1 ±



1 − 42 2

and

=



42 − 1:

3. By letting now r → 0 (and then  → 0) in (24) and (25) or  → 0 in (26) and (27) (or in (18) and

C. Fetecau, C. Fetecau / International Journal of Non-Linear Mechanics 38 (2003) 1539 – 1544

2 0

2

S1 (y) t = 10 s

S2 (y) S3 (y) −10 −17

0 0

(a) 2

0.2

0.4

0.6

0.8

−0.15

1 1

y

2

0 −5

t = 30 s

S3 (y) −10 −12 (b)

0 0

0.2

0.4

0.6

0.8

v1 (y) v2 (y) v3 (y) v4 (y)

2

S1 (y) t = 60 s

S3 (y) −8 (c)

−5

0 0

0.2

0.4

0.6

0.8

y

1 1

Fig. 2. Tangential tension pro les Sxy (y; t) corresponding to relations (21)—curves S1, (22)—curves S2 and (23)—curves S3 for V = 2;  = 0; 0011746 (glycerin), r = 15 and  = 8, 15 and 30, respectively.

(22)) and having in view the entries 5 and 6 of Tables 4 and 5 of [17] we attain to the solutions of Stokes problem for a Navier–Stokes *uid (see [1], Eq. (2.74)) 
 y v(y; t) = V 1 − Erf √ and 2 t 
y2 V √ ; (28) Sxy (y; t) = − exp − 4t t where Erf (·) is the error function of Gauss. 5. Numerical results and conclusions In this paper, we have determined the velocity eld and the associated tangential tension corresponding to

t = 10 s

0 0 0

0.2

0.4

0.6

0.8

y

1 1

2

v1 (y) v2 (y) v3 (y) v4 (y) −0.15 (c)

t = 30 s

0 0 0

(b)

1 0 S2 (y)

1

1

−0.15

1 1

y

2

(a)

S1 (y) S2 (y)

v1 (y) v2 (y) v3 (y) v4 (y)

1543

0.2

0.4

0.6

0.8

y

1 1

2

t = 60 s

1

0 0 0

0.2

0.4

0.6 y

0.8

1 1

Fig. 3. Velocity pro les v(y; t) corresponding to an Oldroyd (curves v1), second grade (curves v2), Maxwell (curves v3) and Navier–Stokes (curves v3) *uid for V = 2;  = 0; 0011746 (glycerin), r = 15 and  = 8.

the *ow of an Oldroyd-B *uid over a suddenly moved *at plate. The obtained solutions, given by (17)–(19) and (21)–(23), contain sine and cosine terms. This indicates that unlike the Navier–Stokes *uid (see solutions (28)), oscillations are set up in the non-steady *ow of an Oldroyd-B *uid. The amplitudes of these oscillations decay exponentially in time, the damping being proportional to exp(−t=2) or exp(−t=). Direct computations show that v(y; t) and Sxy (y; t) satisfy both the associate partial di5erential equations and all imposed initial and boundary conditions. In the special cases when r → 0 or  → 0 these solutions reduce to those corresponding to a Maxwell *uid or a second-grade one, respectively. If both r and  → 0, corresponding to a Navier–Stokes *uid, our solutions take the simple forms (28) in which v(y; t) and

1544

C. Fetecau, C. Fetecau / International Journal of Non-Linear Mechanics 38 (2003) 1539 – 1544

3

0

S1 (y) S2 (y) S3 (y) −10 S4 (y) −20 −20 (a) 1 S1 (y) S2 (y) S3 (y) S4 (y) −12

S1 (y) S2 (y) S3 (y) S4 (y) −8 (c)

0 0

0.2

0.4

0.6

0.8

y

1 1

References

0 t = 30 s

−5 −10 0 0

(b) 1

t = 10 s

0.2

0.4

0.6

0.8

y

1 1

0 t = 60 s −5

0 0

0.2

0.4

0.6 y

0.8

(21), (25), (27) and (28)2 , respectively. For small values of t one can observe both the di5erences between the adequate diagrams and the oscillations corresponding to the non-Newtonian models. In each case v(y; t) and Sxy (y; t) tend to zero for y → ∞.

1 1

Fig. 4. Tangential tension pro les Sxy (y; t) corresponding to an Oldroyd (curves S1), second grade (curves S2), Maxwell (curves S3) and Navier–Stokes (curves S4) *uid for V = 2;  = 0; 0011746 (glycerin), r = 15 and  = 8.

ySxy (y; t) depend on √ y and t only by means of the similarity variable y= t. The solutions corresponding to the steady state, resulting for t → ∞, are the same for all types of *uid. In Fig. 1, the variations of the velocity eld v(y; t), corresponding to relations (17), (18) and (19), are plotted for di5erent values of t. The variations of the associated tangential tensions, corresponding to relations (21), (22) and (23), are given in Fig. 2 for the same values of t and of the material constants. For large values of t the pro les of the velocity as well as those of the tension are closing by. Figs. 3 and 4 display, for comparison, the velocity and tension pro les corresponding to an Oldroyd-B, second grade, Maxwell and Navier–Stokes *uid as they result from relations (17), (24), (26), (28)1 and

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