Hodograph Transformations in Unsteady MHD Transverse Flows

Transcription

1 Applied Mathematical Sciences, Vol. 4, 010, no. 56, Hodograph Transformations in Unsteady MHD Transverse Flows Pankaj Mishra Department of Mathematics, Faculty of Science Banaras Hindu University, Varanasi-1005 U.P., India R. B. Mishra Department of Mathematics, Faculty of Science Banaras Hindu University, Varanasi-1005 U.P., India rbmishra Abstract Hodograph-Legendre transformation have been used to find exact solutions of equation of motion governing the two dimensional unsteady viscous incompressible electrically conducting fluid flows under the presence of transverse magnetic field in porous media. Further various flows corresponding geometries have been investigated. Keywords: Hodograph Legendre transformation, porous media 1 Introduction The equations of motion governed by Newtonian fluid flow are highly non linear hence are quite complex to solve difficulty increases when we consider unsteady flow. Various methods techniques have been used to solve Navier-Stokes equation arising in Newtonian fluid flow among those one of the powerful analytical tool is hodograph transformation. This paper has been devoted to hodograph transformation for solving a system of non-linear partial differential equations governing unsteady plane, incompressible viscous flow of electrically conducting fluid in porous media. We have introduced a Legendre transform function of the stream function recast all equations in terms of this transformed function.

2 78 P. Mishra R. B. Mishra An excellent survey of this with application in non-linear problems has been given by Ames[13]. In steady plane flow this method has been previously applied by Chna Smith [10]. Chna Garg [8] have studied this technique to viscous incompressible orthogonal flow Chna Barron [1, 11] have studied to constantly variably inclined incompressible plane flows. Again Chna Nguyen [9] apply this method in non-newtonian MHD transverse fluid flows. Mishra Singh [5] have studied this technique to steady Plane MHD flows. Further Mishra Thakur [3] have used this method to rotating hydromagnetic flows. Recently this technique have been used by Siddiqui Hayat [1] in unsteady plane viscous flow. Bhatt Shirley [] have applied this method in plane viscous flow in porous medium. The present paper is the generalization of Bhatt Shirley [] by applying transverse magnetic field in unsteady viscous incompressible plane flow in porous medium. For this purpose first the Navier-Stokes equations for unsteady viscous plane MHD transverse flow in porous medium is transformed in steady state form. Further these equations are transformed in the hodograph plane then by using Legendre transformation solutions are obtained in various cases. we feel that our work is more general in MHD flows. Equations of Motion We consider unsteady, plane, incompressible viscous electrically conducting fluid flow through porous medium in the presence of transverse magnetic field. Darcy-Brinkman-Lapwood equations are.v =0, (1 [ ] V ρ t + V. V = P + μ V + μ( H H μ V, ( k (V H+ 1 μ e σ H = H t, (3.H =0, (4 where V is fluid velocity, ρ is constant fluid density, P is fluid pressure, K is permeability of porous medium, H is magnetic field, μ is coefficient of viscosity

3 Transverse MHD flow 783 μ e is magnetic permeability. In this paper we have considered unsteady flow in (X, Y plane applied transverse magnetic field take V =(U (X, Y, t,v (X, Y, t, 0 (5 H =(0, 0,H. (6 Since P = (P H + μ e (7 We get [ ] V ρ t + V. V = P + μ V μ V. (8 K The above equation can be rewritten in (X,Y as under. U X + V Y = 0 (9 [ ρ t + U X + V ] U = P ( Y X + μ X + U μ Y K U (10 [ ρ t + U X + V ] V = P ( Y X + μ X + V μ Y K V (11 Again employing the following transformations. Then we have x =(X Ct, y = Y,u = U C, v = V (1 u x + v y = 0 (13 [ ρ u x + v ] u = p ( y x + μ x + u μ y K u (14

4 784 P. Mishra R. B. Mishra [ ρ u x + v ] v = p ( y y + μ x + v μ y K v (15 u H x + v H ( y ν H H x + H =0, (16 y where ν H = 1 μ eσ. Now Introducing the vorticity function ω = v x u y eliminating p we get ( u ω x + v ω = (ν ω ν y K ω (17 (18 where ν = μ ρ is kinematical coefficient of viscosity. 3 Equation in Hodograph plane Let us consider u = u(x, y v = v(x, y such that (u, v J = 0. (19 (x, y In such a case we interchange the roles of dependent independent variables. Hence following Chna et al. [3, 4, 5, 9], we have the relations. write u x = J y v, u y g x g y = J x v, v x = (g, y (x, y = (g, x (x, y = J y u, v y = J x u, (0 = J (g, y (u, v, = J (x, g (u, v, (1 where x = x(u, v,y = y(u, v g = g(x, y =g(x(u, v, y(u, v = g(u, v is any continuously differentiable function. Thus we can J = J(x, y = (u, v (x, y = [ ] 1 (x, y = j(u, v. ( (u, v

5 Transverse MHD flow 785 Now we can write ω(x, y = ω(u, v, H(x, y = H(u, v. Then equations (17 (18 become. where ν [ (jq,y (u, v j x u + y v = 0 (3 ( x v y = ω (4 u + (x, jq ] 1 ν ω (u, v K j = uq + vq 1 (5 ug 1 + vg ν H [ (jg1,y (u, v + (x, jg ] =0, (6 (u, v (x, ω Q 1 = Q 1 (u, v = (u, v, (ω, y Q = Q (u, v = (u, v (7 G 1 = G 1 (u, v = (H,y (u, v, G = G (u, v = (x, H (u, v. (8 4 Equation in Legendre transform function Equation (9 implies the existence of stream function Ψ satisfying U = Ψ Y, V = Ψ X similarly equation (13 implies existence of ψ such that (9 u = ψ y, v = ψ x (30

6 786 P. Mishra R. B. Mishra Ψ=Ψ(X, Y, t =ψ (X Ct CY + Constt. (31 In a similar manner the equation (3 implies u = y, v where ψ = ψ (x, y L = L (u, v are related by = x, (3 L (u, v =vx uy + ψ (x, y (33 Introducing L (u, v in (3 to (6 we see that equation (3 is identically satisfied the remaining equations are as under. ( L j u + L = ω (34 v ν [ (,jq u (u, v + (,jq v 1 (u, v ] ν K ω j = uq + vq 1. (35 Again [ ( ug 1 + vg ν,jg u 1 H (u, v (,jg v (u, v ] =0, (36 where Q 1 = Q 1 (u, v = (, ω u (u, v, Q ( v = Q (u, v =, ω (u, v (37 G 1 = G 1 (u, v = ( u, H (u, v, G = G (u, v = ( v, H (u, v (38 j = [ ( ] L L 1 u v L. (39 u u Summing up above we have the following theorem.

7 Transverse MHD flow 787 Theorem 4.1 If L(u, v is the Legendre transform of a stream function of the equation of motion (13 to (15 governing the plane unsteady flow of a finitely conducting viscous incompressible fluid. Then L(u, v must satisfy (35 H satisfy (36. Then we have q = u + v, θ = tan 1 ( v u (40 hence u = cos θ q sin θ q θ, v = sin θ q + cos θ q θ. (41 Again we define L (q, θ, ω (q, θ j (q, θ to be the Legendre transform function, vorticity function Jacobean function in (q, θ plane. Now using the relations (F, G (u, v = (F,G (q, θ (q, θ (u, v = 1 (F,G, (4 q (q, θ where F (u, v =F (q, θ, G (u, v =G (q, θ are continuously differentiable functions. Following Bhatt Shirley [], we have the corrolary (4.1 Corollary 4. If L (q, θ is the Legendre transform function of a stream function of the equation of motion (13 to (15 then L (q, θ must satisfy ( ( ν sin θ + cos θ q q θ,j Q 1 cos θ sin θ q q θ,j Q + (q, θ (q, θ ν ω q = q (sin θq K j 1 + cos θq (43 And if H (q, θ is Legendre transform of magnetic field, then equation (18 reduces to q (cos θg 1 + sin θg ( ( ν H cos θ sin θ q q θ,j G 1 sin θ + cos θ q q θ,j G + =0, q (q, θ (q, θ (44 where Q 1,Q,j ω are same as obtained by Chna et al. [6],[9] ( G 1(q, θ = 1 cos θ sin θ q q θ,h q (q, θ ( (45 sin θ + cos θ G (q, θ =1 q q θ,h. q (q, θ

8 788 P. Mishra R. B. Mishra Further for infinitely conducting fluid we have the corollary (4.3 Corollary 4.3 L (q, θ H (q, θ must satisfy (43 cos θg 1 + sin θg = 0 (46 Once a solution of L (q, θ of the system of equation (43, (44 is at h we employ x = sin θ q + cos θ q θ, y = sin θ cos θ q θ q (47 equation (40 to get u(x, y v(x, y in the physical plane. 5 Solutions In this section we consider some special flow problems by taking following examples. Example.1 Let L(u, v =F (u+g(v (48 such that first second derivative of F(u G(v are non-zero. putting this value of L(u, v in (34, (37, (38, we get Now ω = 1 F (u, j = 1 F (ug (v, Q 1 = F (ug (v F (u, Q = G (vf (u, G (v G 1 = F (u H v, G = F (v H u. (49 Now using equation (47, (48, (4 we get [ ( 1 G ν + 1 ( F ] + ν (G + F = v F G + u. (50 G G F F K F G F 3 G 3 If equation (48 defines the Legendre transformation function such that F (u = 0 G (v = 0 then (50 is satisfied only when η =0orG (v+f (u = 0. Since we have taken the porous media, hence η 0 so we must have G (v+f (v = 0. Now F (u =G (v =0 F (u =K 1 G (v =K,

9 Transverse MHD flow 789 where K 1, K are arbitrary constants. Then, G (v+f (u =0 K 1 = K L(u, v =C 1 u + C u + C 3 + D 1 v + D v + D 3 (51 D 1 = C 1. Now using equation (3, we get u = 1 C 1 (y + C, v = 1 D 1 (x D (5 Again by using equation (30 we get the stream function as ψ = (x D (y + C. (53 4C 1 4C 1 The time dependent stream function is given by From equations (49, we have Ψ(X, Y, t = (X Ct D (Y + C. (54 4C 1 4C 1 Further we consider G 1 =C 1 H v, G =C 1 H u. (55 u H C 1 v + v [ H 1 C 1 u ν H H C1 v + 1 C1 ] H =0. (56 u Let H(u, v =F (u+g(v then. Putting H(u, v in equation (55, we get u G (v+ v [ ] 1 F (u ν H {G (v+f (u} =0. (57 C 1 C 1 C1 Differentiating twice with respect to u we get F (u v + ν H F (4 (u =0. C 1 C1 Now the above equation is true for all v if F (u C 1 Hence F (u = ( C u + C 3 u + C 4 Now equation (57 becomes { G C 1 C v C 1 } u = 0 ν H C 1 F (4 (u =0. { C3 v + ν H G + ν } H C C 1 C1 C1 =0

10 790 P. Mishra R. B. Mishra Now this equation is true for all u, if only if, G C 1 C v C 1 =0, C 3 v + ν H G + ν H C C 1 C1 C1 =0. (58 Now solving above equation, we get G(v= ( C v + C 5, C3 = 0 hence H(u, v = C ( u v + C 6, (59 where C 6 = C 5 + C 4. Now using equation (51, we have H(x, y = C [ (y + C (x D 8C1 ] + C 6. (60 Similarly by using transformations (1 magnetic field is as under: H(X, Y = C [ (Y + C (X Ct D 8C1 ] + C 6. (61 Using equations (34,(, we have ω =0. (6 The total pressure by integrating (14,(15 with respect to x y respectively is as under: p = μ C 1 K [xy + C x D y] ρ [ (x D +(y + C ] + D 3 (63 C 1 where D, C, D 3 are arbitrary constants. Now time dependent pressure is given as μ P = C 1 K [(X CtY + C (X Ct D Y ] ρ [ ((X Ct D +(y + C ] + D 3. (64 C 1 Further by using the equation (7, the fluid pressure is P = P μ eh. Hence P = μ C 1 K [(X CtY + C (X Ct D Y ] ρ [ ((X Ct D +(y + C ] + D 3 C 1 1 { μ C [ e (Y + C (X Ct D ] } + C 6. (65 8C 1

11 Transverse MHD flow 791 Example. Here we let L(u, v =vf(u+g(u, (66 such that F (u 0 first second derivative of F(u G(v are not zero. Now putting this value of L(u, v in (34, (37, we get ω = (vf + G F, j = 1 F Q 1 = F F, Q = v (F F 3F +F G 3F G F. (67 Again using these values in equation (35, we get ν [ ( F iv v F 10F F +15 F ( ] 3 G iv 6G + F 3 F 4 F F 4 G F +15 G F F 3 F 3 F 4 ν K (vf + G = v F [ ( F + uv 3 F F F F Equation (68 gives rise to following differential equations ν ( ] G + u 3 G F. (68 F F [F iv F 10F F F +15F 3 1K F F 4 ] = u ( F F 3 3F F 3 F F 3 (69 ν [G iv F 6G F F 4G F F +15G F 1K G F 4 ] = u ( G F 3 3G F F (70 The above system is coupled non-linear partial differential equation in two unknown F G quite complex to solve, so we assume F (u as where A a are constants. F (u = 1 a ln u A KA, (71

12 79 P. Mishra R. B. Mishra Now, with this value of F (u, equation (69 is satisfied if only if A = νb, where b = a 1. So, we must have Ka F (u = 1 u νb ln a Kνb. (7 With help of equations (70 (7 we can write G iv + where c =(7 1 a K. Now, we put G = H, we get H + (6νa u (cνa 3u νa(u νb G + νa(u νb G =0, (73 (6νa u νa(u νb H + (cνa 3u H =0, (74 νa(u νb which is linear differential equation of order two with a regular singular point at u = νb, so using Frobenious method we have solution as under: where A,B are arbitrary constants H(u =Aφ 1 (u νb+bφ (u νb, (75 (77 φ (u νb=(u νb r ( c 0 + c 1(u νb+c (u νb In the above equations r 1, r are roots of Indicial equation ( r + 5 b ( r + c 3b =0, (78 a a c 0, c 1, c...; c 0, c 1, c...; are arbitrary constants determined by the recurrence relation [ c m (r + m +(5 b ] 3b (r + m+(c a a = 1 νa c m 1 [r + m ]. (79 Now after getting H(u as above, we can get G(u from the relation G = H hence L(u, v from equation (38. Further we can get u v from equation (3 steady state stream function ψ from equation (33.

13 Transverse MHD flow 793 Example.3 Let L(u, v =Auv + B ( u v + Cu + Dv + E. (80 Now using equation (3 we get (Ax By (BC + AD (Bx + Ay+(BD + AC u =, v =, (81 (A +4B (A +4B again using equation (34 (39 we get 1 j =, ω =0. (8 4B + A Now using these values, we see that equation (35 is identically satisfied again using these values in (36 we have. where M 1 = ν H ( A 4B (Bv Au H u + M 1 ( H u A +4B H v M = 8ν HAB A +4B. H +(ub + Av v H + M =0, (83 u v Now above equation is a Hyperbolic PDE for infinitely conducting fluid i.e. ν H = 0, it has a solution as under: H = B ( u v + Auv. (84 Further for finitely conducting fluid it has a solution for B = 0 as under: H = C 1 e Au ν H + C where we have the velocity components as; ( x D (u, v = A Now, total fluid pressure is given as under: e Av ν H + C 3, (85, y + C A. (86 p = N 1 (x D N (y + C + N 3, (87 where, N 1 = ρ + μ, N A KA = ρ μ N A KA 3 are arbitrary constants. Further by using the equation (7 (1 the time dependent fluid pressure is given by P = N 1 ((X Ct D N (Y + C + N 3 ν ] H [C 1 e 1 (X Ct D Aν H + C e 1 (Y +C Aν H (88

14 794 P. Mishra R. B. Mishra 6 Conclusion This paper has been devoted to generalize the work of Bhatt Shirley [], by taking unsteady case by applying transverse magnetic field. First the time dependent equations are transformed in steady state form then Hodograph -Legendre transformation have been applied to find exact solutions. Several examples have been considered to illustrate the theory. We expect that by taking other forms of Legendre transform function we can get more exact solutions. ACKNOWLEDGEMENTS. First author is highly thankful to Council of Scientific Industrial Research, Govt. of India, New Delhi-11001, for providing financial support to carry out the present research work. References [1] A.M. Siddiqui, T. Hayat, J. Siddiqui S. Asghar, Exact Solutions of Time-dependent Navier-Stokes Equations by Hodograph-Legendre transformation Method, Tamsui Oxford Journal of Mathematical Sciences, Aletheia University, 4(3, (008, [] Balswaroop Bhatt Angela Shirley, Plane viscous flows in porous medium, Matematicas: Ensenanza Universitaria, Vol. XVI, No. 1, 51-6, Junio(008. [3] Chreshwar Thakur R.B. Mishra, On steady Plane Rotating Hydromagnetic flows, Astrophysics Space science 146( [4] G.B. Jeffery, The Two-Dimensional Steady Motion of a Viscous Fluid, Philos. Mag. Ser. 6.9(1915, pp [5] H.P. Singh R.B. Mishra, Legendre Transformation in steady plane MHD flows of a viscous fluid, Indian J. of Pure Appl., 18(1: , January [6] M.H. Martin, The flow of viscous fluid, I, Arch. Rat. Mech. Anal. 41(1971. [7] M.K. Swaminathan, O.P. Chna K. Sridhar, Hodograph study of transverse MHD flows, Canadian J. of Physics. 61(1983, [8] O.P. Chna M.R. Garg On steady plane magnetohydrodynamic flows with orthogonal magnetic velocity field,int J.Engg.Sci.17(1979, [9] O.P. Chna P.V. Nguyen, Hodograph method in Non-newtonian MHD transverse fluid flows, J. of Engg. Math. 3(1989,

15 Transverse MHD flow 795 [10] O.P. Chna, R.M. Barron A.C. Smith, Rotational plane steady flows of viscous fluid, SIAM. J. Appl. Math. 4( [11] O.P. Chna,R.M. Barron K.T. Chew, Hodograph transformations solutions in variably inclined MHD plane flows, J. Engg. Math., 16,3-43(198. [1] R.M. Barron O.P. Chna, Hodograph transformation solutions in constantly inclined plane flows, J.Eng. Math.15, p11(1981. [13] W.F. Ames, Non-linear partial Differential equations in Engineering, Academic Press, New York(1965. Received: November, 010