(2) Proceedings of the International MultiConference of Engineers and Computer Scientists 2014 Vol II, IMECS 2014, March 12 - 14, 2014, Hong Kong. which maps the physical flow field into an infinite strip in the ( , ) plane. In fact the A and B satisfy:. AY. (i ). Y. ( i 1). (7). yi 1,1 yi ,1 yi 1,1 y y y i 1,2 i 1,2 i ,2 (i ) ( i 1) . . , Y . , Y . . . yi 1, N yi 1, N yi , N . (3). Y. and. (log( B)) 2 A q. , . ( i 1). (4). and q as known functions. of and the system (3) and (4) as previously mentioned is quasi-linear hyperbolic with characteristics parallel to the and axes which maps the physical flow field into an infinite strip in the ( , ) plane. Bearing in mind the freedom available in the stream wise variation of and the cross stream variation of , suitable values of. A can be prescribed along one characteristic and those of B can be prescribed along one characteristic. III. THE NUMERICAL ALGORITHM IN THE DESIGN PLANE Rewriting the partial differential equation that y satisfies i.e. equation (1) as:. E. (i ). gi ,1 yi ,0 g i ,2 . h2 . gi , N yi , N 1 . and. 4 1 0 0 . 1 4 1 A 0 1 4 . . . 1 1 4 . 1 y y (C ) + ( )c (5) C A where C , for problems posed in the design plane c=0, B. On a uniform mesh with. the value of C will vary depending on whether the flow field is irrotational or swirl free etc. Equation (5) will be rewritten as a Poisson equation that is as:. Y. y c 1 2 y log e | C | (1 2 ) 2 C C 1 1 y C C . 2 y . (6). 2 is the usual two dimensional Laplacian operator 2 2 2 2 2. where. 2 y y y , , , C , c) 2 . h .. IV. DIRECT SOLUTION OF THE DIFFERENCE EQUATIONS The matrix-vector equation (7) is ( i 1). AY. (i ). Y. ( i 1). E (i ) (i ). With all of order (NXN), and column vectors Y(i) and E of order N. To solve the vector recurrence relation a speculation is made that the Y(i-1) vector can be related linearly to the Y(i) vector as follows:. Y. ( i 1). B (i ) Y. (i ). K. (i ). (8) (i ). (i ). where the B and the K are at present unknown matrices and column vectors respectively. Substituting (8) into (7) gives. (W (i ) B (i ) A)Y. so that. 2 y g (. E (i ). where. (log( A)) 2 B q. Regarding temporarily. ( i 1). Y. Y. (i ). (i ). E (i ) W (i ) K. (W ( i ) B ( i ) A) 1 E ( i ) Y. (i ). E (i ) Y. ( i 1). ( i 1). (W ( i ) B ( i ) A) 1 ( E (i ) W ( i ) K ) (i ). but where g is a function of the arguments shown as defined by equation (6). Writing in finite difference form using central differences gives:. ISBN: 978-988-19253-3-6 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online). Y. (i ). B ( i 1) Y. ( i 1). K. ( i 1). Thus equating coefficients implies IMECS 2014.
(3) Proceedings of the International MultiConference of Engineers and Computer Scientists 2014 Vol II, IMECS 2014, March 12 - 14, 2014, Hong Kong. ( i 1). B. (W B. 1. A) E. (i ). (W ( i ) B (i ) A) 1 ( E. (i ). (i ). (i ). u y (9). and. K. ( i 1). W (i ) K ) (i ). For i=0 this gives. Y. ( 0). B (1) Y. (1). To determine the. K. (1). Y. K. (1). (10). K , if the first iterate B (1) 0 then (1). (0). equations (9) and (10) for i=1 to M. The. Y. (i ). now calculated starting from right to left (as known) using (M ). V.. B ( M 1) Y. ( M 1). K. vectors are. Y. ( M 1). . (12). Furthermore. u (u.)u . p t. ( M 1). (u ) H. l where the k and l are constants with ky y. representing solid body rotation and l/y the so-called free vortex term respectively. VI. THE FLOW EQUATIONS IN THE PHYSICAL PLANE(Y,,X). Adopting cylindrical polar coordinates with y being the radial coordinate, the circumferential and x the axial coordinate, defining velocity components uy , u and ux with corresponding vorticity components y , ,, x in the direction of increasing y, and x respectively, then the equation of motion with unit density becomes:. Du . p Dt. (11). D is the material derivative. Equation (11) can be Dt. written using well known vector identities as:. ISBN: 978-988-19253-3-6 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online). u 1 ( u ) ( p q 2 ) . Thus t 2 for steady flow Crocco’s form of the equation of motion is obtained, i.e.. Here numerical solutions to inviscid axisymmetric flow with constant vorticity and a swirl velocity will be derived. The axial velocity component ux(y) at inlet will be chosen to be of the form ux(y) =y+, where and are constants chosen such that ux(y1)=u1 and ux(y2)=u2 where y1 represents the inner radius and y2 the outer radius at inlet. The swirl velocity u(y), will be of the form. Where. t x u u ux t x u x u ux x t x. u y. u2 p y y y u u y u uy 0 y y u p uy x y x uy. is. AXISYMMETRIC FLOW IN THE ABSENCE OF BODY FORCES. u ( y ) ky . u y. can be written (once again using an appropriate vector identity as). The matrix and vector sequences are now defined by. Y. ux. where H is the total head defined by H =. (13). 1 p q2 . 2. Calculating the cross product on the left hand side of equation (13), gives. H u x u x y 0 u x y u y x H u y u x x. (14). In addition for axisymmetric flow the vorticity vector becomes. u u u = u y + y x + y x x 1 ( yu ) x y y . . (15). The equation of continuity becomes. .u . ( yu x ) ( yu y ) 0 x y VII. THE DESIGN PLANE COUNTERPARTS. In order to compute numerical solutions in the design plane, expressions are required for the terms A, B and , thus. IMECS 2014.
(4) Proceedings of the International MultiConference of Engineers and Computer Scientists 2014 Vol II, IMECS 2014, March 12 - 14, 2014, Hong Kong. u x u y 1 y ux uy y x y x q (log( y )) s or. but. q2 (log( A)) B . thus Ay =. f ( ) , that is. . C ( ) dC dH y y d d. which is the required expression to be used in calculation of B according to definition (4). If far upstream the flow is assumed to be cylindrical so that all quantities are independent of x, then with unit density the equation of motion and the Stokes’ Stream function give. q2 (log( y )) , B . . using the Stokes’ stream function this becomes. yq . The arbitrary f ( ) n. f ( ) represents the freedom in the cross stream distribution of and choosing f ( ) to be unity everywhere f ( ) can be identified as the usual Stokes function. p p u2 u y 0; 0; ; 0; yu x x y y x y giving. With. u2 C ( ) dC y d (u x2 u2 ) y d 2 d ux y u x ( y ) y . stream function given by. yu y ; yu x x y Equation (12), (circumferential component) gives. 0 ux. ( yu ) ( yu ) uy x y. previously defined. Once. and. dH d. u ( y ) ky . l y. as. has been calculated. upstream it takes this value throughout the ( , ) since as is self evident the expression is independent of . This last expression for is required in the calculation of B and numerical coupling with equation (1) gives the numerical solution in the design plane. VIII. DOWNSTREAM CONDITIONS. Referring to the meridian plane figure 1, it may be deduced that:. x y ux q ;u y q s s ( yu ) 0 s yu C ( ). Downstream a cylindrical flow condition as discussed below will be prescribed. Defining the pressure function H ( ) and the function C ( ) as. ds . In terms of C() the vorticity vector where q = dt. for cylindrical flow radial equilibrium (from equation (12) ) radial component gives. (expression (15)) becomes. u y u x 1 C y + + y y x x 1 C x y y . = u . = yy + + x x, by definition.. An expression for is required as this appears in the expression for B, so using the radial component of equation (14) gives. . ux u. H ( ) . 1 2 p (u x u2 ) and C ( ) yu 2. 1 dp u2 dy y Integrating gives. 1. . ( p p y inner ) . Which gives. u2 dy y y inner. C 2 ( ) dy 3 y y inner. H ( ) as. 1 C 1 H y y u x y. ISBN: 978-988-19253-3-6 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online). IMECS 2014.
(5) Proceedings of the International MultiConference of Engineers and Computer Scientists 2014 Vol II, IMECS 2014, March 12 - 14, 2014, Hong Kong. H ( ) . where. p y inner 1 2 (u x u2 ) 2. K 2(. C 2 ( ) dy y3 y inner. 0. . ) (u2 ,1 ) inner (u2 , 2 ) inner and. d u 0 x , 2. . (18). IX. CALCULATION PROCEDURE The calculation of the downstream radii y2() follow from equation (18) with ux,2 given by equation (17), which can be written as. Therefore. . p 2,inner. with ux,2 in this case given by (17).. 1 1 C 2 C 2 1 dC 2 2 2 y 2 dy dy 2y 2 y inner y y inner . . . . y 22 y 22,inner 2. 1 C 2 ( ) Now dy C 2 d (1 / y 2 ) 3 2 y inner y y inner. H ( ) . p1,inner. u x2, 2 g ( ) K , where. 1 2 p y inner 1 2 ux (u ) y inner 2 2 . 1 dC 2 d y 2 d. g ( ) u x2,1 . 1 1 d (C 2 ) y 2 y 2 d d 2 0 1. (19). Suppose u x ,1 u x ,1 ( ) and u ,1 u ,1 ( ) , where the. In order to calculate the (n+1)th iterate it is known that:. subscript 1 denotes upstream conditions, then u x , 2 u x , 2 ( ) and u , 2 u , 2 ( ) are required as. d ( y 22,outer ) 2 K 0 K g ( ) K. functions of , where the subscript 2 similarly denoting downstream conditions, so that. p 2,inner 1 2 1 2 u x , 2 H ( ) (u , 2 ) inner 2 2 (16). and. 1 d d y 22 y 22,inner 2 u 0 x , 2. . . ( y 22,outer K . . d. u . but. 2. 1 1 dC 2 d 2 0 y1 d. =. . (n). 0. y . ( n) 3 x,2. . ( n 1) 2 2 ,outer ( n 1). K. ( y 22,outer ) ( n ) K (n). (20). from which as can be seen from equation (20) the K(n) must be calculated iteratively with K(0)=0, Once the K(n+1) has been calculated it is introduced into equation (19), giving. . 2. rise to a new (u x , 2 ). Furthermore C ( ) y1u ,1 y 2 u , 2 , and equation (16). ( n 1). which in turn gives a new. ( y x2, 2 ) ( n 1) from equation (18) and the process repeated until some convergence criteria is satisfied.. now gives X.. p1,inner p 2,inner 1 2 1 u x , 2 u x2,1 2 2 1 2 ((u ,1 ) inner (u2 , 2 ) inner ) 2 1 1 1 2 2 d (C 2 ) 2 0 y1 y 2 . In this paper the Dirichlet boundary conditions will be prescribed on the wall boundaries so that it is the radii values, y that are given as a function of on the boundaries. The function chosen to give a y distribution is based on the hyperbolic tangent, choosing y()=Ctanh(a+b)+k where C, a, b and k are constants, applying the conditions that y=yu at =0 and y=yd at = taking a+b=3(arbitrary) and b=3, so that tanh(a+b) 1 and tanh(b) -1, then it follows that. or. u x2, 2 u x2,1 K . 1 1 2 y 2 y 2 d (C ) 2 0 1. ISBN: 978-988-19253-3-6 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online). PRESCRIPTION OF WALL GEOMETRIES.. (17). IMECS 2014.
(6) Proceedings of the International MultiConference of Engineers and Computer Scientists 2014 Vol II, IMECS 2014, March 12 - 14, 2014, Hong Kong. y yu y yu y ( ) d tanh(a b) d 2 2 . problem (21). replacing , by x in equation (21) gives a y(x) distribution. The inner radius is prescribed to be equal to unity in this paper (arbitrary). The geometries produced are shown in figures 2, 3 and 4 respectively.. XI. ALTERNATIVE SOLUTION USING AN INTEGRAL FORMULA BASED ON GREEN’S THEOREM.. without. F ( , ) 1 ,. loss of generality and for interior points:. 2t ( 0 , 0 ) t C. setting. v D ds n. v D g ( R. 2t 2 L , )dd 2 2. (22). i.e., the Green’s function vD satisfies Laplace’s equation on. where is defined by:. Here a second method of solution is derived using an integral formula. Commencing with the generalized form of Green’s theorem for the self adjoint elliptic operator E(t) in normal form given by:. vE (t ) tE. ( A). R. C. where t = y , E (u ) E (u ) where E (u ) is the adjoint of E and v is the fundamental solution to the adjoint equation. In this case the adjoint equation is given by ( A). contour C bounding the surface R is traversed in the counter clockwise sense. For a doubly connected region introducing a singularity at the point (0 , 0 ) (inside or on the contour C) and assuming. v( , ) F ( , ) log e | r | so that the distance r is given by: r ( 0 ) ( 0 ) with. 2. C. vg ( R. determination of the speed q given by equation (2) hence differentiating under the integral sign above with respect to. . v t v ds n n. and gives integral formulae for both. 2. (0 , 0 ) is within C and. the appropriate Plemelj formulae or by indenting the contour at (0 , 0 )) . For the Dirichlet case of boundary of. t ( , ) the. requirement. t t and , . such that:. (0 , 0 ) is on C (the m=1 case can be shown using. condition. vN t vanishing on C. Knowledge of the derivatives n t and are also required for the . 2. with L log e t . Now m = 2 if m=1 if. Which give integral formulae for the square of the radius t, from which the radius y can be determined., above Green’s function v N satisfies the Laplace equation on with. . t L , )dd , m 1,2 2 2 2. R. 2t 2 L , )dd 2 2. 2 1/ 2. F ( , ) analytic, then it can be shown that. mt ( 0 , 0 ) F ( 0 , 0 ) t. t ds n. v N g (. ( A). 2 0 and E(t) = g as defined by equation (6). The. . the Neumann problem. 2t ( 0 , 0 ) v N. v t v ds (v)dd t C n n. 2. 0 M1, 0 N1 and vanishes on C. For. is. that. 2 vD t v D t ( 0 , 0 ) ds t C n n. v D R. v g g D dd . and similarly for. v( , ) 0 on C in addition to v( , ) being harmonic. t . and for the Neumann conditions on t, the requirement is that. v =0 and v ( , ) once again satisfying Laplace’s n. equation. Much literature is available for the Green’s function for the Laplace equation (see Williams ) and need not be mentioned here. Hence for the Dirichlet. 2 vD t v D t ( 0 , 0 ) 2 ds t C n n v D R. ISBN: 978-988-19253-3-6 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online). v g g D dd IMECS 2014.
(7) Proceedings of the International MultiConference of Engineers and Computer Scientists 2014 Vol II, IMECS 2014, March 12 - 14, 2014, Hong Kong. XII. ITERATIVE SOLUTION To convert formula (22) to a system of linear algebraic equations the point t (0 , 0 ) inside C is related to its boundary iterates t. values. (1) i. on. (0 , 0 ) , g. (0) i. C.. To. obtain. the. first. where the summations on the right hand side are performed over j=0,1,…N+1 and j=M+N+3, M+N+4.,,,2M+N+3. Once the first iterate tj has been calculated the field integral containing g is then computed, where the central difference approximation to the second derivative is used, this is then introduced into the right hand side of equation (23) and (2). is set equal to zero, so that. compute the second iterate t . The procedure is repeated until some convergence criteria is satisfied e.g.. 2 ti(1) . j. vD t j s j n j. ( k 1). || p , where is a constant and the p denotes the p-norm (p=1,2 or ). || t i t i (k ). 2 N 2M 4. XIII. CONCLUSIONS As shown, geometries have been produced subject to given upstream and downstream conditions with prescribed Dirichlet boundary conditions. In this case vorticity at inlet has been specified by defining the axial velocity to be of the form ux(y) =y+, and the swirl velocity of the form. Using the trapezoidal rule. ti(1) . 2 N 2M 4. j. 1 vD s j 1 s j 1 t j 4 n j. u ( y ) ky . . ti(1) . 2 N 2 M 4. . the so-called free and forced vortex whirl respectively. The downstream conditions where such that: cylindrical flow was present, Dirichlet boundary conditions were prescribed, however the case with Neumann conditions can be accommodated using the algorithm, in addition so can the case with Robin boundary condition. Further examples of the algorithm with a combination of boundary condition is given in Pavlika . It was found that at most eight iterations were required to achieve an acceptable level of convergence, with the technique accelerated using Aitken’s Method.. K (vD , s ) t j , . j. j. Where K (vD , s ) =. 1 vD s j 1 s j 1 4 n . Using this method there is a simple self-consistency check. i.e. the tj are known upstream and downstream for j=0,1,2,…,N+1 and j=N+M+3,N+M+4,….2N+M+3, hence the first iteration may be written as:. REFERENCES  . . 1KN2 KN3 K N2 1 KN3 KN4 . . KN2 . .. K j t j j Kt j j j . . K j t j j . (1) (1). so that A t. l , where the k and l are constants, defining y. . K2N2M4 tN2 . K2N2M4 tN3 . . . . . . . K2N2M4 t2N2M4 .   . Cousins. J.M., “Special Computational problems associated with axisymmetric flow in Turbomachines. Ph.D Thesis, CNAA 1976 Curle. N., Davies. H.J., “Modern Fluid Dynamics, van Nostrand Reihhold Company, 1971, chapter 1.” Klier. M., “Aerodynamic Design of Annular Ducts”. PhD thesis 1990, chapter 1. Pavlika. V., “Vector Field Methods and the Hydrodynamic Design of Annular Ducts”, chapter VI, University of North London 1995. Pavlika. V., “Vector Field Methods and the Hydrodynamic Design of Annular Ducts”, chapter VIII, University of North London 1995.. Fig. 1. The meridian plane.. b. (i ). ISBN: 978-988-19253-3-6 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online). (23). IMECS 2014.
(8) Proceedings of the International MultiConference of Engineers and Computer Scientists 2014 Vol II, IMECS 2014, March 12 - 14, 2014, Hong Kong. Fig. 4.The geometry and speed distribution (along the top Fig.2. The geometry and speed distribution (along the top boundary) produced given a Swirl velocity. 0.2 and a velocity at inlet given y by u x ( y ) y . u 0.5 y . boundary) produced given a Swirl velocity u . 0.6 and y. an axial velocity at inlet given by u x ( y ) y .. Fig. 3. The geometry and speed distribution (along the top boundary) produced given a Swirl velocity. 0.6 and an axial velocity at inlet given by y u x ( y ) y . u 0.3 y . ISBN: 978-988-19253-3-6 ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online). IMECS 2014.