Q 8The velocity potential  for a two dimensional flow is by (x2 y2) + 3xy. – Calculate: i) the stream function  and ii) the flow rate passing between the stream lines through (1,1) and (1, 2). (July 2013, 2014) Ans: = (x2 y2) + 3xy  – To determine the function  y d   x dx    v  d dx Eq.(2) As per definition of velocity potential () and stream function (); and       u v    y x x y     and (2 3 u    x  y)    (2  3x)     y v y  x  x y Substituting the value of u and v in terms of x and y in equation 2, we obtain  v   (2  3x)dx  (2  3 d dx udy y x y)dy 2 2   3xdx   3 d y x Eq.3 Integrating the equation-3 (partially w.r.t the and w.r.t the ‘x’ ‘dx-term’ ‘y’ ‘dy-term’)  3   3  3 2 2 2    xy 2 x2    xy x  =   4  y2 xy  y   2   2  2 The flow rate passing between the stream lines through (1, 1) and (1, 2). The equation of stream function is given by The value of Point streamline at (1, 1) is obtained by substituting x = 1, y = 1 x2  4 3 4 3 (12 ) 7Units   xy   y2  11 12  2 2 (1,1) The value of Point streamline at (1, 2) is obtained by substituting x = 1, y = 2 x2  4 3 4 2 3 (12 22 ) 15.5Units   xy   y2  1    2 2 (1,2) passing between the stream through (1, 1) and (1, 2) The flow rate (1,2) - (1,1) = (15.5-7)
Q 8The velocity potential  for a two dimensional flow is by (x2 – y2) + 3xy. Calculate: i) the stream function  and ii) the flow rate passing between the stream lines through (1,1) and (1, 2). (July 2013, 2014) Ans:  = (x2 – y2) + 3xy To determine the  function d   x dx   y d  v dx  Eq.(2) As per definition of velocity potential () and stream function ();     u and      v x y  y x    u    (2 x  3 y)   and  (2 y  3x)     v x y y  x  Substituting the value of u and v in terms of x and y in equation 2, we obtain d  v dx  udy  (2 y  3x)dx  (2 x  3 y)dy d  2 y  3xdx  2 x  3 Eq.3 Integrating the equation-3 (partially w.r.t ‘x’ the ‘dx-term’ and w.r.t ‘y’ the ‘dy-term’)  3   3  3 2 2    2 xy  x2    2 xy  y2  = 4 xy  x  y   2   2  2 The flow rate passing between the stream lines through (1, 1) and (1, 2). The equation of stream function is given by The value of Point streamline at (1, 1) is obtained by substituting x = 1, y = 1   4 xy  3 x2  y2   4 11 3 (12 12 )  7Units (1,1) 2 2 The value of Point streamline at (1, 2) is obtained by substituting x = 1, y = 2   4 xy  3 x2  y2   4 1 2  3 (12  22 )  15.5Units (1,2) 2 2 The flow rate passing between the stream through (1, 1) and (1, 2) (1,2) - (1,1) = (15.5-7)
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