Given the components of velocity Ans:  y 3    x 2 2 3 2x u    x  y , v   x  3    2   y y   The condition for Irrorational flow  v   u        x   y  v   2 u     x3 y 3  2 LHS x y  y and   y  x x    2x    x RHS 3   y y   i.e. LHS = (y2 x2) and RHS = (y2 x2) – – Hence the flow is Irrorational The corresponding stream function be obtained by using following relationship ‘’can 3 Eq.1   x      2  v x y y 2 x      y3 2x   x2  y Eq.2  u   y   Integrating Eq.1 with respect to ‘x’  (  y C f Eq.3   x 2     x y x 2 2 12 1 Differentiating Eq.3 with respect to ‘y’ 2   x  y  x  C 2 1 y y C y 3   1 3 y y 4 Integratin g, C   1 12 (assu min  0) g C C y 4 ; C   1
Ans: Given the components of velocity  y 3 2   2 x 3  u    3 2x  x  y ,  v   x  y   2 y    The condition for Irrorational flow  v   u        x   y  v   2 x3  u   y 3  LHS   x  y x x   2 y   and RHS    y y  3 2x  x  y   i.e. LHS = (y2 – x2) and RHS = (y2 – x2) Hence the flow is Irrorational The corresponding stream function ‘’can be obtained by using following relationship   x  v   x  y 2  2 y  3 Eq.1 x     y3  u     2x  x2  y Eq.2 y   Integrating Eq.1 with respect to ‘x’   x y 2  2  x  y  x  C  f ( Eq.3 2 12 1 Differentiating Eq.3 with respect to ‘y’   x 2  y  2 x  C 1 y y C y 3 1   y 3 y 4 Integratin g, C 1   12 y 4 C 1   C ; (assu min g C  0)
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