Determining a component of velocity vector of a fluid flow explicitlyIf (u(y),0,0) are the velocity components of an incompressible Newtonian fluid flow due to a pressure gradient in the X-direction, then u(y) is a(a) linear function of y.(b) quadratic function of y.(c) cubic function of y.(d) constant.I started with Euler dynamical equation as follows: D q → D t = F → − 1 ρ ( ∇ p )where q → = ( u ( y ) , 0 , 0 ) and F → is the external force, ∇ p being pressure gradient. Now as per question we have along X-direction that ∂ u ( y ) ∂ t + u ( y ) ∂ u ( y ) ∂ x + 0. ∂ u ( y ) ∂ y + 0. ∂ u ( y ) ∂ z = 0 − 1 ρ ∂ p ∂ x ⟹ ∂ u ( y ) ∂ t = − 1 ρ ∂ p ∂ x ⟹ u ( y ) = − ∫ ( 1 ρ ∂ p ∂ x ) d t + CC being arbitrary constant. But from here I am unable to conclude whether this is linear or quadratic or cubic or constant. Is my thinking correct here?

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Determining a component of velocity vector of a fluid flow explicitlyIf (u(y),0,0) are the velocity components of an incompressible Newtonian fluid flow due to a pressure gradient in the X-direction, then u(y) is a(a) linear function of y.(b) quadratic function of y.(c) cubic function of y.(d) constant.I started with Euler dynamical equation as follows:            D                        q          →                            D      t        =            F      →        −      1    ρ    (  ∇  p  )where             q      →        =  (  u  (  y  )  ,  0  ,  0  ) and             F      →       is the external force,   ∇  p being pressure gradient. Now as per question we have along X-direction that            ∂      u      (      y      )              ∂      t        +  u  (  y  )            ∂      u      (      y      )              ∂      x        +  0.            ∂      u      (      y      )              ∂      y        +  0.            ∂      u      (      y      )              ∂      z        =  0  −      1    ρ              ∂      p              ∂      x            ⟹              ∂      u      (      y      )              ∂      t        =  −      1    ρ              ∂      p              ∂      x            ⟹    u  (  y  )  =  −  ∫            (            1    ρ              ∂      p              ∂      x                  )        d  t  +  CC being arbitrary constant. But from here I am unable to conclude whether this is linear or quadratic or cubic or constant. Is my thinking correct here?

Lena Gomez · Accepted Answer

Step 1Neglecting body force terms other than gravity which can be absorbed into the pressure field, the x-component of velocity satisfies                    (*)                                          ∂            u                                ∂            t                          +        u                              ∂            u                                ∂            x                          +        v                              ∂            u                                ∂            y                          +        w                              ∂            u                                ∂            z                          =        −                  1          ρ                                      ∂            p                                ∂            x                          +                  μ          ρ                          ∇          2                u        ,            where   μ is the viscosity.The flow is steady, fully developed (u depends only on y), and unidirectional. Consequently the partial derivatives of u with respect to t, x, and z vanish and   v  =  w  =  0. Thus, (*) reduces to                    (**)                                                        d              2                        u                                d                          y              2                                      =                  1          μ                                      ∂            p                                ∂            x                          =                  G          μ                    The pressure gradient G is independent of y. This follows from the y-component of the Navier-Stokes equations which reduces to            ∂      p              ∂      y        =  0Step 2Integrating both sides of (**) twice with respect to y we get the quadratic function  u  (  y  )  =      G          2      μ            y    2    +      C    1    y  +      C    2    ,where       C    1   and       C    2   are integration constants.

If (u(y),0,0) are the velocity components of an incompressible Newtonian fluid flow due to a pressure gradient in the X-direction, then u(y) is a. (a) linear function of y. (b) quadratic function of y. (c) cubic function of y. (d) constant.

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