Complex potential describing an inviscid flowFinding Stagnation PointsStream function, velocity potential/field from complex potentialVelocity potential of flow under rigid diskComplex potential questionFinding stagnation points and stream functionFinding pressure using Bernoulli's TheoremSolutions to Euler's Equations and Potential FlowTo find the velocity potential2-D steady incompressible inviscid flow ProblemInviscid Taylor Couette Flow

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Complex potential describing an inviscid flow


Finding Stagnation PointsStream function, velocity potential/field from complex potentialVelocity potential of flow under rigid diskComplex potential questionFinding stagnation points and stream functionFinding pressure using Bernoulli's TheoremSolutions to Euler's Equations and Potential FlowTo find the velocity potential2-D steady incompressible inviscid flow ProblemInviscid Taylor Couette Flow













0












$begingroup$


I'm working through a homework sheet for a Fluid Mechanics module. The question is given:




Consider the flow described by the complex potential $$w=4z+frac8z.$$



  1. Determine $psi$, $phi$, $u$ and $v$ in plane polar coordinates $(r,theta)$.

  2. Determine the location of the stagnation points.

  3. Show that this complex potential describes an inviscid flow around a solid object, What is the shape of the object?

  4. Sketch the streamlines for the flow outside the object.



My working out so far for the question is:



  1. Let $z=re^itheta$, and therefore
    beginalign
    w&=4re^itheta+frac8re^-itheta \
    &=4r(cos(theta)+isin(theta))+frac8r(cos(theta)-isin(theta)) \
    &=(4r+frac8r)cos(theta)+(4r-frac8r)isin(theta).
    endalign

    Using the Cauchy-Riemann equations, write $w=phi+ipsi$, $phi=(4r+frac8r)cos(theta)$ and $psi=(4r-frac8r)sin(theta)$. Also, we have that $$u=fracpartialphipartial rimplies u=(4-8r^-2)cos(theta)$$ and $$v=frac1rfracpartialphipartialtheta implies v=-(r+8r^-2)sin(theta).$$

  2. Stagnation points are given by $u=0$ and $v=0$. So, from $u=0$, we have that $r^2=2$ or $cos(theta)=0$. Similarly from $v=0$, we have that $r^2=-2$ and $sin(theta)=0$. Since $rinmathbbR$ (as it is a distance), we have that $r^2=2$ (from $u=0$) and $sin(theta))=0$ (from $v=0$). Therefore, the stagnation points occur at $(r,theta)=(sqrt2,0),(sqrt2,pi)$.

From here (ie 3 onwards), I fall down. I think that I should use that $textbfucdottextbfn=0$, but I'm not too sure how to use this information. Should I be using Bernoulli's theorem for pressure? Is there some assumption I am missing?



Any help would be much appreciated!










share|cite|improve this question











$endgroup$
















    0












    $begingroup$


    I'm working through a homework sheet for a Fluid Mechanics module. The question is given:




    Consider the flow described by the complex potential $$w=4z+frac8z.$$



    1. Determine $psi$, $phi$, $u$ and $v$ in plane polar coordinates $(r,theta)$.

    2. Determine the location of the stagnation points.

    3. Show that this complex potential describes an inviscid flow around a solid object, What is the shape of the object?

    4. Sketch the streamlines for the flow outside the object.



    My working out so far for the question is:



    1. Let $z=re^itheta$, and therefore
      beginalign
      w&=4re^itheta+frac8re^-itheta \
      &=4r(cos(theta)+isin(theta))+frac8r(cos(theta)-isin(theta)) \
      &=(4r+frac8r)cos(theta)+(4r-frac8r)isin(theta).
      endalign

      Using the Cauchy-Riemann equations, write $w=phi+ipsi$, $phi=(4r+frac8r)cos(theta)$ and $psi=(4r-frac8r)sin(theta)$. Also, we have that $$u=fracpartialphipartial rimplies u=(4-8r^-2)cos(theta)$$ and $$v=frac1rfracpartialphipartialtheta implies v=-(r+8r^-2)sin(theta).$$

    2. Stagnation points are given by $u=0$ and $v=0$. So, from $u=0$, we have that $r^2=2$ or $cos(theta)=0$. Similarly from $v=0$, we have that $r^2=-2$ and $sin(theta)=0$. Since $rinmathbbR$ (as it is a distance), we have that $r^2=2$ (from $u=0$) and $sin(theta))=0$ (from $v=0$). Therefore, the stagnation points occur at $(r,theta)=(sqrt2,0),(sqrt2,pi)$.

    From here (ie 3 onwards), I fall down. I think that I should use that $textbfucdottextbfn=0$, but I'm not too sure how to use this information. Should I be using Bernoulli's theorem for pressure? Is there some assumption I am missing?



    Any help would be much appreciated!










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      I'm working through a homework sheet for a Fluid Mechanics module. The question is given:




      Consider the flow described by the complex potential $$w=4z+frac8z.$$



      1. Determine $psi$, $phi$, $u$ and $v$ in plane polar coordinates $(r,theta)$.

      2. Determine the location of the stagnation points.

      3. Show that this complex potential describes an inviscid flow around a solid object, What is the shape of the object?

      4. Sketch the streamlines for the flow outside the object.



      My working out so far for the question is:



      1. Let $z=re^itheta$, and therefore
        beginalign
        w&=4re^itheta+frac8re^-itheta \
        &=4r(cos(theta)+isin(theta))+frac8r(cos(theta)-isin(theta)) \
        &=(4r+frac8r)cos(theta)+(4r-frac8r)isin(theta).
        endalign

        Using the Cauchy-Riemann equations, write $w=phi+ipsi$, $phi=(4r+frac8r)cos(theta)$ and $psi=(4r-frac8r)sin(theta)$. Also, we have that $$u=fracpartialphipartial rimplies u=(4-8r^-2)cos(theta)$$ and $$v=frac1rfracpartialphipartialtheta implies v=-(r+8r^-2)sin(theta).$$

      2. Stagnation points are given by $u=0$ and $v=0$. So, from $u=0$, we have that $r^2=2$ or $cos(theta)=0$. Similarly from $v=0$, we have that $r^2=-2$ and $sin(theta)=0$. Since $rinmathbbR$ (as it is a distance), we have that $r^2=2$ (from $u=0$) and $sin(theta))=0$ (from $v=0$). Therefore, the stagnation points occur at $(r,theta)=(sqrt2,0),(sqrt2,pi)$.

      From here (ie 3 onwards), I fall down. I think that I should use that $textbfucdottextbfn=0$, but I'm not too sure how to use this information. Should I be using Bernoulli's theorem for pressure? Is there some assumption I am missing?



      Any help would be much appreciated!










      share|cite|improve this question











      $endgroup$




      I'm working through a homework sheet for a Fluid Mechanics module. The question is given:




      Consider the flow described by the complex potential $$w=4z+frac8z.$$



      1. Determine $psi$, $phi$, $u$ and $v$ in plane polar coordinates $(r,theta)$.

      2. Determine the location of the stagnation points.

      3. Show that this complex potential describes an inviscid flow around a solid object, What is the shape of the object?

      4. Sketch the streamlines for the flow outside the object.



      My working out so far for the question is:



      1. Let $z=re^itheta$, and therefore
        beginalign
        w&=4re^itheta+frac8re^-itheta \
        &=4r(cos(theta)+isin(theta))+frac8r(cos(theta)-isin(theta)) \
        &=(4r+frac8r)cos(theta)+(4r-frac8r)isin(theta).
        endalign

        Using the Cauchy-Riemann equations, write $w=phi+ipsi$, $phi=(4r+frac8r)cos(theta)$ and $psi=(4r-frac8r)sin(theta)$. Also, we have that $$u=fracpartialphipartial rimplies u=(4-8r^-2)cos(theta)$$ and $$v=frac1rfracpartialphipartialtheta implies v=-(r+8r^-2)sin(theta).$$

      2. Stagnation points are given by $u=0$ and $v=0$. So, from $u=0$, we have that $r^2=2$ or $cos(theta)=0$. Similarly from $v=0$, we have that $r^2=-2$ and $sin(theta)=0$. Since $rinmathbbR$ (as it is a distance), we have that $r^2=2$ (from $u=0$) and $sin(theta))=0$ (from $v=0$). Therefore, the stagnation points occur at $(r,theta)=(sqrt2,0),(sqrt2,pi)$.

      From here (ie 3 onwards), I fall down. I think that I should use that $textbfucdottextbfn=0$, but I'm not too sure how to use this information. Should I be using Bernoulli's theorem for pressure? Is there some assumption I am missing?



      Any help would be much appreciated!







      complex-analysis physics fluid-dynamics






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      edited Mar 24 at 0:48









      David M.

      2,188421




      2,188421










      asked Mar 20 at 17:52









      FatsFats

      468




      468




















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          To determine the shape of the obstacle, you just look at the streamlines (they're everywhere tangent to the flow). Since you have a potential flow, it holds that $Delta u = 0$, where $u$ is the (real) velocity. So, the term $muDelta u$ in the Navier-Stokes equation will vanish. Obviously, this doesn't itself imply $mu = 0$ so the fluid itsf may not be inviscid, but the flow should be.






          share|cite|improve this answer









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            $begingroup$

            To determine the shape of the obstacle, you just look at the streamlines (they're everywhere tangent to the flow). Since you have a potential flow, it holds that $Delta u = 0$, where $u$ is the (real) velocity. So, the term $muDelta u$ in the Navier-Stokes equation will vanish. Obviously, this doesn't itself imply $mu = 0$ so the fluid itsf may not be inviscid, but the flow should be.






            share|cite|improve this answer









            $endgroup$

















              0












              $begingroup$

              To determine the shape of the obstacle, you just look at the streamlines (they're everywhere tangent to the flow). Since you have a potential flow, it holds that $Delta u = 0$, where $u$ is the (real) velocity. So, the term $muDelta u$ in the Navier-Stokes equation will vanish. Obviously, this doesn't itself imply $mu = 0$ so the fluid itsf may not be inviscid, but the flow should be.






              share|cite|improve this answer









              $endgroup$















                0












                0








                0





                $begingroup$

                To determine the shape of the obstacle, you just look at the streamlines (they're everywhere tangent to the flow). Since you have a potential flow, it holds that $Delta u = 0$, where $u$ is the (real) velocity. So, the term $muDelta u$ in the Navier-Stokes equation will vanish. Obviously, this doesn't itself imply $mu = 0$ so the fluid itsf may not be inviscid, but the flow should be.






                share|cite|improve this answer









                $endgroup$



                To determine the shape of the obstacle, you just look at the streamlines (they're everywhere tangent to the flow). Since you have a potential flow, it holds that $Delta u = 0$, where $u$ is the (real) velocity. So, the term $muDelta u$ in the Navier-Stokes equation will vanish. Obviously, this doesn't itself imply $mu = 0$ so the fluid itsf may not be inviscid, but the flow should be.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 20 at 19:04









                Gary MoonGary Moon

                92127




                92127



























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