Hydrodynamics

Añadido: 2026-05-14

[00:00:01]Good morning class.

[00:00:02]This is Dr. Tarannum Siddiqui.

[00:00:05]Today I am continuing with chapter number one, kinematics, equation of continuity.

[00:00:11]And we shall discuss some numerical problems today.

[00:00:16]So, taking problem number one, show that the motion Q bar equal to K squared times X J cap plus Y times I cap in bracket and in denominator we have X squared plus Y squared is possible for an incompressible fluid.

[00:00:41]If so, find the equation of a streamline.

[00:00:44]Also, test whether the motion is of potential kind or that is irrotational.

[00:00:53]And so, find the velocity potential also.

[00:00:58]So, starting with uh we have given uh the value of Q bar which can be put in the form of components of velocity along the three axis.

[00:01:11]That is in the form of U, V, and W.

[00:01:14]Um so, comparing uh it with the with its component form.

[00:01:23]So, taking out the components firstly, we have U equal to minus K squared times Y upon X squared plus Y squared.

[00:01:32]And small V is equal to K squared times X upon X squared plus Y squared.

[00:01:39]And W equal to zero.

[00:01:43]Now, uh we have to first show that uh whether the motion is possible or not.

[00:01:50]For this, we have to check the equation of continuity whether this equation is satisfied or not.

[00:01:56]So, taking equation of continuity, which is del U upon del X plus del V upon del Y plus del W upon del Z and sum of all these three partial derivatives must be equal to zero.

[00:02:16]So, starting with substituting the value of U, V, and W in this equation on the left-hand side and simplification um gives us a zero after subbing these three values.

[00:02:34]So, we observe that the equation of continuity here is satisfied for incompressible fluid so that the fluid motion is possible.

[00:02:45]Now, coming to equation of a straight line streamlines.

[00:02:49]We know that the equation of a streamlines are given by dx upon u equal to dy upon v equal to dz upon w.

[00:02:58]So, putting the value of u, v, and w here uh we have uh equation of the streamline. Now, uh we shall solve these equations by taking uh two ratios together. So, firstly considering ratio number one and two we have x dx plus y dy equal to zero or on integration we get x squared plus y squared equal to some constant.

[00:03:29]So, we are taking here the constant a squared.

[00:03:33]So, this equation uh represents a family of circles.

[00:03:40]So, this is the equation of the streamline which is the family of a circle.

[00:03:47]And with the third ratio, we have dz equal to zero which on integration give give us z equal to constant.

[00:03:57]So, this also represent a circle.

[00:04:02]So, these are the equations of the streamlines.

[00:04:06]Now, coming to uh check whether the uh motion is of the potential kind or not.

[00:04:15]For this, we have to check uh what is the value of curl Q.

[00:04:20]So, curl Q is equivalent to del del cross Q bar.

[00:04:25]So, on simplification, we have curl Q equal to zero.

[00:04:31]So, that fluid motion is of potential kind. It is proved. Now, we want to obtain the value of potential, which is uh denoted by phi.

[00:04:43]So, we know that velocity potential Q bar is equivalent to minus del phi, where phi is a potential function.

[00:04:53]Now, Q bar has the components U V W.

[00:04:57]And so, uh are the components of del phi are del phi upon del x, del phi upon del y, and del phi upon del z with minus sign. So, we can compare both sides to get the equation in the form of U equal to del phi upon del x, V {comma} W equal to del phi upon del y minus {comma} W equal to minus del phi upon del z.

[00:05:28]So, we can uh simplify any of these equations to get the potential function. So, taking the first equation, del phi upon del x equal to minus U, and integrating it partially with respect to x to get phi equal to k squared tan inverse X upon Y plus some constant of integration.

[00:05:53]Here the constant of integration will be a function of uh Y and Z.

[00:05:58]This is the required velocity potential function.

[00:06:02]Now taking another example.

[00:06:08]Show that phi equal to X {comma} T {dot} Y {comma} T represent velocity potential of an incompressible fluid in 2D.

[00:06:20]Show that the streamline at time T are given by equation X minus T whole squared minus Y minus T whole squared equal to constant.

[00:06:33]So here we have given the value of phi.

[00:06:39]Now for velocity of the fluid in two-dimensional uh region, U is equivalent to minus double phi upon double X and V is equal to minus double phi upon double Y.

[00:06:55]Now differentiating phi partially with respect to X and Y and putting here, we have U equal to minus Y upon uh minus Y in bracket Y minus T.

[00:07:10]V equal to minus and in bracket X minus T.

[00:07:14]We are assigning equation number one.

[00:07:17]Now uh we consider double U upon double X plus double V upon double Y.

[00:07:24]So we are obtaining these partial derivatives and by adding them, we have a zero.

[00:07:31]Hence the given velocity potential is possible for in incompressible fluid.

[00:07:38]Now obtaining the streamline equation, so we know that the equation of the streamlines are given by dx upon u equal to dy upon v if we have two dimensional only.

[00:07:54]So, putting the value of u and v here.

[00:07:59]Now, we are using here variable separable method.

[00:08:03]So, we have x minus t dx y minus t then dy.

[00:08:10]On integration, we get x minus t whole square equal to y minus t whole square equal to some constant.

[00:08:19]Now, shifting the term y minus t whole square upon two in the left and taking two from left and merging it into the constant we have the final form of result.

[00:08:35]It gives the equation of the streamline uh which are required.

[00:08:40]So, these are the some numeric problem.

[00:08:45]The students, continue practicing the solved example as well as the unsolved questions from exercise uh which are covered till now.

[00:09:00]Uh and that is all for today. In the next class, we shall discuss some more numeric problem.

[00:09:07]Thank you.