So, if we use equation 14 in equation 15, what was equation 14? l m was kappa into y, this was what we said in equation number 14. So, we can simply write and this is equation number 16, very simple. So, for small values of y it can be assumed. So, if the y is very small we can assume that tau is equal to tau not, where tau not is the shear stress at the pipe wall and can be assumed to be a constant. So, at the wall the shear stress is assumed to be constant and equal to tau not.

And therefore, what we can say, if we substitute tau is equal to tau not in equation 16, we can obtain or du / dy, this quantity actually tau not can be written as rho. So, but the catch here is, what is the catch? We have considered small value of y. So, du / dy can be written as, 1 / kappa y and under root tau / rho is rho u *. So, it becomes and this u * under root tau not / rho is the sheer velocity and this has the dimension of velocity.

Then using the boundary conditions, what are the boundary conditions? So, u at y is equal to R, where R is the radius of the pipe. We will get, u is equal to u max. That is what we have seen at the center line of the pipe the velocity is going to be the maximum. So, if we use this boundary condition u at y is equal to R is u max, we can get u is equal to, you know, we put u max here, y will be R and therefore, we can obtain C.

So, this is ln. So, we can put it in form of log. This is simple manipulation, we can get u max minus u / u * equals to 2.5 ln R / y. And, so, this is ln. This is just simple manipulation, we can get u max minus u / u * is equal to 5.75 log to the base 10 R / y. u max minus u is called the velocity defect or velocity defect law, this is velocity defect law. This is just simple, you know, manipulation of these terms here.

So, now we are going to solve one of the problems, problem number 7. And what it says is, the velocity of water. So, what we have learned in this particular lecture is about the turbulent flow and this problem 7 will help you in solving any problem that is based on this particular concept. So, it says the velocities of water through a pipe of diameter 10 centimeter are 4 meters per second and 3.5 meters per second at the center of the pipe and 2 centimeters from the pipe center, respectively. Considering turbulent flow in pipe, determine the sheer stress at the wall. So, we need to determine tau not.

As always what we do we solve, we write given, diameter is given as 10 centimeter, try to always write down in SI units. So, we write 0.1 meter. So, diameter is 10. So, radius is going to be 0.05 meter. u max is given, is given as 4 meters per second, that is, at y is equal to R. And this is also given, u at r is equal to 2 centimeter is given 3.5 meters per second, that is, y is equal to R - r. So, y is going to be 5 - 2 is equal to 3 centimeter.

So, now u max we are using the minus u / u * was 5.75 log R / y. So, substituting the values here, this from here, this equation, 4 - 3.5 divided by u * is equal to 5.75 log base 10 5 / 3. This will give us, u * as 0.392 meters per second. We also know, u is under root tau not / rho or tau not is rho u whole square. Therefore, tau not rho is 1000 and u * we already got, 0.392 whole square.

So, now, the turbulent velocity profile is much fuller compared to the parabolic profile of laminar flow case. Actually this is the flow, this is the true picture, this is a laminar flow that we have seen before. But below is, this is the V average and the velocity fluctuates or deviates from these depending upon the flow condition.

So, this is the V average line. There are several other layers, viscous sublayer, buffer layer, overlap layer and turbulent layer. So, as I told you in the last slide, there are different layers, different layers in turbulent flow.

Now, when it comes to these beds and these regimes, some of the important terms that are there is hydro dynamically rough and smooth boundaries. So, this is the, if you see, there is a term called k. Here, if in here, so, k here is the mean height of the surface irregularities. We talked in the beginning that the turbulence could occur due to the presence of irregularities on the surface.

But when the surface irregularities are much smaller than delta dash, the height of the viscous sublayer, the eddies are unable to reach the surface irregularities when the roughness height is much less. Therefore, we define that boundary as smooth boundary. So, smooth boundary are the one, where the thickness of the viscous sublayer is much larger than the surface irregularities.

When k is much larger than the delta dash, that is, the thickness of viscous sub layer, the irregularities are above the laminar sublayer leading to the interaction of eddies with the surface irregularities and therefore, these are called rough boundaries. From Nikuradse's roughness, k / delta dash if it is less than 0.25. So, these values which we are going to talk about, has been derived from experiments by Nikuradse.

In terms of roughness Reynolds number, so actually, there is something called roughness Reynolds number that is dependent upon k the height of the irregularities. So, in terms of roughness Reynolds number, if this Reynolds number is less than the 4, the boundaries is smooth, if it is more than the 100 then the boundary is rough and if it lies between 4 and 100 the boundaries is transitional.

Now, we will solve one problem about this particular concept. So, the question is, a pipeline carrying water has average height of irregularities projecting from the surface of the boundary of the pipe as 0.15 millimeter. What type of boundary it is? We have to estimate the rough or smooth or transitional boundary. The shear stress at the pipe wall is 4.9 Newton per meter square and the kinematic viscosity is 0.01 Stokes.

So, best is to calculate the roughness Reynolds number Re and that is given as, u k / nu. So, Re is, u is 0.07, k is 0.15 into 10 to the power - 3 and nu is 0.01 into 10 to the power - 4 and that comes to be 10.5. So, as Re* lies between 4 and 400, this implies that the boundary is transitional.