2.1. Problem 7: Shear Stress Determination

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.

In this section, we are introduced to Problem 7, which focuses on calculating the shear stress in turbulent flow through a pipe. We are provided with specific data regarding the velocities of water at two points in the pipe: at the center (4 m/s) and at a point 2 cm from the center (3.5 m/s). The problem specifically requires us to determine the shear stress at the wall's surface of the pipe.

Imagine a garden hose: when you turn on the water, it flows quickly in the center but may flow slower near the edges where it touches the hose. The different speeds represent the varying velocities of water in the pipe, and we can calculate the force (shear stress) acting at the hose's surface.

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.

Here, we convert the given diameter (10 cm) into SI units for easier calculations. The diameter is converted into meters (0.1 m), and we derive the radius as half of the diameter, which is 0.05 m. We also note that the maximum velocity (u max) is at the maximum radius of the pipe and equals 4 m/s, as determined from the problem.

Think of measuring ingredients for a recipe. Just as you might convert cups to liters for clarity, we are converting the pipe's measurements into meters to make our calculations easier and standard.

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.

In this chunk, we apply a logarithmic velocity defect equation to find the shear velocity (u*). By substituting the known values, we rearrange the terms to isolate u*. We find that u* equals 0.392 m/s, which is critical for determining the shear stress.

Imagine you are adjusting the speed on a treadmill. Just like trying to find the right settings when you reduce or increase speed, we are tweaking our variables to get the right velocity we need for further calculations.

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, tau not is coming out to be 153.6 Newton per meter square.

At this point, we use the relationship between shear velocity and shear stress. We square the previously determined shear velocity and multiply it by the fluid's density (1000 kg/m³ for water). This calculation yields a shear stress (tau_not) of approximately 153.6 N/m² at the pipe wall.

Think of pressing your hand flat against a surface. The harder you push (analogous to shear stress), the more resistance you feel. In our calculations, the shear stress indicates how strongly the water is pushing against the wall of the pipe.

So, going back again to the slide, so, what we got was approximately 153 Newton per meter square the sheer stress at the wall.

In summary, after performing the necessary calculations, we conclude that the shear stress at the wall of the pipe is approximately 153 N/m². Thus, our computations successfully followed from the initial problem statement through various steps to arrive at the required result.

Imagine measuring how hard you're pressing against a wall; by understanding this force (shear stress), we can infer how fluid flow interacts with surfaces, much like how we experience friction when sliding our hand against a rough surface.

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