4.4.2. Calculation steps

Now, it is a good idea to solve one problem here, on this particular concept, and the solving the problem will give more understanding now. You know, the question is, determine the average height of roughness for a rough pipe of diameter 10 centimeter when the velocity at point 4 centimeter away from the wall is 40 percent more than the velocity at a point 1 centimeter from the wall.

In this chunk, we introduce a practical problem that illustrates the concepts of turbulent flow in pipes. The problem involves finding the average height of roughness in a specific pipe. It emphasizes the importance of relating classroom theory to real-world applications, especially in hydraulic engineering.

Imagine the roughness of a riverbed. Just as different materials and shapes of stones affect how water flows over them, similarly, the texture of a pipe influences how fluid moves through it. The problem here is akin to observing how water flows differently over a gravel path compared to a smooth concrete sidewalk.

So, diameter it says is 10 centimeters, when the velocity at a point 4 centimeters away is 40 percent more than the velocity at a point 1 centimeter away from the wall. So, what we are going to do? We are going to assume our white screen back again and start solving by writing down what are given.

Here, we specify the key parameters of the problem: the diameter of the pipe (10 cm) and the relationship between the velocities at 4 cm and 1 cm from the wall. This setup is crucial for understanding how to approach the problem systematically. By determining what values we have, we can utilize equations related to fluid dynamics to find the unknowns of the problem.

Think of it like measuring how fast a car moves when it’s close to a wall compared to when it’s further away. If two cars are driving by a wall and you know how fast one is going at 1 meter and that the other speeds up by 40% at 4 meters, you can analyze the differences a wall creates in movement.

Given is, D is equal to 10 centimeters or 10 into 10 to the power minus 2 meters or 0.1 meters. It is given, u at y is equal to 4 centimeters or 4 is equal to 1.4 times u at y is equal to 1 and the surface is rough. This is what we have already been told.

In this section, we express the diameter in standard units (meters) for precision in calculations. We also define the equations based on the relationship between velocities at different points. This process is essential for creating a mathematical model that reflects our earlier problem statement. Writing formulas helps bridge theory and calculation.

Consider this like setting up a chart where you put down the lengths of distances in meters instead of centimeters to avoid conversion issues when measuring the height of buildings. Similarly, properly setting up our known values ensures accuracy in our fluid dynamics calculations.

Therefore, log 1 by k is equal to 0.027, which implies, 1 by k is equal to 1.064 per centimeter, which implies, k is equal to 0.9399 centimeter.

This chunk reveals the steps taken to solve the equation for k, the roughness height of the pipe. By manipulating logarithmic equations, we find the specific roughness height that influences how fluid flows through the rough pipe. Understanding this part is imperative as it ties directly back to the earlier defined relationships in the problem.

Consider measuring the friction on a road's surface. Just like different roads have varying levels of grip (smooth vs. rough), pipes have roughness values that determine how smoothly liquid can pass through them. Here, we calculate the value, k, which will dictate how 'rough' the flow is.

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