1.2. Problem on Laminar Flow between Two Plates

So, the question here is, water is flowing between 2 large parallel plates which are 2 millimeters apart. Determine the maximum velocity, the pressure drop per unit length, that is, dp dx and the sheer stress at the wall of the plate if the average velocity is 0.4 meters per second. And viscosity of water is given as 0.01 poise.

In this problem, we are analyzing the flow of water between two parallel plates, commonly used in fluid dynamics studies. The distance between the plates is given as 2 millimeters, which is equivalent to 0.002 meters. We are asked to calculate the maximum velocity of the flow, the pressure drop over a unit length of the plates (dp/dx), and the shear stress at the wall of the plates. The average velocity of the flowing water is provided as 0.4 meters per second, and the viscosity of water is noted in poise, which is a unit of dynamic viscosity. The viscosity will need to be converted to Pascal-seconds for further calculations.

Think of this scenario as water flowing between two flat surfaces, similar to how toothpaste flows out of a tube when squeezed from both sides. The distance between the tube walls and how fast the toothpaste can flow out can be equated to the dynamics of water between these plates.

Also u max is equal to 1.5 V average. This implies that u max is equal to 1.5 and this is 0.4 from here and that comes out to be 0.6 meters per second.

The maximum velocity (u max) in a laminar flow between two plates can be calculated using the relationship u max = 1.5 * V average. Given the average velocity (V average) is 0.4 meters per second, we multiply this value by 1.5 to find u max, which results in a maximum velocity of 0.6 meters per second.

Imagine a crowded hallway where the average speed of people walking through is 4 meters per minute. Now, in some parts of that hallway, people can move faster, perhaps reaching a speed of 6 meters per minute. This maximum speed corresponds to the concept of u max in our fluid flow scenario.

Substituting the value of u max, that is, 0.8 is equal to minus 1 / 8 into 0.001 and dp dx is what we need to determine and t is 2 millimeters. So, 2 into 10 to the power minus 3 whole square and this way we can obtain dp dx is equal to minus 1200 Newton per meter squared per meter.

To find the pressure drop per unit length (dp/dx), we utilize the equation linking maximum velocity to the pressure gradient. Given that u max equals to 0.6 and using the known values for viscosity and the distance between the plates, we can rearrange the equation to calculate dp/dx, which results in a value of -1200 Newton per square meter per meter (negative indicating a drop in pressure).

This can be visualized as water flowing through a faucet with a filter. The greater the pressure difference created by the faucet, the harder it pushes the water through the filter, much like the calculated pressure drop affects the water flow between two plates.

So, tau not is equal to mu du dy at y is equal to 0. So, tau not is mu du, we know u, so, we can differentiate that and obtain minus 1 / 2 mu dp dx into t.

Shear stress (tau) at the wall of the plates can be calculated using the formula tau = mu * (du/dy). Here, we need to differentiate the velocity profile and evaluate it at the point y=0 (the wall). After substituting the values into the equation alongside the pressure gradient we previously found, this results in a shear stress of 1.2 Newton per square meter at the wall of the plates.

Consider how rubbing your hand against a rough surface creates friction. Similarly, shear stress represents the frictional force between the fluid (water) and the walls of the plates, affecting how smoothly the water flows.

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