2.3. Drag Force Calculation

So, we have been given the thickness of the boundary layer, we have been given the length of the plate, we have been given how much wide is it and what is the U it is given, the viscosity is given.

In this chunk, we are introduced to a specific problem involving calculating the drag force on a plate within a fluid. Here, several specific parameters are given, including the thickness of the boundary layer, the length of the plate, the width of the plate, the fluid's velocity (U), and the viscosity of the fluid. These parameters are essential for solving the drag force calculation, as they affect how the fluid interacts with the plate.

Imagine trying to push your hand through water. The thickness of the water layer around your hand (boundary layer), how long your hand is (length of the plate), how fast you are pushing it (velocity), and how thick the water is (viscosity) will all determine how hard it is to push your hand through the water.

When x is equal to 1.5 meter, Reynolds number at x is going to be ρ Ux / mu.

The Reynolds number is a dimensionless value that helps predict flow patterns in different fluid flow situations. In this segment, we calculate the Reynolds number specifically at a position (x = 1.5 meters) using the formula: Reynolds number = (density × velocity × length) / viscosity. After plugging in the given values, we find that the Reynolds number is 3 × 10^5, which indicates that the flow is laminar—as it is below the turbulent flow threshold.

Think of Reynolds number like a measure of how chaotic water flows around an object. A lower Reynolds number represents smooth, laminar flow, similar to how a calm river flows gently around rocks. As the number increases, the flow starts to get more turbulent, much like a stormy sea crashing against boulders.

For laminar boundary layer delta, we know, 4.64 x under root Re at x and after substituting this x and Re what we get is 0.013 meter.

In fluid dynamics, the thickness of the boundary layer (delta) for laminar flow can be estimated using a specific formula. In this case, it is calculated using the position (x) and the Reynolds number. By substituting the previously calculated Reynolds number into the equation, we find that the boundary layer thickness is 0.013 meters, indicating the region where the fluid velocity transitions from zero at the surface of the plate to the free stream velocity.

Imagine the boundary layer as a layer of syrup spreading on a pancake. The syrup sticks to the pancake (just like a fluid sticks to a surface), and as you go further away from the pancake's surface, the syrup gradually gets thinner until it remains just syrupy liquid (representing the free stream velocity).

For that we have already derived 3 / 2 mu U / Delta and after substituting in the values, you will get 0.0231 Newton per meter square. This is simply substituting in the values.

Here, we derive the equation for the wall shear stress (which contributes to drag) in laminar flow, represented as taou = (3/2) * (mu * U / Delta). Using the values obtained earlier, we calculate taou to be 0.0231 Newton per meter squared. This shear stress is critical for calculating the total drag force acting on the plate as it directly affects how much force is needed to move the plate through the fluid.

Think of the drag force like the tug a swimmer feels while moving through water. The viscosity of the water (like the mu in our formulas) plays a significant role in how much resistance (shear stress) the swimmer feels. If the swimmer is thickly bundled in a wetsuit, they may feel more drag compared to when they swim freely.

And similar, if we put in F D, the drag force that we have derived in the last thing, so it will come out to be 0.114 Newton. You should verify it actually.

In this final chunk, we use our previously calculated shear stress value to determine the drag force (FD) on one side of the plate. It turns out that FD equals 0.114 Newton. Since the plate has two sides exposed to fluid flow, we double this value to find the total drag force acting on the plate, resulting in 0.229 Newton. Understanding how to accurately calculate this total drag is vital, as it informs engineering decisions regarding material choice and design.

If you have a sign flapping in the wind, each side of the sign experiences resistance as the wind hits it. By calculating the drag on just one side first (like we did), and then recognizing that both sides feel similar effects, you can determine how much total force the wind is applying to the sign.

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