Question

Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section), and the power fit of your Trendline equation,calculate the drag coefficient. Solve for it first (see video) and then plug in the values.

Answer 1

Answer:

The  drag coefficient is  $D_z = 1.30512$  

Explanation:

From the question we are told that

     The density of air is  $\rho_a = 1.21 \ kg/m^3$

     The diameter of bottom part is  $d = 0.15 \ m$

The  power trend-line  equation is mathematically represented as

      $F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is  20 m/s

Then

      $F_{\alpha } = 0.9226 * 20^{0.5737}$

       $F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

      $D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where  

     $F_{\alpha }$ is the drag force

      $\rho$ is the density of the fluid

       $v$ is the flow velocity

       A is the area which mathematically evaluated as

       $A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

     $A = 3.142 * \frac{(0.15)^2}{4}$

     $A = 0.0176 \ m^2$

Then

   $D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

   $D_z = 1.30512$