Certainly! I will use the \displaystyle command in all future mathematical expressions.

Pressure Drop Calculation:

Given:

• Duct area: $A_d = 1.4 \times 1.4 = 1.96 \, \text{m}^2$
• Flow speed of air: $v = 10.5 \, \text{m/s}$
• Wire diameter: $D = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m}$
• Screen pitch: $P = 1.7 \, \text{mm} = 1.7 \times 10^{-3} \, \text{m}$
• Density of air: $\rho = 1.225 \, \text{kg/m}^3$

The open area of the screen $A_s$ is calculated as:

$\displaystyle A_s = \left(\frac{P - D}{P}\right)^2 = \left(\frac{1.7 \times 10^{-3} - 0.4 \times 10^{-3}}{1.7 \times 10^{-3}}\right)^2 = \left(\frac{1.3 \times 10^{-3}}{1.7 \times 10^{-3}}\right)^2 = \left(0.7647\right)^2 \approx 0.5856$

The pressure drop $\Delta P$ across the screen is calculated using:

$\displaystyle \Delta P = C_d \cdot \frac{1}{2} \cdot \rho \cdot v^2 \cdot \left(\frac{A_s}{A_d}\right)^2$

Assuming a drag coefficient $C_d$ of 1.2:

$\displaystyle \Delta P = 1.2 \cdot \frac{1}{2} \cdot 1.225 \cdot (10.5)^2 \cdot (0.5856)^2$

$\displaystyle \Delta P = 1.2 \cdot 0.6125 \cdot 110.25 \cdot 0.343$

$\displaystyle \Delta P \approx 28.28 \, \text{Pa}$

Thus, the total pressure drop across the aluminum screen is approximately:

$\boxed{\displaystyle 28.3 \, \text{Pa}}$

If you have any further questions or need additional details, feel free to ask.

Follow-up Questions:

1. How does changing the wire diameter $D$ affect the pressure drop $\Delta P$?
2. How would varying the air density $\rho$ influence the pressure drop?
3. What is the relationship between the drag coefficient $C_d$ and different screen materials?
4. How would increasing the pitch $P$ impact the pressure drop?
5. What simplifications are typically made when using empirical equations for pressure drop across screens?

Tip:

When dealing with flow and pressure calculations, always ensure that the units are consistent across all parameters to avoid errors.