$$P_v=\frac{1}{2}\rho V^2$$

where:

- \(P_v\) is the velocity pressure in Pa

- \(\rho\) is the density of the air in kg/m³

- \(V\) is the velocity of the air in m/s

We can rearrange this equation to solve for the velocity:

$$V=\sqrt{\frac{2P_v}{\rho}}$$

Substituting the given values, we get:

$$V=\sqrt{\frac{2(0.20\text{ in w.g.})(47.88\text{ Pa/in w.g.})}{1.225\text{ kg/m}^3}}$$

$$V= 5.67\text{ m/s}$$

Therefore, the air with a velocity pressure of 0.20 in w.g. moves through the square duct at a velocity of 5.67 m/s.