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

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

Solving for $v$, we get:

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

Substituting the given values, we get:

$$v = \sqrt{\frac{2(0.20 \text{ in wg})(47.88 \text{ Pa/in wg})}{1.204 \text{ kg/m³}}} = 18.5 \text{ m/s}$$

Therefore, the air with a velocity pressure of 0.20 in wg moves through the square duct at a velocity of 18.5 m/s.