Answer
$V_{max}=16.97\frac{m}{s}$
Work Step by Step
Knowing that: $W=Q\Delta P$
The maximun flow rate will be:
$Q_{max}=\frac{W}{\Delta P}=\frac{60W}{50Pa}=1.2\frac{m^3}{s}$
The area of the duct is:
$A=\frac{\pi*D^2}{4}=\frac{\pi*(30cm*\frac{1m}{100cm})^2}{4}=0.0707m^2$
Knowing that: $Q=VA$
Then the highest possible average flow velocity in the duct will be:
$V_{max}=\frac{Q_{max}}{A}=\frac{1.2\frac{m^3}{s}}{0.0707m^2}=16.97\frac{m}{s}$
