L= 12 cm=0.12 m
D=0.27 m
Vol flow rate= Q= 1.3 m3/sec
Area of flow= A = ${\pi}{4} d^2=0.0573 m^2$
Q=$A * v$
Therefore v=22.7 m/sec
$p_f=\frac{fLρ_a V^2}{2m}
=$\frac{4fLρ_a V^2)}{2D}$…….(m=$\frac{D}{4}$ for circular duct)
assuming $ρ_a=1.2 \frac{kg}{m3}$ for standard air
Pressure drop=2.748 Pa
i) Let side be a=0.27 m
Area of flow= A = $a^2=0.0729m^2$
$Q=A *v$
Therefore v=17.83 m/sec
$p_f=\frac{(fLρ_a V^2)}{2m}=\frac{(4fLρ_a V^2)}{2a}$
..$(m=\frac{ab}{2(a+b)}$ for rectangular duct=a/4 for square duct)
assuming $ρ_a=1.2 \frac{kg}{m3}$ for standard air
Pressure drop=1.696 Pa