Subject:- Refrigeration and Air Conditioning

Topic:- Duct Design, Controls & Applications

Difficulty:- Medium

i. When duct is circular of diameter 270 mm

ii. When the duct is 270 mm square section

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