Given: a=500mm=0.5m
b=350mm=0.35m
Q=75/60=1.25 $m^3/sec$
$ρ=1.15 kg/m^3$
Calculating equivalent diameter of circular duct
1) The quantity of air carried in both the cases is same
D=1.265$\frac{ [(a^3 b^3)}{(a+b)]}^{(1/5)}$
D=0.46 m=459.23 mm
2) The velocity of air in both the cases is same
D=$\frac{2ab}{(a+b)}$
D=0.4117 m=411.76 mm
Also, pressure loss $P_f=\frac{f.l.v ̅^2.ρ}{2m}$…(1)
Where $v ̅=\frac{Q}{A}=\frac{1.25}{(0.5×0.35)}=7.14 m/sec, f=0.01, l=100, ρ=1.15 kg/m^3$
Hydraulic mean depth, m =$\frac{(cross-section area of flow)}{(wetted perimeter)}$
=$\frac{(a×b)}{2(a+b)}$ (for a rectangular duct)
=0.1029
Substituting all values in equation (1) we get, $P_f=284.76 Pa$
Dividing by 9.81, $P_f=29.03 mm of W.C. {Since 1\frac{N}{m}^2 =\frac{1}{9.81} mm of W.C.}$
[Tip: In duct designing it is a general practice to write pressures in mm of W.C.]