For the system shown below which made from galvanized iron which Roughness of galvanized iron $=0\mathrm{.}00015m.$ Friction pressure drop $(\frac{p}{L})p\frac{a}{m}$ in main duct $AB$ must be calculate by equal friction method and size duct and air velocity in each branch with using the duct friction chart. Also calculate the air fan power required. Take the followings for air $.$Air density $=1\mathrm{.}2k\frac{g}{m^3},$ air dynamic Viscosity $=18\mathrm{.}178{10}^{-6}Pa.s.$ The pressure drop in the Air Handling Unit $(p_{\text{A}\text{i}\text{r}-\text{H}\text{a}\text{n}\text{d}\text{l}\text{i}\text{n}\text{g}})$ is $375Pa.$ Assume the velocity in the main duct $GH$ is $8\frac{m}{s}$

Fan Power $=Q_i[(\frac{P_{\text{A}\text{i}\text{r}-\text{H}\text{a}\text{n}\text{d}\text{l}\text{i}\text{n}\text{g}}}{}{)}_{\text{A}\text{i}\text{r}-\text{H}\text{a}\text{n}\text{d}\text{l}\text{i}\text{n}\text{g}}+{\displaystyle \underset{?}{\overset{?}{}}}(\frac{1}{}\frac{P}{L}L{)}_{\text{F}\text{r}\text{i}\text{c}\text{t}\text{i}\text{o}\text{n}}+{\displaystyle \underset{?}{\overset{?}{}}}(K\frac{V^2}{2}{)}_{\text{F}\text{i}\text{t}\text{t}\text{i}\text{n}\text{g}}+(\frac{V_{\text{E}\text{x}\text{i}\text{t}}}{2}{)}_{\text{E}\text{x}\text{i}\text{t}}]$

Note: one damper and elbow $90$ in each branch of $(HM,KN)$ and one damper in branch $KL$

Fitting $K\{$ Fan discharge to main duct $0\mathrm{.}3$; Branch from main duct $0\mathrm{.}4$; Expander $0\mathrm{.}75$ ; Reducer $0\mathrm{.}05$ ; Standard $90$ elbow $0\mathrm{.}75$ ; Damper $0\mathrm{.}2$ ; Exit grille $0\mathrm{.}5\}$