Three pipes with details as following are connected in parallel between two points When the total discharge of $0.30 m^3$/sec flows through the system, calculate distribution of discharge and loss between the junctions

Given:

For the three pipes in parallel connected between two points

Q ̇=$0.30 m^3/s$ (Total discharge through the system)

To Find:

$Q ̇_1,Q ̇_2 and Q ̇_3$ (Discharge through each pipe)

$h_l$ (Head loss between the junctions)

Consider the given configuration of three pipes in parallel connected between two points

Now, For pipes in parallel

Q ̇=$Q ̇_1+Q ̇_2+Q ̇_3$

and

$h_l=h_f1=h_f2=h_f3$

Now, Friction loss through a pipe is given by,

$h_f=\frac{flv^2}{2gd}$

(Assuming the Coefficient of friction values given,are the Darcy Weisbach friction factor)

∴$4h_l=(\frac{flv^2}{2gd})_{1}=(\frac{flv^2}{2gd})_{2}=(\frac{flv^2}{2gd})_{3}$

Now , $Q ̇_1=(vA)_{1}$

$Q ̇_2=(vA)_{2}$and

$Q ̇_3=(vA)_{3}$

∴$h_l=(\frac{flQ ̇^2}{2gdA^2})_{1}=(\frac{flQ ̇^2}{2gdA^2})_{2}=(\frac{flQ ̇^2}{2gdA^2})_{3}$

Substituting in the above equation,

∴$h_l=\frac{Q ̇_1^2}{2g} (\frac{0.02(1000)}{(0.2(\frac{π}{4} 0.2^2)^2})=\frac{Q ̇_2^2}{2g} (\frac{0.015(1200)}{0.3(\frac{π}{4} 0.3^2)^2})=\frac{Q ̇_3^2}{2g} (\frac{0.02(800)}{(0.15(\frac{π}{4}0.15^2)^2})$

∴$Q ̇_1^2=\frac{2gh_l}{101321.18}$

∴$Q ̇_1=\sqrt{\frac{2gh_l}{101321.18}}$

Similarly,

∴$Q ̇_2^2=\frac{2gh_l}{12008.44}$

∴$Q ̇_2=\sqrt{\frac{2gh_l}{12008.44}}$

and

$∴Q ̇_3^2=\frac{2gh_l}{341573.31}$

$∴Q ̇_3=\sqrt{\frac{2gh_l}{341573.31}}$

Now,

$Q ̇=0.30=Q ̇_1+Q ̇_2+Q ̇_3$

Substituting in above equation,

$Q ̇=0.30=\sqrt{\frac{2gh_l}{101321.18}}+\sqrt{\frac{2gh_l}{12008.44}}+\sqrt{\frac{2gh_l}{341573.31}}$

Solving for $h_l$

∴$\sqrt{h_l } \{\sqrt{\frac{2g}{101321.18}}+\sqrt{\frac{2g}{12008.44}}+\sqrt{\frac{2g}{341573.31}}\}=0.30$

∴$\sqrt{h_l} \{\sqrt{\frac{2g}{101321.18}}+\sqrt{\frac{2g}{12008.44}}+\sqrt{\frac{2g}{341573.31}}\}=0.30$

∴$\sqrt{h_l} {0.06192}=0.30$

∴$\sqrt{h_l }$=4.8453

$h_l$=23.4772 m of liquid column

…this is the head loss between the junctions.

Now,

∴$Q ̇_1=\sqrt{\frac{2gh_l}{101321.18}}=0.06743 m^3/s$

∴$Q ̇_2=\sqrt{\frac{2gh_l}{12008.44}}=0.1959 m^3/s$

and

∴$Q ̇_3=\sqrt{\frac{2gh_l}{341573.31}}=0.03672 m^3/s$

…which are the required discharges through each pipe.