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## Q. 11.28

The branching system shown in Fig. P11.28a is to be solved by using the Hardy Cross method.

(a) Write the equations for $ΔQ_I, ΔQ_{II}$,W, and G.

(b) Perform two iterations by filling in the blanks in the table shown in Fig. P11.28b.

After two iterations, sketch the hydraulic grade line.

## Verified Solution

$ΔQ_{I}=\frac{-(\pm W_1\pm W_2)-10}{G_1+G_2} , ΔQ_{II}=\frac{-(\pm W_2\pm W_3)-15}{G_2+G_3}$,

$\pm W_i=\overline{R} Q_i\left|Q_i\right| , G_i=2\overline{R} \left|Q_i\right|$.

$\begin{matrix} Loop & Pipe & Q & W & G & Q & W & G & Q \\ I & 1 & −3.5 & −24.50 & 14.00 & −2.46 & −12.10 & 9.84 & −2.30 \\ & 2 & 0 & \underline{ 0} & \underline{0} & +0.36 & \underline{+0.26} & \underline{1.44} & +0.43 \\ & & & Σ 24.50 & 14.00 & & Σ −11.84 & 11.28 & \end{matrix}$

$ΔQ_{I}==\frac{+24.50-10}{14.00} =+1.04 ΔQ_{I}=\frac{+11.84-10}{11.28} =+0.16$

$\begin{matrix} II & 2 & 0 & 0 & 0 & −0.36 & -0.26 & 1.44 & −0.43 \\ & 3 & −3.5 & \underline{−24.5} & \underline{14.00} & −2.82 & \underline{−15.90} & \underline{11.28} & +2.73 \\ & & & Σ −24.50 & 14.00 & & Σ −16.16 & 12.72 & \end{matrix}$

$ΔQ_{II}=\frac{+24.50-15}{14.00}=+0.68 ΔQ_{II}=\frac{+16.16-15}{12.72}$ =+0.09

$H _{Junction} = 50 −\overline{R}_1Q_1^2 = 50 − 2 × 2.3^2 = \underline{39.4}$

$\underline{Q_1 = 2.30}$ (into J), $\underline{ Q_2 = 0.43}$ (into J),  $\underline { Q_3 = 2.73}$ (out of J)