Question 11.8: Two reservoirs open to atmosphere are connected by a pipe 80...

Let the length of pipe upstream of C be $L_{1}$ and that of the downstream be $L_{2}$ (Fig. 11.20a).

It is given $L_{1}+L_{2}=800$ m

Considering the entry, friction and exit losses,

the total loss from A to C $=h_{f 1}=\left(0.5+\frac{0.032 L_{1}}{0.3}\right) \frac{V^{2}}{2 g}$ (11.47)

the total loss from C to B $=h_{f 2}=\left(1+\frac{0.032 L_{2}}{0.3}\right) \frac{V^{2}}{2 g}$

Therefore,

the total loss from A to B $=h_{f}=\left(0.5+\begin{array}{c}0.032 \times 800 \\0.3\end{array}+1\right) \begin{array}{l}V^{2} \\2 g\end{array}$

Applying Bernoulli’s equation between A and B, we have

or $12.5=86.83 \frac{V^{2}}{2 g}$

which gives $V=\sqrt{\frac{12.5 \times 2 \times 9.81}{86.83}}=1.68$ m/s

Applying Bernoulli’s equation between A and C, we have

$\begin{array}{c}p_{ atm } \\\rho g\end{array}=\frac{p_{c}}{\rho g}+\frac{V^{2}}{2 g}+6+h_{f_{1}}$ (11.48)

With the atmospheric pressure

$=\frac{760 \times 13.6}{1000}=10.34$ m of water,

Equation (11.47) becomes

which gives $h_{f 1}=2.99$ m

Using the value of $h_{f 1}$ = 2.99 m, and V = 1.68 m/s in Eq. (11.47) we get

or $0.5+0.107 L_{1}=20.78$

which gives $L_{1}=189.53$ m

Rate of discharge through the pipe

$Q=\frac{\pi}{4}(0.3)^{2} \times 1.68=0.119$ $m ^{3} / s$

The equivalent electrical network of the system is shown in Fig. 11.20b.