Question 11.8: Two reservoirs open to atmosphere are connected by a pipe 80...
Two reservoirs open to atmosphere are connected by a pipe 800 metres long. The pipe goes over a hill whose height is 6 m above the level of water in the upper reservoir. The pipe diameter is 300 mm and friction factor f = 0.032. The difference in water levels in the two reservoirs is 12.5 m. If the absolute pressure of water anywhere in the pipe is not allowed to fall below 1.2 m of water in order to prevent vapour formation, calculate the length of pipe in the portion between the upper reservoir and the hill sumit, and also the discharge through the pipe. Neglect bend losses. Draw the equivalent electrical network system.
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}
=86.83 \frac{V^{2}}{2 g}
Applying Bernoulli’s equation between A and B, we have
\Delta H=h_{f}
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
p_{ atm }=760 mm \text { of } Hg
=\frac{760 \times 13.6}{1000}=10.34 m of water,
Equation (11.47) becomes
10.34=1.2+6+\frac{(1.68)^{2}}{2 \times 9.81}+h_{f 1}
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
\left(0.5+0.107 L_{1}\right) \frac{(1.68)^{2}}{2 \times 9.81}=2.99
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.
