If the separation takes place at 2.8 m of water absolute, find the maximum length of siphon from the upper reservoir to summit. Take f'(friction coefficient)=0.004 and atmospheric pressure = 10.3 m of water.

Solution :

Diagram

Given :

1) $ d=200 \mathrm{~mm}=0.2 \mathrm{~m}$

2) length of Siphon $=600 \mathrm{~m}$

3) Friction factor $(f)=0.016$

4) Accelerdion due to gravity $(g)=9.81\mathrm{~m} / \mathrm{s}^{2}$

5) $P_{1}=P_{3}=10.3$ m atm of water

6) $P_{2}=2.8 \mathrm{~m}$ of water absolute pressure

7) $ v = \rho g=$ constant

Required formulae

1) Bernoullie energy equation $$ \frac{P}{v}+\frac{V^{2}}{2 g}+z=\text { constant } $$ 2) From Darcy weisbach equation $$ h_{f}=\frac{F L Q^{2}}{12.1 d^{5}} $$

Part 1 : find value of Q

Applying bernoulies energy equation at point $1 \& 3$ $$ \begin{aligned} \therefore \frac{P_{1}}{v}+\frac{V_{1}^{2}}{2 g}+z_{1}=\frac{P_{3}}{v}+\frac{V_{2}^{2}}{2 g}+z_{3}+\ h_{f(1-3)} \\ \therefore \quad v=\rho g=\text { constant} . \end{aligned} $$

$$ \begin{aligned} \therefore 10.3+0+z_{1} &=10.3+0+z_{3}+h_{f(1-3} \\ \therefore \quad z_{1}-z_{3} &=h_{f(1-3)}=\frac{\mathrm{fL} Q^{2}}{12.1 \mathrm{~d}^{5}} \\ \therefore \quad 15 &=\frac{0.016 \times 600^{2}}{12.1 \times 0.2^{5}} \\ \therefore \quad Q^{2} &=0.00605 \\ \therefore \quad Q &=77.78\times 10^{-3} \mathrm{~m}^{3} / \mathrm{s} \end{aligned} $$

Part 2 : find max length of Siphon from upper reservoir to summit. $L_{(1-2)}$ Applying bernoullies energy equalion at point 1 & 2

$$ \begin{aligned} \therefore \frac{P_{1}}{v}+\frac{V_{1}{ }^{2}}{2 g}+z_{1} &=\frac{P_{2}}{v}+\frac{V_{2}{ }^{2}}{2 g}+z_{2}+h_{ f(1-2)} \\ \therefore 10.3+0+z_{1} &=2.8+\frac{Q^{2}}{A^{2} g}+z_{2}+h_{f(1-2)}\\ \therefore 2.5-\left(z_{2}-z_{1}\right) &=\frac{Q^{2}}{2 A^{2} g}+\frac{f L_{(1-2)} Q^{2}}{12.1 \times d^{5}} \\ \therefore 7.5-4 &=\frac{(77.78 \times 10^{-3})^{2}}{2 \times(\frac{\pi}{4} \times 0.2^{2})^{2} \times 9.81}+\frac{0.016\times(77.78 \times 10^{-3})^{2}}{12.1 \times 0.2^{5}} \\ \therefore 3.5 &=.3124+0.025 L_{(1-2)}\\ \therefore L_{(1-2)} &=127.506 m \end{aligned} $$

therefore The max length of siphon from upper reservoir to summit is is 127.506 m.