A pipe line 45 m long connects two reservoirs which have a difference of water level of 18 m. For a length of 20 m from the upper tank the pipe diameter is 40 mm and for the remaining part the pipe is 65 mm in diameter. At the change in section a partially open valve having a K value of 0.6 is fitted. Considering all major and minor losses calculate the flow rate through the pipeline. Take f = 0.022 for the both pipes.

Given Data, $$ \begin{array}{l} K_{L}=0.6 \\ f=0.022 \end{array} $$ Applying Energy Equation between 1 and 2 , we get $$ \frac{P_{1} }{\rho g}+\frac{v_{1}^{2}}{2 g}+z_{1}=\frac{P_{2}}{\rho g}+\frac{v_{2}^{2}}{2 g}+z_{2}+h_{f} $$ $P_{1}=P_{2}=P_{atm}$ Thus, $z_{1}-z_{2}=h_{f}=18 \mathrm{~m}$ $$ \begin{array}{l} \\ \Rightarrow 18=\frac{0.5 v_{1}^{2}}{2 g}+\frac{f_{1} L_{1} v_{1}^{2}}{d_{1} \times 2g}+\frac{K_{2} V_{1}^{2}}{2g}+\frac{f_{2} L_{2} V_{2}^{2}}{d_{2} \times 2 g}+\frac{V_{2}^{2}}{2 g} \end{array} $$ Applying continuity principal, $\begin{aligned} \Rightarrow \frac{\pi}{4}(0.04)^{2} \times v_{1} &=\frac{\pi}{4} \times(0.065)^{2} \times v_{2} . \\ V_{1} &=2.64 V_{2} \end{aligned}$ $\Rightarrow \quad 18=\frac{0.5 \times\left(2.64 V_{2}^{2}\right.}{2g}+\frac{0.022 \times 20 \times\left(2.64 V_{2}\right)^{2}}{2 g \times 0.04}+\frac{0.6 \times\left(2.64V_{2})^{2}\right.}{2g}$ $+\frac{0.022 \times 25 \times v_{2}^{2}}{2 g \times 0.065}+\frac{V_{2}{ }^{2}}{2 g}$ $\Rightarrow 18=4.81 \mathrm{~V}_{2}{ }^{2}$ $\Rightarrow \quad V_{2}=1.934 \mathrm{~m} / \mathrm{s}$

Discharge, $Q_{2}=A_{2} V_{2}$

$=\frac{\pi}{4}(0.065)^{2} \times 1.934$

$Q=0.00642 \mathrm{~m}^{3} / \mathrm{s}$