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} …