Question 6.SQ.3: Two large reservoirs with a difference in water level of 27 ......
Two large reservoirs with a difference in water level of 27 m are connected by a pipeline that splits into two branches (as in Fig. 6.7) after a distance of 10 km. Ignoring minor losses, calculate the discharge in each of the three pipelines if the details of the pipelines are:
\begin{array}{cccc}\text { pipeline } & \text { Diameter (m) } & \text { Length (km) } & \lambda \\1 & 0.90 & 10 & 0.04 \\2 & 0.75 & 21 & 0.07 \\3 & 0.60 & 23 & 0.05\end{array}
Applying the Bernoulli equation to a streamline joining the surface of the reservoirs via pipes 1 and 2, ignoring minor losses: Z = {\lambda _{1}L_{1}V_{1}^{2}}/{2gD_{1}} + {\lambda _{2}L_{2}V_{2}^{2}}/{2gD_{2}} where Z = 27 m. Thus:
27 = \left[0.04 \times 10000 \times {V_{1}^{2}}/{19.62} \times 0.9\right] + \left[0.07 \times 21000 \times {V_{2}^{2}}/{19.62} \times 0.75\right]27 = 22.653V_{1}^{2} + 99.898V_{2}^{2} (1)
Applying the Bernoulli equation to a streamline joining the surface of the reservoirs via pipes 1 and 3, ignoring minor losses: Z = {\lambda _{1}L_{1}V_{1}^{2}}/{2gD_{1}} + {\lambda _{3}L_{3}V_{3}^{2}}/{2gD_{3}} where Z = 27 m again. Thus:
27 = 22.653V_{1}^{2} + \left[0.05 \times 23000 \times {V_{3}^{2}}/{19.62} \times 0.60\right] so:
27 = 22.653V_{1}^{2} + 97.689V_{3}^{2} (2)
Subtracting equation (2) from equation (1) gives: 0 = 99.898V_{2}^{2} + 97.689V_{3}^{2}.
Thus V_{3}^{2} = ({99.898}/{97.689})V_{2}^{2} giving V_{3} = 1.011V_{2} (3)
Applying the continuity equation: Q_{1} = Q_{2} + Q_{3} or D_{1}^{2}V_{1} = D_{2}^{2}V_{2} + D_{3}^{2}V_{3}.
Thus 0.9^{2}V_{1} = 0.75^{2}V_{2} + 0.6^{2}V_{3} giving 0.81V_{1} = 0.563V_{2} + 0.36V_{3}. Now substituting for V_{3} from equation (3): 0.81V_{1} = 0.563V_{2} + 0.36(1.011V_{2}) thus 0.81V_{1} = 0.927V_{2} or V_{1} = 1.144V_{2} (4)
Substituting for V1 in equation (1): 27 = 22.653(1.144V_{2})^{2} + 99.898V_{2}^{2}
27 = 129.545V_{2}^{2} thus V_{2} = 0.457 m/s.
Substituting for V_{2} in equation (4) gives V_{1} = 0.523 m/s.
Substituting for V_{2} in equation (3) gives V_{3} = 0.462 m/s.
Thus Q_{1} = (\pi \times {0.9^{2}}/{4}) \times 0.523 = 0.333 {m^{3}}/{s}. Q_{2} = (\pi \times {0.75^{2}}/{4}) \times 0.457 = 0.202 {m^{3}}/{s}.
Q_{1} = (\pi \times {0.6^{2}}/{4}) \times 0.462 = 0.131 {m^{3}}/{s}.