Question 6.SQ.4: Three reservoirs are connected as in Fig. 6.8. Details of th......
Three reservoirs are connected as in Fig. 6.8. Details of the three pipelines are shown below. The elevation of the water surface in the reservoirs A, C and D is constant at respectively 250 mOD, 220 mOD and 190 mOD. The pipelines are long so it can be assumed that friction losses will dominate and that minor losses can be ignored. Assuming that the flow is not controlled by means of valves, calculate the discharge through each of the pipes.
\begin{array}{cccc}\text { pipeline } & \text { length } \mathrm{~(km)} & \text { diameter } \mathrm{~(m)} & \lambda \\1 & 17.5 & 1.2 & 0.05 \\2 & 5.3 & 0.9 & 0.04 \\3 & 6.4 & 0.9 & 0.04\end{array}
Apply Bernoulli from A to C:
Z_{AC} = {\lambda _{1}L_{1}V_{1}^{2}}/{2gD_{1}} + {\lambda _{2}L_{2}V_{2}^{2}}/{2gD_{2}}. Z_{AC} = 250 – 220 = 30 m.
30 = \left[0.05 \times 17500 \times {V_{1}^{2}}/{19.62} \times 1.2\right] + \left[0.04 \times 5300 \times {V_{2}^{2}}/{19.62} \times 0.9\right]30 = 37.164V_{1}^{2} + 12.006V_{2}^{2} which gives: V_{2} = \left(2.499 – 3.095V_{1}^{2}\right)^{{1}/{2}} (1)
Apply Bernoulli from A to D:
Z_{AD} = {\lambda _{1}L_{1}V_{1}^{2}}/{2gD_{1}} + {\lambda _{3}L_{3}V_{3}^{2}}/{2gD_{3}}. Z_{AD} = 250 – 190 = 60 m.
60 = 37.164V_{1}^{2} + \left[0.04 \times 6400 \times {V_{3}^{2}}/{19.62} \times 0.9\right] thus 60 = 37.164V_{1}^{2} + 14.498V_{3}^{2}. This gives: V_{3} = \left(4.139 – 2.563V_{1}^{2}\right)^{{1}/{2}} (2)
Apply the continuity equation: Q_{1} = Q_{2} + Q_{3} so D_{1}^{2}V_{1} = D_{2}^{2}V_{2} + D_{3}^{2}V_{3}.
Thus 1.2^{2}V_{1} = 0.9^{2}V_{2} + 0.9^{2}V_{3} and hence 1.440V_{1} = 0.810V_{2} + 0.810V_{3}.
Thus: V_{1} = 0.563V_{2} + 0.563V_{3} (3)
Substituting for V_{2} and V_{3} in equation (3) from equations (1) and (2):
V_{1} = 0.563\left(2.499 – 3.095V_{1}^{2}\right)^{{1}/{2}} + 0.563\left(4.139 – 2.563V_{1}^{2}\right)^{{1}/{2}} (4)
For a real solution 3.095V_{1}^{2} \lt 2.499 and 2.563V_{1}^{2} \lt 4.193 giving V_{1} \lt 0.9 {m}/{s} and 1.3 {m}/{s}.
Solving equation (4) by trial and error gives V_{1} = 0.895 m / s. Substituting in equation (1) gives V_{2} = 0.141 {m}/{s} and V_{3} = 1.444 {m}/{s} from equation (2). Thus Q_{1} = (\pi \times {1.2^{2}}/{4}) \times 0.895 = 1.012 {m^{3}}/{s}. Q_{2} = (\pi \times {0.9^{2}}/{4}) \times 0.141 = 0.090 {m^{3}}/{s}.
Q_{3} = (\pi \times {0.9^{2}}/{4}) \times 1.444 = 0.919 {m^{3}}/{s}.