Question 6.8: Three large reservoirs are joined by a branching pipeline ex......
Three large reservoirs are joined by a branching pipeline exactly as in Fig. 6.8. The elevation of the water surface in the reservoirs is A = 680 mOD, C = 640 mOD and D = 590 mOD. These levels do not change. Details of the three pipelines are:
\begin{array}{cccc}\text { pipeline } & \text { length } & \text { diameter } & \lambda \\1 & 0.5 \mathrm{~km} & 1.2 \mathrm{~m} & 0.04 \\2 & 0.3 \mathrm{~km} & 0.9 \mathrm{~m} & 0.06 \\3 & 0.4 \mathrm{~km} & 0.6 \mathrm{~m} & 0.05\end{array}Assume that all minor losses are negligible, so only friction losses need be considered. Determine the discharge through each pipeline if the flow is not controlled by valves.
First, apply the Bernoulli equation to a streamline joining A to C assuming only a friction loss:
Z_{AC} = {\lambda _{1}L_{1}V_{1}^{2}}/{2gD_{1}} + {\lambda _{2}L_{2}V_{2}^{2}}/{2gD_{2}}where Z_{AC} = 680 mOD – 640 mOD = 40 m.
40 = \left[0.04 \times 500 \times {V_{1}^{2}}/{19.62} \times 1.2\right] + \left[0.06 \times 300 \times {V_{2}^{2}}/{19.62} \times 0.9\right] 40 = 0.849V_{1}^{2} + 1.019V_{2}^{2}1.019V_{2}^{2} = 40 – 0.849V_{1}^{2}
V_{2} = (39.254 – 0.833V_{1}^{2})^{{1}/{2}} (1)
Now apply the Bernoulli equation to a streamline joining A to D, as above:
Z_{AD} = {\lambda _{1}L_{1}V_{1}^{2}}/{2gD_{1}} + {\lambda _{3}L_{3}V_{3}^{2}}/{2gD_{3}}where Z_{AD} = 680 mOD – 590 mOD = 90 m.
90 = \left[0.04 \times 500 \times {V_{1}^{2}}/{19.62} \times 1.2\right] + \left[0.05 \times 400 \times {V_{3}^{2}}/{19.62} \times 0.6\right] 90 = 0.849V_{1}^{2} + 1.699V_{3}^{2}1.699V_{3}^{2} = 90 – 0.849V_{1}^{2}
V_{3}= (52.972 – 0.500V_{1}^{2})^{{1}/{2}} (2)
The continuity equation for a branching pipeline can be written as Q_{1} = Q_{2} + Q_{3} or: A_{1}V_{1} = A_{2}V_{2} + A_{3}V_{3}
({\pi D_{1}^{2}}/{4})V_{1} = ({\pi D_{2}^{2}}/{4})V_{2} + ({\pi D_{3}^{2}}/{4})V_{3}D_{1}^{2}V_{1} = D_{2}^{2}V_{2} + D_{3}^{2}V_{3}
1.2^{2}V_{1} = 0.9^{2}V_{2} + 0.6^{2}V_{3}
1.440V_{1} = 0.810V_{2} + 0.360V_{3}
V_{1} = 0.563V_{2} + 0.250V_{3} (3)
This completes the hydraulic analysis. The remaining part of the question concerns the solution of the equations. This can be done by trial and error (as below), graphical means or by using appropriate computer software.
Substituting the expressions for V_{2} and V_{3} in equations (1) and (2) into equation (3) gives:
V_{1} = 0.563(39.254 − 0.833V_{1}^{2})^{{1}/{2}} + 0.250 52.972 − 0.500V_{1}^{2})^{{1}/{2}} (4)
This equation must be solved by trial and error. However, from equation (4), for the solution to be real (+ve) then 0.833V_{1}^{2} \lt 39.254 and 0.500V_{1}^{2} \lt 52.972. This gives V_{1} < 6.8 and
10.3 m/s respectively. Try V_{1} = 5.0 m/s in equation (4) and see if the left-hand side (LHS) equals the right-hand side (RHS), as required for a valid solution.
= 2.417 + 1.590 = 4.007 m/s
The RHS = 4.007 not 5.0 m/s so this is not the answer. Repeating the calculation with V_{1} = 4.5 m/s and summarising the results in a table gives:
LHS: Try \begin{matrix} V_{1} = 5.0 {m}/{s} & RHS = 4.007 {m}/{s} & (LHs – RHS) = + 0.993 \\ V_{1} = 4.5 {m}/{s} & = 4.300 {m}/{s} & = + 0.200 \\ V_{1} = 4.3 {m}/{s} & = 4.402 {m}/{s} & = – 0.102 \end{matrix}
Now RHS > LHS so V_{1} lies between 4.3 and 4.5 m/s. V_{1} can be found by interpolation: (LHS – RHS) changes by 0.200 + 0.102 = 0.302 when V_{1} changes by 4.5 – 4.3 = 0.2 m/s.
Therefore 0.102 is equivalent to (0.102/0.302) × 0.2 = 0.068 m/s.
Thus V_{1} = 4.3 + 0.068 = 4.368 m/s.
Check the solution by substituting V_{1} = 4.368 m/s in equation (4):
RHS = 0.563(39.254 − 0.833 \times 4 368^{2})^{{1}/{2}} + 0.250(52.972 – 0.500 \times 4.368^{2})^{{1}/{2}}=2.721 + 1.647 = 4.368 m/s OK
From equation (1):
V_{2} = (39.254 – 0.833V_{1}^{2})^{{1}/{2}} = (39.254 – 0.833 \times 4.368^{2})^{{1}/{2}} = 4.833 m/s
From equation (2):
V_{3}= (52.972 – 0.500V_{1}^{2})^{{1}/{2}} = (52.972 – 0.500 \times 4.368^{2})^{{1}/{2}} = 6.590 m/s
Now apply the continuity equation to the individual pipelines to calculate the discharge:
Q_{1} = A_{1}V_{1} = (π × 1.2²/4) × 4.368 = 4.940 m³/s
Q_{2} = A_{2}V_{2} = (π × 0.9²/4) × 4.833 = 3.075 m³/s
Q_{3} = A_{3}V_{3} = (π × 0.6²/4) × 6.590 = 1.863 m³/s
CHECK
Apply the continuity equation to the pipe branches.
Q_{1} = Q_{2} + Q_{3}From above Q_{1}= 4.940 m³/s
Q_{2} + Q_{3} = 3.075 + 1.863 = 4.938 m³/s OK