Question 6.7: Two reservoirs are connected by a pipeline that splits into ......
Two reservoirs are connected by a pipeline that splits into two branches as shown in Fig. 6.7. The difference in the elevation of the water surface between the upper and lower reservoir, Z, is constant at 150 m. The roughness factor of all the pipes is λ = 0.04. The branch in the pipeline occurs 23 km from the upper reservoir, the lower pipelines being each 20 km long. The diameters of pipelines 1, 2 and 3 are respectively 1.5 m, 0.9 m and 1.0 m. Ignoring minor losses, calculate the discharge through the three pipelines.
The pipeline is long so friction losses will dominate, allowing minor losses to be ignored. Applying the Bernoulli equation to the streamline joining A to B via pipes 1 and 2 gives:
Z = {\lambda _{1}L_{1}V_{1}^{2}}/{2gD_{1}} + {\lambda _{2}L_{2}V_{2}^{2}}/{2gD_{2}}
150 = \left[0.04 \times 23000 \times {V_{1}^{2}}/{19.62} \times 1.5\right] + \left[0.04 \times 20000 \times {V_{2}^{2}}/{19.62} \times 0.9\right]150 = 31.261V_{1}^{2} + 45.305V_{2}^{2} (1)
Now applying the Bernoulli equation to the streamline joining A to B via pipes 1 and 3:
Z = {\lambda _{1}L_{1}V_{1}^{2}}/{2gD_{1}} + {\lambda _{3}L_{3}Q_{3}^{2}}/{2gD_{3}}
The first two terms of this equation are the same as in equation (1) above.
150 = 31.261V_{1}^{2} + \left[0.04 \times 20000 \times {V_{3}^{2}}/{19.62} \times 1.0\right]150 = 31.261V_{1}^{2} + 40.775V_{3}^{2}
The continuity equation provides the third equation needed to solve for three variables:
Q_{1} = Q_{2} + Q_{3} or A_{1}V_{1} = A_{2}V_{2} + A_{3}V_{3}D_{1}^{2}V_{1} = D_{2}^{2}V_{2} + D_{3}^{2}V_{3} (the π/4’s have been cancelled)
1.5^{2}V_{1} = 0.9^{2}V_{2} + 1.0V_{3}2.250V_{1} = 0.810V_{2} + 1.000V_{3} (3)
This effectively ends the hydraulic analysis, the remaining calculations being concerned with solving the equations. This is relatively simple because the first two terms of equations (1) and (2) are identical. Subtracting equation (2) from equation (1) gives:
0 = 45.305V_{2}^{2} – 40.775V_{3}^{2} V_{3}^{2} = 1.111V_{2}^{2}V_{3} = 1.054V_{2} (4)
Substituting for V_{3} in equation (3) gives:
2.250V_{1} = 0.810V_{2} + 1.000(1.054V_{2}) 2.250V_{1} = 1.864V_{2}V_{1} = 0.828V_{2} (5)
Substituting for V_{1} in equation (1) gives:
150 = 31.261(0.828V_{2})^{2} + 45.305V_{2}^{2}150 = 21.432V_{2}^{2} + 45.305V_{2}^{2}
150 = 66.737V_{2}^{2}
V_{2} = 1.449 m/s
From equation (4): V_{3} = 1.054V_{2} = 1.054 × 1.499 = 1.580 m/s
From equation (5): V_{1} = 0.828V_{2} = 0.828 × 1.499 = 1.241 m/s
Therefore: Q_{1} = A_{1}V_{1} = (π × 1.5²/4) × 1.241 = 2.193 m³/s
Q_{2} = A_{2}V_{2} = (π × 0.9²/4) × 1.499 = 0.954 m³/s
Q_{3} = A_{3}V_{3} = (π × 1.0²/4) × 1.580 = 1.241 m³/s
CHECK: Q_{1} = Q_{2} + Q_{3}
Q_{1} = 0.954 + 1.241 = 2.195 m³/s OK