Question 6.SQ.3: Two large reservoirs with a difference in water level of 27 ......

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:

$27 = 22.653V_{1}^{2} + 99.898V_{2}^{2}$  (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 $Q_{1} = (\pi \times {0.9^{2}}/{4}) \times 0.523 = 0.333  {m^{3}}/{s}.  Q_{2} = (\pi \times {0.75^{2}}/{4}) \times 0.457 = 0.202  {m^{3}}/{s}$.

$Q_{1} = (\pi \times {0.6^{2}}/{4}) \times 0.462 = 0.131  {m^{3}}/{s}$.