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I have a little question about this problem I computed the right answer this way

First I applied Bernoulli between the top and the exit

$$\frac{P_{t}}{\rho g}+z_{t}+\frac{v_{t}^2}{2g}-H_{1}-H_2-H_{\small{turbine}}=\frac{P_e}{\rho g}+z_e+\frac{v_e^2}{2g}$$

Where $H_1$ is the loss in the first pipe and $H_2$ in the second.

So

$$\frac{v_e^2}{2g}+H_{1}+H_2+H_{\small{turbine}}=z_t-z_e$$

Replacing all the data from the problem and solving for $Q$ I have:

$$32,276.11161Q^2+1,062,588.184\lambda _1Q^2+24,207,087.07\lambda _2Q^2+\frac{0.04086}{Q}=20 \\ (32,276.11161+1,062,588.184\lambda _1+24,207,087.07\lambda _2)Q^3-20Q+0.04086=0$$

Where $\lambda _1$ and $\lambda _2$ are the loss coefficient in the pipes 1 and 2 respectively.

So now cames the iterative process, first I let $\lambda _1=\lambda _2=0.025$ then

$$664,017.9975Q^3-20Q+0.04086=0$$

Unfortunately this polynomial have three real solutions, One negative and two positive, and I don't know what is the correct $Q$ to choose, I only know that $Q$ must be positive.

I chose the smaller $Q$ positive in all iteration, and after three iterations computing Reynolds and getting $\lambda _{new}$ 1 and 2 from Moody Diagram I got $Q=0.00416 \, \text{m}^3\text{/s} \approx 15 \text{m}^3\text{/h}$ which in fact is the answer of the problem.

Why I not chose the bigger $Q$ positive?

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  • $\begingroup$ Did you check your positive answers to see which one conserves energy? $\endgroup$ – David White Jul 11 at 1:17
  • $\begingroup$ Which term in your equation describes the viscous frictional pressure drops? $\endgroup$ – Chet Miller Jul 11 at 2:03

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