# Wich is the correct flow rate $Q$ solved with iterations?

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?

• Did you check your positive answers to see which one conserves energy? – David White Jul 11 '19 at 1:17
• Which term in your equation describes the viscous frictional pressure drops? – Chet Miller Jul 11 '19 at 2:03