Three reservoirs with 250 mm concrete pipes

Data. $Z_1 = 29$ m, $Z_2 = 129$ m, $Z_3 = 69$ m; $L_1 = 80$ m, $L_2 = 150$ m, $L_3 = 110$ m; $D = 0.25$ m, $k = 0.5$ mm, $\nu = 1.02\times10^{-6}$ m²/s. The friction factor is found from the Colebrook-White equation (no Moody chart), neglecting minor losses and velocity heads at the reservoirs.

$$\frac{k}{D} = \frac{0.5}{250} = 0.0020,\qquad \frac{1}{\sqrt f} = -2\log_{10}\left(\frac{k/D}{3.7} + \frac{2.51}{Re\sqrt f}\right)$$

A starting value comes from the fully rough formula $\dfrac{1}{\sqrt f} = 2\log_{10}\dfrac{D}{k} + 1.14$, giving $f_0 = 0.0234$.

Method

Let $H_J$ be the piezometric head at the junction. For pipe $i$ the head loss is $|Z_i - H_J| = f_i\dfrac{L_i}{D}\dfrac{V_i^2}{2g}$, so $V_i = \sqrt{\dfrac{2gD|Z_i - H_J|}{f_iL_i}}$, with $f_i$ updated from $Re_i = V_iD/\nu$. The correct $H_J$ satisfies continuity at the junction: $\sum Q = 0$.

Reservoir 2 is the highest (129 m) and reservoir 1 the lowest (29 m), so the junction head lies between 29 and 129 m. If $H_J > 69$ m, reservoir 3 receives water; if $H_J < 69$ m, it also feeds the junction. Trying values (each with Colebrook for $f$):

• $H_J = 69$ m: $Q_1 \approx 0.50$ m³/s out, $Q_2 \approx 0.45$ m³/s in, $Q_3 = 0$. Outflow > inflow, so $H_J$ must be lowered below 69 m, and reservoir 3 then supplies the junction.
• Further trials just below 69 m give the balance at $H_J \approx 68.48$ m.

Final solution ($H_J = 68.48$ m)

(Positive $Q$ = into the junction; negative = out of the junction.) Check: $450.4 + 47.9 = 498.3$ l/s in, and $498.3$ l/s out.

Answer: Reservoir 2 delivers $Q_2 = 450.4$ l/s and reservoir 3 delivers $Q_3 = 47.9$ l/s to the junction; both flow on to reservoir 1 with $Q_1 = 498.3$ l/s. Junction piezometric head $H_J = 68.48$ m (about 68.5 m).