Branching pipe system and three-reservoir cases

Branching pipe system

A branching (three-reservoir) pipe system is a set of three pipes that run from three reservoirs at different levels and meet at one junction J. The flow in each pipe is not obvious: the middle reservoir B may either supply or receive water, depending on the piezometric head at the junction $H_J$.

   A (highest)          Z_A
     \                   
      \ pipe 1      B  Z_B
       \         /  pipe 2
        \       /
         +--- J ---- pipe 3 ---- C (lowest) Z_C

Two conditions govern it: (1) at J, continuity $\sum Q = 0$; (2) the energy (head) loss in each pipe is $|Z_i - H_J| = r_iQ_i^2$ (friction; minor losses and velocity heads neglected), with $r_i = \dfrac{8f_iL_i}{\pi^2gD_i^5}$. Because A is the highest and C the lowest reservoir, A always supplies and C always receives. The direction of flow in pipe 2 depends on $H_J$:

• $H_J > Z_B$: B receives water, so $Q_1 = Q_2 + Q_3$.
• $H_J < Z_B$: B supplies water, so $Q_1 + Q_2 = Q_3$.
• $H_J = Z_B$: no flow in pipe 2, and $Q_1 = Q_3$.

Solution procedures (three cases)

Case 1: Reservoir levels and all pipe data given; find the discharges. (Most common)

1. Assume $H_J$ between $Z_C$ and $Z_A$ (first try $H_J = Z_B$).
2. Compute $Q_1 = \sqrt{(Z_A - H_J)/r_1}$, $Q_3 = \sqrt{(H_J - Z_C)/r_3}$, and $Q_2 = \sqrt{|Z_B - H_J|/r_2}$.
3. With $H_J = Z_B$: if $Q_1 > Q_3$, the junction has too much inflow, so B receives and $H_J$ must be raised; if $Q_1 < Q_3$, B supplies and $H_J$ must be lowered.
4. Adjust $H_J$ until $Q_1 \pm Q_2 - Q_3 = 0$ (the correct sign for $Q_2$ as above). Interpolate between two trials.

Case 2: Discharge in one pipe and level/data of the others known; find an unknown reservoir level or the other discharges. Calculate $H_J$ directly from the pipe with known $Q$ (for example $H_J = Z_A - r_1Q_1^2$), then find the other discharges from $H_J$ and the continuity equation, or find the unknown level from the energy equation $Z_B = H_J \pm r_2Q_2^2$.

Case 3: Discharges and levels given; find the unknown pipe diameter (or length). From the known discharges find $H_J$ (using the pipe whose data is fully known), then find the head loss available for the pipe whose diameter is required, $h_f = |Z_i - H_J|$, and solve $D = \left(\dfrac{8f L Q^2}{\pi^2 g\,h_f}\right)^{1/5}$ (with $f$ from Moody/Colebrook if not given).

In every case check that the final flows satisfy continuity at the junction, and that the head at the junction is between the levels of the lowest and highest reservoirs.