--- title: "Branching pipe system and three-reservoir cases" subject: "Hydraulics" chapter: 3 key: "semester_4/hydraulics:branching-pipe-system-three-cases" source: https://www.bctengineer.com/q/semester_4/hydraulics/branching-pipe-system-three-cases site: BCT Engineering --- # Branching pipe system and three-reservoir cases Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks - Asked in: 2071 Bhadra, 6 marks - Video: [Unit:2 | Pipe Networking | Numerical from Three Reservoirs| Types:1 | Prashant YT | BE Civil |](https://www.youtube.com/watch?v=pIXbQd4ZiiM) — Prashant YT - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/branching-pipe-system-three-cases - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3 ## Question What do you understand by branching pipe system? Explain. Describe the solution procedures for three possible different cases of three reservoir problem. ## Short answer A branching (three‑reservoir) system has three pipes meeting at a single junction; the middle reservoir can supply or receive depending on the junction head HJ. Solution cases: (1) Given all data, assume HJ and adjust until continuity holds using hf = rQ²; (2) Given one discharge, compute HJ from that pipe, then other flows or an unknown level; (3) Given flows and levels, find unknown diameter from D = (8fLQ²/π²g hf)^(1/5). ## Answer #### 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 $H_J$. ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. If H_J > Z_B in a three-reservoir system, what happens in pipe 2? A. Pipe 2 reverses the flow of pipe 3 B. B receives water and Q1 = Q2 + Q3 C. B supplies water and Q1 + Q2 = Q3 D. There is no flow Answer: B. If the junction head exceeds B's level, water flows into B, so A's flow splits between B and C. (Key point: Direction of flow in pipe 2) 2. In Case 1 with H_J = Z_B, Q1 > Q3. What is concluded? A. B receives and H_J must be raised B. B supplies and H_J must be lowered C. The solution is found D. C supplies water Answer: A. Too much inflow at J means B must absorb the surplus, which requires a higher H_J. (Key point: Starting trial in Case 1) 3. What energy condition is used for each pipe? A. Z_i = H_J B. H_J = sum of Z_i C. |Z_i - H_J| = Q_i / r_i D. |Z_i - H_J| = r_i Q_i^2 Answer: D. The difference between reservoir level and junction head equals the friction loss r Q^2. (Key point: Condition governing each pipe) 4. In Case 3, how is an unknown diameter found? A. By assuming H_J = Z_B B. By setting D equal to the largest pipe C. From the head loss available, D = (8fLQ^2/(pi^2 g h_f))^(1/5) D. From continuity alone Answer: C. Known discharges give H_J, then h_f = |Z_i - H_J| is solved for D. 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