The Three-Reservoir Problem: Junction Head and Flow Direction
Three tanks, three pipes, one unknown junction head. How to guess the piezometric level, reverse a flow, and why Newton-Raphson beats nested square roots.
Key takeaways
- The unknown is the junction piezometric head H_j.
- Each pipe: Q from ΔH = H_end − H_j and the pipe’s K.
- Continuity: ΣQ into the junction = 0 (no local demand).
- If H_j sits between two reservoir levels, one pipe reverses.
The three-reservoir problem is the first network that is not a straight series or a simple parallel pair. Two (or three) surface levels try to feed a junction. You do not know the junction head, and you may not even know which way the lowest pipe flows until you solve.
Setup
A useful first guess for H_j is the flow-weighted average of the three surface levels, or simply the middle reservoir. Compute each Q, sum them, and raise H_j if the junction has net inflow (too much arriving) or lower it if it has net outflow.
Open solver: Three-reservoir solverThe reversal case
If H_j lands between reservoir 2 and 3, flow in the middle pipe may go either way depending on K. Students often lock the direction from a sketch. Let the sign of (H_i − H_j) choose the direction each iteration.
From three tanks to a network
Add a fourth pipe and a loop and you need Hardy-Cross or Newton-Raphson. The three-reservoir problem is the node equation you will keep using inside those methods.
Open solver: Looped network (next step)Open solver: Pipe K from Darcy fFrequently asked questions
No. There is no loop. It is a single node with three known-head boundaries. You iterate H_j, not a loop ΔQ. The same idea extends to n reservoirs.