In reservoir-system problems, assume a trial junction head and compute flow in each connecting pipe from available head loss. Flow direction is from higher energy head to lower energy head.
The correct junction head satisfies continuity: inflow to the junction equals outflow from the junction.
For three-reservoir problems, one reservoir may supply the junction while the others receive flow depending on elevations.
For closed pipe loops, apply continuity at junctions and require algebraic head loss around each loop to be zero. Iterative Hardy Cross work must keep signs consistent.
A junction receives 0.30 m3/s from reservoir A and sends 0.12 m3/s to reservoir B. Determine the flow to reservoir C.
Answer: The flow to reservoir C is 0.18 m3/s.
Three reservoirs connect to a junction. Reservoir heads are $H_A=90\text{ m}$, $H_B=70\text{ m}$, and $H_C=40\text{ m}$. The pipe coefficients are $Q=0.020\sqrt{h}$ for A, $Q=0.015\sqrt{h}$ for B, and $Q=0.025\sqrt{h}$ for C, with $Q$ in m3/s and $h$ in m. If the junction head is 60 m, check continuity.
Answer: Continuity is not satisfied. The assumed junction head of 60 m gives excess inflow, so the true junction head should be higher.
Reservoirs A and B supply a junction that discharges 0.160 m3/s to reservoir C. If A supplies 0.095 m3/s, find the flow supplied by B and state the governing condition at the junction.
Answer: Reservoir B supplies 0.065 m3/s. The junction head must make continuity true while each pipe head loss matches its reservoir-to-junction head difference.
Three reservoirs are connected to a common junction J by three pipes. Reservoir A has a water surface elevation of 100 m, reservoir B has elevation 80 m, and reservoir C has elevation 50 m. The pipe characteristics expressed as $Q = K\sqrt{h_f}$ (where $h_f$ is in meters and Q in m³/s) are: Pipe A–J: $K_A = 0.025$, Pipe B–J: $K_B = 0.018$, Pipe C–J: $K_C = 0.022$. Find the junction head $H_J$ and the flow in each pipe using successive trials.
Trial 1: Assume $H_J = 70$ m (between B and C elevations).
Positive excess inflow → raise $H_J$. Trial 2: Assume $H_J = 80$ m.
Slightly negative → junction head is just below 80 m. Interpolating between trials 1 and 2:
At $H_J = 79.2$ m: B reservoir barely supplies (or receives) nearly zero flow. A supplies to J, J sends to C.
Answer: Junction head ≈ 79.2 m. Pipe A supplies ≈ 0.115 m³/s, pipe B has nearly zero flow, pipe C receives ≈ 0.115 m³/s. Reservoir B becomes a "dead" reservoir when the junction head equals its surface elevation.
In a three-reservoir system, reservoir A (elevation 120 m) and reservoir B (unknown elevation $H_B$) are connected to junction J (elevation 60 m), and the junction also connects to reservoir C (elevation 40 m). Flow measurements at the junction show that pipe A carries 0.08 m³/s into J, and pipe C carries 0.12 m³/s out of J. The junction piezometric head $H_J$ has been determined by trial to be 75 m. Each pipe has the form $h_f = RQ^2$ where $R_A = 5000$, $R_B = 3200$, $R_C = 4500$. Verify the junction head is consistent with the given flows and find $H_B$.
Verify pipe A: $h_{fA} = R_A Q_A^2 = 5000(0.08)^2 = 32$ m. Available head = $120 - 75 = 45$ m ≠ 32 m — inconsistency. Correct $H_J$ so that $H_J = 120 - 32 = 88$ m. Re-solve:
Continuity: $Q_A + Q_B = Q_C \Rightarrow Q_B = 0.1033 - 0.08 = 0.0233$ m³/s (into J from B).
Answer: The corrected junction head is 88 m. Reservoir B has a surface elevation of approximately 89.74 m, and it contributes only 0.023 m³/s into the junction. Reservoir C receives a combined 0.103 m³/s from both A and B.
A single closed pipe loop has four pipes with head loss expressed as $h_f = RQ^2$. The assumed clockwise flows and pipe resistances are: Pipe 1 (clockwise): Q₁ = 0.06 m³/s, R₁ = 1000. Pipe 2 (clockwise): Q₂ = 0.04 m³/s, R₂ = 800. Pipe 3 (counter-clockwise): Q₃ = 0.03 m³/s, R₃ = 1200. Pipe 4 (counter-clockwise): Q₄ = 0.05 m³/s, R₄ = 600. Apply one Hardy Cross correction and find the corrected discharges.
Take clockwise as positive. Compute $h_f = RQ|Q|$ (signed) and $|2RQ|$ for each pipe:
Corrected flows: $Q_1' = 0.06 - 0.00728 = 0.0527$ m³/s; $Q_2' = 0.04 - 0.00728 = 0.0327$ m³/s; $Q_3' = 0.03 + 0.00728 = 0.0373$ m³/s; $Q_4' = 0.05 + 0.00728 = 0.0573$ m³/s.
Answer: After one Hardy Cross correction: Q₁ = 52.7 L/s, Q₂ = 32.7 L/s, Q₃ = 37.3 L/s, Q₄ = 57.3 L/s. Iterate further for convergence. The large initial imbalance (2.30 m) requires 2–3 more iterations.
Additional board-style practice items for this topic.
The three reservoirs A, B and C are connected by pipelines 1, 2 and 3, respectively, which meet at junction D as shown in the figure. The elevation of reservoir A is at 300 m, while that of C is at elevation 277 m. The discharge flowing towards B is 0.60 m3/sec. The length diameter and frictiom factors for the pipes are as shown in the accompanying table.
Compute the rate of flow in or out reservoir A, in liters/sec.
Compute the rate of flow in reservoir C, in liters/sec.
Compute the elevation of reservoir B in meters.
Solution pending in psadquestions/q346.json.
In the figure shown, reservoir A is the source of water supply and is at elevation 150m, B is the junction at elevation 91.46 m, C is a town at elevation 30.49 m. with 25000 inhabitants, D is another town at elevation 15.24 m. with a population of 30000 inhabitants. Length AB = 15,240m, BC is 9150m, BD = 6100m. Determine the size of the pipes if the consumption is 150 liters per capita per day. For the pipes, friction factor f = 0.02.
Determine the required diameter of pipe AB.
Determine the required diameter of pipe BC.
Determine the required diameter of pipe BD.
Part 1.
The total demand is:Part 2.
Demand for town C:Part 3.
Demand for town D: