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Reservoir Junction Head

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.

$$h_f=H_{reservoir}-H_j$$$$Q=C\sqrt{h_f}\quad \text{or solve from Darcy-Weisbach/Hazen-Williams}$$

Continuity at a Pipe Junction

The correct junction head satisfies continuity: inflow to the junction equals outflow from the junction.

$$\sum Q_{in}=\sum Q_{out}$$

For three-reservoir problems, one reservoir may supply the junction while the others receive flow depending on elevations.

Network Loop Reminders

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.

$$\sum h_L=0\quad \text{around a closed loop}$$

Junction Flow Balance

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.

$$Q_A=Q_B+Q_C$$$$Q_C=0.30-0.12=0.18\text{ m}^3/\text{s}$$

Answer: The flow to reservoir C is 0.18 m3/s.

Three-Reservoir Trial Junction Head

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.

$$Q_A=0.020\sqrt{90-60}=0.1095\text{ m}^3/\text{s}\quad \text{into junction}$$$$Q_B=0.015\sqrt{70-60}=0.0474\text{ m}^3/\text{s}\quad \text{into junction}$$$$Q_C=0.025\sqrt{60-40}=0.1118\text{ m}^3/\text{s}\quad \text{out of junction}$$$$\sum Q_{in}-\sum Q_{out}=0.1095+0.0474-0.1118=0.0451\text{ m}^3/\text{s}$$

Answer: Continuity is not satisfied. The assumed junction head of 60 m gives excess inflow, so the true junction head should be higher.

Two Supply Reservoirs to a Common Demand

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.

$$Q_A+Q_B=Q_C$$$$Q_B=0.160-0.095=0.065\text{ m}^3/\text{s}$$

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.

Problem: Three-Reservoir System — Full Iterative Solution

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).

$$Q_A = 0.025\sqrt{100-70} = 0.025(5.477) = 0.1369 \text{ m}^3/\text{s (into J)}$$ $$Q_B = 0.018\sqrt{80-70} = 0.018(3.162) = 0.0569 \text{ m}^3/\text{s (into J)}$$ $$Q_C = 0.022\sqrt{70-50} = 0.022(4.472) = 0.0984 \text{ m}^3/\text{s (out of J)}$$ $$\sum Q_{in} - \sum Q_{out} = 0.1369 + 0.0569 - 0.0984 = +0.0954 \text{ m}^3/\text{s}$$

Positive excess inflow → raise $H_J$. Trial 2: Assume $H_J = 80$ m.

$$Q_A = 0.025\sqrt{20} = 0.1118, \quad Q_B = 0.018\sqrt{0} = 0 \text{ (no flow from B)}$$ $$Q_C = 0.022\sqrt{30} = 0.1205 \text{ m}^3/\text{s}$$ $$\sum Q_{in} - \sum Q_{out} = 0.1118 - 0.1205 = -0.0087 \text{ m}^3/\text{s}$$

Slightly negative → junction head is just below 80 m. Interpolating between trials 1 and 2:

$$H_J \approx 70 + (80-70)\frac{0.0954}{0.0954+0.0087} = 70 + 10(0.916) = 79.2 \text{ m}$$

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.

Problem: Finding an Unknown Reservoir 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:

$$h_{fA} = 120 - H_J = 5000(0.08)^2 = 32 \text{ m} \Rightarrow H_J = 88 \text{ m}$$ $$h_{fC} = H_J - 40 = 88 - 40 = 48 \text{ m}$$ $$Q_C = \sqrt{\frac{48}{4500}} = 0.1033 \text{ m}^3/\text{s (out of J)}$$

Continuity: $Q_A + Q_B = Q_C \Rightarrow Q_B = 0.1033 - 0.08 = 0.0233$ m³/s (into J from B).

$$h_{fB} = R_B Q_B^2 = 3200(0.0233)^2 = 1.735 \text{ m}$$ $$H_B = H_J + h_{fB} = 88 + 1.735 = 89.74 \text{ m}$$

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.

Problem: Hardy Cross — Single Loop Pipe Network Correction

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:

$$\sum h_f = R_1Q_1^2 + R_2Q_2^2 - R_3Q_3^2 - R_4Q_4^2$$ $$= 1000(0.06)^2 + 800(0.04)^2 - 1200(0.03)^2 - 600(0.05)^2$$ $$= 3.60 + 1.28 - 1.08 - 1.50 = +2.30 \text{ m}$$
$$\sum 2R|Q| = 2[1000(0.06) + 800(0.04) + 1200(0.03) + 600(0.05)]$$ $$= 2[60 + 32 + 36 + 30] = 2(158) = 316$$
$$\Delta Q = -\frac{\sum h_f}{\sum 2R|Q|} = -\frac{2.30}{316} = -0.00728 \text{ m}^3/\text{s}$$

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.