07-Str-B11 · May 2013

Question 1 of 6: Two-Reservoir Supply to a Single Demand Node

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations — May 2013 — 07-Str-B11 Hydraulic Engineering. Three-hour, CLOSED-BOOK examination; one 8.5 × 11 in aid sheet (both sides) and one approved Casio or Sharp calculator are permitted. The paper prints six questions of 20 marks each and instructs the candidate to complete any five; where a question has more than one part, the parts carry equal marks. Candidates are urged to state any interpretive assumptions with their answer. All six questions are worked below, because this set is intended as a study resource.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, pipe networks, pump-system curves and quasi-steady reservoir routing; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning's equation, compound and divided channel sections, and the specific-energy and momentum treatment of channel obstructions; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American design practice for transmission mains, minimum service pressures and pump selection; Henderson, F.M., Open Channel Flow (Macmillan) — surges, hydraulic jumps and the unsteady response of a channel to a sudden blockage.

Notation and conventions used throughout. The paper supplies the SI Hazen-Williams form $Q = 0.278\,C\,D^{2.63}S^{0.54}$ with $Q$ in m³/s, $D$ in metres and $S = h_f/L$; it is used exactly as printed. Local losses and velocity head are neglected, as instructed in Note 6, so the hydraulic grade line (HGL) and the energy grade line coincide and "pressure head at a node" means $p/\gamma = \mathrm{HGL} - z$. Water properties are $\rho = 1000$ kg/m³ and $\nu = 1.31\times10^{-6}$ m²/s.

Question 1: Two-Reservoir Supply to a Single Demand Node (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two elevated reservoirs feed a single junction through short 200 mm mains, and the junction draws a modest morning demand.

Given data — Question 1
Quantity	Symbol	Value
Water level, Reservoir 1	$z_1$	96 m
Water level, Reservoir 2	$z_2$	91 m
Demand withdrawn at the node	$Q_d$	20 L/s = 0.020 m³/s
Length, Reservoir 1 to node	$L_1$	200 m
Length, node to Reservoir 2	$L_2$	300 m
Internal diameter, both pipes	$D$	200 mm = 0.200 m
Hazen-Williams coefficient, both pipes	$C$	120

Find. The discharge carried by each of the two mains and the pressure head at the demand node at 7 am.

Figure 1-S. Computed steady state at 7 am. The HGL falls monotonically from Reservoir 1 through the junction to Reservoir 2: the 20 L/s demand is far too small to draw water out of Reservoir 2, so the second main runs as a transfer line and refills it at 31.2 L/s.

Approach. Take the HGL elevation at the junction as the single unknown, write Hazen-Williams for each main with its head loss measured from its own reservoir level, impose continuity at the node, and solve the resulting one-equation problem.

Reduce Hazen-Williams to a single pipe conductance. Both mains share $C$ and $D$, so the leading group is common to them: $$K = 0.278\,C\,D^{2.63} = 0.278(120)(0.200)^{2.63} = 0.4841\ \text{m}^3\text{/s}$$ Each main then obeys $Q = K\,(h_f/L)^{0.54}$, and inverting gives the head loss for a known discharge, $h_f = L\,(Q/K)^{1/0.54} = L\,(Q/K)^{1.852}$.
Test the direction of flow in the second main before assuming it. Figure 1 as printed sketches the HGL sagging below both reservoir levels, which would mean both reservoirs feed the node. Checking that assumption against the numbers, if Reservoir 1 alone supplied the whole 20 L/s then $$h_{f,1} = 200\left(\frac{0.020}{0.4841}\right)^{1.852} = 0.55\ \text{m} \qquad\Rightarrow\qquad H = 96 - 0.55 = 95.45\ \text{m}$$ which sits 4.45 m above Reservoir 2. The junction therefore cannot be a low point on the grade line, and the second main must carry water from the node into Reservoir 2.
Write continuity at the junction with the corrected direction. Inflow from Reservoir 1 equals the demand plus the outflow to Reservoir 2, so with $91\ \text{m} \lt H \lt 96\ \text{m}$ $$K\left(\frac{96-H}{200}\right)^{0.54} = 0.020 + K\left(\frac{H-91}{300}\right)^{0.54}$$ Solving this single non-linear equation by bisection on $H$ gives $$\boxed{H = 92.87\ \text{m}}$$
Back-substitute for the two discharges. The head losses follow immediately as $h_{f,1} = 96 - 92.87 = 3.125$ m over 200 m and $h_{f,2} = 92.87 - 91 = 1.875$ m over 300 m, so $$Q_1 = 0.4841\left(\frac{3.125}{200}\right)^{0.54} = 0.05124\ \text{m}^3\text{/s} = 51.2\ \text{L/s}$$ $$Q_2 = 0.4841\left(\frac{1.875}{300}\right)^{0.54} = 0.03124\ \text{m}^3\text{/s} = 31.2\ \text{L/s}$$ Continuity closes exactly, since $51.24 - 31.24 = 20.00$ L/s is the stated demand.
Convert the junction HGL to a pressure head. With local losses and velocity head neglected the pressure head is the vertical distance from the pipeline up to the HGL, which is precisely what Figure 1 labels $H$. Taking the pipeline invert as the datum to which the 96 m and 91 m water levels are referred, $$\frac{p}{\gamma}\bigg|_{\text{node}} = H - z_{\text{node}} = 92.87 - 0 = \boxed{92.9\ \text{m} \approx 911\ \text{kPa}}$$ If the junction actually sits at some elevation $z_{\text{node}}$ above that datum, subtract it directly; the HGL elevation of 92.87 m is unaffected.

Final results — Question 1
Quantity	Value
HGL elevation at the demand node	92.87 m
Flow in pipe 1 (Reservoir 1 to node)	51.2 L/s, into the node
Flow in pipe 2 (node to Reservoir 2)	31.2 L/s, out of the node, refilling Reservoir 2
Head loss, pipe 1 / pipe 2	3.13 m / 1.87 m
Pressure head at the demand node	92.9 m above the pipeline, about 911 kPa

Check — the printed figure and the printed data disagree, and the data win. Figure 1 shows the HGL sagging to a low point at the junction, the classic picture in which both sources feed the demand. At 20 L/s that cannot happen: the 200 m main from Reservoir 1 loses only 0.55 m while carrying the entire demand, which leaves the junction 4.5 m above Reservoir 2. The sketch is best read as a generic schematic, or as a different and much larger demand hour. The arithmetic above is the state that actually satisfies both continuity and the Hazen-Williams law at 7 am, and flagging the contradiction is exactly the assumption statement Note 1 of the paper invites.

Topic. Steady flow at a two-source pipe junction, the two-reservoir special case of the classical three-reservoir problem.

Key relations. Hazen-Williams in SI, $Q = 0.278\,C\,D^{2.63}(h_f/L)^{0.54}$, inverted as $h_f = L(Q/K)^{1.852}$ with $K = 0.278\,C\,D^{2.63}$; nodal continuity $\sum Q_{\text{in}} = \sum Q_{\text{out}}$; and $p/\gamma = \mathrm{HGL} - z$ when velocity head is neglected.

Why this works. A junction fed by fixed-level reservoirs has exactly one unknown, the HGL elevation there, because every pipe discharge is a monotonic function of that elevation. Raising the junction head reduces the inflow from every reservoir above it and increases the outflow to every reservoir below it, so the net nodal imbalance decreases strictly as the head rises and there is one and only one root. That monotonicity is what makes simple bisection unconditionally reliable here, and it is also the reason the direction of flow in each pipe emerges from the answer rather than having to be assumed in advance.

Common pitfall. Assuming both reservoirs supply the node because the figure is drawn that way, obtaining a negative discharge, and then quietly forcing it positive. The second classic error is applying Hazen-Williams with the diameter in millimetres: the SI form printed on this paper requires metres, and the resulting factor of $1000^{2.63}$ is not a subtle mistake. Always test the no-flow head of each candidate pipe first.

Source. Mays, Water Resources Engineering, 3rd ed., Ch. 4 (pipe flow and the Hazen-Williams equation) and Ch. 5 (pipe networks, nodal head method); Chin, Water-Resources Engineering, 3rd ed., Sections 2.3 to 2.5.

← Paper overview Next: Question 2 →

Quantity	Symbol	Value
Water level, Reservoir 1	\(z_1\)	96 m
Water level, Reservoir 2	\(z_2\)	91 m
Demand withdrawn at the node	\(Q_d\)	20 L/s = 0.020 m³/s
Length, Reservoir 1 to node	\(L_1\)	200 m
Length, node to Reservoir 2	\(L_2\)	300 m
Internal diameter, both pipes	\(D\)	200 mm = 0.200 m
Hazen-Williams coefficient, both pipes	\(C\)	120