07-Str-B11 · May 2013

Question 4 of 6: Quasi-Steady Simulation of Two Draining Elevated Tanks

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 4: Quasi-Steady Simulation of Two Draining Elevated Tanks (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 cylindrical elevated tanks discharge through equal mains to a common node where a half-open valve releases water to atmosphere, and the tanks draw down as the simulation proceeds.

Given data — Question 4
Quantity	Symbol	Value
Initial water level, Tank 1	$Z_1(0)$	96 m
Initial water level, Tank 2	$Z_2(0)$	89 m
Tank diameter (both)	$D_t$	5 m
Plan area of each tank	$A_t$	$\pi(5)^2/4 = 19.635$ m²
Valve discharge coefficient (half open)	$C_v$	0.13 m^5/2/s
Initial steady-state valve discharge	$Q_v(0)$	400 L/s = 0.400 m³/s
Pipe length (both)	$L$	300 m
Pipe internal diameter (both)	$D$	250 mm = 0.250 m
Hazen-Williams coefficient (both)	$C$	110
Simulation time step	$\Delta t$	10 s

Find. The pressure head at the demand node and the discharge in each of the two pipes at the initial condition and after each of the first three 10-second time steps.

Figure 4-S. Initial steady state. Both tanks feed the node, the HGL sags to 69.25 m there, and the valve passes 400 L/s to atmosphere under a pressure head of 9.47 m.

Approach. Establish the initial state by reconciling the valve law with the pipe network, then advance in time by holding the flows constant over each 10-second step while the tank levels fall, and re-solving the steady network at the start of each new step.

Check — the elevation of the pipeline is not printed, and the three givens fix it. Figure 2 measures the node head $H$ from the pipeline up to the HGL but never labels the pipeline elevation, so it must be treated as an unknown $z_p$. The valve law fixes the node pressure head at $H_0 = (0.400/0.13)^2 = 9.47$ m, while the two mains, working from the printed tank levels, can only deliver 400 L/s when the HGL at the node stands at 69.25 m. Those two statements are consistent only if $z_p = 69.25 - 9.47 = 59.78$ m, and with that single value every printed datum — both tank levels, both pipe descriptions, the valve coefficient and the 400 L/s initial flow — is satisfied simultaneously. That elevation is adopted below. Had the pipeline instead been placed at datum, the valve law would demand a node head of 9.47 m at which the mains would deliver about 798 L/s, which is twice the stated initial flow; the printed data would then be self-contradictory.

Set up the pipe conductance and the two governing laws. Both mains are identical, so $$K = 0.278(110)(0.250)^{2.63} = 0.7980\ \text{m}^3\text{/s}$$ and each carries $Q_i = K\left[(Z_i - \mathrm{HGL})/300\right]^{0.54}$. The valve, discharging to atmosphere, obeys $Q_v = C_v\sqrt{H}$ with $H = \mathrm{HGL} - z_p$ the pressure head immediately upstream of it.
Fix the initial condition and the pipeline elevation. From the valve law at $t = 0$, $$H_0 = \left(\frac{Q_v}{C_v}\right)^2 = \left(\frac{0.400}{0.13}\right)^2 = 9.467\ \text{m}$$ and requiring the two mains to deliver exactly 0.400 m³/s from the printed tank levels, $$0.7980\left[\left(\frac{96-\mathrm{HGL}}{300}\right)^{0.54} + \left(\frac{89-\mathrm{HGL}}{300}\right)^{0.54}\right] = 0.400 \quad\Rightarrow\quad \mathrm{HGL}_0 = 69.248\ \text{m}$$ $$\boxed{z_p = 69.248 - 9.467 = 59.78\ \text{m}}$$
Split the initial flow between the two mains. Substituting $\mathrm{HGL}_0 = 69.248$ m back into each pipe equation gives the head losses $96 - 69.248 = 26.752$ m and $89 - 69.248 = 19.752$ m, hence $$Q_1 = 0.7980\left(\frac{26.752}{300}\right)^{0.54} = 0.21634\ \text{m}^3\text{/s} = 216.3\ \text{L/s}$$ $$Q_2 = 0.7980\left(\frac{19.752}{300}\right)^{0.54} = 0.18366\ \text{m}^3\text{/s} = 183.7\ \text{L/s}$$ and the sum is 400.0 L/s, matching the given valve discharge exactly.
Write the quasi-steady stepping rule. Under the quasi-steady assumption the pipe flows are governed at every instant by the steady equations above, and only the tank levels carry memory. Volume conservation in each cylindrical tank over one step gives $$Z_i(t + \Delta t) = Z_i(t) - \frac{Q_i(t)\,\Delta t}{A_t}, \qquad A_t = \frac{\pi (5)^2}{4} = 19.635\ \text{m}^2$$ so with $\Delta t = 10$ s the level drop is $Q_i \times 10/19.635 = 0.5093\,Q_i$ metres per step.
Advance to $t = 10$ s. The levels fall by $0.5093(0.21634) = 0.1102$ m and $0.5093(0.18366) = 0.0935$ m, giving $Z_1 = 95.890$ m and $Z_2 = 88.907$ m. Re-solving the nodal continuity equation $$0.7980\left[\left(\frac{Z_1-\mathrm{HGL}}{300}\right)^{0.54} + \left(\frac{Z_2-\mathrm{HGL}}{300}\right)^{0.54}\right] = 0.13\sqrt{\mathrm{HGL} - 59.78}$$ gives $\mathrm{HGL} = 69.217$ m, hence $\boxed{H = 9.436\ \text{m}}$, with $Q_1 = 216.0$ L/s, $Q_2 = 183.3$ L/s and $Q_v = 399.3$ L/s.
Advance to $t = 20$ s. Repeating the level update from the $t = 10$ s flows gives $Z_1 = 95.780$ m and $Z_2 = 88.813$ m; the same nodal equation then returns $\mathrm{HGL} = 69.186$ m, so $H = 9.405$ m, $Q_1 = 215.7$ L/s, $Q_2 = 183.0$ L/s and $Q_v = 398.7$ L/s.
Advance to $t = 30$ s. One further step gives $Z_1 = 95.670$ m and $Z_2 = 88.720$ m, from which $\mathrm{HGL} = 69.155$ m, $H = 9.374$ m, $Q_1 = 215.3$ L/s, $Q_2 = 182.7$ L/s and $Q_v = 398.0$ L/s. This completes the three requested steps.
Sanity-check the rate of change. The node head falls by a steady 0.031 m and the valve discharge by 0.66 L/s per step, so after 30 s the system has drifted only 0.8 per cent from its initial state. That slow drift is the justification for the quasi-steady assumption itself: the tanks hold enough water that inertial and elastic effects in the mains, which act over seconds, have long since died out relative to the minutes over which the tanks empty. Tank 1 drains faster than Tank 2 because it starts 7 m higher, so the two levels converge; the simulation would eventually reach a state in which Tank 2 stops contributing and, later still, begins to be refilled by Tank 1.

Final results — Question 4, quasi-steady simulation with $\Delta t$ = 10 s
Time (s)	$Z_1$ (m)	$Z_2$ (m)	HGL at node (m)	Pressure head $H$ (m)	$Q_1$ (L/s)	$Q_2$ (L/s)	$Q_v$ (L/s)
0 (initial)	96.000	89.000	69.248	9.467	216.3	183.7	400.0
10 (step 1)	95.890	88.907	69.217	9.436	216.0	183.3	399.3
20 (step 2)	95.780	88.813	69.186	9.405	215.7	183.0	398.7
30 (step 3)	95.670	88.720	69.155	9.374	215.3	182.7	398.0
Pipeline and node elevation $z_p$ = 59.78 m throughout; tank plan area $A_t$ = 19.635 m².

Topic. Quasi-steady (rigid-column, extended-period) simulation of a small distribution system with storage, and the reconciliation of a valve characteristic with a pipe network.

Key relations. Valve discharging to atmosphere, $Q_v = C_v\sqrt{H}$; Hazen-Williams for each main, $Q_i = K\left[(Z_i-\mathrm{HGL})/L\right]^{0.54}$; nodal continuity $Q_1 + Q_2 = Q_v$; tank storage $A_t\,\mathrm{d}Z_i/\mathrm{d}t = -Q_i$, integrated explicitly as $Z_i^{k+1} = Z_i^{k} - Q_i^{k}\Delta t/A_t$; and $H = \mathrm{HGL} - z_p$.

Why this works. A quasi-steady or extended-period simulation exploits a separation of time scales. Pressure waves cross a 300 m main in well under a second and the water column accelerates over a few seconds, whereas a 19.6 m² tank losing 0.2 m³/s changes its level by only a centimetre per second. Over any interval long compared with the first time scale and short compared with the second, the pipes may be treated as being in instantaneous steady state while the tanks provide all the memory. The result is a sequence of independent steady network solutions coupled only through the storage update, which is why the whole simulation reduces to solving one scalar equation per time step. The scheme is explicit first-order Euler, so its error falls linearly with the time step; halving $\Delta t$ would halve the small lag introduced by holding the flows constant across each step.

Common pitfall. Confusing the HGL elevation with the pressure head at the node, which is the distinction the whole valve calculation turns on. A close second is updating the tank levels with the new flows rather than the flows that actually applied during the step just completed, which makes the scheme implicit without any of the stability benefit. A third is forgetting that a half-open valve has its own fixed coefficient in this formulation, so that the outflow varies only with the square root of the node head and not with the tank levels directly.

Source. Mays, Water Resources Engineering, 3rd ed., Ch. 5 (network simulation and extended-period simulation); Chin, Water-Resources Engineering, 3rd ed., Sections 2.5 and 2.7 (network analysis, storage reservoirs and valves).

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Quantity	Symbol	Value
Initial water level, Tank 1	\(Z_1(0)\)	96 m
Initial water level, Tank 2	\(Z_2(0)\)	89 m
Tank diameter (both)	\(D_t\)	5 m
Plan area of each tank	\(A_t\)	\(\pi(5)^2/4 = 19.635\) m²
Valve discharge coefficient (half open)	\(C_v\)	0.13 m^5/2/s
Initial steady-state valve discharge	\(Q_v(0)\)	400 L/s = 0.400 m³/s
Pipe length (both)	\(L\)	300 m
Pipe internal diameter (both)	\(D\)	250 mm = 0.250 m
Hazen-Williams coefficient (both)	\(C\)	110
Simulation time step	\(\Delta t\)	10 s