07-Str-B11 · December 2015

Question 4 of 6: Quasi-Steady Simulation of Two Tanks Feeding a Valved Demand Node

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

Notes on this paper

Paper format. National Exams, December 2015 — 07-Str-B11, Hydraulic Engineering. Three hours, closed book; one 8.5 × 11 inch aid sheet (both sides) and any non-communicating calculator are permitted. Six questions of twenty marks each; candidates complete any five, and where a question has parts the parts carry equal weight. The solutions below work all six, so the paper can be used whichever five a reader chooses.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, series/parallel equivalent pipes, looped-network analysis and extended-period simulation; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, service-pressure criteria, control-valve characteristics and gutter/roadway hydraulics; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation and compound cross-sections; Henderson, F.M., Open Channel Flow (Macmillan) — the Saint-Venant equations, wave celerity and the kinematic/diffusion/dynamic hierarchy; Transportation Association of Canada, Geometric Design Guide for Canadian Roads — crossfall, curb reveal and roadway drainage conventions.

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. Writing $K = 0.278\,C\,D^{2.63}$ inverts it to the head-loss form $h_f = L\,(Q/K)^{1.852}$, the exponent $n = 1.852$ being the value the paper itself quotes in its loop-correction note. Following Note 6, local losses and velocity head are neglected, 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 is taken as $\rho = 1000$ kg/m³ with $\nu = 1.31\times10^{-6}$ m²/s.

Question 4: Quasi-Steady Simulation of Two Tanks Feeding a Valved 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 cylindrical elevated tanks, each 5 m in diameter, connected by 300 m pipes to a common demand node whose valve discharges freely to atmosphere. As the tanks drain, their levels fall and the flows change — but slowly enough that each instant can be treated as a steady-state network solve.

Given data — Question 4
Quantity	Symbol	Value
Tank diameter (both)	$D_t$	5 m → plan area $A_t$ = 19.635 m²
Initial level, Tank 1	$H_1(0)$	96 m
Initial level, Tank 2	$H_2(0)$	89 m
Pipe length, each tank to node	$L$	300 m
Pipe diameter	$D$	250 mm = 0.250 m
Hazen-Williams coefficient	$C$	110
Valve discharge constant	$E_v$	0.10 m^5/2/s
Valve opening (half open)	$\tau$	0.50
Initial steady flow through the valve	$Q_v(0)$	300 L/s = 0.300 m³/s
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 end of each of the first three 10-second time steps.

Figure 4.1 — The supply system. Both tanks drain into a common node; the HGL falls from each tank surface to the node, and the pressure head $H$ there drives the free discharge through the valve.

Approach. Quasi-steady (extended-period) simulation alternates two operations. At each instant, the tank levels are treated as fixed reservoirs and a steady network solve returns the node head and the two pipe flows; those flows are then held constant over the time step and used to lower the tank levels by continuity. Repeating gives the transient. The initial 300 L/s is used first to fix the valve datum, which the paper does not otherwise supply.

Establish the pipe resistance and the tank area. Both pipes share $C = 110$, $D = 0.250$ m as in Question 3, but are 300 m long: $$K = 0.79803\ \mathrm{m^3/s}, \qquad R = \frac{300}{0.79803^{1.852}} = 455.60$$ so $h_f = 455.60\,Q^{1.852}$ in each pipe. Each tank has plan area $A_t = \tfrac{\pi}{4}(5)^2 = 19.635$ m².
Fix the node head at $t = 0$ from pipe continuity. Both pipes deliver to the same node, so they share a common HGL $h$ there, and the two flows must sum to the 300 L/s passing the valve: $$K\left(\frac{96-h}{300}\right)^{1/1.852} + K\left(\frac{89-h}{300}\right)^{1/1.852} = 0.300$$ Solving gives $h = 78.72$ m, and back-substituting, $$\boxed{Q_1(0) = 170.9\ \mathrm{L/s}, \qquad Q_2(0) = 129.1\ \mathrm{L/s}}$$ Tank 1 supplies the larger share because it stands 7 m higher; the split is not proportional to the head difference because of the 1.852 exponent.
Fix the valve datum from the given initial discharge. The valve is half open, so $\tau = 0.5$, and it discharges to atmosphere, so $H_{d/s} = 0$. The valve equation then gives the pressure head at the node directly: $$H(0) = \left(\frac{Q_v}{\tau E_v}\right)^2 = \left(\frac{0.300}{0.5 \times 0.10}\right)^2 = 6.00^2$$ $$\boxed{H(0) = 36.0\ \mathrm{m}}$$ Since the HGL at the node is 78.72 m and its pressure head is 36.0 m, the node itself sits at $z = 78.72 - 36.00 = 42.72$ m. That elevation is never printed on the paper; the stated initial discharge is what determines it, and it stays fixed for the rest of the simulation.
Write the governing equations for a general time step. With $C_v = \tau E_v = 0.05$ m^5/2/s and $z = 42.72$ m, the node balance at any instant is one equation in the single unknown $h$: $$K\left(\frac{H_1-h}{300}\right)^{0.53996} + K\left(\frac{H_2-h}{300}\right)^{0.53996} = C_v\sqrt{h-z}$$ and the tank levels then march forward by continuity over the step: $$H_1(t+\Delta t) = H_1(t) - \frac{Q_1\,\Delta t}{A_t}, \qquad H_2(t+\Delta t) = H_2(t) - \frac{Q_2\,\Delta t}{A_t}$$ This is the explicit (forward-Euler) form of the quasi-steady scheme: the flows computed at the start of a step are held constant across it.
Advance to $t = 10$ s. Using the $t = 0$ flows, the level drops are $\Delta H_1 = 0.1709 \times 10/19.635 = 0.087$ m and $\Delta H_2 = 0.1291 \times 10/19.635 = 0.066$ m, giving $H_1 = 95.913$ m and $H_2 = 88.934$ m. Re-solving the node balance at these levels, $$h = 78.662\ \mathrm{m} \;\Longrightarrow\; H = 78.662 - 42.718 = 35.94\ \mathrm{m}$$ $$Q_1 = 170.7\ \mathrm{L/s}, \qquad Q_2 = 129.0\ \mathrm{L/s}, \qquad Q_v = 299.8\ \mathrm{L/s}$$

Advance to $t = 20$ s and $t = 30$ s. Repeating the same two operations twice more produces the table below. Because the head losses are small compared with the 36 m of pressure head at the node, the system responds gently: over 30 seconds the discharge falls by only 0.7 L/s.

Quasi-steady simulation, $\Delta t$ = 10 s
$t$ (s)	$H_1$ (m)	$H_2$ (m)	Node HGL (m)	Node pressure head $H$ (m)	$Q_1$ (L/s)	$Q_2$ (L/s)	$Q_v$ (L/s)
0	96.000	89.000	78.718	36.00	170.9	129.1	300.0
10	95.913	88.934	78.662	35.94	170.7	129.0	299.8
20	95.826	88.869	78.607	35.89	170.6	129.0	299.5
30	95.739	88.803	78.551	35.83	170.4	128.9	299.3

Check the physics of the trend. Tank 1 falls 0.261 m over the three steps while Tank 2 falls only 0.197 m, so the 7.000 m gap between them has narrowed to 6.936 m. That is the expected behaviour: the higher tank works harder and therefore empties faster, and the two levels converge. The node HGL falls monotonically, the pressure head with it, and the valve discharge follows through $Q_v = 0.05\sqrt{H}$. Nothing in the simulation approaches a reversal — both tanks continue to supply the node throughout, since the node HGL of 78.6 m remains far below both tank levels.

Check: the valve datum is inferred, not given. The paper supplies the tank water-surface elevations (96 m and 89 m) but no elevation for the demand node, while simultaneously specifying the initial valve discharge (300 L/s). These are consistent only if the node sits at $z$ = 42.72 m, and that is the value adopted here. The alternative reading — that the quoted 0.10 m^5/2/s is already the half-open coefficient, so that $Q_v = 0.10\sqrt{H}$ — gives $H(0)$ = 9.00 m and $z$ = 69.72 m. Note that the pipe flows, the node HGL and the tank-level history are identical under either reading, because both are calibrated to the same 300 L/s; only the reported pressure head shifts, from 36.0 m to 9.0 m. The reading used above follows the paper’s own valve equation from Question 2, in which $\tau$ and $E_s$ appear as separate factors.

Final results — Question 4
Quantity	$t$ = 0 s	$t$ = 10 s	$t$ = 20 s	$t$ = 30 s
Pressure head at the demand node (m)	36.00	35.94	35.89	35.83
Flow in pipe 1, from Tank 1 (L/s)	170.9	170.7	170.6	170.4
Flow in pipe 2, from Tank 2 (L/s)	129.1	129.0	129.0	128.9
Valve discharge (L/s)	300.0	299.8	299.5	299.3
Tank 1 water level (m)	96.000	95.913	95.826	95.739
Tank 2 water level (m)	89.000	88.934	88.869	88.803

Topic. Extended-period (quasi-steady) simulation of a network with storage; explicit level-marching by tank continuity.

Key relations. Node continuity $\sum Q_{\mathrm{in}} = Q_{\mathrm{valve}}$ with a common nodal HGL; pipe head loss $h_f = R\,Q^{1.852}$; free-discharge valve $Q_v = \tau E_v\sqrt{H}$; tank continuity $\mathrm{d}H/\mathrm{d}t = -Q/A_t$, discretised as $H^{k+1} = H^{k} - Q^{k}\Delta t/A_t$.

Why this works. The quasi-steady assumption says that the water in the pipes reaches its steady distribution far faster than the tank levels change, so the inertia of the pipe columns can be dropped and the momentum equation reduces to the ordinary steady head-loss law. The justification is a comparison of time scales: the pipes here are 300 m long and the tanks have a drawdown time scale of hours, so the pipe transit and acceleration times of seconds are negligible by comparison. The only genuinely time-dependent equation left is the tank storage balance, which is a simple first-order ODE integrated explicitly. This is exactly how EPANET and comparable packages run an extended-period simulation, and the same caveat applies: the time step must be short enough that the flows do not change appreciably within it. Here 10 s changes the flows by roughly 0.1 %, so the explicit scheme is comfortably accurate.

Common pitfall. Updating both tanks with the same flow, or with half the valve discharge each — the tanks drain at different rates precisely because they stand at different levels, and using $Q_v/2$ destroys the convergence of the two levels that is the interesting feature of the result. A second pitfall is confusing the nodal HGL with the nodal pressure head; the valve equation is driven by pressure head above the outlet, not by the HGL on the survey datum.

Source. Mays, Water Resources Engineering, 3rd ed., Ch. 4 and Ch. 5 (network simulation, storage tanks and extended-period analysis); Chin, Water-Resources Engineering, 3rd ed., §2.5 and §7.3 (storage-node continuity, EPANET-style extended-period simulation).

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Quantity	Symbol	Value
Tank diameter (both)	\(D_t\)	5 m → plan area \(A_t\) = 19.635 m²
Initial level, Tank 1	\(H_1(0)\)	96 m
Initial level, Tank 2	\(H_2(0)\)	89 m
Pipe length, each tank to node	\(L\)	300 m
Pipe diameter	\(D\)	250 mm = 0.250 m
Hazen-Williams coefficient	\(C\)	110
Valve discharge constant	\(E_v\)	0.10 m^5/2/s
Valve opening (half open)	\(\tau\)	0.50
Initial steady flow through the valve	\(Q_v(0)\)	300 L/s = 0.300 m³/s
Simulation time step	\(\Delta t\)	10 s