07-Str-B11 · May 2016

Question 3 of 6: Transmission Main with a Throttling Valve

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

Notes on this paper

Paper format. National Exams, May 2016 — 07-Str-B11, Hydraulic Engineering. Three hours, closed book; one 8.5 × 11 inch aid sheet (both sides may be used) and any non-communicating calculator are permitted. The paper prints six questions of equal value, and a candidate completes any five of them; where a question has parts, the parts also carry equal value. All six are worked below, so the set is useful whichever five a reader chooses.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, branched and looped network analysis, control valves and extended-period simulation; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, maximum-day-plus-fire-flow loading and service-pressure criteria; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation, specific energy and channel transitions; Henderson, F.M., Open Channel Flow (Macmillan) — critical flow, controls and surface-profile classification; White, F.M., Fluid Mechanics (7th ed., McGraw-Hill) — the pipe force balance, the friction-factor definition and laminar film flow.

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 the cover-page 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 (the cover page prints the exponent without its minus sign, which is an obvious typesetting slip — $1.31\times10^{6}$ m²/s is not a physical viscosity).

Source note — the Question 4 schematic is not reproduced in the examination paper. Question 4 refers to “the figure below”, but no figure is printed between Question 4 and Question 5 on page 3 of the paper, and page 4 opens with the continuation of Question 5. The network is therefore assumed to be a single loop: a source reservoir feeding node N1 through pipe P1 and node N2 through pipe P2, with pipe P3 closing the loop between N1 and N2. That topology is adopted here and is stated explicitly in the Given.

Question 3: Transmission Main with a Throttling Valve (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. A single pipe joins two reservoirs, with a throttling valve 4,000 m along it and 1,000 m short of the downstream end.

Given data — Question 3
Quantity	Symbol	Value
Total pipeline length	$L$	5,000 m (4,000 m + 1,000 m about the valve)
Hazen-Williams coefficient	$C$	110
Inner diameter	$D$	1,067 mm = 1.067 m
Upstream reservoir level	$h_A$	105 m
Valve discharge constant	$E_s$	0.35 m^5/2/s
Discharge, parts (a) and (b)	$Q$	1 m³/s
Head loss across the valve, part (a)	$\Delta H_v$	5 m
Valve setting, part (c)	$\tau$	0.3

Find. The valve opening parameter that produces the stated loss at the stated discharge; the downstream reservoir level that the resulting energy balance implies; and the new discharge after the valve is throttled to $\tau = 0.3$ with the downstream level held fixed.

Figure 3.1 — The transmission main and its hydraulic grade line for the steady state of parts (a) and (b). The grade line falls at a constant friction slope along the pipe and drops vertically by 5 m across the valve; the downstream reservoir surface is whatever level closes the balance.

Approach. Part (a) is a direct inversion of the supplied valve equation. Part (b) adds the pipe friction loss to the valve loss and subtracts the total from the upstream level. Part (c) is the same energy balance read backwards: with both reservoir levels and the valve setting fixed, the available head is shared between a friction loss and a valve loss that both grow with discharge, so the discharge is found by solving a single nonlinear equation.

Part (a) — invert the valve equation for the opening parameter. The valve equation as printed is $Q = \tau E_s \sqrt{H_{u/s} - H_{d/s}}$, in which the radicand is the head loss across the valve itself. With $Q = 1$ m³/s and a 5 m loss, $$\tau = \frac{Q}{E_s\sqrt{\Delta H_v}} = \frac{1}{0.35\sqrt{5}} = \frac{1}{0.78262}$$ $$\boxed{\tau = 1.278}$$ The value exceeds unity, which simply reflects how this paper has scaled $E_s$: $\tau$ here is a dimensionless multiplier on a reference capacity rather than a fractional opening bounded by one. What matters for the rest of the question is the direction of travel, and part (c) closes the valve further by lowering $\tau$ to 0.3.
Part (b) — establish the pipe resistance. Before the energy balance can be closed, the friction loss over the whole 5,000 m must be evaluated. With $K = 0.278\,C\,D^{2.63}$, $$K = 0.278 \times 110 \times (1.067)^{2.63} = 30.58 \times 1.185967 = 36.267\ \text{m}^3\text{/s}$$ so that $$h_f = 5000\left(\frac{Q}{36.267}\right)^{1.852} = 6.4678\,Q^{1.852}$$ At the stated discharge of 1 m³/s this evaluates to $h_f = 6.468$ m. The mean velocity is $V = Q/A = 1/0.8942 = 1.118$ m/s, a sensible transmission-main figure, and the friction slope is 1.29 m per kilometre.
Part (b) — close the energy equation between the two reservoir surfaces. Both surfaces are at atmospheric pressure and at rest, so the entire difference in level is consumed by the two losses in series: $$h_A - h_B = h_f + \Delta H_v$$ $$h_B = 105 - 6.468 - 5 = 93.532$$ $$\boxed{h_B = 93.53\ \text{m}}$$ The split is instructive: 6.47 m of the 11.47 m total is spent on 5 km of pipe and 5.00 m on a single valve, so the partially closed valve is already doing more than the whole pipeline to control the flow.
Part (c) — set up the balance for the throttled valve. Closing the valve to $\tau = 0.3$ leaves the reservoirs where they are, so the available head is unchanged at $105 - 93.532 = 11.472$ m, but it must now be shared differently. Rearranging the valve equation for its head loss at a given discharge, $$\Delta H_v = \left(\frac{Q}{\tau E_s}\right)^{2} = \left(\frac{Q}{0.3 \times 0.35}\right)^{2} = \left(\frac{Q}{0.105}\right)^{2}$$ and adding the friction loss from Step 2 gives one equation in one unknown: $$6.4678\,Q^{1.852} + \left(\frac{Q}{0.105}\right)^{2} = 11.472$$ Both terms increase with $Q$, so the left-hand side is strictly increasing and the root is unique.
Part (c) — solve for the throttled discharge. Because the valve term dominates, a good first estimate ignores friction: $Q \approx 0.105\sqrt{11.472} = 0.3556$ m³/s. Substituting that back gives $h_f = 0.955$ m, leaving 10.52 m for the valve and a revised $Q = 0.3405$ m³/s; one further pass gives 0.3416, and the iteration has converged. Solving the equation numerically confirms $$\boxed{Q = 0.342\ \text{m}^3\text{/s}\ \ (342\ \text{L/s})}$$ with the head split as $h_f = 0.885$ m in the pipe and $\Delta H_v = 10.586$ m across the valve, which sum to the 11.472 m available.
Interpret the control behaviour. Throttling $\tau$ from 1.278 to 0.3, a factor of 4.26, has cut the discharge only from 1.000 to 0.342 m³/s, a factor of 2.92. The valve is not a proportional device: as it closes it takes over more and more of the available head, and once it holds nearly all of it the discharge varies as $\tau$ itself. Here the valve’s share of the loss has risen from 44 per cent to 92 per cent, and the mean pipe velocity has fallen to 0.382 m/s. That is the classic argument against flow control by throttling on a long main — almost all of the available energy ends up being destroyed in the valve seat rather than used.

Final results — Question 3
Quantity	Result
Pipe resistance	$h_f = 6.468\,Q^{1.852}$ (SI)
(a) Valve parameter at the 5 m loss	τ = 1.278
(b) Friction loss over 5,000 m at 1 m³/s	6.468 m
(b) Downstream reservoir level	93.53 m
(c) Discharge with τ = 0.3	0.342 m³/s (342 L/s)
(c) Head split, pipe / valve	0.885 m / 10.586 m

Topic. Series analysis of a pipeline with a throttling control valve, and the valve characteristic expressed as an equivalent orifice.

Key relations. The valve law $Q = \tau E_s\sqrt{\Delta H_v}$, equivalently $\Delta H_v = (Q/\tau E_s)^2$; Hazen-Williams as $h_f = L(Q/K)^{1.852}$; and the reservoir-to-reservoir energy equation $h_A - h_B = \sum h_{loss}$, which for two elements in series is $h_f + \Delta H_v$.

Why this works. Everything rests on the fact that the two reservoir surfaces are fixed-head boundaries: the pressure is atmospheric and the velocity negligible at both, so the total energy difference between them is exactly the sum of the losses along whatever path joins them. Losses in series simply add, and each is a known increasing function of the single discharge, which makes the system determinate. The two functions have different exponents — 1.852 for the pipe, 2 for the valve — so the head is shared in a proportion that depends on the operating point, and the equation must be solved rather than inverted term by term. Any monotone root-finder converges immediately because the total loss increases strictly with discharge.

Common pitfall. The commonest error in part (c) is to reuse the 5 m valve loss from part (a), which belongs to the old setting and the old discharge; the valve loss is a variable, not a datum. The second is to interpret $H_{u/s} - H_{d/s}$ in the valve equation as the difference between the two reservoir levels rather than the drop across the valve alone, which double-counts the friction loss. A third is to force $\tau$ to lie between zero and one and conclude that part (a) is wrong; the constant $E_s$ supplied here is simply not scaled to make $\tau$ a fractional opening.

Source. Mays, Water Resources Engineering 3e, Ch. 4 and Ch. 9 (valves and system curves); Chin, Water-Resources Engineering 3e, §2.4 and §7.5 for control-valve characteristics.

← Question 2 Next: Question 4 →

Quantity	Symbol	Value
Total pipeline length	\(L\)	5,000 m (4,000 m + 1,000 m about the valve)
Hazen-Williams coefficient	\(C\)	110
Inner diameter	\(D\)	1,067 mm = 1.067 m
Upstream reservoir level	\(h_A\)	105 m
Valve discharge constant	\(E_s\)	0.35 m^5/2/s
Discharge, parts (a) and (b)	\(Q\)	1 m³/s
Head loss across the valve, part (a)	\(\Delta H_v\)	5 m
Valve setting, part (c)	\(\tau\)	0.3

Quantity	Result
Pipe resistance	\(h_f = 6.468\,Q^{1.852}\) (SI)
(a) Valve parameter at the 5 m loss	τ = 1.278
(b) Friction loss over 5,000 m at 1 m³/s	6.468 m
(b) Downstream reservoir level	93.53 m
(c) Discharge with τ = 0.3	0.342 m³/s (342 L/s)
(c) Head split, pipe / valve	0.885 m / 10.586 m