Question 2 of 6: Transmission Main with an In-Line Control Valve
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 2: Transmission Main with an In-Line Control Valve (20 marks)
Given. A single 450 mm main, 5,000 m long, running between two fixed-level reservoirs with a throttling valve 4,000 m along its length (Figure 2 of the exam paper).
Given data — Question 2
Quantity
Symbol
Value
Pipeline length (total)
\(L\)
5,000 m (4,000 m upstream of the valve, 1,000 m downstream)
Inner diameter
\(D\)
450 mm = 0.450 m
Hazen-Williams coefficient
\(C\)
120
Upstream reservoir level
\(h_A\)
105 m
Downstream reservoir level, part (a)
\(h_B\)
95 m
Downstream reservoir level, part (b)
\(h_B\)
85 m
Valve discharge constant
\(E_s\)
0.45 m5/2/s
Valve opening ratio, part (a)
\(\tau\)
0.70
Find. (a) the steady discharge with \(\tau = 0.70\) and a 10 m reservoir difference; (b) the opening ratio \(\tau\) that holds that same discharge once the reservoir difference widens to 20 m, together with the physical argument for whether the valve opens or closes.
Figure 2.1 — Transmission main and hydraulic grade line for part (a). The HGL falls linearly through the 4,000 m of pipe upstream of the valve, steps down by the valve loss, then falls again through the remaining 1,000 m to the downstream reservoir surface.
Approach. The reservoir difference is shared between pipe friction over the full 5,000 m and the local loss across the valve. Writing both losses in terms of the single unknown \(Q\) gives one non-linear equation to solve for part (a); part (b) reverses the logic, fixing \(Q\) and solving the valve equation for \(\tau\).
Establish the pipe resistance. Using the same inversion as in Question 1,
$$K = 0.278 \times 120 \times (0.450)^{2.63} = 33.36 \times 0.122444 = 4.0848\ \mathrm{m^3/s}$$
so that friction over the whole main is \(h_f = 5000\,(Q/4.0848)^{1.852}\). Because the valve is a lumped local loss, it makes no difference to the friction term whether it sits 4,000 m or 400 m along the pipe; the split only matters when the head at the valve itself is wanted.
Write the valve loss in terms of the same unknown. Squaring the supplied valve equation gives the head the valve destroys:
$$\Delta H_{\mathrm{valve}} = H_{u/s} - H_{d/s} = \left(\frac{Q}{\tau E_s}\right)^{2} = \left(\frac{Q}{0.7 \times 0.45}\right)^{2} = \left(\frac{Q}{0.315}\right)^{2}$$
a conventional orifice-type law in which loss grows with the square of discharge.
Impose the energy balance between the two reservoir surfaces. Neglecting velocity head and entry/exit losses per Note 6, the entire elevation difference is consumed by the pipe and the valve:
$$h_A - h_B = h_f + \Delta H_{\mathrm{valve}} \;\Longrightarrow\; 105 - 95 = 5000\left(\frac{Q}{4.0848}\right)^{1.852} + \left(\frac{Q}{0.315}\right)^{2} = 10\ \mathrm{m}$$
This is one equation in one unknown, non-linear because the two terms carry different exponents.
Solve for the discharge. Trial values bracket the root quickly: at \(Q = 0.14\) m3/s the two losses total 9.87 m, and at \(Q = 0.142\) m3/s they total 10.13 m. Interpolating and refining gives
$$\boxed{Q = 0.141\ \mathrm{m^3/s} = 141\ \mathrm{L/s}}$$
which answers part (a). The corresponding mean velocity is \(V = 0.141/0.15904 = 0.886\) m/s, entirely typical of a transmission main.
Check how the 10 m is actually shared. At \(Q = 0.14096\) m3/s the friction loss is \(h_f = 9.80\) m and the valve loss is only \(\Delta H_{\mathrm{valve}} = (0.14096/0.315)^2 = 0.20\) m, and \(9.80 + 0.20 = 10.00\) m as required. The heads immediately either side of the valve are therefore
$$H_{u/s} = 105 - 4000\left(\tfrac{0.14096}{4.0848}\right)^{1.852} = 97.16\ \mathrm{m}, \qquad H_{d/s} = 95 + 1000\left(\tfrac{0.14096}{4.0848}\right)^{1.852} = 96.96\ \mathrm{m}$$
Even at \(\tau = 0.7\) the valve is barely working: the pipeline itself is doing 98 % of the throttling. That observation is what makes part (b) behave the way it does.
Reason out the direction of the change in part (b). Dropping \(h_B\) from 95 m to 85 m doubles the driving head to 20 m while leaving the pipeline unchanged. If the valve setting were left at \(\tau = 0.7\), the extra head would simply accelerate the flow. To hold \(Q\) at its part-(a) value the system must dissipate an extra 10 m of head, and since the pipe friction is fixed once \(Q\) is fixed, every metre of that extra dissipation has to be taken by the valve. A valve destroys more head only by presenting a smaller opening, so
$$\boxed{\text{the valve must be closed further } (\tau \text{ decreases})}$$
Compute the new opening ratio. Holding \(Q = 0.14096\) m3/s keeps \(h_f = 9.80\) m, so the valve must now absorb
$$\Delta H_{\mathrm{valve}} = (105 - 85) - 9.80 = 10.20\ \mathrm{m}$$
Rearranging the valve equation for the opening ratio,
$$\tau = \frac{Q}{E_s\sqrt{\Delta H_{\mathrm{valve}}}} = \frac{0.14096}{0.45\sqrt{10.20}} = \frac{0.14096}{1.4371}$$
$$\boxed{\tau = 0.098}$$
The valve goes from 70 % of its fully-open characteristic to under 10 % — a drastic movement, and a direct consequence of Step 5: because the valve was contributing so little loss at \(\tau = 0.7\), it has to be throttled almost shut before it can dominate the energy budget. The valve loss rises fiftyfold, from 0.20 m to 10.20 m, while the flow does not change at all.
Check: operational note. A setting of \(\tau \approx 0.10\) puts the valve deep into the range where its characteristic curve is steep and poorly conditioned, where cavitation is likely given the 10 m head break across it, and where seat erosion is rapid. In practice a designer confronted with this duty would not throttle a single main valve to 10 %; the head would be broken across a dedicated pressure-reducing station, a multi-stage or anti-cavitation trim, or an orifice plate in series with a more modestly closed valve. The 0.098 answer is the correct response to the question as posed, but it is a diagnosis of a badly-matched control element, not a recommended set-point.
