Question 3 of 6: Transmission Main Throttled by an In-Line Valve
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations — May 2014 — 07-Str-B11 Hydraulic Engineering. Three-hour, CLOSED-BOOK examination; one 8.5 × 11 in aid sheet (both sides) and any non-communicating 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 submit a clear statement of any interpretive assumptions with the answer paper. 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, branched and looped network analysis, transmission mains and valve characteristics; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, fire-flow criteria, minimum service pressures and gutter hydraulics; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation and the momentum treatment of the hydraulic jump; Henderson, F.M., Open Channel Flow (Macmillan) — specific force, conjugate depths and jump classification; White, F.M., Fluid Mechanics (7th ed., McGraw-Hill) — the control-volume force balance that links wall shear stress to the Darcy friction factor.
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. 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 3: Transmission Main Throttled by an In-Line Valve (20 marks)
Given. A single 5,000 m transmission main between two reservoirs, with a throttling valve 4,000 m from the upstream end (Figure 3 of the paper).
Given data — Question 3
Quantity
Symbol
Value
Pipeline length (total)
\(L\)
5,000 m (4,000 m upstream of the valve)
Inner diameter
\(D\)
1,167 mm = 1.167 m
Hazen-Williams coefficient
\(C\)
100
Upstream reservoir water level
\(h_A\)
105 m
Valve discharge constant
\(E_s\)
0.35 m5/2/s
Steady discharge, part (a) and (b)
\(Q\)
0.92 m3/s
Head loss across the valve, part (a)
\(\Delta h_v\)
6 m
Valve setting, part (c)
\(\tau\)
0.3
Find. (a) the valve opening parameter \(\tau\); (b) the downstream reservoir level \(h_B\); and (c) the discharge once \(\tau\) is reduced to 0.3 with \(h_B\) held fixed.
Figure 3.1 — Energy budget for the throttled transmission main at \(Q = 0.92\) m3/s. The reservoir difference is spent on 4.27 m of pipe friction distributed along the full 5,000 m plus a 6.0 m step loss concentrated at the valve; the valve step is drawn at its true position, 4,000 m from the upstream reservoir.
Approach. The valve equation is a local orifice-type law that fixes the loss across the valve for a given \(\tau\) and \(Q\), while Hazen-Williams fixes the distributed loss along the main; the reservoir difference must equal their sum, and each part of the question rearranges that single energy statement for a different unknown.
Rearrange the supplied valve equation for the opening parameter. The head difference in the valve equation is the loss across the valve itself, \(\Delta h_v = H_{u/s} - H_{d/s} = 6\) m, so
$$\tau = \frac{Q}{E_s\sqrt{\Delta h_v}} = \frac{0.92}{0.35\sqrt{6}} = \frac{0.92}{0.35 \times 2.4495} = \frac{0.92}{0.8573} = \boxed{1.073}$$
This is the answer the printed data give.
Compute the distributed friction loss in the main. Hazen-Williams applies over the entire 5,000 m; the valve’s position along the main affects where the HGL steps down but not the total friction:
$$K = 0.278 \times 100 \times (1.167)^{2.63} = 27.8 \times 1.5010 = 41.73\ \mathrm{m^3/s}$$
$$h_f = L\left(\frac{Q}{K}\right)^{1.852} = 5000\left(\frac{0.92}{41.73}\right)^{1.852} = 5000 \times 8.532\times10^{-4} = 4.27\ \mathrm{m}$$
The main is generously sized: 0.92 m3/s in a 1.167 m pipe is only 0.86 m/s, so friction over 5 km costs barely 4 m.
Close the energy equation on the downstream reservoir. The upstream level must supply the friction loss, the valve loss and the downstream level:
$$h_A = h_B + h_f + \Delta h_v \quad\Rightarrow\quad h_B = 105 - 4.27 - 6.00 = \boxed{94.73\ \mathrm{m}}$$
Ten and a quarter metres of the available head is being consumed, of which the valve throttles away well over half.
Set up the energy equation for part (c) with \(\tau\) reduced. Closing the valve raises its loss, which reduces \(Q\), which in turn reduces the friction loss — so the two loss terms must be solved together. With \(h_B\) held at 94.73 m the available head is fixed:
$$h_A - h_B = 105 - 94.73 = 10.27\ \mathrm{m}$$
$$\underbrace{5000\left(\frac{Q}{41.73}\right)^{1.852}}_{\text{pipe friction}} + \underbrace{\left(\frac{Q}{\tau E_s}\right)^{2}}_{\text{valve, from } Q = \tau E_s\sqrt{\Delta h_v}} = 10.27, \qquad \tau E_s = 0.3 \times 0.35 = 0.105$$
The valve term is quadratic in \(Q\) and the friction term has exponent 1.852, so the equation is solved numerically.
Iterate to the discharge. Neglecting friction as a first estimate gives \(Q \approx 0.105\sqrt{10.27} = 0.336\) m3/s; substituting that back to evaluate the friction term and re-solving converges in three passes:
$$Q_1 = 0.336 \rightarrow h_f = 0.66 \rightarrow Q_2 = 0.325 \rightarrow h_f = 0.62 \rightarrow Q_3 = 0.326\ \mathrm{m^3/s}$$
$$\boxed{Q = 0.326\ \mathrm{m^3/s}}$$
At the converged answer the valve absorbs 9.65 m and the pipe only 0.63 m, confirming that a valve this far closed, not the main, is what controls the system.
Sanity-check the result against part (a). Reducing \(\tau\) from 1.073 to 0.3, a factor of 3.6, cuts the discharge from 0.92 to 0.326 m3/s, a factor of 2.8. The reduction is less than proportional because throttling shifts head away from pipe friction and onto the valve, leaving the valve a larger differential to work with — the classic reason a throttling valve is a poor flow controller near its fully-open position and a good one when nearly shut.
Check: the part (a) answer exceeds unity. The printed data give \(\tau = 1.073\). In the usual convention \(\tau\) is a dimensionless relative opening with \(\tau = 1\) at the fully-open position, so a “partially closed” valve returning \(\tau > 1\) means the quoted \(E_s = 0.35\ \mathrm{m^{5/2}/s}\) has been calibrated at a slightly more closed reference setting than the state described, or that the 6 m loss has been rounded. The value is reported as the data give it because the method, and parts (b) and (c), are unaffected: only the product \(\tau E_s\) enters the calculations. If a strictly physical opening is required, note the inconsistency in the answer booklet and proceed with \(\tau E_s = 0.857\ \mathrm{m^{5/2}/s}\).
