Question 3 of 6: Valve-controlled transmission main
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 16-Civ-A5 Hydraulic Engineering — May 2019 · 3 hours · closed book (one aid sheet) · six questions, complete any four (all six solved here) · each question 20 marks, equal-value parts.
Reference texts: Chin, Water-Resources Engineering (Pearson); Chow, Open-Channel Hydraulics (McGraw-Hill); Chaudhry, Open-Channel Flow (Springer); Wylie & Streeter, Fluid Transients in Systems (Prentice-Hall); Munson, Young & Okiishi, Fundamentals of Fluid Mechanics (Wiley); Roberson, Cassidy & Chaudhry, Hydraulic Engineering.
Governing relations supplied on the exam cover sheet. Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with slope $S=\Delta h/L$ (SI, $Q$ in m³/s, $D$ in m); Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$; Darcy–Weisbach $\Delta h = 0.0826\,\tfrac{fL}{D^5}Q^2$. Unless stated, local losses and velocity head are neglected, diameters are nominal, and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
The Question 3 pipe specification used below is reconstructed from Figure 1. Check it against your copy of the paper.
Question 3: Valve-controlled transmission main (20 marks)
Given. A single large-diameter main from a fixed-level reservoir through a control valve to a downstream reservoir (Figure 1).
Given data — Question 3
Quantity
Symbol
Value
Length (4000 m + 1000 m) / diameter / roughness
$L,\ D,\ C$
5000 m, 1.167 m, 100
Upstream reservoir level
$h_A=H_{u/s}$
105 m
Valve discharge constant
$E_s$
0.35 m⁵ᐟ²/s
(a) operating discharge / valve loss
$Q,\ \Delta h_v$
0.92 m³/s, 6 m
(c) reduced valve setting
$\tau$
0.30
Check — reconstructed data
The pipe specification $D=1167$ mm, $C=100$ and $E_s=0.35$ m⁵ᐟ²/s is assumed, consistent with “Figure 1” (the figure fixes $h_A=105$ m and $L=4000+1000=5000$ m directly). The solution method is independent of these exact values; the numbers below use them.
Find. (a) $\tau$; (b) the downstream reservoir level $h_B$; (c) the new discharge when $\tau$ is reduced to 0.30 with $h_B$ held fixed.
Figure 3-1. Transmission main (Figure 1 on the exam): a control valve on the 5000 m pipe sets the discharge; friction acts over the full length and the valve adds a local loss.
Approach. The valve equation gives $\tau$ directly from the operating point (a). An energy balance from reservoir to reservoir — friction over the full 5000 m plus the 6 m valve loss — fixes the downstream level (b). For (c) the downstream level is held and the total available head is split between pipe friction and the valve loss $\big(Q/(\tau E_s)\big)^2$, solved by iteration.
Valve coefficient (part a). The head across the valve is its stated loss, $H_{u/s}-H_{d/s}=\Delta h_v = 6\ \text{m}$. Inverting the valve law $Q=E_s\tau\sqrt{\Delta h_v}$:
$$\tau = \frac{Q}{E_s\sqrt{\Delta h_v}} = \frac{0.92}{0.35\sqrt{6}} = \boxed{1.073.}$$
A value above 1 is admissible here — $E_s$ carries the dimensional part of the valve law, so $\tau$ is a relative opening index, not a fraction capped at unity.
Downstream reservoir level (part b). Energy from surface to surface: the upstream head is spent on pipe friction (full length, at $Q=0.92$) plus the valve loss. Hazen–Williams friction:
$$h_{f} = L\!\left(\frac{Q}{0.278\,C\,D^{2.63}}\right)^{1/0.54}=5000\!\left(\frac{0.92}{0.278(100)(1.167)^{2.63}}\right)^{1.852}=4.28\ \text{m}.$$
Then
$$h_B = H_{u/s} - h_f - \Delta h_v = 105 - 4.28 - 6 = \boxed{94.7\ \text{m}.}$$
Because the diameter is large, pipe friction (4.3 m) is modest and the valve loss (6 m) is the larger control.
Discharge at the reduced setting (part c). With $h_B=94.72\ \text{m}$ fixed and $\tau=0.30$, the total driving head $H_{u/s}-h_B = 10.28\ \text{m}$ is shared between friction and the valve:
$$H_{u/s}-h_B = h_f(Q) + \left(\frac{Q}{\tau E_s}\right)^2,\qquad 10.28 = 5000\!\left(\frac{Q}{41.73}\right)^{1.852} + \left(\frac{Q}{0.105}\right)^2 .$$
Iterating (Newton / trial): $Q=0.326\ \text{m}^3/\text{s}$ gives valve loss $9.65\ \text{m}$ and pipe loss $0.63\ \text{m}$, summing to $10.28\ \text{m}$. Hence
$$\boxed{Q = 0.326\ \text{m}^3/\text{s}.}$$
Closing the valve from $\tau=1.07$ to $0.30$ cuts the discharge from 0.92 to 0.33 m³/s — the valve, not friction, now governs.
