Question 3 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: National Exams — 98-Civ-A5 Hydraulic Engineering, December 2013 · 3 hours, closed book (one 8.5×11 aid sheet). Six questions; any five constitute a complete paper — all six are solved here as a study resource. All parts of a question are of equal value; ‘cms’ = m³/s.
Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe systems, valves, network analysis); Chow, Open-Channel Hydraulics (Manning uniform flow, gutter flow); Crowe, Elger & Roberson, Engineering Fluid Mechanics (laminar film flow, energy/continuity). Exam-supplied relations used throughout: Hazen–Williams $Q=0.278\,C\,D^{2.63}\,S^{0.54}$ with $S=h_f/L$ (SI), Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, Darcy–Weisbach, and total dynamic head $\text{TDH}=H_s+H_f$. Unless stated, local losses and velocity head are neglected and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
The Hazen–Williams relation is used in inverted form to get the friction loss carried by a known pipe discharge $Q$ (SI, $Q$ in m³/s):
Given. One long main with an in-line valve; energy is dissipated by pipe friction (full 5,000 m) plus the valve.
Given data (Question 3)
Upstream level $h_A$
105 m
Pipe $L$ / $D$ / $C$
5,000 m / 1,067 mm / 110
Valve constant $E_s$
0.35 m$^{5/2}$/s
Valve law
$Q=\tau E_s\sqrt{H_{u/s}-H_{d/s}}$
Find. (a) $\tau$ at the given operating point; (b) downstream level $h_B$; (c) discharge when $\tau=0.3$.
Figure 3. Upstream reservoir, 5,000 m main with an in-line valve (4,000 m + 1,000 m), downstream reservoir.
Approach. The valve law gives $\tau$ directly from the observed discharge and valve headloss; an energy balance from A to B (friction + valve) fixes $h_B$; then holding $h_B$ and $\tau=0.3$, the same balance is solved iteratively for the new discharge.
(a) Valve coefficient. The headloss across the valve is $H_{u/s}-H_{d/s}=5$ m at $Q=1$ m³/s:
$$\tau=\dfrac{Q}{E_s\sqrt{\Delta h_{\text{valve}}}}=\dfrac{1}{0.35\sqrt{5}}=\boxed{1.28.}$$
(A value above unity simply reflects the chosen units of $E_s$; it is the calibration constant for this valve at that opening.)
(b) Pipe friction at $Q=1$ m³/s (Hazen–Williams, full length):
$$h_{f,\text{pipe}}=5000\left(\dfrac{1}{0.278(110)(1.067)^{2.63}}\right)^{1/0.54}=6.47\ \text{m}.$$
Energy balance A→B: $h_A-h_B=h_{f,\text{pipe}}+\Delta h_{\text{valve}}$, so
$$\boxed{h_B = 105-6.47-5 = 93.5\ \text{m}.}$$
(c) Re-solve with $\tau=0.3$, $h_B=93.5$ m. The available head is $h_A-h_B=11.47$ m, split between pipe friction and the (now tighter) valve, whose loss is $\Delta h_{\text{valve}}=\big(Q/(\tau E_s)\big)^2=\big(Q/0.105\big)^2$:
$$5000\left(\dfrac{Q}{0.278(110)(1.067)^{2.63}}\right)^{1/0.54}+\left(\dfrac{Q}{0.105}\right)^2 = 11.47.$$
Iterate. The valve term dominates. Solving numerically gives
$$\boxed{Q = 0.342\ \text{m}^3/\text{s}\ (342\ \text{L/s}),}$$
with the valve now absorbing $\approx 10.6$ m and the pipe only $\approx 0.9$ m. Closing the valve from $\tau=1.28$ to $0.3$ cuts the discharge from 1.00 to 0.34 m³/s.
