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 2018 · 3 hours · closed book (one aid sheet) · six questions, complete any five (all solved here) · each question 20 marks, equal-value parts.
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 $\Delta h = 0.0826\,\tfrac{fL}{D^5}Q^2$; total dynamic head $TDH=H_s+H_f$. Unless stated, local losses and velocity head are neglected, diameters are nominal, and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
Question 3: Valve-controlled transmission main (20 marks)
Given. Main $L=5000\ \text{m}$, $C=110$, $D=0.45\ \text{m}$; upstream head $H_u=105\ \text{m}$; downstream head $H_d=70\ \text{m}$ (part a) and $30\ \text{m}$ (part b); system discharge $Q=0.7\ \text{m}^3/\text{s}$; valve equation $Q=\tau E_s\sqrt{H_u-H_d}$.
Find. (a) $E_s$ at $\tau=0.5$; (b) the $\tau$ needed to pass the same 0.7 m³/s when the downstream reservoir is lowered to 30 m.
Figure 3. Water transmission system: upstream reservoir, valve at 4,000 m, downstream reservoir at 1,000 m beyond.
Approach. The valve equation supplied by the exam relates the system discharge directly to the driving head $H_u-H_d$ (the reservoir head difference). Solve it for $E_s$ in (a) with the given $\tau$ and $Q$, then invert for $\tau$ in (b) with the new downstream head.
Solve the valve equation for $E_s$ (part a). Driving head $H_u-H_d=105-70=35\ \text{m}$:
$$E_s=\frac{Q}{\tau\sqrt{H_u-H_d}}=\frac{0.7}{0.5\sqrt{35}}.$$
$$\boxed{E_s\approx0.237\ \text{m}^{5/2}/\text{s}.}$$
Invert for $\tau$ at the new downstream head (part b). Now $H_u-H_d=105-30=75\ \text{m}$, same $Q=0.7$ and $E_s=0.237$:
$$\tau=\frac{Q}{E_s\sqrt{H_u-H_d}}=\frac{0.7}{0.237\sqrt{75}}.$$
$$\boxed{\tau\approx0.34.}$$
Lowering the downstream reservoir increases the driving head, so the same flow is passed with the valve more closed ($\tau$ drops from 0.50 to 0.34).
Check / model note. If one instead tried to split the 35 m between pipe friction and valve loss, the Hazen–Williams friction for $0.7\ \text{m}^3/\text{s}$ in this 5,000 m / 450 mm / $C{=}110$ main is $h_f\approx224\ \text{m}$ — far more than the 35 m available (velocity $\approx4.4\ \text{m/s}$). The stated operating point is therefore only self-consistent under the exam’s lumped valve model, in which the reservoir head difference drives the valve equation and distributed friction is not separately resolved (per cover-sheet NOTE 1, negligible local losses/velocity head). Both parts are answered on that basis.
