Question 3 of 6: Valve-controlled transmission main
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2018 — 16-Civ-A5, Hydraulic Engineering. Closed book (one aid sheet; approved calculator); 3 hours; six questions, candidates complete any five — all six are solved here as a study resource. Each question 20 marks; parts of equal value.
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 negligible, diameters are nominal, and the fluid is water ($\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$).
Question 3: Valve-controlled transmission main (20 marks)
Given. A single transmission main ($L=5000\ \text{m}$, $C=140$, $D=1.167\ \text{m}$) with an in-line control valve; upstream reservoir $h_A=105\ \text{m}$; valve constant $E_s=0.45\ \text{m}^{5/2}/\text{s}$; valve law $Q=\tau E_s\sqrt{\Delta h_{valve}}$.
Given data
Quantity
Value
Length $L$ / diameter $D$ / roughness $C$
5000 m / 1.167 m / 140
Upstream reservoir $h_A$
105 m
Valve constant $E_s$
0.45 m$^{5/2}$/s
Part (a) discharge / valve headloss
0.92 m³/s / 6 m
Part (c) valve setting $\tau$
0.3
Find. (a) $\tau$ at the partially-closed setting; (b) the downstream reservoir level $h_B$; (c) the discharge when $\tau=0.3$ with $h_B$ fixed.
Figure 3. Transmission main with an in-line control valve (4,000 m upstream + 1,000 m downstream of the valve).
Approach. The valve law gives $\tau$ directly from the valve headloss. The downstream level follows from an energy balance $h_A-h_B=h_{f,pipe}(Q)+\Delta h_{valve}$ over the full 5,000 m of main plus the valve. For (c), hold $h_B$, drop $\tau$, and solve the same balance — now with $\Delta h_{valve}=(Q/\tau E_s)^2$ — iteratively for $Q$.
Part (a): $\tau$ from the valve law. With $\Delta h_{valve}=6\ \text{m}$,
$$\tau=\frac{Q}{E_s\sqrt{\Delta h_{valve}}}=\frac{0.92}{0.45\sqrt{6}}\;\Rightarrow\;\boxed{\tau\approx 0.835.}$$
Pipe friction at $Q=0.92$. With $a=0.278(140)(1.167)^{2.63}=38.92$ and $k_{pipe}=L/a^{1/0.54}=2.676$,
$$h_{f,pipe}=k_{pipe}\,Q^{1.852}=2.676(0.92)^{1.852}=2.29\ \text{m}.$$
Part (b): downstream level by energy balance. Total loss = pipe friction + valve loss:
$$h_A-h_B=h_{f,pipe}+\Delta h_{valve}=2.29+6.00=8.29\ \text{m},$$
$$\boxed{h_B=105-8.29\approx 96.7\ \text{m}.}$$
Part (c): new discharge at $\tau=0.3$. With $h_B$ fixed, the available head is still $h_A-h_B=8.29\ \text{m}$, now split between pipe friction and the tighter valve:
$$k_{pipe}\,Q^{1.852}+\left(\frac{Q}{\tau E_s}\right)^{2}=8.29,\qquad \frac{1}{(0.3\cdot0.45)^2}=54.87.$$
Solving $2.676\,Q^{1.852}+54.87\,Q^{2}=8.29$ iteratively gives
$$\boxed{Q\approx 0.378\ \text{m}^3/\text{s}}\quad(\Delta h_{valve}=7.85\ \text{m},\ h_{f,pipe}=0.44\ \text{m}).$$
Check / where the head goes. Because the main is a large 1,167 mm diameter, pipe friction is small (2.3 m at 0.92 m³/s) and the valve dominates the head budget. Closing the valve from $\tau=0.835$ to $0.30$ cuts the discharge from 0.92 to 0.38 m³/s — a $\sim\!59\%$ reduction — exactly the throttling behaviour expected when valve loss, not friction, controls the line.
