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 — 98-Civ-A5 Hydraulic Engineering, May 2014. 3 hours, closed book (one aid sheet). Six questions of equal value (20 marks each); candidates complete any five, and all six are solved here as a study resource.
Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe networks, valves, pumps); Chow, Open-Channel Hydraulics (Manning uniform flow, hydraulic jump, gutter flow); Crowe, Elger & Roberson, Engineering Fluid Mechanics (pipe force balance, wall shear). 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 $\Delta h=0.0826\,\tfrac{fL}{D^5}Q^2$, and $\text{TDH}=H_s+H_f$. Unless stated, local losses and velocity head are neglected and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
Question 3: Valve-Controlled Transmission Main (20 marks)
Given. $L=5000\ \text{m}$, $C=100$, $D=1.167\ \text{m}$, upstream head $h_A=105\ \text{m}$, $E_s=0.35\ \text{m}^{5/2}/\text{s}$; part (a) state $Q=0.92\ \text{m}^3/\text{s}$ with valve loss $\Delta h_v=6\ \text{m}$; part (c) $\tau=0.3$.
Find. (a) τ of the partly closed valve; (b) downstream reservoir level $h_B$; (c) discharge when $\tau=0.3$ with $h_B$ held.
Figure 2 (Q3). Transmission main from the upstream reservoir through an in-line control valve to the downstream reservoir; 4,000 m upstream and 1,000 m downstream of the valve (5,000 m total).
Approach. The valve equation gives τ directly from the state in (a); a full energy balance from reservoir to reservoir (pipe friction over the entire 5,000 m plus the valve loss) gives $h_B$; and holding $h_B$ with a smaller τ makes the same balance a single nonlinear equation for $Q$.
τ from the valve equation (a). The valve head drop is $\Delta h_v=6\ \text{m}$, so $\tau=\dfrac{Q}{E_s\sqrt{\Delta h_v}}=\dfrac{0.92}{0.35\sqrt{6}}=\boxed{1.073}$. (τ carries the units of $1/E_s$ absorbed, and can exceed 1 for a nearly open valve.)
Pipe friction at 0.92 m³/s. Hazen–Williams over the full length: $h_{f}=L\left(\dfrac{Q}{0.278\,C\,D^{2.63}}\right)^{1/0.54}=4.28\ \text{m}$.
Energy balance to the downstream reservoir (b). $h_A - h_B = h_f + \Delta h_v \Rightarrow h_B = 105 - 4.28 - 6 = \boxed{94.72\ \text{m}}$.
Re-close the valve, hold $h_B$ (c). The available head is fixed at $h_A-h_B=105-94.72=10.28\ \text{m}$, now split between pipe friction and the valve: $h_f(Q)+\left(\dfrac{Q}{\tau E_s}\right)^{2}=10.28$, with $\tau=0.3$.
Solve for $Q$ (c). The valve term $\left(Q/(0.3\times0.35)\right)^2$ now dominates; iterating gives $\boxed{Q=0.326\ \text{m}^3/\text{s}}$ (valve loss 9.65 m, pipe friction 0.63 m, summing to 10.28 m).
