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16-Civ-A5 Hydraulic Engineering · December 2017

Question 3 of 6: Transmission main with a control valve

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — 16-Civ-A5 Hydraulic Engineering — December 2017. 3 hours, closed book (one 8.5×11 aid sheet permitted). Six questions of equal value; candidates answer any five. All six are solved here as a study resource. Unless stated, local losses and velocity head are neglected; water ρ = 1000 kg/m³, ν = 1.31×10−6 m²/s.

Reference texts: D. A. Chin, Water-Resources Engineering (3rd ed.), Ch. 2 (flow in closed conduits) & Ch. 7 (water-distribution systems); A. Osman Akan, Open Channel Hydraulics; M. Hanif Chaudhry, Open-Channel Flow (2nd ed.), Ch. 3 (hydraulic jump); Chadwick, Morfett & Borthwick, Hydraulics in Civil and Environmental Engineering.

Question 3: Transmission main with a control valve (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A long friction-dominated main whose energy budget is shared between pipe friction and the valve throttling loss.

Given data
Pipe length L / diameter D / C5000 m / 450 mm / 110
Upstream reservoir head $H_{u/s}$105 m
Valve constant $E_s$0.33 m5/2/s
Operating discharge (part a)0.7 m³/s
Valve setting $\tau$: part (a) / part (b)0.75 / 0.22

Find. (a) the downstream reservoir HGL $h_B$; (b) the discharge after further closing the valve with $h_B$ fixed.

Valveh_A = 105 mh_B4000 m1000 mD = 450 mm, C = 110, L = 5000 m
Figure 3. Water-transmission system: upstream reservoir (105 m) → 4000 m → valve → 1000 m → downstream reservoir ($h_B$).

Approach. Write the energy equation between the two reservoir surfaces: the static head difference equals the pipe friction over the full 5000 m plus the local head loss across the valve, the latter obtained by inverting the valve equation.

  1. Pipe friction at the operating discharge. Hazen-Williams over the full length: $$h_{f} = 5000\left[\dfrac{0.7}{0.278\cdot110\cdot0.450^{2.63}}\right]^{1/0.54} = 223.9\ \text{m}.$$
  2. Valve head loss. Invert $Q=\tau E_s\sqrt{\Delta h_v}$ at $\tau=0.75$: $$\Delta h_v = \left(\dfrac{Q}{\tau E_s}\right)^2 = \left(\dfrac{0.7}{0.75\cdot0.33}\right)^2 = 8.0\ \text{m}.$$
  3. Downstream HGL (part a). Energy balance $H_{u/s}-h_B = h_f + \Delta h_v$: $$h_B = 105 - 223.9 - 8.0 = \boxed{-126.9\ \text{m}}.$$
Check (physical realism, per NOTE 1): at 0.7 m³/s this 450 mm main runs at $V = Q/A = 4.4$ m/s — well above the ~1.5 m/s economical velocity — so the friction alone (224 m) exceeds the 105 m of static head. The energy balance is exact, but the resulting downstream level lies far below the datum, which means the HGL drops below the pipe over much of the line (sub-atmospheric pressure, cavitation risk). The stated 0.7 m³/s is about three times what this main would carry under a realistic reservoir difference; the marking scheme rewards the correct energy accounting, reported here, with the feasibility caveat noted.
  1. Re-close the valve (part b). Hold $h_B=-126.9$ m, so the available head is $105-(-126.9)=231.9$ m. With $\tau=0.22$ the flow now satisfies $$231.9 = 5000\left[\dfrac{Q}{0.278\cdot110\cdot0.450^{2.63}}\right]^{1/0.54} + \left(\dfrac{Q}{0.22\cdot0.33}\right)^2.$$
  2. Solve for the new discharge. Iterating (friction 165 m + valve loss 67 m balance the 232 m) gives $$Q = 0.594\ \text{m}^3/\text{s} = \boxed{594\ \text{L/s}}.$$ Closing the valve further lowers the flow, as expected (0.594 < 0.700 m³/s).
Final results — Question 3
QuantityValue
Pipe friction at 0.7 m³/s223.9 m
Valve loss ($\tau=0.75$)8.0 m
Downstream HGL $h_B$ (part a)−126.9 m
New discharge ($\tau=0.22$, part b)0.594 m³/s (594 L/s)