Question 2 of 6: Transmission main with in-line valve
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2017 · 16-Civ-A5 Hydraulic Engineering.
Three hours; closed book (one aid sheet). Six questions of equal value; candidates answer any five.
All six are solved here as a study resource. Take water ρ = 1000 kg/m³,
ν = 1.31×10−6 m²/s, g = 9.81 m/s²; local losses and velocity head are neglected (Note 6).
Reference texts (subject):
L.W. Mays, Water Resources Engineering (Wiley) — pipe flow, Hazen-Williams, distribution networks, pumps.
V.T. Chow, Open-Channel Hydraulics (McGraw-Hill) — normal & critical depth, hydraulic jump, specific energy, unsteady flow.
Munson, Young & Okiishi, Fundamentals of Fluid Mechanics — momentum/energy principles.
Question 2: Transmission main with in-line valve (20 marks)
Check (data reconciliation): the prose gives an upstream level of
102 m and a length of 4,500 m, while Figure 1 labels $h_A = 105$ m and dimensions
$4{,}000 + 1{,}000 = 5{,}000$ m. Following the paper’s Note 1 (state assumptions) and the convention
that the written data governs, the prose values (102 m, 4,500 m) are adopted. Also, part a) supplies the
valve setting $\tau=0.75$ but no operating discharge, and the downstream level is the unknown — one
energy equation in two unknowns. The operating point is therefore anchored on an economic transmission-main velocity
$V \approx 1.5$ m/s ($Q_a \approx 0.24\ \text{m}^3/\text{s}$); the method below is independent of that choice, and
part b) follows self-consistently from the resulting $h_B$.
Given. A single main carries flow from the upstream reservoir through an in-line control
valve to the downstream reservoir; friction is by Hazen-Williams and the valve loss follows the stated valve law.
Given data
Quantity
Symbol
Value
Length (adopted)
L
4,500 m
Hazen-Williams coefficient
C
110
Diameter
D
0.450 m
Upstream level (adopted)
$h_A$
102 m
Valve constant
$E_s$
0.33 m5/2/s
Valve settings
$\tau$
0.75, then 0.22
Find. (a) the downstream reservoir HGL $h_B$; (b) the discharge when $\tau$ is reduced to 0.22 with $h_B$ held fixed.
Transmission main with an in-line control valve (Figure 1). Head is lost to pipe friction over the full length plus a local loss at the valve.
Approach. Write the energy balance from the upstream to the downstream reservoir as
pipe friction plus valve loss, $h_A - h_B = h_f(Q) + \Delta h_v$, with $\Delta h_v = \left(Q/(\tau E_s)\right)^2$ from
the valve law; evaluate at the anchored $Q$ for (a), then invert for $Q$ at the new $\tau$ for (b).
Friction resistance of the main. Writing Hazen-Williams as $h_f = k\,Q^{1.852}$ with
$k = L\,[0.278\,C\,D^{2.63}]^{-1.852}$:
$$k = \frac{4500}{\left[0.278(110)(0.450)^{2.63}\right]^{1.852}} = 390.2\ \text{s}^{1.852}\text{m}^{-4.556}.$$
Anchor the operating discharge (part a). With $A = \tfrac{\pi}{4}(0.45)^2 = 0.159\ \text{m}^2$
and an economic $V = 1.5$ m/s, $Q_a = VA = 0.239\ \text{m}^3/\text{s}$. The pipe friction is then
$$h_f = 390.2\,(0.239)^{1.852} = 27.44\ \text{m}.$$
Valve loss at $\tau = 0.75$. Inverting the valve law, $H_{u/s}-H_{d/s} = \left(Q/(\tau E_s)\right)^2$:
$$\Delta h_v = \left(\frac{0.239}{0.75\times 0.33}\right)^2 = 0.93\ \text{m}.$$
Close the valve to $\tau = 0.22$ (part b). Hold $h_B = 73.6$ m, so the available head is
$h_A - h_B = 28.4\ \text{m}$, split between friction and the (now larger) valve loss:
$$390.2\,Q^{1.852} + \left(\frac{Q}{0.22\times 0.33}\right)^2 = 28.4.$$
Solve for the discharge. Iterating (Newton / trial) gives
$$Q = \boxed{0.203\ \text{m}^3/\text{s}}, \qquad \Delta h_v = 7.8\ \text{m}.$$
The valve loss is no longer negligible, yet because this long, small-bore main is friction-dominated,
throttling from $\tau = 0.75$ to $0.22$ trims the discharge by only about 15%.
