Reservoir Discharge Through a Pipe Network: 21-Mat-A2 Materials Transport Phenomena, May 2017, Q1

Question 1: Reservoir Discharge Through a Pipe Network (25 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.

Quantity	Symbol	Value
Net elevation drop, reservoir surface to pipe exit	$z$	$150-50=100\ \text{ft}$
Pipe diameter	$D$	6 in $=0.5\ \text{ft}$
Total pipe length	$L$	$250+50+150=450\ \text{ft}$
Kinematic viscosity of water	$\nu$	$1\times10^{-5}\ \text{ft}^2/\text{s}$
Pipe roughness	$\epsilon$	$0.00015\ \text{ft}$ ($\epsilon/D=3\times10^{-4}$)
Entrance loss coefficient	$K_{ent}$	0.25
Elbow loss coefficient (each of 2)	$K_{elb}$	0.90

Find. The volumetric flow rate $Q$ discharged at Point 2.

Fig. 1 — pipe run from the reservoir free surface B, through the entrance (1), two 90° elbows and 450 ft of 6 in. pipe, to the atmospheric exit (2). The vertical leg rises, so the exit stands 50 ft above the entrance and only 100 ft below the free surface.

Approach. Write the mechanical-energy (Bernoulli-with-losses) balance between the reservoir free surface B ($V\approx0$, $P=P_{atm}$) and the exit at Point 2 ($P=P_{atm}$), which is driven purely by the 100 ft net elevation head (the free surface stands 150 ft above the entrance at Point 1, but the vertical leg rises, putting Point 2 50 ft above Point 1); because the Darcy friction factor $f$ depends on the still-unknown velocity through $Re$, solve the energy balance and the Colebrook correlation (the algebraic form of the supplied Moody chart) simultaneously by iteration.

Energy balance, B → 2. With both ends open to atmosphere and $V_B\approx0$, the elevation drop supplies the exit kinetic energy plus all friction and minor losses: $$z=\frac{V^2}{2g}\left[1+K_{ent}+2K_{elb}+f\frac{L}{D}\right]\;\Rightarrow\;100=\frac{V^2}{2(32.2)}\Big[1+0.25+2(0.90)+f\frac{450}{0.5}\Big]=\frac{V^2}{64.4}\big[3.05+900f\big]$$
Colebrook check on the Moody chart. For commercial steel at $\epsilon/D=3\times10^{-4}$ the friction factor is read (or, equivalently, computed from the Colebrook equation the chart is built from) as a function of $Re=VD/\nu$: $$\frac{1}{\sqrt f}=-2\log_{10}\left(\frac{\epsilon/D}{3.7}+\frac{2.51}{Re\sqrt f}\right)$$ Guessing $f_0\approx0.0157$ (near the fully-turbulent asymptote for this $\epsilon/D$) and iterating steps 1–2 to convergence:
Converged solution. $$f=0.0157,\qquad V=\sqrt{\frac{2(32.2)(100)}{3.05+900(0.0157)}}=19.37\ \text{ft/s},\qquad Re=\frac{VD}{\nu}=\frac{19.37(0.5)}{1\times10^{-5}}=9.69\times10^{5}$$ which lies in the complete-turbulence zone of the chart for $\epsilon/D=3\times10^{-4}$, confirming $f=0.0157$ is self-consistent (no further iteration changes it).
Flow rate. With $A=\tfrac{\pi}{4}D^2=\tfrac{\pi}{4}(0.5)^2=0.1963\ \text{ft}^2$: $$\boxed{Q=VA=19.37(0.1963)=3.80\ \text{ft}^3/\text{s}\ \;(=1707\ \text{gpm}=0.108\ \text{m}^3/\text{s})}$$

Quantity	Value
Darcy friction factor, $f$	0.0157
Velocity, $V$	19.37 ft/s
Reynolds number, $Re$	$9.69\times10^5$
Volumetric flow rate, $Q$	3.80 ft³/s (1707 gpm)

Check

Assumes turbulent-flow kinetic-energy correction $\alpha=1$ at the exit, negligible entrance velocity at the free surface, and that the two 90° elbows are the only fittings besides the entrance (i.e. no additional valve/tee losses at the unlabelled corner points). The driving head is the net 100 ft from the free surface B down to the exit, not 200 ft. The printed figure places the free surface 150 ft above the pipe centreline at Point 1 and shows the 50 ft vertical leg rising between the two elbows, so Point 2 sits 50 ft above Point 1 and $150-50=100$ ft below B; neither horizontal leg changes elevation. Reading that leg as a drop ($z=200$ ft) would inflate the answer by a factor of $\sqrt2$, to $Q=5.41$ ft³/s.

Concept & Theory

Topic. Steady incompressible pipe flow with friction and minor (fitting) losses, closed by the Moody/Colebrook friction-factor correlation.

Key relations. Mechanical energy balance $z=\dfrac{V^2}{2g}\big[1+\sum K+f\tfrac{L}{D}\big]$; Darcy–Weisbach friction loss $h_f=f\tfrac{L}{D}\tfrac{V^2}{2g}$; Colebrook equation $\tfrac{1}{\sqrt f}=-2\log_{10}\!\big(\tfrac{\epsilon/D}{3.7}+\tfrac{2.51}{Re\sqrt f}\big)$; $Re=VD/\nu$.

Why this works. Because both ends of the system are open to atmosphere and the reservoir is large (negligible surface velocity), the entire elevation head is dissipated by exit kinetic energy and by friction/fitting losses — a single algebraic energy balance in the unknown $V$ closes once $f(Re)$ is supplied by the Moody relation, requiring iteration only because $Re$ itself depends on $V$.

Common pitfall. Forgetting the "$+1$" exit kinetic-energy term in the bracket (the jet still carries kinetic energy as it leaves to atmosphere), or mistaking the sense of the vertical leg. Here it rises, so its 50 ft subtracts from the 150 ft reservoir head ($z=100$ ft) instead of adding to it; the full 450 ft of pipe still enters $fL/D$ either way, because only the elevation difference between the two end states enters $z$.

Source. Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer, Ch. 6 (mechanical energy balance, friction factor, Moody chart).

Question 1 of 5: Reservoir Discharge Through a Pipe Network

Question 1: Reservoir Discharge Through a Pipe Network (25 marks)