A2 — Discharge from a reservoir through a pipe system: 23-Chem-B1 Transport Phenomena, December 2019, Q2

Question 2: A2 — Discharge from a reservoir through a pipe system (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
Pipe inside diameter (6 in)	$D$	$0.5\ \text{ft}$
Total pipe length ($250+50+150$)	$L$	$450\ \text{ft}$
Net head: surface above entrance (150 ft) less the 50-ft rise to the outlet	$\Delta z$	$100\ \text{ft}$
Kinematic viscosity	$\nu$	$1\times10^{-5}\ \text{ft}^2/\text{s}$
Roughness height	$\epsilon$	$0.00015\ \text{ft}\ (\epsilon/D=3\times10^{-4})$
Minor losses: entrance + 2 elbows	$\sum K$	$0.25+2(0.90)=2.05$

Find. The volumetric discharge $Q=V\cdot\tfrac{\pi}{4}D^2$, requiring the pipe velocity $V$ from a mechanical-energy balance with a Reynolds-dependent friction factor.

Fig. A2: Gravity-driven pipe flow. The net 100-ft elevation head (150 ft to the entrance less the 50-ft rise) is spent on the exit velocity head plus friction and minor (entrance + two-elbow) losses. Friction factor and velocity are coupled, so the solution is iterative on the Moody chart / Colebrook equation.

Approach. Apply the steady mechanical-energy (Bernoulli-with-losses) balance between the reservoir surface and the free discharge; both are at atmospheric pressure, and the reservoir velocity is negligible. Collect friction and minor losses, then iterate between the velocity and the Reynolds-dependent Darcy friction factor until they agree.

Energy balance, surface 1 to jet 2. With $P_1=P_2=P_{atm}$ and $V_1\approx0$, the head is charged against the exit kinetic energy plus all losses: $$\Delta z=\frac{V^2}{2g}+\left(f\frac{L}{D}+\sum K\right)\frac{V^2}{2g}=\left(1+f\frac{L}{D}+\sum K\right)\frac{V^2}{2g}.$$ The leading “$1$” is the velocity head carried out by the free jet.
Insert the geometry. The 150-ft dimension runs from the free surface to the pipe centreline at the entrance, and the pipe then rises 50 ft to the outlet run, so $\Delta z=150-50=100$ ft. With $L/D=450/0.5=900$ and $\sum K=2.05$, $$100=\bigl(1+900f+2.05\bigr)\frac{V^2}{2g},\qquad g=32.2\ \text{ft/s}^2.$$ So $V=\sqrt{\dfrac{2g\,\Delta z}{3.05+900f}}=\sqrt{\dfrac{6440}{3.05+900f}}.$
First pass — seed from the Moody chart. For $\epsilon/D=3\times10^{-4}$ at $Re\sim10^6$ the Moody chart gives about $f\approx0.0155$ (the fully-rough limit is $0.0149$). Then $3.05+900(0.0155)=17.0$, giving $V=\sqrt{6440/17.0}=19.5\ \text{ft/s}$ and $$Re=\frac{VD}{\nu}=\frac{19.5\times0.5}{1\times10^{-5}}=9.7\times10^{5}.$$
Update $f$ from the Colebrook equation. At $Re=9.7\times10^{5}$, $\epsilon/D=3\times10^{-4}$, $$\frac{1}{\sqrt f}=-2\log_{10}\!\left(\frac{\epsilon/D}{3.7}+\frac{2.51}{Re\sqrt f}\right)\ \Rightarrow\ f=0.0157.$$ The bracket becomes $3.05+900(0.0157)=17.17$ and $V=\sqrt{6440/17.17}=19.4$ ft/s; a further pass leaves $f$ unchanged, so the velocity is converged at $\boxed{V\approx19.4\ \text{ft/s}.}$
Volumetric flow rate. $$Q=V\cdot\frac{\pi}{4}D^2=19.37\times\frac{\pi}{4}(0.5)^2=19.37\times0.1963\ \Longrightarrow\ \boxed{Q\approx3.80\ \text{ft}^3/\text{s}.}$$ Equivalently $Q\approx1710\ \text{gpm}\approx0.108\ \text{m}^3/\text{s}$.

Quantity	Result
Relative roughness	$\epsilon/D=3\times10^{-4}$
Darcy friction factor	$f\approx0.0157$
Pipe velocity	$V\approx19.4\ \text{ft/s}$
Reynolds number	$Re\approx9.7\times10^{5}$ (fully turbulent)
Volumetric discharge	$Q\approx3.80\ \text{ft}^3/\text{s}\ (\approx1710\ \text{gpm})$

Check — free-jet velocity head vs. exit loss

The discharge is a free jet to atmosphere, so the exit velocity head $V^2/2g$ is retained as kinetic energy (the “$1$” in the bracket) rather than added as an exit-loss coefficient $K_{ex}=1$; both bookkeeping conventions give the identical numeric bracket here. Read the figure carefully: the 150-ft dimension is measured from the free surface to the entrance centreline (Point 1), and the vertical run between the elbows rises 50 ft to the outlet run. The static head is therefore the net surface-to-outlet difference of 100 ft; taking 150 ft would overstate $Q$ by about 23 %. Every foot of pipe (450 ft) still contributes friction.

Concept & Theory

Topic. Single-pipe gravity discharge: the iterative “find the flow” class of pipe-network problem.

Key relations. Extended Bernoulli $\Delta z=\bigl(1+fL/D+\sum K\bigr)V^2/2g$; Darcy–Weisbach friction $h_f=f\frac{L}{D}\frac{V^2}{2g}$; Colebrook $\frac1{\sqrt f}=-2\log_{10}\!\bigl(\frac{\epsilon/D}{3.7}+\frac{2.51}{Re\sqrt f}\bigr)$; $Re=VD/\nu$; $Q=\frac\pi4 D^2V$.

Why this works. Head is fixed but $V$ and $f$ are mutually dependent, so the problem is solved by iteration: guess $f$ (fully-rough Moody value is an excellent seed at high $Re$), get $V$, update $Re$ and $f$, repeat. Convergence is fast because at $Re\sim10^6$ the Colebrook curve is nearly flat in $Re$ — the friction factor is dominated by roughness, so one correction suffices.

Common pitfall. Omitting the exit velocity head (the “$1$”), double-counting it as both KE and an exit loss, or mixing Darcy and Fanning factors ($f_{Darcy}=4f_{Fanning}$). Also, using pipe length instead of net elevation for the static head — the two horizontal runs contribute friction but no elevation change. Here the elbow run also rises, so the static head is 150 − 50 = 100 ft, not the 150 ft dimensioned beside the reservoir.

Source. Welty, Wicks, Wilson & Rorrer, Momentum, Heat and Mass Transfer, Ch. 13–14 (pipe friction, Moody chart, minor losses); Perry’s Handbook, fluid-flow section (Colebrook correlation).

Question 2 of 6: A2 — Discharge from a reservoir through a pipe system

Question 2: A2 — Discharge from a reservoir through a pipe system (25 marks)