18-Env-A2 Hydrology and Municipal Hydraulics Engineering, December 2018, Q3

Problem 3 (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.

(i) Plastic pipe: flow rate, Reynolds number, head loss (6 marks)

Given.

Quantity	Value
Length, $L$	1000 m
Diameter, $d$	500 mm = 0.500 m
Full-flow velocity, $V$	4 m/s
Pipe material	plastic (PVC), $\varepsilon\approx1.5\times10^{-6}\ \text{m}$
Kinematic viscosity, $\nu$ (water, 20 °C)	$1.004\times10^{-6}\ \text{m}^2/\text{s}$

Find. (a) $Q$ in m³/min; (b) $Re$ and flow type; (c) $H_f$ in m.

Approach. $Q$ from continuity; classify with $Re$; solve Colebrook–White for $f$ and apply Darcy–Weisbach.

Part (a) — flow rate. $$A=\frac{\pi d^2}{4}=\frac{\pi(0.500)^2}{4}=0.1963\ \text{m}^2\qquad Q=VA=(4)(0.1963)=0.7854\ \text{m}^3/\text{s}=\boxed{47.12\ \text{m}^3/\text{min}}$$
Part (b) — Reynolds number. $$Re=\frac{Vd}{\nu}=\frac{(4)(0.500)}{1.004\times10^{-6}}=\boxed{1.99\times10^{6}}$$ Since $Re\gg4000$, the flow is turbulent.
Part (c) — friction head loss. Relative roughness $\varepsilon/d=3.0\times10^{-6}$ (hydraulically smooth plastic). Solving Colebrook–White iteratively gives $f=0.0105$. Applying Darcy–Weisbach: $$H_f=f\frac{L}{d}\frac{V^2}{2g}=(0.0105)\left(\frac{1000}{0.500}\right)\frac{(4)^2}{2(9.81)}=\boxed{17.11\ \text{m}}$$

Quantity	Value
Flow rate, $Q$	47.12 m³/min
Reynolds number, $Re$	1.99×10&sup6; (turbulent)
Friction head loss, $H_f$	17.11 m

(ii) Trapezoidal grass channel: uniform-flow discharge and Reynolds number (7 marks)

Given.

Quantity	Value
Normal depth, $y$	5 m
Base width, $b$	4 m
Side slope (H:V)	1:4, i.e. $z=0.25$
Bed slope, $S_o$	3% = 0.03
Manning's $n$ (short grass, well maintained)	0.030

Find. (a) Discharge $Q$; (b) Reynolds number and flow type.

Approach. Trapezoidal geometry gives $A$, $P$, $R$; Manning's equation gives $Q$; open-channel $Re$ uses the hydraulic radius, $Re=VR/\nu$.

Part (a) — geometry and discharge. $$A=(b+zy)y=(4+0.25\times5)(5)=\boxed{26.25\ \text{m}^2}\qquad P=b+2y\sqrt{1+z^2}=4+2(5)\sqrt{1.0625}=14.31\ \text{m}\qquad R=\frac{A}{P}=1.835\ \text{m}$$ $$Q=\frac{1}{n}AR^{2/3}S_o^{1/2}=\frac{1}{0.030}(26.25)(1.835)^{2/3}(0.03)^{1/2}=\boxed{227.1\ \text{m}^3/\text{s}}$$
Part (b) — Reynolds number. $$V=\frac{Q}{A}=\frac{227.1}{26.25}=8.65\ \text{m/s}\qquad Re=\frac{VR}{\nu}=\frac{(8.65)(1.835)}{1.004\times10^{-6}}=\boxed{1.58\times10^{7}}$$ Since open-channel flow is turbulent for $Re\gtrsim12{,}500$ (hydraulic-radius Reynolds number), this flow is strongly turbulent.

Quantity	Value
Flow area, $A$	26.25 m²
Hydraulic radius, $R$	1.835 m
Discharge, $Q$	227.1 m³/s
Reynolds number, $Re$	1.58×10&sup7; (turbulent)

Check: a 1:4 (H:V) side slope is unusually steep for a grass-lined channel (most vegetated channels use $z\ge2$ for mowability/stability); it is taken as printed on the exam, consistent with this subject's established convention of solving with the literal given data and flagging an unusual value rather than silently "correcting" it.

(iii) Specific energy over a bed rise (7 marks)

Fig. 3 — Longitudinal profile of the channel bed rising by $\Delta z$ between Section 1 (upstream, depth $Y_1$) and Section 2 (crest, depth $Y_2$); same trapezoidal cross-section ($b=4$ m, $z=0.25$) as part (ii).

Given. $Q=10\ \text{m}^3/\text{s}$, $Y_1=2.5$ m, bed rise $\Delta z=0.5$ m, same trapezoidal section ($b=4$ m, $z=0.25$); frictional losses negligible.

Find. Depth $Y_2$ at the crest.

Approach. With no friction loss, total energy head is conserved along the channel, so the specific energy at the crest is $E_2=E_1-\Delta z$ (specific energy is referenced to the LOCAL bed). Check the crossing is not choked ($E_2\ge E_{\min}$ at critical depth) before solving the subcritical root of $E_2=Y_2+Q^2/(2gA(Y_2)^2)$.

Approach conditions at Section 1. $$A_1=(4+0.25\times2.5)(2.5)=11.56\ \text{m}^2\qquad V_1=\frac{10}{11.56}=0.865\ \text{m/s}\qquad E_1=Y_1+\frac{V_1^2}{2g}=2.5+0.038=\boxed{2.538\ \text{m}}$$
Check the bump does not choke the flow. Solving $Q^2T/(gA^3)=1$ numerically for the critical depth of this trapezoidal section gives $Y_c=0.845$ m, with minimum specific energy $E_{\min}=Y_c+A_c/(2T_c)=1.248$ m. Since $E_2=E_1-\Delta z=2.538-0.5=2.038\ \text{m} > E_{\min}=1.248\ \text{m}$, the crossing is not choked and a subcritical depth $Y_2$ exists at the crest.
Solve for $Y_2$ on the subcritical branch ($Y_2>Y_c$). Solving $E_2=Y_2+Q^2/(2gA(Y_2)^2)=2.038$ m numerically (subcritical root): $$\boxed{Y_2=1.973\ \text{m}}$$ As expected for subcritical approach flow over a bump, the depth drops from $Y_1=2.5$ m to $Y_2=1.973$ m as the bed rises — the free surface itself dips by $Y_1-(Y_2+\Delta z)=2.5-2.473=0.027$ m over the crest.

Quantity	Value
Specific energy at Section 1, $E_1$	2.538 m
Critical depth, $Y_c$ (this $Q$)	0.845 m
Depth at crest, $Y_2$	1.973 m

Topic: Closed-pipe turbulent flow; Manning uniform flow in a trapezoidal channel; specific-energy analysis of a channel bed rise (bump).

Key relations: Darcy–Weisbach $H_f=f(L/d)(V^2/2g)$; Manning $Q=(1/n)AR^{2/3}S^{1/2}$; specific energy $E=y+Q^2/(2gA^2)$; critical-flow condition $Q^2T/(gA^3)=1$.

Why this works: for the pipe, Colebrook–White gives the friction factor consistent with the pipe's own relative roughness and $Re$, rather than assuming a fixed value; for the channel, Manning's equation is dimensionally an empirical friction law analogous to Darcy–Weisbach but written in terms of hydraulic radius; and specific energy is conserved (not total energy alone) because it is measured relative to the LOCAL bed, so a bed rise directly reduces the energy available to sustain flow, which is why a subcritical channel's depth drops (not rises) over a bump.

Common pitfall: using $4\nu$ scaling instead of the open-channel $Re=VR/\nu$ convention; and, for the bump, forgetting to check $E_2\ge E_{\min}$ before solving — if the bed rise were large enough to choke the flow, no subcritical $Y_2$ would satisfy the equation and the upstream depth would have to rise instead.

Source: Chow, Open-Channel Hydraulics, uniform-flow and specific-energy chapters; Davis & Cornwell, Introduction to Environmental Engineering, pipe-flow chapter.

Question 3 of 7

Problem 3 (20 marks)

(i) Plastic pipe: flow rate, Reynolds number, head loss (6 marks)

(ii) Trapezoidal grass channel: uniform-flow discharge and Reynolds number (7 marks)

(iii) Specific energy over a bed rise (7 marks)