Question 4 of 7

Problem 4 (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) Corrugated steel pipe — flow rate, Reynolds number, head loss (6 marks)

Given.

Find. (a) $Q$ in m³/min; (b) $Re$ and flow regime; (c) head loss $H_f$.

Approach. Continuity for $Q$; $Re=VD/\nu$ for the regime; Colebrook–White (fully rough, high-$Re$ regime) for the friction factor, then Darcy–Weisbach for $H_f$.

1. Flow rate. $A=\dfrac{\pi}{4}D^2=\dfrac{\pi}{4}(0.6)^2=0.2827\ \text{m}^2$. $$Q = VA = 3\times0.2827 = 0.848\ \text{m}^3/\text{s} = \boxed{50.9\ \text{m}^3/\text{min}}$$
2. Reynolds number and regime. $$Re = \frac{VD}{\nu} = \frac{3\times0.6}{1.00\times10^{-6}} = \boxed{1.8\times10^{6}}$$ $Re\gg4000$, so the flow is unambiguously turbulent; at this Reynolds number and relative roughness the flow is in the fully-rough regime (friction factor essentially independent of $Re$).
3. Friction factor and head loss. Relative roughness $\varepsilon/D=45/600=0.075$. Solving Colebrook–White, $$\frac{1}{\sqrt f}=-2\log_{10}\!\left(\frac{\varepsilon/D}{3.7}+\frac{2.51}{Re\sqrt f}\right)$$ at $Re=1.8\times10^6$ gives $f\approx0.087$ (fully-rough asymptote). Applying Darcy–Weisbach, $$H_f = f\,\frac{L}{D}\,\frac{V^2}{2g} = 0.087\times\frac{1200}{0.6}\times\frac{3^2}{2(9.81)} = \boxed{80\ \text{m}}$$

(ii) Grass-lined trapezoidal channel — discharge and Reynolds number (7 marks)

Given.

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

Approach. Trapezoidal geometry for $A$, $P$, $R$; Manning's equation for uniform-flow velocity and discharge; hydraulic-radius-based Reynolds number, $Re=4VR/\nu$, for the regime check.

1. Section properties. $$A=(b+zy)y=(5+3\times4)(4)=\boxed{68\ \text{m}^2}$$ $$P=b+2y\sqrt{1+z^2}=5+2(4)\sqrt{10}=30.3\ \text{m}, \qquad R=A/P=\boxed{2.24\ \text{m}}$$
2. Manning's equation. $$V=\frac{1}{n}R^{2/3}S_0^{1/2}=\frac{1}{0.030}(2.24)^{2/3}(0.03)^{1/2}=\boxed{9.90\ \text{m/s}}$$ $$Q = VA = 9.90\times68 = \boxed{673\ \text{m}^3/\text{s}}$$
3. Reynolds number. $$Re = \frac{4VR}{\nu} = \frac{4\times9.90\times2.24}{1.00\times10^{-6}} = \boxed{8.9\times10^{7}}$$ $Re\gg2000$, so the flow is unambiguously turbulent.

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

Given. Same trapezoidal channel ($b=5$ m, $z=3$); $Q=15\ \text{m}^3/\text{s}$; upstream normal depth $Y_1=3$ m; bed rise $\Delta z=0.6$ m; negligible friction loss between the two sections.

Find. Depth of flow $Y_2$ over the crest of the bed rise.

Approach. Specific energy is conserved (no friction loss, no change in total head other than the bed-elevation rise): $E_1=E_2+\Delta z$. Before solving for $Y_2$, verify the crossing is not choked by checking the available energy above the upstream value against the minimum specific energy (critical depth) for this discharge.

1. Upstream specific energy. $A_1=(b+zY_1)Y_1=(5+9)(3)=42\ \text{m}^2$, $V_1=Q/A_1=15/42=0.357\ \text{m/s}$. $$E_1 = Y_1+\frac{V_1^2}{2g} = 3+\frac{0.357^2}{2(9.81)} = \boxed{3.007\ \text{m}}$$
2. Check for choking — critical depth and minimum specific energy. Solving $\dfrac{Q^2T}{gA^3}=1$ for the trapezoidal section ($T=b+2zY$) gives $Y_c=0.82\ \text{m}$, $E_{min}=1.13\ \text{m}$. The maximum bed rise the flow can climb without choking is $$\Delta z_{max} = E_1-E_{min} = 3.007-1.126 = 1.88\ \text{m}$$ Since $\Delta z = 0.6\ \text{m} < \Delta z_{max}=1.88\ \text{m}$, the crossing is not choked and a subcritical $Y_2$ exists.
3. Solve for $Y_2$ from $E_2=E_1-\Delta z$. $$E_2 = 3.007-0.6 = 2.407\ \text{m} = Y_2+\frac{Q^2}{2g\left[(b+zY_2)Y_2\right]^2}$$ Solving iteratively (subcritical root, $Y_2>Y_c$): $$Y_2 = \boxed{2.39\ \text{m}}$$