Question 6 of 7

Problem 6 (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) Uniform flow in the trapezoidal channel — discharge and Reynolds number

Given.

Find. Discharge $Q$ and Reynolds number/flow regime.

Approach. Trapezoidal geometry gives area $A$, wetted perimeter $P$ and hydraulic radius $R$; Manning's equation gives velocity and hence $Q$; the open-channel Reynolds number then uses $R$ as the characteristic length.

1. Geometry. $$A = (b+zy)y = (12+0.25\times4)(4) = 52.0\ \text{m}^2$$ $$P = b + 2y\sqrt{1+z^2} = 12 + 2(4)\sqrt{1+0.25^2} = 20.25\ \text{m}$$ $$R = \frac{A}{P} = \frac{52.0}{20.25} = 2.57\ \text{m}$$
2. (a) Manning velocity and discharge. $$V = \frac{1}{n}R^{2/3}S_0^{1/2} = \frac{1}{0.020}(2.57)^{2/3}(0.03)^{1/2} = 16.2\ \text{m/s}$$ $$Q = VA = 16.2 \times 52.0 = \boxed{845\ \text{m}^3/\text{s}}$$
3. (b) Reynolds number. Using the hydraulic radius as the open-channel characteristic length, $$Re = \frac{VR}{\nu} = \frac{16.2\times2.57}{1.00\times10^{-6}} = \boxed{4.16\times10^{7}}$$ This is far above the open-channel turbulent threshold ($Re \gtrsim 2000$), so the flow is turbulent.

(ii) Gradually varied flow — alternate depth over a bed rise (8 marks)

Given. Same trapezoidal channel ($b=12\ \text{m}$, $z=0.25$) as part (i); $Q = 20\ \text{m}^3/\text{s}$; upstream depth $Y_1=3\ \text{m}$; bed rise $\Delta Z = 1.0\ \text{m}$ over the 10 m reach; frictional losses negligible.

Find. Downstream depth $Y_2$.

Approach. With no friction loss, total energy head is conserved between the two sections, but the datum (bed) rises by $\Delta Z$, so the specific energy must fall by exactly $\Delta Z$: $E_1 = E_2 + \Delta Z$. Solve $E_2 = y_2 + \dfrac{Q^2}{2gA(y_2)^2}$ for $y_2$ by trial.

1. Upstream specific energy. $A_1=(12+0.25\times3)(3)=38.25\ \text{m}^2$, $V_1 = 20/38.25 = 0.523\ \text{m/s}$. $$E_1 = y_1 + \frac{V_1^2}{2g} = 3 + \frac{0.523^2}{19.62} = 3.014\ \text{m}$$
2. Target downstream specific energy. $$E_2 = E_1 - \Delta Z = 3.014 - 1.0 = 2.014\ \text{m}$$
3. Solve for $y_2$ (subcritical branch, continuous with the subcritical approach flow). Trial values of $A(y_2)=(12+0.25y_2)y_2$ bracket the target: at $y_2=1.98$ m, $E \approx 2.013$ m, converging to $$\boxed{y_2 \approx 1.98\ \text{m}}$$ (check: $E(y_2)=2.0139\ \text{m}$, matching $E_2$).

(iii) Limitation of Stokes' Law for large grains (6 marks)

$$\omega = \frac{D^2(\rho_s-\rho_w)g}{18\mu}$$

Stokes' Law is derived assuming purely laminar, creeping flow around a settling sphere (particle Reynolds number $Re_p = \omega D/\nu \ll 1$), where drag varies linearly with velocity and the drag coefficient follows $C_d = 24/Re_p$. This assumption holds well for small, fine particles (silts and fine sands, $D$ well under about 0.1 mm) settling slowly enough that the flow around them stays laminar. For large grains ($D>0.01\ \text{m}$, i.e. coarse sand and gravel), the settling velocity itself becomes large enough that the particle Reynolds number rises into the hundreds or thousands, the flow around the grain separates and a turbulent wake forms, and drag transitions to the Newtonian/turbulent regime where $C_d$ approaches a roughly constant value (of order 0.4–1) rather than continuing to fall as $1/Re_p$. Because Stokes' Law's linear (laminar) drag law does not capture this much larger turbulent drag, it significantly over-predicts the fall velocity for large grains; in practice, an empirical or transitional formulation (e.g. Newton's drag law for the fully turbulent range, or a composite relation such as Rubey's or Van Rijn's equation that blends the laminar and turbulent regimes) must be used instead to match observed settling velocities for coarse material.