Trapezoidal canal section (0.5 m bottom, 45° sides) flowing at 1.5 m depth; the wetted cross-section is converted to an equivalent hydraulic diameter for a Moody-chart friction analysis.
Find. The elevation (bed) drop over 8 km needed to sustain uniform flow at 1.5 m depth and 3 m³/s.
Approach. "General pipe flow relations" means treating the open channel as an equivalent pipe: compute the flow area and wetted perimeter at the design depth, form the hydraulic (equivalent) diameter $D_e=4A/P$, find the Reynolds number and relative roughness, read/compute the Darcy friction factor $f$ from the Moody relation (Colebrook), then apply the Darcy–Weisbach head-loss formula — which, for UNIFORM open-channel flow, equals the required bed elevation drop.
Flow area and wetted perimeter at $y=1.5$ m. At 45° side slope the water surface width is $b_0+2y=0.5+2(1.5)=3.5$ m, so
$$A=\frac{b_0+(b_0+2y)}{2}y=\frac{0.5+3.5}{2}(1.5)=\boxed{3.00\ \text{m}^2}$$
Each sloped side has wetted length $y\sqrt{2}=2.121$ m, so
$$P=b_0+2(y\sqrt2)=0.5+2(2.121)=4.743\ \text{m}$$
Hydraulic (equivalent) diameter and velocity.
$$D_e=\frac{4A}{P}=\frac{4(3.00)}{4.743}=\boxed{2.530\ \text{m}}\qquad V=\frac{Q}{A}=\frac{3}{3.00}=1.000\ \text{m/s}$$
Reynolds number and relative roughness. Using $\nu_{water}=\mu/\rho=1.0\times10^{-3}/1000=1.0\times10^{-6}\ \text{m}^2/\text{s}$,
$$Re=\frac{D_eV}{\nu}=\frac{(2.530)(1.000)}{1.0\times10^{-6}}=2.53\times10^6\qquad e/D_e=\frac{0.001}{2.530}=3.95\times10^{-4}$$
Friction factor from the Colebrook/Moody relation. Solving $\dfrac{1}{\sqrt{f}}=-2\log_{10}\!\left(\dfrac{e/D}{3.7}+\dfrac{2.51}{Re\sqrt{f}}\right)$ iteratively at $Re=2.53\times10^6$, $e/D=3.95\times10^{-4}$ gives
$$f=\boxed{0.0161}$$
Required elevation drop over 8 km. The Darcy–Weisbach head loss equals the bed elevation drop needed to sustain uniform flow,
$$\Delta z=h_L=f\left(\frac{L}{D_e}\right)\!\frac{V^2}{2g}=(0.0161)\left(\frac{8000}{2.530}\right)\!\frac{(1.000)^2}{2(9.81)}=\boxed{2.59\ \text{m over the 8 km run}}$$
