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Clemente Gotelli
Bundaleer open channel carrying water toward the aqueduct

Bundaleer concrete channel. Photo credit: denisbin.

Prismatic channel hydraulics solver

Compute open-channel hydraulics for trapezoidal, rectangular, triangular, and circular sections: normal and critical depth, discharge, velocity, slope, Manning’s n, width/diameter, or side slopes. Uses Manning for gradually varied flow plus the Froude condition for critical depth.

Channel cross-section diagram

Intermediate calcs (normal depth)

Flow area $A_n$ (m²)
--
Wetted perimeter $P_n$ (m)
--
Top width $T_n$ (m)
--
Hydraulic radius $R_n$ (m)
--
Hydraulic depth $D_n$ (m)
--

Output (normal depth)

Depth $y_n$ (m)
--
Velocity $V_n$ (m/s)
--
Froude number $F_n$
--

Intermediate calcs (critical depth)

Flow area $A_c$ (m²)
--
Wetted perimeter $P_c$ (m)
--
Top width $T_c$ (m)
--
Hydraulic radius $R_c$ (m)
--
Hydraulic depth $D_c$ (m)
--

Output (critical depth)

Depth $y_c$ (m)
--
Velocity $V_c$ (m/s)
--
Froude number $F_c$
--

Formulas

Trapezoidal Channel (asymmetric slopes $z_1$, $z_2$):

$$A = b y + \tfrac{1}{2}(z_{1}+z_{2}) y^{2}, $$

$$P = b + y\sqrt{1+z_{1}^{2}} + y\sqrt{1+z_{2}^{2}},$$

$$T = b + (z_{1}+z_{2}) y$$

Rectangular Channel:

$$A = b y, \quad P = b + 2 y, \quad T = b$$

Triangular Channel (asymmetric slopes $z_1$, $z_2$):

$$A = \tfrac{1}{2}(z_{1}+z_{2}) y^{2},$$

$$P = y\sqrt{1+z_{1}^{2}} + y\sqrt{1+z_{2}^{2}},$$

$$T = (z_{1}+z_{2}) y$$

Circular Channel:

$$\theta = 2 \cos^{-1}\left(\frac{D - 2y}{D}\right), \quad A = \frac{D^{2}}{8}(\theta - \sin\theta), $$

$$P = \frac{D\theta}{2}, \quad T = D\sin\left(\frac{\theta}{2}\right)$$

Common Equations (all channels):

$$R = \frac{A}{P}, \quad D = \frac{A}{T}, \quad Q = \frac{k}{n} \, A \, R^{2/3} \, S^{1/2},$$

$$V = \frac{Q}{A}, \quad F = \frac{V}{\sqrt{g D}}$$

Critical depth condition:

$$F = 1 \Rightarrow \frac{Q^{2} T}{g A^{3}} = 1$$