Discharge of triangular river from surface velocities

Data: triangular section, top width $T = 10$ m, maximum depth $D=3$ m at 4 m from one side (the left side). Surface velocities: 1.5 m/s at 3.0 m and 1.8 m/s at 7.0 m from the same side. Mean velocity in a vertical $\approx 0.85\times$ surface velocity (usual surface-to-mean coefficient for floats and surface readings).

  0     3   4   5   7         10 m
  +-----v---v---v---v---------+  water surface
   \    |   |   |   |       /
     \  |   |   |   |     /
       \|   |   |   |   /
        \   |3 m|   | /
          \_ V __|_/
          deepest point at x = 4 m

Depths

Left side (0 to 4 m): depth $= 3\times\dfrac{x}{4}$. Right side (4 to 10 m): depth $= 3\times\dfrac{10-x}{6}$.

• At $x=3$ m: $d_1 = 3\times3/4 = 2.25$ m.
• At $x=7$ m: $d_2 = 3\times(10-7)/6 = 1.50$ m.

Areas

The total area is $\tfrac12\times10\times3=15$ m$^2$. The section is divided at $x = 5$ m (midway between the two verticals), where depth is $3\times5/6 = 2.5$ m.

• Left part (vertical at 3 m): $A_1 = \tfrac12(4)(3) + \tfrac12(3+2.5)(1) = 6 + 2.75 = 8.75$ m$^2$.
• Right part (vertical at 7 m): $A_2 = 15 - 8.75 = 6.25$ m$^2$.

Mean velocities

$$v_1 = 0.85\times1.5 = 1.275\ \text{m/s},\qquad v_2 = 0.85\times1.8 = 1.53\ \text{m/s}$$

Discharge

$$Q = v_1A_1 + v_2A_2 = 1.275\times8.75 + 1.53\times6.25 = 11.156 + 9.563 = 20.72\ \text{m}^3/\text{s}$$

Answer: Discharge $\approx 20.7$ m$^3$/s.