Stream discharge from current meter revolutions (V = 0.51N + 0.03)

The mid-section method is used: each vertical represents a segment extending halfway to the neighbouring verticals, so $\Delta Q_i = w_i\,y_i\,V_i$ with $w_i = (x_{i+1} - x_{i-1})/2$. Velocity is from the rating equation $V = 0.51N + 0.03$, with $N$ = revolutions/second at 0.6 depth (gives mean velocity).

Computation

(At the edges, x = 0 and 12 m, depth and velocity are zero, so they add nothing.)

Sample: at x = 5 m, $N = 112/150 = 0.7467$ rev/s, $V = 0.51 \times 0.7467 + 0.03 = 0.4108$ m/s, $\Delta Q = 2.0 \times 2.5 \times 0.4108 = 2.054$ m³/s.

$$Q = \sum \Delta Q_i = 6.24\ \text{m}^3/\text{s}$$

Answer: Discharge ≈ 6.24 m³/s.

(Subramanya's textbook version gives the end verticals a width $\bar{W}_1 = (W_1 + W_2/2)^2/(2W_1) = 2.0$ m instead of 1.5 m, which gives Q ≈ 6.45 m³/s. Either method is accepted if stated.)