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Stream discharge from current meter revolutions (v = 0.55N + 0.04)

Short answer

Mid-section method: N from revolutions/duration, V=0.55N+0.04, segment widths per spacing. Total discharge summed from w y V equals 10.99 m³/s.

The data pertaining to a stream-gauging operation at a gauging site are given below. The rating equation of the current meter is v = (0.55 N + 0.04) m/s. Calculate the discharge in the stream.

Dist from left edge (m)01.04.07.010.011.0
Depth (m)01.52.53.52.40
Revolution of current meter at 0.6d040601201250
Duration of observations (s)01001001501500

The mid-section method is used: each vertical represents the strip extending halfway to the adjacent verticals, wi=(xi+1−xi−1)/2w_i = (x_{i+1} - x_{i-1})/2, and ΔQi=wi yi Vi\Delta Q_i = w_i\,y_i\,V_i. Velocity at 0.6 depth (taken as mean velocity) is from V=0.55N+0.04V = 0.55N + 0.04, with NN = revolutions per second.

Computation

x (m)Width wiw_i (m)Depth (m)N (rev/s)V (m/s)Area (m²)ΔQ (m³/s)
1(4 − 0)/2 = 2.01.540/100 = 0.4000.26003.00.7800
4(7 − 1)/2 = 3.02.560/100 = 0.6000.37007.52.7750
7(10 − 4)/2 = 3.03.5120/150 = 0.8000.480010.55.0400
10(11 − 7)/2 = 2.02.4125/150 = 0.8330.49834.82.3920
Total25.810.987

Sample: at x = 10 m, V=0.55×0.8333+0.04=0.4983V = 0.55 \times 0.8333 + 0.04 = 0.4983 m/s; ΔQ=2.0×2.4×0.4983=2.392\Delta Q = 2.0 \times 2.4 \times 0.4983 = 2.392 m³/s.

Q=0.780+2.775+5.040+2.392=10.99 m3/sQ = 0.780 + 2.775 + 5.040 + 2.392 = 10.99\ \text{m}^3/\text{s}

Answer: Discharge ≈ 10.99 m³/s.

(If Subramanya's end-segment rule Wˉ=(W1+W2/2)2/(2W1)\bar{W} = (W_1 + W_2/2)^2/(2W_1) is used for the first and last verticals, their widths become 3.125 m and Q ≈ 12.77 m³/s.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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