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Discharge from multi-point current meter data

Short answer

Mid-section method with v = 0.3N+0.05, mean velocity from two-point (0.2d,0.8d) or single-point (0.6d) readings, total discharge is 20.61 m³/s. Q ≈ 20.6 m³/s.

The following data were collected for a stream at a gauging station. Compute the discharge. Rating equation of current meter: V=0.3N+0.05V = 0.3N + 0.05.

Distance from one end of water surface (m)Depth, d (m)at 0.6d Rev.at 0.6d Sec.at 0.2d Rev.at 0.2d Sec.at 0.8d Rev.at 0.8d Sec.
31.41250
63.338522355
95.040583054
129.048603458
155.434523050
183.835523054
211.81850

Method: mid-section method. Velocity from v=0.3N+0.05v = 0.3N + 0.05 (NN in rev/s). Where the depth is small (1.4 m, 1.8 m) one reading at 0.6 d is the mean velocity; at other verticals two readings are used: vm=v0.2+v0.82v_m=\dfrac{v_{0.2}+v_{0.8}}{2}.

The verticals are 3 m apart, so each strip width is 3 m. The banks are assumed at 0 m and 24 m (zero depth).

Dist. (m)Depth (m)Revolutions / timevmv_m (m/s)Width (m)Area (m2^2)qq (m3^3/s)
31.40.6d: 12/50 = 0.2400.122034.200.512
63.30.2d: 0.731, 0.8d: 0.418 rps → 0.269, 0.1750.222339.902.201
95.00.2d: 0.690, 0.8d: 0.556 rps → 0.257, 0.2170.2368315.003.552
129.00.2d: 0.800, 0.8d: 0.586 rps → 0.290, 0.2260.2579327.006.964
155.40.2d: 0.654, 0.8d: 0.600 rps → 0.246, 0.2300.2381316.203.857
183.80.2d: 0.673, 0.8d: 0.556 rps → 0.252, 0.2170.2343311.402.671
211.80.6d: 18/50 = 0.3600.158035.400.853
Total89.1020.610

Mean velocity =0.231= 0.231 m/s.

Answer: Discharge Q≈20.6 m3/sQ \approx 20.6\ \text{m}^3/\text{s}.

✓ Answer checked by Claude Sonnet 5.5

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