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Stream flow from weir and float measurements

Short answer

(a) Effective length 0.6−0.1×2×0.18=0.564 m, H^1.5=0.0764, Q=1.84×0.564×0.0764=0.0793 m³/s (79 L/s). (b) Surface velocity 20/25=0.8 m/s, mean 0.85×0.8=0.68 m/s, area 1.2×0.4=0.48 m², Q=0.48×0.68=0.326 m³/s (326 L/s).

(a) A rectangular sharp-crested weir of crest length 0.6 m with two end contractions has a measured head of 0.18 m. Find the flow using Francis' formula Q=1.84 (L−0.1nH)H3/2Q = 1.84\,(L - 0.1nH)H^{3/2} (SI units). (b) In a rectangular channel of width 1.2 m and mean water depth 0.40 m, a float takes 25 s to travel 20 m. Taking the mean velocity as 0.85 times the surface velocity, find the flow.

(a) Weir

Here L=0.6L = 0.6 m, n=2n = 2, H=0.18H = 0.18 m.

Effective length: L−0.1nH=0.6−0.1(2)(0.18)=0.564L - 0.1nH = 0.6 - 0.1(2)(0.18) = 0.564 m

H3/2=0.181.5=0.0764H^{3/2} = 0.18^{1.5} = 0.0764

Q=1.84×0.564×0.0764=0.0793 m3/sQ = 1.84 \times 0.564 \times 0.0764 = 0.0793\ \text{m}^3/\text{s}

That is 79 litres per second.

(b) Float

Surface velocity:

vs=2025=0.80 m/sv_s = \frac{20}{25} = 0.80\ \text{m/s}

Mean velocity:

v=0.85×0.80=0.68 m/sv = 0.85 \times 0.80 = 0.68\ \text{m/s}

Area: A=1.2×0.40=0.48A = 1.2 \times 0.40 = 0.48 m².

Q=Av=0.48×0.68=0.326 m3/sQ = A v = 0.48 \times 0.68 = 0.326\ \text{m}^3/\text{s}

Answer: (a) Q=0.0793Q = 0.0793 m³/s (79 L/s); (b) Q=0.326Q = 0.326 m³/s (326 L/s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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