The approximate discharge over a 4 m long rectangular weir (with suppressed end contractions) with head over the crest as 0.36 m is

4.2 litres/s

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2.4

m^{3} / s

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6.1 litres/s

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1.6

m^{3} / s

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Solution

The correct option is D 1.6 $m^{3} / s$
$Q = \frac{2}{3} C_{d} \sqrt{2 g} L_{e} H^{3 / 2}$

$= 1.84 L_{e} H^{3 / 2} (L o n g r e c t a n g u l a r w e i r)$

$L_{e} = (L - 0.1 n H)$

$= 4 - 0.1 \times 2 \times 0.36$

$= 3.928 m$

$∴ Q = 1.84 \times 3.928 \times {(.36)}^{3 / 2} = 1.56 m^{3} / s$

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The approximate discharge over a 4 m long rectangular weir (with suppressed end contractions) with head over the crest as 0.36 m is

The correct option is D 1.6 m3/sQ=23Cd√2gLeH3/2 =1.84LeH3/2(Long rectangular weir) Le=(L−0.1nH) =4−0.1×2×0.36 =3.928m ∴Q=1.84×3.928×(.36)3/2=1.56m3/s

The correct option is D 1.6 $m^{3} / s$
$Q = \frac{2}{3} C_{d} \sqrt{2 g} L_{e} H^{3 / 2}$

$= 1.84 L_{e} H^{3 / 2} (L o n g r e c t a n g u l a r w e i r)$

$L_{e} = (L - 0.1 n H)$

$= 4 - 0.1 \times 2 \times 0.36$

$= 3.928 m$

$∴ Q = 1.84 \times 3.928 \times {(.36)}^{3 / 2} = 1.56 m^{3} / s$