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The critical depth of flow in a most economical triangular channel section for a discharge of 1 m3/sec is given by
- a)
metre - b)
metre - c)
metre - d)
metre
Correct answer is option 'C'. Can you explain this answer?
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The critical depth of flow in a most economical triangular channel sec...
Critical depth= (2Q^2/(gm^2))^(1/5) for triangular channel
m= side slope= tan 45 degree= 1 for most economical channel
Q =1 m^3/s
yc= (2*1^2/(9.8*1^2))^(1/5)= (1/4.9)^(1/5)
m= side slope= tan 45 degree= 1 for most economical channel
Q =1 m^3/s
yc= (2*1^2/(9.8*1^2))^(1/5)= (1/4.9)^(1/5)
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