Question
Download Solution PDFA triangular open channel has a vertex angle of 90 degrees and carries flow at a critical depth of 0.30 m . The discharge in the channel is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
For triangular open channels, the discharge at critical flow can be found using the critical flow condition:
\(Q_c=\frac{8}{15}\sqrt {g} zy_c^{2.5}\)
where, z = side slope of channel, yc = critical depth, g = 9.81 m/s2
Calculation:
Given: For 90 degrees, z = 1
yc = 0.30 m
\(Q_c=\frac{8}{15}\times \sqrt {9.81}\times 1\times0.3^{2.5}\)
\(Q_c=0.11m^3/s\)
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