Water Resources and Environmental Hydraulics · Environmental Hydraulics · 1 mark · Numerical answer
A rectangular channel of width 5 m carries $15\,\mathrm{m^3/s}$. For $g=9.81\,\mathrm{m/s^2}$,find the critical velocity in m/s, rounded to three decimal places.
Discharge per unit width is $q=Q/b=15/5=3\,\mathrm{m^2/s}$. At critical flow $Fr=1$,so $v_c^2=gy_c$ and $q=v_cy_c$. Eliminating depth gives $v_c=(gq)^{1/3}=(9.81\cdot3)^{1/3}\approx\mathbf{3.087}$ m/s.
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