Water flows in a 6-m-wide rectangular channel with a depth of 0.55 m and a flow rate of 12 m3/s.

Determine the critical depth of the flow.

Express your answer in m.

Respuesta :

Answer:

[tex]Y_{c}=0.7415m[/tex]

Explanation:

General Formula for calculating the critical Depth is:

[tex]Y_{c}=\frac{V^{2} }{g*A_{c}^{2}}[/tex]        Eq(1)

where:

V is the volume flow rate

g is gravitational acceleration i.e 9.81 m/s^2

A_c is the  critical area

In case of Rectangular channel:

[tex]A_{c} =w*y_{c}[/tex]

where:

w is the width

In case of Rectangular channel Eq (1) will become:

[tex]Y_{c}=(\frac{V^{2} }{g*w^{2} } )^{\frac{1}{3} }[/tex]

[tex]Y_{c}=(\frac{12^{2} }{9.81*6^{2} } )^{\frac{1}{3} }[/tex]

[tex]Y_{c}=0.7415m[/tex]

Actual depth i.e Y < Critical depth i.e Y_c

Flow is Supercritical