Practice Question: Sub and Supercritical Flow

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Solution

Solution: \(Subcritical\)

 

 

\(b\ =\ 5\ ft\)

\(Q\ =\ 100\frac{ft^3}{sec}\)

\(n=0.013\)

\(S=0.001\)

\[Q=vA\]

\[v=\frac{1.486}{n}R^{{2}/{3}}_hS^{{1}/{2}}\]

\[A=by\]

\[R_h=\frac{by}{b+2y}=\frac{y}{1+{2y}/{b}}\]

\[Q=\frac{1.486}{n}\ \frac{by^{{5}/{3}}}{{\left(1+{2y}/{b}\right)}^{{2}/{3}}}S^{{1}/{2}}\]

\[y={\left(\frac{nQ}{1.486\times b\sqrt{S}}\right)}^{{3}/{5}}{\left(1+{2y}/{b}\right)}^{{2}/{5}}\]

\[h=2.79\ {\left(1+{2y}/{5}\right)}^{{2}/{5}}\]

by iteration; \(y=4.12\ ft\)

 

 Froude Number is given by:

 

\[F_r=\frac{v}{\sqrt{gy}}\]

\[v=\frac{1.486}{n}R^{{2}/{3}}_hS^{{1}/{2}}\]

\[R_h=\frac{by}{b+2y}=\frac{5\times 4.12}{5+2\times 4.12}=1.55\ ft\]

\[v=\frac{1.486}{0.013}\times {1.55}^{{2}/{3}}\times \ {0.001}^{{1}/{2}}=4.85\frac{ft}{sec}\]

\[F_r=\frac{4.85}{\sqrt{32.2\times 4.12}}=0.42<1\]The flow is Subcritical

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