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Chapter 4: Problem 125

Liquid falls vertically into a short horizontal rectangular open channel of width \(b\). The total volume flow rate, \(Q\). is distributed uniformly over area \(b L\). Neglect viscous effects. Obtain an expression for \(h_{1}\) in terms of \(h_{2}, Q,\) and \(b\) Hint: Choose a control volume with outer boundary located at \(x=L\). Sketch the surface profile, \(h(x)\). Hint: Use a differential control volume of width \(d x\).

Short Answer

Expert verified
The expression for \(h_{1}\) in terms of \(h_{2}, Q,\) and \(b\) is \(h_{1} = h_{2} + \frac{QL}{b}\)

Step by step solution

01

Representation of control volume and differential control volume

The control volume chosen is a rectangular volume of length \(L\), width \(b\), and height varies from \(h_{1}\) at \(x=0\) to \(h_{2}\) at \(x=L\). This control volume includes the liquid in the channel. Next, choose a differential control volume at distance \(x\) from where the water falls, having dimensions \(dx\), \(b\), and \(h(x)\), where \(dx\) is an infinitesimally small length.
02

Apply the Principle of Conservation of Mass

Considering steady state condition, the principle of conservation of mass implies the inflow rate is equal to outflow rate for the differential control volume. The inflow rate is \(\rho Q\) and the outflow rate is \(\rho (Q + b h(x) dx)\), where the extra term represents the outflow from \(x+dx\). Setting these two to be equal, we get \(\rho Q = \rho (Q + b h(x) dx)\). Simplifying this equation, gives the differential equation \(b h(x)= -Q dx\). Solving this differential equation gives \(h(x) = \frac{-Q}{b}x + C\), where \(C\) is the integration constant.
03

Determine the integration constant

At \(x=L\), \(h(x)=h_{2}\). Substituting these values in the equation \(h(x) = \frac{-Q}{b}x + C\), we get \(h_{2} = \frac{-Q}{b}L + C\). Solving for \(C\) gives \(C=h_{2}+\frac{QL}{b}\).
04

Obtain the expression for \(h_{1}\)

At \(x=0\), \(h(x)=h_{1}\). After substituting the values obtained for \(C\) and \(x\) into the equation \(h(x) = \frac{-Q}{b}x + h_{2}+\frac{QL}{b}\), we get the final expression for \(h_{1}\): \(h_{1} = h_{2}+\frac{QL}{b}\).

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