We can solve this problem by applying Eq. 15.13 between a point upstream where the velocity, width, and depth are V₁, w₁, and y₁, and a point downstream where the velocity, width, and depth are V₂, w₂, and y₂ as shown in Figure 15.18. Writing Eq. 15.13 between these points gives

\frac{1}{2g}\left[\frac{V_1y_1w_1}{y_2w_2} \right]^2+y_2= \frac{V^2_1}{2g} +y_1

Multiplying by y^2 _2 and rearranging we obtain the cubic equation

y^3_2-y^2_2\left[\frac{V^2_1}{2g}+y_1\right]\frac{1}{2g} \left[\frac{V_1y_1w_1}{w_2} \right]^2=0

After inserting the data, we have y^3_2-y^2_2[1.062 ft] + 0.1104 ft = 0, which can be solved to obtain the three solutions: −0.286 ft, 0.412 ft, and 0.936 ft. The negative root can be immediately discarded on physical grounds. Next we calculate the downstream velocities and corresponding Froude numbers for each of the positive roots. For y₂ = 0.412 ft we find:

V_2=\frac{V_1y_1w_1}{y_2w_2}=\frac{(2\ \mathrm{ft} /s)(1\ \mathrm{ft} )(4\ \mathrm{ft} )}{(0.412\ \mathrm{ft} )(3\ \mathrm{ft} )} =6.47\ \mathrm{ft} /s

and

Fr_2=\frac{V_2}{\sqrt{gy_1} }=\frac{6.47\ \mathrm{ft} /s}{\sqrt{(32.2\ \mathrm{ft} /s^2)(1\ \mathrm{ft} )}}=1.14

For y₂ = 0.936 ft we find:

V_2=\frac{V_1y_1w_1}{y_2w_2}=\frac{(2\ \mathrm{ft} /s)(1\ \mathrm{ft} )(4\ \mathrm{ft} )}{(0.936\ \mathrm{ft} )(3\ \mathrm{ft} )} =2.85\ \mathrm{ft} /s

and

F_2=\frac{V_2}{\sqrt{gy_1} }=\frac{2.85\ \mathrm{ft} /s}{\sqrt{(32.2\ \mathrm{ft} /s^2)(1\ \mathrm{ft} )}}=0.50

Calculating the upstream Froude number, we find

Fr_1=\frac{V_1}{\sqrt{gy_1} }=\frac{2\ \mathrm{ft} /s}{\sqrt{(32.2\ \mathrm{ft} /s^2)(1\ \mathrm{ft} )}}=0.352

hence the flow is subcritical. Since the depth of a subcritical for flow must decrease in a contraction, the correct depth downstream of the contraction is 0.936 ft and the corresponding flow speed is 2.85 ft/s.