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## Q. 15.7

The ﬂow of a stream from a steep slope to a horizontal section results in a hydraulic jump. The depth in the horizontal section is 1 m and the velocity is 16 m/s. The stream is 10 m wide. Determine the change in depth, velocity, and Froude number across the jump, and the power dissipated in the jump.

## Verified Solution

Figure 15.24 can serve as a sketch for this problem. The ﬁrst step in solving any hydraulic jump problem is to ﬁnd the upstream Froude number:

$Fr_1=\frac{V_1}{\sqrt{gy_1}}=\frac{16\ m/s}{\sqrt{\left(9.81\ m/s^2\right)(1\ m)} } =5.1$

Then from Eq. 15.28 the depth ratio across the jump is determined to be

$\frac{y_2}{y_1}=\frac{1}{2}\left(\sqrt{1+8Fr^2_1}-1\right)=\frac{1}{2}\left(\sqrt{1+8(5.1)^2}1\right)=6.7$

Thus, since y1 = 1 m we ﬁnd y2 = 6.7 m. The velocity ratio is given by V1/V2 = y2/y1 = 6.7, so V2 = (16 m/s)/(6.7) = 2.4 m/s. The downstream Froude number is now easily calculated to be

$Fr_2=\frac{V_2}{\sqrt{gy_2}}=\frac{2.4\ m/s}{\sqrt{\left(9.81\ m/s^2\right)(6.7\ m)} } =0.3$

The power dissipated in the jump is found from Eq. 15.26b: P = ρghLQ. We can obtain the head loss by using Eq. 15.30 as follows:

$\frac{h_L}{y_1}=\frac{1}{4}\frac{[(y_2/y_1)-1]^3}{y_2/y_1}=\frac{1}{4} \frac{[6.7-1]^3}{6.7}=6.9$

so hL = 6.9y1 = 6.9(1 m) = 6.9 m. The power dissipated in the jump is then
P = ρghL(wy1V1) = (1000 kg/m3)(9.81 m/s2)(6.9 m )[(10 m)(1 m)(16 m/s)]

= 1.1 × 107 (N-m)/s

This is over 10 megawatts of power. If you are wondering where this enormous amount of energy goes, it is converted into heat. The heat capacity of water is so large however, that the temperature rise is minimal. Perhaps you can estimate the temperature rise?