Question 12.11: A rectangular channel, 6.1 m wide, carries 11.32 m^3/s and d...

(i)   V_{1}=6.1 \mathrm{~m} / \mathrm{s}

q_{1}   (the rate of discharge per unit width) = 11.32/6.1

=1.86 \mathrm{~m}^{3} / \mathrm{sm} width

Therefore, h_{1}=q_{1} / V_{1}=1.86 / 6.1=0.305 \mathrm{~m}

And   \mathrm{Fr}_{1}=V_{1} /\left(g h_{1}\right)^{1 / 2}=6.1 /(9.81 \times 0.305)^{1 / 2}=3.53

using

Eq. (12.46),

\frac{h_{2}}{h_{1}}=\frac{1}{2}\left[\left(1+8 \mathrm{Fr}_{1}^{2}\right)^{1 / 2}-1\right]                       (12.46)

from which          h_{2}=1.38 \mathrm{~m}

Hence, the height of the hydraulic jump = 1.38 – 0.305 = 1.075 m using

Eq. (12.48)

h_{j}=\frac{\left(h_{2}-h_{1}\right)^{3}}{4 h_{1} h_{2}}                   (12.48)

Loss of head in the hydraulic jump h_{j}=\frac{(1.07)^{3}}{4 \times 0.305 \times 1.38}

= 0.73 m

Therefore the loss of total energy per second \rho g Q h_{j}