Discharge from hydraulic jump depths 1.5 m to 4 m

Given: conjugate depths of the jump $y_1=1.5$ m and $y_2=4$ m, crest length $L=120$ m.

Step 1: Froude number before the jump

From the jump equation

$$\frac{y_2}{y_1}=\frac12\left(\sqrt{1+8Fr_1^2}-1\right)\Rightarrow Fr_1^2=\frac{1}{2}\frac{y_2}{y_1}\left(\frac{y_2}{y_1}+1\right)$$ $$Fr_1^2=\tfrac12(2.667)(3.667)=4.889\ \Rightarrow\ Fr_1=2.211$$

Step 2: Velocity and unit discharge

$$V_1=Fr_1\sqrt{gy_1}=2.211\times\sqrt{9.81\times1.5}=8.48\ \text{m/s}$$ $$q=V_1y_1=8.48\times1.5=12.72\ \text{m}^3/\text{s/m}$$

(Check: $q=\sqrt{\dfrac{g\,y_1y_2(y_1+y_2)}{2}}=12.72$ m³/s/m.)

Step 3: Discharge over the spillway

$$Q=qL=12.72\times120=1526.7\ \text{m}^3/\text{s}$$

Answer: discharge over the spillway $Q\approx1527$ m³/s. The energy lost in the jump is $\dfrac{(y_2-y_1)^3}{4y_1y_2}=0.651$ m.