The loss of energy in a hydraulic jump formed in a rectangular channel is given by (Symbols have the usual meanings)

- 1
  
  $∆ E = \frac{{(y_{2} - y_{1})}^{3}}{4 y_{1} y_{2}}$
- 2
  
  $∆ E = \frac{v Q {(y_{2} - y_{1})}^{3}}{8 y_{1} y_{2}}$
- 3
  
  $∆ E = \frac{v Q {(y_{2} - y_{1})}^{3}}{75 x 4 y_{1} y_{2}} - H P$
- 4
  
  $∆ E = \frac{{(y_{1} - y_{2})}^{3}}{4 y_{1} y_{2}}$

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