Problem Related to Hydraulic Jump

Problem 1. A rectangular channel 6m wide and discharge 1200 l/s of water into a 6m wide apron with zero slope with a mean velocity of 16 m/s. What is the height of hydraulic jump? Exam Name: BWDB 2016

Solution: 
We know that,
              Hydraulic jump, h_f = h₂ - h₁
              Here, V₁ = Q/( bh₁ )
                        h₁ =  (1200* 10^-3) / (6 * 16)
                             =  0.0125

             And, F₁ =  V₁ / √ (gh₁)
                           = 16 / √ (9.81* 0.0125 )
                                =  45.69
                      h₂ = (1/2) * h₁ {( √ 1  + 8F₁) - 1}
                          = (1/2) * 0.0125 {( √ 1  + 845.69²) - 1}
                          = 0.80

                      h_f = h₂-h₁ 
                          = 0.80 - 0.012
                          = 0.79 m
Answer: 0.79 m

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Problem 2. The sequent depth ration of a hydraulic jump in a rectangular channel is 16.48. Find Frude number at the beginning of jump and type of jump.
Exam Name: BWDB 2014

Solution: 
We know,
            h_2/  h₁ = (1/2) {( √ 1  + 8F₁) - 1}
         ∴ 16.48 =  (1/2) {( √ 1  + 8F₁) - 1}
            ⇒    F₁= 12, which is greater than 9

So that jump is strong.
Answer: 12

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