Hydraulic jump question in open channel flow

[Aurora09]

Hello, I am stuck on the following problem:

Water flows in a rectangular channel with depth of 0.40 m, width of 1.0 m and velocity of 6 m/s. Calculate the alternate depth. 

Here is how the solution manual solved it:

E = y1 + v2/2g

Plugging in the values above will give you 2.235 m for the initial energy E. Then to solve for y2, they set 2.235 = y2 + Q^2/(2gA^2) in which A equals width (1.0m) times y2.

Plugging in the values gives you y2 = 2.17 m. 

Now here is where my confusion is kicking in:  I am trying to solve for y2 using the equation 19.93(a) from the CERM book but I end up getting 1.52 m. Am I not supposed to use this equation? 

 

[Acedeala]

@Aurora09

Please read the paragraph below equation 19.94 on p. 19-25 of the CERM (16th Edition). It states that CONJUGATE DEPTHS ARE NOT THE SAME AS ALTERNATIVE DEPTHS!!! Alternative Depths process and examples are on page 19-17 to 19-18 of CERM (16th Edition). In Water Resources, skimming through the paragraphs after equations provides insight. In fact, the many equations have assumptions/conditions and you will not be aware unless you read it (or someone told you) in your review notes, text book, reference manual, and etc. On the exam they will trick you this way, by adding a word which force you to use an alternate equation from the norm. Do not be discourage, this just forces you to really understand concepts. I hope this helps.