WR/Env Six Minute Solutions P. 15

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DanHalen

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The problem states:

A hydraulic jump occurs in a trapezoidal channel with a 4.2 m bottom width and 1:1 side slopes. The flow in the channel is 38 m3/sec and the water depth upstream of the jump is 0.74 m. What is the water depth downstream of the jump?

(A) 0.69 m

(B) 0.74 m

C) 2.3 m

(D) 3.3 m

The solution is fairly straight forward but I don't get the same answer. I keep getting 0.74 m and the answer should be 3.34 m. I use the Solve function on my calculator and it keeps giving me the same answer. I haven't been able to find an errata for this problem so I want to see if anyone else gets the same answer as me. This is what I'm using in my Casio FX-115:

(3.66*0.74)+((382)/(9.81*3.66))=X2(2.1+0.33X)+((382)/(9.81*(4.2X+X2)))

Hit solve and X = 0.74 m

 
I dont even know the equation for a trapezoidal channels hydraulic jump. The rectangular jump equation is straight forward.

I would guess that if the equation isn't in CERM it wont be on the exam.

 
Last edited by a moderator:
The solution begins by solving for the upstream and downstream areas in a trapezoidal channel. The formula is: A = bd+(d2)/(tan(theta)).

A1 = (4.2*0.74)+(0.742)/(tan(45o)) = 3.66 m2

A2 = (4.2*d2)+d22

Next, they use the specific force equation to solve for d2.

Ad = (d2(3b+(2d)/(tan(theta))))/(6)

so,

A1d1 = (0.742((3*4.2)+(2*0.74)/(tan(450)))/(6) = 1.29 m3

A2d2 = (d22((3*4.2)+(2*d2)/(tan(45o)))/(6) = d22(2.1+0.33d2)

A1d1+(Q2)/(gA1) = A2d2+(Q2)/(gA2)

1.29+(382)/(9.81*3.66) = d22(2.1+0.33d2)+(382)/(9.81(4.2d2+d22))

Solve for d2........ = 3.3 m

I found my mistake in working through this. I was forgetting to solve for A1d1 and was using (3.66*0.74) instead of 1.29.

 
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