I am trying to work through the six-minute solution problems, WR and got stuck on #10 breadth. I think using the Continuity equation (A1*V1 = A2*V2) is enough to solve the problem (I arrive at .14), but my answer doesn't match with any of the answers in the provided solution (which is .28). I don't understand why my answer would differ in the back using the momentum equation... Please help me solve this mystery!!! I am desperate!
Thanks!
Please provide the problem.
Here is the problem:
A concrete-lined open channel is used to convey storm water runoff along a roadway. The roadway and channel make an abrupt transition from 12% slope to 2% slope, which causes a hydraulic jump to form. The channel is a triangular cross section and 1-to-1 side slopes. The upstream water depth is 10 cm. What is the water depth in the downstream channel section with the 2% slope?
I am not taking the credit for this answer. Somebody explained it 4yrs ago on this board. It still not very clear to me, but that what I found:
"
The conjugate depth method is the typical approach for determining the height of the hydraulic jump. The formula you most likely have seen is derived for a rectangular cross section and will not work for a circular, triangular, or other geometrical cross-section.
I have taken the exam before and I can say with some confidence that you would not be required to derive equations in order to solve the problem. Remember, you only have approximately 6 minutes per problem.
*IF* you wanted to derive the conjugate depth equation for a geometry other than rectangular, you utilize the following three equations:
Momentum
Energy
Continuity
So, for the triangular cross section you start with:
H = z + D + v^2/2g -- Bernoulli
However, you are going to set your datum to the channel bottom, so your equation looks more like:
H = D + v^2/2g
We know Q = vA from the continuity equation, which yields
H = D + Q^2/2gA (**) from simple substitution
Now, if you follow the King and Brater text, you can apply the following:
A = z*D; such that z = side slope and D = Depth and
Let x = D/H
Make those substitutions into H = D + Q^2/2gA, yields:
x^4 - x^5 = Q^2/(2*g*z^2*H^5)
Table 8.2 in the text is dedicated to providing values of x for values of equal energy, D = x*H. "