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!

# Six Minute Solutions - Water Resources Breadth #10

Started by
ckjy
, Mar 04 2011 02:40 AM

5 replies to this topic

### #1

Posted 04 March 2011 - 02:40 AM

### #2

Posted 05 March 2011 - 12:31 AM

Continuity and momentum need to be applied, rather than just continuity due to the initial velocity of the water and the resulting hydraulic jump. In essence, the water will eventually lower to a depth of 0.14 m, but where the slope of the channel changes.

### #3

Posted 05 March 2011 - 05:47 AM

QUOTE (sac_engineer @ Mar 4 2011, 04:31 PM) <{POST_SNAPBACK}>

Continuity and momentum need to be applied, rather than just continuity due to the initial velocity of the water and the resulting hydraulic jump. In essence, the water will eventually lower to a depth of 0.14 m, but where the slope of the channel changes.

Thanks for much for your response! I applied both continuity and momentum and reached the result of 0.28. However, if I plug that back into the continuity equation it doesn't balance, however. The momentum equation as used in the solution doesn't take into account the second slope - S2 isn't even used in the problem. How can that be?

### #4

Posted 05 March 2011 - 09:06 PM

QUOTE (ckjy @ Mar 3 2011, 09:40 PM) <{POST_SNAPBACK}>

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 help me solve this mystery!!! I am desperate!

Thanks!

Please provide the problem.

### #5

Posted 05 March 2011 - 11:19 PM

QUOTE (IlPadrino @ Mar 5 2011, 01:06 PM) <{POST_SNAPBACK}>

QUOTE (ckjy @ Mar 3 2011, 09:40 PM) <{POST_SNAPBACK}>

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 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?

**Edited by ckjy, 05 March 2011 - 11:21 PM.**

### #6

Posted 10 March 2011 - 03:26 PM

QUOTE (ckjy @ Mar 5 2011, 07:19 PM) <{POST_SNAPBACK}>

QUOTE (IlPadrino @ Mar 5 2011, 01:06 PM) <{POST_SNAPBACK}>

QUOTE (ckjy @ Mar 3 2011, 09:40 PM) <{POST_SNAPBACK}>

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. "

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