Hydraulic Jump Equations, Non-Rectangular Channels

[Timmy!]

OK, so I'm the 52 year-old Electrical PE who is taking the Civil PE this April in WR (think Kevin Bacon getting paddled in "Animal House": "Thank you sir, may I have another"!!!)

The CERM readily gives the equation to calculate the hydraulic jump for a rectangular channel. I was solving a problem for hydraulic jump with a triangular channel [six-Minute Solutions, WR, #10 Breadth] and was told that I need to derive the hydraulic jump equation for a triangular channel from the momentum equation.

Ain't got time for that horseshit during the exam. Anyone got a list of hydraulic jump equations for trapezoidal, triangular and circular channel cross-sections?

 

[Guest]

Anyone got a list of hydraulic jump equations for trapezoidal, triangular and circular channel cross-sections?

First, I would like to say excellent taste in movies. I am still on double-secret probation from the 'other' forum :"the other board": I have to watch out for

Now for your question.

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.

This is the easiest geometrical shape - it gets a lot more difficult from. If you have specific questions, you can post follow-on questions and I can try to answer.

Regards,

JR

 

[Timmy!]

OK, so I'm the 52 year-old Electrical PE who is taking the Civil PE this April in WR (think Kevin Bacon getting paddled in "Animal House": "Thank you sir, may I have another"!!!)
The CERM readily gives the equation to calculate the hydraulic jump for a rectangular channel. I was solving a problem for hydraulic jump with a triangular channel [six-Minute Solutions, WR, #10 Breadth] and was told that I need to derive the hydraulic jump equation for a triangular channel from the momentum equation.

Ain't got time for that horseshit during the exam. Anyone got a list of hydraulic jump equations for trapezoidal, triangular and circular channel cross-sections?

 

[Timmy!]

Thanks, jregieng. I'll take that advice to heart, and also thanks for leading me along the paths of righteous for Civil Engineering's sake by giving me the tip of momentum/energy/continuity.

Really, I must be a sick puppy. I absolutely love studying this stuff. Honest, I do. I'd take the Mechanical PE exam if the State of Arizona would let me.

 

[Guest]

Thanks, jregieng. I'll take that advice to heart, and also thanks for leading me along the paths of righteous for Civil Engineering's sake by giving me the tip of momentum/energy/continuity.
Really, I must be a sick puppy. I absolutely love studying this stuff. Honest, I do. I'd take the Mechanical PE exam if the State of Arizona would let me.

Glad I could help !!

I am a sick bastard too !! I like studying this stuff so much, I am STILL in school :lmao:

I just passed the Oct 2006 Civil PE Exam WR Depth. It took me 4 times, but I finally got it right :thumbs: Please feel free to continue posting questions!

Best of luck in your preparations.

JR

 

[Cheese]

This page is helping me - nice charts.

http://www.fhwa.dot.gov/engineering/hydrau...6/hec14ch06.cfm

PS - Glad to have found this website - great resource.

Cheese

 

[ktulu]

Cheese-

That's a nice link...thanks for the information.....

ktulu

 

[Cheese]

I have two questions about hydraulic jump.

1) Why is the froude # calculated differently when finding hydraulic jump. I have

F=V/(gycos(tan-1S)^.5) for rectangular channels (S=slope)

and

F=V/(gy)^.5 when calculating for hydraulic jump in rectangular channels.

Is the 1st equasion for uniform, steady flow only. Obviously the flow is inturupted when there is a jump.

2) I have two formula for energy disipated h=y1+v1^2/2g-y2-v2^2/2g

and

h=(y2-y1)^3 / (4y1*y2)

Are these used differently, they seem to produce different answers.

Please help.

Thanks,

Cheese

 

[Guest]

Cheese --

In response to your questions:

1. The second equation is the 'basic' equation for the Froude Number. However, I think you have an error in the formula. The bottom term should be the square root of the product of the gravitational acceleration and the LENGTH (L) of interest, not the DEPTH .

The first equation is a modification of the second 'basic' equation accounting for the slope of the channel (first equation assumes level channel bottom). In this case, if you are not provided the horizontal length along the channel, then you must find that horizontal length based on the trigonometry of the problem (Think triangle with respect to depth from water surface and legth of channel bottom).

I can pull out my references and elaborate further if necessary.

Both equations assume uniform (water surface level parallel to channel bottom) and steady-state (not varying over time).

2. The first equation that you provided is the 'general' equation for energy loss based on the energy equation.

The second equation is the special case when the cross-section is rectangular. Therfore, if you assume:

i. The channel cross section is rectangular;

ii. The rectangular width is b; and

iii. The discharge per unit width is q = Q/b

Then the hL equation simplifies to the 2nd equation you provided. Or, in other words, the second equation is the specific form for a rectangular channel.

I can expand on this as necessary as well.

I hope this helps. Let me know if I need to add detail.

JR

 

[Cheese]

Thanks JR. That makes a lot of sence.

I now understand the y = Dh = hydraulic depth = Flow area / top water surface length

Is that how you understand it?

Freude # = V/(g*Dh) (general form).

Another quicky, is characteristic length the same as hydraulic diameter = 4A/wetted perimeter.

 

[Guest]

Cheese,

In response to your comments:

I now understand the y = Dh = hydraulic depth = Flow area / top water surface length. Is that how you understand it?

Freude # = V/(g*Dh) (general form).

This is correct

Another quicky, is characteristic length the same as hydraulic diameter = 4A/wetted perimeter.

I want to be careful in answering this one. I have two questions for you:

1. What are you calling characteristic length ?

2. In the relationship above you say hydraulic diameter. Do you mean hydraulic radius ?

In general, hydraulic radius = cross-sectional area / wetted perimeter. I noted you have 4A in the numerator of your relationship. I am thinking this is probably a special case - where did you get that relationship from?

These are very good questions - these concepts will most likely be tested on the exam in some form.

JR

 

[Cheese]

JR

What I think I think I know.

Hydraulic Radius - used frequently for open channel flow - mannings eqn specifically. Rh = (Cross Sectional flow area) / (Wetted perimeter)

the hydraulis diameter is NOT = 2 * Hydraulic Radius. Hydraulic Diameter (Dh) is entirely different. Hydraulic diameter is used frequently in pressure pipes specifically when finding Renolds #.

Re = V/(Dh*fluid viscosity)

where

Dh = 4*(Cross Sectional flow area) / (Wetted Perimeter)

Dh = 2*R = D for full flowing circular pipes only

It would appear that hydraulic diameter = 4* hydraulic radius ??

hydraulic depth is used when determining Froude # and when comparing depths (congugate method) determing critical flow and flow in wiers, etc.

y = (Cross Sectional flow area) / top width of water

rectangular channel....

area = channel width*flow height

top water lenght = channel width

thus

y = flow height for rectangular channel (not the length) for square channels

I guess I don't know what the characteristic lenght is.... I thought is was sonomous (the same as) hydraulic diameter.

Sorry for the rambling....

Anybody see any problems with my thoughts here?

And miles to go before I sleep..........

Cheese

 

[Cheese]

In above I have expressed Re incorrectly

Re = VDh/viscosity

sorry...

by the way my definition for Dh is from

http://en.wikipedia.org/wiki/Hydraulic_diameter

 

[Guest]

Cheese,

Looking over your musings (no problem - we all need to talk out loud every now and then) it seems that you have everything in order.

I would add a word of caution - I haven't used hydraulic diameter (Dh) when approaching solutions to my problems. I have always used hydraulic radius. There are some emperical relationships (e.g. Manning) that specifically use hydraulic radius in the application of solving for velocity and conveyance (flow). My word of caution is to make sure you are clear about each terms' meaning. That's why I asked you about hydraulic diameter.

Again, looking over your musings, you appear to be grasping the concepts. Please post again if you have any questions or problems.

JR