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Use of Manning equation when a pipe runs full

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Tremolo

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Feb 26, 2003
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Can the Manning equation be used for case where a pipe is running full of water?

For example, in open channel flow where the channel flow runs through a culvert pipe, if the conditions are such that the culvert pipe is running full, can the Manning equation still be used? Are there any restrictions or limitations to the use of the Manning equation in this situation?

Thanks.


 
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Manning's will only give a snpashot of full capacity, due to area and friction. It can not account for HW/TW control, or indicate pressure head. So, as BigInch said, yes you will get a capacity. But, you will not see roadway topping, or backwater.

I always run a culvert analysis for an open, stand alone culvert in a channel. I always check all losses with Bernoulli's for a system.

Engineering is the practice of the art of science - Steve
 
Our concern is whether or not a 21” culvert will have sufficient capacity to prevent flooding of nearby property. I was trying to verify an engineering calculation and put together a bounding scenario for maximum culvert capacity, as follows:

1. Storm water flows freely through an open ditch until it reaches the culvert. The culvert is 120’ long, 21” corrugated steel pipe running through a railroad embankment.

2. Assume culvert is the primary restriction to flow, therefore culvert capacity controls total flow rate.

3. Neglect any back pressure/ head water on the discharge side of the culvert.

4. The maximum water level prior to flooding is 6.13 feet above the bottom of the culvert. Use this water level for any slope/head calculation.

5. Assume the culvert pipe runs full.

I applied the Bernoulli equation and Manning’s equation and got similar results:

BERNOULLI

Volumetric flow = q = A*sqrt(2*g*h/K)

A = (21/12)^2*pi/4 = 2.4 ft^2
K = 1 + fL/D + Kentrance + Kexit = 1+ .02*120/(21/12) + 0.5 + 1.0 = 3.9
h = water level to pipe centerline = 6.13 – 21/12/2 = 5.255 ft

q = 2.4*sqrt(2*32.2*5.255/3.9) = 22.4 cfs

MANNING

Volumetric flow = q = (1.49/n)* R^2/3 * S^1/2 * A

n = 0.014 (consistent with previous engineering calculation)
R = hydraulic radius = 1/4*(21/12) = 0.4375 ft
S = slope = h/Leq
Leq = equivalent length = L + (Kentrance + Kexit)*D/f = 120 + (0.5 + 1.0)*21/12*.02
= 251 ft
S = 5.255 / 251 = 0.0209 ft/ft

q = (1.49/0.014) * 0.4375^2/3 * 0.0209^1/2 * 2.4 = 21.3 cfs


So the Bernoulli equation and the Manning equation give similar results (22.4 and 21.3 cfs, respectively). This is a good thing. I am not familiar with the Churchill equation.

The original engineering calculation neglected the entrance and exit losses on the culvert, so I presented this calculation to the engineer. Then he said I shouldn’t be using Manning for a pipe running full. I felt pretty good about the calculation, since there was good agreement between the methods.

This is an important issue because the original calculation shows a culvert capacity that exceeds the storm water peak flow rate for a 100 year storm event. Hence, no flooding would be expected. My calculations show that the culvert capacity would be exceeded for even a 10 year storm event.

Any comments you may have are welcome. Thanks for the input so far.
 
BigInch: Thanks for the link. I downloaded the spreadsheet and am looking at it.

dicksewrrat: I have to double check, but I thought the culvert was corrugated steel pipe. The Ohio Dept of Transportation (ODOT) culvert design manual suggests n=.024 for 24" type A corrugated steel pipe. The HEC-RAS computer code users manual lists n=.016 for 24" corrugated pipe. I will have to look into this some more.

Thanks for the input.
 
0.024 is the Manning's used mainly for Annularly-wound CMP. They don't make small diameter (your 21") pipe Annularly anymore, but I think they used to. .016-.018 is more typical for design/analysis of Helically-wound CMP. Manufactures claim as low as 0.012, but I wouldn't use that.

You can easily look in it and see; Helical is the style where the corrugations "corkscrew" from one end to the other (looks liek a DNA Helix). Annular is where the corr's are symetrically arranged through out (looks like rings around the inside).

Your 0.014 would be a poorly-jointed/slightly corroded RCP, or conservative, but OK, for a Smooth-Lined pipe, including possibly a Bituminuous-coated CMP, in excellent shape.

Signature under construction, sorry about the mess - Steve
 
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