1 / 11

Rock Creek Canal, Montana

Rock Creek Canal, Montana. n = 0.06 d 50 = 200 mm. http://wwwrcamnl.wr.usgs.gov/sws/fieldmethods/Indirects/nvalues/. At each cross section R, S, and U were measured and Mannings equation was rearranged to solve for n n = u m (R 2/3 )(S 1/2 ) U.

kiayada-olson
Download Presentation

Rock Creek Canal, Montana

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Rock Creek Canal, Montana n = 0.06 d50 = 200 mm http://wwwrcamnl.wr.usgs.gov/sws/fieldmethods/Indirects/nvalues/

  2. At each cross section R, S, and U were measured and Mannings equation was rearranged to solve for n n = um (R2/3)(S1/2) U

  3. Mannings roughness coefficients (n) for flood plains are usually much greater than the channel.

  4. Example Application of Manning’s Equation Example R = 1.37 ft S = 10 ft /1000ft n = 0.05 U = um(R2/3)(S1/2) = 1.49 (1.37)2/3 (10/1000)1/2= 3.7 ft/sec n 0.05 Channel discharge and average velocity Q = A • U Q = discharge A = cross sectional area U = average flow velocity Example: U = 3.7 ft/sec, A = 17 ft2 Q = A • U = (17 ft2) ( 3.7 ft/sec) = 62.9 ft3/sec

  5. Estimated rating curve for an ungaged channel: Determine channel bed geometry For several assumed flow stages (water surface elevations) calculate wetted perimeter, cross sectional area and hydraulic radius estimate Manning’s n for different stages calculate water flow velocity, U = um (R2/3)(S1/2)/ n calculate discharge Q = A • U plot Q vs water surface elevations to determine flow depth at various values of Q Estimated Discharge Assumed Stage

  6. Convex Routing Method UF = 1.67 • [ um (Y2/3)(S1/2)] n UF = velocity of flood wave Y = water depth  R Slope = S n = Mannings roughness Routing Coefficient, CX = t / T* t = time step T* = X /UF = time of flood wave travel through river reach length X X = River reach length CX = t = t • UF= t • 1.67 • [ um (Y2/3)(S1/2)] T* X n X Error in text page 431: “… reduce the speed of the flood wave and decrease the value of CX…”

  7. Bankfull flow – maximum discharge that can be conveyed by a channel. If a flood is equal to or less than bankfull, and if hydraulic resistance (n) is very small, the velocity of a flood wave (UF) may be as large as 1.67 times the average velocity of the water: UF = 1.67 U A flood wave can move faster than the flood water because prior to peak flow, at the leading edge of the flood wave, the water surface slope is greater than the channel slope: Y > Z xx But, several factors can cause UF to be less than 1.67 U such as: flood plain storage, hydraulic resistance (n), flat slopes, shallow water

More Related