Chapter 4: Culvert Hydraulics Jam Aleem Nawaz Amanda Yancoskie - PowerPoint PPT Presentation

Chapter 4: Culvert Hydraulics Jam Aleem Nawaz Amanda Yancoskie

Table of Contents ● Introduction ● Types ● Inlet Control ● Outlet Control ● Tutorial Example ● Problem Statement ● Questions

4.5 Tutorial Example

Problem Statement Several circular, concrete culverts that have square-edged entrances with headwalls (n=0.013, k e =0.5) must be used to carry 54 m 3 /s. The culverts are 20 meters long, on a 5.5 percent slope with and upstream and downstream invert elevations of 261.0 meters and 259.9 meters, respectively. The maximum allowable headwater elevation is 268 meters, and the elevation of the tailwater is 257.4 meters. a) How many culverts should be used if the diameter of the culverts is 1950mm? b) What is the headwater depth? c) Is the system flowing under inlet or outlet control? d) What is the velocity at the culvert exit? e) Construct a performance curve showing the headwater elevation as a function of discharge. Consider both inlet and outlet control.

When you start CulvertMaster, you should be prompted with the Welcome to Bentley CulvertMaster dialog. Select Create New Project .

Navigate to the folder where you want to save your project, enter a filename, and click Save .

The Project Setup wizard will appear. Add a project title, project engineer, and appropriate comments, and click Next .

In this example we will not be defining a rainfall table or a rainfall equation, so select Next .

We need to check the unit system before creating an element , so select the None radio button and Finished to exit.

Select Options from the Tools pulldown menu.

Change the unit system to System International . Click OK to exit the dialog.

Click the Quick Culvert Calculator button on the left hand side of the window.

Provide a title for the worksheet and click OK .

Click the drop-down arrow next to the Solve For: data entry field and Select Discharge from the list.

Enter: Maximum Allowable HW: 268 m Tailwater Elevation: 257.4 m Invert Upstream: 261 m Invert Downstream: 259.9 m Length: 20 m Shape: Circular Material: Concrete Size: 1950 mm Mannings: 0.013 Entrance: square edge w/headwall Ke: 0.5 Note: You can change the units by double-clicking the units label next to the variable.

The default value for the number of culvert barrels is 1 . Run the model using one culvert barrel by clicking the Solve button.

The Discharge value that is calculated is displayed in the Exit Results part of the worksheet. Run the model several times, increasing the number of culvert barrels until the calculated discharge is ≥ 54 m 3 /s.

a) How many culverts should be used if the diameter of the culverts is 1950 mm? Answer: 3 culverts with 1950 mm diameter will carry 61 m 3 /s

b) What is the head water depth? Answer: The headwater depth is equal to the difference between the Computed Headwater and the Inverted Upstream : 268.0 m - 261.0 m = 7.0 m

c) Is the system flowing under inlet or outlet control? Answer: The system is flowing under inlet control, because the inlet control elevation is greater then the outlet control. Inlet: 268.00 m ＞ Outlet: 266.34 m

d) What is the velocity at the culvert exit? Answer: 8.19 m/s.

e) Construct a performance curve showing the headwater elevation as a function of discharge. Consider both inlet and outlet control. Click the Output button.

Select the Plot Curves button.

Under the Compute heading, select Inlet Control HW Elev. (m) and Outlet Control Elev. (m) . Click the drop-down arrow next to Vary: and select Discharge (m 3 /s) . Set the graph scale to, Minium: 0.0 m 3 /s Maximum: 100 m 3 /s Increment: 20 m 3 /s Click the Refresh button.

The graph shows the performance curve for the three-barrel culvert set.

Any Questions?