manning's formula

Department of Transport and Main Roads Appendix 8B Road Drainage Manual Worked Examples

Appendix 8B Worked Examples

8B

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March 2010

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8B

Appendix 8B Amendments – Mar 2010

Revision Register Issue/

Rev

No.

Reference

Section

Description of Revision Authorised

by

Date

1 - Initial Release of 2nd Ed of manual. Steering Committee

Mar 2010


Manning’s Formula Example 1

This example describes the process to determine the flow rate, the velocity of a flow and the state of flow in a stream based on discussion in Section 8.4.

The example commences after the stream data (such as cross section, terrain, condition of channel and stream profile to determine site bed slope) has been gathered (refer Chapter 4).

The task for this example is, given the stream data and height of flow (see diagram below), determine the velocity of flow in the channel, the flow rate and state of flow (subcritical / critical / supercritical flow).

Stream Data

Bed slope about the site is 0.8%

Ht of channel bed is 110.60 m

Channel is regular and considered a little rough with a lot of trees and weeds along the banks.

sides are 1 on 1

Ht = 111.80 m

2.5m Solution

8B We need to determine the velocity of flow using Manning’s formula first, then the flow rate using the fundamental equation and finally, determine Froude’s number to describe the state of flow. (Refer Sections 8.4.3, 8.4.4 and 8.4.5)

Manning’s equation is:

nSR

V5.0667.0

Where R is the hydraulic radius, determines as cross sectional area of flow (A) divided by the wetted perimeter (P). Also, S is the slope of the energy line which we don’t have, therefore we can use the bed slope (So) to approximate S.

Step 1. Calculate the cross sectional area of the flow.

A = 1.22 + 1.2 x 2.5 = 4.44 m2

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Step 2. Calculate the wetted perimeter.

P = (1.22 + 1.22) x 2 + 2.5 = 5.89 m

Step 3. Calculate the hydraulic radius.

R = A / P = 4.44 / 5.89 = 0.75 m

Step 4. Now determine an appropriate Manning’s roughness coefficient.

Using Table 8.4.3(b), Manning’s n Values for Natural Channels, we can see that our channel, being the main channel and regular in shape, places us in the top portion of the table and within Section 1. The trees and weeds would suggest (e), the range 0.06 – 0.08. Now, with the bed being a little rough, a value n = 0.07 (in the middle of range) is considered appropriate.

Step 5. All variables have now been determined, therefore calculate velocity.

07.0008.075.0 5.0667.0 V

V = 1.06 m/s

Step 6. Using fundamental equation Q = V.A, we can now determine the flow rate in the channel.

8B

= 4.71 m3 /s

Q = 1.06 x 4.44

Step 7. To determine the ‘state of flow’, we calculate Froude’s number.

3gABQFr

We have determined that Q = 4.71m3/s, A = 4.44m2 and g is acceleration due to gravity (taken as 9.81m/s2), therefore we need to calculate B, the width of flow across the surface.

B = 1.2 + 2.5 + 1.2 = 4.9 m

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Therefore:

344.481.99.471.4

rF

Fr = 0.36

Froude’s number is below 1.0, therefore the flow is subcritical.

End of Example

8B

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8B



This example describes the process to determine the depth and velocity of flow based on a known discharge / flow rate in a stream, based on discussion in Section 8.4.

The example commences after the stream data (such as cross section, terrain, condition of channel and stream profile to determine site bed slope) has been gathered (refer Chapter 4) and the flow rate (as determine using Rational Method) has been determined (refer Chapter 5).

The task for this example is, given the stream data and flow rate (see diagram below), determine the depth and velocity of flow in the channel.

Stream Data

Discharge / flow rate = 17.86 m3/s


Ht of channel bed is 65.10 m

Max depth of flow is 2.0 m

Manning’s n = 0.06.

8B

d = ?? m

sides are 1 on 1

4m

Solution

To solve for d, we need to use Manning’s formula and develop a Stage-Discharge curve.

nSR

V5.0667.0

A Stage-Discharge curve plots discharge against depth of flow. Therefore several iterations using Manning’s formula are required for several depths of flow.

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8B-5


Step 1. Using the maximum channel depth of 2.0m, calculate stream velocity and flow rate.

Calculate the cross sectional area of the flow, wetted perimeter and hydraulic radius:

A = 12.00 m2, P = 9.66 m therefore R = 1.24 m

Now,

smV /11.206.0

012.024.1 5.0667.0

Using Q/VA, Q=25.33 m3/s

This flow is greater than the known discharge therefore we know that the channel can easily carry the flow.

Step 2. Now, using the same method, re-calculate stream velocity and flow rate for several lesser depths (suggest using even increments).

Depth A P R Velocity Discharge

1.50 8.25 8.24 1.00 1.83 15.07

1.00 5.00 6.83 0.73 1.48 7.42

0.50 2.25 5.41 0.42 1.02 2.29

Step 3. Now draw the Stage-Discharge curve for this site / channel (refer next page).

8B

Step 4. From the curve, we can now read of the flow depth for our design flow of 17.86m3/s.

Q = 17.86 m3/s, therefore d = 1.62 m

Step 5. Now we can use the depth to calculate flow area, then Q=V.A to determine the average flow velocity.

A = 1.622 + 1.62 x 4 = 9.10 m2

17.86 m3/s = V x 9.10 m2

V = 1.96 m/s

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Stage-Discharge Curve

0

5

10

15

20

25

30

0 0.5 1 1.5 2 2.5

Flow Depth (m)

Flo

w R

ate

(m3/

s)

8B

End of Exercise

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8B



This example describes the process to determine the flow rate, the average velocity of a flow in a compound stream based on discussion in Section 8.4.

The example commences after the stream data (such as cross section, terrain, Manning’s n and stream profile to determine site bed slope) has been gathered (refer Chapter 4).

The task for this example is, given the stream data and height of flow (see diagram below), determine the average velocity of flow in the channel and the flow rate.

Stream Data


d = 1.2m

8B

Solution

To solve for Q, we need to use Manning’s formula for each sub section of stream:

nSR

V5.0667.0

After calculating V for each sub-section, use Qtotal = VA x AA + VB x AB + VC x AC to determine total flow rate.

Step 1. For sub-section A, calculate VA using Manning’s equation.

Calculate the cross sectional area of the flow, wetted perimeter and hydraulic radius for sub-section A are:

AA = (1.22 /2) + 1.2 x 4.0 = 5.52 m2

PA = (1.22 + 1.22) + 2.5 = 5.70 m

sides are 1 on 1

d = 2.1m d = 1.0m n values A = 0.07

4m

2m

3mA B

CB = 0C = 0.0

.035 6

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It is important to remember that that water - water boundary between sub-sections A & B does not contribute any length to the wetted perimeter.

RA = A / P = 4.44 / 5.89 = 0.97 m

Now,

smVA /25.107.0

008.097.0 5.0667.0

Using Q/VA , QA = 6.90 m3/s

Step 2. For sub-section B, calculate VB using Manning’s equation.

AB = 7.39 m2

PB = 4.83 m

RB = 1.53 m

Now,

smVB /40.3035.0

008.0531.1 5.0667.0

Using Q/VA, QB = 25.09 m3/s

8B

Step 3. For sub-section C, calculate VC using Manning’s equation.

AC = 3.50 m2

PC = 4.41 m

RC = 0.79 m

Now,

smVC /28.106.0

008.0793.0 5.0667.0

Using Q/VA, QC = 4.47 m3/s

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8B

Step 4. Now we can calculate:

Qtotal = 6.90 + 25.09 + 4.47 = 36.46 m3/s

and

Atotal = 5.52 + 7.39 + 3.50 = 16.41 m2

therefore

Vavg = 36.46 / 16.41 = 2.22 m/s

End of Exercise

manning's formula

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