Traffic and Highway Engineering
Traffic and Highway Engineering
5th Edition
ISBN: 9781305156241
Author: Garber, Nicholas J.
Publisher: Cengage Learning
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Chapter 16, Problem 20P
To determine

Whether the given culvert will operate under inlet or outlet control for the given conditions.

Expert Solution & Answer
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Answer to Problem 20P

The given culvert is acceptable and inlet control governs.

Explanation of Solution

Given:

The diameter of concrete circular box is 6.5ft.

The angle of inclination of flared walls is 45°.

The 50 years flow rate is 200ft3/sec.

The design headwater elevation is 105ft.

Elevation of stream bed at face of invert is 99.55ft

Tail water depth is 4.75ft

Approximate length of culvert is 200ft

Slope of stream is 1.5%.

The value of n is 0.012.

The value of k is 0.5.

Formula used:

The flow rate per width is given by,

  q=QD ....... (I)

Here, Q is the discharge and D is the diameter of circular box.

The required headwater is given by,

  HWi=(HWD)(depthofculvert) ....... (II)

Here, HWD is the headwater depth at culvert face.

The design headwater depth is given by,

  HWd=ELhdELsd ....... (III)

Here, ELhd is the design headwater elevation and ELsf is the elevation at the stream bed at the face.

The formula for fall is given by,

  Fall=HWiHWd ....... (IV)

The culvert invert elevation is given by,

  Culvert invert elevation=ELsfFall ...... (V)

The depth from outlet invert to hydraulic line is given by,

  ho=(dc+d)2 ...... (VI)

Here, dc is the critical depth and D is the depth of box.

The outlet water elevation is given by,

  ELo=(Culvert invert elevationS0)(Approximatelengthofculvert) ...... (VII)

Here, S0 is the slope.

The required outlet headwater elevation is given by,

  ELho=ELo+H+ho ...... (VII)

Here, H is the total head loss.

Calculation:

The flow rate per width is calculated as,

Substitute 200ft3/sec for Q and 6.5ft for D in equation (I).

  q=200ft3/sec6.5ft=30.8ft2/sec

Consider Figure 16.17, "Headwater Depth for Concrete Pipe Culverts with Inlet Control" from the book "Traffic and Highway Engineering", for a depth of 6.5ft and a discharge of 200ft3/s, the headwater depth of the culvert face HWD, is 0.88.

The required headwater is calculated as,

Substitute 0.88 for HWD and 6.5ft for depth of culvert in equation (II).

  HWi=0.88×6.5ft=5.72ft

The design headwater depth is calculated as,

Substitute 105ft for ELhd and 99.55ft for ELsf in equation (III).

  HWd=105ft99.55ft=5.45ft

The fall is calculated as,

Substitute 5.45ft for HWd and 5.72ft for HWi in equation (IV).

  Fall=5.72ft5.45ft=0.27ft

The culvert invert elevation is calculated as,

Substitute 99.55ft for ELsf and 0.27ft for HWd in equation (V).

  Culvert invert elevation=99.55ft0.27ft=99.28ft

The depth from outlet invert to hydraulic line is calculated as,

Substitute 3.75ft for dc and 6.5ft for D in equation (VI).

  ho=(3.75ft+6.5ft)2=5.13ft

The tail water is not greater so take ho=5.13ft.

The outlet water elevation is calculated as,

Substitute 99.05ft for culvert invert elevation, 0.015 for S0 and 200ft for approximate length of culvert in equation (VII).

  ELo=[99.05ft(0.015)(200ft)]=96.05ft

For circular culvert with n=0.012 the total head loss is 1.05ft.

The required outlet headwater elevation is calculated as,

Substitute 96.05ft for ELo, 5.13ft for ho and 1.05ft for H in equation (VIII).

  ELho=96.05ft+1.05ft+5.13ft=102.46ft ...... (VII)

The required outlet headwater elevation is less than the design headwater design so the given culvert is acceptable and inlet control governs.

Conclusion:

Therefore, the given culvert is acceptable and the inlet control governs.

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