Chapter 16, Problem 19P

To determine

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

Answer to Problem 19P

The given culvert is acceptable and the inlet control governs.

Explanation of Solution

Given:

The dimension of concrete box is $5\text{ft}\times 5\text{ft}$.

The angle of inclination of flared walls is $45°$.

The $50$ years flow rate is $200{\text{ft}}^3/\text{sec}$.

The design headwater elevation is $105\text{ft}$.

Elevation of stream bed at face of invert is $99.55\text{ft}$

Tail water depth is $4.75\text{ft}$

Approximate length of culvert is $200\text{ft}$

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,

  $q=\frac{Q}{B}$ ....... (I)

Here, $Q$ is the discharge and $B$ is the width.

The required headwater is given by,

  $H{W}_{\text{i}}=\left(\frac{HW}{D}\right)\left(\text{depth}\text{of}\text{culvert}\right)$ ....... (II)

Here, $\frac{HW}{D}$ is the headwater depth at culvert face.

The design headwater depth is given by,

  $H{W}_{\text{d}}=E{L}_{\text{hd}}-E{L}_{\text{sd}}$ ....... (III)

Here, $E{L}_{\text{hd}}$ is the design headwater elevation and $E{L}_{\text{sf}}$ is the elevation at the stream bed at the face.

The formula for fall is given by,

  $\text{Fall}=H{W}_{\text{i}}-H{W}_{\text{d}}$ ....... (IV)

The culvert invert elevation is given by,

  $\text{Culvert invert elevation}=E{L}_{\text{sf}}-\text{Fall}$ ...... (V)

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

  $h_{\text{o}}=\frac{\left(dc+d\right)}{2}$ ...... (VI)

Here, $d_{\text{c}}$ is the critical depth and $D$ is the depth of box.

The outlet water elevation is given by,

  $E{L}_{\text{o}}=\left(\text{Culvert invert elevation}-S_0\right)\left(\text{Approximate}\text{length}\text{of}\text{culvert}\right)$ ...... (VII)

Here, $S_0$ is the slope.

The required outlet headwater elevation is given by,

  $E{L}_{\text{ho}}=E{L}_{\text{o}}+H+h_{\text{o}}$ ...... (VII)

Here, $H$ is the total head loss.

Calculation:

The flow rate per width is calculated as,

Substitute $200{\text{ft}}^3/\text{sec}$ for $Q$ and $5\text{ft}$ for $B$ in equation (I).

  $\begin{array}{c}q=\frac{200{\text{ft}}^3/\text{sec}}{5\text{ft}}\\ =40{\text{ft}}^2/\text{sec}\end{array}$

Consider Figure 16.16, "Headwater Depth for Inlet Control, Rectangular Box Culverts, Flared Wingwalls $18°$ to $33.7°$ and $45°$ with Beveled Edge at Top of Inlet" from the book "Traffic and Highway Engineering", for a height of $5\text{ft}$ and flow rate per width of $40\text{CFS}/\text{ft}$, the headwater depth of the culvert face $\frac{HW}{D}$, is $1.19$.

The required headwater is calculated as,

Substitute $1.19$ for $\frac{HW}{D}$ and $5\text{ft}$ for depth of culvert in equation (II).

  $\begin{array}{c}H{W}_{\text{i}}=1.19\left(5\text{ft}\right)\\ =5.95\text{ft}\end{array}$

The design headwater depth is calculated as,

Substitute $105\text{ft}$ for $E{L}_{\text{hd}}$ and $99.55\text{ft}$ for $E{L}_{\text{sf}}$ in equation (III).

  $\begin{array}{c}H{W}_{\text{d}}=105\text{ft}-99.55\text{ft}\\ =5.45\text{ft}\end{array}$

The fall is calculated as,

Substitute $5.45\text{ft}$ for $H{W}_{\text{d}}$ and $5.95\text{ft}$ for $H{W}_{\text{i}}$ in equation (IV).

  $\begin{array}{c}\text{Fall}=5.95\text{ft}-5.45\text{ft}\\ =0.50\text{ft}\end{array}$

The culvert invert elevation is calculated as,

Substitute $99.55\text{ft}$ for $E{L}_{\text{sf}}$ and $0.50\text{ft}$ for $H{W}_{\text{d}}$ in equation (V).

  $\begin{array}{c}\text{Culvert invert elevation}=99.55\text{ft}-0.50\text{ft}\\ =99.05\text{ft}\end{array}$

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

Substitute $3.7\text{ft}$ for $d_{\text{c}}$ and $5\text{ft}$ for $D$ in equation (VI).

  $\begin{array}{c}{h}_{\text{o}}=\frac{\left(3.7\text{ft}+5\text{ft}\right)}{2}\\ =4.35\text{ft}\end{array}$

The tail water is greater so take $h_{\text{o}}=4.75\text{ft}$.

The outlet water elevation is calculated as,

Substitute $99.05\text{ft}$ for culvert invert elevation, $0.015$ for $S_0$ and $200\text{ft}$ for approximate length of culvert in equation (VII).

  $\begin{array}{c}E{L}_{\text{o}}=\left[99.05\text{ft}-\left(0.015\right)\left(200\text{ft}\right)\right]\\ =96.05\text{ft}\end{array}$

For $n=0.012$ the total head loss is $2.2\text{ft}$.

The required outlet headwater elevation is calculated as,

Substitute $96.05\text{ft}$ for $E{L}_{\text{o}}$, $4.75\text{ft}$ for $h_{\text{o}}$ and $2.2\text{ft}$ for $H$ in equation (VIII).

  $\begin{array}{c}E{L}_{\text{ho}}=96.05\text{ft}+2.2\text{ft}+4.75\text{ft}\\ =103\text{ft}\end{array}$ ...... (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.