Question

8 A reinforced concrete rectangular box culvert has the following properties: D=1 m, b=1m, L=40 m, n=0.012, S=0.002. The inlet is square-edge on three edges and has a headwater parallel to the embankment. The tailwater depth = 0.6 m. Determine the headwater depth (in m) when the culvert is flowing partially full at Q=2 m3/s.* 15 (6 Points) O 7.35 2.99 0 3.56 01.40 01.24

Answer 1

discharge in the culvert =2m3/s

the area of the rectangular culvert of D= 1 and b = 1 = 1X1 = 1m2

thevelocity of water V =Q/A ( by continuity eqn)

V=2/1 =2m/s

Hw Q CO :)? +Y – 0.55 D

Where:
HW is headwater depth at the culvert entrance
Q is discharge through the culvert =2 m2/s
A is full open area of the culvert 1m2
D is culvert rise =1m
S is the slope of the culvert barrel s =0.002

c and Y are inlet control regression coefficients for submerged conditions.

c=0.5 and y =1

Hw 2 = 0.50 1x10.5)2 +1 - (0.5X0.002) = 2.99 D

Hw = dx2.99=1x2.99=2.99m

or

n HDS-5 design methodology, outlet control is determined assuming that the culvert is flowing full. The headwater due to outlet control is found from Equation 5, which is an energy balance between the upstream and downstream ends of the culvert.

Equation 5:

n HDS-5 design methodology, outlet control is determined assuming that the culvert is flowing full. The headwater due to outlet control is found from Equation 5, which is an energy balance between the upstream and downstream ends of the culvert.

Equation 5:

Where:
HW is headwater depth above the inlet invert (m);
EL0 is the elevation of the culvert invert at the outlet;
H0 is the governing tailwater (fm);
hL is head loss through the culvert (m).

HW = ELO + Ho+h

To find the governing tailwater, H0, the critical depth in the culvert must first be determined.

Where:
TW is the tailwater at the downstream end of the culvert (feet);
DC is critical depth in the culvert (feet);
D is culvert diameter or rise (feet).

Dc+D Ho = max(TW, 2

$H_{O}=max(0.6,\frac{0.33+1}{2}=0.665m$

The head loss through the culvert, hL , is found by considering all losses, including entrance losses, exit losses, and friction losses. Manning's equation is rearranged to quantify friction losses.
$h_{L}=\frac{V^{2}}{2g}(K_{x} +\frac{29n^{2}L}{R^{1.33}}+K_{e})$
Where:
Kx is an exit loss coefficient;
n is Manning's roughness coefficient;
L is the length of the culvert (feet);
R is the hydraulic radius of the culvert (feet);
Ke is an entrance loss coefficient;
V is velocity in the culvert (feet per second);
g is the gravitational constant (feet per second per second).

$h_{L}=\frac{V^{2}}{2g}(K_{x} +\frac{29n^{2}L}{R^{1.33}}+K_{e})$

$h_{L}=\frac{2^{2}}{2X9.81}(1+\frac{29X0.012^{2}X40}{0.25^{1.33}}+1)=0.6229m$

$H_{W}=EL_{o}+H_{o}+h_{L}$

Where:
HW is headwater depth above the inlet invert (m);
EL0 is the elevation of the culvert invert at the outlet;=1m
H0 is the governing tailwater (m);=0.665m
hL is head loss through the culvert (m).=0.663m

To find the governing tailwater, H0, the critical depth in the culvert must first be determined. The critical depth is then

the option 2 Hw =2.99m can be choosen as the answer