## Water Slope

Consider water is fully flowing through a 0.5 ft diameter pipe at a 5% slope. If the pipe’s roughness coefficient is 0.02, what is the volumetric flow in cubic feet per second?

##
__
__**Hint**

**Hint**

Manning’s equation:

$$$Q=\frac{K}{n}AR_{H}^{2/3}S^{1/2}$$$

where
$$Q$$
is the volumetric discharge,
$$A$$
is the cross sectional flow area,
$$K=1.486$$
for USCS units,
$$n$$
is the roughness coefficient,
$$R_H$$
is the hydraulic radius, and
$$S$$
is the slope.

##
__
__**Hint 2**

**Hint 2**

To find hydraulic radius:

$$$R_H=\frac{A}{P}$$$

where
$$A$$
is the area, and
$$P$$
is the perimeter.

Manning’s equation:

$$$Q=\frac{K}{n}AR_{H}^{2/3}S^{1/2}$$$

where
$$Q$$
is the volumetric discharge,
$$A$$
is the cross sectional flow area,
$$K=1.486$$
for USCS units,
$$n$$
is the roughness coefficient,
$$R_H$$
is the hydraulic radius, and
$$S$$
is the slope. To find hydraulic radius:

$$$R_H=\frac{A}{P}$$$

where
$$A$$
is the area, and
$$P$$
is the perimeter. Thus, the hydraulic radius is:

$$$R_H=\frac{(\pi/4)(D^2) }{(\pi D)}=\frac{D}{4}=\frac{0.5ft}{4}=0.125\:ft$$$

To find the volumetric discharge:

$$$Q=\frac{1.486}{0.02}\cdot \frac{\pi}{4}(0.5)^2(0.125)^{2/3}\sqrt{5/100}$$$

$$$Q=74.3\cdot \frac{\pi}{4}(0.25)(0.25)\sqrt{0.05}=0.82\:ft^3/s$$$

$$$0.82\:ft^3/s$$$