## Water Tank

Consider the below water tank with water height H = 8 meters and a nozzle cross-sectional area A = 6 cm^2. Assuming there are no minor losses in the discharge, and a common coefficient of discharge of 0.6, what is the tank’s discharge velocity?

##
__
__**Hint**

**Hint**

$$$Q=CA(2gh)^{1/2}$$$

where
$$Q$$
is the volumetric flow rate,
$$C$$
is the coefficient of discharge,
$$A$$
is the cross sectional area of flow,
$$g$$
is the acceleration due to gravity, and
$$h$$
is the height of the fluid above the orifice.

##
__
__**Hint 2**

**Hint 2**

$$$Q=vA$$$

where
$$Q$$
is the volumetric flow rate,
$$v$$
is the fluid’s velocity, and
$$A$$
is the cross sectional area of flow.

For an orifice discharging fluid freely into the atmosphere:

$$$Q=CA(2gh)^{1/2}$$$

where
$$Q$$
is the volumetric flow rate,
$$C$$
is the coefficient of discharge,
$$A$$
is the cross sectional area of flow,
$$g$$
is the acceleration due to gravity, and
$$h$$
is the height of the fluid above the orifice.

Since
$$Q=vA$$
where
$$v$$
is the fluid’s velocity and
$$A$$
is the cross sectional area of flow, we can combine the two equations into one and rearrange to solve for velocity:

$$$vA=CA(2gh)^{1/2}$$$

$$$v=C(2gh)^{1/2}=(0.6)\sqrt{2(9.8m/s^2)(8m)}$$$

$$$v=(0.6)\sqrt{156.8m^2/s^2)}=(0.6)(12.52m/s)=7.5\:m/s$$$

7.5 m/s