## Discharging Orifice

Consider the shown water tank with a nozzle cross-sectional area of 10 cm^2 and a discharge velocity of 5 m/s. Assuming there are no minor losses in the discharge and a common coefficient of discharge of 0.2, what is the water line’s height from the nozzle?

##
__
__**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, it’s possible to combine the two equations into one and rearrange to solve for height:

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

$$$v=C(2gh)^{1/2}=(0.2)\sqrt{2(9.8m/s^2)}(h)^{0.5}=5m/s$$$

$$$\sqrt{2(9.8m/s^2)h}=\frac{5m/s}{0.2}=25m/s$$$

$$$2(9.8m/s^2)h=(25m/s)^2=625m^2/s^2$$$

$$$h=\frac{625m}{2(9.8)}=31.89m=32\:m$$$

32 m