## Jet Velocity

A 0.5 m tall water cooler jug is flipped upside down to freely flow water onto the ground through the jug's opening. If the jug was 1 m above the floor when flipped and losses are neglected, what is the jet velocity when the jug is half full of water?

##
__
__**Hint**

**Hint**

$$$\frac{P_1}{\rho g}+z_1+\frac{v_{1}^{2}}{2g}=\frac{P_2}{\rho g}+z_2+\frac{v_{2}^{2}}{2g}+h_f$$$

where
$$h_f$$
is head loss,
$$P_1$$
and
$$P_2$$
are the pressures at sections 1 and 2,
$$v_1$$
and
$$v_2$$
are the average fluid velocities at the sections,
$$z_1$$
and
$$z_2$$
are the vertical distances from a datum to the sections (the potential energy),
$$\rho$$
is the fluid density, and
$$g$$
is gravity.

##
__
__**Hint 2**

**Hint 2**

In our situation, the following are negligible or cancel out:

$$P_1=0$$
since the water jug is not internally pressurized → zero gauge

$$v_1=0$$
since the tank is relatively large, meaning no noticeable movement at the water line

$$P_2=0$$
since it is the pressure at the opening → atmospheric, which is zero gauge

$$z_2=0$$
since there is no potential energy (height difference) at the exit point, if we're finding the speed at the exit point (the jug suspension is there to throw you off)

$$h_f =0$$
since the problem statement defined head loss as negligible

The energy equation for steady incompressible flow with no shaft device is

$$$\frac{P_1}{\rho g}+z_1+\frac{v_{1}^{2}}{2g}=\frac{P_2}{\rho g}+z_2+\frac{v_{2}^{2}}{2g}+h_f$$$

where
$$h_f$$
is head loss (friction effect),
$$P_1$$
and
$$P_2$$
are the pressures at sections 1 and 2,
$$v_1$$
and
$$v_2$$
are the average fluid velocities at the sections,
$$z_1$$
and
$$z_2$$
are the vertical distances from a datum to the sections (the potential energy),
$$\rho$$
is the fluid density, and
$$g$$
is gravity.

In our situation, the following are negligible or cancel out:

$$P_1=0$$
since the water jug is not internally pressurized → zero gauge

$$v_1=0$$
since the tank is relatively large, meaning no noticeable movement at the water line

$$P_2=0$$
since it is the pressure at the opening → atmospheric, which is zero gauge

$$z_2=0$$
since there is no potential energy (height difference) at the exit point, if we're finding the speed at the exit point (the jug suspension is there to throw you off)

$$h_f =0$$
since the problem statement defined head loss as negligible

Thus,

$$$z_1=\frac{v_{2}^{2}}{2g}$$$

$$$v_{2}^{2}=z_1(2g)\rightarrow v_2=\sqrt{z_1(2g)}$$$

$$$v_2=\sqrt{(0.25m)(2)(9.8m/s^2)}=\sqrt{4.9m^2/s^2}=2.2\:m/s$$$

2.2 m/s