If the hole has a diameter of 3.00 mm, what is the height, h, of the water level in the tank?

Question

A jet of water squirts out horizontally from a hole near the bottom of the tank in Figure P9.45.

http://www.webassign.net/sf/p9_45.gif

Answer 1

The static pressure (P) of the water in the tank is

P = p*g*h

where p is the density of water, g is the gravitational acceleration, and h is the height of water. When the water exits the hole at the bottom of the tank it has kinetic energy per unit volume given by

K = 0.5*p*v^2

where v is the exit speed.

The jet travels a horizontal distance L hitting the ground.

L = v*t

where t is the travel time, which can be solved for by knowing how high the tank is above the ground. The water falls through a distance Y under the influence of gravity since there is no y-component to the initial velocity.

Y = 0.5*g*t^2.

Solving for t yields

t = sqrt(2*Y/g)

which gives a speed of

v = L/t = L*sqrt(0.5*g/Y).

Equating the fluid potential energy per unit volume (pressure) to the kinetic energy we have

p*g*h = 0.5*p*v^2

canceling out p and substituting v gives

g*h = 0.5*L^2*0.5*g/Y

Solving for h, the height of the fluid, yields

h = L^2/(4*Y)

Y = 1 m according to the picture provided. Of course, this is a dynamic problem as fluid exits the hole the height of the fluid will change as will the exit speed and impact distance. The height here is for the initial jet of water.

The links below provide more information regarding pressure, Torricelli, and Torricelli's law.