# Application of Bernoulli's theorem

Problem 1 : Here is a cup with two holes (see attachment). The cup is filled with water to the top and the water is coming out of the holes. Assuming suitable values for heights, radius's etc. please answer the following questions :

1) Velocity with which the stream of water comes out.

2) Acceleration of the stream of water.

3) Distance and time after which the water stream hits the ground.

4) How long it will take for the tank to drain water above the hole.

5) Pressure of water coming out of the hole.

Problem 2 : Answer the same questions but for one hole instead of two holes.

(See attached file for full problem description with diagrams)

© BrainMass Inc. brainmass.com March 4, 2021, 6:51 pm ad1c9bdddfhttps://brainmass.com/physics/acceleration/application-bernoullis-theorem-65527

#### Solution Preview

Please see the attached file.

SOLUTION

One hole problem

R

x

dx H = 3 m

h1 R = 1 m

h1 = 0.5 m

H vH r1 = 2 cm

L1

The fig. shows a circular drum open from the top, of radius R and height H filled with water up to the top. It has one hole of radius r1 at a depth of h1 from the top.

a) Velocity with which the stream of water comes out

Let us apply Bernoulli's equation (P + hρg + ½ ρv2 = Constant) to the water surface in the drum and the water just coming out of the hole.

Ps + hsρg + ½ ρ(vs)2 = PH + hHρg + ½ ρ(vH)2 .........(1)

Where Ps = Pressure on the surface = Atmospheric pressure = Patm

hs= Height of the surface = H

vs= Velocity of the water surface = 0 (As the water comes out of the hole, the water surface in the drum also comes down. However, as the radius of the drum is very large as compared to the radius of the hole, the water surface comes down with a very small velocity as compared to the velocity of the stream coming out from the hole. Hence vs can be taken as 0)

PH = Pressure at the hole = Patm (As the hole is open to atmosphere, the pressure just out side the hole is atmospheric pressure)

hH= Height of the hole = H - h1

vH = Velocity of the water at the hole.

Equation (1) now becomes : Patm + Hρg = Patm + (H-h1)ρg + ½ ρ(vH)2

____ _______

Solving we get : vH = √2gh1 = √2x9.8x0.5 = 3.13 m/s

The direction of the velocity vector is horizontal at the instant water just emerges out of the hole.

b) Acceleration of the stream of water

As the water comes out of the hole, its velocity is horizontal. Neglecting the air friction, the water is subjected to only the force of gravity acting downwards. Hence, water will have acceleration equal to g (acceleration due to gravity) in the down wards direction. However, as there is no force acting on the water in the horizontal direction, there is no horizontal acceleration (it means the horizontal velocity of water as it just comes out of the hole remains constant).

c) Distance and time after which the water stream hits the ground

Under the influence of gravity the water stream behaves like a projectile thrown ...

#### Solution Summary

These problems on streams of water coming out of holes in the side of the container are solved using continuity equation, Bernoulli's theorem.