# Question Video: Kinematics of Pressure Induced Fluid Flow

A container of water has a cross-sectional area of 0.100 m². A piston sits on top of the water, as shown. There is a spout located 0.150 m from the bottom of the tank, open to the atmosphere. And a stream of water exits the spout. The cross-sectional area of the spout is 7.00 × 10⁻⁴ m². What is the speed of the water as it leaves the spout? How far from the spout does the water hit the floor? Ignore all friction and dissipative forces.

06:24

### Video Transcript

A container of water has a cross-sectional area of 0.100 meters squared. A piston sits on top of the water, as shown. There is a spout located 0.150 meters from the bottom of the tank, open to the atmosphere. And a stream of water exits the spout. The cross-sectional area of the spout is 7.00 times 10 to the negative fourth meters squared. What is the speed of the water as it leaves the spout? How far from the spout does the water hit the floor? Ignore all friction and dissipative forces.

We can call the speed of the water as it leaves the spout 𝑣 sub two. And the distance from the spout the water hits the floor we’ll label 𝑑. Since this example involves a flowing fluid and all friction and dissipative forces are ignored, that means this is an opportunity to apply Bernoulli’s equation.

This relationship tells us that the pressure plus half the fluid density times its speed squared plus its density times 𝑔 times its altitude ℎ is constant throughout any point in our fluid system.

Looking at our diagram, we’ll want to pick two points over which to apply Bernoulli’s equation. We’ll pick point number one to be at the location right beneath our piston. And point number two is right at the spout of our outlet.

Applying Bernoulli’s equation to our scenario, we write that 𝑝 one plus one-half 𝜌 𝑣 one squared plus 𝜌𝑔 times ℎ one is equal to 𝑝 two plus one-half 𝜌 𝑣 two squared plus 𝜌𝑔 ℎ two. We’ll treat 𝑔, the acceleration due to gravity, as exactly 9.8 meters per second squared.

Looking at this long expression, let’s see if there’re any terms we can eliminate. At point one, the speed of our fluid is zero. It won’t be in motion. So one-half 𝜌 times 𝑣 one squared goes to zero. We can also condense a bit by simplifying our 𝜌𝑔 ℎ one and 𝜌𝑔 ℎ two terms. If we subtract 𝜌𝑔 ℎ two from each side, then we can combine our 𝜌 times 𝑔 terms.

Since we want to solve for 𝑣 two, the speed of the water as it leaves the spout, we can do that algebraically now. When we do, we find that 𝑣 two is the square root of two over 𝜌 times the quantity 𝑝 one minus 𝑝 two plus 𝜌 times 𝑔 times the quantity ℎ one minus ℎ two.

In the problem statement, we’re told that 𝜌 is equal to 1000 kilograms per cubic meter. But we’ll want to solve for 𝑝 one minus 𝑝 two and ℎ one minus ℎ two. When we consider the pressure term, we know that 𝑝 one consists of two pressures. The pressure due to the atmosphere plus the pressure due to the piston comprise 𝑝 one.

On the other hand, 𝑝 two, which we’re told is exposed to the atmosphere, consists solely of atmospheric pressure. This means that 𝑝 one minus 𝑝 two is equal to atmospheric pressure plus pressure due to the piston minus the same amount of atmospheric pressure. So that term cancels out. So then what is the pressure due to the piston?

If we recall that pressure is equal to force over area, we know that the pressure due to the piston will be the pressure created by the force of gravity divided by the cross-sectional area of our cylinder. That area, which we’re told in the problem statement is 0.100 meters squared, we can call 𝐴 one. So the pressure due to the piston and therefore 𝑝 one minus 𝑝 two is equal to the piston’s mass, given as 20.0 kilograms, multiplied by 𝑔 divided by 𝐴 one. We insert this expression for 𝑝 one minus 𝑝 two in our expression for 𝑣 two.

And now we look to solving for ℎ one minus ℎ two. Because we’re given the heights at these two points we’ve chosen, 0.500 meters and 0.150 meters, ℎ one minus ℎ two is simply the difference between them, or 0.350 meters. Plugging that into our expression for 𝑣 two, we’re now ready to insert values for our other variables.

When we plug in for 𝜌, 𝑚, 𝑔, 𝐴 one, and ℎ one minus two and enter these values on our calculator, we find that 𝑣 two is 3.28 meters per second. That’s the speed at which water leaves the spout. Then if we call 𝑑 the horizontal distance that the water travels before it hits the ground, that’s what we want to solve for next.

We knew that as the water falls, its speed in the horizontal direction, which is 𝑣 two, will be constant over its descent. That means if we can solve for the time it takes for the water to fall from the spout down to the ground, when we multiply that time by 𝑣 two, that will give us 𝑑. We’re not able to solve for that time using motion in the horizontal direction. But we can solve for it using motion in the vertical.

Because the acceleration of the water once it leaves the spout is constant, that means the kinematic equations of motion describe how it moves. In particular, we can use the kinematic equation, which says that distance is equal to initial speed times time plus one half 𝑎 times time squared.

In our case, we can write that ℎ two, the height of the spout, is equal to one-half 𝑔 times 𝑡 squared. We don’t have an initial speed term because, initially, our speed in the vertical direction is zero.

Rearranging to solve for 𝑡, we see it’s equal to the square root of two ℎ two over 𝑔. Plugging in for these values and entering this expression on our calculator, the time 𝑡 it takes for the water to fall from the spout to the ground is approximately 0.175 seconds. Recalling that, for a constant speed, distance traveled equals speed times time, we can write that 𝑑 is equal to 𝑣 two times 𝑡. Plugging in for these values, when we calculate 𝑑, we find that, to three significant figures, it’s 0.574 meters. That’s how far from the end of the spout the water reaches the ground.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.