All AP Physics 2 Resources
Example Question #38 : Fluid Dynamics
Water is flowing at a rate of through a tube with a diameter of 1m. If the pressure at this point is 80kPa, what is the pressure of the water after the tube narrows to a diameter of 0.5m?
We need Bernoulli's equation to solve this problem:
The problem statement doesn't tell us that the height changes, so we can remove the last term on each side of the expression, then arrange to solve for the final pressure:
We know the initial pressure, so we still need to calculate the initial and final velocities. We'll use the continuity equation:
Rearrange for velocity:
Where is the cross-sectional area. We can calculate this for each diameter of the tube:
Now we can calculate the velocity for each diameter:
Now we have all of the values needed for Bernoulli's equation, allowing us to solve:
Example Question #111 : Fluids
Suppose that a huge tank 50m high and filled with water is open to the atmosphere and is hit with a bullet that pierces one side of the tank, allowing water to flow out. The hole is 2m above the ground. If the hole is very small in comparison with the size of the tank, how quickly will the water flow out of the tank?
There is not enough information
To begin with, it will be necessary to make use of Bernoulli's equation:
For the situation described in the question stem, we'll designate the top of the container as point 1, and the hole where water is flowing out as point 2.
To begin simplifying things, it's important to realize a few things. First, both points are open to the atmosphere. Therefore, the term on each side of the above equation is equal to 1atm and thus can cancel out. Secondly, since the size of the hole on the side of the tank is so small compared to the rest of the tank, the velocity of the water at point 1 is nearly equal to 0. Hence, we can cancel out the term on the left side of the equation. Thus far, we have:
Dividing everything by , we obtain:
Example Question #1 : Bernoulli's Equation
Below is a picture of a pipe with a fluid running through it.
If the initial velocity of the fluid is , the final velocity is , and the initial pressure is (with an unchanging potential energy), what is the value of the final pressure?
None of the other answers is correct
The equation relating fluid pressure, kinetic energy, and potential energy from state to state is known as the Bernoulli equation, and is as follows:
Our potential energy is the same, so we can remove that part from the equation.
is the density, which in our case is equal to 1, so it doesn't change anything here.
We have the values for the initial pressure, initial velocity, and final velocity, so we can rearrange our equation to equal final pressure.
Now, we can plug in our values.
Therefore, the final pressure is equal to .
Example Question #41 : Fluid Dynamics
A house is to be designed to withstand hurricane-force winds. The maximum wind velocity is . The surface area of the roof is . If the density of air is , how much force must the roof supports be able to withstand?
This is solved using Bernoulli's equation and the definition of pressure. First choose the "Bernoulli points", one just inside the roof where the air is still (Point A) and one just outside where the air is moving (Point B). This will allow us to eliminate many of the terms:
Since the air is still inside,
Since our points are at the same height, the terms cancel. Rearrange:
Since we only care about the difference in pressure inside and out.
Plug in known values and solve for the difference in pressure.
Force is pressure times area:
Example Question #41 : Fluid Dynamics
A kite boarder is using a kite to generate a force on a windy day. The area of the kite is . The wind speed is . The density of air is . If the kite is designed such that the air is stationary on the inner surface, how much force can the kite boarder expect the kite to generate?
Use Bernoulli's equation to find the pressure difference on the two sides of the kite. Call point A the inner surface (where the air is still) and point B the outer surface (where the wind is at full speed).
Since the points are at the same height, the terms cancel. Since the air is still at point A, the term is zero. Since we only care about the difference in pressure on the two sides of the kite, solve for :
Net force is pressure difference times area:
Example Question #42 : Fluid Dynamics
At a dock, a metal plate is completely submerged and attached to an underwater wall. The side of the metal plate is exposed to the ocean, and to the flow of water caused by tides. The plate has dimensions of and . If the current has a speed of at maximum tidal flow, how much force will the water exert on the metal plate?
The moving water on one side of the metal plate has a lower pressure than the still water on the other side, resulting in a force. We start by writing the Bernoulli equation:
We choose our two "Bernoulli points" to make the problem as easy as possible. Take point A to be on the side where the water is still, and point B on the side where the water is moving. If we make them at the same height, the terms can be subtracted from both sides. Since the water is still at point A, the velocity term on the right-hand side of the equation is zero. Rearrange to find the difference in pressure from side A to side B:
This is saying that the pressure at point B is less than the pressure at point A. Using the definition of pressure, find the resulting force:
Example Question #43 : Fluid Dynamics
A venturi is a T-shaped tube in which the vertical tube is in water. A high-speed stream of air is forced through the horizontal tube. As a result, water rises in the vertical tube, as shown in the given figure. If air is forced through the horizontal tube at , how high will the water rise in the vertical tube?
The water in the vertical tube will be pushed down
The water in the vertical tube will not change its height
We will use Bernoulli's equation to solve this. We must do this twice: once for the air and once for the water. The central principle here is that the moving stream of air has a lower pressure than still air. In this problem we will ignore the atmospheric pressure since it is applied at the tube ends and at the surface of the water outside the vertical tube. For the air, choose our two "Bernoulli points": point A is just outside the horizontal tube and point B is just inside. Write the equation:
The heights are the same, so they cancel out of the equation. The air is still at point A, so the velocity term is zero for the left side. Finally, as mentioned, we care only about the difference in pressure:
This is saying that the pressure inside the tube is below the pressure outside.
Now we solve for the water. Our points will be on the surface outside the vertical tube (point A) where the pressure is one atmosphere, and inside the vertical tube at the surface of the risen column (point B). Write the equation:
Since the water is no longer moving, the terms containing are equal to zero. Rearrange:
Put in numbers and solve for the height difference:
Example Question #45 : Fluid Dynamics
To which of the following fluid situations does Bernoulli's principle apply?
Incompressible flow with internal friction
Incompressible, steady flow with no internal friction
Compressible, steady flow
Incompressible, steady flow
Incompressible, steady flow with no internal friction
Since Bernoulli's principle is derived from the work-energy theorem, it is a requirement that the flow be incompressible, steady, and without internal friction. Otherwise, energy would be lost to these outlets and the equation would no longer be applicable.
Example Question #46 : Fluid Dynamics
Suppose that a fluid with a density of flowing through a horizontal pipe at a speed of has a pressure of 100000Pa. If this fluid then starts flowing through the pipe at a speed of , what is the new pressure that this fluid exerts?
In the question stem, we're told that a fluid with a density of moves through a pipe at a speed of and has a pressure of . We're then told that the same fluid begins to move through the pipe at a new speed of , and we're asked to determine what the new pressure will be.
In order to answer this question, we'll need to make use of Bernoulli's equation. This equation essentially tells us that the pressure, kinetic energy, and potential energy of a moving fluid is constant. Or, put another way:
Alternatively, since we know the sum of these values is constant, we can relate the sum of these values at one instant to the sum of these values at another instant.
Furthermore, since we're told that the fluid remains flowing in a horizontal pipe, the height of the fluid does not change. Therefore, we can cancel out the potential energy terms on both sides.
Next, if we define the initial conditions as instant 1, and the final conditions as instant 2, then the term we are trying to solve for is .
Then, plugging in the values we have allows us to obtain the answer:
Example Question #47 : Fluid Dynamics
Water is flowing through a hose. It comes out of the tap at a pressure of, velocity of , and height of . It leaves the nozzle at a velocity of and a height of . What is the pressure of the water when it leaves the hose?
To solve this problem, we will use Bernoulli's equation, a simplified form of the law of conservation of energy. It applies to fluids that are incompressible (constant density) and non-viscous.
Bernoulli's equation is:
Where is pressure, is density, is the gravitational constant, is velocity, and is the height.
In our question, state 1 is at the tap and state 2 is at the nozzle. Input the variables from the question into Bernoulli's equation: