Fluid Dynamics - AP Physics 2
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As the velocity of a fluid increases, what happens to the pressure? Assume both states of the fluid are at the same height and the pipe has a constant diameter.
As the velocity of a fluid increases, what happens to the pressure? Assume both states of the fluid are at the same height and the pipe has a constant diameter.
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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. Since the height of the pipe is constant, we can eliminate the height term leaving us with:
The velocity is increasing,
making the velocity term negative. We are subtracting some amount from 
, therefore 
, meaning the pressure is decreasing.
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
The velocity is increasing,
Suppose a physician is considering the flow of blood through a blood vessel. Dissolved within the blood are several important gasses, such as oxygen. If there is a blood clot that is partially obstructing one of the blood vessels, how will the blood's partial pressure of oxygen change as it is flowing past this clot?
Suppose a physician is considering the flow of blood through a blood vessel. Dissolved within the blood are several important gasses, such as oxygen. If there is a blood clot that is partially obstructing one of the blood vessels, how will the blood's partial pressure of oxygen change as it is flowing past this clot?
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In this question, we're presented with a scenario in which blood is flowing through a blood vessel with a clot. We're asked to determine how the partial pressure of oxygen changes as the blood flows pass the clot. To answer this, we'll need to understand the concepts of the continuity equation, Bernoulli's equation, and partial pressures.
Since the blood is flowing past an obstruction, and because the volume flow rate is the same everywhere, we can use the continuity equation to show the relationship between velocity and area.
What this equation shows is that, when the volume flow rate is constant, the velocity of the fluid is inversely proportional to the area in which it is flowing. When the blood passes through the narrow space caused by the obstruction, it temporarily speeds up and thus has a higher velocity.
Next, we can use the Bernoulli equation to show how the velocity is related to pressure.
In this case, we can assume that the height does not change, and thus we can take the potential energy component out of the equation. What we are left with is an inverse relationship between velocity and pressure. Thus, as the velocity increases, the pressure decreases.
The next thing to do is relate total pressure with partial pressure. Remember that partial pressure refers to the proportion of pressure that a certain gas contributes to the overall pressure when there is a mixture of gasses. For instance, the partial pressure of oxygen in the atmosphere is roughly 
, meaning that oxygen is responsible for contributing of the overall pressure of the atmosphere. The expression for partial pressure is as follows.
Again, what this means is that the partial pressure of a certain gas is dependent on both the fraction of that gas out of all other gasses present, and also the total pressure. Even though the proportion of oxygen will remain the same as blood flows pass the clot, the total pressure in that region decreases. Thus, the partial pressure of oxygen in this region will decrease as well.
All in all, let's summarize what happened by bringing it all together. As the blood flows through a narrower space due to the obstruction, its velocity increases. As a result of the increase in velocity, the pressure lessens. And as the pressure decreases, so too does the partial pressure of any of the gasses within that region, including that of oxygen.
In this question, we're presented with a scenario in which blood is flowing through a blood vessel with a clot. We're asked to determine how the partial pressure of oxygen changes as the blood flows pass the clot. To answer this, we'll need to understand the concepts of the continuity equation, Bernoulli's equation, and partial pressures.
Since the blood is flowing past an obstruction, and because the volume flow rate is the same everywhere, we can use the continuity equation to show the relationship between velocity and area.
What this equation shows is that, when the volume flow rate is constant, the velocity of the fluid is inversely proportional to the area in which it is flowing. When the blood passes through the narrow space caused by the obstruction, it temporarily speeds up and thus has a higher velocity.
Next, we can use the Bernoulli equation to show how the velocity is related to pressure.
In this case, we can assume that the height does not change, and thus we can take the potential energy component out of the equation. What we are left with is an inverse relationship between velocity and pressure. Thus, as the velocity increases, the pressure decreases.
The next thing to do is relate total pressure with partial pressure. Remember that partial pressure refers to the proportion of pressure that a certain gas contributes to the overall pressure when there is a mixture of gasses. For instance, the partial pressure of oxygen in the atmosphere is roughly
Again, what this means is that the partial pressure of a certain gas is dependent on both the fraction of that gas out of all other gasses present, and also the total pressure. Even though the proportion of oxygen will remain the same as blood flows pass the clot, the total pressure in that region decreases. Thus, the partial pressure of oxygen in this region will decrease as well.
All in all, let's summarize what happened by bringing it all together. As the blood flows through a narrower space due to the obstruction, its velocity increases. As a result of the increase in velocity, the pressure lessens. And as the pressure decreases, so too does the partial pressure of any of the gasses within that region, including that of oxygen.
An incompressible fluid flows through a pipe. At location 1 along the pipe, the volume flow rate is 
. At location 2 along the pipe, the area halves. What is the volume flow rate at location 2?
An incompressible fluid flows through a pipe. At location 1 along the pipe, the volume flow rate is
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When the area halves, the velocity of the fluid will double. However, the volume flow rate (the product of these two quantities) will remain the same. In other words, the volume of water flowing through location 1 per second is the same as the volume of water flowing through location 2 per second.
When the area halves, the velocity of the fluid will double. However, the volume flow rate (the product of these two quantities) will remain the same. In other words, the volume of water flowing through location 1 per second is the same as the volume of water flowing through location 2 per second.
Consider the case in which a fluid is flowing through a cylindrical tube. Which of the following would be expected to not increase the volume flow rate?
Consider the case in which a fluid is flowing through a cylindrical tube. Which of the following would be expected to not increase the volume flow rate?
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For this question, we need to consider the flow of a fluid through a cylindrical pipe. We then need to determine which parameter, when changed, will decrease the flow rate of fluid through the pipe.
First, let's consider a few variables that are implicated in fluid flow. In fact, the flow of a fluid through a pipe (or a connection of pipes) is analogous to a circuit and Ohm's law. For example, the expression for Ohm's law is as follows:
From this expression, we can see that the voltage difference (driving force for movement of charge) is equal to the current (flow of charge) multiplied by the resistance (which impedes the flow of charge). Similarly, the flow of a fluid through a pipe can be expression as follows:
Where 
is equal to the pressure difference, 
is equal to flow rate, and 
is equal to resistance. This expression takes the same form as Ohm's law. The pressure (driving force for movement of the fluid) is equal to the flow rate (movement of fluid) multiplied by the resistance (which, again, impedes the flow of the fluid).
Making this comparison between Ohm's law and flow rate should help make it easier to remember. But to answer the question, let's rearrange the expression slightly.
In this form, we can see that as the pressure difference is increased, so too does the flow rate increase. Moreover, flow rate is higher when resistance is lower. We need to keep these things in mind when determining what will reduce flow rate. Since increasing the pressure difference increases the flow rate, we can rule this answer choice out.
Next, let's see what things contribute to resistance. While the equation for resistance is quite complicated, we can remember a few generalities. First, the resistance of flow is inversely related to the radius of the pipe. In other words, as the radius increases, there is more room for the fluid to flow through, and thus the flow rate increases.
When the length of the tube increases, the resistance also increases. Think of it this way: it's easier to move a given amount of water through a straw than it is a garden hose. A big reason for this is that the increased length of the tube provides more area to come into contact with the moving fluid, which thus allows more opportunity for friction between the fluid and the walls of the pipe.
Fluids flow fastest when their flow is laminar. Turbulent flow is characterized as disordered and involves "wasting" energy through the movement of liquid molecules in directions other than the main direction of flow, which also impedes other fluid molecules and causes them to bump into the walls of the container/other fluid molecules irregularly, ultimately disrupting flow.
Finally, increasing the viscosity of the fluid also increases its resistance, and thus decreases its flow rate. This is because viscosity is a measure of the frictional interactions between molecules of the fluid itself. Think of it this way: it's harder to get something like syrup (very viscous) to flow as opposed to water (much less viscous). Thus, when we decrease the viscosity of a fluid, the resistance to flow decreases, and flow rate increases.
For this question, we need to consider the flow of a fluid through a cylindrical pipe. We then need to determine which parameter, when changed, will decrease the flow rate of fluid through the pipe.
First, let's consider a few variables that are implicated in fluid flow. In fact, the flow of a fluid through a pipe (or a connection of pipes) is analogous to a circuit and Ohm's law. For example, the expression for Ohm's law is as follows:
From this expression, we can see that the voltage difference (driving force for movement of charge) is equal to the current (flow of charge) multiplied by the resistance (which impedes the flow of charge). Similarly, the flow of a fluid through a pipe can be expression as follows:
Where
Making this comparison between Ohm's law and flow rate should help make it easier to remember. But to answer the question, let's rearrange the expression slightly.
In this form, we can see that as the pressure difference is increased, so too does the flow rate increase. Moreover, flow rate is higher when resistance is lower. We need to keep these things in mind when determining what will reduce flow rate. Since increasing the pressure difference increases the flow rate, we can rule this answer choice out.
Next, let's see what things contribute to resistance. While the equation for resistance is quite complicated, we can remember a few generalities. First, the resistance of flow is inversely related to the radius of the pipe. In other words, as the radius increases, there is more room for the fluid to flow through, and thus the flow rate increases.
When the length of the tube increases, the resistance also increases. Think of it this way: it's easier to move a given amount of water through a straw than it is a garden hose. A big reason for this is that the increased length of the tube provides more area to come into contact with the moving fluid, which thus allows more opportunity for friction between the fluid and the walls of the pipe.
Fluids flow fastest when their flow is laminar. Turbulent flow is characterized as disordered and involves "wasting" energy through the movement of liquid molecules in directions other than the main direction of flow, which also impedes other fluid molecules and causes them to bump into the walls of the container/other fluid molecules irregularly, ultimately disrupting flow.
Finally, increasing the viscosity of the fluid also increases its resistance, and thus decreases its flow rate. This is because viscosity is a measure of the frictional interactions between molecules of the fluid itself. Think of it this way: it's harder to get something like syrup (very viscous) to flow as opposed to water (much less viscous). Thus, when we decrease the viscosity of a fluid, the resistance to flow decreases, and flow rate increases.
A baseball is thrown at a catcher with a high velocity and the ball passes right by the glove of the catcher. Which of the following scenarios will occur because of this?
A baseball is thrown at a catcher with a high velocity and the ball passes right by the glove of the catcher. Which of the following scenarios will occur because of this?
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As the baseball passes by the glove, the air surrounding the glove increases in velocity. The increase in air velocity will cause a decrease in pressure. The decrease in pressure causes the ball to be pulled towards the glove.
Therefore the correct answer is that the ball will be pulled toward the glove because of the decreased air pressure.
As the baseball passes by the glove, the air surrounding the glove increases in velocity. The increase in air velocity will cause a decrease in pressure. The decrease in pressure causes the ball to be pulled towards the glove.
Therefore the correct answer is that the ball will be pulled toward the glove because of the decreased air pressure.
Methanol is traveling through a cylindrical tube with a diameter of 
. What is the maximum velocity of methanol that maintains laminar flow?
Methanol is traveling through a cylindrical tube with a diameter of
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We will use the expression for Reynold's number for this problem:
Where 
is the hydraulic diameter, and for a cylindrical tube, 
Rearranging for velocity:
For laminar flow: 
Plugging in our values into the expression, we get:
Methanol has a relatively low viscosity, so it only has a very small range for laminar flow and quickly becomes turbulent
We will use the expression for Reynold's number for this problem:
Where
Rearranging for velocity:
For laminar flow:
Plugging in our values into the expression, we get:
Methanol has a relatively low viscosity, so it only has a very small range for laminar flow and quickly becomes turbulent
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?
Water is flowing at a rate of
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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:
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
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:
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?
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?
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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:
And rearranging:
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
Dividing everything by
And rearranging:
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?
Below is a picture of a pipe with a fluid running through it.
If the initial velocity of the fluid is
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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 
.
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.
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
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?
A house is to be designed to withstand hurricane-force winds. The maximum wind velocity is
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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:
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
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:
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?
A kite boarder is using a kite to generate a force on a windy day. The area of the kite is
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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:
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
Net force is pressure difference times area:
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?
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
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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:
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
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
Put in numbers and solve for the height difference:
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?
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
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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:
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
This is saying that the pressure at point B is
To which of the following fluid situations does Bernoulli's principle apply?
To which of the following fluid situations does Bernoulli's principle apply?
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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.
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.
Water is flowing through a horizontal pipe. Water enters the left side of the pipe at a pressure 
and a velocity of 
. It leaves the pipe at a pressure of 
. What is the velocity of the water when it leaves the pipe?
Water is flowing through a horizontal pipe. Water enters the left side of the pipe at a pressure
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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 entrance of the pipe and state 2 is at the exit of the pipe. Both states of the pipe are at the same height because the pipe is horizontal. Input the variables from the question into Bernoulli's equation and solve for the final velocity:
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
In our question, state 1 is at the entrance of the pipe and state 2 is at the exit of the pipe. Both states of the pipe are at the same height because the pipe is horizontal. Input the variables from the question into Bernoulli's equation and solve for the final velocity:
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?
Suppose that a fluid with a density of
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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:
In the question stem, we're told that a fluid with a density of
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:
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?
Water is flowing through a hose. It comes out of the tap at a pressure of
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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:
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
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:
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 pressure of 
and a height of . What is the velocity of the water when it leaves the hose?
Water is flowing through a hose. It comes out of the tap at a pressure of
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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:
Simplify and solve for the final velocity.
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
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:
Simplify and solve for the final velocity.
Water is flowing through a horizontal pipe. Water enters the left side of the pipe at a pressure 
and a velocity of 
. It leaves the pipe at a velocity of 
. What is the pressure of the water when it leaves the pipe?
.
Water is flowing through a horizontal pipe. Water enters the left side of the pipe at a pressure
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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 entrance of the pipe and state 2 is at the exit of the pipe. Both states of the pipe are at the same height because the pipe is horizontal. Input the variables from the question into Bernoulli's equation and solve for the final pressure:
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
Water flows through a tube with a diameter of 2m at a rate of 
. What is the velocity of the water?
Water flows through a tube with a diameter of 2m at a rate of
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The velocity of the water can be determined from the following formula:
We need to calculate the volumetric flow rate and the cross-sectional area. For the flow rate:
Rearrange to solve for volumetric flow rate:
Next, calculate cross-sectional area:
Now we can solve for the velocity:
The velocity of the water can be determined from the following formula:
We need to calculate the volumetric flow rate and the cross-sectional area. For the flow rate:
Rearrange to solve for volumetric flow rate:
Next, calculate cross-sectional area:
Now we can solve for the velocity: