Ap Physics 2 Bernoulli and Continuity Lab

All AP Physics 2 Resources

Water is flowing at a rate of $2 \frac{m^3}{s}$ 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?

$\rho_{water} = 1.0\frac{kg}{l}$

Correct answer:

76.2kPa

Explanation:

We need Bernoulli's equation to solve this problem:

$P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2$

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:

$P_2 = P_1 + \frac{1}{2}\rho (v_1^2-v_2^2)$

We know the initial pressure, so we still need to calculate the initial and final velocities. We'll use the continuity equation:

$\dot{V} = vA$

Rearrange for velocity:

$v = \frac{\dot{V}}{A}$

Where is the cross-sectional area. We can calculate this for each diameter of the tube:

$A_1 = \frac{\pi d^2}{4} = \frac{\pi(1m)^2}{4} = \frac{\pi}{4}m$

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.5m)^2}{4} = \frac{\pi}{16}m$

Now we can calculate the velocity for each diameter:

$v_1 = \frac{2\frac{m^3}{s}}{\frac{\pi}{4}m} = \frac{8}{\pi}\frac{m}{s}$

$v_2 = \frac{2\frac{m^3}{s}}{\frac{\pi}{16}m} = \frac{32}{\pi}\frac{m}{s}$

Now we have all of the values needed for Bernoulli's equation, allowing us to solve:

$P_2 = (80,000Pa)+ \frac{1}{2}(1000 \frac{kg}{m^3})(\frac{8}{\pi}-\frac{32}{\pi})$

$P_2 = (80,000Pa)+ (500)(\frac{-24}{\pi}) = 76.2kPa$

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?

Possible Answers:

There is not enough information

$112.8\frac{m}{s}$

$21.69\frac{m}{s}$

$30.67\frac{m}{s}$

$940.8\frac{m}{s}$

Correct answer:

$30.67\frac{m}{s}$

Explanation:

To begin with, it will be necessary to make use of Bernoulli's equation:

$P_{1}+\frac{1}{2} \rho v^{2}_{1}+\rho gh_{1}=P_{2}+\frac{1}{2} \rho v^{2}_{2}+\rho gh_{2}$

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 $\frac{1}{2} \rho v^{2}_{1}$ term on the left side of the equation. Thus far, we have:

$\rho gh_{1}=\frac{1}{2} \rho v^{2}_{2}+\rho gh_{2}$

Dividing everything by $\rho$ , we obtain:

$gh_{1}=\frac{1}{2} v^{2}_{2}+ gh_{2}$

And rearranging:

$\frac{1}{2} v^{2}_{2}= gh_{1}-gh_{2}$

$\frac{1}{2} v^{2}_{2}= g(h_{1}-h_{2})=g\Delta h$

$v_{2}=\sqrt{2g\Delta h}=\sqrt{2(9.8\frac{m}{s^{2}})(50m-2m)}=30.67\frac{m}{s}$

Below is a picture of a pipe with a fluid running through it.

Bernoullipipe

If the initial velocity of the fluid is $5\frac{m}{s}$ , the final velocity is $10\frac{m}{s}$ , and the initial pressure is 110 kPa (with an unchanging potential energy), what is the value of the final pressure?

$\rho=1\frac{kg}{m^3}$

Possible Answers:

None of the other answers is correct

72.5kPa

147.5kPa

35kPa

185kPa

Correct answer:

72.5kPa

Explanation:

The equation relating fluid pressure, kinetic energy, and potential energy from state to state is known as the Bernoulli equation, and is as follows:

$P_1 + \frac{1}{2}\rho v_1^2 +\rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 +\rho gh_2$

Our potential energy is the same, so we can remove that part from the equation.

$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

$\rho$ is the density, which in our case is equal to 1, so it doesn't change anything here.

$P_1 + \frac{1}{2} v_1^2 = P_2 + \frac{1}{2}v_2^2$

We have the values for the initial pressure, initial velocity, and final velocity, so we can rearrange our equation to equal final pressure.

$P_1 + \frac{1}{2} v_1^2 &= P_2 + \frac{1}{2}v_2^2$

$P_1 + \frac{1}{2}v_1^2-\frac{1}{2}v_2^2&= P_2$

$P_1 + \frac{1}{2}(v^2_1-v_2^2)&= P_2$

Now, we can plug in our values.

$P_2&=P_1 + \frac{1}{2}(v^2_1-v_2^2)$

$P_2=110 +\frac{5^2+10^2}{2}$

$P_2= 110 + ({-}37.5)$

P_2=72.5

Therefore, the final pressure is equal to 72.5kPa .

A house is to be designed to withstand hurricane-force winds. The maximum wind velocity is $v=88\frac{m}{s}$ . The surface area of the roof is A=450 m . If the density of air is $\rho=1.029 \frac{kg}{m^3}$ , how much force must the roof supports be able to withstand?

Correct answer:

F=1.8*10^6N

Explanation:

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:

$P_A + \frac{1}{2}\rho v_A^2+\rho g h_A=P_B + \frac{1}{2}\rho v_B^2+\rho g h_B$

Since the air is still inside, $\frac{1}{2}\rho v_A^2 =0$

Since our points are at the same height, the $\rho g h$ terms cancel. Rearrange:

$P_A-P_B=\frac{1}{2}\rho v_B^2$

Since we only care about the difference in pressure inside and out.

$\Delta P = \frac{1}{2}\rho v_B^2$

Plug in known values and solve for the difference in pressure.

$\Delta P=\frac{1}{2}* 1.029\frac{kg}{m^3}* \left(88\frac{m}{s}\right)^2=3984Pa$

Force is pressure times area:

$F=\Delta P*A$

$F=3984Pa* 450m^2* 10^6\,N$

A kite boarder is using a kite to generate a force on a windy day. The area of the kite is A=4m^2 . The wind speed is $v=9\frac{m}{s}$ . The density of air is $\rho = 1.29\frac{kg}{m^3}$ . 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?

Correct answer:

F=209N

Explanation:

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).

$P_A + \frac{1}{2}\rho v_A^2+\rho g h_A=P_B + \frac{1}{2}\rho v_B^2+\rho g h_B$

Since the points are at the same height, the $\rho g h$ terms cancel. Since the air is still at point A, the $\frac{1}{2}\rho v_A^2$ term is zero. Since we only care about the difference in pressure on the two sides of the kite, solve for P_A - P_B :

$P_A -P_B = \frac{1}{2}\rho v_B^2 = \frac{1}{2}* 1.29\frac{kg}{m^3}* \left(9\frac{m}{s}\right)^2=52.25Pa$

Net force is pressure difference times area:

$F=\Delta P *A$

F=52.25Pa* 4m^2=209N

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 height=2.3m and width=19m . If the current has a speed of $v=3\frac{m}{s}$ at maximum tidal flow, how much force will the water exert on the metal plate?

Correct answer:

F=1.97*10^5N

Explanation:

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:

$P_A+\frac{1}{2}\rho v_A^2+\rho g h_A=P_B+\frac{1}{2}\rho v_B^2+\rho g h_B$

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 $\rho g h$ 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:

$P_A-P_B=\frac{1}{2}\rho v_B^2=\frac{1}{2}*1000\frac{kg}{m^3}*\left(3\frac{m}{s}\right)^2=4500Pa$

This is saying that the pressure at point B is 4500Pa less than the pressure at point A. Using the definition of pressure, find the resulting force:

F=PA=4500Pa* 2.3m*19m=1.97*10^5N

Venturi

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 $v=7\frac{m}{s}$ , how high will the water rise in the vertical tube?

Possible Answers:

The water in the vertical tube will not change its height

The water in the vertical tube will be pushed down

h=4.52cm

h=3.22 cm

h=1.23cm

Correct answer:

h=3.22 cm

Explanation:

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:

$P_A+\frac{1}{2}\rho v_A^2+\rho g h_A=P_B+\frac{1}{2}\rho v_B^2+\rho g h_B$

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:

$P_A-P_B=\frac{1}{2}\rho v_B^2=\frac{1}{2}* 1.29 * 7^2=31.6Pa$

This is saying that the pressure inside the tube is 31.6Pa 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:

$P_A+\frac{1}{2}\rho v_A^2+\rho g h_A=P_B+\frac{1}{2}\rho v_B^2+\rho g h_B$

Since the water is no longer moving, the terms containing are equal to zero. Rearrange:

$P_A-P_B=\rho g h_B -\rho g h_A = \rho g(h_B-h_A)$

Put in numbers and solve for the height difference:

$31.6Pa=1000\frac{kg}{m^3}*9.8\frac{N}{kg}*(h_B-h_A)$

$h_B-h_A =3.22*10^{-3}m=3.22cm$

To which of the following fluid situations does Bernoulli's principle apply?

Possible Answers:

Compressible, steady flow

Incompressible flow with internal friction

Incompressible, steady flow

Incompressible, steady flow with no internal friction

Correct answer:

Incompressible, steady flow with no internal friction

Explanation:

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.

Suppose that a fluid with a density of $1000\: \frac{kg}{m^{3}}$ flowing through a horizontal pipe at a speed of $5\: \frac{m}{s}$ has a pressure of 100000Pa. If this fluid then starts flowing through the pipe at a speed of $10\: \frac{m}{s}$ , what is the new pressure that this fluid exerts?

Correct answer:

62500Pa

Explanation:

In the question stem, we're told that a fluid with a density of $1000\: \frac{kg}{m^{3}}$ moves through a pipe at a speed of $5\: \frac{m}{s}$ and has a pressure of $100,000\:Pascals$ . We're then told that the same fluid begins to move through the pipe at a new speed of $10\: \frac{m}{s}$ , 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:

$P+\frac{1}{2}\rho v^{2}+\rho gh=Constant$

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.

$P_{1}+\frac{1}{2}\rho v_{1}^{2}+\rho gh_{1}=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gh_{2}$

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.

$P_{1}+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2}$

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 $P_{2}$ .

$P_{2}=P_{1}+\frac{1}{2}\rho v_{1}^{2}-\frac{1}{2}\rho v_{2}^{2}$

$P_{2}=P_{1}+\frac{1}{2}\rho (v_{1}^{2}-v_{2}^{2})$

Then, plugging in the values we have allows us to obtain the answer:

$P_{2}=100000 Pa+\frac{1}{2}(1000\: \frac{kg}{m^{3}})((5\: \frac{m}{s^{2}})^{2}-(10\: \frac{m}{s^{2}})^{2})$

$P_{2}=62500Pa$

Correct answer:

$P_{2}=116.9kPa$

Explanation:

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: $P_{1}+\frac{\rho v_{1}^{2}}{2}+\rho gh_{1}=P_{2}+\frac{\rho v_{2}^{2}}{2}+\rho gh_{2}$

Where is pressure, $\rho$ 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:

$120*10^{3}+\frac{1000* (7)^{2}}{2}+1000*9.8*0=P_{2}+\frac{1000* (4)^{2}}{2}+1000*9.8*2$

$120*10^{3}+24.5*10^{3}+0=P_{2}+8*10^{3}+19.6*10^{3}$

$144.5*10^{3}=P_{2}+27.6*10^{3}$

$116.9*10^{3}=P_{2}=116.9kPa$

All AP Physics 2 Resources

Report an issue with this question

If you've found an issue with this question, please let us know. With the help of the community we can continue to improve our educational resources.

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice ("Infringement Notice") containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys' fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner's agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

garciahicad1961.blogspot.com

Source: https://www.varsitytutors.com/ap_physics_2-help/bernoulli-s-equation

Ap Physics 2 Bernoulli and Continuity Lab

All AP Physics 2 Resources

All AP Physics 2 Resources

Report an issue with this question

DMCA Complaint

0 Response to "Ap Physics 2 Bernoulli and Continuity Lab"

Post a Comment

Iklan Atas Artikel

Iklan Tengah Artikel 1

Iklan Tengah Artikel 2

Iklan Bawah Artikel