Fluids - AP Physics 2

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Question

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?

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

Answer

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:

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

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

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Question

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?

Answer

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:

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

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

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

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Question

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

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 (with an unchanging potential energy), what is the value of the final pressure?

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

Answer

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.

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

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

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

Therefore, the final pressure is equal to .

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Question

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 . If the density of air is $\rho=1.029 \frac{kg}{m^3}$ , how much force must the roof supports be able to withstand?

Answer

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:

$F=\Delta P*A$

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Question

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 $v=9\frac{m}{s}$ . 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?

Answer

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 :

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

Net force is pressure difference times area:

$F=\Delta P *A$

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Question

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?

Answer

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 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:

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

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

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Question

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 $v=3\frac{m}{s}$ at maximum tidal flow, how much force will the water exert on the metal plate?

Answer

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 less than the pressure at point A. Using the definition of pressure, find the resulting force:

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Question

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

Answer

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.

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Question

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?

Answer

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 . 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:

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

$P_{2}=62500Pa$

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Question

Water is flowing through a hose. It comes out of the tap at a pressure of, velocity of $v=7\frac{m}{s}$ , and height of . It leaves the nozzle at a velocity of $v= 4\frac{m}{s}$ and a height of . What is the pressure of the water when it leaves the hose?

$\rho_{water}=1000\frac{kg}{m^3}$

Answer

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:

$12010^{3}+\frac{1000 (7)^{2}}{2}+10009.80=P_{2}+\frac{1000* (4)^{2}}{2}+10009.82$

$12010^{3}+24.510^{3}+0=P_{2}+810^{3}+19.610^{3}$

$144.510^{3}=P_{2}+27.610^{3}$

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

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Question

Water is flowing through a hose. It comes out of the tap at a pressure of, velocity of $v=10\frac{m}{s}$ , 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?

$\rho_{water}=1000\frac{kg}{m^3}$

Answer

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:

$7510^{3}+\frac{1000 (10)^{2}}{2}+10009.80=11510^{3}+\frac{1000 v_{2}^{2}}{2}+10009.82$

Simplify and solve for the final velocity.

$7510^{3}+5010^{3}+9.810^{3}=11510^{3}+\frac{ v_{2}^{2}}{2}10^{3}+19.610^{3}$

$134.810^{3}=134.610^{3}+\frac{ v_{2}^{2}}{2}10^{3}$

$0.210^{3}=\frac{ v_{2}^{2}}{2}*10^{3}$

$0.4= v_{2}^{2}$

$v_{2}=0.632\frac{m}{s}$

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Question

Water is flowing through a horizontal pipe. Water enters the left side of the pipe at a pressure and a velocity of $v=9.5\frac{m}{s}$ . It leaves the pipe at a velocity of $v= 12\frac{m}{s}$ . What is the pressure of the water when it leaves the pipe?

$\rho_{water}=1000\frac{kg}{m^3}$ .

Answer

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 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:

$10010^{3}+\frac{1000 (9.5)^{2}}{2}+10009.80=P_{2}+\frac{1000* (12)^{2}}{2}+10009.80$

$10010^{3}+45.12510^{3}=P_{2}+7210^{3}$

$P_{2}=145.12510^{3}-7210^{3}=73.12510^{3}=73.125kPa$

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Question

Water is flowing through a horizontal pipe. Water enters the left side of the pipe at a pressure and a velocity of $v=9.5\frac{m}{s}$ . It leaves the pipe at a pressure of . What is the velocity of the water when it leaves the pipe?

$\rho_{water}=1000\frac{kg}{m^3}$

Answer

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 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:

$10010^{3}+\frac{1000 (9.5)^{2}}{2}+10009.80=8510^{3}+\frac{1000 v_{2}^{2}}{2}+10009.80$

$10010^{3}+45.12510^{3}=8510^{3}+\frac{v_{2}^{2}}{2}10^{3}$

$145.12510^{3}=8510^{3}+\frac{v_{2}^{2}}{2}10^{3}$

$60.12510^{3}=\frac{v_{2}^{2}}{2}*10^{3}$

$v_{2}=5.48\frac{m}{s}$

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Question

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.

Answer

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. Since the height of the pipe is constant, we can eliminate the height term leaving us with:

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

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

The velocity is increasing, $v_{1}^{2}< v_{2}^{2}$ making the velocity term negative. We are subtracting some amount from $P_{1}$ , therefore $P_{2}< P_{1}$ , meaning the pressure is decreasing.

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Question

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?

Answer

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.

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

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.

$P_{gas}=X_{gas}P_{total}$

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.

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Question

Water flows through a tube with a diameter of 2m at a rate of $800 \frac{kg}{s}$ . What is the velocity of the water?

$\rho_{water} = 1\frac{kg}{l}$

Answer

The velocity of the water can be determined from the following formula:

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

We need to calculate the volumetric flow rate and the cross-sectional area. For the flow rate:

$\dot m = \dot V \cdot \rho_{water}$

Rearrange to solve for volumetric flow rate:

$\dot V = \frac{\dot m}{\rho_{water}} = \frac{800\frac{kg}{s}}{1000\frac{kg}{m^3}} = 0.8 \frac{m^3}{s}$

Next, calculate cross-sectional area:

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

Now we can solve for the velocity:

$v = \frac{0.8\frac{m^3}{s}}{\pi m^2} = 0.25 \frac{m}{s}$

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Question

Suppose that water flows from a pipe with a diameter of 1m into another pipe of diameter 0.5m. If the speed of water in the first pipe is $5: \frac{m}{s}$ , what is the speed in the second pipe?

Answer

To find the answer to this question, we'll need to use the continuity equation to determine the flow rate, which will be the same in both pipes.

$v_{1}A_{1}=v_{2}A_{2}$

We'll also need to calculate the area of the pipe using the equation:

$A=\pi r^{2}$

$A_{1}=\pi r_{1}^{2}=\pi (0.5\m)^{2}=0.25\pi m^{2}$

$A_{2}=\pi r_{2}^{2}=\pi \left ( 0.25 m\right )^{2}=0.0625\pi m^{2}$

Solve the combined equation for and plug in known values to find the velocity of the water through the second pipe.

$v_{2}=\frac{v_{1}A_{1}}{A_{2}}$

$v_{2}=\frac{(5 \frac{m}{s})(0.25\pi m^{2})}{0.0625\pi m^{2}}=20 \frac{m}{s}$

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Question

A diameter garden hose with a diameter of 3cm sprays water travels through a hose at $1\frac{m}{s}$ . At the end of the garden hose, the diameter reduces to 2cm. What is the speed of the water coming out at the end?

Answer

Use the continuity equation for incompressible fluids.

The cross sectional area of the garden hose at both ends are circular regions. Rewrite the equation replacing areas with the formula for an area of a circle and solve for the velocity at the second point.

$\left(\frac{\pi d_1^2}{4}\right)v_1 =\left(\frac{\pi d_2^2}{4}\right)v_2$

$v_2 = \frac{v_1 d_1^2}{d_2^2}= \frac{1\frac{{m}}{{s}}(3^2)}{2^2}= \left(\frac{3}{2}\right)^2=2.25 \frac{{m}}{{s}}$

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Question

An civil engineer is designing the outflow of a pond. The pond has a radius of , and the maximum sustained rainfall rate is $v=2.1*10^{-5}\frac{m}{s}$ , about 3 inches per hour. If the engineer makes the outflow with a cross-sectional area of , what maximum velocity will the outflow of water have during a heavy rainstorm if the surface level of the pond does not change?

Answer

This is a volume flow rate problem. Because the water is an incompressible fluid, we can apply the flow rate equation:

$A_{1}* v_{1}=A_{2}* v_{2}$

Find the surface area of the pond:

$A_{1}=\pi r^2 = 3.14* 200^2 =1.3* 10^5m^2$

Substitute into the flow rate equation:

$1.310^5 2.110^{-5}=0.5 v_2$

$v_2=5.56\frac{m}{s}$

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Question

An incompressible fluid flows through a pipe. At location 1 along the pipe, the volume flow rate is $10 \frac{m^3}{s}$ . At location 2 along the pipe, the area halves. What is the volume flow rate at location 2?

Answer

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.

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