Flow Rate - AP Physics 2

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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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Question

Water is flowing through a pipe of radius $0.5\textup{ m}$ at a velocity of $2\ \frac{\textup{m}}{\textup{s}}$ . The piper then narrows to a radius of $0.23\textup{ m}$ . Determine the new velocity

Answer

Initial volume rate must equal final volume rate:

$\pi r_i^2v_i=\pi r_f^2v_f$

Solving for

$v_f=\frac{r_i^2v_i}{r^2_f}$

Plugging in values:

$v_f=\frac{.5^2*2}{.23^2}$

$v_f=9.45\frac{m}{s}$

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Question

What is the volumetric flow rate of oil in a diameter pipe? The velocity of the oil is $5\frac{m}{s}$ .

Answer

The volumetric flow rate of fluid is found using the equation:

Where is the velocity of the fluid and is the cross-sectional area of the space through which the fluid is flowing.

In this problem the cross-section of the pipe is a circle. The area of the cross-section is:

$a=\pir^{2}=\pi\left ( \frac{12}{2} \right )^{2}\approx113.04m^2$

The volumetric flow rate is:

$Q=va=5113.04=565.2\frac{m^3}{s}$

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Question

What is the volumetric flow rate of ethanol flowing through a square pipe with sidelength 4m? The velocity of the ethanol is $7\frac{m}{s}$ .

Answer

The volumetric flow rate of fluid is found using the equation:

Where is the velocity of the fluid and is the cross-sectional area of the space through which the fluid is flowing.

In this problem the cross-section of the pipe is a square. The area of the cross-section is:

The volumetric flow rate is:

$Q=va=716=112\frac{m^3}{s}$

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Question

Molasses has a volumetric flow rate of $6\frac{cm^3}{s}$ . What volume of molasses has flowed after ?

Answer

The volumetric flow rate of fluid can be defined using the equation:

$Q=\frac{V}{t}$

Where is the volume of the fluid and is the time the fluid is flowing. To solve this problem, time must be converted from minutes to seconds:

$t=1min \times \frac{60sec}{1min}=60sec$

Using the volumetric flow rate equation we find that the volume of molasses is:

$Q=\frac{V}{t}\Rightarrow V=Qt$

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Question

Water is flowing through a diameter pipe at $8\frac{m}{s}$ . Oil is flowing through a $1m\times1m$ square pipe at $5\frac{m}{s}$ . Which has the higher volumetric flow rate?

Answer

The volumetric flow rate of fluid is found using the equation:

Where is the velocity of the fluid and is the cross-sectional area of the space through which the fluid is flowing. In this problem the cross-section of the water pipe is a circle. The area of the cross-section is:

$a=\pir^{2}=\pi\left ( \frac{7}{2} \right )^{2}\approx 38.465m^2$

The volumetric flow rate is:

$Q=va=838.465=307.72\frac{m^3}{s}$

The cross-section of the oil pipe is a square. The area of the cross-section is:

The volumetric flow rate is:

$Q=va=51=5\frac{m^3}{s}$

The water pipe has the larger volumetric flow rate.

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Question

A pipe narrows from a diameter to a diameter. What is the velocity of the fluid when it exits the pipe (at the end) if it entered the pipe at $15\frac{m}{s}$ ?

Answer

The volumetric flow rate of fluid is found using the equation:

Where is the velocity of the fluid and is the cross-sectional area of the space through which the fluid is flowing. Use the continuity equation, we see that $Q_{in}=Q_{out}$ , therefore

$v_{in}a_{in}=v_{out}a_{out}$

In this problem, the cross-section of the pipe is a circle, which is

$a_{in}=\pir^{2}=\pi\left ( \frac{30}{2} \right )^{2}\approx706.5m^2$

The area of the exit cross-section is:

$a_{out}=\pir^{2}=\pi\left ( \frac{15}{2} \right )^{2}\approx176.71m^2$

Plug in these variables into the continuity equation and solve:

$15706.5=v_{out}176.71$

$v_{out}=59.97\frac{m}{s}$

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Question

Fluid enters a pipe at $24\frac{m}{s}$ and exits the pipe at $48\frac{m}{s}$ . What is the diameter of the pipe exit if the entrance has a diameter?

Answer

The volumetric flow rate of fluid is found using the equation:

Where is the velocity of the fluid and is the cross-sectional area of the space through which the fluid is flowing.

Use the continuity equation, we see that $Q_{in}=Q_{out}$ , therefore:

$v_{in}a_{in}=v_{out}a_{out}$

In this problem,

The cross-section of the pipe is a circle, which is:

$a_{in}=\pir^{2}=\pi\left ( \frac{26}{2} \right )^{2}\approx530.93m^2$

Plugging our variables into the continuity equation gives us

$24530.93=48a_{out}$

$a_{out}=265.46m^2$

$r_{out}=\sqrt{\frac{a}{\pi}}=\sqrt{\frac{265.46}{\pi}}=9.19m$

$d_{out}=2r_{out}=29.19=18.38m$

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Question

Fluid flows through a pipe whose diameter goes from to . How are flow in the two parts of the pipe related?

Answer

Based on the continuity equation, both parts of the flow must have the same volumetric flow rate.

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Question

Water is flowing through a pipe of radius $0.25\textup{ m}$ at a velocity of $4\ \frac{\textup{m}}{\textup{s}}$ . The pipe then narrows to a radius of $0.18\textup{ m}$ . Determine the new velocity.

Answer

Initial volume rate must equal final volume rate

$.5\pi r_i^2v_i=.5\pi r_f^2v_f$

Solving for :

$v_f=\frac{r_i^2v_i}{r^2_f}$

Plugging in values:

$v_f=\frac{.25^2*4}{.18^2}$

$v_f=7.72\ \frac{\textup{m}}{\textup{s}}$

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Question

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?

Answer

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:

$\Delta V=iR$

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:

$\Delta P=QR$

Where $\Delta P$ 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.

$Q=\frac{\Delta P}{R}$

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.

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Question

A tank is completely full of water to the height of . On the side of the tank, at the very bottom a small hole is punctured. With what velocity does water flow though the hole at the bottom of the water tank?

Answer

The equation for determining the velocity of fluid through a hole is as follows:

$v=\sqrt{2gh}$

This equation is actually derived from Bernoulli's principle. The is for velocity, the is the acceleration due to gravity and is the height. We solve for velocity by substituting for the values:

$v=\sqrt{2(10\frac{m}{s^{2}})(5m)}=\sqrt{100}= 10\frac{m}{s}$

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Question

How can the velocity of fluid through a pipe be increased?

Answer

By decreasing the diameter of the pipe we increase the volume flow rate, or the velocity of the fluid which passes through the pipe according to the continuity equation.

Increasing or decreasing the length of the pipe has no effect on fluid velocity. Therefore the correct answer is to decrease the diameter of the pipe.

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Question

A syringe has a cross-sectional area of and the needle attached to the syringe has a cross-sectional area of . The fluid in the syringe is pushed with a speed of $.1\frac{m}{s}$ , with what velocity does the fluid exit the needle opening?

Answer

The correct answer is $10 \frac{m}{s}$ because the cross-sectional area of the syringe is times larger than the needle opening. Therefore, the velocity will be larger as well.

$v= 100 \cdot 0.1 \frac{m}{s} = 10\frac{m}{s}$

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Question

A pipe has fluid flowing through it. Which of the following situations will occur if a section of the pipe is compressed resulting in a small area?

Answer

The velocity of the fluid in the compressed section will increase and the The pressure of the fluid in the compressed section will decrease. Therefore the correct answer is: More than one of these is true.

When the area of a pipe decreases the fluid velocity increases, and an increase in fluid velocity results in the decrease of pressure.

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Question

Water is flowing through a horizontal cylindrical tube. By what factor does the velocity change by if the circumference of the tube doubles?

Answer

We can model the volumetric flow through the tube as the following expression:

$\dot{V}=vA$

Where:

$A= \pi r^2$

so:

$\dot{V}=v\pi r^2$

Applying this to the first and second scenario, we get:

$\dot{V_1}=v_1\pi r_1^2$

$\dot{V_2}=v_2\pi r_2^2$

According to the law of continuity, we know that the volumetric flow through the tube (when neglecting friction and assuming that it is horizontal) is constant. Therefore, we can say:

$v_1\pi r_1^2=v_2\pi r_2^2$

Rearranging for the ratio $\frac{v_2}{v_1}$ :

$\frac{v_2}{v_1}=\frac{r_1^2}{r_2^2}$

From the problem statement, we are told that the circumference is doubled. Thus, we know that the radius of the tube doubles as well:

Plugging this into the expression, we get:

$\frac{v_2}{v_1}=\frac{r_1^2}{(2r_1)^2}$

$\frac{v_2}{v_1}=\frac{1}{4}$

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