Fluid Dynamics

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AP Physics 2 › Fluid Dynamics

Questions 1 - 10

Pipe has radius $R_{1}$ , and pipe has radius $R_{2}$ . The two pipes are connected. In order for the speed of water in pipe to be times as great as the speed in pipe , what must be $\frac{R_{2}}{R_{1}}$ ?

$\frac{\sqrt{2}}{4}$

$2\sqrt{2}$

$\frac{1}{8}$

$\frac{1}{4}$

Explanation

The continuity equation says that the cross sectional area of the pipe multiplied by velocity must be constant. Let be the water speed in pipe .

$\pi R_{1}^2v=\pi R_{2}^2(8v)$

$\frac{R_{2}^2}{R_{1}^2}=\frac{1}{8}$

$\frac{R_{2}}{R_{1}}=\frac{\sqrt{2}}{4}$

$\frac{\sqrt{2}}{4}$

$2\sqrt{2}$

$\frac{1}{8}$

$\frac{1}{4}$

Explanation

The continuity equation says that the cross sectional area of the pipe multiplied by velocity must be constant. Let be the water speed in pipe .

$\pi R_{1}^2v=\pi R_{2}^2(8v)$

$\frac{R_{2}^2}{R_{1}^2}=\frac{1}{8}$

$\frac{R_{2}}{R_{1}}=\frac{\sqrt{2}}{4}$

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?

Water pipe

Oil pipe

Both pipes have the same volumetric flow rate

It is impossible to determine without knowing the density of the fluids

Explanation

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:

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

The volumetric flow rate is:

$Q=v*a=8*38.465=307.72\frac{m^3}{s}$

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

The volumetric flow rate is:

$Q=v*a=5*1=5\frac{m^3}{s}$

The water pipe has the larger volumetric flow rate.

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?

Water pipe

Oil pipe

Both pipes have the same volumetric flow rate

It is impossible to determine without knowing the density of the fluids

Explanation

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

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

The volumetric flow rate is:

$Q=v*a=8*38.465=307.72\frac{m^3}{s}$

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

The volumetric flow rate is:

$Q=v*a=5*1=5\frac{m^3}{s}$

The water pipe has the larger volumetric flow rate.

Suppose two pipes made out of an identical material have the same length and the same liquid flowing through them. If pipe A has a cross-sectional diameter that is twice as great as pipe B, how does the flow rate in pipe A differ from the flow rate in pipe B?

The flow rate in pipe A is greater than pipe B by a factor of

The flow rate in pipe B is greater than pipe A by a factor of

The flow rate in pipe A is greater than pipe B by a factor of

The flow rate in pipe B is greater than pipe A by a factor of

The flow rate is the same in both pipes

Explanation

For this question, we're asked to consider two pipes. Each pipe has the same length, is made out of the same material, and has the same fluid moving through it. The only difference is the cross-sectional diameter of these pipes. We're asked to find how the flow rate will differ between the two pipes.

In order to solve this question, we'll need to use Poiseuille's equation.

$Q=\frac{\pi \Delta P r^{4}}{8\eta l}$

This equation tells us that the volume flow rate is directly proportional to two things: the pressure gradient between the ends of the pipe and the radius of the pipe raised to the fourth power. Moreover, the volume flow rate is inversely proportional to the viscosity of the fluid and also to the length of the pipe.

Now we need to ask the question - which of the variables in this equation is different in pipes A and B? Both pipes have the same length. Since each pipe has the same fluid moving through it, the viscosity will also be the same. Furthermore, we can assume that the pressure gradient at the end of each pipe is the same. The only thing that's left is the radius.

We're told that the diameter of pipe A is twice as great as pipe B. Since the radius is just half of the diameter, this also means that the radius of pipe A is twice as great as pipe B. Because the volume flow rate is dependent on the radius of the pipe raised to the fourth power, we can see that doubling the radius will result in a fold difference in the volume flow rate. Thus, the flow rate in pipe A will be times the flow rate in pipe B.

The flow rate in pipe A is greater than pipe B by a factor of

The flow rate in pipe B is greater than pipe A by a factor of

The flow rate in pipe A is greater than pipe B by a factor of

The flow rate in pipe B is greater than pipe A by a factor of

The flow rate is the same in both pipes

Explanation

In order to solve this question, we'll need to use Poiseuille's equation.

$Q=\frac{\pi \Delta P r^{4}}{8\eta l}$

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?

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

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

$v_{2}=1\frac{m}{s}$

$v_{2}=10\frac{m}{s}$

$v_{2}=0.5\frac{m}{s}$

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:

$75*10^{3}+\frac{1000* (10)^{2}}{2}+1000*9.8*0=115*10^{3}+\frac{1000* v_{2}^{2}}{2}+1000*9.8*2$

Simplify and solve for the final velocity.

$75*10^{3}+50*10^{3}+9.8*10^{3}=115*10^{3}+\frac{ v_{2}^{2}}{2}*10^{3}+19.6*10^{3}$

$134.8*10^{3}=134.6*10^{3}+\frac{ v_{2}^{2}}{2}*10^{3}$

$0.2*10^{3}=\frac{ v_{2}^{2}}{2}*10^{3}$

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

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

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

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

$v_{2}=1\frac{m}{s}$

$v_{2}=10\frac{m}{s}$

$v_{2}=0.5\frac{m}{s}$

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:

$75*10^{3}+\frac{1000* (10)^{2}}{2}+1000*9.8*0=115*10^{3}+\frac{1000* v_{2}^{2}}{2}+1000*9.8*2$

Simplify and solve for the final velocity.

$75*10^{3}+50*10^{3}+9.8*10^{3}=115*10^{3}+\frac{ v_{2}^{2}}{2}*10^{3}+19.6*10^{3}$

$134.8*10^{3}=134.6*10^{3}+\frac{ v_{2}^{2}}{2}*10^{3}$

$0.2*10^{3}=\frac{ v_{2}^{2}}{2}*10^{3}$

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

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

Turbulent flow is characterized by __________.

irregular and chaotic flow

low flow rates

regular streamlines

All of these

Explanation

Turbulent flow is caused by sufficiently high flow rates, and is characterized by chaotic and irregular flow patterns. Streamlines represent the curves tangent to the point of the direction of flow; thus, if a flow is turbulent the streamlines will be chaotic as well.

Turbulent flow is characterized by __________.

irregular and chaotic flow

low flow rates

regular streamlines

All of these

Explanation

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