AP Physics 2 - Flow Rate

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 .

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

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

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

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.

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?

$20 \frac{m}{s}$

$10 \frac{m}{s}$

$5 \frac{m}{s}$

$2.5 \frac{m}{s}$

Explanation

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.

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

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

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

Decrease the diameter of the pipe

Decrease the length of the pipe

Increase the density of the fluid

Increase the length of the pipe

Increase the diameter of the pipe

Explanation

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.

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?

$10 \frac{m}{s}$

$100 \frac{m}{s}$

$20 \frac{m}{s}$

$200 \frac{m}{s}$

$150 \frac{m}{s}$

Explanation

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:

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

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.

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

None of these

$v_f=6.68\ \frac{\textup{m}}{\textup{s}}$

$v_f=4.58\ \frac{\textup{m}}{\textup{s}}$

$v_f=9.81\ \frac{\textup{m}}{\textup{s}}$

Explanation

Initial volume rate must equal final volume rate

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

Solving for :

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

Plugging in values:

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

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

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 $15\frac{m}{s}$ ?

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

$v_{out}=100\frac{m}{s}$

$v_{out}=15\frac{m}{s}$

$v_{out}=30.15\frac{m}{s}$

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. 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

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

The area of the exit cross-section is:

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

Plug in these variables into the continuity equation and solve:

$15*706.5=v_{out}*176.71$

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

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?

$10 \frac{m}{s}$

$100 \frac{m}{s}$

$1 \frac{m}{s}$

$5 \frac{m}{s}$

$50 \frac{m}{s}$

Explanation

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.

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

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?

The velocity of the fluid in the compressed section will increase

The pressure of the fluid in the compressed section will increase

The velocity of the fluid in the compressed section will decrease

More than one of these is true

Explanation

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.

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

$\frac{1}{4}$

$\frac{1}{2}$

Explanation

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:

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