Chapter 14: Fluid Mechanics

Adria Updike

13 Chapter 14: Fluid Mechanics

Section 14.1: Fluids

There are three main phases of matter we explore in this chapter. A solid is a material with a definite shape and volume, with atoms close together and chemically bound. A liquid has fewer bonds between nearby atoms and can change its shape, but not its volume. Liquids are incompressible; they cannot be forced into a smaller volume than they normally inhabit. A gas has no bonds between nearby particles, no definite volume and takes the shape of its container. Gasses can be compressed. Both gasses and liquids are consider fluids; solids are not fluids.

Materials in a definite shape, volume, or container can be characterized by a density equal to the mass of the material divided by its volume, as we see in Equation 1. Density is represented by the Greek letter $\rho$ (rho) and has units of kg/m $^3$ .

$\rho = \frac{m}{V}$

Section 14.2: Pressure

Textbook Section 14.1: Fluids, Density, and Pressure

Textbook Section 14.2: Measuring Pressure

When a fluid is contained in a volume, the particles in the fluid move around and hit each other and the sides of the container. The pressure of a fluid in a container or on an object is due to the particles in the fluid colliding with the sides of the container or object. As they hit the object, they elastically rebound and transfer momentum. We measure pressure in units of Pascals (Pa) which are equal to one Newton per square meter, and it is calculated as we see in Equation 2 by the force of the collisions divided by the area over which they come into contact.

$p = \frac{F}{A}$

We feel a constant air pressure on our bodies as particles in the air run into us. At sea level, the air pressure is known as 1 atmosphere (1 atm) of pressure, which is equal to 101,325 Pa. This air pressure includes the weight of all the air above you in our atmosphere pushing down on you. If you go to a higher altitude, the air pressure decreases because the column of air above your head gets smaller. Similarly, if you dive under water (or another fluid), the deeper you go, the more water pressure you have pushing down on you (the more fluid is located above your head). To determine the pressure $p$ at some height $h$ above or below the water, you would use the following equation.

$p = p_0 + \rho \, g \, h$

In the above equation, $p$ is the new pressure at some height $h$ , $p_0$ is a known pressure at a height defined as zero, $g$ is the local gravitational constant, and $\rho$ is the density of the fluid producing the pressure. If you take $p_0$ to be air pressure at sea level, $p$ gets lower as you go up (use $g$ = -9.81 m/s $2$ ) and $p$ gets higher if you go down or under a fluid. Remember to draw a picture of the situation you are dealing with and determine ahead of time if your new pressure $p$ should be higher or lower than your original pressure $p_0$ to make sure your answer makes sense.

Section 14.3: Pascal’s Principle

Textbook Section 14.3: Pascal’s Principle and Hydraulics

Pascal’s Principle states that when you change the pressure on one part of an enclosed fluid, the change in pressure is applied to all parts of the enclosed fluid equally, as well as the walls of the container. Because pressure is equal to force over area, if the same pressure is applied on two different areas, two different forces are applied, as we see in the following equation.

$\frac{F_1}{A_1} = \frac{F_2}{A_2}$

This is the principle behind the hydraulic lift. Applying a small force on a small area connected by a fluid to a large area results in a large force acting on the large area.

Section 14.4: Buoyancy

Textbook Section 14.4: Archimede’s Principle and Buoyancy

We saw that with air pressure, the air above you pushes down on you. However, the air to the side and below you (if you’re suspended in air somehow) also push on you in other directions. The force of a fluid pushing on you opposite the force of gravity is called the buoyancy force. Buoyancy is the force that pushes back when you try to push a balloon under the water, and the reason why you float in a pool of water. The buoyancy force $\vec{F_B}$ is equal to the weight of the displaced fluid. If you completely submerge an object in a fluid, it displaces an amount of fluid equal to its volume. However, if an object is floating in a fluid but not completely submerged, it instead displaces a smaller volume of fluid. The volume of fluid displaced is equal to the volume of the object that is under the surface of the fluid. When solving buoyancy problems, draw yourself a free body diagram for the object in question and make sure to keep track of the different densities, masses, and volumes correctly.

Section 14.5: Fluid Dynamics

Textbook Section 14.5: Fluid Dynamics

Textbook Section 14.6: Bernoulli’s Equation

Fluid dynamics refers to fluids that are moving in some fashion. For this introduction to fluid mechanics, we will only consider ideal fluids (no internal friction, know as viscosity, no turbulent motion, and an incompressible fluid). A fluid moving through a pipe that cannot be compressed means that the same amount of fluid must go past a certain point in a certain amount of time (say, every second). This does not mean that the fluid always has the same velocity — if the pipe suddenly became far more narrow, a fluid moving at the same velocity everywhere would only move part of the total amount past a certain point per second. This principle, known as the continuity equation (see equation below) actually means that a fluid passing through a more compressed space must move faster in that smaller space to transport the same amount of fluid in a second.

$A_1\,v_1 = A_2\,v_2$

In the equation above, the velocity $v_1$ of the fluid passing through a pipe of cross-sectional area $A_1$ changes to $v_2$ as the pipe changes to a cross-sectional area of $A_2$ .

When you have a change in height of the tube, the pressure can change as well as the velocity and energy of the fluid. Bernoulli’s Equation (equation below) is similar to a conservation of energy equation for fluids, describing how these variables are related.

$p_1 + \frac{1}{2} \rho v_1^2 + \rho g y_1 = p_2 + \frac{1}{2} \rho v_2^2 + \rho g y_2$

In Class Group Problem 14.1:

Airplane windows are made up of three layers with air gaps in between — an outer glass layer, a middle glass layer with a small hole in it, and an inner plastic layer which mostly serves to keep passengers from damaging the middle layer.

When flying at a normal cruising altitude of 30,000 feet, the pressure outside the airplane has dropped from 14.7 PSI (lbs/in $^2$ ) at sea level to 4.30 PSI. The drastically decreased air pressure would make it impossible for humans to breathe in enough oxygen to stay conscious, so the airplane cabin is pressurized to the equivalent of being at 7,000 feet (11 PSI).

Find the pressure inside and outside the plane in Pascals at a typical cruising altitude.

The hole in the middle of the glass pane of the windows allows the pressure to equalize between the plastic and glass layers so that the outer glass window supports most of the force that results due to the imbalance in pressures. The hole also keeps moisture from forming between the panes. If the average surface area of an airplane window is 115 in $^2$ , what net force is exerted on the window, and which way does this force point (in or out of the plane)?

If the outer window breaks under the force, the inner glass layer can support the pressure difference until the plane is able to descend to a safer altitude. The small hole, which is now a leak, is easily compensated for by the pressure system of the plane.

The difference in air pressure between the ground and cabin when flying causes the same kind of force (although not as large) on your inner ear, causing pain until you can cause your ears to “pop” by draining fluid out the Eustachian tube into your nose and equalizing the pressure inside your head with the cabin pressure. Small children (and people flying with head colds) have trouble doing this, which is the cause of most of the crying kids on planes.

If this happens to you, do not attempt to fix the problem by blowing your nose! This can cause your eardrums to rupture, leaving you with much more serious problems. Instead, close your mouth, hold your nose, and turn your head as far as you can to the side while swallowing. Continue this back and forth until your ears pop naturally.

In Class Group Problem 14.2:

A swimmer in a pool ( $\rho_{water}$ = 1.0 g/cm $^3$ ) dives from just above the surface (at sea level) to 3.5 m below the surface of the water. What is the pressure on the swimmer in atm below the surface?

In Class Group Problem 14.3:

What force do you need to apply to area 2 of a hydraulic lift with areas $A_1$ = 0.35 m $^2$ and $A_2$ = 0.0067 m $^2$ if you want to lift your 2,000 lb car a total of 2.0 meters in the air?

In Class Group Problem 14.4:

The manometer below is filled with water. If the right side is open to air pressure, $y_1$ = 3.5 cm, and $y_2$ = 0.75 cm, what is the pressure $P$ ?

The density of water is 1,000 kg/m $^3$ at room temperature. The density of mercury is 13.56 g/cm $^3$ . You want to make a barometer capable of measuring atmospheric pressure at sea level. How long does the tube need to be if you make it out of water? How about if you make it out of mercury?

In Class Group Problem 14.5:

You drop a spherical, latex balloon ( $m_{balloon}$ = 5.0 g, radius $r$ = 7.0 cm) filled with air from a height of 3.0 m. How long does it take to reach the ground if the density of air is $\rho_{air}$ = 1.225 kg/m $^3$ ?

In Class Group Problem 14.6:

You have been hired to lift a statue out of the ocean. The statue has a mass of 500 kg and is made entirely out of silver (10.49 g/cm $^3$ ). The density of seawater is 1030 kg/m $^3$ . The statue is lifted at a constant velocity.

What is the tension in the cable supporting the statue when the statue is completely under the water?

What is the tension in the cable when the statue is completely above the water and still moving at a constant velocity? Ignore the buoyancy due to air.

In Class Group Problem 14.7:

At the end of the movie Titanic, Rose lies on a piece of furniture, floating in the ocean, and Jack drowns. A common objection posed to this scene is that there’s room on the furniture for Jack as well, so why didn’t she let him get on it with her?

Let’s assume that the piece of furniture has an area of 1.5 m $^2$ , a thickness of 5 cm, and is made of pine ( $\rho_{pine}$ = 350 kg/m $^3$ ). Assuming Rose has a mass of 46 kg and Jack has a mass of 58 kg, could they have both managed to stay afloat on the raft together?

In Class Group Problem 14.8:

Water is flowing through the spherical pipe system below. The opening on the left has a radius of 2.0 cm; the opening on the right has a radius of 6.0 cm. If the water is moving at 3.8 m/s through the pipe on the left, how fast is it moving through the right-hand end of the pipe?

In Class Group Problem 14.9:

Water enters a house through a pipe at ground level with an inside diameter of 2.0 cm at $p_1$ = 4.0 $\times 10^5$ Pa. In the bathroom 5.0 m above the ground, the pipe has narrowed to 1.0 cm diameter. If the initial flow speed when the water entered the house was 1.5 m/s, find the final flow speed in the bathroom, the water pressure in the bathroom, and the volume flow rate in the bathroom.

In Class Group Problem 14.10:

You have been hired by Nintendo to do the physics for a new Mario Brothers level. This underwater level includes large stone blocks falling on Mario from above. The blocks begin to fall when Mario first steps directly beneath them, and are cubic with each side 1.70 m long in Mario’s world. The blocks will fall from a height of 3.50 m (at the base of the block) above the ground, and Mario is 1.50 m tall. If the density of the blocks is 2,350 kg/m $^3$ and the density of water is 1,000 kg/m $^3$ , what is the minimum speed at which Mario must be running to avoid getting hit in the head by a block? Ignore drag.

In Class Group Problem 14.11:

Gaseous helium has a density of $\rho_H$ = 0.1786 kg/m $^3$ and air has a density of $\rho_{air}$ = 1.225 kg/m $^3$ . You have decided to recreate the balloon scene from “Up” with yourself (62 kg), your lawn chair (3.0 kg), and a bunch of helium-filled balloons 12 inches in diameter.

How many balloons do you need to attach to your lawn chair to keep yourself and the lawn chair at a stable altitude without sinking or rising? You have already determined the balloons each have an un-inflated mass of 4.0 g.

In Class Group Problem 14.12:

At your next family gathering at the lake, you want to fool your cousins into thinking you can walk on water. You design a piece of wood so that when you, a 50 kg person, walk on it, the wood is entirely submerged but you are entirely out of the water (just the bottom of your feet get wet). The wood is 2.0 m long and wide and 37 cm thick. What density of wood do you need to use for your trick to work?

Practice Exam Question 14.1: