Ch. 8 Fluids and Elasticity of Solids (still learning)

Hydrostatics (fluids at rest)

-Fluids are substances that can flow

-Both liquids and gases are fluids (because both liquids and gases are able to flow)

There are distinctions between liquids and gases. At the molecular level, the substance in the liquid phase is similar to one in the solid phase, in that the molecules are close to, and interact with one another. The molecules in a liquid are able to move around a little more freely than those in a solid, in which the molecules typically only vibrate around fixed positions. By contrast, the molecules of a gas are not constrained and fly around in a chaotic swarm, with hardly any interaction.

Filling of a container

On a macroscopic level, there is another distinction between liquids and gases. If you pour a certain volume of a liquid into a container of a greater volume, the liquid will occupy its original volume. whatever the shape and size of the container. However, if you introduce a sample of gas into a container, the molecules will fly around and fill the entire container.

Density and Specific Gravity

The density is the amount of mass contained in a unit volume. In SI units, density is usually expressed in kg/m³ or g/cm³.

Equation for density

Density= mass/ volume

Density of water

There is one substance whose density you should memorize: The density of liquid water is taken to be 1000 kg/m³. or 1 g/cm³. (Another useful version of the same value: 1 kg/L, where L stands for a liter; a liter is 1000 cm³.

Specific Gravity

Specific gravity (poorly named) is a unitless number tells how dense something is compared to water:

formula for specific gravity

specific gravity = density of substance/density of water

sp. gr. = density/ density of water

The density of solids, liquids and gases

The density of a solid remains constant. For solids, density doesn’t change much with surrounding pressure or temperature. For example, the density of marble is pretty close to 2700 kg/m³ under most conditions.

The density of a liquid remains constant. Liquids behave the same way: the density of water is pretty close to 1000 kg/m³ under all conditions at which it’s a liquid.

The density of a gas changes. However, the density of a gas changes markedly with pressure and temperature.

Example 8.1

Turpentine has a specific gravity of 0.9. What is the density of this liquid?

1. given the specific gravity (specific gravity = density/ density of water)

2. find the density of the liquid (rearrange the equation of specific gravity to give you the density)

3. density= specific gravity x density of water = 0.9 × 1000 kg/m³= 900 kg/m³.

Come back to example 8.2, 8.3

Force of Gravity for Fluids

It is handy to know how to find the force of gravity acting on the fluid itself or objects that are immersed in the fluid. In previous chapters, we have used F_grav= mg. However, with fluids, it is more difficult to remove a portion of fluid from a tank, place it on a scale and find its mass. Using the relationship between mass, volume and density, we can redefine the magnitude of F_grav for fluids questions:

F_grav= mg= pVg

make sure the density (p) and volume (V) describe the properties of the correct fluid.

Pressure

1. place an object in a fluid

2. the fluid exerts a force on the object

if we look at how the force is distributed over any small area of the objects surface, we have the concept of pressure.

Pressure Formula

P= F⊥/A

P: pressure

F: force

A: area

The subscript ⊥, meaning perpendicular, indicates that pressure is defined as the magnitude of the force, acting perpendicular to the surface, divided by the area. We don’t need to worry about this, because on the MCAT, at any given point in a fluid the pressure is the same in all directions, meaning the force does not depend on the orientation of the force.

Pressure is a scalar quantity

-although the formula for pressure involves “force”, pressure is a scalar quantity, because the perpendicular force is the same for all orientations of surface.

Unit of pressure

The unit of pressure is N/m², which is called a pascal (abbreviated Pa). Because 1N is a pretty small force and 1 m² is a pretty big area, 1 Pa is very small. Often, you’ll see pressure expressed in kPa (or even in MPa). For example, at sea level, normal atmospheric pressure is about 100 kPa.

Suppose we have a:

1. tank of water with a lid.

2. suspended from a lid is a string attached to a thin metal sheet

3. (picture on top of page 221)

The weight of the water above the metal sheet produces a force that pushes down on the sheet. If we divide this force by the area of the sheet, w/A, we get the pressure, due to the water, on the sheet. The formula for calculating this pressure depends on the density of the fluid in the tank (P_fluid), the depth of the sheet (D), and the acceleration due to gravity (g).

P= p_fluidDg

P: pressure

p: density

D: the depth of sheet

g: acceleration due to gravity

Hydrostatic Gauge Pressure (using the water in tank situation on page 221)

Hydrostatic gauge pressure is the pressure exerted by a fluid at rest due to the weight of the fluid column above it, measured relative to local atmospheric pressure.

P_gauge= p_fluidgD

This formula gives the pressure due to only the fluid (in this case, the water) in the tank. This is called hydrostatic gauge pressure. It’s called hydrostatic, because the fluid is at rest, and gauge pressure means that we don’t take the pressure due to the atmosphere into account. If there were no lid on the water tank, then the water would be exposed to the atmosphere, and the total pressure at any point in the water would be equal to the atmospheric pressure pushing down on the surface plus the pressure due to the water (that is, the gauge pressure). So, below the surface, we’d have

P_total= P_atm+ P_gauge

Another Situation

If the tank were closed to the atmosphere, but there we a layer of gas above the surface of the water, then the total pressure at a point below the surface would be the pressure of the gas pushing down at the surface plus the gauge pressure: P_total=P_gas+ P_gauge.

come back to page 222-224

Buoyancy and Archimedes’ Principle

Buoyancy is the upward force exerted by a fluid (liquid or gas) that opposes the weight of a partially or fully immersed object.

1. Let’s place a wooden block in our tank of water.

2. Since the pressure on each side of the block depends on its average depth (the deeper you go, the more pressure exerted), we see that there’s more pressure on the bottom of the block than on the top of it.

3. Therefore, there’s a greater force pushing up on the bottom of the block than there is on the top of it. The forces due to the pressure on the other four sides (left and right, front and back) cancel out, so the net fluid force on the block is upward. This net upward fluid force is called the bouyant force, denoted by F_Buoy.

Archimedes’ Principle

We can calculate the magnitude of the buoyant force using Archimede’s principle:

The magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.

When an object is 1. partially or 2. completely submerged in a fluid, the volume of the object that’s submerged, which we call V_sub, is the volume of the fluid displaced.

there is a picture on page 225

By multiplying V_sub by the density of the fluid, we get the mass of the fluid displaced; then, multiplying this mass by g gives us the weight of the fluid displaced. So, here’s Archimedes’ principle as a mathematical equaiton:

F_B= p_fluidV_subg

When an object floats, its submerged volume is just enough to make the buoyant force it feels balance its weight. That is, for a floating object, we always have w_object=F_bouy. If an object’s density if p_object and its volume is V, its weight will be p_object V_object g. The buoyant force ti feels is p_fluidV_subg. Setting these equal we get:

Floating Object in Equilibrium on Surface

come back to this

Pascal’s Law

Pascal's law states that a change in pressure applied to an enclosed, incompressible fluid at rest is transmitted equally and undiminished to every point in the fluid and to the walls of its container

Pascal’s law is a statement about fluid pressure. It says that a confined fluid will transmit an externally applied pressure change to all parts of the fluid and the walls of the container without loss of magnitude. In less formal language, if you squeeze a container of fluid, the fluid will transmit your squeeze perfectly throughout the container. The most important application of Pascal’s law is to hydraulics.

Example of Pascal’s law

Consider a simple hydraulic jack consisting of two pistons resting above two cylindrical vessels of fluid that are connected by a pipe.

1. If you push down on one piston, the other piston will rise.

Let’s make this more precise. Let F₁ be the magnitude of the force you exert down on one piston (whose cross-sectional area is A₁) and let F₂ be the magnitude of the force that the other piston (cross-sectional area A₂) exerts upward as a result.

there is a picture on page 230

1. Pushing down on the left-hand piston with a force F₁ introduces a pressure increase of F₁/A₁.

2. Pascal’s law tells us that this pressure change is transmitted, without loss of magnitude, by the fluid to the other end. Since the pressure change at the other piston is F₁/A₁, we have, by Pascal’s law,

F₁/A₁=F₂/A₂

Solving this equation for F₂, we get

F₂=A₂/A₁ (F₁)

So, if A₂ is greater than A₁ (as it is in the figure), then the ratio of the areas, A₂/A₁ will be greater than 1, so F₂ will be greater than F₁; that is, the output force, F₂, is greater than your input force, F₁. This is why hydraulic jacks are useful; we end up lifting something very heavy (a car for example) by exerting a much smaller force (one that would be insufficient to lift the car if it were just applied directly to the car).

This seems too good to be true, doesn’t this violate some conservation law? No, since there’s no such thing as a “Conservation of Force” law. However, there is a price to be paid for the magnification of the force. Let’s say 1. you push the left-hand piston down by a distance d₁, and that 2. the distance the right-hand piston moves upward is d₂. Assuming the fluid is incompressible, whatever fluid you push out of the left-hand cylinder must appear in the right-hand cylinder. Since volume is equal to cross-sectional area times distance (v= Ad), the volume of the fluid you push out of the left-hand cylinder is A₁d₁, and the extra volume of fluid that appears in the right-hand cylinder is A₂d₂.

There is a picture on page 231

But these volumes have to be the same, so A₁d₁. Solving this equation for d₂, we get

d₂=A₁/A₂

If the area of the right-hand piston (A₂) is greater than the area of the left-hand piston (A₁), the ratio A₁/A₂ will be less than 1, so d₂ will be less than d₁. In fact, the decrease in d is the same as the increase in F. For example, if A₂ if five times larger than A₁, then F₂ will be five times greater than F₁, but d₂ will only be one-fifth of d₁. We can now see that the product of F and d will be the same for both pistons:

F₂d₂=(A₂/A₁ F₁) x (A₁/A₂ d₁)= F₁d₁

Recall that the product of F and d is the amount of work done. What we have shown is that the work you do pushing the left-hand piston down is equal to the work done by the right-hand piston as it pushes upward. Just as when we discussed simple machines and mechanical advantage in Chapter 6, we can’t cheat when it comes to work. True, we can do the same job with less force, but we will always pay for that by having to exert the smaller force through a greater distance. This is the whole idea behind all simple machines, not just a hydraulic jack.

Surface Tension

The reason a drop falls from the faucet into the sink is because of surface tension

1. the spherical shape is distorted by the pull of gravity

2. the force, due to surface tension, causing the drop to cling to the head of the faucet is overwhelmed by the increasing weight of the drop.

The strength of the surface tension depends on the particular liquid and is determined by the coefficient of surface tension, γ, which is the force per unit length.

γ=F/ 2L

defines the surface tension of a liquid film

surface tension (gamma) is the force per unit length

Why is it 2L instead of L? A thin liquid film suspended in the air has two distinct surfaces—a top surface and a bottom surface

8.2 Hydrodynamics: fluid in motion

Flow rate and continuity equation

Flow rate, f, is the volume of fluid that passes a particular point per unit time.

ex: how many liters of water per minute are coming out of a faucet.

The SI unit for flow rate is m³/s.

Flow rate equation

f=Av

to find the flow rate, we need to multiply the cross-sectional area of that pipe at any point, A, by the average speed of the flow, v, at that point.

Be careful not to confuse flow rate with flow speed

1. flow rate: tells us how much fluid flows per unit time

2. flow speed: tells us how fast the fluid moves

There’s a difference between saying a hose ejects 4 liters of water every second (flow rate) and saying the water leaves the hose at a speed of 4m/s (flow speed).

If a pipe is carrying an incompressible liquid (that is, the density of the fluid remains constant), then a flow rate must be the same everywhere along the pipe. Choose any two points in a flow tube carrying a liquid. Point 1 and Point 2. If there aren’t any sources or sinks between these points (i.e., no leaks and no additional liquid), then the liquid that flows by Point 1 must also flow by Point 2, and vice versa. In other words, f1=f2, or since f=Av, we get A₁v₁=A₂v₂ (this is the continuity equation)

Continuity Equation

A₁v₁=A₂v₂

The continuity equation tells us when a fluid has constant density (incompressible fluid), the volume flow rate remains constant throughout the pipe, when the tube narrows, the flow speed will increase, and if the tube widens, the flow speed will decrease.

There is a picture on page 233

1. Incompressible Fluid (Constant Density)

When fluid density (ρ) does not change—common for liquids like water—the volume flow rate remains constant throughout the pipe:
A₁V₁=A₂V₂

• A₁, A₂: Cross-sectional areas of the pipe sections

• V₁, V₂: Fluid velocities at each section

• Meaning: If the pipe gets narrower, the fluid must speed up; if the pipe gets wider, the fluid slows down

Example 8-12,8-13,8-14

Bernoulli’s Equation

Bernoulli’s equation applies to ideal fluid flow. A fluid must satisfy to following four requirements in order to be considered an ideal fluid:

1. The fluid is incompressible.

This works very well for liquids, gases are quite compressible, but it turns out that we can use the Bernoulli equation for gases provided the pressure changes are small.

2. There is negligible viscocity

Viscosity is the force of cohesion between the molecules in a fluid; think of it as internal friction for fluids. For example, maple syrup is more viscous than water, and there’s more resistance to a flow of maple syrup than to a flow of water. (While Bernoulli’s equation give good results when applied to a flow of water, it would not give good results if it were applied to a flow of maple syrup)

3. The flow is laminar

In a tube carrying a flowing fluid, a streamline is just what it sounds like: a “line” in the stream (look at the figure). If we were to inject a drop of dye into a clear glass pipe carrying, say, water, we’d see a streak of dye in the pipe, indicating a streamline. The entire flow is called streamline (as an adjective) or laminar if the individual streamlines don’t cross. When the flow is laminar, the fluid flows smoothly through the tube.

take a picture of the page of streamlines.

The opposite of streamline flow is turbulent flow. In this case, the flow is not smooth; it is chaotic (unpredictable). Turbulence is characterized by whirlpools and swirls (vortexes). At high enough speeds, all real fluids experience turbulent flow, and no simple equation can be applied to such a flow.

4. the flow rate is steady

That is, the value of f is constant. If we’re analyzing the water flowing through a garden hose connected to a faucet sticking out of the side of the house, turn the faucet handle to a particular setting and then leave it there. The flow rate through the hose must be steady while we’re taking our measurements.

If these conditions hold- (1) the fluid is incompressible, (2) the flow is smooth (laminar), (3) there’s no friction (viscocity), and (4) the flow rate is stready- then total mechanical energy will be conserved. Bernoulli’s equation is the statement of conservation of total mechanical energy for ideal fluid flow. On the MCAT, you will often be told to consider a fluid to be ideal, allowing you to use Bernoulli’s equation.

Bernoulli’s Equation

P₁+1/2ρv₁²+pgy₁= P₂+1/2ρv₂²+ρgy₂

In this equation, ρ is the density of flowing fluid. P₁ and P₂ give the pressures at any two points along a streamline within the flow, v₁ and v₂ give the flow speeds at these points, and y₁ and y₂ give the heights of these points above some chosen horizontal reference level.

there is a picture on page 236

Although the equation may look complicated, notice that the two sides are the same, except all the subscripts on the left-hand side are 1’s while all the subscripts on the right-hand side are 2’s. Also, each 1/2pv²₁ term looks very much like the kinetic energy (sometimes it’s referred to as kinetic energy density), and each term ρgy looks very much like gravitational potential energy. So, just take the equation you already know for conservation of total mechanical energy, KE₁+PE₁=KE₂+PE₂, change the m’s to ρ’s, add P to both sides and you’ve got Bernoulli’s equation.

there is a picture on page 236.

Although the equation may look complicated, notice that the two sides are the same, except all the subscripts on the left-hand side are 1’s while all the subscripts on the right-hand side are 2’s. Also, each 1/2pv₁² term looks very much like the kinetic energy (sometimes it’s referred to as kinetic energy density), and each term ρgylooks very much like graviational potential energy. So, just take the equation you already known for conservation of total mechanical energy. KE₁+PE₁=KE₂+PE₂, change the m’s to ρ’s, add P to both sides, and you’ve got Bernoulli’s equation.

Look at the picture on page 236.

stopped here