1. Lecture 21 Goals: Chapter 15, fluids Assignment
HW-8 due Tuesday, Nov 15
Wednesday: Read through Chapter 16 1

2. SI unit for pressure is 1 Pascal = 1 N/m2
Fluids
Another parameter
Pressure (force per unit area)
P=F/A
SI unit for pressure is 1 Pascal = 1 N/m2
The atmospheric pressure at sea-level is
Now, for fluids, there is also another very important quantity that we usually talk about, and that is the pressure. Pressure is the ratio of the force to the area that the force is exerted.So we divide the force to the area.
The pressure is a scalar quantity, so it does not have a direction. However, the force exerted by a fluid is perpendicular to the surface.
The SI unit for pressure is N/m^2, which is also called pascal.
There are a lot of different units for pressure, and you should be familiar with some of them. One of the most important units is the atmospheric pressure at sea level. So the pressure of the gas that forms the atmosphere, mostly oxygen and nitrogen molecules, at sea level, is what we call 1 atmospheres, which is roughly 10^5 pascals which is 14.7 pounds/in^2, so at every inch^2, there is a a force that is equivalent to the weight of 14.7 pounds of mass
1 atm = x105 Pa
= 1013 mbar
= 760 Torr
= 14.7 lb/ in2 (=PSI)

3. Incompressible fluids (liquids)
What is the pressure at the bottom of the container?
F=Mg=ρVg
F=ρAyg
Pressure=F/A=ρyg y
P=ρgy
One of the key ideas is that , when you go deeper in a liquid, the pressure increases. So, in this slide we will make this observation more quantitative.
Let’s consider a liquid that fills a cup of area A, and depth y, and let’s ask the question, what is the pressure that is applied to the bottom surface of the cup.
Now the liquid is not going anywhere, so it is at equilibrium, which means that, the bottom surface here must be carrying all the weight of the liquid. So the total force on the bottom surface must be M g where M is the total mass of the liquid. Now M is rho V where rho is the density.
Area=A

4. What if there is outside gas?
Pressure=P0
F=P0A+Mg
P0A
P=P0+ρgy y
Area=A

5. P1=P0+ρgy1
Pressure=P0
P2=P0+ρgy2 y1 y2
P2-P1=ρg(y2-y1)
Area=A

6. What is the pressure 10m down?
P=P0+ρgy
=P0+(1000 kg/m3)(10 m/s2) (10 m)
=P0+105 N/m2
= approximately 2 atm
Home exercise: what is the pressure 6 miles down?
1) Home exercise: what is the pressure 6 miles down the ocean, basically the deepest place in the ocean? There are only one or two vessels that can go this far

7. Consider the open, connected container shown below
Consider the open, connected container shown below. How would the two heights compare?
y1<y2
y1=y2
y1>y2
The next key idea is that in an open connected container, the fluid level will rise to the same height. Let’s consider a container that looks something like this, so this side has a much larger area compared to this side.
In such a container, the fluid level will be the same on both side. The reason for this is that, if the fluid levels were different, let’s say that this side had a higher level, then from the pressure expression that we derived, the pressure at the bottom of this container would be higher than at the bottom of this container. Which means that, the forces from left and right on this middle region would have been different, so this middle region would not be in equilibrium and there would be flow. And the flow would continue until the pressure was equalized on both sides which means that until each side reaches the same height. y1 y2

8. Pressure vs. Depth
In a connected liquid, the pressure is the same at all points through a horizontal line. p
1) So if we draw a horizontal line here, regardless of these funny shapes of these containers, if we draw a horizontal line, we would have exactly the same pressure p at all of these points. And the arguments for seeing this are very similar to the one that I just made, if they weren’t at the same pressure, then the system would not be at static equilibrium and there would be flow.

9. Pressure Measurements: Barometer
Invented by Torricelli
A long closed tube is filled with mercury and inverted in a dish of mercury
The closed end is nearly a vacuum
Measures atmospheric pressure as
1 atm = m (of Hg)
P0=ρgh

10. Archimedes’ Principle
Suppose we weigh an object in air (1) and in water (2).
How do these weights compare?
W2? W1
a) W1 < W2
b) W1 = W2
c) W1 > W2

11. F1 F2 Buoyancy F2=P2 Area F1=P1 Area F2-F1=(P2-P1) Area
=ρg(y2-y1) Area
=ρ g Vobject
=weight of the fluid
displaced by the
object y1 F1 y2
The reason for this is that due to the buoyant force. When an object is immersed in a liquid, there is a force that tends to push the object up, and the object will either float or sink depending on how the buoyant force compares with the weight of the object.
Now why is there a force that pushes the object up. This is directly related to the fact that the pressure goes up as we go further down. So in this slide we are going to derive that. Let’s consider an object that is immersed in a liquid. For simplicity, we will consider a simple rectangular object, but the argument holds for objects of arbitrary shape Let’s call the pressure at the bottom of the object P2…. F2

12. ρobject Vobject< ρfluid Vobject ρobject < ρfluid
Float or sink?
If we immerse the object completely in the liquid:
weight of the object < bouyant force
float
ρobject Vobject< ρfluid Vobject
float
ρobject < ρfluid
float
How does a steel ship float?
ρsteel < ρwater
overall density of the ship < ρwater
none of the above
It is easier to float in a fluid with a larger density. This is the reason why it is easier to float in the sea compared to the pool, simply because salty water has slightly larger density and that density makes a difference.
It would, for example, be very easy to float in liquid mercury. You actually can walk on liquid mercury, you would only sink slightly higher than your knees.

13. Vimmersed ρfluid =Vobject ρobject
Float
If the object floats, then we can find the portion of the object that will be immersed in the fluid
FB=mg
Vimmersed ρfluid g =Vobject ρobject g FB
Vimmersed ρfluid =Vobject ρobject
1) This principle is responsible for 90% of the icebergs being in the water and only 10% being visible over the sea. You can easily show that by using the density of the ice and the density of the water, I will leave it as a home exercise for you to show that.

14. P=P0+ρgy Pascal’s Principle Pressure=P0 y
Any change in the pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel.
Pressure=P0
1) Because the liquid is not compressible, the pressure is that is applied to the fluid is transmitted to every portion of the fluid. We have already kind of derived this. We argued that if we have an outside pressure P0, the pressure at a distance y inside the liquid is P=P0+rhogy. We derived this from the requirement of static equilibrium. So this pressure P0 is felt everywhere in the liquid. And it is transmitted to the walls of the container. y
P=P0+ρgy

15. Pascal’s Principle in action: Hydraulics, a force amplifier
Consider the system shown:
A downward force F1 is applied to the piston of area A1.
This force is transmitted through the liquid to create an upward force F2.
Pascal’s Principle says that increased pressure from F1 (F1/A1) is transmitted throughout the liquid.
P1 = P2
F1 / A1 = F2 / A2
A2 / A1 = F2 / F1
Energy conservation!!

16. To describe fluid motion, we need something that describes flow:
Fluid dynamics
To describe fluid motion, we need something that describes flow:
Velocity v
Ideal fluid model:
Incompressible fluid.
No viscosity (no friction).
Steady flow
So far, we have discussed fluids at static equilibrium, now we will start our discussion of fluid dynamics, where there will be fluid-flow, fluid-motion.
As we discussed many times during this course, whenever we have motion, one of the most important kinematic quantities is the velocity of the motion. So here we will associate a velocity to fluid flow, keeping in mind that different portions of the fluid might be flowing with different velocities.
Fluid flow is a very complicated subject. When the flow is turbulant in particular, there are almost no analytical techniques, and the problems are almost always analyzed by large-scale numerical calculations.
So to make the problem more tractable, we will make simplifying assumptions, which are called the ideal fluid model:
These conditions are called laminar flow.

17. Types of Fluid Flow
When the flow is turbulent, you typically observe very erratic flow pattern, a lot of circular motion, and also the flow will not be steady that is at some point in time, you might see a flow pattern shown here, and at a little later time, the pattern may be quite different. So this is not what we will be considering in this course.
So the flow that we will be considering, laminar flow, will have smooth velocity vectors, that don’t change with time as shown here.

18. Keep track of a small portion of the fluid:
Streamlines
Keep track of a small portion of the fluid:
When we discuss fluid-flow, we frequently draw lines that are called streamlines, and in this slide I would like to introduce them.
So basically, we concentrate on a small volume of the fluid and we follow it through as it flows.
So these are the velocity vectors, and in this example, this small volume of the fluid is moving here, and then from here, to here
If we connect the trajectory of this small volume, we call that a streamline.
If we go though the same procedure at other positions of the fluid, we get the streamlines that characterize the flow.
For laminar flow, the streamlines do not cross
The velocity vector at each point is tangent to the streamlines

19. A1v1=A2v2 A1v1 : units of m2 m/s = volume/s
Continuity equation A 1 2 v
A1v1 : units of m2 m/s = volume/s
A2v2 : units of m2 m/s = volume/s
Now, one of the most important equations in fluid flow is the continuity equation, which you can view as mass conservation.
2) Considers cross sectional area A1 and assume that the flow has a uniform speed v1 at this first point
3) Similarly, at a later point consider cross sectional area A2 with velocity v2
4) Now if we multiply the area with the velocity, we get the volume of fluid per second that is flowing through the first point
5) Similarly, if we do that at the second point, we would get the volume of fluid per second that is flowing through the second point.
6) Now, whatever is coming in must be coming out.
7) To find the mass flow rate just multiply by the density, this is really a statement of mass conservation.
A1v1=A2v2