Physics Notes on Fluids

8/2/2019 Physics Notes on Fluids

1/10

Physics 7A Discussion Notes

Chapter 13: Fluids

Aaron Alpert

August 2010

1 Introduction to Fluids.

1.1 Fluid Movement

Figure 1: Kinematics of a Fluid.

Kinematics. In particulate motion, each atom had the same displacement, velocity, andacceleration. In rigid body motion, the relative position of the particles was constant, so themotion of any one particle could be described by the translation of a point and the rotationof the rigid body. Here, every particle moves with different velocity, and the relative positionof the particles is not fixed. In order to better imagine fluid motion, we imagine a cubeof fluid, as seen in figure 1. The cube of fluid can translate (i.e., every particle is displacedby the same vectorial amount). The cube of fluid can rotate through an angle about anaxis. The rotation is referred to as the vorticity. While mathematically involved, it can be

conceptually related to the circulationhow much the fluid is circling around a certain point.(Imagine a rotating merry-go-round. The circulation would be analagous to how much thehorses circle around the center.) You have observed vorticity many times when water spinsaround in a drain or a tornado swirls the air.

1


2/10

Figure 2: Viscosity in Couette Flow.

Deformation. Unlike in a solid, our cube of fluid can become deformed. It will not holdits shape if subjected to forces. Deformation that changes the side length of the cube butdoes not change its right angles can be referred to as elongation or stretching. Deformationthat changes the angles of the cube can be called shear.

Types of Flow. Flows are classified as either laminar or turbulent. With laminar flow,each particle of fluid traces out a line. The particles in front and behind it lie on the samelines, and the lines do not cross. These are called stream lines. However, the flow can becomechaotic, meaning the particles go everywhere, crossing paths, and so on. The classic exampleis a cigarette. Near the cigaratte, the smoke flows in neat orderly lines (laminar), but faraway from the flow, the smoke begins to disperse (turbulent). When we get to problems, wewill only consider laminar cases.

Figure 3: A Cigarette Displays Laminar and Turbulent Flow.

1.2 Fluid Forces.

Because fluids can deform in so many ways, we need to consider not just in what directiona forces acts, but where it acts. Pressure and viscous forces act on the surface of the fluid;body forces, like gravity and magnetic forces, act on the volume of the fluid. As shown in

2


3/10

Figure 4: Forces Act in Both a Certain Direction and on a Certain Place.

figure 4, the red (viscous) and blue (pressure) forces act on the surface of the box of fluid ina certain direction, while the green (body, e.g., gravity) force acts on the entire volume ofth box of fluid. For this reason, it is very difficult to analyze fulids using forces.

Pressure. The force that acts to stretch the cube of fluid is called pressure. Pressure is anormal forceit can only act perpendicular to a surface. An amazing result is the pressureis the same in all directions (up, down, right, left) at a certain spatial point. We relate the

force acting on a surface area with the pressure as follows.

P =dF

dA F

A(1)

The SI units of pressure are Newtons per meter, also called Pascal and notated P a. Othercommonly used units are millimeters of mercury (mmHg, also called torr, based on old mer-cury barometers), standard atmospheres (atm), other SI units (kP a, barwhich is 10P a),or pounds per square inch (psi). Typical atmospheric pressure is 101 .3 kP a (or, if you prefer,760 mmHg, 29.9 inHg, 1 atm, 1.013 bar, or 14.7psi).

Pressure as many important applications in driving the translation of a fluid. In a pipe,for example, the pressure difference along the pipe means that there is a force driving thefluid forward. In your home, the city pressurizes the water, which reaches your home afterhaving lost a lot of its pressure, and comes out your faucet. The reason that the water goesfrom the city to your home and not the other way is the pressure difference in the watermain.

Also, the pressure can change the volume of the box. Since the box would have the samemass but a different volume, the pressure is capable of changing the density. (Think aboutwhat happens when a scuba diver exhales. The bubble increases in size as it rises.) However,in this class, we will almost exclusively consider incompressible fluids. Density will be, forus, a constant.

Viscosity. The force that acts to shear the cube is called viscosity. It is analagous tofriction, and acts very similarly, in many ways. (For one, it removes energy from the system.)In Couette flow, a viscous fluid is placed between two plates separated by H, and fluid flowslike sheets gliding over each other. The bottom plate is stationary, and a person applies aforce to the top plate. The force divided by the area of the plate is the shear on the plate.The plate will travel at a certain speed v, and the fluid at the plate boundary will match

3


4/10

Figure 5: Definition of Viscosity.

the velocity of the plate. Assuming the velocity varies linearly, the velocity u at any heighth above the bottom plate is

u =

y

Hv du

dy =

v

H (2)

The shearing force is proportional to the change (gradient) in velocity by in the equation = du

dy. The constant of proportionality is called viscosity and has units of 1poise =

1 P a s. Since du/dy = v/H and the shear at the top of the plate is just F/A,

F

A=

v

H = F H

Av(3)

Note that this only applies to so-called Newtonian fluids. Many other common fluids (frompaint to blood to cornstarch-water mixture) exhibit other properties.

2 Hydrostatics

2.1 Hydrostatic Pressure

Imagine you are scuba diving 10 m below the surface of the water. You hold out your hand,which is about 60 cm2, and you imagine a large, hand-shaped column of water rising above it.The cross sectional area of the water is 60 cm2, its height is 10 m, making the volume of thatcolumn of water 10m60cm2. If water has a constant density of 1000 kg/m3, then the mass ofthe column of water is V = (1000kg/m3)(10m)(60cm2). The downward force of that wateris mg, or (1000kg/m3)(10m)(60cm2)(9.8m/s2). So what about the pressure on your hand?

It is the force divided by the area of your hand, or (1000kg/m3

)(10m)(9.8m/s2

) = 98 kP a.In general, the pressure on any surface submerged in a fluid (near the surface of earth) is

P = P0 + gy (4)

where P0 is the pressure at the surface, is the density of the fluid, and y is the distancebelow the surface. For most of our situautions, the pressure at the surface P0 is atmospheric

4


5/10

pressure, 101.3kP a. Thus, in water, every 10 meters of descent adds roughly 1 atmosphereof pressure.

Example: Hoover Dam. Hoover Damp is 221 meters tall and 379 meters wide (at thetop). When it is full, what is the force (per unit width) acting on the wall? What is the

torque (per unit width) acting at the base of the dam?

Figure 6: Pressure on Hoover Dam.

Answer: Hoover Dam. In this case, we can ignore the air pressure, as its more or lessbalanced out by the air on the other side of the dam. Thus, the pressure is P = gy. If weimagine a little height y, the force is P A = gyy(379). We can integrate this to find

F = 379

2210

gydy = 379(1000)(9.8)

2210

y dy = 379(1000)(9.8)y2

2

2

0

21 = 2.39 108 N (5)

The torque from that force is r F. The distance from the base to the little height y isactually 221

y, so the torque is actually rF

= (221

y)gyy(379).

= 379

2210

gy(221 y) dy = 379(1000)(9.8)2210

221y y2 dy

= 379(1000)(9.8)

221y2

2 y

3

3

20

21 = 6.69 1014 N m (6)

2.2 Archimedes Principle and Buoyant Force.

Consider a solid mass partially or totally submerged in fluid. If that particle we removed

and replaced by fluid, the surrounding fluid would not care that the solid had replaced.Regardless of what is there, the matter will feel the same force. When the fluid is there, weassume it to be in static equilibrium. Thus, the forces must be zero. There is a gravitationalforce down of mg and an equal and opposite buoyant force. When the solid mass replacesthe fluid, the solid mass feels that same buoyant force. Thus,

FB = fV g (7)

5


6/10

where f is the fluids density and V is the volume that is under the water.

Example: Its Drafty in Here. A barge is a square boat. Suppose the floor of theboat is 6 meters by 8 meters. The barge is 200 kg and is carrying a load of mass m. Howdoes the draft depend on m? (Draft is the distance below the surface of the water that the

boat floats.)

Answer: Its Drafty in Here. Call the draft d. The volume of the boat below the surfaceis (6)(8)d = 48d. Thus, the buoyant force is H2O(48d)g. However, the loaded boat has amass of 200 + m and experiences gravity of (200 + m)g. The boat is in static equilibrium,so the buoyant force equals the graviational force.

H2O(48d)g = (200 + m)g d =200 + m

48H2O= .0042 +

m

48000meters

3 Fluid Dynamics3.1 Conservation of Mass

We now allow the fluid to move. First of all, the mass must be conserved. Consider apipe. Whatever flows into the pipe must either flow out of it or stay in it. We will considersteady state systems in which mass does not accumulate in the pipe. In a situation likethis, we need to find out what mass flows into the pipe per unit time (i.e., mass flux), whichwe will call m. First, a small mass ofm moves into the pipe in some amount of time t.The amount of mass is V. Now, the small volume V has a cross-sectional area A, whichmoves a small distance x, making the small amount of mass A x. The rate of this mass

is m = m/t = Axt , which can be limited down to show

m = Av (8)

where v is the flow velocity. The quantity m is constant throughout the pipe. If the densityis constant (incompressible), then Av is a constant.

3.2 Bernoullis Theorem.

Derivation. The governing F=ma equation for a fluid is called the Navier-Stokes Equa-tion. It is a differential equation which has yet to be solved. In fact, were not even completely

sure it has a solution or if that solution is unique, though we think it does.1

Dv

Dt= P + ( ) + f (9)

1The Clay Mathematics Institute has designated it one of their seven Millenium Problems, and if you

do manage to solve the Navier-Stokes equation, you can earn the $1 million prize.

6


7/10

Because this is a differential equation, it describes what happens to a little box of fluid,so instead of mass we have the density , which is mass per volume. Thus, the first termacts sort of like mass times acceleration. Then, the right-hand side is a description of theforces. P is the pressure, and represents the derivative in all three directions. Its the3D extension of ddx . The middle term is the viscous force. The is the viscosity, and

describes how fast the fluid is deforming. For our analysis, we will ignore viscosity, andthis term is zero. This is, by the way, a very big assumption. The last term, f describesother forces, such as gravity or electromagnetic forces acting on the body of the fluid. Forour purposes, we will consider only the force of gravity (per unit volume): mgV = g. Weobtain something that looks like this (although, this is a bit fluffy to be sure):

a = dPds g (10)

When we first learned about energy, we integrated F = ma with respect to displacement:

F dx =

madx. The left side was defined as work, and the right side as the change in

kinetic energy: W = 12

mv2 = KE, which is the work-kinetic energy theorem. Now, ourfluid is displaced along a path C, which is called the streamline. If we measure the lengthalong the streamline s, then ds is an infinitesimal distance along the path. We integrate theNavier-Stokes Equation with respect to ds.

C

a ds =

C

dPds

ds +

C

g ds (11)

The first term is similar to integrating

madx, which gives us the kinetic energy. The secondterm is easy, using the Fundamental Theorem of Calculus. For the third term, we have toremember that gravitational energy only responds to changes in the vertical direction, z.

Altogether, we obtain the following equation.

P +1

2v2 + gz = B = constant (12)

P is the pressure, v is the velocity, z is the height, and is the density. This is BernoullisEquation, which is essentially an energy equation. In order to be valid, there are severalconditions.

Constant density (incompressible)

No viscosity (inviscid)

B is only constant along a given streamline

Steady flow (the flow doesnt change over timeif the flow is unsteady, then B is afunction of time)

Although this puts some serious limits on the types of problems we can do, it also permitsa good number of problems.

7


8/10

Figure 7: Pitot Tube.

Example: Pitot Tube. Modern aircraft use pitot tubes to measure the airspeed. A smalltube extends from inside the aircraft, has a little fluid-filled dip, and a narrowing tip thatpoints into the direction. If the area at point 1 is half that of the area at the other labelledpoints, the density of air is a, the density of the fluid is f, and the height difference of thefluid is h, what is the planes airspeed and what is the velocity of the air at point 2?

Answer: Pitot Tube. The height differences are small enough to neglect gz. Outsidethe airplane, the pressure is Patm and the velocity is v, the airspeed. The constant fromBernoullis equation is Patm +

1

2av

2. At point 3, the air cannot move, so the velocity goesto zero. However, we are along the sam streamline, so the Bernoulli term will be the same.

Patm +1

2av

2 = P3

Since the other end of the pitot tube opens up to the airplane, the pressure on the other end

of the fluid is Patm. Imagine what is happening at the dotted line shown in figure 7. Thatline is at the same height as point 3, so it must be experiencing the same pressure, P3. Thatpoint feels the atmospheric pressure from the plane P3, plus the additional pressure of thefluid on top of it, fgh. Then,

Patm + fgh = P3 h = P3 Patmfg

We can plug in the value we know for P3 and solve.

h =

Patm +1

2av

2

Patm

fg =

1

2av

2

fg v =2hfg

a

As the flow passes from point 1 and expands to point 2, mass is conserved, so we can findthe velocity at point 2 as

A1v1 = A2v2 v2 = v A1A2

=v

2

8


9/10

We just found the airspeed, so v2 =1

2

2hfg

a.

Example: A Problem Even Torricelli Would Find Draining. This is a variation on

a classic Physics 7A problem... There is a tower of height H and radius R, which is filledwith water. A small hole of diameter a is put near the bottom of the tower. What is thevelocity of the water exiting the hole as a function of time t? In this example, you may notignore the fact that the height of the water is decreasing. You may assume the density ofwater and the atmospheric pressure P0 are constants. [Hint: Once you find the appropriatedifferential equation, the height of the water is a quadratic function of time.]

Figure 8: Water Draining Out of a Tower.

Answer: A Problem Even Torricelli Would Find Draining. We imagine a streamlinebeginning at the surface of the water and following along the path shown as a red line infigure 8 until it exits the tower. At the surface of the water, the Bernoulli equation tells us

P +1

2v2 + gz = P0 + gz(t) =B(t) (13)

As the water exits the tower, it obtains the ambient (atmospheric) pressure, so we can equatethe two points on the red streamline.

P +1

2v2 + gz = P +

1

2v2 = B(t) = P0 + gz(t) v =

2gz (14)

In other words, the speed of water leaving the tube is the same as the speed of an objectthat falls z. However, we must be able to express z as a function of time, and z decreases

as the water drains. What is the rate at which water leaves the tower? If water leaves at aspeed of

2gz through a hole of area a2, then the rate change in volume is

V = Av = (a2)(

2gz) (15)

The volume of water in the tower is V = Ah = (R2)z, which we can differentiate to findthe rate at which water is lost. V = (R2)z. We can set these two equations equal (the

9


10/10

rate at which water is leaving the tank = the rate at which the volume inside the tank isdecreasing).

V = a2

2gz = R2z z2 a2

2g

R2z = 0 (16)

Since we are told that the solution is a quadratic function of time, we know that z(t) =

c2t2 + c1t + c0, which we plug into the differential equation. Since we know that the wateris initially at height H, z(0) = c0 = H.

4c22t2 + 4c2c1t + c

2

1 a

2

2g

R2

c2t2 + c1t + H

= 0 (17)

Matching coefficients tells us that c1 =aR

(2g)1/4

H and c2 =a22g

4R2, giving us a final

expression for z(t). We can plug that in to the expression v =

2gz.

v(t) = 2ga2

2g

4R2t2 +

a

R(2g)1/4

Ht + H

1/2

= a

R(2)1/4(g)5/4t + 2g

H

2

(18)

These notes 2010 by Aaron Alpert.

10

Physics Notes on Fluids

Documents

Transcript of Physics Notes on Fluids