Lecture 10: Fluids

Don't Panic

---

Introduction

---

Fluids are liquids and gases.

They flow, hence fluids.

They cannot resist a shear force much. One that tries to change their shape, but not volume.

But liquids are relatitively dense and incompressible. The atoms and molecules are ``touching'', and like to stick together, but flow over each other.

The pressure of fluids is caused by trying to squush the atoms: they resist.

In gases, the particles (atoms and/or molecules in ordinary gases) are far apart and fly around pretty freely.

Pressure is caused collisions.

The particles move because they have kinetic energy which is measured by temperature

---

Ideal Gas Law

---

Most (ordinary) gases obey the ideal gas law to some approximation:

    P = n_(particle density) * k_(boltzmann's constant) * T_(Kelvin) .

    Pressure increases linearly with n_(particle density).

    Pressure increases linearly with T_(Kelvin).

k_(boltzmann) = 1.380 6504(24)*10**(-23) J/K.

But remarkably the ideal gas law does not depend on the mass of the particles.

This means that a gas of lower particle mass will be less dense in (mass) density than a gas higher particle mass when the n_(particle density) and T_(Kelvin) are the SAME.

This in combination with Arhimedes principle explains ``lighter-than-air'' aircraft using hydrogen (molecular hydrogen [H_2]) and helium (atomic helium since helium forms no known molecules: it is an inert gas): i.e., balloons (aircraft) and airships (dirigibles).

But one can also use hot air one can also create ``lighter-than-air'' aircraft.

This is done only for balloons I believe.

The air is heated by a burner below the balloon bottom which is open and above the basket.

The hot air has the same pressure as the surrounding air, but lower density.

Higher T_(Kelvin) compensates for lower n_(particle density).

---

Archimedes Principle

---

Archimedes (c. 287--212 BCE): He has his principle:

An object partially or wholly immersed in a fluid is acted on by an upward buoyant force equal to the weight of the fluid it displaces.

               F_buoyant= rho_(fluid)*V_(displaced)*g    upward  .

      

Spectacular examples of the buoyant force include:
1. balloons These have been around since the 18th century and have been used in war since the French Revolutionary War and the American Civil War. The latter given a realistic depiction in Mysterious Island.
2. airships (dirigibles): which included the famous LZ 129 Hindenburg.
3. icebergs.
4. submarines.

Prof. Chris's Galileo thermometer.

Now for some equations on the board.

               Recall

               F_buoyant= rho_(fluid)*V_(displaced)*g  upward  .


               For a floating object

               in the vertical direction we have from 

               Newton's 2nd law (F=ma),

               ma = 0 = sum_i F_i =  F_buoyant - F_gravity

               0 = rho_(fluid)*V_(displaced)*g - mg

               0 = rho_(fluid)*V_(displaced)*g - rho_(object)*V*g .
              
               Now g cancels out

               and one finds

               rho_(fluid)*V_(displaced) = rho_(object)*V

               or

               V_(displaced)          rho_(object)
               -------------     =    ----------
                    V                 rho_(fluid)

               For example, 

               ice has density 0.9167 g/cm**3 at 0 degrees C 
               and 
               liquid water has density 0.9998 g/cm**3 degrees C at the same
               temperature. 

               Thus

               V_(displaced)          rho_(object)              0.9
               -------------     =    ----------   = approx ---------  = 0.9
                    V                 rho_(fluid)               1

               So ice floats with about 90 % of its volume submerged.

               The values change a bit for sea water which has salt dissolved in it.

      

Question: What happens if V_(displaced)/V > 1?
1. The object floats on the surface.
2. The object floats in the middle of the fluid.
3. Well actually this can't happen. The object can only displace as much fluid as its own volume. But if rho_(object) > rho_fluid, the object sinks.



Answer 3 is right.

Question: What happens if V_(displaced)/V = 1?
1. The object floats on the surface.
2. The object floats anywhere it is initially placed.
3. The object sinks to the bottom.



Answer 2 is right.

This is the ideal case.

Of course, fluid density varies with depth as we will see below, so the object rise/sink to the location where the surrounding density is just equal to its own and stay there.

Or perturbations (motions in the fluid) might send it to the surface or bottom or have it wander around in depth.

Question: Steel is denser than water, but steel boats float. Why?
1. This whole discussion is hogwash.
2. They don't float. Wooden boats float. We've never left the age of sail.
3. Steel is denser than water, but the average density of a steel boat is less than water since most of the interior is air.



Answer 3 is right.

But don't replace the interior air with water.

The Titanic tried that.

---

Pascal's Principle

---

Pascal (1623--1662): Pascal had his principle too:

A change in the pressure of an enclosed incompressible fluid is conveyed undiminished to every part of the fluid and the surfaces of its container.

This principle is one of the bases of hydraulics.

The conveying is at of order of the sound speed in the fluid which means delay times are usually negligible in applications: e.g., hydraulic machinery.

No equations on the board.

---

Bernoulli's Equation

---

Daniel Bernoulli (1700--1782) and Leonhard Euler (1707--1783) winged it up.

We will derive the Bernoulli's equation for incompressible fluids.

But there is a version for gases too.

Some board work.

The result is simple:

 
    p + (1/2)*rho*v**2 + rho*g*y = constant

       where 

       p is pressure,

       rho is (mass density),

       v is speed,

       g=9.8 N/kg is the gravitational field at the Earth's surface
                  (which is the only location we are considering here),

       y is height measured upward

       and the constant depends on the zero-point for chosen for y.

    

Bernoulli's equation applies along stream lines which are just the path a tiny test particle in the fluid would travel along.

Remember we have derived the equation only for steady-state flows.

Also we have derived it assuming energy conservation which is absolutely true---but we havn't accounted for all possible energy transformations.

Loses to viscosity (the fluid counterpart of friction) will convert some energy to waste heat.

So the Bernoulli's equation is an approximation to some degree.

We have only derived Bernoulli's equation for incompressible fluids, but there is a version for compressible fluids (i.e., gases too).

That version is similar.

Thus, Bernoulli's equation explains Bernoulli lift.

Question: Take a some single sheet of paper in your fingers with your fingers on either side of one of narrow ends. Hold this end just BELOW your lips and blow a strong gust.
1. Nothing happens, because you've blown too hard.
2. Nothing happens, because you've blown too softly and you've never succeeded in blowing up a balloon in your life.
3. You spit.
4. The instructions are unintelligible.
5. The paper rises.



Answer 5 is right if all went well.

The paper rises because you've created a high-speed, low-pressure zone above the paper.

This is the Bernoulli lift which is part of aerodynamic lift by which airplanes fly.

Of course, if you put the paper above your lips and blow the paper rises too. This time it is the reaction lift reaction lift which is the other part of aerodynamic lift.

The blown air is deflected down by the paper, but for every force there is an equal and opposite force and so the air pushes up on the paper too.

How a carburetor works.

---

Pressure, Depth, and Gravity

---

Why can't we snorkel to any depth?

See Snorkeling

But we can scuba dive.

Jacques Cousteau (1910--1997) and the Undersea World of Jacques Cousteau.

Two derivations of the pressure-depth relation?

1. Take the Bernoulli's equation for incompressible fluids.

       
           p + (1/2)*rho*v**2 + rho*g*y = constant
   
           If there is no flow (i.e., we have a static case), v = 0. 
   
           Change y to -y, so that y is measured downward, and one obtains
   
           p = rho*g*y + constant  
   
           or 
   
           p = rho*g*y + p_0 where p_0 is the pressure at y=0.
          
           In the case of surface open to the air,
   
           p = rho*g*y + p_air where y is depth measured from the surface. 
   
       

   This derivation is limited to incompressible fluids and seems a bit unbelievable since we derived it from result that depended on motion.

2. Consider a fluid of any kind at rest (i.e., a static fluid).

   The fluid could be in any convoluted container of any sort.

   But we are assuming it is on the Earth's surface or just a few kilometers up so that g=9.8 N/kg can be considered constant.

   We will assume incompressible fluids for simplicity though a similar derivation using calculus holds for compressible fluids.

   On any horizontal plane the pressure is constant, or else there would be motion.

   Take down as positive.

   Divide the fluid into thin horizontal layers.

   In each layer i (where we count i downward), consider a little cylinder of area A and depth Delta y_i=y_i-y_(i-1).

   i=0 is a reference layer where depth $y=0$.

   The cylinders are not stacked on top of each other. They can be anywhere in their layer.

   
               The gravity force on a cylinder is A*(Delta y_i)*rho*g.
            
               The pressure forces are 
   
               (p_(i-1) - p_i)A
   
               From Newton's 2nd law (F=ma), we find
   
               ma = 0 = A*Delta_y*rho*g + (p_(i-1) - p_i)A  
   
               which leads to
   
               p_i-p_(i-1) = (Delta y_i)*rho*g 
   
               or
   
               If we sum all the equations from $i=0$ to $i$, we find
   
               p_i-p_0= sum_i (Delta y_i)*rho*g  
   
               or
                
               p_i= rho*g*y_i + p_0.
   
               If we go to the continuum limit (i.e., let all the
               layers approach zero thickness),
   
               p= rho*g*y + p_0 which is just our previous result.
   
            

As an example, let's consider water with surface open to the air.

   p = rho*g*y + p_air  = approximately ( 1000*10*y + 10**5 ) Pascals

     = ( 10**4*y + 10**5 ) Pascals

     = ( 0.1*y + 1 ) atmospheres  where y is measured in meters.

   

Every time you go down another 10 meters, the pressure increases by one atmosphere.

Humans can do free-diving (i.e., diving without breathing apparatus) to considerable depths (record is 244 meters with fins and/or weights).

This is because our bodies adjust to the pressure change among other things.

But snorkeling can't be done to deeper than about 40 centimeters (15 inches) of tube.

   For air rho = approximately 1.2 kg/m**3 at sea level 
   Wikipedia:  air density. 
  
   The same formula holds for air as for water, but with a density
   about 1000 times smaller.  

   p = rho*g*y + p_air  = approximately ( 1.2*10*y + 10**5 ) Pascals

     = ( 12y + 10**5 ) Pascals

     = ( 0.001*y + 1 ) atmospheres  where y is measured in meters.

   

The air pressure down a snorkel tube will not keep pace with water pressure as you descend.

The snorkeler has trouble inhaling below 40 centimeters because the lungs won't expand against the higher pressure of water outside.

If you are free-diving, you are not breathing externally (you are in a state of apnea) and you don't need to expand the lungs.

Still going to 244 meters ( Wikipedia: free-diving records) where water pressure is 24 times surface pressure is pretty incredible.

Scuba diving uses high pressure air adjusted to the depth.

---

Water Towers and Plumbing

---

Water towers give us pressure in our tap water.

In static cases, the pressure-depth relation holds.

In moving cases, the Bernoulli equation which also holds in static cases too of course.

      P/rho + (1/2)v**2 + gy = constant

      At the top of the water tower :  P = P_air.

      At an open tap:  P = P_air.

      At the tap:  y << y_(water tower).

      At the top of the water tower:  v is small. 

      At at the tap:  v is large. 

      Of course, really, pumps and valves and the like 
      are used adjust things so that the tap speed isn't a drip nor a jet.