Monday, November 9, 1998 Chapter 9: Archimedes’ principle compressibility bulk modulus fluids & Bernoulli’s equation

Archimedes had this whole process figured out some 2000 years ago! He said, A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid. So, the cork naturally float with just the right portion of its volume under the water’s surface so that the buoyant force upward from the water equals the gravitational force.

If we have a cork with density of 0.8 g/cc, what fraction of its volume will be below the surface in a pool of water when it reaches equilibrium? Let’s look at the free body diagram for our cork. FgFg FBFB The gravitational force:

FgFg FBFB The buoyant force is given by the weight of the displaced water. If we have a cork with density of 0.8 g/cc, what fraction of its volume will be below the surface in a pool of water when it reaches equilibrium? Now, set this equal to the gravitational force...

If we have a cork with density of 0.8 g/cc, what fraction of its volume will be below the surface in a pool of water when it reaches equilibrium? FgFg FBFB Remarkable!

What else happens to our cork while it’s completely submerged? After all, there’s a pressure on all its surfaces, acting inward, from the water which surrounds it... P P P P The water exerts a pressure force that tries to compress the cork (i.e., reduce the volume of the cork).

P P P P We call this type of stress a volume stress and define it simply as This type of stress is caused by a change in pressure. The response of an object to an increase in pressure around its side is to… DECREASE ITS VOLUME!

P P P P You can now probably guess how we’ll define the volume strain The way an object responds to a volume stress is again simply a property of the material from which the object is made. We call this property the bulk modulus:

P P P P Often this characteristic of materials is tabulated as the inverse of the bulk modulus, and known as the compressibility:

Seawater has a bulk modulus of 23 X N/m 2. Find the change in the volume which 1024 kg occupies at a depth where the pressure is 800 atm. Use a surface density of seawater of 1024 kg/m 3.

We’re now going to spend some time examining the behavior of liquids as they flow or move through pipes, the atmosphere, the ocean,... Let’s trace out the motion of a given piece or parcel of water as if flows through a channel. These lines, which tell us where a parcel has been and in which direction it is going, are called trajectories.

If the flow is in a condition known as steady state (not varying) then the trajectories are the same as the streamlines. The streamlines tell us the instantaneous direction of motion of a parcel in a flow, whereas the trajectories trace out exactly where the parcel has been. Real flows often result in turbulence -- a condition in which the flow becomes irregular.

turbulent region Real flows are also often viscous. Viscosity describes the internal “friction” of a fluid, or how well one layer of fluid slips past another.

To simplify our problems, we’re going to study the behavior of a class of fluids known as “ideal” fluids. These fluids have the following set of properties: 1) The fluid is nonviscous (no internal friction) 2) The fluid is incompressible (constant density) 3) The fluid motion is steady (velocity, density and pressuer at each point remain constant) 4) The fluid moves without turbulence.

This is really just a conservation of mass argument. It says that if I put in 5 g of water each second at the left end of my hose, then under steady-state flow conditions, I must get out 5 g of water each second at the right end of the hose. 5 g/s

For ideal fluids in steady-state (unchanging) flows, this must be true regardless of the shape of the hose. For instance, I could have a hose that’s narrower at the left end where the fluid enters the hose than it is at the right end where fluid leaves. 5 g/s Nevertheless, the mass entering at the left each second must equal the mass exiting at the right.

v1v1 v2v2 A1A1 A2A2 The mass entering at the left side is given by And similarly, the mass leaving at right is

v1v1 v2v2 A1A1 A2A2 These two quantities must be equal, leaving us with the relationship

An ideal fluid flows through a pipe of cross-sectional area A. Suddenly, the pipe narrows to half it’s original width. What happens to the speed of the flow in the pipe? So, the cross-sectional area goes down by a factor of 4. That means that the velocity must go up by a factor of 4!

In examining flows through pipes in the Earth’s gravitational field, Bernoulli found a relationship between the pressure in the fluid, the speed of the fluid, and the height off the ground of the fluid. The sum of the pressure ( P ), the kinetic energy per unit volume (0.5  v 2 ), and the potential energy per unit volume (  gy) has the same value at all points along a streamline.

There’s a nice derivation of this in the book…so I won’t derive it here…but let’s apply this to a problem.

A large water tower is drained by a pipe of cross section A through a valve a distance 15 m below the surface of the water in the tower. If the velocity of the fluid in the bottom pipe is 16 m/s and the pressure at the surface of the water is 1 atm, what is the pressure of the fluid in the pipe at the bottom? Assume that the velocity of the fluid in the tank is approximately 0 m/s downward. A 15 m v = 16 m/s v = 0 m/s

A 15 m v = 16 m/s v = 0 m/s Let’s look at the conditions at the top of the tower: = 248 kPa This must match the conditions for the pipe at the bottom of the tower...

A 15 m v = 16 m/s v = 0 m/s P kPa = 248 kPa P = 120 kPa Let’s look at the conditions in the pipe at the bottom of the tower: