Download The four phses of matter are solid, liquid, gas, and plasma

Chapter 10
LECTURE NOTES
FLUIDS
The four phases of matter are solid, liquid, gas, and plasma. Liquids and gasses are known as
fluids.
Define density  to be the mass per unit volume of a substance measured in units of kg/m3.
Thus  = m/V. The density of water at 4C is 1g/cm3 = 1000 kg/m3.
Define specific gravity SG as the ratio of the density of a substance to the density of water. This
quantity is dimensionless and is numerically equal to the density of the substance g/cm3. Thus
SG =  substance/  water.
Define P as the perpendicular component of the force exerted by a fluid per unit area. The units
of pressure are pascals (Pa) where 1 Pa = 1 N/m2. Pressures in the British system are often given
in lb/ft2 or lb/in2 (psi). Note 1psi = 6900Pa. We write P = F/A.
For fluids at rest, we observe that at any point in a fluid the pressure is the same in all directions.
The force due to fluid pressure acts perpendicularly to any area in contact with it.
The pressure at depth h in a fluid of constant density is P =  gh. This is due to the weight of the
overlying fluid. The pressure is the same at all points of the same depth. Also the change in
pressure is given by  P =  g  h for changes in depth  h. Both formulas assume constant
density which is true for liquids (which are nearly incompressible) but not gasses.
Note: For every 1m difference in height, the change in pressure (for water) is  P = (1000kg/m3)
(9.80 m/s2) 1m = 9,800 Pa.
Air Pressure
Air is a fluid and the weight of the atmosphere above us exerts air pressure PA = 101,300 Pa =
14.7 lb/in2 = 1 atm. Atmospheric pressure is also measured in bars where 1 bar = 100,000 Pa.
When we measure pressure with a gauge we really are measuring the pressure above or over
atmospheric pressure. The absolute or total pressure P is the sum of atmospheric pressure P A and
the gauge pressure PG. If a tire gauge says your tires are inflated to 29.4 psi = 202.6 kPa, the
absolute pressure P is PA (14.7psi) + PG (29.4psi) = 44.1psi (or 3 atm).
A barometer can demonstrate the pressure of air. Air pressure will push a fluid like water up a
tube if there is less than atmospheric pressure at the top of the tube (straws work like this).
Chapter 10
LECTURE NOTES
P=0
If there is a vacuum at the top of the tube, then
the pressure of the air balances the pressure of
the water column when PA =  fluid gh . For
water, h = 10.3m or 34 ft.
PA
How can trees taller than this draw up water to
their leaves?
Pascal's Principle
Pressure applied to a confined fluid increases the pressure throughout by the same amount. This
is how a hydraulic lift operates. A small force pushes a piston into a small cylinder; this produces
pressure P in the fluid which is transmitted to the lift where a larger area cylinder lifts a great
weight.
A
F
Fin
 out or Fout   out
A in A out
 A in
where MA is the mechanical advantage of this machine.
Since by Pascal's Principal Pin = Pout we have

 Fin = MA • Fin

Measuring Pressure
A manometer is a U-shaped tube filled with a fluid like mercury. One end is open to the air -the other connects to the source whose pressure is being measured. The difference in fluid levels
in the two arms of the U is a measure of the gauge pressure. We write P = PA +  gh; if the h is
lower on the side of the tube open to air, it is negative; if higher, positive.
In many cases, we simply report the height of the fluid h in mm of mercury. 1mm-Hg = 133Pa
using  Hg = 13.6 g/cm3. The previously mentioned barometer was invented by Toricelli. The
torr = 1mm-Hg is named in his honor. A column of mercury 76cm = 760mm high is supported
by 1 atm pressure so 1 atm = 760 torr.
Buoyancy
Fluids exerts an upward buoyant force . In water, this force almost balances the weight of fish or
swimmers. The buoyant force is due to pressure increasing with depth; the greater pressure on
the lower surface of an object results in a net upward force.
In a fluid of density  , an object with upper surface at depth h1 and lower surface at depth h2
feels a buoyant force equal to the weight of the fluid it displaces. Pupper surface =  g h1 and the
Chapter 10
LECTURE NOTES
force this exerts is F1 =  g h2 A. Thus Fbuoyant = F2 - F1 = A g (h2 - h1) =  • g • volume = mass
of fluid displaced by volume of object • g.
This is known as Archimedes' principle: The buoyant force on a body immensed in fluid is equal
to the weight of the fluid displaced by the body.
This principle also applies to floating objects. A solid object floats if  object <  fluid. The
buoyant force upward balances the gravitational force downward of  fluid Vdisplaced g =
Vdisplaced  object

= SG of object if the fluid is water.
 objectVobject g which gives
Vobject
 fluid
Look up the density of ice; what part of an ice cube will be under water when dropped in a glass
full of water?
Note: A hydrometer indicates SG of a liquid by how far it sinks.
Note: In a helium filled balloon, the buoyant force equals the weight of displaced air. The
balloon floats if this is greater than the weight of the balloon, helium, and payload.
Fluids in Motion
We study streamline or laminar flow (smooth flow at low speed in a less viscous fluid) versus
turbulent flow (characterized by eddies, occurring at higher speeds or more viscous fluids).
The equation of continuity states that the mass flowing into a system must equal the mass
flowing out (assuming no sources or sink) per unit time.
This reduces to the equation A1 v1 = A2 v2 where A1 and A2 are the cross sectional areas through
which the fluid enters and leaves at speeds v1 and v2.
Note: When the area decreases, the velocity increases. Putting your finger partly over the end of
a garden hose decreases the area and increases the speed of the water.
Bernoulli's Equation
Daniel Bernoulli, member of a celebrated Swiss family, discovered that a fluid's velocity is high
when its pressure is low. If a fluid at one point has pressure P1 velocity v1, and is at height h1, at
1
1
another point P2, v2, y2, then P1 + v 12  gy 1  P2  v 22  gy
2
2
Example
A hole is punched a distance h beneath the surface of a can, open to the air above and filled with
water. How fast will the water exit the hole?
Chapter 10
LECTURE NOTES
Air pressure is above the open top of the can and the hole. The velocity v1 is the fluid in the can
1
is low. Pair + 0 + gy top  Pair  v 22  gy hole and v 2  2gy top y hole   2gh , just as if the
2
water fell through the height h.
Question: If holes are punched at different heights beneath the top of the can, which jet of water
leaking out of the can will travel farthest?