Fluids
• Fluids flow – conform to shape of container
– liquids OR gas
Density
• Defined as:
M
r
V
• Or:
M  rV
• Specific gravity = r /r water at 4°C
– r water at 4°C = 1.0 g/cm3 = 1000 kg/m3
– Specific gravity has no units
• density of air (1 atm, 0ºC): 0.00129 g/cm3 = 1.29 kg/m3
Example 1
Find the volume of 20,000 N weigh water.
Pressure as Function of Depth
Static Fluid
(Fluid at rest)
P2A = P1A + mg
P2 = P1 + ρgh
For a given fluid, all
points at same depth
below surface are at
the same pressure.
Example2: Deepest Fish
(a) Deepest fish ever sighted was on the floor of the Mariana's trench
at 11,500m depth. What is the pressure there?
(b) What would be the inward force on a 20cm diameter circular of a
submarine wall at that depth?
Clicker!
Which liquid container will have the highest pressure (at the black dot locations)?
(1)A,
(4) D,
(2) B,
(3) C,
(5) all the same.
Clicker!
Consider the four points in the lake. The points are all at the same depth below the
surface, but the depth of the bottom of the lake underneath the points varies. At
which point is the pressure the greatest?
(1)A,
(4) D,
(2) B,
(3) C,
(5) all the same.
Pressure only depends on depth of point where pressure is
being measured.
Clicker!
Two beakers are filled with fluid. One is filled with water. The other is filled with a
mixture of oil (specific gravity 0.8) and water to the same level. Which beaker has
the greatest pressure at the bottom of the beaker.
(1) The Water beaker
(2) The Oil and Water beaker
(3) Both the same
Pressure at depth where oil and water meet is lower in the oil and
water mixture (oil less dense).
Increase in pressure from this level is same, since water in both
columns.
Clicker!
The following bulbs are connected by a fluid
filled tube. The fluid is blue. How do the
pressures in the two bulbs, P1 and P2, compare?
(1) P1 > P2
P1
P2
A
(2) P1 = P2
(3) P1 < P2
Pressure at level B greater
than at level A in fluid
B
Measuring Pressure Using a Column of Fluid
Height of column and density of fluid
determine pressure difference.
Manometer
Po
P
h
• Absolute Pressure, P
• Gauge Pressure = P – Po
Barometer
P=0
• Po = Atmospheric Pressure
= 1.01 x 105 Pa
h
P  Po  r gh
Pressure sometimes given in inches of Mercury (Hg)
30.2”
Po
Blood Pressure
120 mm of Mercury/80mm of Mercury
Gauge pressure (relative to atmosphere)
Clicker!
As you travel up a mountain, your ears pop. The figure below is a pitiful attempt to
show the ear drum (in red). As you travel up a mountain, what will the ear drum do
between 'pops'?
(1) Bow towards the middle ear
(2) Hold its shape
(3) Bow towards the outside
Pressure outside decreases.
Example 3
A manometer is made of a solution that has a specific gravity of 4.15 and is used
to measure the pressure of gas in a container as shown. The reading in
manometer shows 650 mm.
(a) What is the gauge pressure?
(b) What is the absolute pressure?
Clicker!
Hydraulic Car Lift
The area of the left piston is 10mm2. The area of the right piston is 10,000mm2.
What force must be exerted on the left piston to keep the 10,000-N car on the right
at the same height?
(1) 10N
(2) 100N
A1 = 10
mm2
A2 = 10,000
mm2
(3) 10,000N
(4) 106N
(5) 108N
P = F1/A1 = F2/A2
F1 = (A1/A2) F2
F1 = (10 mm2 / 10,000 mm2) * 10,000N
Clicker!
5N
When an object is submerged in a fluid, there is a buoyant force.
An object of 7N weigh submerged in a fluid is hanged by a rope,
which has a tension of 5 N. What is the buoyant force?
(1) 1 N downward
(4) 2 N upward
(2) 1 N upward
(5) 5 N downward
(7) 12 N downward
FB
(3) 2 N downward
(6) 5 N upward
(8) 12 N upward
7N
Just another force
Σ FY = 0
5N – 7N + FB = 0
FB = 7N – 5N
Floating versus Completely Submerged
FB
• Floating:
FB = W
VDISPLACED< VOBJECT
ρOBJ < ρFLUID
FB = ρFLUID VDISPLACED g
W = ρOBJ VOBJECT g
• Completely Submerged
VDISPLACED=VOBJECT
ρOBJ > ρFLUID
FB = ρ g VOBJECT
FG
Clicker!
Object B is larger than A. If we put both objects into water, which of the following
statement is true?
(1) Both A, and B will immerse into water.
A
2000 kg/m^3
B
500 kg/m^3
(2) A will completely immerse, B will float.
(3) B will completely immerse, A will float.
(4) Both will float.
(5) None of the above is correct.
Density of A > density of water
Density of B < density of water
Example 4: If ice has a density of 920 kg/m3, what percentage of an
iceberg sticks out above the waterline?
If the mass of the ice is 20 kg, what is the volume of the ice that is sticking out of
water?
FBUOYANT = Weight
ρw g VDISPLACED = ρICE g VICE
ρw g VSUBMERGED = ρICE g VICE
VSUB / VOBJ = ρICE / ρw
VSUB / VOBJ = 0.92
So fraction not submerged is 1-0.92 or 0.08, 8%
Example 5
An object of mass 1500 kg with a density of 2000 kg/m3 is put into water.
(a) Will this object float?
(b) What will be the weight of the object in water?
Clicker!
Fill two identical glasses with water up to the rim. Now place ice in one (water will
overflow into the sink). Now consider the two glasses, one with water, one with
water and ice. Which weighs more?
(1) The glass without ice cubes
(2) The glass with ice cubes
(3) The two weigh the same
Ice less dense, but occupies more volume.
Each cube displaces the weight of the cube in water.
Think of the following: two beakers filled to the edge with
water. Add an ice cube. The weight of the fluid lost over the
edge equals the weight of the ice cube.
Clicker!
An incompressible fluid is flowing through a pipe. At which point is the fluid
traveling the fastest?
(6) All points have the same speed
Smallest cross-sectional area, highest speed
Bernoulli's Equation – Conservation of Energy
One way to look at it:
WNC =
F*d
Divide by volume: ΔPressure
ΔKE +
ΔPE
(1/2)mv2
mgh
(1/2)ρv2
ρgh
P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv22 + ρgy2
If height not changing much, higher speed leads to lower
pressure
Clicker!
An incompressible fluid flows through a pipe. Compare the speed of the fluid at
points 1 and 2.
(1) Greater at 1
(2) Greater at 2
(3) Both the same
(4) Not enough information
Area is smaller at 2, therefore fluid must travel faster in order
to conserve the volume flow rate.
Clicker!
An incompressible fluid flows through a pipe. Compare the pressure at points 1 and
2.
(1) Greater at 1
(2) Greater at 2
(3) Both the same
(4) Not enough information
P + (1/2) ρv2 + ρgh = constant
(1/2) ρv2 is larger (higher v due to smaller A)
ρgh is larger (higher)
Therefore pressure must decrease at point 2
P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv22 + ρgy2
Example 6
Calculate the lift force of an airplane wing with a surface area of 12.0m2. Assume
the air above the wing is traveling at 70.0 m/s and the air below the wing is
traveling at 60.0m/s. Assume the density of the air to be constant at 1.29 kg/m3.
FLIFT = FUP - FDOWN
= PBELOW A – PABOVE A
= (ΔP) A
From Bernoulli, find (ΔP) assuming Δy negligible.
PA + (1/2)ρvA2 + ρgyA = PB + (1/2) ρvB2 + ρgyB
(ΔP) = PB – PA = ((1/2)ρvA2 – (1/2)ρvB2) + (ρgyA – ρgyB)
(assume ΔPE negligible – if yA-yB = 10cm, only 1.26 Pa)
ΔP = (1/2) (1.29)(702 – 602 ) = 838.5 Pa
FLIFT = (838.5Pa)(12m2) = 1.01 x104 N
The Water Tank
Water is leaving a tank at a speed of 3.0 m/s. The tank
is open to air on the top. What is the height of the
water level above the spigot?
Assume the area of the tank is much larger than the
area of the spigot.
P1 + (1/2)ρv12 + ρgy1 = P2 + (1/2)ρv22 + ρgy2
P0 + (1/2)ρv12 + 0 = P0 + (1/2)ρv22 + ρgh
Since A2 >> A1, v1 >> v2
(1/2)ρv12 = (essentially 0) + ρgh
h = 0.459m