Transcript Lecture 10

```Liquids and Gasses
Matter that “Flows”
Density and Specific Gravity
 Density is mass per unit volume
kg
 SI unit is
m3
m

V
g
 But this is also used
cm 3
Specific Gravity
 Ratio of density to the
density of water at 4oC.
 Density of water at 4oC=1
g/cm3
 The 4oC is important since
water’s density changes
with temperature
 It has nothing to do with
gravity
 Relative density would be a
better term
Pressure
 A static fluid will exert a
force perpendicular to
any surface it is in
contact with
 Pressure is defined as
Force per unit Area
F
p
A
 Note this is scalar
 SI unit of
N/m2=pascal=pa
Atmospheric Pressure
 The big volume of air
above us exerts pressure
downwards.
 Pressure at sea level
pa  1atm  1.013x10 pa
5
 Changes depending on
elevation, temperature,
weather condition etc.
Pressure at depth
 Assume that  is
constant all throughout
the liquid.
 Fluid is incompressible
 Force is from weight of
liquid
F m g Vg Ahg
p



A
A
A
A
p  hg
Pressure at depth
 If fluid is compressible
(gasses) density changes
with height and equation
is only valid for very thin
slices.
Pascal’s Law
 Equations independent of
Area, only height matters.
 Pressure is the same at any
two points at the same
level of the fluid.
 If pressure is applied to
one end it is distributed
throughout the fluid
and to the walls of the
container
Pascal’s Law Application
 Hydraulic Lift
F1 F2
p

A1 A2
 However, volume of the
liquid remains the same
V1  V2
A1x1  A2 x2
Hydraulic Lift
F1 A1 x2


F2 A2 x1
F1x1  F2 x2
 Work done by F1 = Work
done by F2
Example
 A car lift used in a service station uses compressed air
to exert a force on a small piston with a circular cross
section of radius 5.00cm. The pressure is transmitted
by a liquid to a second piston of radius 15.0 cm. What
force must the compressed air exert to lift a car
weighing 13300N? What air pressure produces this
force?
Example
F1 F2
p

A1 A2
A1
r12
F1 
F2  2 F2
A2
r1
52
F1  2 13300
15
F1  1480N
 Note: no need to convert to
meters since units cancel
F1 F2
p

A1 A2
1480N
p
 188000Pa
2
 (0.05)
Absolute Pressure and Gauge
Pressure
 Absolute Pressure
 Total Pressure within a
vessel
 Gauge Pressure
 Excess Pressure above
atmospheric pressure
 Gauge Pressure – Ex. A
tire shows 220kPa
 Absolute Pressure of tire
is 220kPa+1atm=320kPA
 Flat tire actually has 1
atm of pressure inside
Pressure Gauges - Manometer
 Many different devices
used to measure
pressure
 Manometer – consists of
a U-shaped tube filled
with a liquid of known
density. It is open in one
end and the other end is
connected to a container
filled with gas whose
pressure to be measured.
Pressure Gauges - Manometer
 Bottom of tube has same
pressure
p  gy1  patm  gy2
p  patm  g ( y2  y1 )
p  patm  gh
 Gauge pressure is
proportional to height
difference
Pressure Gauges - Barometer
 Mercury barometer –
consists of filled tube of
mercury closed on one
end and the other end is
inverted into a dish of
mercury.
 Pressure in the closed
end of the tube can be
approximated to be zero.
Pressure Gauges - Barometer
patm  gh  p0
patm  gh
 mmHg or Torr, is
another common unit of
measure of pressure. But
it is affected by
temperature (density of
Hg changes) as well as
gravity.
Buoyancy
 Bodies immersed in
water will weigh less
 Bouyant Force is exerting
an upward force
 Assume a submerged
object with uniform
cross section area A
Bouyancy
B  FB  FT
B  PB A  PT A  A( PB  PT )
B  A( ghB  ghT )
B  Ag (h)
B  Vg
B  mg
Bouyancy
 Archimedes’ Principle-
When a body is
immersed in a fluid, the
fluid exerts an upward
force on the body equal
to the weight of the fluid
displaced by the body.
Example
 Archimedes was
determine if the crown of
the king was pure gold. He
solved this by first
weighing the crown in air,
and then weighing the
crown when submerged in
water. Suppose the scale
6.84N in water, what is the
Example
F  0
0  T1  m g
T1  m g
1   wVg
1
1
V

 w g 1000(9.8)
T1  T2  B   wVg
V  1.02x10 4 m3
mc
7.84
c 

4
V
(1.02x10 )9.8
7.84  6.84   wVg
 c  7840kg m
1   wVg
 gold  19300kg m
0  T2  B  m g
T2  m g  B
3
3
Bouyancy
 An object will float if
B  Wobject
lVobject g  mobject g
lVobject g  objectVobject g
l  object
Fluid Flow – Fluid Dynamics
 Complicated, but can be
simplified with certain
assumptions
 Ideal Fluid
 Incompresible- same
density throughout
 Not viscous - has no
internal friction
 Minimal turbulence

Fluid Flow
 Assume the flow is
m
flow 
t
V Al
flow 

t
t
flow  Av
 Flow rate is constant
1 A1v1  2 A2v2
Fluid Flow
1 A1v1  2 A2v2
 Since incompresible
A1v1  A2v2
 Continuity Equation
Example
 A water hose of 2.50cm diameter is used by a gardener
to fill a 30.0L bucket. The gardener notes that it takes
1.00min to fill the bucket. A nozzle with an opening of
cross-sectional area 0.500cm2 is attached and held so
water is projected horizontally from a point 1.00m
above ground. Over what horizontal distance can the
water be projected?
Example
Flow  A1v1
3
30L 0.03m
3
m

 0.0005 s
min
60s
3
m
0.0005 s  Av
m3
0.0005 s
m
v

10
.
0
s
2
0.00005m
Example
v  10.0 m s
1 2
y   gt
2
t   2 / g   2 /(9.8)(1)
t  0.45s
R  vxt  10(0.45)
R  4.5m
Bernoulli’s Equation
 Pressure in a fluid
depends on velocity and
height
 Assuming ideal fluid,
fluid at certain cross
section will do work on
other parts of the fluid.
W1  F1d1  p1 A1d1
W2   F2 d 2   p2 A2 d 2
Bernoulli’s Equation
Wg   m gh   m g( y2  y1 )
W  W1  W2  Wg
W  p1 A1d1  p2 A2 d 2  m g( y2  y1 )
K  p1 A1d1  p2 A2 d 2  m g( y2  y1 )
1 2
1 2
m v1  p1 A1d1  m gy1  m v2  p2 A2 d 2  m gy2
2
2
1 2
1 2
v1  p1  gy1  v2  p2  gy2
2
2
Example
 Water enters a house
through a pipe with an
inside diameter of 2.0 cm
at an absolute pressure of
4.0x105 Pa. A 1.0cm
second floor bathroom 5.0
m above. When the flow
speed at the inlet pipe is 1.5
m/s, find the flow speed,
pressure and volume flow
in the bathroom.
Example
A1v1  A2 v2
A1
v2 
v1
A2
 (1) 2
m
v2 
1
.
5

6
s
2
 (0.5)
1 2
1 2
v1  p1  gy1  v2  p2  gy2
2
2
1
1
2
5
(1000)(1.5)  4.0 x10  0  (1000)6 2  p2  (1000)(9.8)(5)
2
2
p2  3.34x105 Pa
Example
Flow  A1v1
Flow   (0.5 x10 2 ) 2 6
 4 m3
Flow  4.7 x10
s
Problems
 Giancoli 10-20
 In working out Pascal’s principle, Pascal showed
dramatically how force can be multiplied with fluid
pressure. He placed a long thin tube of radius
r=0.30cm, vertically into a wine barrel of radius
R=21cm. He found that when the barrel was filled with
water and the tube filled to a height of 12m, the barrel
burst. Calculate (a) the mass of water in the tube? (b)
the net force exerted on the barrel lid just before
rupture.
Serway 14-20
 A U tube of uniform cross-secitonal area, open to the
athmosphere, is partially filled with mercury. Water is
then poured into both arms. If the equilibrium
configuration fo the tube is as shown with h2=1.00cm,
find h1.
Young and Friedman 14.31
 A cubical block of wood, 10 cm on a side, floats in the
interface of oil and water with its lower surface, 1.50cm
below the interface. The density of oil is 790kg/m3. (a)
what is the gauge pressure at the upper face of the
block? (b) What is the gauge pressure at the lower face
of the block? (c) What are the mass and density of the
block?
Young and Friedman 14.46
 A golf course sprinkler system discharges water from a
horizontal pipe at the rate of 7200cm3/s. At one point
in the pipe, where the radius is 4.00cm, the absolute
pressure is 2.40x105 Pa. At a second point in the pipe,
the water passes through a connection where the
radius is 2.00cm. What is the absolute pressure as it
flows through the constriction?
```