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Unformatted text preview: Liquids CHAPTER 13
o Bulk modulus
o Compressibility Gases 8
4 To begin with ... some important definitions ...
, i.e., ρ =
• Pressure in a fluid
o Hydraulic lift
o Hydrostatic paradox
• Measurement of pressure
o Manometers and barometers
• Buoyancy and Archimedes Principle
o Apparent weight
• Fluids in motion
o Bernoulli’s equation Units: ⇒ kg ⋅ m−3 PRESSURE: Force
, i.e., P =
A Dimension: ⇒ [M]
[L][ T]2 Units: ⇒ N ⋅ m−2 ⇒ Pascals (Pa) 1 definitions (continued) ... BULK MODULUS:
Dimension: ⇒ B= ( ∆P
V ) [M]
[L][ T]2 Units: ⇒ N ⋅ m−2
Compressibility ⇒ B −1 DISCUSSION PROBLEM [13.1]: ∆P What, approximately, is the mass of air in this room if ∆V the density of air ⇒ 1.29 kg ⋅ m−3 ?
∆P ∆P V ∆P Gases are easily compressed (B ⇒ very small).
Liquids and solids much less compressible. 2 Pressure at a depth in a fluid ... Atmospheric pressure: Po = 1 × 105 Pa
∴ Force on the ceiling from floor of room above ⇒ pressure × area ≈ 1 × 105 × (8 × 15) ≈ 1.2 × 107 N . Po Po h Area = A w = mg P Why doesn’t it collapse under that weight ... ?
Because pressure operates equally in all directions! Why? r
F At equilibrium ∑ Fy = 0 at the lower surface,
i.e., PA = Po A + mg .
But m = ρV = ρAh . When air molecules “bounce” off ∴ P = Po + ρgh . the walls they produce an impulse: F∆t = ∆p ⇒ pressure
Since the molecules are traveling
with equal speeds in all directions
... the pressure is the same ! What’s the pressure in water at a depth of 10m, say? ρgh = 1 × 103 × 9.81 × 10 ≈ 105 Pa .
Po = Pat ≈ 1.01 × 105 Pa .
∴ P ≈ 2Pat. 3 The pressure difference is
∆r Question 1: A balloon has a radius of 10cm. By how
much does the radius change if the balloon is pushed
down to a depth of 10m in a large tank of water? (The
bulk modulus of air is 2 × 105 N ⋅ m−2 .) ∆ P = hρg
But B =
V ( ∴ ) ∆V ∆P hρg
B For an air-filled balloon at a
depth of 10m (with B = 2 × 105 N ⋅ m−2 ), we have
∆V 10 × 1 × 10 3 × 9.81
2 × 105 (i.e., 50%). Assuming the balloon is spherical,
V = πr 3 and ∆V = 4πr 2 × ∆r .
So, for an initial radius r ⇒ 0.1m, we find
= 0.017 m.
3 4 Hydraulic lift:
What’s the difference in the air pressure from the
ceiling to the floor in this room ... ? Area A1 F1 Area A 2 The room height is ~ 3m so the pressure difference is:
∆P = hρg = 3 × 1.29 × 9.81 ≈ 38Pa
5 ≈ 3.8 × 10
Po 1 × 10 same pressure i.e., negligible. If a force F1 is applied to the left hand piston, the
additional pressure, P1 = 1 , is transmitted through the
A1 Pascal’s principle ...
Po + ∆P If additional pressure ( ∆P ) is
applied, it is transmitted h2 through the whole fluid: h1
P1 P2 F2 h ∴ P1 = Po + h 1ρg + ∆P and
P2 = Po + h 2ρg + ∆P .
Blaise Pascal (1623-1662) whole fluid. Therefore, on the surface of the right hand
piston, P2 = 2 = P1.
∴ 1 = 1 , i.e., F2 = 2 F1. A1 F2 A2 Wow ... the force is amplified!!
Mechanical advantage 5 Area A1 F1 Area A 2 ∆x 2
∆x1 Here’s something surprising ...
no matter the shape of a vessel,
the pressure depends only on the
vertical depth. It is known as the
hydrostatic paradox . F2 Get a larger force OUT than you put IN ? Too good to be
No, not really, because, to do work (like lift something
heavy) the force F2 is applied through a distance ∆x 2.
But by conservation of energy
F2 ∆x 2 = F1∆x1 ⇒ ∆x1 = 2 ∆x 2 .
So, although F1 < F2, it is applied through a greater
distance ∆x1 > ∆x 2 .
• lifts Po Po Po Po h Po + ρgh • dentist’s chair
• hydraulic brake systems 6 We can use these ideas to measure pressure :
Po P=0 Barometer P=0
h Po ≈ 1.01 × 105 Pa h y2 P Po P
Po = ρgh , i.e., h = o
ρg y2 Using water: Po ρ = 1 × 103 kg ⋅ m−3 h y1
Po y1 Manometer
P + ρgy1 = Po + ρgy 2 Po This is absolute pressure 1.01 × 105
~ 10 m.
1 × 103 × 9.81 Using mercury: Barometer
0 + ρgy 2 = Po + ρgy1 ∴h = ρ = 13.6 × 103 kg ⋅ m−3
∴h = 1.01 × 105
~ 0.76 m
13.6 × 103 × 9.81 “Standard pressure” is 760mm of Hg.
i.e., P − Po = ρgh .
This is the Gauge pressure i.e., Po = ρgh .
Atmospheric pressure 7 If the object is floating then ... w = B.
Buoyancy and the concept of upthrust B B
Vs w = mg Vs w = mg VL Weight of object is w = mg = ρs Vs g. If VL is the
volume submerged, then the weight of liquid displaced is Archimedes Principle : when an object is partially or ρL VL g . But according to Archimedes principle, this is wholly immersed in a fluid, the fluid exerts an upward equal to the upthrust (B). force ... upthrust ... (or buoyant force, B) on the object,
which is equal to the weight of fluid displaced. ρ
∴ ρs Vs g = ρL VL g ⇒ VL = s Vs .
Example: what volume of an iceberg is submerged? Submerged: w > B
Weight of object w = mg = ρs Vs g
Weight of liquid displaced = ρL Vs g
If ρs > ρL
the object will sink. ρs = 0.92 × 103 kg ⋅ m−3. ρL = 1.03 × 103 kg ⋅ m−3.
0.92 × 103
∴ L= s =
Vs ρL 1.03 × 103
i.e., 89% of an iceberg is submerged!
What ... you don’t believe me ... 8 Question 2: A 0.5kg piece of copper is suspended from a
spring scale and is fully submerged in water. What is the
reading (in N) on the spring scale.
(The density of copper is 9.0 × 103 kg ⋅ m−3.) 9 T (apparent weight)
B T Question 3: A beaker containing water is placed on a
w = mg = Vsρs g balance; its combined mass reading on the balance is
1.200kg. In (a) below, a copper block is hanging freely
from a spring scale, which has a mass reading of 0.20kg. Identify the forces acting on the block. If the copper block is totally immersed in the water, as At equilibrium ∑ Fy = T + B + ( − w ) = 0 . ρ
∴ T = w − B = ρs Vs g − ρL Vs g = ρs Vs g1 − L . ρs shown in (b), what are the readings on the balance and
the spring scales?
(The density of copper is 9.0 × 103 kg ⋅ m−3.) “True weight” = mg .
1 × 103
ρs 9.0 × 103 ∴ T = 0.5 × 9.81 × (1 − 0.111) = 4.36 N 0.20kg
?? (0.444kg). In air instead of water ...
3 = 0.143 × 10 .
9.0 × 10 1.200kg ?? (a) (b) ∴ T = mg(1 − 0.143 × 10 −4 ) = 0.9998 mg . 10 Initially, the spring scale registers the weight of the The upthrust B is the force on the water on the block ; by copper block and the
balance registers the on the balance will increase by 0.178kg, i.e., it will read beaker). When the ?? opposite force on the water. Consequently, the reading weight of (water + 0.20kg Newton’s 3rd law the block must exert an equal and 1.378kg when the bock is suberged. block is lowered into
?? (a) (b) the water, the water So, when you dip a teabag into your cup, the weight of exerts an upward the teabag is reduced, but the weight of the cup (and buoyant force (the 1.200kg contents) is increased! upthrust) on the copper block. The upthrust is equal to
the weight of water displaced, i.e., B = ρw Vs g , which
will reduce the initial reading on the spring scale T to ( T − B). Now
Vs = s =
3 = 2.22 × 10 m .
ρs 9.0 × 10 ( ) ∴ T − B = 0.20 − 1 × 103 × 2.22 × 10 −5 g
= 0.178 g ,
i.e., the reading on the spring scale will be 0.178kg. Image from Paul Hewitt’s Conceptual Physics website:
http://www.arborsci.com/ConceptualPhysics/ 11 Fluids in motion; mass continuity
Area A1 Area A 2
v1 v 2 ∆t v 1∆t We assume the fluid is incompressible, i.e., a liquid, so
there is no change in density from 1 → 2. Then, the mass
and volume must be conserved from 1 → 2 as the liquid
flows down the pipe, v2
Area A1 Area A 2
v 2 ∆t v 1∆t If the density changes (from ρ1 ⇒ ρ2) then, since mass is
i.e., V1 = V2 . Then in time ∆t , we have
A1v1∆t = A2 v 2 ∆t , i.e., A1v1 = A2 v 2 ⇒ constant.
This is called the continuity equation for an conserved we have ...
m1 = m2 i.e., ( A1v1∆t )ρ1 = ( A2 v 2 ∆t )ρ2
∴ A1v1ρ1 = A2 v 2ρ2
This is the mass continuity equation. incompressible liquid. The conserved quantity ... Area × velocity
( m3 ⋅ s−1 ) .
... is called the volume flow rate dt 12 Bernoulli’s equation Area A 2 The continuity equation in everyday life ...
 The garden hose: v2 If you squeeze the end of a
A1 Area A1 y2 garden hose the area is
v1 reduced and so the water y1 v1 velocity increases, since
A2 v2 A1v1 = A2 v 2 A i.e., v 2 = 1 v1. A2 If you don’t restrict the pipe too much, the volume flow
rate remains constant, which means you will fill a bucket
in the same time whether the end of the hose is restricted We state Bernoulli’s equation without proof. It relates
the pressure (P), elevation (y) and speed (v) of an
incompressible fluid in steady (i.e., non-turbulent) flow
down a pipe,
P1 + ρgy1 + ρv12 = P2 + ρgy 2 + ρv 22 ,
i.e., P + ρgy + ρv 2 = constant.
2 or not!
 Lanes at highway tolls:
~ See useful notes on web-site ~ “potential energy” “kinetic energy” Note, if fluid is at rest v1 = v 2 = 0 .
∴ P2 − P1 = ρg( y 2 − y1 ) ⇒ ∆P = ρg∆y ,
a result we obtained earlier. 13 A2
2 Question 4: A large tank of water has an outlet a
distance h below the surface of the water. (a) What is h
A1 the speed of the water as it flows out of the hole? (b)
What is the distance x reached by the water flowing out y1 x of the hole? (c) What value of h would cause the water
to reach a maximum value of x? You may assume that
the tank has a very large diameter so the level of the
water remains constant. (a) We start with Bernoulli’s equation
P1 + ρgy1 + ρv12 = P2 + ρgy 2 + ρv 22 .
But, P2 = P1 = Po since both the hole and the top of the
tank are at atmospheric pressure. Then
∴ ρv12 = ρg( y 2 − y1 ) + ρv 22 = ρgh + ρv 22.
x However, if A2 >> A1, then v1 >> v 2 ; in fact, we were
told that v 2 = 0. Then
ρv12 = ρgh ⇒ v1 = 2 gh .
(Does this seem familiar?) 14 A2
y2 1 (c) To find the maximum value of x, we take A1 y1 x and set (b) To find x we model the water leaving the hole as a
projectile. The time it takes for a volume element of
water to exit the hole and reach the ground is given by,
y12 = v yi t − gt 2 ,
but v yi = 0 .
∴t = x=2 2 y1
g (y1y2 − y12 ) dx
= 0. (Note that we were told that the water
dy1 level remains constant, so y 2 is constant.)
= (2) y1y 2 − y12 2 ( y 2 − 2 y1 ) = 0.
dy1 2 ( ) This equation is satisfied if ( y 2 − 2 y1 ) = 0,
i.e., when y1 = y 2.
Thus, the hole should be halfway between the bottom of
the tank and the surface of the water. The “range” is x = v xi t , but v xi = v1.
∴ x = v1
= 2 hy1 = 2
g ( (y1y2 − y12 )). 15 2
h d 1 Question 5: A large tank of water has an outlet a
distance d below the surface of the water. The outlet is
curved so that the exiting water is directed vertically
upward. Prove that the vertical height achieved by the
water, h, is equal to the depth of the outlet below the
surface of the water, d. Using Bernoulli’s equation we have
P1 + ρgy1 + ρv12 = P2 + ρgy 2 + ρv 22 .
But P1 = P2 = Po . Also, if the speed of an element of
water is v1, then
v 22 − v12 = 2( − g) h . But v 2 = 0
h ∴ v1 = 2 gh . d Substituting in Bernoulli’s equation we obtain
ρv12 = ρg( y 2 − y1 ) ⇒ ρ(2 gh ) = ρgd ,
since ( y 2 − y1 ) = d .
∴ d = h. 16 Why does a tarpaulin bulge outwards on a fast moving
1 Stationary Moving fast Take regions 1 and 2 under and above the tarpaulin,
respectively. Bernoulli’s equation tells us:
P1 + ρgy1 + ρv12 = P2 + ρgy 2 + ρv 22 ,
but y1 ≈ y 2 and v1 = 0.
∴ P1 − P2 = ρv 22 i.e., P1 > P2.
2 F2 = P2 ∆A 1 F1 = P1 ∆A ∴ F1 > F2 Pressure is reduced when a fluid is in motion!
Also explains why roofs lift off during hurricanes!
~ See brain busters for more examples ~ 17 ...
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