CHAPTER 13

FLUIDS

• Density ! Bulk modulus ! Compressibility

• Pressure in a fluid ! Hydraulic lift ! Hydrostatic paradox

• Measurement of pressure ! Manometers and barometers

• Buoyancy and Archimedes Principle ! Upthrust ! Apparent weight

• Fluids in motion ! Continuity ! Bernoulli’s equation

To begin with ... some important definitions ...

DENSITY:

MassVolume

, i.e., ρ =

mV

Dimension: ⇒

[M][L]3

Units: ⇒ kg ⋅m−3

PRESSURE: ForceArea

, i.e., P =

FA

Dimension: ⇒

[M][L][T]2

Units: ⇒ N ⋅m−2 ⇒ Pascals (Pa)

FLUIDS

Liquids

Gases

Question 13.1: What, approximately, is the mass of air

in this room if the density of air is 1.29 kg ⋅m−3 ?

Mass = Volume × Density

Volume of room ≈ 15 m ×10 m × 3 m

∴M ≈ 450 m3 ×1.29kg ⋅m−3 = 580 kg

(over half-a-ton!)

Question 13.2: How does the mass of a medium sized apples compare with the mass of air in a typical

refrigerator? the density of cold air is ~ 1.3 kg/m3.

The weight of a medium apple is

~ 1 N, so the mass of a medium apple is ~ 0.1 kg.

A typical refrigerator has a capacity of ~ 18 ft3.

∴18 ft3 ⇒18 × (0.305 m)3 = 0.51 m3.

But 1 m3 of air has a mass of 1.3 kg, so the mass of air in

the refrigerator is

0.51×1.3 kg = 0.66 kg,i.e., approximately the mass of 6 apples!

We don’t notice the weight of air because we are immersed in air ... you wouldn’t notice the weight of a bag of water if it was handed to you underwater would you?

definitions (continued) ...

BULK MODULUS: B =

ΔPΔV

V( )Dimension:

⇒

[M][L][T]2

Units: ⇒ N ⋅m−2

Compressibility ⇒ B−1

Gases are easily compressed (B ⇒ very small).Liquids and solids much less compressible.

V

ΔP

ΔV

ΔP

ΔP

ΔP

Atmospheric pressure: Po = 1×105Pa

∴ Force on the ceiling from floor of room above is

pressure × area ≈1×105 × (15 m ×10 m)

≈ 1.2 ×107 N.

Why doesn’t it collapse under that weight ... ?Because pressure operates equally in all directions!

When air molecules “bounce” off the walls they produce an impulse:

FΔt = Δp ⇒ pressureSince the molecules are traveling with equal speeds in all directions ... the pressure is the same !

! F Why?

Pressure at a depth in a fluid ...

Imagine a cylindrical body of the fluid with its top face at the surface of the fluid. At equilibrium there is no net force acting on the surfaces of the cylinder.

∴ Fy∑ = 0 at the lower face,

i.e., P × A = P! × A + mg.

But the mass of fluid in the cylinder is m = ρV = ρAh.

∴P = P! + ρgh .

What’s the pressure in water at a depth of 10m, say?

ρgh = 1×103 kg/m3 × 9.81 m/s2 ×10 m ≈105Pa .

P! = Pat ≈1.01×105Pa .

∴P ≈ 2Pat.

P!

P

w = mg h

Area = A

P!

Question 13.3: A balloon has a radius of 10 cm. By how much does the radius change if the balloon is pushed down to a depth of 10 m in a large tank of water? (The

bulk modulus of air is 2 ×105N ⋅m−2.)

The pressure difference is

ΔP = hρg

But B =

ΔPΔV

V( ).

∴

ΔVV

=ΔPB

=hρgB

.

For an air-filled balloon at a depth of 10m, we have

ΔVV

=10 m ×1×103 kg/m3 × 9.81 m/s2

2 ×105 N/m2 ≈ 0.5 (50%).

Assuming the balloon is spherical,

V =

43πr3 so ΔV = 4πr2 × Δr.

∴

ΔVV

=3Δr

r≈ 0.5,

so, for an initial radius r = 0.1 m,

Δr ≈

0.5r3

= 0.017m (1.7 cm).

h

Δr

rWhat’s the difference in the air pressure from the ceiling to the floor in this room ... ?

The room height is ~ 3m so the pressure difference is:

ΔP = hρg = 3 ×1.29 × 9.81 ≈ 38Pa

∴

ΔPP!

≈38

1×105 ≈ 3.8 ×10−4 (0.038%)

i.e., negligible.

Pascal’s principle ...

If additional pressure ( ΔP) is applied, it is transmitted through the whole fluid:

∴P1 = P! + h1ρg + ΔP

and

P2 = P! + h2ρg + ΔP.

Blaise Pascal (1623-1662)

h1

P1 h2

P2

P! + ΔP

Hydraulic lift:

If a force F1 is applied to the left hand piston, the

additional pressure, P1 =

F1A1

, is transmitted through the

whole fluid. Therefore, on the surface of the right hand

piston, P2 =

F2A2

= P1.

∴

F1F2

=A1A2

, i.e., F2 =

A2A1

⎛

⎝ ⎜

⎞

⎠ ⎟ F1.

Wow ... the force is amplified!! Mechanical advantage

h

F1Area A1 Area A2

F2

same pressure

Get a larger force OUT than you put IN? Too good to be true?

No, not really, because, to do work (like lift something heavy) the force F2 is applied through a distance Δx2.

But by conservation of energy

F2Δx2 = F1Δx1 ⇒ Δx1 =

A2A1

Δx2.

So, although F1 < F2, it is applied through a greater

distance Δx1 > Δx2.

Examples:• lifts• dentist’s chair• hydraulic brake systems

F1Area A1 Area A2

F2 Δx1

Δx2

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.

P! + ρgh

h

P! P! P! P!

We can use these ideas to measure pressure:

P

y1

y2

h

P!

h y2

y1

P = 0

P! P!

Manometer Barometer

P + ρgy1 = Po +ρgy2 0 + ρgy2 = Po +ρgy1

This is absolute pressure

i.e., P − P! = ρgh . i.e., P! = ρgh .This is the Gauge pressure Atmospheric pressure

Barometer

P! = ρgh , i.e., h =

P!ρg

P! ≈ 1.01×105Pa

Using water:

ρ = 1×103kg ⋅m−3

∴h =

1.01×105

1×103 × 9.81~ 10m.

Using mercury:

ρ = 13.6 ×103kg ⋅m−3

∴h =

1.01×105

13.6 ×103 × 9.81~ 0.76m

‘Standard pressure’ is 760mm of Hg.

h

P = 0

P! P!

DISCUSSION PROBLEM 13.1:

The drawing shows two pumps, #1 and #2 to be used for pumping water from a very deep well (~30 m deep) to ground level. Pump #1 is submerged in the water at the bottom of the well; the other pump, #2, is located at ground

level. Which pump, if either, can be used to pump water to ground level?

A: Both pumps #1 and #2.B: Pump #1.C: Pump #2.D: Neither pump #1 nor pump #2.

#1

#2

Buoyancy and the concept of upthrust

Archimedes Principle : when an object is partially or wholly immersed in a fluid, the fluid exerts an upward force ... upthrust ... (or buoyant force, B) on the object, which is equal to the weight of fluid displaced.

Submerged: w > B

Weight of object w = mg = ρsVs g

Weight of liquid displaced = ρLVs g

If ρs > ρL

the object will sink.

w = mg

B

Vs

If the object is floating then ... w = B.

Weight of object is w = mg = ρsVs g. If VL is the

volume submerged, then the weight of liquid displaced is

ρLVL g. But according to Archimedes principle, this is

equal to the upthrust (B).

∴ρsVs g = ρLVL g ⇒ VL =

ρsρL

Vs.

Example: what volume of an iceberg is submerged?

ρs = 0.92 ×103kg ⋅m−3.

ρL = 1.03×103kg ⋅m−3.

∴

VLVs

=ρsρL

=0.92 ×103

1.03×103 = 0.89,

i.e., 89% of an iceberg is submerged!What ... you don’t believe me ...

w = mg

B

Vs VL

Question 13.4: On Earth, an ice cube floats in a glass of water with 90% of its volume below the level of the water. If we poured ourselves a glass of water on the Moon, where the acceleration due to gravity is about

1

6th

of that on Earth, and dropped in an ice cube, how

much of the ice cube would be below the level of the water?

When an ice cube floats, the weight of the ice cube ( = ρsVsg)

equals the weight of the water displaced ( = ρLVLg), which is

proportional to VL, the volume of

the ice cube below the surface.

∴ρLVLg = ρsVsg,

i.e., VL =

ρsρL

Vs,

which is independent of g. So, the volume submerged would remain the same!

Since both the weight of an object, which is floating, and the weight of a fluid it displaces are proportional to the local value of g, the submerged volume does not depend on g.

Question 13.5: A piece of copper of mass 0.50 kg is suspended from a spring scale. If it is fully submerged in water, what is the reading (in N) on the spring scale?

(The density of copper is 9.0 ×103kg ⋅m−3.)

Identify the forces acting on the block.At equilibrium Fy∑ = T + B + (−w) = 0.

∴T = w − B = ρsVsg −ρLVsg = ρsVsg 1−

ρLρs

⎛

⎝ ⎜

⎞

⎠ ⎟ .

True weight = mg . Upthrust

But

ρLρs

=1×103 kg/m3

9.0 ×103 kg/m3 = 0.111.

∴T = 0.5 × 9.81× (1− 0.111) = 4.36 N (0.444kg).In air instead of water ...

ρairρs

=1.29 kg/m3

9.0 ×103 kg/m3 = 0.143×10−4.

So, the mass of the block would be 0.9998mg less than in vacuum (0.50 kg).

T

w = mg = Vsρsg B

T (apparent weight)

Question 13.6: A beaker containing water is placed on a balance; its combined mass reading on the balance is 1.200 kg. In (a) below, a copper block is hanging freely from a spring scale, which has a mass reading of 0.20 kg. If the copper block is totally immersed in the water, as shown in (b), what are the readings on the balance and the spring scales?

(The density of copper is 9.0 ×103kg ⋅m−3.)

1.200kg

??

0.20kg

??

(a) (b)

(a) Initially, the spring scale registers the weight of the copper block and the balance registers the weight of (water + beaker). (b) When the block is lowered into the water, the water exerts an upward buoyant force (the upthrust) on the copper block. Then

T2 + B = mg, i.e., T2 = mg −B,

where B is the upthrust, which is equal to the weight of water displaced, i.e., B = ρwVsg. The submerged

volume is

Vs =

msρs

=0.20 kg

9.0 ×103 kg/m3 = 2.22 ×10−5m3

1.200kg

0.20kg

(a)(a)

T1

mg

??

?? (b)

mg

T2 B

∴T2 = m −ρwVs( )g

= 0.20 kg −1×103 kg/m3 × 2.22 ×10−5m3( )g = 0.178g,

i.e., the reading on the spring scale is 0.178 kg.

The upthrust B is the force on the water on the block; by Newton’s 3rd law the block must exert an equal and opposite force on the water. Consequently, the reading on the balance will increase by 0.178 kg, i.e., it will read 1.378 kg when the bock is suberged.

So, when you dip a teabag into your cup, the weight of the teabag is reduced, but the weight of the cup (and contents) is increased!

Image from Paul Hewitt’s

Conceptual Physics website:

http://www.arborsci.com/ConceptualPhysics/

Question 13.7: A barge, loaded with steel cannisters, is floating in a lock. If the cargo is thrown into the water, what happens to the level of water at the side of the lock? Does it rise, stay the same, or fall?

With cargo on the barge, as in (a), the weight of water displaced is equal to the weight of the barge plus the cargo, i.e.,

ww (a) = wb + ws = wb + NVsρsg,

where wb is the weight of the barge, N is the number of

blocks, Vs is the volume of each block and ρs their

density. When the blocks are thrown into the water, as in (b), the weight of water displaced is equal to the weight of the barge plus the weight of the water displaced by the blocks, i.e.,

ww (b) = wb + NVsρwg.

Since ρw < ρs, ww (b) < ww (a), i.e., less water is

displaced in (b) than in (a). Since the volume of the lock is the same, the depth of water is less in (b) than in (a), i.e., the water level falls.

(a) (b)

Question 13.8: A block of balsa wood with a rock tied to it floats in water. When the rock is on top, exactly one-half of the wood is submerged below the water line. If the block is turned over so that the rock is now underneath and submerged, what can you say about the amount of the block below the water line? Is it less than one-half of the block, one-half of the block, more than one-half of the block.

By Archimedes principle, the rock plus wood displaces its combined weight of water whether the rock is on top or underneath. So, the weight and volume of water displaced is the same in both cases. If the volume of the block is Vb and the volume of the rock is Vr, when the

rock is on top the volume of water displaced is 0.5Vb. If

the fraction of the wood submerged when the rock is underneath is xVb , then, since the volume of water

displaced with the rock underneath is the same as with the rock on top,

0.5Vb = xVb + Vr,

i.e., x < 0.5, so less than one-half of the wood is submerged.

Fluids in motion; mass continuity

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,

i.e., V1 = V2.

Then in time Δt , we have

A1v1Δt = A2v2Δt ,

i.e., A1v1 = A2v2 ⇒ constant.

This is called the continuity equation for an incompressible liquid. The conserved quantity ...

Area × velocity

... is called the volume flow rate

dVdt

(m3 ⋅ s−1)⎛ ⎝ ⎜

⎞ ⎠ ⎟ .

v1Δt v2Δt

Area A1 Area A2

v1 v2

If the density changes (from ρ1 ⇒ρ2) then, since mass is

conserved we have ...

m1 = m2

i.e., A1v1Δt( )ρ1 = A2v2Δt( )ρ2

∴A1v1ρ1 = A2v2ρ2

This is the mass continuity equation.

v1Δt v2Δt

Area A1 Area A2

v1 v2

The continuity equation in everyday life ...

[1] The garden hose:

If you squeeze the end of a garden hose the area is reduced and so the water velocity increases, since

A1v1 = A2v2

i.e., v2 =

A1A2

⎛

⎝ ⎜

⎞

⎠ ⎟ v1.

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 or not!

[2] Lanes at highway tolls:

~ See useful notes on web-site ~

v2

v1 A1

A2

Bernoulli’s equation

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 +

12ρv1

2 = P2 + ρgy2 +12ρv2

2,

i.e., P + ρgy +

12ρv2 = constant.

Note, if fluid is at rest v1 = v2 = 0.

∴P2 − P1 = ρg(y2 − y1)⇒ ΔP = ρgΔy,

a result we obtained earlier.

“potential energy” “kinetic energy”

Area A1

Area A2

v1

v2

y2

y1

h y2

y1

x

Question 13.9: A large tank of water has an outlet a distance h below the surface of the water. (a) What is the speed of the water as it flows out of the hole? (b) What is the distance x reached by the water flowing out 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 +

12ρv1

2 = P2 + ρgy2 +12ρv2

2.

But, P2 = P1 = P! since both the hole and the top of the tank are at atmospheric pressure. Then

∴

12ρv1

2 = ρg(y2 − y1) +12ρv2

2 = ρgh +12ρv2

2.

However, if A2 >> A1, then v1 >> v2; in fact, we were told that v2 = 0. Then

12ρv1

2 = ρgh , i.e., v1 = 2gh .

(Does this seem familiar?)

h y2

y1

x

1

2 A2

A1

(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,

y1

2 = vyit −12

gt2,

but vyi = 0, as the water emerges horizontally.

∴ t =

2y1g

.

The “range” is x = vxit , but vxi = v1, i.e., in projectile

motion it remains constant.

∴x = v1

2y1g

= 2 hy1 .

Also, substituting for h we find x = 2 y1y2 − y12( ) .

h y2

y1

x

1

2 A2

A1 (c) To find the maximum value of x, we take

x = 2 y1y2 − y12( )

and set

dxdy1

= 0. (Note: we were told that the water level

remains constant, so y2 is constant.)

∴

dxdy1

=12(2) y1y2 − y1

2( )−12(y2 − 2y1) = 0.

This equation is satisfied if (y2 − 2y1) = 0,

i.e., when y1 =

12

y2.

Thus, the hole should be halfway between the bottom of the tank and the surface of the water.

Question 13.10: 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.

h d

Using Bernoulli’s equation we have

P1 + ρgy1 +

12ρv1

2 = P2 + ρgy2 +12ρv2

2.

But P1 = P2 = P!. Also, if the initial speed of an element

of water is v1, then

v22 = v1

2 = 2(−g)h.

But v2 = 0.

∴v1 = 2gh .

Substituting in Bernoulli’s equation we obtain

12ρv1

2 = ρg(y2 − y1), i.e., 12ρ(2gh) = ρgd ,

since (y2 − y1) = d, then

d = h.

h d 1

2