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: How does the mass of air inside a

typical refrigerator compare with the mass of a medium

size apple? The density of cold air is ~ 1.3 kg ⋅m−3.

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

In the previous chapter we defined the Young’s modulus

and the shear modulus. He is another modulus. 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?

Atmospheric pressure P! = 1×105 Pa.

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!

Pressure at a depth in a fluid

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

ρgh = (1×103 kg ⋅m−3)(9.81 m/s2)(10 m) ≈105 Pa.P! = Pat ≈1.05×105 Pa.

∴P ≈ 2Pat .

Question 13.2: 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. Assume the

balloon remains spherical. (The bulk modulus of air is

) 2 ×105 N ⋅m−2.

The pressure difference is

But

For an air-filled balloon at a depth of

10 m, we have

Since the balloon is spherical,

So, for an initial radius

ΔP = hρg.

B = ΔPΔV

V( ) .

∴ΔV

V= ΔP

B= hρg

B.

∴ΔV

V= (10 m)(1×103 kg ⋅m−3)(9.81 m/s2)

2 ×105 N ⋅m−2≈ 0.5 (50%).

V = 43πr3, i.e., ΔV = 4πr2Δr.

∴ΔVV

= 3Δrr≈ 0.5,

r = 0.1 m,

Δr ≈ 0.5r

3= 0.017 m (1.7 cm).

What’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

What’s the pressure difference from the ceiling to the floor

in a typical room?

Assume a room height of 3 m, so the pressure difference is

i.e., negligible.

ΔP = hρg = (3 m)(1.29 kg ⋅m−3)(9.81 m/s2) ≈ 38 Pa.

∴ΔPP

≈ 38 Pa

1×105 Pa≈ 3.8×10−4 (0.038%).

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

Hydraulic lift

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

P! P! P! P! P!

h

P! + hρg

Hydrostatic Paradox

No matter the shape of a vessel, the pressure depends

only on the vertical depth.

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

Measurement of pressure

Barometer

P! = hρg, i.e., h =P!ρg

.

P ≈1.01×105 Pa.

Using water:

Using mercury:

ρ = 1×103 kg ⋅m−3,

∴h = 1.01×105 Pa

(1×103 kg ⋅m−3)(9.81 m/s2)∼10 m.

ρ = 13.6 ×103 kg ⋅m−3,

∴h = 1.01×105 Pa

(13.6 ×103 kg ⋅m−3)(9.81 m/s2)∼ 0.76 m.

‘Standard atmospheric pressure’ is defined

as 760 mm of Hg.

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

DISCUSSION PROBLEM

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

Buoyancy and Archimedes Principle 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

Example: What fraction of an iceberg is submerged?

i.e., 89% of an iceberg is submerged!

If you don’t believe it …

ρs = 0.92 ×103 kg ⋅m−3 and ρL = 1.03×103 kg ⋅m−3.

∴VLVs

=ρsρL

= 0.92 ×103 kg ⋅m−3

1.03×103 kg ⋅m−3= 0.89,

Question 13.3: On Earth, an ice cube floats in a glass of

water with about 90% of its volume below the level of the

water. If we poured ourselves a glass of water on the

Moon, where the acceleration of gravity is about 16% 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 13.4: A block of copper with mass 0.50 kg is

suspended from a spring scale. When it 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.

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)

ρLρs

= 1×103 kg ⋅m−3

9 ×103 kg ⋅m−3= 0.111.

∴T = (0.5 kg)(9.81 m/s2)(1− 0.111) = 4.36 N (0.444 kg).

In air instead of water:

So, the mass of the block would be 71.5 mg less than in

vacuum (0.50 kg).

ρairρs

= 1.29 kg ⋅m−3

9 ×103 kg ⋅m−3= 1.43×10−4.

Question 13.5: A beaker containing water is placed on top

of a weighing scale and the reading is 1.200 kg. In (a), a

copper block is hanging freely from a spring scale, which

has a mass reading of 0.200 kg. When the copper block is

totally immersed in the water, as shown in (b), what are

the readings on the two scales?

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

Vs =

msρs

= 0.20 kg

9.0 ×103 kg ⋅m−3= 2.22 ×10−5 m.

(a) Initially, spring scale registers the weight of the copper

block and the weighing scale registers the weight of the

beaker plus water. (b) When the block is lowered into the

water, the water exerts a buoyant force, B, (upthrust) on

the copper block. Then,

B is equal to the weight of water displaced, i.e.,

where the submerged volume, i.e., the volume of the

copper block, is

T2 + B = mg, i.e., T2 = mg − B.

B = ρwVsg,

∴T2 = (m −ρwVs)g

= (0.20 kg) − (1×103 kg ⋅m−3)(2.22 ×10−5 m3( )(9.81 m/s2)

= 1.74 N,i.e., the reading on the spring scale is 0.178 kg.

The upthrust B is the force of 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 weighing scale will increase by 0.178 kg, i.e., it will

read 1.378 kg when the block is submerged.

∴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 13.6: A barge, loaded with steel canisters is

floating in a closed lock. If the cargo is thrown over the

side, what happens to the level of water in the lock?

Does it rise, stay the same, or fall?

With the cargo on board, as in (a), the weight of water

displaced is equal to the weight of the barge plus cargo,

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

canisters, is the volume of each canister and their

density. When the canisters 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

canisters,

Since i.e., less water is displaced

in (b) than in (a). The volume of water is unchanged, so the

depth of water is less in (b) compared with (a), i.e., the

water level falls.

i.e., ww (a) = wb + wc = wb + NVcρcg,

wb

Vc ρc

i.e., ww (b) = wb + NVcρwg.

ρw < ρc , ww (b) < ww (a),

Fluids in motion

Consider an incompressible fluid (a liquid) flowing down a

tube of non-uniform size. In time the mass of fluid in

the left-hand shaded volume is and the mass

of fluid in the right-hand volume is Note

that because the fluid is incompressible, the density

remains constant. If the flow is steady, the mass that

crosses must equal that crossing

This expression is called the continuity equation for an

incompressible fluid. The conserved quantity, is

called the volume flow rate, Q

Δt,

m1 = ρA1v1Δt

m2 = ρA2v2Δt.

A1 A2, i.e., m1 = m2.

∴A1v1Δt = A2v2Δt, i.e., A1v1 = A2v2 ⇒ constant.

A × v,

(m3/s).

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

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 speed of the water from a faucet

increases as it falls because of gravity.

The continuity equation tells us that the

cross sectional area will decrease as the

speed increases.

The speed of the water from a

garden hose increases as you reduce

the area by putting your thumb over

the end of the hose. So, the water

“squirts” further.

Other examples include lanes at highway tolls (increasing

the number of lanes in an attempt to maintain traffic flow).

Question 13.7: A garden hose with an inside diameter of

16 mm fills a 10 liter bucket in 20 s.

(a) What is the speed of the water out the end of the

hose?

(b) What diameter nozzle would increase the speed by

a factor of two?

(c) How long would it take to fill the same bucket with

the nozzle referred to in part (b)?

(a)  The volume flow rate is

(b)  Since remains constant, if v is increased by a

factor of 2, then A must be reduced by a factor of 2. But

so the radius must be reduced by a factor of

So, the nozzle diameter should be 11.3 mm.

(c) Since Q, the volume flow rate, i.e., the volume of water

delivered each second, remains constant, it takes the same

time (20 s) to fill the bucket with the nozzle as without the

nozzle.

Q = 10 L20 s

= 10(1×10−3 m3)20 s

= 5.0 ×10−4 m3/s.

v = QA

= Q

πr2= 5.0 ×10−4 m3/s

π(0.008 m)2= 2.5 m/s.

Q = Av

A ∝ r2 2.

Question 13.8: A large tank of water has an outlet a

distance below the surface of the water. Initially,

the tank is filled with water to a depth

(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?

You may assume the tank has a very large diameter so the

level of the water remains constant. Also, you can model

the water leaving the hole as a projectile.

h = 3.0 m

y2 = 4.0 m.

A1

A2

(a)  We apply Bernoulli’s equation to points and .

But since both the hole and the surface of the

water in the tank are at atmospheric pressure.

Since then

(As though in free fall!)

P1 +ρgy1 +

12ρv1

2 = P2 +ρgy2 +12ρv2

2.

P1 = P2 = P!

∴ 1

2ρv1

2 = ρg(y2 − y1) + 12ρv2

2 = ρgh + 12ρv2

2.

v2 = 0,

12ρv1

2 = ρgh, i.e., v1 = 2gh

= 2(9.81 m/s2)(3.0 m) = 7.67 m/s.

A1

A2

(b) If we model the water leaving the hole as a projectile,

then the time to reach the ground is given by

but as the water emerges horizontally.

The range is where which remains

constant in projectile motion.

y1 = vyit −

12

gt2,

vyi = 0,

∴ t =

2y1g

=2(y2 − h)

g= 2(1.0 m)

(9.81 m/s2)= 0.45 s.

x = vxit, vxi = v1,

∴x = v1t = (7.67 m/s)(0.45 s)

= 3.46 m.

Question 13.9: A siphon is a device for transferring a

liquid from one container to another. The tube must be

filled with liquid to start the siphon.

(a) Derive an expression for the speed that water would

flow through the tube.

(b) What is the pressure at the highest section of the

tube?

(a) We apply Bernoulli’s equation

to the surface of the liquid in the

left hand container and the liquid

at C. Then

where the velocity at the surface

as the surface area of the container is much greater

than the area of the tube.

So the velocity depends only on the height difference

between the surface of the liquid in the reservoir and the

drain point.

(b) Applying Bernoulli’s equation to the surface in the

reservoir and the point B we find

P! +12ρvs

2 + 0

= P! +12ρvc

2 − hcρg,

vs = 0

∴ 1

2vc

2 = hcg, i.e., vc = 2ghc .

P! = PB + hbρg + 12ρvB

2 .

∴PB = P! − hbρg − 12ρvB

2 .

But, since the tube has a constant

cross sectional area,

Therefore, substituting for

we find

Note that if

then which represents the longest length for the

siphon tube. With water as the liquid,

vA = vB = vC.

vB = vC = 2ghc ,

PB = P! − (hb + hc )ρg.

(hb + hc ) =

P!ρg

,

PB = 0,

(hb + hc ) ≈10 m.