14-1

Note 14 Fluid DynamicsSections Covered in the Text: Chapter 15, except 15.6

To complete our study of fluids we now examinefluids in motion. For the most part the study of fluidsin motion was put into an organized state by scientistsworking a generation after Pascal and Torricelli. Weshall see here that much of the physics of fluids isencapsulated in the statement of Bernoulli’s principle.

The study of fluid flow was driven by the demandsof the industrial revolution. It was vital for progressand profit that the movement of water, steam and oilthrough pipes, and the movement of aerofoils throughthe air, were understood mathematically. Today, thephysics of fluids in motion is of special interest in thevarious branches of the environmental and lifesciences.

Fluids in MotionThere are two types of fluid motion called laminar flowand turbulent flow.

A fluid will execute laminar flow when it is moving atlow velocity. The particles of the fluid follow smoothpaths that do not cross, and the rate of fluid flowremains constant in time. This is the easiest type offlow to describe mathematically.

A fluid will execute turbulent flow when it is movingabove a certain critical velocity. Strings of vorticesform in the fluid, resulting in highly irregular motion.A white water rapid is a good example. A systemmoving irregularly is difficult to describe mathemat-ically, so we shall not be concerned with turbulentflow here.

An Ideal FluidSince a fluid is in general a complicated medium todescribe mathematically, even in laminar flow, weshall assume for simplicity that the fluid is ideal. Byideal we mean

1 The fluid is moving in laminar flow; viscous forcesbetween adjacent layers are negligible.

2 The flow is steady, that is, the flow rate does notchange with time.

3 The fluid has a uniform density and is thus incom-pressible.

4 The flow is irrotational, that is, the angular momen-tum about any point is zero; in common parlancethe fluid does not “swirl”.

Flow RateWhen a fluid occupies a pipe of cross sectional area Aand flows with average speed v, the rate of flow Q isgiven by

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Q = Av . …[14-1]

The units of Q are m3.s–1. These are units of volume.s–1,so flow rate is the same as volume rate.

The paths of particles in a fluid moving with laminarflow are called streamlines. Streamlines never crossone another (Figure 14-1).

Figure 14-1. Particles in a fluid moving with laminar flowfollow streamlines that do not cross.

The Equation of ContinuityA fluid moving in laminar flow in a flow tube (thatmay be a pipe) can be shown to satisfy a simple rela-tionship. Consider an ideal fluid flowing through atube of variable cross-section (Figure 14-2). In anelapsed time ∆t, a volume A1v1∆t of fluid crosses areaA1, and a volume A 2v2∆t crosses area A2. Since thefluid is incompressible and the streamlines do notcross the volume of fluid crossing A1 must equal thevolume of fluid crossing A2, so A1v1∆t = A2v2∆t, fromwhich it follows that A1v1 = A2v2 or Q 1 = Q2. Thismeans that the flow rate

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Q = Av = const . …[14-2]

An important consequence of eq[14-2] is that if thecross sectional area of the flow tube is reduced atsome point, then the flow speed increases. Eq[14-2] is

Note 14

14-2

known as the equation of continuity.1

Figure 14-2. Illustration of the equation of continuity. Theflow tube of a fluid is shown at two positions 1 and 2. At notime does fluid enter or leave the flow tube.

The equation of continuity can be applied to explainthe various rates of blood flow in the body. Bloodflows from the heart into the aorta from which itpasses into the major arteries; these branch into thesmall arteries (arterioles), which in turn branch intomyriads of tiny capillaries. The blood then returns tothe heart via the veins. Blood flow is fast in the aorta,but quite slow in capillaries. When you cut a finger(capillary) the blood oozes, or flows very slowly. Wecan show this by means of a numerical example.

1 The equation of continuity can be thought of as a statement ofthe conservation of fluid. As the fluid flows through the pipe thevolume of fluid remains constant; it neither increases nor decreases.

Example Problem 14-1Speed of Blood Flow in Capillaries

The radius of the aorta is about 1.0 cm and the bloodflowing through it has a speed of about 30.0 cm.s–1.Calculate the average speed of the blood in thecapillaries using the fact that although each capillaryhas a diameter of about 8.0 x 10–4 cm, there areliterally billions of them so that their total crosssection is about 2000 cm2.

Solution:From the equation of continuity the speed of blood inthe capillaries is

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v2 =v1A1A2

=0.30(m.s−1) × 3.14 × (0.010)2(m2)

2.0 ×10−1(m2)

= 5.0 x 10–4 m.s–1

or about 0.5 mm.s–1. This is a very low speed.

Bernoulli’s EquationDaniel Bernoulli (1700-1782), a Swiss mathematicianand scientist, lived a generation after Pascal andTorricelli. The equation he derived is a more generalstatement of the laws and principles of fluids we haveexamined thus far.

Bernoulli allowed for the flow tube to undergo apossible change in height (Figure 14-3). Considerpoints 1 and 2. Let point 1 be at a height y1 and let v1,A1 and p1 be the speed of the fluid, cross sectional areaof the tube and pressure of the fluid at that point.Similarly let v2, A2 and p 2 be the same variables atpoint 2. The actual system is the volume of fluid in theflow tube.

In an elapsed time ∆t the amount of fluid crossing A1is ∆V1 = A1v1∆t and the amount of fluid crossing A2 is∆V2 = A2v2∆t. But from the equation of continuity, A1v1= A2v2. So the volume of fluid crossing either area isthe same; let us simply write it as

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ΔV = AvΔt .

Fluid is moved in the flow tube as the result of thework done on the fluid by the surrounding fluid (theenvironment). The net work W done on the fluid inthe elapsed time ∆t is

Note 14

14-3

Figure 14-3. Illustration of Bernoulli’s equation.

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W = F2 − F1( )Δx = F2 − F1( )ΔVA

= p2 − p1( )ΔV .

This work goes into achieving two things:

1 changing the kinetic energy of the fluid betweenthe two points by the amount:

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ΔK =ρΔV2

v2

2 − v12( ) . …[14-3]

2 changing the gravitational potential energy of thefluid between the two points by the amount mg∆hor

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ΔU = ρΔVg y2 − y1( ) . …[14-4]

Since

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W = p2 − p1( )ΔV = ΔK + ΔU ,

we have, by substituting eqs[14-3] and [14-4]:

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p1 − p2( )ΔV

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=ρΔV2

v2

2 − v12( ) + ρΔVg y2 − y1( ) .

Dividing through by ∆V we obtain the general form ofBernoulli’s equation:

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p1 +ρ2

v1

2 + ρgy1 = p2 +ρ2

v2

2 + ρgy2 .

…[14-5]

We can put this equation into the simpler form:

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p +ρ2

v 2 + ρgy = const . …[14-6]

Many of the “principles” and “laws” we have seencan be shown to be special cases of Bernoulli’s equa-tion. We shall consider a number of them.

Many homes and buildings in the colder climates areheated by the circulation of hot water in pipes. Even ifthey are not explicitly aware of it, architects mustensure that their designs conform to Bernoulli’sprinciple in order to avoid system failure. Let usconsider an example.

Example Problem 14-2Application of Bernoulli’s Principle

Water circulates throughout a house in a hot waterheating system. If the water is pumped at a speed of0.50 m.s–1 through a pipe of diameter 4.0 cm in thebasement under a pressure of 3.0 atm, what will bethe flow speed and pressure in a pipe of diameter 2.6cm on the second floor 5.0 m above?

Solution:Let the basement be level 1 and the second floor belevel 2. We can obtain the flow speed by applying theequation of continuity, eq[14-2]:

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v2 =v1A1A2

= 0.50(m.s−1) π(0.020m)2

π(0.013m)2

= 1.2 m.s–1.

To find the pressure p2 we use Bernoulli’s equation,eq[14-5]:

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p2 = p1 + ρg y1 − y2( ) +12ρ v1

2 − v22( ) .

Substituting p1 = 3.0 x 105 Pa, ρwater = 1000 kg.m–3, g =9.8 m.s –2, y1 = 0, y2 = 5.0 m, v1 = 0.50 m.s–1, v2 = 1.2 m.s–1

we getp2 = 2.5 x 105 Pa.

Thus p2 < p1. Obviously, the pipe on the second floorof the house must be able to withstand less pressurethan the pipe in the basement.

Eqs[14-5] and [14-6] have the look of conservation ofenergy expressions including work, because that, ineffect, is what they are.

Note 14

14-4

1 Torricelli’s TheoremConsider the container filled with fluid in Figure 14-4.A distance h below the surface of the fluid a smallhole allows fluid to escape. What is the velocity of theoutflowing fluid?

h

Figure 14-4. A container of fluid with a small hole a distanceh below the surface allowing fluid to escape.

Applying Bernoulli’s equation we have

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p +ρ2

v 2 + ρgy = const .

Assuming a container of normal laboratory size, theatmospheric pressure p is essentially the same at itstop and bottom, so we can cancel p and write to agood approximation

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ρ2

v 2 + ρgy = const .

But if the container is not vanishingly small then thefluid velocity is essentially zero at the top (at y = 0). Sothe velocity v of the outflowing fluid (at y = –h) isgiven by

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ρ2

v 2 + ρg(−h) = 0 ,

from which it follows that

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v = 2gh . …[14-7]

This is what is known as Torricelli’s theorem. Note thatthe velocity is independent of the fluid density. Notetoo that our treatment here neglects the effect of fluidviscosity (that would otherwise reduce the speed). 2

2 Eq[14-5] is the same expression as obtained for the final velocityof an object dropped from rest at a height h (Note 09).

Example Problem 14-3Applying Torricelli’s Theorem

A glass container of height 1.0 m is full of water. Asmall hole appears on the side of the container at thebottom (as shown in Figure 14-4). What is the speed ofthe water flowing out the hole?

Solution:This is a straightforward application of Torricelli’stheorem and eq[14-7]. The speed of the water is

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v = 2gh = 2 × 9.80(m.s−1) ×1.0(m) = 4.43 m.s–1

The value measured would be less than this if thefluid (for example, shampoo) had a significantviscosity.

2 A Fluid at RestFor a fluid at rest (static fluid) the speed v in eq[14-6]is zero. Bernoulli’s equation then reduces to

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p + ρgy = const .

This is just Pascal’s law; see eq[13-3] in Note 13. Thepoint to be made here is that Bernoulli’s equation is amore general statement of the physics of fluids than isPascal’s law.

3 Pressure in a Flowing FluidWe studied hydrostatic pressure in Note 13. We canshow that in a moving fluid the pressure is depen-dent on velocity.

A fluid flowing along a horizontal level experiencesa constant gravitational potential (y = constant). Ber-noulli’s equation therefore becomes

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p +ρ2

v 2 = const . …[14-8]

This means that as the speed v of the fluid increasesthe pressure p must fall. This result forms theprinciple of operation of many practical devices. Oneis the Venturi tube, which we consider next.

Note 14

14-5

The Venturi TubeThe Venturi tube (Figure 14-5) is a device that is usedto measure the flow rate Q of a fluid. It consists of twosections of different cross sectional areas A1 and A 2that are known with good precision. The difference inpressure in the fluid in the two sections (p1 – p2) ismeasured with a built-in manometer.

Figure 14-5. The Venturi tube.

From Bernoulli’s equation with y2 = y1 we have

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p1 − p2 =ρ2

v2

2 − v12( ) .

Substituting the equation of continuity to eliminate v2we can rearrange and solve for v1. The flow rate Q istherefore given by

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Q = A1v1 = A1A22 p1 − p2( )ρ A1

2 − A22( )

. …[14-9]

Since the density of the fluid ρ is also known, as arethe areas A1 and A2, Q can be calculated once (p1 – p2)is measured with the manometer.

Example Problem 14-4Using a Venturi Tube

A Venturi tube is used to measure the flow of water. Ithas a main diameter of 3.0 cm tapering down to athroat diameter of 1.0 cm. The pressure difference p1 –p2 is measured to be 18 mm Hg. Calculate the velocityv1 of the fluid input and the flow rate Q.

Solution:The pressure difference in Pa is, using the conversionexpression eq[13-4]:

p1 – p2 = 18 mm Hg ≡ 1.333 x 105 (Pa.m–1)x18x10–3 (m)

= 24.0 x 102 Pa.

Therefore from eq[14-9] the speed of the input fluid is

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v1 = A22 p1 − p2( )ρ A1

2 − A22( )

= 24.6 cm.s–1.

Multiplying by A1 we obtain the flow rate Q:

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Q = π(1.5cm)2 × 24.6(cm.s−1) =174 cm3.s–1

These results, too, would be quite in error if the fluidhad a non-negligible viscosity. The viscosity of waterhas a negligible effect in this example.

The AerofoilBernoulli’s equation helps to explain why an airplaneis equipped with wings to help it stay in the air. Anairplane wing is an example of an aerofoil (Figure 14-6). In a moving stream of air, the air travels morequickly over the top surface of an aerofoil than over thebottom surface. According to Bernoulli’s principle thepressure on the top surface is less than the pressure onthe bottom surface, contributing to a net upward forcecalled aerodynamic lift. When an airplane is flying atconstant altitude and speed, the upward aerodynamiclift balances the downward gravitational force andprevents the plane from falling. This is sometimescalled the Bernoulli effect.

A glider moving on an air track is a kind of aerofoil.On the top surface of a glider the pressure is atmos-pheric pressure. In the space between the bottom

Note 14

14-6

surface and the surface of the air track the pressure ofthe air is higher than atmospheric. The difference inpressure accounts for the aerodynamic lift that keepsthe glider from contacting the air track surface andgrinding to a halt.

Figure 14-6. Streamlines around an aerofoil.

Example Problem 14-5Aerodynamic Lift

An airplane has a mass of 2.0 x 106 kg and the airflows past the lower surface of the wings at 100 m.s–1.If the wings have a surface area of 1200 m2, how fastmust the air flow over the upper surface of the wing ifthe plane is to stay in the air? Consider only theBernoulli effect.

Solution:The wing is a surface moving horizontally throughthe air. The effect is the same as if the wing were stat-ionary and the air were flowing horizontally over the

wing. From eq[14-5] we can write

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pabove +ρair2

vabove

2 = pbelow +ρair2

vbelow

2

…[14-10]

where “above” and “below” refer to the airplanewing. The difference in pressure accounts for the aero-dynamic lift, that is, the difference in pressure is equalto the resultant force per unit area on the wing. Themagnitude of force must just equal the weight of theairplane. Thus

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pbelow − pabove =FA

=2.0 ×106(kg)× 9.80(m.s−2)

1200(m2)

= 16,300 Pa.

Rearranging eq[14-10] we can write

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vabove2 =

pbelow − paboveρ /2

+ vbelow

2 .

Taking the density of air as the value at the Earth’ssurface, i.e., 1.29 kg.m–3 we have

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vabove2 =

16300(1.29 /2)

+104 .

Evaluating and taking the square root we have finally

vabove = 190 m.s–1.

Thus the air must flow over the upper surface of thewing nearly twice as fast as past the lower surface.

Note 14

14-7

To Be Mastered

• Definitions: laminar flow, turbulent flow, equation of continuity

• Physics of: Bernoulli’s Equation

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p1 +ρ2

v1

2 + ρgy1 = p2 +ρ2

v2

2 + ρgy2• Physics of: the Venturi tube• Physics of: the aerofoil

Typical Quiz/Test/Exam Questions

1. State Benoulli’s law relating pressure p, velocity v and density ρ in a moving fluid.

2. Sketch a Venturi tube, labelling the important features.

3. Explain briefly what a Venturi tube is used for, and how it is used.

4.

5.6.

7.

8.