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Chapter 3

Pressure

Our first topic as we begin our detailed exploration of fluid mechanics is the pressure that

prevails whether or not a fluid is bounded by a container. We will quickly discover that fluidpressure, p, acts as a force potential, i.e., the gradient of the pressure, p, is a force (per unitvolume) that acts to accelerate and/or balance other forces acting on a fluid particle. Roughly

speaking, pressure is analogous to voltage in electromagnetic theory. If it's constant, the fluid

is at some base state. If it changes, then something happens.

We will develop two particularly interesting formulas in this chapter, viz., the hydrostatic

relation for pressure and Bernoulli's equation. The hydrostatic relation shows why pressure

increases with depth in the ocean and decreases with altitude in the atmosphere. It can also be

used to measure the pressure at any point in a stationary fluid indirectly by simply measuring

the height of a column of fluid. Bernoulli's equation, which gives an explicit relation between

velocity and pressure for a moving fluid, is one of the most famous and widely used equations

in fluid-flow theory. The section on Bernoulli's equation includes a simple experiment you

can perform at home.

Even in the absence of fluid motion, the presence of a gravitational field gives rise to someinteresting phenomena. On the one hand, we learned in Chapter 1 that pressure acts equally

in all directions at a given point in a fluid. On the other hand, the gravitational field causes

a pressure gradient, and we find that pressure increases with depth in a fluid. There is no

contradiction as the former observation applies to a single point, while gravitational effects

are felt over a finite distance.

After developing and applying the hydrostatic relation and Bernoulli's equation, the con-

cluding sections of this chapter focus on fluid-pressure variation in a gravitational field for

a motionless, or static fluid. For the obvious reason, general engineering literature describes

this topic as fluid statics.

We will develop a method for computing forces and moments on both planar and curved

surfaces. With a clever balance of forces, we will replace the problem of integrating a varying

pressure over a general curvilinear surface with a simple set of algebraic operations. We

will accomplish this by deriving formulas for a point force and moment arm that give the

mechanical equivalent of the force and moment due to the distributed pressure. This method

is useful for determining the loads on a structure such as a dam or a flow-control valve.

We conclude the chapter with a discussion of buoyancy and a famous result known as

Archimedes' Principle. This principle says the buoyancy force on a floating or submerged

body is equal to the weight of the displaced fluid. The section on buoyancy includes another

home experiment.

81

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82 CHAPTER 3. PRESSURE

3.1 Pressure on an Infinitesimal Element

Before proceeding, it is worthwhile to pause and derive a very useful mathematical relationship

that we will make use of from time to time in developing the basic equations governing the

behavior of fluids. Consider the differential element shown in Figure 3.1. The element is a

rectangular parallelepiped and the pressure, p, acts on its six faces. We assume that the fluidelement's center lies at the point (x,y,z) and that the lengths of the three sides are x, yand z in the x, y and z directions, respectively.

x

y

z

i

j

k

xy

z

p(x- 12

x , y, z)

p(x+ 12

x , y, z)

p(x, y- 12

y, z) p(x, y+ 12

y, z)

p(x,y,z- 12

z)

p(x,y,z+ 12

z)

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Figure 3.1: Pressure acting on an infinitesimally small fluid element centered at the point

(x,y,z). i, j and k are unit vectors in the x, y and z directions, respectively.

We can calculate the net force on the fluid element by integrating the pressure over the

surface bounding the fluid element. We indicate this symbolically as follows.

F = Sp n dS (3.1)There are three key points that require discussion to explain the notation.

1. The `O' joining the integrals denotes closed-surface integral, i.e., we must integrate

over the entire surface that bounds the fluid element. For the infinitesimal element in

Figure 3.1, this means

S

p n dS =6

m=1

Am

p n dAm

where the six Am are the areas of the six faces of the element.

2. The vector n is the outer unit normal, i.e., a unit vector normal to the bounding surface

that points outward. In the present case we have

Face at x+ 12

x: n = i, Face at x- 12

x: n = i

Face at y+ 12

y: n = j, Face at y- 12

y: n = j

Face at z+ 12

z: n = k, Face at z- 12

z: n = k

where i, j and k are unit vectors in the x, y and z directions, respectively.

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3.1. PRESSURE ON AN INFINITESIMAL ELEMENT 83

3. We require a minus sign because the pressure pushes in the opposite direction of n.

So, for our infinitesimal element, we have

F = A

p i dydz Face at x+ 1

2x

A

p (i) dydz Face at x- 1

2x

A

pj dxdz Face at y+ 1

2y

A

p (j) dxdz Face at y- 1

2y

A

p kdxdy Face at z+ 1

2z

A

p (k) dxdy Face at z- 1

2z

(3.2)

Because we are dealing with infinitesimally small quantities, we can treat the pressure as being

constant on each face. Our result will be exact in the limit as x, y and z approach zero.Thus,

F p(x + 12x,y,z) p(x 12x,y,z) i yzp(x, y + 1

2y, z) p(x, y 1

2y, z)

j xz

p(x,y,z + 1

2z) p(x,y,z 1

2z)

k xy (3.3)

Next, we rearrange terms to arrive at

F p(x + 1

2x,y,z) p(x 1

2x,y,z)

xi xyz

p(x, y + 1

2y, z) p(x, y 1

2y, z)

yj xyz

p(x,y,z + 1

2z) p(x,y,z 1

2z)

zkxyz (3.4)

Finally, as x, y and z approach zero, each line in Equation (3.4) contains a partialderivative of the pressure. Defining the differential volume, V, according to

V = xyz (3.5)

we conclude that

F

p

xi +

p

yj +

p

zk

V (3.6)

The quantity in square brackets is the familiar gradient of p, which we write symbolically as

p =p

xi +

p

yj +

p

zk (3.7)

Therefore, we have proven that, for an infinitesimally small fluid element,

S

p n dS p V for V 0 (3.8)

In words, Equation (3.8) tells us that the net force on a differential fluid element due to the

pressure is equal to the product of the element's volume and the pressure gradient within the

element. We will use this result in the next section in order to derive a key result for fluids at

rest.

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3.2 The Hydrostatic Relation

For a fluid at rest in a gravitational field, the only forces acting are gravity and the pressure

within the fluid. In the absence of gravity, the pressure in the fluid would obviously be the

same at all points. Thus, the presence of a gravitational field must give rise to a variation in

pressure. In this section, we will determine how the pressure varies.

gkdV

Bounding surface S

Volume Vn

dS

p

xy

z

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Figure 3.2: Fluid volume in a gravitational field.

Consider the arbitrary volume V bounded by a closed surface S shown in Figure 3.2. Asnoted in the preceding section, if n is the outer unit normal to the differential surface element

dS, the net pressure force exerted by the surroundings on the surface is the closed-surfaceintegral of p n (the minus sign makes it an inward directed force). That is, the net pressureforce acting on the volume is

Pressure Force =

S

p n dS (3.9)

Now, the density of the fluid contained in the volume is and the gravitational accelerationvector is given by gk where k is a unit vector in the z direction. So, the weight of the fluidis the volume integral of gk, i.e.,

Weight =

V

gkdV (3.10)

Since the volume is at rest, balancing forces yields

S

p n dS+

V

gkdV = 0 (3.11)

Equation (3.11) applies to stationary volumes of arbitrary shape and size. We will make use of

this integral relationship in Section 3.10 when we address buoyancy. Because of the generality

of this equation, we can apply it to an infinitesimal fluid element such as the one depicted in

Figure 3.1. We can use this fact to determine how the pressure behaves at an arbitrary point

within the volume V. Clearly, as the volume size approaches zero, we haveV

gkdV gkV as V 0 (3.12)

where is an averaged value for the fluid within V. In the limit V 0, the value of approaches the value at the point we are interested in. Recalling Equation (3.8), the forcebalance simplifies to

p V + gkV = 0 as V 0 (3.13)

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3.2. THE HYDROSTATIC RELATION 85

Dividing through by V yields

p = gk (3.14)

Thus, in component form, we have shown that,

p

x=

p

y= 0 and

p

z= g (3.15)

In addition to demonstrating that a stationary fluid in a gravitational field has a pressure

gradient parallel to the direction of gravitational acceleration, we have also shown that pressure

is a function only of depth. This means p = p(z) and we can replace Equation (3.15) withthe following ordinary differential equation.

dp

dz= g (3.16)

We can integrate immediately for fluids such as liquids in which density is constant. The

resulting pressure variation is

p + gz = constant (3.17)

We call Equation (3.17) the hydrostatic relation. It tells us that pressure increases linearly

with depth (remember that negative values of z correspond to increasing depth).The physical meaning of the hydrostatic relation becomes obvious if we regard the pressure

as a force potential. This is sensible since, by definition, a force potential is a function whose

gradient is a force vector (the standard convention is to include a minus sign). In this sense

pressure is analogous to voltage, whose gradient is proportional to electrostatic force. As

demonstrated in the last section, the net force on an infinitesimal fluid element is pV.Consequently, p is force per unit volume and p is thus the pressure-force potential per unitvolume.

Similarly, the term gz is potential energy per unit volume. Thus, Equation (3.17) tellsus that the sum of the pressure-force potential per unit volume and the potential energy per

unit volume is constant. In other words, the hydrostatic relation is a mechanical-energyconservation principle for a motionless fluid.

Example 3.1 At what depth in a 20o C fresh-water lake is the pressure twice the pressure of the

atmosphere?

Solution. Since the pressure at the lake surface (z = 0) is atmospheric, the value of \constant" inEquation (3.17) is pa. Thus, the depth at which p = 2pa is given by

2pa + gz = pa

Solving for z yields

z = pag

The density of water at 20o C is = 998 kg/m3, g = 9.807 m/sec2 and atmospheric pressure ispa = 1.01 105 Pa. Therefore,

z = 1.01 105 kg/(m sec2)

(998 kg/m3)(9.807 m/sec2)= 10.32 m

where we make use of the fact that 1 Pa = 1 kg/(msec2). Thus, the pressure is 2 atm at a depthof 10.32 m.

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Figure 3.3: Because of the increase in pressure with depth, which can cause the bends, sport

diving is done no deeper than 130 feet. The hydrostatic relation tells us the pressure is nearly

5 atm at this depth. [Photograph courtesy of Barbara Wilcox]

The hydrostatic relation tells us that the pressure at 130 feet in the ocean is 72.8 psi (note

that the density of seawater is 2.0 slugs/ft3). This high a pressure is sufficient to cause nitrogen

to dissolve in a diver's bloodstream, leading to impaired judgment and a painful condition

known as the bends if the diver returns to the surface too rapidly. Impaired judgment is very

noticeable at a depth of 130 feet, which is the established maximum depth for sport diving.

Using no breathing equipment, divers have exceeded this depth and risen to the surface rapidly

with no ill effects. However, their pulse rates have gone so low at these great depths as to be

in a state close to death|this practice is not recommended!

3.3 Atmospheric Pressure Variation

The pressure in the Earth's atmosphere varies in a more complicated manner than the simple

linear relation in Equation (3.17). The reason for this is the following. Atmospheric air

behaves like a perfect gas so that its density is given by

=p

RT(3.18)

Hence, Equation (3.16) assumes the following form.

dp

dz=

g

RTp (3.19)

If we knew the variation of temperature T with altitude, integration of Equation (3.19)would be straightforward. There is a simple model for the atmosphere over the United States

known as the U. S. Standard Atmosphere [U. S. Government Printing Office (1974)]. This

model represents average conditions in the United States at 40o N latitude (e.g., New York

City). In the U. S. Standard Atmosphere, the region from the Earth's surface (z = 0) upto z = 11.0 km (6.84 miles) is called the troposphere and the temperature decreases linearlywith altitude according to

T = T0 z for 0 z 11.0 km (3.20)

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3.3. ATMOSPHERIC PRESSURE VARIATION 87

z (km)

0

10

20

30

40

50

60

200 220 240 260 280 300T (K)

T = T1

T = T0troposphere

stratosphere

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Figure 3.4: Temperature in the U. S. Standard Atmosphere.

where the coefficient is the lapse rate, and T0 is the average surface temperature. Theregion from z = 11.0 km (6.84 mi) to z = 20.1 km (12.5 miles) is called the stratosphere.The temperature in this idealized model is constant in the stratosphere, and denoted by T1.Figure 3.4 shows the temperature variation in the U. S. Standard Atmosphere, with the tropo-

sphere and stratosphere clearly indicated. Above the stratosphere, temperature increases in a

nontrivial manner. Table 3.1 lists values of , T0 and T1 in SI and USCS units.

Table 3.1: Properties of the U. S. Standard Atmosphere

Property SI Units USCS Units

6.50 K/km 18.85o R/miT0 288 K (15o C) 518.4o R (59o F)T1 218 K (55o C) 392.4o R (67o F)

Tropospheric Pressure Variation. Focusing first on the troposphere, combining Equa-

tions (3.19) and (3.20) yieldsdp

p=

g dz

R(T0 z)(3.21)

Integrating, we find that the pressure varies according to

p = p0

1

z

T0

g/(R)for 0 z 11.0 km (3.22)

where p0 = 101 kPa (14.7 psi) is the pressure at sea level. The exponent g/(R) in Equa-tion (3.22) is approximately 5.26. Note that the pressure falls to 22.5 kPa (3.28 psi) at the

upper boundary of the troposphere, which is the approximate altitude at which modern airliners

fly.

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Stratospheric Pressure Variation. The integration is even easier in the stratosphere since

temperature is constant. The pressure varies as follows.

p = p1 exp g(z z1)RT1 for 11.0 km z 20.1 km (3.23)The pressure p1 = 22.5 kPa (3.28 psi) follows from insisting that Equations (3.22) and (3.23)yield the same pressure at the interface between the troposphere and stratosphere. Thus, the

pressure (and density) fall off exponentially in the stratosphere, and we sometimes refer to this

as an exponential atmosphere.

Example 3.2 The cabin pressure in a modern airliner at cruise altitude is typically about 12 psi.

As the airplane descends to land, low pressure air is trapped inside your ears. This is what causes

the \popping" sensation you experience when you yawn and allow the pressure to equilibrate.

Estimate the altitude in the atmosphere at which this pressure prevails.

Solution. We can use Equation (3.22) to compute the altitude in the U. S. Standard Atmosphere.Solving for z, we find

z =T0

1

p

p0

R/g=

518.4o R

18.85o R/mi

1

12.0 psi

14.7 psi

1/5.26= 1.04 mi

Inversion

z

z1

T1 T0 T

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Figure 3.5: Temperature variation over Los Angeles.

The atmospheric temperature variation near Los Angeles (and many other large cities)

deviates from the U. S. Standard Atmosphere in a significant manner. Specifically, there is

a region in the troposphere, typically at an altitude of about 0.5 km (0.3 mi), known as

the inversion layer, in which temperature increases with increasing altitude. This layer is

present because the Los Angeles area is almost completely enclosed by high mountains. As

air descends from the mountains, the sun heats it and creates a warm layer that rises above

cooler air blowing in from the Pacific Ocean. This is the primary mechanism that creates the

temperature inversion as in Figure 3.5, with the heavier cool air trapped near the surface. The

inversion layer puts a \lid" on the area that traps surface emissions responsible for smog. The

prevailing winds in the Los Angeles area are unable to relieve the pollution problem because

of this lid. Rather, they merely move the smoggy air from one part of the region to another.

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3.4. MANOMETRY 89

3.4 Manometry

The hydrostatic relation tells us that a change in depth is directly proportional to the corre-

sponding change in pressure. This is the basis of the mercury barometer invented by Torricelli

in 1644. As shown in Figure 3.6, a barometer consists of a glass tube closed at one end with

the open end immersed in a container filled with mercury. The tube is initially filled with

mercury and turned upside down when inserted in the container. Mercury vapor fills the open

space above the column of mercury. However, because the vapor pressure of mercury is very

small (Subsection 1.4.4 and Table A.9), there will be a near vacuum in the top of the tube.

Mercury

pa pah

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Figure 3.6: Mercury barometer.

Since pressure in the tube above the mercury, p, is negligible, the hydrostatic relation tells us

p + ghHg = pa = hHg =pa p

g

pag

(3.24)

Thus, for atmospheric pressure, pa = 1 atm = 101 kPa, the height, hHg , is

hHg =101000 N/m

213550 kg/m

3

9.807 m/sec2 = 760 mm (3.25)

The hydrostatic relation is also the basis of a pressure-measurement device known as the

U-Tube Manometer, an example of which is shown in Figure 3.7. To measure pressure in

a channel filled with water (at Point 4), we attach a U-shaped tube containing mercury. Thetube is open to the atmosphere at Point 1. The hydrostatic relation tells us that the pressure at

Point 2 equals the pressure at Point 1 plus the weight of the column of mercury above. Hence,

p1 + Hggh = p2 (3.26)

Similarly, the pressure at Point 3 is equal to the pressure at Point 4 plus the weight of the

column of water above, i.e.,

p4 + H2OgL = p3 (3.27)

L

h

z = 0

Hg

H2O

1

h2h3h

4h

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Figure 3.7: U-Tube manometer.

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Finally, since z3 = z2, the hydrostatic relation also tells us that p3 = p2. Consequently, wecan equate the left-hand sides of Equations (3.26) and (3.27), which yields the following.

p4 p1 = (Hgh H2OL)g(3.28)

The implication of Equation (3.28) is clear. Since the U-Tube is open to the atmosphere,

then p1 is the pressure of the atmosphere. The density of mercury and water are knownquantities. Thus, by simply measuring L and the height of the mercury column, h, we caninfer the pressure in the water channel.

When a manometry configuration involves layers or columns of different fluids, pressure

can be computed by repeated use of the hydrostatic relation, with one important word of

caution. Since Equation (3.17) follows from Equation (3.15) only when density is constant,

we must stay within the same fluid when applying Equation (3.17) . The following example

illustrates how this is done when several fluids are present.

Example 3.3 Consider the arrangement shown in the figure below, which is designed to monitor

the pressure at the interface of the water and oil in the tank. To make the problem a little moreinteresting, we assume some air has been trapped in the attached manometer. The measured heights

are L1 = 100 mm, L2 = 450 mm and L3 = 400 mm. Determine the pressure difference p1 p0,and how much of a difference ignoring the contribution from the air makes.

L1

L2 L3

Oil

H2O Hg

Air

1g 2g

3g

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Solution. Working first in the column of mercury, we have

p3 = p0 + HggL3

Next, applying the hydrostatic relation in the column of air yields

p3 = p2 + AirgL2

Finally, for the column of water, we have

p1 = p2 + H2OgL1

We thus have three equations for the three unknown pressures,p1, p2 andp3. A little straightforwardalgebra shows that the pressure at the oil-water interface is as follows.

p1 = p0 + (HgL3 + H2OL1

AirL2)g

Using densities corresponding to a temperature of 20o C, we find that the pressure difference, p1p0,is 54.127 kPa. Since L1, L2 and L3 are all of the same order of magnitude, we can neglect thecontribution from the trapped air because Air Hg, H2O . Therefore,

p1 p0 + (HgL3 + H2OL1)g

This equation indicates that p1 p0 is 54.133 kPa|a difference of 0.01% from the value above.

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3.5. BERNOULLI'S EQUATION 91

3.5 Bernoulli's Equation

While there are many important applications in which a fluid is motionless, the \real fun"

begins when a fluid moves. That is, fluids are in motion for any vehicle that moves in the

atmosphere, the ocean or the laboratory. The Earth's atmosphere is in continuous motion,

often bringing cool breezes and sometimes bringing violent storms. Blood flows throughout

your body on a continuous basis. The complicated motions involved are rich with interesting

phenomena that have fascinated mankind for many centuries.

We can reasonably ask if the hydrostatic relation [Equation (3.17)], which is an energy

conservation principle, has any relevance for a fluid that is moving. While the answer is

essentially no (there are simple applications where it does, most notably for constant velocity),

our knowledge of basic physics tells us that total energy is conserved. By simply replacing

the hydrostatic relation with an equation that includes the fluid's kinetic energy, we arrive at

one of the best known equations in fluid-flow theory.

Recall the discussion immediately following Equation (3.17) where we noted that the

hydrostatic relation represents conservation of mechanical energy per unit volume. Thus, to

include kinetic energy, we must determine the appropriate way to represent kinetic energy perunit volume. If and u are the fluid density and velocity, respectively, at a given point, thenthe kinetic energy of a fluid particle with volume V is 1

2Vu u and the corresponding

kinetic energy per unit volume is 12

u u (this quantity is also referred to as the dynamicpressure or dynamic head). Therefore, in the presence of a gravitational field, we replace

the hydrostatic relation by

p +1

2 u u + gz = constant (3.29)

This equation is one of the most famous in fluid mechanics literature and we call i t Bernoulli's

equation. It tells us that the sum of the pressure, p, kinetic energy per unit volume, 12 u u,and potential energy per unit volume, gz, is constant. Clearly, the hydrostatic relation is thelimiting form of Bernoulli's equation for zero velocity.

Although we have arrived at this equation heuristically, it can be rigorously derived. How-

ever, this requires first developing the differential equations governing fluid motion, which wewill do in Chapter 9. When we derive Bernoulli's equation we will find that several conditions

must be satisfied in order for the equation to be valid. There are five conditions and they are

as follows.

1. Viscous effects must be negligible;

2. The flow must not be changing with time, i.e., it must have (u/t = 0);

3. The flow must be incompressible ( = constant);

4. The fluid must be subject only to a conservative forces, i.e., only forces that can be

defined in terms of a force potential (f= V where V is a force potential);

5. The flow must be irrotational, i.e., the fluid velocity must have zero curl (u = 0).

The first four conditions are straightforward and require no further elaboration. The last

condition is more subtle and deserves a bit more discussion. With some analysis, we can show

that the curl of a fluid particle's velocity is proportional to the particle's angular-rotation rate

about its center of gravity. In Chapter 9, we will see explicitly why Bernoulli's equation holds

only in a restricted way when a fluid particle rotates and tumbles as it moves.

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AN EXPERIMENT YOU CAN DO AT HOME

Contributed by Prof. Peter Bradshaw, Stanford University

To observe one of the implications of Bernoulli's equation, you willneed a paper napkin, a thin sheet of paper such as toilet paper, or even

a sheet of notebook paper. Grasp the paper in each hand and hold it up

to your mouth. Blow gently, being careful that you blow only on the

top side. It may take a couple practice tries, but when done correctly,

the paper will rise to a horizontal orientation. You might even want to

try all of the types of paper noted above to observe that you must blow

harder for the more massive sheets.

From Bernoulli's equation, it should be clear that by creating a

moving stream of air over the top of the paper, the pressure decreases.

Thus, you create a suction on the top surface that causes the paper to

rise.

Example 3.4 Consider a large tank with a small hole a distance h below the surface. A jet of fluid issues from the hole with velocity U. You may assume that, as is generally true for thin jets,the surrounding air impresses atmospheric pressure, pa , throughout the jet. Assuming the tankdiameter, D, is very large compared to the diameter of the small hole, d , determine U.

p = pa

z = 0

U

p = pa

h

ho

D

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Solution. Bernoulli's equation tells us that

p +1

2 u u + gz = constant

The tank is open to the atmosphere so that the pressure at the top of the tank is also pa. Since thetank diameter, D, is very large compared to the diameter of the small hole, d, we can ignore thevelocity of the fluid at the top of the tank. Hence, we can use Bernoulli's equation to relate a point

at the top of the tank to a point in the jet to show that

pa + g(h + ho) = pa +1

2 U2

+ gho

Note that we can set the origin, z = 0, anywhere we wish. This is true since the potential energyappears on both sides of this equation, so that only the difference in potential energy between the

surface and the jet, gh, matters. Thus, the velocity in the jet is given by

U =

2gh

8/4/2019 EFM-Ch3

13/52

3.5. BERNOULLI'S EQUATION 93

In Example 3.4, we used Bernoulli's equation to relate conditions at two specified points in

the flow. Often it is helpful to determine the \constant" in Bernoulli's equation by evaluating

each term at a point in the flow where all terms are known. Typically, we seek a point that

lies very far from solid boundaries where the flow is uniform. The most important point

about using Bernoulli's equation in this way is that, because we have selected one universal

reference point, the resulting equation applies at every point in the flow.

The following example illustrates how to implement Bernoulli's equation in this manner.

It involves flow with what we call a free surface, i.e., a flow with an interface between a

liquid and a gas. As in the preceding example, we make use of the fact that the surrounding

air impresses atmospheric pressure, pa, throughout the jet of fluid issuing from the tube.

Example 3.5 Consider water flowing with uniform velocity, U1, as shown in the figure. The waterenters a uniform-diameter tube at some point below the surface. Determine the velocity of the water

leaving the tube at a height z2 above the surface. Also, determine the pressure in the primarystream of fluid.

z

x

Air

Water,

U1

U2

z = 0

z2

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Solution. Clearly, one point where we know everything about the flow is at the free surface very far

upstream. Because of the interface with the air, the pressure must be atmospheric so that p = pa.Letting the free surface lie at z = 0, the potential energy is zero. Since the flow is uniform farupstream, we also know that the velocity is U1. Thus, we have

p +1

2 u u + gz = pa +

1

2 U21

Hence, since p = pa in the jet of fluid issuing from the tube, we find

pa +1

2 U22 + gz2 = pa +

1

2 U21

wherefore

U2 =

U21 2gz2Provided we are not too close to either the tube or the bottom, we expect the flow to be uniform

with velocity u = U1 i. Then, Bernoulli's equation becomes

p +1

2U21 + gz = pa +

1

2U21 = p = pa gz

Thus, in the moving stream, the pressure satisfies the hydrostatic relation. Close to the tube, the

velocity will deviate from the freestream value and, correspondingly, the pressure will depart from

the hydrostatic relation. In a real fluid, viscous effects would result in nonuniform velocity near the

bottom, which would also cause the pressure to differ from the hydrostatic value.

8/4/2019 EFM-Ch3

14/52


3.6 Velocity-Measurement Techniques

We can use Bernoulli's equation to infer velocity from a pressure measurement. This is useful

because pressure is fundamentally easier to measure than velocity. To understand how thisis done, we must first introduce the notion of a stagnation point. Then, we discuss two

measurement devices known as the Pitot tube and the Pitot-static tube.

3.6.1 Stagnation Points

Figure 3.8 illustrates ideal (i.e., frictionless) two-dimensional flow past a cylinder. The figure

includes several contours that we call streamlines (we will discuss streamlines in detail in

Section 4.4). These are contours that are everywhere parallel to the flow velocity, which

means fluid particles move along the streamlines. Since the contours are parallel to the

velocity there is no flow across (normal to) streamlines. Because there is no flow across a

solid boundary, the cylinder surface is a streamline. We call the streamline coincident with

the x axis upstream and downstream of the cylinder the dividing streamline. This streamline

splits at the front of the cylinder so that half of the fluid moves over the cylinder and halfmoves below. The streamlines rejoin at the back of the cylinder.

x, u

y, v

StagnationPoints,

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Figure 3.8: Ideal flow past a cylinder.

We have a very special situation at these two points. Specifically, we have the intersection

of two perpendicular streamlines, namely, the dividing streamline and the cylinder surface.

Now, since streamlines are parallel to the velocity, necessarily the vertical velocity component,

v, on the dividing streamline must be zero approaching the cylinder. Also, very close to thefront of the cylinder we must have zero horizontal velocity, u, on the surface (see Figure 3.8)in order to have flow tangent to the surface. Thus, at the point where the streamlines collide,

u = 0 (Stagnation Point) (3.30)

So, we have two points on the cylinder where the velocity vanishes, and we refer to these

points as stagnation points.

3.6.2 Pitot Tube

The Pitot tube is one of the simplest devices based on Bernoulli's equation that provides

an indirect measurement of velocity. Figure 3.9 illustrates flow in the immediate vicinity

of a Pitot tube placed a distance d below the surface in a flowing stream of water. For

8/4/2019 EFM-Ch3

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3.6. VELOCITY-MEASUREMENT TECHNIQUES 95

precise measurements, the tube must have a very small diameter such as that characteristic of

a hypodermic needle. As shown, the water fills the Pitot tube up to a point a distance h abovethe free surface. Because the fluid in the tube cannot move, there must be a stagnation point at

the tube's entrance below the surface. As we will now show, this device permits determining

the velocity by a simple measurement of the distance the fluid rises above the surface. It is

thus the analog, for moving fluids, of the U-Tube manometer discussed in Section 3.4.

Air

Water,

U1

Stagnation Point

h

d

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