Fluid MechanicsChapter 1 – EUects of pressure

last edited September 10, 2017

1.1 Motivation 291.2 Concept of pressure 29

1.2.1 The direction of pressure 291.2.2 Pressure on an inVnitesimal volume 30

1.3 Pressure in static Wuids 311.3.1 Fluid statics 311.3.2 Shear in static Wuids 311.3.3 Pressure and depth 311.3.4 Pressure distribution 331.3.5 Hydrostatic pressure 331.3.6 Atmospheric pressure 34

1.4 Wall pressure forces and buoyancy 351.4.1 Pressure forces on plane surfaces 351.4.2 Buoyancy and Archimedes’ principle 37Buoyancy and Archimedes’ principle 37

1.5 Exercises 39

These lecture notes are based on textbooks by White [13], Çengel & al.[16], and Munson & al.[18].

1.1 Motivation

Video: pre-lecture brieVng forthis chapter

by o.c. (CC-by)https://youtu.be/4MVn1PoxGqY

In Wuid mechanics, only three types of forces apply to Wuid particles: forcesdue to gravity, pressure, and shear. This chapter focuses on pressure (we willaddress shear in chapter 2), and should allow us to answer two questions:

• How is the eUect of pressure described and quantiVed?

• What are the pressure forces generated on walls by static Wuids?

1.2 Concept of pressure

We approached the concept of pressure in the introductory chapter with thenotion that it represented force perpendicular to a given Wat surface (eq. 0/15),for example a Wat plate of area A:

p ≡F⊥A

(1/1)

To appreciate the concept of pressure in Wuid mechanics, we need to gobeyond this equation.

1.2.1 The direction of pressure

An important concept is that in continuum mechanics, the Wat surface isimaginary. More precisely, a Wuid is able to exert pressure on not only onsolid surfaces, but also upon and within itself. In this context, we need to 29

rework eq. 1/1 so that now pressure is deVned as perpendicular force perarea on an inVnitesimally small surface of Wuid:

p ≡ limA→0

F⊥A

(1/2)

Equation 1/2 may appear unsettling at Vrst sight, because as A tends to zero,F⊥ also tends to zero; nevertheless, in any continuous medium, the ratio ofthese two terms tends to a single non-zero value.

This brings us to the second particularity of pressure in Wuids: the pressureon either side of the inVnitesimal Wat surface is the same regardless of itsorientation. In other words, pressure has no direction: there is only one (scalar)value for pressure at any one point in space.

Thus, in a Wuid, pressure applies not merely on the solid surfaces of itscontainer, but also everywhere within itself. We need to think of pressure asa scalar property Veld p(x ,y,z,t ) .

1.2.2 Pressure on an inVnitesimal volume

While pressure has no direction, it may not have the same value everywherein a Wuid, and so the gradient (the space rate of change) of pressure may notbe null. For example, in a static water pool, pressure is uniform in the twohorizontal directions, but it increases along with depth.

Instead of a Wat plate, let us now consider an inVnitesimally small cube withinthe Wuid (Vg. 1.1). Because the cube is placed in a scalar Veld, the pressureexerting on each of its six faces may be diUerent. The net eUect of pressurewill therefore have three components: one for each pair of opposing faces.Further down in this chapter, we will express this eUect with the help of thegradient of pressure ~∇p, a vector Veld. For now, it is enough to notice thatpressure itself is quantiVed as a (one-dimensional) scalar Veld, and the eUectof pressure on a submerged volume as a (three-dimensional) vector Veld.

Figure 1.1 – The pressure on each face of an inVnitesimal volume may have adiUerent value. The net eUect of pressure will depend on how the pressure varies inspace. These changes are labeled dp��i in each of the i = x ,y,z directions.

Vgure CC-0 o.c.

30

1.3 Pressure in static Wuids

1.3.1 Fluid statics

Fluid statics is the study of Wuids at rest, i.e. whose velocity Veld ~V iseverywhere null and constant:

~V = ~0∂~V∂t = ~0

(1/3)

We choose to study this type of problem now, because it makes for a con-ceptually and mathematically simple case with which we can start analyzingWuid Wow and calculate Wuid-induced forces.

1.3.2 Shear in static Wuids

We will see in chapter 2 that shear eUorts can be expressed as a functionof viscosity and velocity: ~τ = µ ∂ui∂x j

~i (2/16). We will also see that for mostordinary Wuids (termed Newtonian Wuids) the viscosity µ is independent fromthe velocity distribution.

For now, it is enough to accept that in a static Wuid (one in which ∂ui∂x j = 0

since ~V(x ,y,z,t ) = ~0), there cannot be any shear ~τ . Static Wuids are unable togenerate shear eUort.

This is a critical diUerence between solids and Wuids. For example:

• A static horizontal beam extruding from a wall carries its own weightbecause it is able to support internal shear; but this is impossible for astatic Wuid.

• A solid brick resting on the ground exerts pressure on its bottom face,but because of internal shear, the pressure on its sides is zero. On thecontrary, a static body of water of the same size and weight cannotexert internal shear. Because pressure has no direction, the body exertspressure not only on the bottom face, but also laterally on the sides ofits container.

1.3.3 Pressure and depth

We now wish to express pressure within a Wuid as a function of depth. Let usconsider, inside a static Wuid reservoir, a small element of Wuid. We choosean element with Vnite horizontal surface S but inVnitesimal height dz, asshown in Vg. 1.2.

The element’s geometry is such that all of the lateral forces (which are due topressure only) compensate each other. In the vertical direction, we are leftwith three forces, which cancel out because the element is not accelerating:

~Fp bottom + ~Fp top +m~д = ~0

pbottomS − ptopS −mд = 0

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Figure 1.2 – A Wuid element of (Vnite) horizontal surface S and inVnitesimal height dzwithin a static Wuid.

Vgure CC-0 o.c.

If we now rewrite the bottom pressure as pbottom = ptop + dp, we obtain:

(ptop + dp)S − ptopS −mд = 0

dp =m

Sд (1/4)

This equation 1/4 expresses the fact that the inVnitesimal variation of pres-sure dp between the top and bottom surfaces of the element is due only to itsown weight. If we evaluate this variation along the inVnitesimal height dz,we obtain:

dpdz=

m

S dzд =

m

dVд

dpdz= ρд (1/5)

Thus, equation 1/5, which is obtained from Newton’s second law applied toa static Wuid, links the vertical variation in pressure to the Wuid’s density ρand the local gravitational acceleration д. This equation is so useful that it issometimes called “the fundamental equation of Wuid statics”.

If our coordinate system is not aligned with the vertical direction (and thuswith gravity), we will have to adapt equation 1/5 and obtain a more generalrelationship:

∂p∂x = ρдx∂p∂y = ρдy∂p∂z = ρдz

(1/6)

∂p

∂x~i +∂p

∂y~j +∂p

∂z~k = ρ~д (1/7)

We can simplify the writing of this last equation by using the mathematicaloperator gradient (see also Appendix A2 p.217), deVned as so:

~∇ ≡ ~i∂

∂x+ ~j∂

∂y+ ~k∂

∂z(1/8)

32

And thus, equation 1/7 is more elegantly re-written as:

~∇p = ρ~д (1/9)

This equation is read “the gradient of pressure is equal to the density timesthe gravitational acceleration vector”. It expresses concisely a very importantidea: in a static Wuid, the spatial variation of pressure is due solely tothe Wuid’s own weight.

In chapter 4, we shall see that eq. 1/9 is a special case in a much more potentand complex relation called the Navier-Stokes equation (eq. 4/37 p.95). Fornow, it suXce for us to work with just pressure and gravity.

1.3.4 Pressure distribution

An immediate consequence stemming from equation 1/9 is that in a staticWuid (e.g. in a glass of water, in a swimming pool, in a calm atmosphere),pressure depends solely on height. Within a static Wuid, at a certain altitude,we will measure the same pressure regardless of the surroundings (Vg. 1.3).

Figure 1.3 – Pressure at a given depth (or height) in a static Wuid does not depend onthe environment. Here, as long as the Wuid remains static, pA = pB = pC = pD.

Figure CC-0 o.c.

1.3.5 Hydrostatic pressure

How is pressure distributed within static liquid water bodies? The densityof liquid water is approximately constant: ρwater = 1 000 kgm−3. In a waterreservoir, if we align the z-direction vertically downwards, equation 1/9becomes:

~∇p = ρ~дdpdz= ρд(

dpdz

)water

= ρwater д(dpdz

)water

= 1 000 × 9,81 = 9,81 · 103 Pam−1 = 9,81 · 10−2 barm−1

(1/10)

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Therefore, in static water, pressure increases by approximately 0,1 bar/mas depth increases. For example, at a depth of 3m, the pressure will beapproximately 1,3 bar (which is the atmospheric pressure plus ∆z × dp

dz ).

1.3.6 Atmospheric pressure

The density ρair of atmospheric air is not uniform, since —unlike for water—it depends strongly on temperature and pressure. If we model atmosphericair as a perfect gas, once again orienting z vertically downwards, we canexpress the pressure gradient as:(

dpdz

)atm.= ρair д = p

1T

д

R(1/11)

This time, the space variation of pressure depends on pressure itself (and it isproportional to it). A quick numerical investigation for ambient temperatureand pressure (1 bar, 15 ◦C) yields:(

dpdz

)atm. ambient

= 1 · 105 ×1

288,15×9,81287

= 11,86 Pam−1 = 1,186 · 10−4 barm−1

This rate (approximately 0,1mbar/m) is almost a thousand times smaller thanthat of water (Vg. 1.4).

Since the rate of pressure change depends on pressure, it also varies withaltitude, and the calculation of pressure diUerences in the atmosphere is alittle more complicated than for water.

If we focus on a moderate height change, it may be reasonable to considerthat temperature T , the gravitational acceleration д and the parameter R (gasconstant) are uniform. In this admittedly restrictive case, equation 1/11 can

Figure 1.4 – Variation of pressure as a function of altitude for water and air at thesurface of a water reservoir. The gradient of pressure with respect to altitude isalmost a thousand times larger in water than in air.

Figure CC-0 o.c.

34

be integrated as so:

dpdz=

д

RTcst.p∫ 2

1

1pdp =

д

RTcst.

∫ 2

1dz

lnp2p1=

д

RTcst.∆z

p2p1= exp

[д∆z

RTcst.

](1/12)

Of course, air temperature varies signiVcantly within the atmosphere (atmoderate altitudes the change with altitude is approximately −6 K km−1).Adapting equation 1/12 for a uniform temperature gradient (instead of uni-form temperature) is the subject of exercise 1.8 p.43.

In practice, the atmosphere also sees large lateral pressure gradients (whichare strongly related to the wind) and its internal Wuid mechanics are com-plex and fascinating. Equation 1/12 is a useful and convenient model, butreVnements must be made if precise results are to be obtained.

1.4 Wall pressure forces and buoyancy

1.4.1 Pressure forces on plane surfaces

Video: when pressure-inducedforces in static Wuids matter: 24hours of heavy tonnage transitthrough the MiraWores locks inPanama. Can you quantify theforce applying on a single lockdoor?

by Y:XyliboxFrance (styl)https://youtu.be/ke4Rf9FrHnw

When the pressure p exerted on a Wat surface of area S is uniform, theresulting force F is easily easily calculated (F = pcst.S). In the more generalcase of a three-dimensional object immersed in a Wuid with non-uniformpressure, the force must be expressed as a vector and obtained by integration:

~Fpressure =

∫Sd~F =

∫Sp ~n dS (1/13)

where the S-integral denotes an integration over the entire surface;and ~n is a unit vector describing, on each inVnitesimal surface element dS , thedirection normal to the surface.

Equation 1/13 is easily implemented in software algorithms to obtain numeri-cally, for example, the pressure force resulting from Wuid Wow around a bodysuch as an aircraft wing. In our academic study of Wuid mechanics, however,we will restrict ourselves to the simpler cases where the surface is perfectlyWat (and thus ~n is everywhere the same). Equation 1/13 then becomes:

Fpressure =

∫SdF =

∫Sp dS (1/14)

for a Wat surface.

What is required to calculate the scalar F in eq. 1/14 is an expression of pas a function of S . In a static Wuid, this expression will be given by eq. 1/9.Typically in two dimensions x and y we re-write dS as dS = dx dy and wemay then proceed with the calculation starting from

Fpressure =

"p(x ,y) dx dy (1/15)

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Such a calculation gives us the magnitude of the pressure force, but not itsposition. This position can be evaluated by calculating the magnitude ofmoment generated by the pressure forces about any chosen point X. Thismoment ~MX, using notation shown in Vg. 1.5, is expressed as:

~MX =

∫Sd ~MX =

∫S~rXF ∧ d~F =

∫S~rXF ∧ p ~n dS (1/16)

where ~rXF is a vector expressing the position of each inVnitesimal surface relative topoint X.

Much like eq. 1/13 above, this eq. 1/16 is easily implemented in softwarealgorithms but not very approachable on paper. In the simple case where westudy only a Wat surface, and where the reference point X is in the same planeas the surface, eq. 1/16 simpliVes greatly and we can calculate the magnitudeMX as:

MX =

∫SdMX =

∫SrXF dF =

∫SrXF p dS (1/17)

for a Wat surface, with X in the plane of the surface.

Once both Fpressure and MX pressure have been quantiVed, the distance RXF

between point X and the application point of the net pressure force is easilycomputed:

RXF =MX pressure

Fpressure(1/18)

While we restrict ourselves to simple cases where surfaces are Wat, theprocess above outlines the method used by computational Wuid dynamics(cfd) software packages to determine the position of forces resulting fromWuid pressure on any arbitrary solid object.

Figure 1.5 – Moment generated about an arbitrary point X by the pressure exertedon an arbitrary surface (left: perspective view; right: side view). The vector ~n is aconvention unit vector everywhere perpendicular to the inVnitesimal surface dSconsidered.

Figure CC-0 o.c.

36

1.4.2 Buoyancy and Archimedes’ principle

Video: playing around with anair pump and a vacuum cham-ber

by Y:Roobert33 (styl)https://youtu.be/ViuQKqUQ1U8

Any solid body immersed within a Wuid is subjected to pressure on its walls.When the pressure is not uniform (for example because the Wuid is subjectedto gravity, although this may not be the only cause), then the net force dueto Wuid pressure on the body walls will be non-zero.

When the Wuid is purely static, this net pressure force is called buoyancy.Since in this case, the only cause for the pressure gradient is gravity, thenet pressure force is oriented upwards. The buoyancy force is completelyindependent from (and may or may not compensate) the object’s weight.

Since it comes from equation 1/9 that the variation of pressure within a Wuidis caused solely by the Wuid’s weight, we can see that the force exerted onan immersed body is equal to the weight of the Wuid it replaces (that is tosay, the weight of the Wuid that would occupy its own volume were it notthere). This relationship is sometimes named Archimedes’ principle. Theforce which results from the static pressure gradient applies to all immersedbodies: a submarine in an ocean, an object in a pressurized container, andof course, the reader of this document as presently immersed in the earth’satmosphere.

Figure 1.6 – Immersion in a static Wuid results in forces that depend on the body’s vol-ume. They can evidenced by the removal of the Wuid (for example in a depressurizedsemi-spherical vessel).

Figure CC-0 o.c.

37

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