skmm 2313 mechanics of fluids 1 - fkm.utm.myabuhasan/content/teaching/skmm2313/... · skmm 2313...

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School of Mechanical EngineeringMechanics of Fluids & Engineering Computing

SKMM 2313 Mechanics of Fluids 1Pressure Distribution in a Fluid

« An excerpt (mostly) from White (2011) »

Abu Hasan ‘ABDULLAH

September 2018

ibn.abdullah@ddress c b n a 2018 SKMM 2313 Mechanics of Fluids 1 Pressure Distribution 0 / 40

Outline

1 Pressure and Pressure Gradient

2 Pressure Force on a Fluid Element

3 Equilibrium of a Fluid Element

4 Hydrostatic Pressure Distributions

5 Application to Manometry

6 Hydrostatic Forces on Plane Surfaces

7 Hydrostatic Forces in Layered Fluids

8 Buoyancy and Stability

9 Pressure Distribution in Rigid-Body Motion

10 Bibliography


Pressure and Pressure Gradient

Normal stress on any plane through afluid element at rest is equal to aunique value called the fluid pressure p,taken positive for compression bycommon convention.

Consider a small wedge of fluid at restof size ∆x by ∆z by ∆s and depth binto the paper. There is no shear bydefinition, but we postulate that thepressures px, pz, and pn may bedifferent on each face. Figure 1: Equilibrium of a small wedge of fluid

at rest.

Summation of forces must equal zero (no acceleration) in both the x and zdirections.∑

Fx = 0 = pxb∆z− pnb∆s sin θ∑Fz = 0 = pzb∆x − pnb∆s cos θ − 1

2γb∆x∆z(1)



The geometry of the wedge is such that

∆s sin θ = ∆z ∆s cos θ = ∆x (2)

and substitution into Eq.(1) and rearrangement give

px = pn pz = pn + 12γ∆z (3)

These relations illustrate two important principles of the hydrostatic, or shear-free,condition:

1 there is no pressure change in the horizontal direction, and2 there is a vertical change in pressure proportional to the density, gravity, and depth

change.

In the limit as the fluid wedge shrinks to a “point” z→ 0 and Eqs. (3) become

px = pz = pn = p (4)

Since θ is arbitrary, we conclude that the pressure p at a point in a static fluid isindependent of orientation.



In the case of a moving fluid, the pressure is defined as the average of the threenormal stresses σii on the element

p = − 13 (σxx + σyy + σzz) (5)

Minus sign on LHS because a compression stress is taken as −ve whereas p is +ve.

The great majority of viscous flows have negligible viscous normal stresses, soEq.(5) is rarely needed.


Pressure Force on a Fluid Element

Pressure causes no net force on a fluidelement unless it varies with space.

Figure 2 shows pressure acting on thetwo x faces of the element:

p = p(x, y, z, t) (6)

Net force in the x-direction is given by

dFx = pdydz−(

p +∂p∂x

dx)

dy dz

= −∂p∂x

dx dy dz

(7)

Figure 2: Net x force on an element due topressure variation.

In like manner the net force dFy involves −∂p/∂y, and the net force dFz concerns−∂p/∂z.


Pressure Force on a Fluid Element

Total net-force vector on the element due to pressure is

dFpress =

(−i∂p∂x− j

∂p∂y− k

∂p∂z

)dx dy dz (8)

Denoting f as the net force per unit element volume, Eq.(8) can be rewritten as

fpress =dFpress

dx dy dz= −

(i∂

∂x+ j

∂

∂y+ k

∂

∂z

)p = −∇p (9)

Thus it is not the pressure p but the pressure gradient ∇p (which is a surface force)that is causing a net force which must be balanced by

body forces (e.g. gravity, e.m.f.),surface forces (e.g. viscous stresses gradients), orsome other effect in the fluid.


Equilibrium of a Fluid Element

Body forces due to electromagnet or gravity could be acting on the entire mass of theelement. The gravity force (i.e. weight) of the element is

dFgrav = ρg dx dy dz or fgrav = ρg (10)

where g is acceleration of gravity; on earth g ≈ 9.807 m/s2.

Surface force due to the gradient of the viscous stresses is assumed to be present,given below without derivation,

fvisc = µ

(∂2V∂x2 +

∂2V∂y2 +

∂2V∂z2

)= µ∇2V (11)

The total vector resultant of these three forces—pressure, gravity, and viscousstress—must either keep the element in equilibrium or cause it to move withacceleration a. From Newton’s law, we have∑

f = fpress + fgrav + fvisc = −∇p + ρg + µ∇2V = ρa (12)


Equilibrium of a Fluid Element

In this chapter, V and a are known, we need to solve for p. See next slide fornomenclature of the pressure terminology.

Note: In later chapters, V, a and p are all unknown.

We rewrite Eq.(12) as

∇p = ρ(g− a) + µ∇2V = B(x, y, z, t) (13)

where B is a short notation for the vector sum on the RHS.If V and a = dV/dt, time, density and viscosity are known, we can solve Eq.(13) forp(x, y, z, t) by direct integration.

We could use Eq.(13) in at least four special cases:1 Flow at rest or at constant velocity—Section 2.3 [White (2002)].2 Rigid-body translation and rotation—Section 2.9 [White (2002)].3 Irrotational motion (∇× V ≡ 0)—Section 4.9 [White (2002)].4 Arbitrary viscous motion—Section 6.4 [White (2002)].

Cases 1 and 2 will be treated in this chapter.


Equilibrium of a Fluid ElementAbsolute, Atmospheric, Gauge and Vacuum Pressures

NomenclatureAbsolute Pressure ( pabs or p) The actual pressure at a given position is called the absolutepressure, and it is measured relative to absolute vacuum (i.e., absolute zero pressure).Gauge Pressure ( pgage) Gage pressure is the pressure relative to the atmosphericpressure. In other words, how much above or below is the pressure with respect to theatmospheric pressure.Vacuum Pressure ( pvac) Pressures below atmospheric pressure are called vacuumpressures and are measured by vacuum gauges that indicate the difference between theatmospheric pressure and the absolute pressure.Atmospheric Pressure ( patm) The atmospheric pressure is the pressure that an areaexperiences due to the force exerted by the atmosphere, easily measured using mercurybarometer. For engineering calculations patm used is the pressure at sea level, i.e.patm = 101350 Pa.

These four pressures are relatedthrough

pgage = pabs − patm

pvac = patm − pabs

pabs = patm + pgage Figure 3: Absolute, gauge, and vacuumpressure readings.


Hydrostatic Pressure Distributions

If the fluid is at rest or at constant velocity, a = 0 and fvisc = 0. Eq.(12) for thepressure distribution reduces to

∇p = ρg (14)

This is hydrostatic distribution and is correct for all fluids at rest, regardless of theirviscosity, because the viscous term vanishes.

A fluid in hydrostatic equilibrium will align its constant-pressure surfaceseverywhere normal to the local-gravity vector:

g = −gk (15)

where g = 9.807 m/s2. The maximum pressure increase will be in the direction ofgravity i.e. “down”.

For these coordinates Eq.(14) has the components

∂p∂x

= 0∂p∂y

= 0∂p∂z

= −ρg = −γ (16)


Hydrostatic Pressure Distributions

Eq.(16) shows that p is independent of x and y. Hence the partial derivative ∂p/∂zcan be replaced by the total derivative dp/dz, and the hydrostatic condition reducesto

dpdz

= −γ =⇒ p2 − p1 = −∫ 2

1γ dz (17)

which is the solution to the hydrostatic problem.

In Figure 4points a, b, c, and d are at equaldepths in water and therefore haveidentical pressures.Points A, B, and C are also at equaldepths in water and have identicalpressures higher than a, b, c, and d.Point D has a different pressure fromA, B, and C because it is not connectedto them by a water path. Figure 4: Hydrostatic-pressure distribution.


Hydrostatic Pressure DistributionsHydrostatic Pressure in Liquids

Since density is constant in liquid hydrostatic calculations, Eq.(17) integrates to

p2 − p1 = −γ(z2 − z1) or z2 − z1 = −p2

γ− p1

γ(18)

with the first form used in most problems; γ is the specific weight of the fluid andp/γ is the pressure head of the fluid whose SI unit is m.

Exercise: Work through Example 2.1 (White, 2011, p. 71).


Hydrostatic Pressure DistributionsMercury Barometer

Mercury (Hg) barometer, Figure 5,measures atmospheric pressure. Ituses Eq.(18).

It’s a tube filled with Hg andinverted while submerged in areservoir. Hg has extremely smallvapour pressure at roomtemperatures (0.16 Pa at 20◦C); thiscauses a near vacuum in the closedupper end.

Atmospheric pressure pa forces amercury column to rise a distance hinto the tube, the upper mercurysurface is at p1 ≈ 0.

Figure 5: Mercury barometer.

With p1 = 0 at z1 = h and p2 = pa at z2 = 0, Eq.(18) reduces to

pa − 0 = −γHg(0− h) or h =pa

γHg(19)


Hydrostatic Pressure DistributionsHydrostatic Pressure in Gases

Gases are compressible, with ρ nearly proportional to p. Thus ρ must be consideredas a variable in Eq.(17) if the integration carries over large p changes.We introduce the perfect-gas law p = ρRT in Eq.(17)

dpdz

= −γ = −ρg = − pRT

g

then separate the variables and integrate between points 1 and 2:∫ 2

1

dpp

= lnp2

p1= − g

R

∫ 2

1

dzT

(20)

The integral over z requires an assumption about the temperature variation T(z).Case 1: We assume isothermal atmosphere

T = T0 (21)

and integration of Eq.(20) lead to

p2 = p1 exp[−g(z2 − z1

RT0

](22)



Case 2: We assume earth’s mean atmospheric temperature drops off linearly with zwithin the troposphere (0 < z ≤ 11000 m)

T ≈ T0 − Bz (23)

Introducing Eq.(23) into Eq.(20)

lnp2

p1= − g

R

∫ 2

1

dzT0 − Bz

(24)

and integrating yields

p = pa

(1− Bz

T0

)g/(RB)

and ρ = ρ0

(1− Bz

T0

)g/(RB)−1

(25)

where g/RB = 5.26 (air), ρ0 = 1.2255 kg/m3, p0 = 101350 Pa.

By international agreement these standard values are assumed to apply:T0 = 288.15 K= 15◦ C and B = 0.00650 K/m.



The linear approximation from Eq.(18)

δp = −ρgδz =⇒ p2 − p1 = −γ(z2 − z1)

is satisfactory for liquids, which are nearly incompressible. For gases, it isinaccurate unless z is very small.

Note: Calculation of p through binomial expansion of Eq.(18) incurs anerror of less than 1% if z < 200 m.


Application to ManometryColumn of Multiple Fluids

A change in elevation z2 − z1 of aliquid is equivalent to a change inpressure (p2 − p1)/γ. Thus a staticcolumn of one or more liquids or gasescan be used to measure pressuredifferences between two points. Such adevice is called a manometer.

If multiple fluids are used, the densityin the formula is changed as we movefrom one fluid to another. Figure 6illustrates the use of the formula withcolumn of multiple fluids.

Figure 6: Pressure changes through a columnof multiple fluids.

Total change p5 − p1 is addition of successive changes p2 − p1, p3 − p2, p4 − p3 andp5 − p4.

p5 − p1 = −γo(z2 − z1)− γw(z3 − z2)− γG(z4 − z3)− γM(z5 − pz) (26)


Application to ManometryA Simple Manometer

Figure 7 shows a simple U-tube openmanometer that measures the gaugepressure pA relative to the atmosphere,pa.

The chamber fluid ρ1 is separated fromthe atmosphere by a second, heavierfluid ρ2.

We first apply the hydrostatic formula,Eq.(18) from A down to z1.

Figure 7: Simple open manometer.

Note: Note that we can then go down to the bottom of the U-tube and backup on the right side to z1 , and the pressure will be the same, p = p1.

Instead, we “jump across” and then up to level z2:

pA + γ1|zA − z1| − γ2|z1 − z2| = p2 = pa (27)



Application to ManometryPascal’s Law

Another physical reason that we can “jump across” at section 1 is due to thePascal’s law or the principle of transmission of fluid-pressure.

Pascal’s law Any two points at the same elevation in a continuous mass ofthe same static fluid will be at the same pressure.

This law is applied in the hydraulic lift,Figure 8, to multiply the force, whichfor the two pistons implies

p1 = p2

This allows the lifting of a heavy loadwith a small force, as in an autohydraulic lift, but of course there canbe no multiplication of work, W, so inan ideal case with no frictional loss:

W1 = W2

See http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html

Figure 8: Hydraulic lift.


Application to ManometryMultiple-fluid Manometer

Figure 9 shows a multiple-fluid manometer for finding the difference in pressurebetween two chambers A and B.

Figure 9: Multiple-fluid manometer.

We compute four pressure differences while making three jumps:

pA − pB = (pA − p1) + (p1 − p2) + (p2 − p3) + (p3 − pB)

= −γ1(zA − z1)− γ2(z1 − z2)− γ3(z2 − z3)− γ4(z3 − zB)(28)



Hydrostatic Forces on Plane Surfaces

Many design of containment structures requires computation of hydrostatic forces onsolid surfaces adjacent to the fluid; these forces relate to the weight of fluid bearingon the surface.

Figure 10 shows a plane panel ofarbitrary shape completelysubmerged in a liquid making anarbitrary angle θ with the horizontalfree surface so that depth variesover the panel surface. If h is depthto elemental area dA of the plate,Eq.(18) yields pressure p = pa + γh.

To derive formulae involving theplate shape we need to establish

an xy coordinate system in theplane of the plate with the origin atits centroid,a dummy coordinate ξ down fromthe surface in the plane of the plate.

Figure 10: Hydrostatic force and centre ofpressure on an arbitrary plane.



Notice that in Figure 10, depth h = ξ sin θ and slant distance from the surface tothe centroid of elemental area dA is

ξCG =1A

∫ξ dA

The total hydrostatic force on one side of the plate is given by

F =

∫p dA =

∫(pa + γh)dA = paA + γ

∫h dA (29)

Since θ is constant along the plate, Eq.(29) becomes

F = paA + γ sin θ∫ξ dA = paA + γ sin θ ξCG A (30)

Straight down from the surface to the plate centroid is depth hCG = ξCG sin θ, thus

F = paA + γhCG A = (pa + γhCG)A = pCGA (31)



Eq.(31) can be visualized physically inFigure 11 as the resultant of a linearstress distribution over the plate area.

This simulates combined compressionand bending of a beam of the samecross section.

To balance the bending-momentportion of the stress, the resultantforce F acts NOT through the centroidbut through the centre of pressure CPtoward the high-pressure side.

Figure 11: Hydrostatic pressure on a planesurface.

To find the coordinates (xCP, yCP), we sum moments of the elemental force pdAabout the centroid and equate to the moment of the resultant F.



To compute yCP, we equate

F yCP =

∫yp dA =

∫y(pa + γξ sin θ)dA = γ sin θ

∫yξdA

The term∫

paydA vanishes by definition of centroidal axes and introducingξ = ξCG − y, we obtain

F yCP = γ sin θ(ξ

∫ydA−

∫y2dA

)= −γ sin θIxx

where again∫

ydA = 0 and Ixx is the area moment of inertia of the plate area aboutits centroidal x axis, computed in the plane of the plate.

Substituting for F from Eq.(31) results in

yCP = −γ sin θIxx

pCGA(32)



The determination of xCP is exactly similar:

F xCP =

∫xp dA =

∫x[pa + γ(ξCG − y) sin θ] dA = −γ sin θ

∫xy dA = −γ sin θIxy

Substituting for F from Eq.(31) results in

xCP = −γ sin θIxy

pCGA(33)

For positive Ixy, xCP is negative because the dominant pressure force acts in thethird, or lower left, quadrant of the panel. If Ixy = 0, usually implying symmetry,xCP = 0 and the center of pressure lies directly below the centroid on the y axis.



Hydrostatic Forces on Plane SurfacesArea Moment of Inertia & Product Moment of Area

The area moment of inertia is a property of a two-dimensional plane shape whichcharacterizes its deflection under loading, a.k.a. second moment of area or secondmoment of inertia. It has dimensions of L4.

Not to be confused with the usual (mass) moment of inertia (which characterizesthe angular acceleration undergone by a solids when subjected to a torque and hasdimensions of ML2)

The area moment of inertia about the x-axis is defined by

Ix = Ixx =

∫y2 dx dy =

∫y2 dA

. . . while more generally, the product moment of area a.k.a. product of inertia isdefined by

Ixy =

∫x y dx dy =

∫x y dA

Ixy of an area symmetrical about an axis will cancel out each other and becomeszero, and the integral will reduces to zero also.


Hydrostatic Forces on Plane SurfacesArea Moment of Inertia & Product Moment of Area

Figure 12: Ixx and Ixy for: (a) rectangle, (b) circle, (c) triangle, and (d) semicircle.


Hydrostatic Forces on Curved Surfaces

The resultant pressure force on a curved surface is most easily computed byseparating it into horizontal and vertical components.

Figure 13 shows a curved surface and a free-body diagram of the column of fluidcontained in the vertical projection above it.

Figure 13: Hydrostatic force on a curved surface: (a) submerged curved surface; (b)free-body diagram of fluid above the curved surface.

Forces FH and FV are exerted by the surface on the fluid column.


Hydrostatic Forces on Curved Surfaces

On the upper part of the column bcde, the horizontal components F1 exactlybalance and are not relevant to the discussion.

F1,vertical side cd = F1,vertical side be

On the lower, irregular portion of fluid abc, summation of horizontal forces showsthat force FH due to the curved surface is exactly equal to force FH on the verticalleft side of the fluid column, which can be computed using Eq.(9), based on avertical projection of the area of the curved surface.

FH,vertical side ac = FH,curved side ab

Summation of vertical forces on the fluid free body is

FV = W1 + W2 + Wair (34)

Thus the calculation of FV involves little more than finding centres of mass of acolumn of fluid.

Exercise: Work through Examples 2.8 & 2.9 (White, 2011, pp.87–89).


Hydrostatic Forces in Layered Fluids

If the fluid is layered with different densities, as in Figure 14, a single formulacannot solve the problem because the slope of the linear pressure distributionchanges between layers, i.e.

(total force on the plate)6=(pressure at the centroid)× (plate area)

Different formula applies separately toeach layer, and the remedy is tocompute and sum the separate layerforces and moments.

F =∑

Fi =∑

pCG,i Ai (35)

Figure 14: Hydrostatic forces on a surfaceimmersed in a layered fluid.


Hydrostatic Forces in Layered Fluids

The centroid of the plate portion in each layer can be used to locate the centre ofpressure on that portion:

yCP,i = −ρig sin θi Ixx,i

pCG Aiand xCP,i = −ρig sin θi Ixy,i

pCG Ai(36)

Exercise: Work through Example 2.10 (White, 2011, p.90).


Buoyancy and StabilityArchimedes’ Principle

Archimedes’ Laws of Buoyancy:1 a body immersed in a fluid experiences a vertical buoyant force equal to the weight of the

fluid it displaces,2 a floating body displaces its own weight in the fluid in which it floats,

Figure 15: Two different approaches to the buoyant force on an arbitrary immersed body.

are shown in Figure 15—the body lies between upper curved surface 1 & lowercurved surface 2.


Buoyancy and StabilityBuoyant Force

There two ways to calculate buoyant force FB acting on the body.Method 1: Vertically, the body experiences a net upward force

FB = FV(2)− FV(1) = (fluid weight above 2)− (fluid weight above 1)

= weight of fluid equivalent to body volume(37)

Method 2: We sum the vertical forces on elemental vertical slices through theimmersed body, assuming fluid has uniform specific weight:

FB =

∫body

(p2 − p1) dAH = −γ∫

(z2 − z1) dAH = (γ)(body volume) (38)

Buoyant force FB passes through the centre of volume of the displaced body, i.e. itscentre of mass computed as if it had uniform density. This point through which FB

acts is called the centre of buoyancy.Eqs.(38) and (39) are identical results and equivalent to Archimedes’ law 1.Eq.(38) can be generalized to a layered fluid (LF) by summing the weights of eachlayer of density ρi displaced by the immersed body:

(FB)LF =∑

ρig (displaced volume)i (39)


Buoyancy and StabilityFloating Bodies

Floating bodies are a special case—a portion of the body is submerged, theremainder poking up out of the free surface—see Figure 16.

Eq.(38) is modified to apply to the submerged volume:

FB = (γ)(displaced volume) = floating-body weight = W (40)

FB and W are collinear since there can be no net moments for static equilibrium.

Figure 16: Static equilibrium of a floating body.

Eq.(40) is the mathematical equivalent of Archimedes’ law 2.

Exercise: Work through Example 2.11 (White, 2011, pp.93–94).


Buoyancy and StabilityStability

Figure 17 outlines the basic principle of the static stability calculation.

(a) The basic floating position is calculated from Eq.(40). The body’s centreof mass G and centre of buoyancy B are computed.

(b) The body is tilted a small angle ∆θ, and a new waterline is establishedfor the body to float at this angle. The new position B of the center ofbuoyancy is calculated. A vertical line drawn upward from B intersectsthe line of symmetry at a point M, called the metacenter, which isindependent of ∆θ for small angles & MG is metacentric height.

(c) If M is above G (positive MG):a restoring moment is present &the original position is stable.If M is below G (negative MG):body is unstable and willoverturn if disturbed.Stability increases withincreasing MG .

Figure 17: Static equilibrium of a floatingbody.


Buoyancy and StabilityStability

Metacentric height is a property of the cross section for the given weight, and itsvalue gives an indication of the stability of the body. Its computation can be veryinvolved if a body has varying cross section and draft, such as a ship.

A less involved computation requires area moment of inertia of the waterline areaabout the axis of tilt.

Figure 18: A floating body tilted through a small angle ∆θ.

In Figure 18, y axis of the body is assumed to be a line of symmetry. Tilting thebody a small angle θ then submerges the small wedge Obd and uncovers an equalwedge cOa.


Buoyancy and StabilityStability Related to Waterline Area

The new position B of the centre of buoyancy is calculated as the centroid of thesubmerged portion aObde of the body:

x̄VaObde =

∫cOdea

x dV +

∫Obd

x dV−∫

cOax dV

= 0 +

∫Obd

x (L dA)−∫

cOax (L dA)

= 0 +

∫Obd

x L(x tan θ dx)−∫

cOax L(−x tan θ dx)

= tan θ∫

waterlinex2dAwaterline

= IO tan θ

(41)

where IO is the area moment of inertia of the waterline footprint of the body aboutits tilt axis O.Note:

The first integral vanishes because of the symmetry of the original submerged portioncOdea.The remaining two “wedge” integrals combine into IO when we notice that Ldx equalsan element of waterline area.


Buoyancy and StabilityStability Related to Waterline Area

Thus we determine the desired distance from M to B:

x̄tan θ

= MB =IO

Vsubmerged= MG + GB (42a)

or

MG =IO

Vsubmerged− GB (42b)

Exercise: Work through Example 2.12 (White, 2011, p.96).

Note:Determining the floating stability of a buoyant body of irregular shape is difficult, even toan expert. Such bodies may have two or more stable positions.A floating iceberg (average density 1025 kg/m3) in seawater (density is 1025 kg/m3)has approximately 900/1025 ≈ 7

8 of its volume lies below the waterline. It mayoverturn after becoming unstable when its shape changes due to melting.


Pressure Distribution in Rigid-Body Motion

In rigid-body motion, all particles are in combined translation and rotation, andthere is no relative motion between particles.

With no relative motion, there are no strains or strain rates, so that the viscousterm in Eq. (12) vanishes, leaving a balance between pressure, gravity, and particleacceleration:

∇p = ρ(g− a) (43)

Fluids can rarely move in rigid-body motion unless restrained by confining wallsfor a long time.

Water inside a tank in a car that starts a constant acceleration would begin to sloshabout, and that sloshing would damp out very slowly until finally the particles ofwater would be in approximately rigid-body acceleration.


Bibliography

1 FRANK M. WHITE (2011): Fluid Mechanics, 7ed, McGraw-Hill (ISBN 0073529346)2 MERLE C. POTTER, DAVID C. WIGGERT, BASSEM RAMADAN & TOM I-P. SHIH (2012):

Mechanics of Fluids, 4ed, Cengage Learning (ISBN-13: 978-0-495-66773-5)3 BRUCE R. MUNSON, THEODORE H. OKIISHI, WADE W. HUEBSCH & ALRIC P. ROTHMAYER

(2013): Fundamentals of Fluid Mechanics, 7ed, John Wiley & Sons (ISBN 9781118116135)4 YUNUS A. CENGEL & JOHN M. CIMBALA (2014): Fundamentals of Fluid Mechanics, 6ed,

McGraw-Hill (ISBN 9780073380322)5 JOHN F. DOUGLAS, ET AL. (2013): Fluid Mechanics, 5ed, Pearson (ISBN 9780131292932)6 DONALD F. ELGER, ET AL. (2013): Engineering Fluid Mechanics, 5ed, John Wiley & Sons

(ISBN 9781118164297)7 PHILIP J. PRITCHARD & JOHN C. LEYLEGIAN (2011): Fox and MacDonald’s Introduction to

Fluid Mechanics, 8ed, John Wiley & Sons (ISBN 9780470547557)


. . . must end

. . . and I end my presentation with two supplications

my Lord! increase me in knowledge(TAA-HAA (20):114)

O Allah! We ask You for knowledge that is of benefit(IBN MAJAH)

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