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8/13/2019 Fluid Mecanics Nusif Grundgl Engl

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5 Basic Equations of Fluid Mechanics

5.1 General Considerations

Fluid mechanics considerations are carried out in many domains, especiallyin the fields of engineering. Below there is a list which clearly indicates thefar-reaching application of fluid-mechanics knowledge. The list lends itself toshow the importance of fluid mechanics. While it was usual in the past to

carry out special fluid-mechanics considerations on each of the below-listeddomains, today one strives increasingly at developing and introducing gen-eralized ways of consideration that are applicable unreservedly to all below-cited domains. This makes it necessary to derive the basic laws formulatedfor the application of solving flow problems so generally that they fulfill therequirements for broadest applicability indicated in the below-cited list. Theobjective of the derivations in this section is to formulate the conservationlaws for mass, momentum, energy, chemical species etc. such that they canbe applied to all the flow problems occurring in the following domains:

Heat exchanger, cooling and drying technology

Reaction technology and reactor layout

Aerodynamics of vehicles and aero planes Semiconductor-crystal production, thin-film technology, vapor-phase sep-

aration processes

Layout and optimization of pumps, valves and nozzles

Usage of flow aggregates like bent pipes, junctions etc.

Development of measuring instruments and production of sensors

Ventilation, heating and air-conditioning techniques, function of labora-tory vents

Problem solutions in roof ventilation and flows around buildings

Production of electronic components, micro-systems analysis engineering

Layout of stirrer systems, propellers and turbines

Sub-domains of biomedicine and medical engineering

Layout of baking ovens, melting furnaces as well as other fire places

Development of engines, catalyzers and exhaust systems


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118 5 Basic Equations of Fluid Mechanics

Combustion and explosion processes, energy generation, environmentalengineering

Sprays, atomizing and coating technologies

Concerning the formulation of the basic equations of fluid mechanics it iseasy to formulate the conservation equations for mass, momentum, energyand chemical species for a fluid element, i.e. to derive in the Lagrange formof the equations. In this way the derivations can be represented in an easilycomprehensible way and it is possible to build the derivations upon the basicknowledge of physics. Derivations of the basic equations in the Lagrange formare usually followed by transformation considerations whose aim is to striveat local formulations of the conservation equations and to introduce fieldquantities into the mathematical representations, i.e. the Euler form of theconservation equations is sought for solutions of flow-mechanical problems.This requires to express temporal modifications of substantial quantities astemporal modifications of field quantities, which makes it necessary to partly

repeat in this section the considerations cited in chapter 2 or to complementthem in a deepening way.

Figure 5.1: Division of a fluid in fluid elements

The considerations to be carried out apart from the assumption that at acertain point in time t = 0 the mass of a fluid is subdivided in fluid elementsof the mass m i.e.

M=

m.

Here for each fluid element m is to be chosen large enough to make theassumption m = const possible with sufficient precision in spite of themolecular structure of the matter, it also allows to assign a arbitrary thermo-dynamical and fluid-mechanics properties (x(t), t) = (t)for a fluidelement and with a satisfactory precision for fluid-mechanics considerations.

The statement (x, t), with x = x(t), expresses that the thermo-

dynamical or fluid-mechanics property, which is assigned to the consideredfluid element and thus only represents a substantial quantity, is only a func-tion of time. This property of the element is changing with time, also due tothe motion of the fluid element and for the description of this modification it


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5.2 Mass Conservation (Continuity Equation) 119

is important that one follows the mass m, i.e. knows x(t) and also takesit into consideration as known. It is assumed that this motion of particlesis constant and unequivocal, i.e. that the considered fluid element does notsplit up during the considerations of its motion. The fluid pertaining to the

considered fluid element at the moment in time t = 0 remains also at all latermoments in time. This signifies that it is not possible for two different fluidelements to take the same point in space at an arbitrary time: x(t) =xL(t)for = L.

When a fluid element is at the position xi, at the time t i.e. xi = (x(t))iat the timet, then the substantial thermodynamic property or fluid property(t) is equal to the field quantity at the point xi at time t:

(t) =(xi, t) when (x(t))i =xi at timet (5.1)

For the temporal change of a quantity (t) results (see also chapters 2 and3):

ddt =

t +

xi dxi

dt

(5.2)

With ( dxi/dt)= (Ui)= Ui holds:

ddt

=D

Dt =

t +Ui

xi(5.3)

The operator (xi, t) applied to the field quantity D/Dt = /t+Ui/xiis often defined as the substantial derivatve and will be applied in the subse-quent derivations. Significance of individual terms are:

/t= (/t)xi = change with time at a fixed location,partial differentiation with respect to time

d/dt = total change with time (for a fluid element),

total differentiation with respect to timefor e.g. for a fluid when = = const i.e. the density is constant, then itholds:

ddt

= D

Dt = 0 or

t = Ui

xi(5.4)

When at a certain point in space /t ()xi = 0 indicate of stationary condi-tions, i.e. the field size(xi, t) is stationary and thus has no time dependency.On the other hand d()/dt= D/Dt= 0, is (t) =(xi, t) = const. i.ethe field is independent of space and time.

5.2 Mass Conservation (Continuity Equation)

For fluid-mechanics considerations a closed fluid system can always befound, i.e. a system for whose total mass holdsM=const. This is easily seenfor a fluid mass, which is stored in a container. For fluid setups as shown in


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Fig. 5.2, control volumes can always be defined, within which the systemictotal mass can be stated as constant. If necessary these control volumes cancomprise the whole earth.

When one subdivides the fluid mass within the considered system in fluidelements with the sub-masses m, then it holds true for the temporalchange of the total mass:

0 = dM

dt =

d

dt

(m) =

d

dt(m) (5.5)

Figure 5.2: : Different Fluid setups and aero foil within control volumes whereM= const

This equation expresses that the mass conservation in the total fluid systemis preserved when each individual fluid element conserves its mass m. Withthis the balance equation for the mass conservation can be stated as follows

in Lagrange notation:d (m)

dt = 0 (5.6a)

The basic molecular structure of matter and thermal motion connected withit indicates that to fulfil the above relation absolutely, it is necessary thatm 0 The consideration carried out in this book therefore requires thatall the m are considered as finite but nevertheless as very small. In Fig.5.3 a fluid element with position coordinates (xi) is shown. 1

The estimation ofm is referred to considerations that are listed in chapter3, where it is shown what dimensions a volume of an ideal gas has to havein order to define sufficiently clearly e.g. a density of the gas within thevolume. The considerations carried out there would have to be repeated here

in order to guaranteem= const, in spite of the molecular structure of thematter. With the choice of m = const the conditions are set to carry out

1see the considerations carried out in chapter 3.2


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5.2 Mass Conservation (Continuity Equation) 121

continuum-mechanics considerations for the motion of fluids, although thefluids show a molecular structure.

In this context it is often referred to the continuum considerations in whichusually fluid-mechanics considerations are carried out. Strictly speaking thismeans that the properties of the molecules, especially their transport prop-erties, can only be introduced in fluid-mechanics considerations in integralform.

Figure 5.3: m= const, condition for the mass of a fluid element

Because of the above explanations the mass conservation can be stated asfollows:

dM

dt = 0 und

dmdt

= 0 (5.6b)

The above considerations confirm that it is very simple to formulate the mass-conservation law in Lagrange variables. Working practically with the law ofmass conservation, however, requires its representation in field quantities, i.e.the Lagrange form of mass-conservation law has to be brought into the Eulerform.

Transformed into Euler variables (i.e. into field quantities) one obtains from(5.6a) for the mass conservation:

0 = d

dt(m) =

d

dt(V) =

d (V)

dt I

+ Vd ()

dt II

. (5.7)

For term I in equation (5.7) using = and (x (t))i = xi at the time t:

yields (according to equation 4.89):

d (V)

dt =V

Uixi

. (5.8)


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5.3 Second Newtonian Law (Momentum Equation) 123

This relation expresses for a fluid in the thermodynamic sense, = const, i.eif the fluid density is constant, then it has to be thermodynamically incom-pressible.

1

DDt

= Uixi

= 0 = DPDt DT

Dt with = 0 and = 0 (5.16)

whereandare:

= 1

v

v

P

T

= 1

P

T

= isothermal compressibility coefficient

= 1

v

v

T

P

=1

T

P

= thermal expansion coefficient

Thus the continuity equation holds in one of the two forms listed below:

t +

(Ui)

xi= 0 (compressible flows) (5.17)

Uixi

= 0 (incompressible flows) (5.18)

5.3 Second Newtonian Law (Momentum Equation)

The derivations of the equations of momentum for the three coordinate di-rections j = 1, 2, 3 in fluid mechanics, the second Newtonian law to a fluidelement, i.e. the Lagrange formulation of the equation of momentum is cho-

sen. For a fluid element it is stated that the time derivative of the momentumin the j -direction is equal to the sum of the external forces acting in this di-rection on the fluid element, plus the molecular-dependent input per unittime. The forces can be stated as inertia forces caused by gravitation forces(Mj) and electromagnetic forces2 as well as the surface forces caused bypressure (Oj). After addition of a temporal change of momentum fed inby the molecular movement, the equation of motion can be formulated asfollows:

d(Jj)dt

=

(Mj) Inertia forces

+

(Oj) surface forces

+

d

dt (JM)j

molecular-dependentmomentum input

(5.19)

Here, as stated in Fig. 5.4 , (Jj)= m(Uj)

2(the latter are not taken into consideration in the following)


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Figure 5.4: For the derivation of momentum equations

The fluid element acts like a rigid body since it does not change its stateof motion, i.e. its momentum, when no inertia or surfaces forces act on thefluid element and a molecular-dependent momentum input fails to appear.However, when forces are present, or when a momentum has input, the fluidelement changes its momentum in accordance with the above-stated relation(5.19). It represents the Lagrange form of the equations of momentum (j =1, 2, 3) of fluid mechanics.

In order to conserve the Euler form of the equation of momentum, it isimportant to express each of the terms contained in equation (5.19) in fieldquantities. For the left side of the equation (5.19) can be written as:

d(Jj)dt

= d

dt[m(Uj)] =m

d ((Uj))

dt + (Uj)

d ((m))

dt (5.20)

Because of the universal mass conservation in formulation (5.6a), the lastterm in equation (5.20) ) is equal to zero:

d(Jj)dt

=md ((Uj))

dt =m

Ujt

+UiUjxi

(5.21)

This can be written as follows:m= V= Vapplying= when:(x(t))i =xi at the time t,

d(Jj)dt

=V

Ujt

+UiUjxi

(5.22)

Owing to the above derivations it is possible to state the left side of theequation of momentum (5.19)) in field quantities. For the right side the below-cited considerations can be carried out. The inertia forces can be represented

in an easy way by the acceleration vector {gj}, i.e. by its components gj.

(Mj) = Mass forces acting on a fluid element

The mass forces acting on a fluid element can be stated by means of the


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5.3 Second Newtonian Law (Momentum Equation) 125

Mass forces:

Figure 5.5: Mass force on a fluid element

acceleration {gj} = {g1, g2, g3} acting per measuring unit. The mass force

acting on a fluid element in j -direction can be stated as follows:(Mj)= (m) gj =Vgj (5.23)

Even when only gravitational acceleration is present, depending on the posi-tion of the coordinate system, several components ofgj may not be equal tozero

Surface forces:

Figure 5.6: Considerations concerning surface force on a fluid element

(Oj) = Surface forces on a fluid element

Fluids as they are treated in this book, i.e. fluids (e.g. water) and gases (e.g.air) are characterized by the way they can apply surface forces on a fluid

element only by the molecular pressure. The pressure force acting on a fluidelement is calculated as the difference of the forces acting on the areas thatstand vertically on the considered axes. It holds since


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5.4 The Navier-Stokes Equations 127d(JM)j

dt

= ijxi

V (5.27)

When one inserts all these derived relations (5.22), (5.23), (5.25) and (5.27)

in (5.19) after division byV

the equation of momentum of fluid mechanicsin directionj results, i.e. for j = 1, 2, 3 three equations result:

Ujt

+UiUjxi

= gj

P

xj

ijxi

(5.28)

As in this equation the volume of the fluid element V appearing in allterms was eliminated, the equation of momentum hold per unit volume. Themomentum equation in the three coordinate directions results:

U1t

+U1U1x1

+U2U1x2

+U3U1x3

=

P

x1

11x1

21

x2

31x3

+g1

U2t +U1 U2x1 +U2 U2x2 +U3 U2x3 = Px2 12x1 22x2 32x3 +g2

U3t

+U1U3x1

+U2U3x2

+U3U3x3

=

P

x3

13x1

23

x2

33x3

+g3

(5.29)

From the fluid mechanics point of view only for ideal fluids ij = 0,

For fluids in generalij = 0 but for ideal fluids in terms of fluid mechanicsij = 0. Hence the following forms of the momentum equations can be stated:

Ujt

+UiUjxi

= P

xj

ijxi

+gj (viscous Fluide) (5.30)

Ujt

+UiUjxi

=

P

xj+gj (ideal Fluide) (5.31)

5.4 The Navier-Stokes Equations

In the above equation (5.30) the molecular-dependent momentum inputij occurring per surface and unit time, is an unknown term, i.e. it isintroduced formally in derivations, without details being considered, as itcan be formulated for several fluids. When one takes into consideration the

symmetry of the term ij i.e. |ij |= |ji |, holds, one notices/ascertains thatthere are the following unknowns in the equations:

U1, U2, U3, P , 11, 12, 13, 22, 23, 33= 10 unknowns


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Figure 5.8: Momentum input due to flow through the plane Fj

For these unknowns there are only four partial differential equations, thecontinuity equation and three equations of momentum, i.e. an incompleteequation system exists that does not permit the solution of flow problems.It is therefore necessary to state additional equations, i.e. to express the

unknown terms ij physically well-founded, as functions of Uj/xi. This isdone below for ideal gases, as their properties are known to a large extentfrom considerations in physics. From the below-stated derivations, relationsfor ij = f(Uj/xi) result, that are valid also for non-ideal gases and fora whole class of fluids whose molecular momentum-transport properties canbe classified as Newtonian. Thus the derived relations ij are valid farbeyond ideal gases and represent in this book the basic equations to describethe molecular-dependent momentum transport in Newtonian fluids.

When one considers a fluid element, as shown in Fig. 5.8 one can see thatthe momentum j fed in direction i by a velocity field and can be stated asfollows:

Iij =UiUjFi (5.32)

Assuming that the instantaneous velocity components are composed of thevelocity of the fluid flow Ui and the share of the molecular motion ui oneobtains:

Ui Uj =(Ui+ui)(Uj + uj)=(UiUj + uiUj+ ujUi+uiuj)

(5.33)

By time averaging, one obtains for the time-averaged total momentum changeof the fluid element:

Ui Uj =[UiUj

I

+ uiUj

II

+ ujUi

III

+ uiuj

IV

] (5.34)

The total momentum input consists of four terms that can be interpretedphysically as follows:


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5.4 The Navier-Stokes Equations 129

Term I: Momentum inputj in direction in i due to the velocity field ofthe Fluid.

Term II: Momentum inputj in direction i due to the molecular motionin i-direction.

Term III: Momentum inputj in direction i due to the molecular motionin j -direction.

Term IV: For i =j is uiuj = 0 as the molecular motion inthe three coordinate directions - are not correlated; for i = jresults the pressure treated in chapter 3

The molecular motion is distinguished by the presence of free path lengthswith finite dimensions, i.e. l = 0, the time averages uiUj and ujUi are un-equal to zero. In order to calculate these contributions to ij considerationson the molecular momentum transport, ideal gases are recommended. Forthe number of molecules moving in direction xi and passing the plane A inFig. 5.9 in the timet when x1= x2= x3=a it can be written:

zi = 16

na2uit (5.35)

Wheren is equal to the number of molecules per unit volume, a2 the volumeof areaFianduithe mean velocity of the molecules in directioni. Connectedto zi, a mass transport throughFi, can be stated as follows:

mzi = 1

6(mn)

a2uit (5.36)

wherem represents the mass of a molecule and thus can be set mn = .

Figure 5.9: Momentum input in xidirection with the molecular velocity ui

When one considers now the auxiliary planes in Fig. 5.9 located to Fi in

the distance l and introduced for the derivations, in which the mean flowfield owns the velocity components Uj(xi+ l) andUj(xi l). In positive andnegative i-directions, jdirectional momentum input and discharge can bestated s follows:


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i+ij = +zimUj(xi l) Momentum input over areaFi

iij =zimUj(xi+l) Momentum discharge over areaFi(5.37)

Therefore the sum of the molecular-dependent input and discharge results:iij =zim[Uj(xi l) Uj(xi+l)] (5.38)

or with zi inserted from equation (5.35):

iij =1

6(mn)

a2uit[Uj(xi l) Uj(xi+l)] (5.39)

The momentum flow per unit area and time can be obtained by Taylor seriesexpansion of velocity at around xi by neglecting the higher order terms:

IIij = 1

a2iijt

= 1

6ui

Uj(xi)

Ujxi

l Uj(xi) Ujxi

l

, (5.40)

so that for Term II in equation (5.34) results:

IIij = 1

3uil

Ujxi

= Ujxi

. (5.41)

Analogous to this it can be carried out considerations onIIIij where forzj canbe written:

zj = 1

6na2ujt (5.42)

Figure 5.10: Momentum input in xjdirection with molecular velocity ujandfluid velocityUi

In accordance with term III in equation (5.34) a momentum input j-results

which can be stated as follows:i+ij =zjmUi(xj l)

iij =zjmUi(xj+ l)(5.43)


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5.4 The Navier-Stokes Equations 131

or.

iij =zjm[Ui(xj l) Ui(xj+ l)] (5.44)

Analogous to the derivations in equations (5.38) to (5.41) yields:

IIIij = 1

3(ujl)

Uixj

= Uixj

(5.45)

For reasons of symmetry ij =ji , so that ui = uj has to hold, i.e. the meanvelocity field of the molecules is isotropic (no preferred velocity direction), sothat it can be written:

ij =IIij+

IIIij =

Ujxi

+Uixj

for i =j (5.46)

This is the total momentum inputij for = const, i.e. when d/dt(V

) = 0,the thermodynamic state equation for an ideal fluid is assumed. For =const, an additional term needs to be added which is conditioned by thevolume increase of a fluid element. For the volume increase of a fluid elementat pointxi and time t, (see section 4):

d(V)

dt = (V)

Uixi

(5.47)

For the corresponding surface increase holds:

d(F)

dt =

2

3(F)

Uixi

(5.48)

With surface increase an increased momentum input results:

ij = +2

3ij

Ukxk

(5.49)

This term has to be added to obtain general formulas of the total momentuminput per unit time and unit area for ideal gases. It can be stated as follows:

ij =

Ujxi

+Uixj

+

2

3ij

Ukxk

(5.50)

When one considers this relation forij, the basic equations of fluid mechanicscan be stated as follows:Continuity equation:

t +

(Ui)

xi= 0 (5.51)


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Momentum equations:(j = 1, 2, 3)

Ujt

+UiUjx

i = P

xj

ijx

i

+gj (5.52)

For Newtonian fluids

ij =

Ujxi

+Uixj

+

2

3ij

Ukxk

(5.53)

With ij from (5.53) there exist equations for the 6 unknown terms ijinthe momentum equations. The four equations, one continuity equation andthree Navier-Stokes equations, contain five unknowns: P,, Uj , so that anincomplete system of partial differential equations exists still. With the aidof the thermal energy equation and the thermodynamic state equation validfor the considered fluid, it is possible to obtain a complete system of partial

differential equations that permits general solutions for flow problems, wheninitial and boundary conditions are present.

For = const and = const, using 2Ui

xixj=

2Uixjxi

= xj

Uixi

= 0 can be

stated:

Continuity equation: Ui

xi= 0 (5.54)

Navier-Stokes equations (j = 1, 2, 3):

Ujt

+UiUjxi

=

P

xj+

2Ujx2i

+gj

This system of equations comprises four equations for the four unknownsP,U1, U2, U3. In principle, it can be solved for all flow problems to be investi-gated, when suitable initial and boundary conditions are given. For thermo-dynamic ideal fluids, i.e. = const, a complete system of partial differentialequations exists with the continuity equation and the momentum equations,which can be used for solutions of flow problems.

5.5 Mechanical Energy Equation

In many domains in which fluid-mechanic considerations are carried out,the mechanical energy equation is employed which can be derived from the

momentum equation j-For this purpose one multiplies equation (5.52) withUj :

Uj

Ujt

+UiUjUjxi

= Uj

P

xj Uj

ijxi

+Ujgj (5.55)


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5.5 Mechanical Energy Equation 133

This equation can be transcribed as follows:

t 1

2U2j

+Ui

xi 1

2U2j

= (P Uj)

xj+P

Ujxj

(ijUj)

xi+ij

Ujxi

+gjUj

(5.56)

The equation states how the kinetic energy of a fluid element is changing ata location due to energy-production and dissipation terms that occur on theright side of the above equation (5.56). In order to discuss the significance ofthe different terms, the following modification of the last term is carried out,introducing a potential G from which the gravity is derived:

gj = G

xj gjUj =

G

xjUj (5.57)

Thus, employing G

t = 0 :

gjUj =

G

t +Uj

G

xj

=

DG

Dt (5.58)

The combined equations (5.56) and (5.58) ) yield for the temporal change ofthe kinetic and potential energy of a fluid element:

D

Dt

1

2U2j +G

=

(P Uj)

xj I

+ PUjxj II

(ijUj)

xi III

+ ijUjxi IV

(5.59)

where the terms I to IV have the following physical significance:


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Term I: This term describes the difference between input anddischarge of pressure energy. Here it is referred to theconsiderations on ideal gases, in the framework of whichwas shown thatP= 1

3u2, i.e. expresses an energy per unit

volume. Therefore the following can be said:

(P Uj(xi)) = Input of pressure energy per unitarea

(P Uj(xi+xi)) = Discharge of pressure energy perunit area

Taylor series expansion and forming the difference yieldsfor the energy per unit volume:

P Uj(xi)

P Uj(xi) +

(P Uj)

xj+

(P Uj)

xj(5.60)

Term II: Considering:

Ujxj= 1V

d(V)dt (5.61)

the term:

PUjxj

= P

V

d(V)

dt (5.62)

proves to be the work done during expansion, occurring perunit volume .

Term III: When taking into consideration that ij repre-sents the molecular-dependent momentum transportper unit area and unit time into a fluid element:

(ijUj)

xj = difference from the molecular-dependent

input and discharge of the kinetic energy ofthe fluid

Term IV: The term ijUjxi

describes the dissipation of mechanical

energy into heat.The above presentations show that the mechanical energy equation can bederived from the j momentum equation by multiplication with Uj. It is tomake sure that no independent equation and should be employed along withthe momentum equation for the solution of fluid-mechanical problems.

A special form of the mechanical energy equation is the Bernoulli equation

which can be derived from the general form of the mechanical energy equa-tion:

D

Dt

1

2U2j +G

=

P

xjUj

ijxi

Uj (5.63)


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5.6 Thermal Energy Equation 135

Forij = 0 and P

t= 0, as well as = const. Holds:

D

Dt 12 U2j +G=

Pt +Uj

P xj =

DP

Dt (5.64)

D

Dt

1

2U2j +

P

+G

= 0

1

2U2j +

P

+G= const. (5.65)

This form of mechanical energy equation can be employed in many engineer-ing applications for an imputation estimate of flow processes.

5.6 Thermal Energy Equation

The derivations in chapter 5 showed that the mechanical energy equation is

derivable from the momentum equation, so that both equations have to beconsidered as not being independent from the another. From the derivationsresulted the equation in the following form:

D

Dt

1

2U2j

=

(P Uj)

xj+P

Ujxj

(ijUj)

xi+ij

Ujxi

+gjUj (5.66)

When one sets up the equation for the total energy balance, the considerationstated below results, which starts from the entire internal kinetic and poten-tial energy of a fluid element and considers its evolution as a function of time:

d

dtm 1

2U2j +e+G

= m

d

dt[. . . ] + [. . . ]

dmdt

For the temporal change of the total energy of a fluid element results withm= const, d.h.

ddt

(m) = 0:

d

dt

m

1

2U2j +e+G

= m

D

Dt

1

2U2j +e+G

This is the total energy change which has to be considered concerning thederivation of the total energy equation.

The change of the total energy of the fluid element can emanate from theheat conduction, which yields the following inputs minus the discharges ofheat:

qixi

V= Energy input per unit time by heat conduction


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can originate from the convective transport of pressure energy :

xj(P Uj)V= Input of pressure energy through convection

and by the input of kinetic energy due to molecular transport into the fluidelement:

xi(ijUj)V= molecular-dependent input of kinetic energy

The following total energy balance is thus resulting:

VD

Dt

1

2U2j +e+G

=

qixi

V (P Uj)

xjV

(ijUj)

xiV, (5.67)

as V= 0 follows:

D

Dte+ 12 U2j +G= qixi (P Uj)xj (ijUj)xi . (5.68)

When one deducts from this the mechanically derived parts, i.e. by subtract-ing the equation (5.68) from the equation for the mechanical energy, givenhere once again:

D

Dt

1

2U2j +G

=

(P Uj)

xj+P

Ujxj

(ijUj)

xi+ij

Ujxi

, (5.69)

one obtains the thermal energy equation:

De

DtI=

qixi II

PUjxj II I

ijUjxi IV

(5.70)

Term I : Temporal change of the internal energy of a fluid per unit volume.Term II : Heat supply per time and unit area .Term III : Work done per unit volume and unit time.Term IV : Irreversible transfer of mechanical energy into heat, per unit

volume and unit time.

Considering the energy equation from technical thermodynamics

dq= de+Pdv dldiss (5.71)

and the sign convention is usual in technical thermodynamics, that the energyto be dissipated by a fluid element has to be regarded as negative, it results:

dedt

= De

Dt;

dqdt

= 1

qixi

; Pdv

dt =

1

P

Ujxj

and dldiss

dt =

1

ij

Ujxi

(5.72)


21/42


22/42


De

Dt =

qixi

PUixi

ijUixi

(5.80)

Equation for the total energy:

D

Dt

1

2U2j +G+e

=

qixi

xj(P Uj)

xi(ijUj)

=2T

x2i

xj(P Uj)

xi(ijUj)

(5.81)

From this final relation the Bernoulli equation can be derived which is oftenused for fluid-mechanical considerations:

Ideal Fluid: (= const): no heat conduction and viscous dissipation

DDt 1

2U2j +G= Uj Pxj = Ui

P

xi(5.82)

For a steady flow:

=0

t

1

2U2j +G

+Ui

xi

1

2U2j +G

= Ui

P

xi(5.83)

xi

1

2U2j +G+

P

= 0

or after integration:

1

2U2j +G+

P

= const. (5.84)

Ideal Gas: P/= RT, no heat conduction and neglecting viscous dis-sipation as well as of the potential energy

D

DT

1

2U2j +e

=

xj(P Uj) =

xi(P Ui) (5.85)

For steady flow:

xi

1

2U2j +e

=

xi(P Ui) = P

Uixi

UiP

xi(5.86)

From the continuity equation follows for steady flows:

Uixi

= Ui xi

(5.87)

Inserted consideration ofe = cT


23/42

5.7 Basic Equations in different Coordinate Systems 139

xi

1

2U2j +e

=

xi

1

2U2j

+c

T

xi=

P

xi

P

xi(5.88)

xi 12 U2j = P2 xi 1 Pxi c TxiIntroducing:

T

xi=

P

R2

xi+

1

R

P

xi(5.89)

xi

1

2U2j

=

P

2

xi

1 +

cR

1

P

xi

1 +

cR

=

1

xi

P

(5.90)

xi 1

2U2j +

1 P

= 0 1

2U2j +

1 P

= const (5.91)

5.7 Basic Equations in different Coordinate Systems

5.7.1 Continuity Equation

The derivations carried out for the continuity equation inCartesian coordinates result in:

t +

(Ui)

xi= 0 (5.92)

or

t +

(U1)

x1+

(U2)

x2+

(U3)

x3= 0 (5.93)

For = const: U1

x1+

U2x2

+ U3x3

= 0

In cylindrical coordinates (r,,z) with (Ur, U, Uz) results the followingequation:

t +

(Ur)

r +

1

r

(U)

+

(Uz)

z +

Urr

= 0 (5.94)

and for = const equation reduces to:

Urr

+1

r

U

+ Uz

z +

Urr

= 0 (5.95)


24/42


In spherical coordinates (r,,) the continuity equation can be stated asshown below for (Ur, U, U) :

t +

1

r2

r (r2

Ur) +

1

r sin

(Usin ) +

1

r sin

(U) = 0 (5.96)

Here the below-mentioned coordinates in the derivations of the relations wereemployed.

Cylindrical coordinates

Figure 5.11: Coordinate systems and transformation equations for cylindricalcoordinates

The use of cylindrical coordinates in the derivations of the basic equationsleads to the metric coefficients introduced in section 2.10 for the transforma-tion of the equations

hr = 1; h= r; hz = 1

for the general continuity equation:

(Ui)

xi=

1

r

r(rUr) +

1

r

(U) +

(Uz)

z , (5.97)

or for the continuity equation with = const:

Urr

+1

r

U

+ Uz

z +

Urr

= 0 (5.98)

Analogous to the above derivations of the continuity equation in cylindricalcoordinates one obtains for spherical coordinates:

hr = 1; h =r; h = r sin (5.99)


25/42


Spherical coordinates

Figure 5.12: Coordinate systems and transformation equations for spherical co-ordinates

and thus for

(Ui)

xi=

1

r2

r(r2Ur) +

1

r sin

(Usin ) +

1

r sin

(U) (5.100)

the continuity equation in spherical coordinates with = const results:

1

r2

r(r2Ur) +

1

r sin

(Usin ) +

1

r sin

(U) = 0 (5.101)

5.7.2 Navier-Stokes Equations

Analogous to the transformation of the continuity equation in cylindricaland spherical coordinate systems, the different terms of the Navier-Stokesequations can also be transferred, which can be stated in cartesian coordinatesas follows for Newtonian fluids:

=DUj

Dt =

Ujt

+UiUjxi

=

P

xj+

xi

Ujxi

+Uixj

2

3ij

Ukxk

+gj (5.102)

Written out forj = 1, 2, 3:

DU1Dt

=P

x1+

x1 2

U1x1

2

3( U)

+

x2

U1

x2+ U2

x1

+

x3

U3

x3+ U1

x1

+g1

(5.103)


26/42


DU2Dt

=P

x2+

x1

U2x1

+ U1x2

+

x2

2

U2x2

2

3( U)

+

x3

U3x3

+ U2x1

+g2

(5.104)

DU3

Dt =

P

x3+

x1

U3x1

+ U1x3

+

x2

U3x2

+ U2x3

+

x3

2

U3x3

2

3( U)

+g3

(5.105)

where it holds U= Ukxk

Momentum Equations in Cartesian Coordinates

- Momentum equations with ijterms:

x1Component: U1t

+U1 U1x1

+U2 U1x2

+U3 U1x3

= Px1

11x1

+21

x2+

31x3

+g1

(5.106)

x2Component:

U2t

+U1U2x1

+U2U2x2

+U3U2x3

=

P

x2

12x1

+22

x2+

32x3

+g2

(5.107)

x3Component:

U3t

+U1U3x1

+U2U3x2

+U3U3x3

=

P

x3

13x1

+23

x2+

33x3

+g3

(5.108)

- Navier-Stokes equations for and equally constant:

x1Component: U1t

+U1U1x1

+U2U1x2

+U3U1x3=

P

x1

+

2U1x21

+2U1

x22+

2U1x23

+g1

(5.109)


27/42


x2Component:

U2t

+U2U2x1

+U2U2x2

+U3U2x3

=

P

x2

+2U2x2

1

+2U2

x22

+2U2

x23 +g2

(5.110)

x3Component:

U3t

+U1U3x1

+U2U3x2

+U3U3x3

=

P

x3

+

2U3x21

+2U3

x22+

2U3x23

+g3

(5.111)

Momentum Equations in Cylindrical Coordinates

- Momentum equations with ijterms:

rComponent:

Urt

+UrUrr

+U

r

Ur

U2r

+UzUrz

=

P

r

1

r

r(rrr) +

1

r

r

r +

rzz

+gr

(5.112)

Component:

U

t +Ur

Ur

+U

r

U

+UrU

r +Uz

Uz

= 1

r

P

1

r2

r(r2r) +

1

r

+z

z +g

(5.113)

zComponent:

Uzt

+UrUzr

+U

r

Uz

+UzUzz

=

P

z

1

r

r(rrz ) +

1

r

z

+ zz

z

+gz

(5.114)

- - Navier-Stokes equations for and equally constant:

rComponent: Urt +Ur Urr + Ur Ur U2

r +UzUrz

(5.115)=

P

r +

r

1

r

r(rUr)

+

1

r22Ur2

2

r2U

+2Ur

z2

+gr


28/42


Component:

U

t +Ur

Ur

+U

r

U

+UrU

r +Uz

Uz

(5.116)

= 1

r

p

+

r 1

r

r(rU) +

1

r22U2

+ 2

r2Ur

+2U

z2 +g

zComponent:

Uzt

+UrUzr

+U

r

Uz

+UzUzz

(5.117)

= P

z +

1

r

r

r

Uzr

+

1

r22Uz2

+2Uz

z2

+gz

Momentum Equations in Spherical Coodinates

- Momentum equations with ijterms

rComponent: Ur

t +Ur

Ur

r +

U

r

Ur

+

U

r sin

Ur

U2 +U

2

r

=

P

r

1

r2

r(r2rr) +

1

r sin

(rsin ) +

1

r sin

r

+

r

+gr

(5.118)

Component:

Ut

+UrUr

+U

r

U

+ Ur sin

U

+UrU

r

U2cot

r = 1

r

p

1

r2

r

(r2r) + 1

r sin

(sin ) + 1

r sin

+r

r

cot

r

+g

(5.119)

Component:

U

t +Ur

Ur

+U

r

U

+ Ur sin

U

+UUr

r

+UU

r cot

=

1

r sin

p

1

r2

r

r2r

+

1

r

+ 1

r sin

+r

r +

2cot

r +g

(5.120)

Navier-Stokes Equations for and Equally Constant


29/42


rComponent:

Urt

+UrUrr

+U

r

Ur

+ Ur sin

Ur

U2 +U

2

r

= P

r +2Ur 2r2 Ur

2

r2

U

2

r2Ucot

2

r2 sin

U

+gr (5.121)

Component:

Ut

+UrUr

+U

r

U

+ Ur sin

U

+UrU

r

U2cot

r

=

1

r

P

+

2U+

2

r2Ur

U

r2 sin2

2cos

r2 sin2

U

+g (5.122)

Component:

U

t +Ur

Ur

+U

r

U

+ Ur sin

U

+UUr

r

+UU

r cot

=

1

r sin

P

+

2U

U

r2 sin2

+ 2

r2 sin2

Ur

. + 2cos

r2 sin2

U

+g (5.123)

In these equations:

2 = 1

r2

r r2

r+ 1

r2 sin

sin

+ 1

r2 sin2 2

2 (5.124) Components of the Molecular Momentum Transport Tensor in

Cartesian Coordinates:

11=

2

U1x1

2

3( U)

; 22=

2

U2x2

2

3( U)

; (5.125)

33=

2

U3x3

2

3( U)

.

12

= 21

= U1x2 +U2

x1 ; 23= 32= U2

x3+

U3

x2 ;(5.126)

31= 13=

U3x1

+ U1x3

.


30/42


above being employed

( U) =U1x1

+U2x2

+U3x3

= Ukxk

(5.127)

Components of the Molecular Transport Tensor inCylindrical Coordinates:

rr =

2

Urr

2

3( U)

; (5.128)

=

2

1

r

U

+Ur

r

2

3( U)

; zz =

2

Uzxz

2

3( U)

r = r =

r

r

Ur

+

1

r

Ur

; (5.129)

z =z = Uz +1r Uz ; zr =rz = Uzr + Urz .above being employed

( U) = 1

r

r(rUr) +

1

r

U

+Uz

z (5.130)

Components of the Molecular Momentum Transport Tensor inSpherical Coordinates:

rr =

2

Urr

2

3( U)

(5.131)

= 21r U + Urr 23 ( U) (5.132) =

2

1

r sin

U

+Ur

r +

Ucot

r

2

3( U)

(5.133)

r =r =

r

r

Ur

+

1

r

Ur

(5.134)

= =

sin

r

Usin

+

1

r sin

U

(5.135)

r =r =

1

r sin

Ur

+r

r Ur

(5.136)

( U) = 1

r2

r(r2Ur) +

1

r sin

(Usin ) +

1

r sin

U

(5.137)


31/42

5.8 Special Forms of the Basic Equations 147

Dissipation FunctionijUjxi

=:

Cartesian coordinates:

= 2U1x12

+ U2x22

+U3x32 + U2x1 + U1x2

2

+

U3x2

+ U2x3

2+

U1x3

+ U3x1

2

2

3

U1x1

+U2x2

+U3x3

2(5.138)

Cylindrical coordinates:

= 2

Urr

2+

1

r

U

+Ur

r

2+

Uzxz

2

+ r r U

r +1

r

Uz

2

+ 1rUz

+U

xz 2

+ Urxz +Uz

r 2

2

3

1

r

r(rUr) +

1

r

U

+ Uzxz

2(5.139)

Spherical coordinates:

= 2

Urr

2+

1

r

U

+Ur

r

2 1

r sin

U

+Ur

r +

Ucot

r

2

+

r

r

Ur

+

1

r

Ur

2+

sin

r

Usin

+

1

r sin

U

2

+ 1r sin Ur

+r

r U

r 2

2

3

1

r2

r(r2Ur) +

1

r sin

(Usin ) +

1

r sin

U

2(5.140)

The above equations can be solved in connection with the initial and bound-ary conditions describing the actual flow problems.

5.8 Special Forms of the Basic Equations

Due to the multitude of fluid-mechanical considerations special forms havecrystallized from the equations treated in the preceding chapters, some ofwhich shall be derived and explained in this chapter. These are the vortex-power equation (Wirbelstrkengleichung), the Bernoulli equation and the


32/42


Crocco equation, that have already been treated before. The derivations willmoreover treat the Kelvin theorem as a basis for explanations of its physicalsignificance. The objective of the considerations is to bring out clearly theprerequisites under which the special forms of the basic equations are valid.

Only the simplified treatments of flow problems that are sought with thespecial forms of the equations lead to secure results.

5.8.1 Transport equation for vortex power

The vortex poweri is a property of the flow field which can be employed ad-vantageously in considerations of rotating fluid motions. It can be computedfrom the velocity field as follows:

k = U= ijkUjxi

=

Ujxi

Uixj

(5.141)

For a fluid with the properties = const and = const, the Navier-Stokesequation can be written in the following way:

Ujt

+UiUjxi

=

P

xj+

2Ujx2i

+gj (5.142)

or in vector form:U

t +

U

U

=

1

P+2U+ g (5.143)

Considering that this vector form of the Navier-Stokes equation can also bewritten as:

U

t +

1

2U U

U

U

=

1

P+2U+ g (5.144)

When one applies the operator (. . . ) . to each of the terms appearing inthe above equation one obtains:

t (U ) =2 (5.145)

Making use of the relation valid for vectors:

(U ) = U( ) ( U) (U )+ ( )U

where = 0 as the divergence of the rotation of each vector is equal tozero, and where at the same time = const U = 0 holds owing to thecontinuity equation. When one introduces all this into the above equations,the transport equation for the vortex power reads:


33/42


t + ( U) = ()U+ 2 (5.146)

or in tensor notation:

DjDt

= j

t +Ui

jxi

=jUixj

+2jx2i

(5.147)

The equation (5.147) does not contain the pressure term, from this it isapparent that the vortex-power field can be determined without knowledgeof the pressure distribution. To be able to compute the pressure, one formsthe divergence of the Navier-Stokes equation and obtains for gj = 0:

2

x2i

P

= 2j +Uj

2Ujx2i

1

2

2U2jx2i

(5.148)

Thus yields the Poisson equation for the computation of the pressure. Fortwo-dimensional flows, for which the vortex-power vector stands vertical onthe flow plane, ( )U = 0. The transport equation for the vortex powertherefore reads:

jt

+Uijxi

=2jx2i

(5.149)

5.8.2 The Bernoulli Equation

The general momentum equation is transferred into the Euler equations, byassuming an ideal fluid to derive the Bernoulli equation, i.e:

DUj

Dt = Ujt +Ui Ujxi = Pxj +gj (5.150)Multiplying this equation by Uj, one obtains the mechanical energy equationvalid for dissipation-free fluid flows:

D

Dt

1

2U2j

=

t

1

2U2j

+Ui

xi

1

2U2j

= Uj

P

xj+gjUj.

(5.151)When one introduces the potential field G for the presentation ofgj so as:

gj = G

xj(5.152)

the last term of the equation(5.151) reads:

gjUj = UjG

xj=

DG

Dt +

G

t, (5.153)


34/42


one obtains for G

t = 0:

D

Dt 1

2U2j +G= Uj

P

xj. (5.154)

Considering that it holds:

D

Dt

P

=

t

P

+Uj

xj

P

(5.155)

and that moreover the following conversions are possible:

t

P

=

P

t

P

t, (5.156)

Uj

xj

P

= Uj

P

xj

P

Uj

xj, (5.157)

then it can be written:

D

Dt P = Pt +Uj Pxj P DDt . (5.158)From the equation (5.154) and (5.158):

D

Dt

1

2U2j

+G

= Uj

P

xj=

D

Dt

P

+

P

t

P

D

Dt, (5.159)

or after conversion of some terms:

D

Dt

1

2U2j

+

P

+G

=

P

t

P

D

Dt, (5.160)

or

D

Dt

1

2U2j

+

P

+G

=

P

t +

Ujxj

P. (5.161)

For stationary pressure fields Pt = 0, and for = const, the Bernoulliequation can be stated as follows:

1

2U2j +

P

+G=

1

2U2j +

P

xjgj = const (5.162)

The above derivations make clear under which conditions the well-knownBernoulli equation (5.162) holds.

From the above derivations, general form of the mechanical energy equationby including dissipative flow field can be written in other form:

D

Dt 1

2U2j

+P

+G= Pt +P

Uj

xj+

xi(ij

Uj

) ij

Uj

xi(5.163)

left side of this form of the mechanical energy contains all terms of theBernoulli equation.


35/42


5.8.3 Crocco Equation

The Crocco equation is a special form of the momentum equation which showsin an impressive manner how purely fluid-mechanical considerations can be

supplemented by thermodynamical insights. The Crocco equation connectsthe vortex power of a flow field to the entropy of the considered fluid. Itcan be shown from this equation that isotropic flows are free from rotationand vice versa under certain conditions. So when one recognizes a flow fieldto be isentrope, the simplified rotation-free flow fields considerations can beapplied.

For the derivation of the Crocco equation one starts from the Navier-Stokesequation, as it is stated in equation (5.144) supplemented by = 0, i.e. oneintroduces an ideal fluid into the considerations, to neglect inertia forces.

U

t +

1

2U U

U

U

=

1

P (5.164)

In section 3.6 it was shown that it holds:

T ds= de+P d= de+Pd

1

(5.165)

With e= hn P/ the following relation holds:

dh d

P

= Pd

1

+Tds (5.166)

because of d

P

= P d

1

+

1

dP it holds:

1dP= Tds dh (5.167)

This relation can also be written:

1

P =Ts h (5.168)

equation (5.168) is inserted in equation (5.164) to yield:

U

t +

1

2U U

U

U

= Tsh (5.169)

For stationary adiabatic processes the thermal energy equation can be written

in the following form:

Dh

Dt =

DP

Dt (5.170)

From the momentum equation it follows further:


36/42


D

Dt

1

2UU

= UP (5.171)

Thus:

D

Dt

h+

1

2UU

=

DP

Dt UP (5.172)

D

Dt

h+

1

2UU

=

DP

Dt (5.173)

equation (5.173) is inserted in (5.164) under stationary flow conditions toyield:

U +Ts=

h+

1

2UU

(5.174)

If flow is considered along a flow line, then (h+ 1/2 U U) is a vector vertical

to the considered flow line. U is also a vector and lies also vertical to the

flow line. ThusTsas well lies vertical to the fluid motion along a flow line,therefore it can be stated:

Unn+T ds

dn =

d

dn

h+

1

2UU

(5.175)

when

h+ 12

UU

is constant along a flow field, then ddn

h+ 1

2UU

= 0 and

thus:

Unn+T ds

dn = 0 (5.176)

Ifn = 0 then ds/dn = 0, thus rotation-free flows are isentropic and viceversa. If the flow is assumed to be stationary and in the absence of viscos-

ity,the inertial forces turn out to be zero.

5.8.4 Further Forms of the Energy Equation

The close connection between fluid mechanics and thermodynamics becomesclear from different forms of the energy equation, summarized in the fol-lowing table, as introduced by Bird, Steward & Lightfoot [5.1] the notationadapted in this book are as follows


37/42


Symbols Explanation Dimension

cp Heat capacity at constant pressure, L2/(T t2)

per mass unit

cv Heat capacity at constant volume, L2/(T t2)

per mass unit

etotal Total energy of the fluid, per M L2/t2

unit mass

e Internal energy, per unit mass M L2/t2

g,gi External mass acceleration L/t2

G Potential energy, potential ofG M L

2

/t

2

Symbols Explanation Dimension

h Enthalpy ML2/t2

P Pressure field M/(Lt2)

q, qi Heat flow per unit area M/t3

T Absolute temperature TU, Ui Velocity field L/t

V Volume L3

xi Cartesian coordinates L

Thermal expansion coefficient 1/T

Fluid density field M/L3

, ij Molecular momentum transport M/(Lt2)

Mass conservation (Continuity Equation)


38/42


39/42


Equation of Motion (Momentum Equation)

Special Equations in vector and Notes/observationsForm Tensor Notation

Imposed DU

Dt = P [ ] +g For = 0

convection one obtains

DUjDt

= Pxj

ijxi

+gj Euler equations

Free DU

Dt = [ ] gT This equation

convection comprises approximation

DUjDt

= ijxi

gjT by Boussinesque-assumptions

Energy Equations

Special Equations in vector and tensor notation Notes/Form observations

Written

for DetotalDt

= ( q) ( U) ( [ U]) Exact onlyetotal = e+ for G time

+12

U2

+G DetotalDt

= qixi

(PUi)xi

(ijUj)

xiindependent

D(e+ 1

2

U2)

Dt = ( q) ( U)

( [ U]) +( U g)

e+ 12

U2

D(e+ 1

2U2i)

Dt = qi

xi (PUi)

xi (ijUj)

xi+Uigi


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Special Equations in vector and Notes/observationsForm tensor notation

D 1

2U

2

Dt = ( U P) ( U [ ])

+( U g)12

U2

D 1

2U2i

Dt = Ui

Pxi

Uiijxj

+Uigi

DeDt

= ( q) P( U) ( : U) the PTerm containinge is zero

DeDt

= qixi

PUixi

ijUixj

for DDt

= 0

DhDt

= ( q) ( : U) + DPDt

h

DhDt

= qixi

ijUixj

+ DPDt

Written cvDTDt

= ( q) T(PT

)( U) For an ideal

for ( : U)cv and T Gas (

PT

) = PT

cvDTDt

= qixi T(PT

)(Uixi

) ijUixj

Written cpDTDt

= ( q) + ( ln Vln T

)DPDt

For an ideal

four ( : U)cp and T Gas (

ln Vln T) = 1

cpDTDt

= qixi

+ (ln Vln T

)DPDt ij

Uixj

5.9 Transport Equation for Chemical Species

In many domains of engineering science investigations of fluids with chemical

reactions are required which make it necessary to enlarge the considerationscarried out to-date. It is necessary to state the basic equations of fluid me-chanics for the different chemical components:


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5.9 Transport Equation for Chemical Species 157

Local modification of the mass of thechemical component A

At

V

Modification of the mass of component Aby inflow and outflow

xi

A(UA)i V

Production of the chemical component Aby chemical reaction

rAV

Thus yields a mass balance:

At

V=

xi[A(UA)i] V+rAV (5.177)

and the equation for the mass conservation for the chemical component A ofa fluid results as:

At

+ xi

[A(UA)i] =rA (5.178)

For a chemical component B, as a consequence of equal considerations:

Bt

+

xi[B(UB)i] =rB (5.179)

The addition of these equations yields:

t +

(Ui)

xi= 0 (5.180)

i.e. the total mass conservation equation for a mixture of different componentsis equal to the continuity equation for a fluid which consists of one chemicalcomponent only. By considering Ficks law of diffusion it can be stated:

At

+

xi(AUi) =

xi

DAB

(CA/C)

xi

+rA (5.181)

For = const and DAB = const, one obtains:

At

+

=0 A

Uixi

+UiAxi

=DAB2Axi

+rA (5.182)

or expressed in terms of concentration, CA

DCADt

= CA

t +Ui C

A

xi

= DAB

2CAxi2

+RA (5.183)

with rA= A RA, see Bird, Stewart, Lightfoot [5.1].


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5.10 Literature

[5.1]Bird, R.B., Stewart, W.E. & Lightfoot, E.N., Transport Phenomena, JohnWiley & Sons Inc., New York, 1960

[5.2]Spurk, J.H., Stromungslehre - Einfuhrung in die Theorie derStromungen, Springer Verlag Berlin, 4. Aufl., 1996

[5.3]Brodkey, R.S., The Phenomena of Fluid Motions, Dover Publications,Inc., New York, 1967

[5.4]Sherman, F.S., Viscous Flow, McGraw-Hill Book Co. Singapore, 1990

[5.5]Schlichting, H., Boundary Layer Theory, 6 Edition, McGraw Hill BookCompany New York, 1968

fluid mecanics nusif grundgl engl

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