Equations of Motion - Fluid Dynamics- Part 1- Classical Fluid Dynamics

to be a tensor, its elements should satisfy the following transformation rule:

q^′_ij= αikαjlqkl. Using this information and

(a) applying equation (1.2.10) for a direction n = e^′_ialong the x^′_i-axis, prove that the stress tensor (1.2.6) is a real tensor;

(b) taking into account that the velocity components transform as V_i^′ = αij, Vi= αjiV_j^′,

prove that the rate-of-strain tensor (1.4.26) is also a real tensor.

1.6 Equations of Motion

The differential equations describing fluid motion may be deduced from conservation of mass, momentum, and energy. The mass conservation law has been already used to formulate the Lagrangian continuity equation (1.4.6). We shall now do it using Eulerian variables.

1.6.1 Continuity equation in Eulerian variables

In order to formulate the mass conservation law in Eulerian variables, we shall consider an arbitrary regionD fixed in space (see Figure 1.25a) with fluid passing through D as time t increases. We denote the surface surrounding regionD by S and the external unit normal to S by n. The fluid mass contained in regionD at instant t is given by the integral

m(t) = ZZZ

ρ(t, r) dτ, (1.6.1)

where dτ is a volume element inD.

Notice that time t plays the role of a parameter in the integral (1.6.1). Since region D does not change with time, the differentiation of (1.6.1) involves variation of the integrand only. We have

dm dt =

ZZZ

∂ρ

∂t dτ, (1.6.2)

which represents the rate of change of mass inside regionD.

Provided that there are no sources or sinks of fluid inside D, this mass variation can be due only to the influx of fluid through the surface S. The mass flux through a small element ds of the surface S is given by

ρ V· n ds.

It equals the local density ρ multiplied by the volume of the slanted parallelepiped (see Figure 1.25b) swept by the fluid passing through ds per unit time. The area of the base of the parallelepiped is ds and its ribs are|V| long and have the same direction as the velocity vector V.

The total mass flux through the entire surface S is ZZ

ρV· n

ds. (1.6.3)

Since n is the external normal, the integral (1.6.3) represents the outflow from region D. So the mass conservation law, applied to D, can be written as

ρV· n

ds =−dm dt .

Applying Gauss’s divergence theorem to the left-hand side of this equation, we have ZZZ

div(ρV) dτ =−dm

dt. (1.6.4)

Combining (1.6.2) with (1.6.4) results in ZZZ

∂ρ

∂t + div(ρV)

dτ = 0. (1.6.5)

Let us suppose that both ρ and V have continuous derivatives. Then, by virtue of the arbitrariness of regionD, the integrand in (1.6.5) should be zero everywhere in the flow field, and so

∂ρ

∂t + div(ρV) = 0. (1.6.6)

S ds

x y

(a) Region D used in derivation of the mass con-servation law.

|V|

(b) Zoomed surface element ds and cylindri-cal region filled by the fluid passing through ds per unit time.

Fig. 1.25: Geometrical layout used in the derivation of the mass conservation law.

1.6. Equations of Motion 53 Indeed, if the function

Φ =∂ρ

∂t + div(ρV)

did not vanish (say, it was positive) at least at one point in the flow, then, because Φ is continuous, it would be positive in a small vicinity of this point. Hence the integral (1.6.5) taken over such a vicinity would also be positive.

Equation (1.6.6) is the continuity equation in Eulerian variables. In Chapters 2 and 3, we will be dealing with so-called incompressible fluid flows. These are flows where the fluid density ρ remains constant, and the continuity equation (1.6.6) reduces to

divV = 0. (1.6.7)

1.6.2 Momentum equation

We shall now formulate the fluid-dynamic version of Newton’s Second Law. When applied to a solid body, Newton’s Second Law is written as

ma = F. (1.6.8)

Here m is the mass of the body, a its acceleration, and F the force applied to the body.

The following three remarks should be made concerning equation (1.6.8). First, this equation is valid if a body’s motion is analysed in the framework of an inertial coordinate system. Second, if more than one force is applied to the body, then the vector F in (1.6.8) must be interpreted as the resultant force that is equal to the vector sum of all forces acting on the body. Third, if the body is not ‘small’ and its different parts experience different accelerations, then equation (1.6.8) serves to determine the acceleration a of the mass centre. A convenient idealisation is based on the assumption that the body size is small compared with the characteristic path traced by the body during the time of observation, in which case the body may be thought as a material point. In fluid dynamics, the role of material points is played by the fluid particles.

Let us consider an assemblage of N material points. Each element obeys Newton’s Second Law

dVi

dt = Fi. (1.6.9)

Here suffix i is used to enumerate the elements in the assemblage, and the acceleration ai of the ith element is written via the velocity derivative dVi/dt. Since the mass mi

does not vary with time, equation (1.6.9) may also be written as dKi

dt = Fi, (1.6.10)

with Ki= miVi being the momentum of the ith material point.

The force Fiis represented as a superposition of internal and external forces:

Fi= XN j=1

Fij+ Fie, (1.6.11)

where Fij is the force exerted by the element with number j upon the element with number i, and Fie is the external force produced by any physical agents outside the system under consideration.

Substitution of (1.6.11) into (1.6.10) and summation over all the elements in the assemblage yields the momentum equation

dt = R, (1.6.12)

where K is the momentum of the entire material system,

K= XN

i=1

miVi, (1.6.13)

and R is the resultant external force,

R= XN i=1

Fie.

Internal forces obviously cancel in the course of summation owing to Newton’s Third Law, which states that an action and reaction are equal and opposite: Fij =−F^ji.

Now we shall apply equation (1.6.12) to a moving fluid. To express the derivative dK/dt on the left-hand side of (1.6.12) in terms of the fluid-dynamic variables, we have to keep in mind that the momentum equation (1.6.12) is valid for a material system consisting of the same elements; no exchange of matter between the system under consideration and the surrounding medium is allowed.

LetD be again an arbitrary region in an inertial coordinate system. The surface surroundingD we will denote as before by S and the external unit normal to S by n (see Figure 1.26). However, now we shall treat this region differently. We choose an arbitrary instant t and ‘mark’ all the fluid particles that happen to be insideD at this instant. Then we follow these fluid particles as time increases; considered together,

S ds

x y

Fig. 1.26: Fluid body deforming with time.

1.6. Equations of Motion 55 they form a fluid body, to which we shall apply the momentum equation (1.6.12). The momentum (1.6.13) of the body is expressed by the integral

K(t) = ZZZ

ρ(t, r)V(t, r) dτ. (1.6.14)

Direct differentiation of (1.6.14) with respect to t involves differentiation of the integrand ρV as well as analysis of the fluid body deformation,¹⁰which is significantly more intricate. This may be avoided if, before differentiating K(t), a transformation of integration variables is performed in (1.6.14) with new variables chosen in such a way that the region of integration ceases to depend on time t. A proper choice is obviously provided by the Lagrangian position vector function

r= r(t, r0), (1.6.15)

which relates the ‘initial location’ r0of a fluid particle at time t0to its current location r at time t. Formula (1.6.15) may obviously be considered as a transformation of variables from (x, y, z) to (x0, y0, z0) and vice versa. Using (1.6.15) in (1.6.14) results in

whereD⁰ is the region occupied by the fluid body at the initial time t0 and dτ0 is a volume element fromD⁰.

It follows from the continuity equation in Lagrangian variables (1.4.6) that Hence, (1.6.16) may be rewritten as

K(t) =

Here both the initial density ρ0 and initial regionD⁰ are independent of t. Therefore differentiation of (1.6.17) yields

Taking into account that the expression in the square brackets is the acceleration of a fluid particle (1.4.10), we can further write

10The shape of the body at the next instant t + dt is depicted in Figure 1.26 by the dashed line.

It remains to return to the original integration variables (x, y, z), and we will have dK

dt = ZZZ

ρDV

Dt dτ. (1.6.18)

We shall now calculated the resultant force R acting on the fluid inside regionD.

As we know, all the forces acting on a moving fluid may be subdivided into two classes:

body forces and surface forces. With f (r, t) denoting the body force upon a unit mass, the force acting on a volume element dτ insideD is calculated as

ρ(r, t)f (r, t) dτ,

and the entire body force acting upon the fluid contained in regionD proves to be ZZZ

ρ(r, t)f (r, t) dτ. (1.6.19)

Let us now turn to the surface forces. According to (1.2.3), an element ds of the surface S surrounding regionD experiences a force (see Figure 1.27)

dPn= pnds. (1.6.20)

The total force acting upon the fluid inD through S is calculated as

Rs= ZZ

pnds. (1.6.21)

S ds

x y

Fig. 1.27: Calculation of the surface force.

1.6. Equations of Motion 57 Adding (1.6.21) to (1.6.19), we obtain the resultant force R that should be used on the right-hand side of equation (1.6.12). Using for the left-hand side the formula (1.6.18), we arrive at the following integral form of the momentum equation:

ZZZ

In order to express this equation in a differential form, we need to convert the integral (1.6.21) for the surface force Rs into a volume integral. When performing this task, it is convenient to use index notation, with (x, y, z) denoted as (x1, x2, x3).

Taking into account that, according to (1.2.10),¹¹ pn = nipi,

the projection of the vector equation (1.6.21) on the xα-axis may be written as Rs

then the right-hand side of (1.6.23) may be expressed via the scalar product of A and the normal unit vector n. We have

Now Gauss’s divergence theorem may be used, leading to Rs

11Here we again use the well-known summation convention according to which terms containing a repeated suffix are to be regarded as summed over all three possible values of the suffix, i.e.

nipi=

i=1

nipi.

Since equation (1.6.24) is valid for all α = 1, 2, 3 , it may be written in vector form as Rs=

ZZZ

divP dτ. (1.6.25)

The vector divP is referred to as the divergence of the stress tensor (1.2.6); its com-ponents are written as

divP

Substitution of (1.6.25) into (1.6.22) yields ZZZ

and we can conclude that in a region of smooth variation of the fluid-dynamic functions ρ, p, and V, the following equation holds:

ρDV

Dt = ρf + divP, (1.6.27)

which represents the sought differential form of the momentum equation.

1.6.3 The energy equation

To derive the energy equation we shall return to the First Law of Thermodynamics (1.3.11) and apply it to the fluid body contained at time t inside regionD as shown in Figure 1.26. Instead of the work performed by the body, we will consider an opposite quantity, the work performed by the forces acting upon the body. Correspondingly, we shall write equation (1.3.11) in the form

dt = W + Q, (1.6.28)

where E is the energy of the fluid body, W the work performed per unit time by the forces acting on the body, and Q the rate of heat transfer to the body.

In a fluid in motion, the energy E consists of the internal energy of fluid particles and their kinetic energy. The internal energy per unit mass is e. Hence the internal energy of a fluid particle occupying a volume element dτ is eρ dτ . The kinetic energy of the fluid particle is ¹₂V²ρ dτ . This means that the entire energy E of the fluid body occupying regionD at time t may be written as

E(t) =

1.6. Equations of Motion 59 The derivative of this function is calculated in the same way as the derivative of the momentum K(t) given by (1.6.18). We have

dE dt =

ZZZ

ρD Dt

e +V²

dτ. (1.6.29)

Let us now calculate the work W on the right-hand side of (1.6.28). It is known that the work is given by the scalar product F· δr of the force F acting on a body, and the distance δr travelled by the body. Thus the work per unit time will be F· V.

The work performed by the body force acting on a fluid particle is ρ f· V dτ , and therefore the entire work of the body forces acting upon regionD is

Wb= ZZZ

ρ f· V

dτ. (1.6.30)

The surface force acting on an element ds of surface S (see Figure 1.27) is given by equation (1.6.20). The work performed by this force is pnds· V

. Integrating over the entire surface S surrounding the fluid body in regionD yields

Ws= ZZ

pn· V

ds. (1.6.31)

The second term on the right-hand side of equation (1.6.28) represents the heat rate Q. We shall assume that the only physical process that leads to heat transfer is the heat conduction due to the temperature variation in the flow field.¹² The heat conduction vector q is known to be proportional to the temperature gradient, i.e.

q=−κ∇T,

where κ is a positive constant called the heat conductivity coefficient. It depends on the temperature only, and is related to the dynamic viscosity coefficient, µ, such that the quantity

P r =µcp

κ , (1.6.32)

called the Prandtl number, can be treated as a constant for a given fluid. In particular, for air at ‘room temperature’ and atmospheric pressure, P r≈ 0.713.

With n being the external normal to the surface S (see Figure 1.27), the heat transfer towards the fluid body contained in regionD is calculated as

Q =− Z Z

q· n ds =

κ∇T · n

ds. (1.6.33)

Substituting (1.6.29), (1.6.30), (1.6.31), and (1.6.33) into (1.6.28), we arrive at the

12In hypersonic flows, the gas temperature might become sufficiently high to provoke an additional process, namely radiation of heat.

following integral form of the energy equation:

In order to convert the surface work integral (1.6.31) into a volume integral, we again use formula (1.2.10). We have

pn· V = (nⁱpi)· V = nⁱ pi· V

= nipijVj. Hence, if we introduce a vector A = (A1, A2, A3) such that

Ai= pijVj, i = 1, 2, 3, (1.6.35) then the integral (1.6.31) may be written as

Ws=

Now we can use Gauss’s divergence theorem. We have ZZ

Here it is taken into account that the vector A, whose components are defined by equation (1.6.35), is equal to the product of the stress tensor (1.2.6) and the velocity vector V.

Applying Gauss’s divergence theorem also to the heat transfer integral on the right-hand side of equation (1.6.34), we can express this equation in the form

ZZZ

Taking into account that the regionD is arbitrary, we deduce that the integrand must be zero. We have

ρD The kinetic energy term on the left-hand side of (1.6.37) may be calculated using the momentum equation (1.6.27). Indeed, scalar multiplication of (1.6.27) with the velocity vector V yields

ρD

which, when subtracted from (1.6.37), results in the following form of the energy equation:

ρDe

Dt = div(PV) − V · div P + div(κ∇T ). (1.6.38)

In document Fluid Dynamics- Part 1- Classical Fluid Dynamics (Page 63-73)