Baumgarten

54 3 Basic Equations

Fig. 3.4. One-dimensional pipe flow with heat and work

	Q Ws	³	§ p			u2	·	&	&
			U ¨		e		gz ¸	u	ndA .	(3.28)
				U		2
			©				¹
		Ain Aout

Since the flow is one-dimensional, p, Υ, u, and e are uniform over A1 and A2. Further on, the velocity is also normal to all surfaces where fluid crosses, resulting in

§ p		2		u2	·	§ p			u2
Q Ws ¨			e2	2	gz2 ¸ U2u2 A2	¨	1	e1	1
	U	2		2			U		2
©					¹	©	1

·

gz1 ¸ U1u1 A1 . (3.29)

¹

Using the continuity equation Υ u A = Υ u A = m for one-dimensional steady flow, we obtain

Q

W

§

p

u2

· §

p

u2

s

¨

2

e

2

gz

2

¸

¨

1

e

1

m m

U2

2

2

U1

1

2

©

¹

©

·

gz1 ¸ . (3.30)

¹

Introducing the enthalpy h = e + p/Υ yields

Q

W

§

u2

·

§

u2

s

¨

h

2

gz

2 ¸

¨

h

1

m

m

2

2

1

2

©

¹

©

·

gz1 ¸ . (3.31)

¹

If the energy transfer caused by the left hand side of Eq. 3.30 is zero and the internal energy of the fluid does not change (isothermal flow), the energy equation becomes equal to the Bernoulli equation (Sect. 3.1.3.2):

p	2		u2			p		u2
			2	gz		1		1	gz .	(3.32)
U			2		2 U			2
	2								1
					1

3.1.3 Conservation Equations for Multi-Dimensional Flows

In order to describe three-dimensional compressible and non-isothermal flows, the differential conservation equations of mass (one equation), momentum (three equations), and energy (one equation) are used to calculate the three unknown ve-

3.1 Description of the Continuous Phase	55

locity components u1, u2, u3 as well as the two thermodynamic variables pressure p and temperature T. The remaining thermodynamic variables and transport properties density Υ, enthalpy h (or internal energy e), dynamic viscosity µ, and heat conductivity Ο, which appear in the final forms of the conservation equations, are given by the state relations Υ =Υ (p, T), h = h (p, T), µ = µ (p, T) and Ο = Ο (p, T), which must be known in order to complete the system of equations. In the following, the differential conservation equations will be derived.

3.1.3.1 Conservation of Mass

The continuity equation in differential form can be derived from a mass balance at an infinitesimal volume element dV = dx1dx2dx3, as shown in Fig. 3.5. Due to the Eulerian approach, the coordinate system as well as the control volume are fixed in space. Thus, the volume element is solely a control volume that is passed by the flow without resistance, and that is used only in order to quantify the amount of mass inside the volume at a given time t. The mass will increase if the inflow exceeds the outflow and will decrease in the opposite case. In both cases, the mass density inside the control volume will change. The total mass balance yields

w

dx1dx2dx3 U

dmx1

dmx2 dmx3 .

wt

The mass balance in the x1-direction gives

dmx1

mx1

mx1

dx2dx3 Uu1 x dx2dx3 Uu1 x dx .

x

x

dx

1

1 1

1

1

1

Using a Taylor series for the second term on the right hand side,

Uu1

Uu1

w

Uu1

dx1 ,

x1 dx1

x1

wx

x1

1

the mass balance in the x1-direction finally reads

dx2dx3

w

Uu1

dx1 .

dmx

wx1

1

x1

(3.33)

(3.34)

(3.35)

(3.36)

The other coordinate directions are treated accordingly. Because the mass balances hold true for any position (x1, x2, x3) of the control volume inside the threedimensional flow field, the indices are omitted in the following. Using the mass balances in Eq. 3.33 and eliminating the constant volume dx1dx2dx3 finally yields the general form of the continuity equation

wU		w	Uu1	w	Uu2	w	Uu3 0 ,	(3.37)
wt		wx		wx		wx
	1		2		3

which can also be written as

56 3 Basic Equations

Fig. 3.5. Mass fluxes entering and exiting the control volume
		wU		div Uu&			0 ,	(3.38)
		wt
or as
	wU				w
						Uui	0 ,	(3.39)
		wt			wx
					i

if the Einstein notation is used. If the product rule is applied to the second term in Eq. 3.39, the continuity equation can be transformed into

	DU	U	wui	0 ,	(3.40)
	Dt		wx
			i

where DΥ /Dt is the substantial derivate of Υ.

If the fluid density is constant (incompressible flow), the continuity equation reduces to

	&		&	wui
	u	div	u		0 .	(3.41)
				wx
				i

3.1.3.2 Conservation of Momentum

The momentum conservation equations can be derived in a manner similar to the mass conservation equation. Fig. 3.6 shows the corresponding infinitesimal control volume, which is again fixed in space and passed by the flow without resistance. The momentum equation is based on the principle that the temporal change

wM&		w	Uu& dx1dx2dx3	(3.42)
wt		wt

3.1 Description of the Continuous Phase	57

Fig. 3.6. Momentum fluxes entering and exiting the control volume

Fig. 3.7. Normal and shear stresses

of the momentum of the fluid inside the control volume at time t equals the sum of all momentum fluxes entering the volume (due to mass entering the volume) minus the sum of all momentum fluxes exiting the control volume, plus the sum of all external forces, which are forces due to static pressure, friction inside the fluid, and gravitation (body forces).

First of all, the momentum fluxes due to mass transport shall be derived. As an example, Fig 3.6 shows the momentum fluxes entering and leaving the control volume through the two faces that are perpendicular to the x1-axis. Because the mass flows through both of these faces have velocity components in all three coordinate directions, momentum also having components in all three directions is transported. This becomes obvious regarding the mass flow entering the control volume through the left face: due to the velocity components u1, u2, and u3, the mass flow transports momentum Υu1,Υu2, and Υu3, which enters the control volume with an effective velocity (perpendicular to the surface) u1. For this reason, the momentum fluxes at the left face in Figure 3.6 are dx2dx3Υu1u1, dx2dx3Υu2u1,

58 3 Basic Equations

and dx2dx3Υu3u1. The momentum fluxes through the remaining five faces are treated accordingly.

Next, the normal and shear stresses, as shown in Fig. 3.7, have to be considered. The meaning of the indices of the tensions is as follows: the first index characterizes the face of the control volume. For example, if the unit vector, which is normal to the surface, is in x1-direction, the first index is x1. The second index indicates the direction of the force caused by the tension. The force due to static pressure p is always directed&into the control volume.

Finally, the body force f = (f1, f2, f3) must be considered. This is usually the gravitational force, whose components are f1 = 0, f2 = 0, f3 = -Υg, see Fig. 3.7.

In contrast to the continuity equation, the momentum is not a scalar but a vector. For this reason, one resulting equation for each of the Cartesian dimensions x1, x2, and x3 has to be written. For the x1-dimension, the equation reads

w

Uu1 dx1dx2dx3

§

§

w

Uu1u1

··

¨Uu1u1

¨Uu1u1

dx1 ¸¸dx2dx3

wt

wx

©

©

1

¹¹

§

§

w Uu1u2

··

¨Uu1u2 ¨Uu1u2

dx2 ¸¸dx1dx3

wx2

©

©

¹¹

§

§

w Uu1u3

··

¨Uu1u3 ¨Uu1u3

dx3 ¸¸dx1dx2

wx3

©

©

¹¹

§

§

wW

x1 x1

dx

·

·

dx dx

¨

W

x1 x1

W

¸

¸

¨

x1 x1

wx1

1

2 3

©

©

¹

¹

§

§

wW

x2 x1

dx

·

·

dx dx

¨

W

x2 x1

W

¸

¨

x2 x1

wx2

2 ¸

1 3

©

©

¹

¹

§

§

wW

x3 x1

dx

··

dx dx

¨

W

x3 x1

W

¸

¨

x3 x1

wx3

3 ¸

1 2

©

©

¹

¹

§ p

§ p

wpx1

dx

··dx dx dx dx dx f .

¨

x1

¨

x1

wx1

1

¸

¸

2 3

1 2 3 1

©

©

¹

¹

(3.43)

The static pressure p has the same value in all three coordinate directions, and the indices can be omitted. Eq. 3.43 can be simplified:

w Uu1

w Uu1u1

w Uu1u2

w Uu1u3

wt

wx

wx

wx

1

2

3

wp

+

wWx1 x1

wWx2 x1

wWx3 x1

f .

wx1

wx1

wx2

wx3

1

(3.44)

Applying the product rule to the left hand side of Eq. 3.44 yields:

3.1 Description of the Continuous Phase	59

w Uu1

w Uu1u1

w Uu1u2

w Uu1u3

U

wu1

u

wU

wt

wx

wx

wx

wt

1 wt

w Uu1

1

2

Uu2

3

w Uu3

wu

w

wu

wu

u

Uu

1

u

Uu

1

u

Uu

3

wx

wx

2 wx

wx

3 wx

1

1 wx

1

1

1

1

2

2

3

3

U

§

wu1

u

wu1

u

2

wu1

u

wu1

·

¨

1

wx1

wx2

3

¸

© wt

wx3 ¹

u

§wU

w Uu1

w Uu2

w Uu3

·.

¨ wt

1

wx

wx

wx

¸

©

1

2

3

¹

(3.45)

The last term of Eq. 3.45 contains the left hand side of the continuity equation and is equal to zero. Hence, Eq. 3.44 can be rewritten:

U

Du1

U

§

wu1

u

wu1

u

2

wu1

u

wu1

·

Dt

¨

1

wx1

wx2

3

¸

© wt

wx3 ¹

wp

+

wWx1 x1

wWx2 x1

wWx3 x1

f .

wx1

wx1

wx2

wx3

1

The corresponding equations for the x2- and the x3-directions are

Du

2

wp

wWx x

wWx x

wWx x

U

1

2

2

2

3 2

f2

,

Dt

wx2

wx1

wx2

wx3

Du

wp

wWx x

wWx x

wWx x

U

3

1

3

2

3

3 3

f3 .

Dt

wx3

wx1

wx2

wx3

(3.46)

(3.47)

(3.48)

A much shorter way of writing these three ( j = 1, 2, 3) momentum equations is achieved using the Einstein notation:

U

Du j

U

§wu j

u

wu j ·

wp

wWij

f

.

(3.49)

¨

¸

Dt

wt

i wx

wx

j

wx

j

©

i

¹

i

The second term on the right hand side includes the components of the stress tensor

	§W	x1 x1	W	x2 x1
	¨
TW			Wx2 x2
	¨Wx1 x2
	¨		Wx2 x3
	©Wx1 x3

	W	x3 x1	·
			¸
	Wx3 x2			,	(3.50)
			¸
	Wx3 x3		¸
			¹

which accounts for momentum transfer due to friction. For Newtonian fluids (µ ζ f(Ω)), the well-known linear relation between shear stress and velocity gradient for

60 3 Basic Equations

one-dimensional flow has been extended for use in three-dimensional flows (Stokes’ hypothesis (1845), e.g. [13, 12, 20]):

	§ wu			wu j	·	2		wu
Wij	µ ¨	i			¸ Gij		µ	i	.	(3.51)
		wx		wx		3		wx
	©	j		i	¹			i

The stress tensor is symmetric, which means that Ωij = Ωji. In Eq. 3.51, Γij is the Kronecker symbol (Γij = 1 if i = j and Γij = 0 otherwise). For non-Newtonian fluids, the relation between stress tensor and velocity is usually much more complex and has to be defined by a set of partial differential equations. This subject is just at the beginning of being explored. However, most of the fluids of interest can be treated as Newtonian fluids.

The so-called Navier-Stokes equations are the complete equations of motion for a viscous Newtonian fluid. Using Eq. 3.51 in Eq. 3.49 finally yields

Du j

wp

w

§

§ wu

i

wu j

2 wu

i

··

U

¨µ ¨

Gij

¸¸

f j .

Dt

wx

j

wx

wx

j

wx

3

wx

i ©

©

i

i ¹¹

(3.52)

As an example, the first of the three Navier-Stokes equations (x1-direction) reads:

U

Du1

U

§

wu1

u

wu1

u

2

wu1

u

wu1

·

Dt

¨

wt

1

wx1

wx2

3

¸

©

wx3 ¹

wp

w §

§wu

wu

2 §wu

wu

2

wu

3

···

¨µ ¨

1

1

¨

1

¸¸¸

wx1

3

wx2

wx1 © © wx1

wx1

© wx1

wx3 ¹¹¹

w §

§wu

2

wu

··

w

§

§wu

wu

··

µ

1

¨

µ

3

1

¸¸

f .

¨

¨

¸¸

¨

wx3

1

wx2 ©

© wx1

wx2 ¹¹

©

© wx1

wx3 ¹¹

(3.53)

For incompressible flows, the first and the last term inside the brackets on the right hand side of Eq. 3.52 cancel out (Eq. 3.41), such that Eq. 3.52 becomes

Du j

wp

w2u j

U

µ

f j .

(3.54)

Dt

wx

j

wx2

i

In a coordinate-free vector form, Eq. 3.54 reads

Du&

§wu&

&

&·

2

&

&

U

U ¨

u u ¸

p µ

u f .

(3.55)

Dt

© wt

¹

In frictionless flows, there are no shear stresses, and the second term on the right hand side of Eq. 3.52 vanishes completely. This form of the momentum equation is the so-called Euler equation. If the flow is also steady, a special form of the momentum equation, the Bernoulli equation, can be derived by integrating between two points along a streamline (coordinate: s)

							3.1 Description of the Continuous Phase 61
	2	wu	2 w u2 / 2			u22 u12	2 1 wp					(3.56)
	³u ws ds ³			ws	ds	2	³ U ws ds g z2 z1 ,
	1		1				1

where z is the elevation above an arbitrary datum. For incompressible flow, the Bernoulli equation reads:

u2	u2		p	2	p
2	1				1	g z2	z1 0 .	(3.57)
	2
					U

3.1.3.3 Conservation of Energy

In order to derive the energy equation, the temporal change of the energy contained in an infinitesimal volume element is regarded, Fig. 3.8. The total energy,

	§			u&		2	·
E	¨Ue U						¸dx dx dx ,		(3.58)
			2
t	¨						¸	1 2 3
	©						¹

consists of the internal energy e [J/kg] and the kinetic energy. The velocity term in Eq. 3.58 can also be expressed as _ u& _2 = u& · u& = u12 + u22 + u32. The change of energy inside the control volume is equal to the sum of energy fluxes dEm/dt entering and exiting the control volume due to mass entering or exiting, plus the sum of the energy fluxes dQ/dt due to heat conduction, plus the sum of the work per unit time done by pressure (dWp /dt), viscous forces (dWΩ /dt), and body forces (dWg /dt), plus energy supply dWs /dt due to shaft work, radiation, or chemical processes inside the volume (e.g. combustion). Hence, the energy equation reads

Fig. 3.8. Convective energy fluxes through a control volume

62 3 Basic Equations

w

§

u&

2

·

E

Q W

W

W

W

¨Ue U

¸dx dx dx

p

g

.

(3.59)

2

wt ¨

¸

1 2 3

m

W

s

©

¹

Remember that energy is not a vector, but a scalar. Hence, there is only one equation, as in the case of the continuity equation.

First, the energy flux due to mass flow will be derived. Fig. 3.8 shows the energy fluxes entering and exiting the control volume. In the x1-direction, the resulting energy flux entering the volume is

E

E

E

m,x1 x1 dx1

§

u&

2 ·

m,x1

m,x1

¨Ue

U

¸u dx dx

2

x1

¨

¸

1

2

3

©

¹

§§

&

2

·

w

§§

&

2

· ·

·

¨¨Ue

U

u

¸u

¨

¨Ue U

u

¸u

¸dx

¸dx dx

2

wx

2

¨

¨

¸

1

¨

¨

¸ 1 ¸

1 ¸

2 3

©

©

¹

1

©

©

¹

¹

¹

w

§

§

&

2

·

·

¨

¨Ue

U

u

¸u

¸ dx dx dx .

wx

2

¨

¨

¸ 1 ¸

1 2 3

1

©

©

¹

¹

(3.60)

The energy fluxes in the two other coordinate directions are derived accordingly. The resulting energy flux entering the volume due to mass flow is

ª

§§

&

2

·

·

§§

&

2

·

·

«

w

¨¨Ue U

u

¸

w

¨¨Ue

u

¸

E

m

«wx

2

¸u

wx

U

2

¸u

¨¨

¸

1

¸

¨¨

¸

2 ¸

¬

1

©©

¹

¹

2

©©

¹

¹

w

§

§

&

2

· ·º

ª§

&

2

·

&

º

¨

u

¸»dx dx dx =- «¨Ue

u

»dx dx dx .

¨Ue U

¸u

U

¸u

wx

2

2

¨

¨

¸

3

¸»

1 2 3

Ǭ

¸

»

1 2 3

3

©

©

¹

¹¼

©

¹

¼

(3.61)

Next, the energy fluxes due to heat conduction must be derived. Assuming that the heat transfer to the element is given by Fourier’s law,

&

O T ,

or

O

wT

,

(3.62)

q

q j

wxj

where

2

q

is in [W/m ], the heat flux in the x1-direction is

Qx1

ª

wT

§

wT

w §

wT ·

·º

« O

¨

O

¨

O

¸dx1

¸»dx2 dx3 ,

(3.63)

wx1

wx1

«

©

wx1 ©

wx1 ¹

¹»

¬

¼

and the resulting heat flux due to heat conduction can be expressed as

3.1 Description of the Continuous Phase	63

Q

ª w

§

wT ·

w

§

wT ·

w

§

wT ·º

«

¨O

¸

¨O

¸

¨O

¸»dx1dx2dx3

wx2

wx3

wx3

(3.64)

«wx1

©

wx1 ¹

©

wx2 ¹

©

¹»

¬

&

O T

dx dx dx .

¼

qdx dx dx

1 2

3

1

2 3

Next, the relations for the work done by pressure as well as normal and shear stresses are discussed. At each face of the control volume, two shear stresses, one normal stress and the static pressure, have to be accounted for, see Fig. 3.7. At first, only the two opposing faces normal to the x1-axis shall be regarded. The rate of work done to the element per unit time due to static pressure is

Wp,x1

ª

§

w pu1

·º

w pu1

«pu1

¨ pu1

dx1 ¸

»dx2dx3

dx1dx2dx3

,

wx1

wx1

«

©

¹»

¬

¼

and the work done due to viscous forces is

ª

§

w Wx1 x1 u1

·

º

«

»

W

«

W

u

¨W u

dx ¸

»

dx dx

wx

W ,x1

x1 x1 1

¨ x1 x1

1

1 ¸

2 3

«

©

1

¹

»

¬

w Wx1 x2 u2

¼

ª

§

·º

«

W

u

¨W u

dx

¸»dx dx

wx1

«

x1 x2

2

¨ x1 x2

2

1

¸»

2 3

«

©

¹»

¬

¼

ª

§

·º

«

W

u

¨W u

w Wx1 x3 u3

dx ¸

»dx dx

«

x1 x3

3

¨ x1 x3

3

wx1

1 ¸

»

2 3

«

©

¹

»

¬

w Wx1 x1 u1

w Wx1 x2 u2

w Wx1 x3 u3

¼

= ¨§

¸·dx dx dx .

¨ wx1

wx1

wx1

¸

1 2 3

©

¹

(3.65)

(3.66)

In Eq. 3.66, the work per unit time, which is the product of velocity and force, is given a positive sign if velocity (directed in positive coordinate direction) and force (directed by the corresponding normal or shear stress, Fig. 3.7) have the same direction, and a negative sign in the case of opposite directions.

The expressions for the other two coordinate directions are derived accordingly. The resulting expressions for the work done per unit time by pressure and viscous forces are

§w pu1

w pu2

w pu3 ·

&

Wp

¨

¸dx1dx2dx3

pu

dx1dx2dx3

,

(3.67)

wx

wx

wx

©

1

2

3

¹

and