Momentum Heat Mass Transfer MHMT9 Mechanical, internal energy and enthalpy balance and heat...

Momentum Heat Mass TransferMHMT9

Mechanical, internal energy and enthalpy balance and heat transfer. Fourier´s law of heat conduction. Fourier-Kirchhoff ´s equation. Steady-state heat conduction. Thermal resistance (shape factor).

Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010

Energy balances. FK equations

sourceDt

D

Power of forcesMHMT9

When analysing energy balances I recommend you to imagine a fixed control volume (box) and forces acting on the outer surface

sdFdu

ds Power of surface force [W] dF u

Kinetic energy balanceMHMT9

Duu u p u u fDt

2

dissipation ofreversiblework done by work done by mechanicalexpansionkinetic energy pressure forces viscous forces energyacting at surface to heat

12 ( ) ( ) :

D upu p u u

Dt

work of external forces

f u

Kinetic energy balance follows directly from the Cauchy’s equation for velocities (kinetic energy is square of velocities, so that it is sufficient to multiply the Cauchy’s equation by velocity vector; scalar product results to scalar energy)

and this equation can be rearranged (how? see the next slide) to the final form of the kinetic energy transport

Kinetic energy balanceMHMT9

irreversible dissipation ofoverall work done by reversible change viscous forces work to heatviscous forces to kinetic energy

( ) ( ) :u u

12 k k

kk

u uuu

t t

12 k k

kk m m

m m

u uuu u u

x x

k kk

k k k

u p upu p

x x x

( )2

km m km mm km

k k k

m km km m k

k k m

m kmkm mk

k

u uu

x x x

u u u

x x x

u

x

Proof follows from the previous equation rewritten to the index notation

( )k k kmk m k m k k

m k k

u u pu u u u u f

t x x x

When considering different forms of energy transport equations it is important to correctly interpret decomposition of energies to reversible and irreversible parts

overall work done by conversion adiabatic expansionpressure forces to kinetic energy

( ) ( ) ( )pu p u p u

Dissipation of kinetic energyMHMT9

U=ux(H)y

1 1 1( )

2 2 2x

xy yx x y y x

uu u

y

1( ( ) )

2Tu u

1: : ( ( ) ) :

2Tu u u

This identity follows from the stress tensor symmetry

Rate of deformation tensor

: : xy yx yx xy xyu

Example: Simple shear flow (flow in a gap between two plates, lubrication)

Example: DissipationMHMT9

Gap width H=0.1mm, U=10 m/s, oil M9ADS-II at 00C

=3.4 Pa.s, =105 1/s, =3.4.105 Pa, = 3.4.1010 W/m3

At contact surface S=0.0079 m2 the dissipated heat is 26.7 kW !!!!

U=10 m/sy

H=0.1 mm

D=5cm

L=5cm

Rotating shaft at 3820 rpm

Generated heating power in a gap between rotating shaft and casing

Internal energy balanceMHMT9

The equation of kinetic energy transport is a direct consequence of Cauchy’s equations and does not bring a new physical information (this is not a new “law”). Such a new law is the equation of internal energy balance which represents the first law of thermodynamics stating, that the internal energy increase is determined by the heat delivery by conduction and by the mechanical work. This statement can be expressed in form of a general transport equation for =uE

specificinternalenergy

Eu q

P ( )

dissipated heat generatedreversible mechanical by volumetric expansion energy (internal ohmic or(gas cools down friction ) microwave heatingwhen expanded)

: gp u Q

Heat flux by conduction

( ): gEDuq p u Q

Dt

Internal energy uE is defined as the sum of all energies (thermal energy, energy of phase changes, chemical energy) with the exception of kinetic energy u2/2 and this is the reason why not all mechanical work terms are included in the transport equation and why the reaction heat is not included into the source term Q(g).


Interpretation using the First law of thermodynamics

du = dq - p dv

( ): gEDuq p u u Q

Dt

Heat transferred by conduction into FE Expansion cools

down working fluid

This term is zero for incompressible fluid

Dissipation of mechanical energy to heat by viscous friction

u

::


Remark: Confusion exists due to definition of the internal energy itself. In some books the energy associated with phase changes and chemical reactions is included into the production term Q(g) (see textbook Sestak et al on transport phenomena) and therefore the energy related to intermolecular and molecular forces could not be included into the internal energy. This view reduces the internal energy only to the thermal energy (kinetic energy of random molecular motion).

Example: Consider exothermic chemical reaction proceeding inside a closed and thermally insulated vessel. Chemical energy decreases during the reaction (energy of bonds of products is lower than the energy of reactants), no mechanical work is done (constant volume) and heat flux is also zero. Temperature increases. As soon as the internal energy is defined as the sum of chemical and thermal energy, the decrease of chemical energy is compensated by the increased temperature and DuE/Dt=0.

Total energy balanceMHMT9

The internal energy transport equation is in fact the transport of total energy (this is expression of the law of energy conservation) from which the transport equation for the kinetic energy is subtracted

2 ( )

heat delivered energy commingwork done by work done byby conduction from ousidetotal energy [J/kg] pressure viscous forces

(electric heat)

1( ) ( ) ( )

2g

E

Du u q pu u Q

Dt

The term is potential of conservative external volumetric forces (for example potential energy in gravitational field gh).

Enthalpy balanceMHMT9

Thermal engineers prefer balancing of steady continuous systems in terms of enthalpies instead of internal energies. The transport equation follows from the equation for internal energy introducing specific enthalpy h=uE+pv (v is specific volume)

( )( ) : gEDu Dh pv q p u Q

Dt Dt

( ) ( )D p Dh p D Dp Dh Dph p u

Dt Dt Dt Dt Dt Dt

( ): gDh Dpq Q

Dt Dt

giving the final form of the enthalpy balance

This enthalpy balance (like all the previous transport equations for different forms of energy) is quite general and holds for compressible/incompressible fluids or solids with variable transport properties (density, heat capacity, …)

Fourier Kirchhoff equationMHMT9

Feininger


Primary aim of the energy transport equations is calculation of temperature field given velocities, pressures and boundary conditions. Temperatures can be derived from the calculated enthalpy (or internal energy) using thermodynamic relationship

( )

it is necessary to includealso reaction and phase this term is zero forchanges enthalpies (andincompressible liquid andelereduces to Dp/Dt for ideal gas.

( ) : Rp p

DT v Dpc T q QDt T Dt

ctric heat as previously)

giving the transport equation for temperature

The reason why reaction enthalpy and enthalpy of phase changes had to be included into the source term Q(R) is the consequence of limited applicability of thermodynamic relationships between enthalpy and temperature (for example DT/Dt=Dp/Dt=0 during evaporation but Dh/Dt>0).

Dt

Dp

T

vTv

Dt

DTc

Dt

Dhpp ))((


Diffusive (molecular) heat flux is proportional to the gradient of temperature according to the Fourier’s law

( )( ) : Rp

DTc T QDt

Tq

where is thermal conductivity of material.

Fourier Kirchhoff equation for temperature field reads like this

As soon as thermal conductivity is constant the FK equation is

2 ( ): Rp

DTc T QDt

Material [W/(m.K)] a [m2/s]

Aluminium Al 200 80E-6

Carbon steel 50 14E-6

Stainless steel 15 4E-6

Glas 0.8 0.35E-6

Water 0.6 0.14E-6

Polyethylen 0.4 0.16E-6

Air 0.025 20E-6

Thermal and electrical conductivities are similar: they are large for metals (electron conductivity) and small for organic materials. Temperature diffusivity a is closely related with the thermal conductivity

Memorize some typical values: pca

Thermal conductivity of nonmetals and gases increases with temperature (by about 10% at heating by 100K), at liquids and metals usually decreases.

Thermal conductivity MHMT9

Let us consider special case: Solid homogeneous body (constant thermal conductivity and without internal heat sources). Fourier Kirchhoff equation for steady state reduces to the Laplace equation for T(x,y,z)

Boundary conditions: at each point of surface must be prescribed either the temperature T or the heat flux (for example q=0 at an insulated surface).

Solution of T(x,y,z) can be found for simple geometries in an analytical form (see next slide) or numerically (using finite difference method, finite elements,…) for more complicated geometry.

2

2

2

2

2

220

z

T

y

T

x

TT

2

2

22

10 ( )

10 ( )

T Tr

x r r rT

rr r r

The same equation written in cylindrical and spherical

coordinate system (assuming axial symmetry)

Conduction - stationaryMHMT9

cylinder

sphere

Steady transversal temperature profile in a plate is described by FK equation which reduces to

1D conduction (plate)MHMT9

22

20

TT

x

with a general solution (linear temperature profile) 21)( cxcxT

h

Tw1

Tw2

Tf1

x

Differential equations of the second order require two boundary conditions (one BC in each point on boundary)

1. BC of the first kind (prescribed temperatures)

2. BC of the second kind (prescribed flux q0 in one point)

3. BC of the third kind (prescribed heat transfer coefficient)

2 1 1( ) ( )w w w

xT x T T T

h

0 02( ) w

q q hT x x T

1 2 1 2( )( ) f w f wT T hT TT x x

h h

these boundary conditions with fixed values are called Dirichlet boundary conditions

the boundary conditions of the second kind is called Neumann’s boundary condition

the boundary condition of the third kind is called Newton’s or Robin’s boundary condition

Steady radial temperature profile in a cylinder and sphere (for fixed temperatures T1 T2 at inner and outer surface)

2 1 1 2 2 11 1 2 1 2

2 1 2 1

2 1 1 2 2 2 1 11 2 1 2 1̀ 2

2 1 2 1

ln ln, T=c ln c

ln / ln /

c, T=- c ( )

T T T R T RTr c r c cr R R R R

R R T R T RTr c c T T c

r r R R R R

Sphere (bubble)

cylinder

R1

R2

1D conduction (sphere and cylinder)MHMT9

Knowing temperature profiles it is possible to calculate heat flux q [W/m2] and the heat flow Q [W]. The heat flow can be also specified as a boundary condition (thus the radial temperature profile is determined by one temperature T0 at radius R0 and by the heat flow Q)

1 1 2

0 1 0 2

1

, T=c ln ,

T =c ln

Q=-2 r 2

Tr c r cr

R c

TL L c

r

1D conduction (cylinder-heating power)MHMT9

For cylinder with thermal conductivity and for specified heat flow Q related to length L, the logarithmic temperature profile can be expressed in the following form

r

R

L

QTrT 0

0 ln2

)(

R0,T0

L

Q

Positive value Q>0 represents a line heat source, while negative value heat sink.

Temperature distribution is a solution of a linear partial differential equation (Laplace equation) and therefore is additive. It means that any combination of simple solutions also satisfies the Laplace equation and represents some stationary temperature field. For example any previously discussed solution of potential flows (flow around cylinder, sphere, see chapter 2) represents also some temperature field (streamlines are heat flux lines, and lines of constant velocity potential are isotherms). Therefore the same mathematical techniques (conformal mapping, tables of complex functions w(z) describing dipoles, sources, sinks, circulations,…) are used also for solution of

temperature fields

electric potential field

velocity potential field

concentration fields

The principle aim is thermal resistance, electrical resistance… Thus it is possible to evaluate for example the effect of particles (spheres) or obstacles (cylinders) to the resistivity of inhomogeneous materials.

2D conduction (superposition)MHMT9

2D conduction (superposition)MHMT9

As an example we shall analyse superposition of two parallel linear sources/sinks of heat Q

Temperature at an arbitrary point (x,y) is the sum of temperatures emitted by source Q and sink -Q

Q

S

SQ r

r

L

QT

r

R

r

R

L

QTyxT ln

2)ln(ln

2),( 0

000

Lines characterised by constant values k=rS/rQ are isotherms.

22

22

2

22

)(

)(

yhx

yhx

r

rk

Q

S

This equation describes a circle of radius R and with center at position m 2

2 2

2 1,

1 1

kh kR m h

k k

rQ

h

y

x

T(x,y)

Sink -Q

rS

Source Qh

T1

T2

x

y

Resulting temperature field describes for example the following cases: (see also the next slide)

T1T2

x

y

m

R

Knowing temperature field and thermal conductivity it is possible to calculate heat fluxes and total thermal power Q transferred between two surfaces with different (but constant) temperatures T1 a T2

TR

TTQ 21

RT [K/W] thermal resistance

L

RRRT 2

/ln 12)(1

2

2

1

1

hh

SRT

h1 h2

S

T1 T2

h

S1

S2

T1 T2

L

R1

R2

T1 T2

L

R1

T1

T2

h

Q

2211 SS

hRT

L

RhRT 2

/2ln

Serial Parallel Tube wall Pipe burried under surface

In this way it is possible to express thermal resistance

of windows, walls, heat transfer surfaces …

Thermal resistanceMHMT9

EXAMMHMT9

Energy transport

What is important (at least for exam)MHMT9

2

dissipation ofreversiblework done by work done by mechanicalexpansionkinetic energy pressure forces viscous forces energyacting at surface to heat

1( ) ( ) ( ) :2

Du pu p u u

Dt

work of external forces

f u

Kinetic energy

( ): gEDuq p u Q

Dt

Internal energy (by subtraction kinetic energy, see also 1st law duE=dq-dw)

2 ( ): Rp

DTc T QDt

Fourier Kirchhoff

Conservation of all forms of energy

2 ( )1( ) ( ) ( )

2g

E

Du u q pu u Q

Dt


2

2

2

2

2

220

z

T

y

T

x

TT

2

2

22

10 ( )

10 ( )

T Tr

x r r rT

rr r r

Steady state heat conduction (cartesian, cylindrical, spherical coordinates)

1 2

12

T=c ln

cT=-

r c

cr

1D temperature profiles (cartesian, cylindrical, spherical coordinates)


Thermal resistance

TR

TTQ 21

L

RRRT 2

/ln 12

)(1

2

2

1

1

hh

SRT

Serially connected plates

Cylinder

Momentum Heat Mass Transfer MHMT9 Mechanical, internal energy and enthalpy balance and heat...

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Transcript of Momentum Heat Mass Transfer MHMT9 Mechanical, internal energy and enthalpy balance and heat...