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).
Internal energy balanceMHMT9
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
::
Internal energy balanceMHMT9
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
Fourier Kirchhoff equationMHMT9
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 ))((
Fourier Kirchhoff equationMHMT9
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
What is important (at least for exam)MHMT9
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)
What is important (at least for exam)MHMT9
Thermal resistance
TR
TTQ 21
L
RRRT 2
/ln 12
)(1
2
2
1
1
hh
SRT
Serially connected plates
Cylinder