Download Momentum Heat Mass Transfer

MHMT9
Momentum Heat Mass Transfer
D
     source
Dt
Energy balances. FK
equations
Mechanical, internal energy and enthalpy balance and
heat transfer. Fourier´s law of heat conduction. FourierKirchhoff ´s equation. Steady-state heat conduction.
Thermal resistance (shape factor).
Rudolf Žitný, Ústav procesní a
zpracovatelské techniky ČVUT FS 2010
MHMT9
Power of forces
When analysing energy balances I recommend you to imagine a fixed control
volume (box) and forces acting on the outer surface
ds
Power of surface force [W] dF  u

 
dF  ds 
u
MHMT9
Kinetic energy balance
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)
Du
u
 u p  u     u f
Dt
and this equation can be rearranged (how? see the next slide) to the final form of
the kinetic energy transport
1 2
D u
 2   ( pu )  p u   ( u )   :    f u
Dt
dissipation of
reversible
work of
work done by
work done by
kinetic energy
pressure forces
acting at surface
expansion
viscous forces
mechanical
energy
to heat
external
forces
MHMT9
Kinetic energy balance
Proof follows from the previous equation rewritten to the index notation
uk
uk
 km
p
 uk (
 um
)  uk
 um
  uk f k
t
xm
xk
xk
um
1

uk uk
u
p uk p
uk
2
uk

p k
 uk

xk
xk
xk
t
t
1
 uk uk
u
 uk um k   u m 2
xm
xm
 km um km
u

  km m 
xk
xk
xk

um km  km um uk

(

)
xk
2 xk xm

um km
  km  mk
xk
When considering different forms of energy transport equations it is important to
correctly interpret decomposition of energies to reversible and irreversible parts
 ( u )
 (  ) u 
overall work done by
viscous forces
reversible change
to kinetic energy
 ( pu )  (p) u 
overall work done by
pressure forces
conversion
to kinetic energy
 :
irreversible dissipation of
viscous forces work to heat
p( u )
adiabatic expansion
MHMT9
Dissipation of kinetic energy
1
 : u   : (u  (u )T )   : 
2
1
  (u  (u )T )
2
This identity follows from
the stress tensor symmetry
Rate of deformation
tensor
Example: Simple shear flow (flow in a gap between two plates, lubrication)
1
1 ux 1
 xy   yx  ( xu y   y ux ) 
 
2
2 y 2
U=ux(H)
y
 : u   :    xy  yx   yx  xy   xy
MHMT9
Example: Dissipation
Generated heating power in a gap between rotating shaft and casing
Rotating shaft at
3820 rpm
D=5cm
L=5cm
U=10 m/s
H=0.1 mm
y
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 !!!!
MHMT9
Internal energy balance
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
Pq
  uE
specific
internal
energy
Heat flux by
conduction
   p u 
reversible
expansion
(gas cools down
when expanded)
 :
dissipated
mechanical
energy (internal
friction )

Q( g )
heat generated
by volumetric
ohmic or
microwave heating
DuE

  q  p u   :   Q ( g )
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).
MHMT9
Internal energy balance
Interpretation using the First
du
=
dq
-
law of thermodynamics
p dv
DuE

   q  p  u   : u  Q ( g )
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
MHMT9
Internal energy balance
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.
MHMT9
Total energy balance
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
D
1 2
 (uE  u   )   q   ( pu )   ( u )  Q ( g )
Dt
2
heat delivered
energy comming
work done by
work done by
total energy [J/kg]
by conduction
pressure
viscous forces
from ouside
(electric heat)
The term  is potential of conservative external volumetric forces (for example
potential energy in gravitational field gh).
MHMT9
Enthalpy balance
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)
DuE
D

(h  pv)   q  p u   :   Q ( g )
Dt
Dt
D
p
Dh p D Dp
Dh
Dp
 (h  )  



 p( u ) 
Dt

Dt  Dt Dt
Dt
Dt

giving the final form of the enthalpy balance
Dh
Dp
(g)

  q 
 :   Q
Dt
Dt
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, …)
MHMT9
Fourier Kirchhoff equation
Feininger
MHMT9
Fourier Kirchhoff equation
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
Dh
DT
v
Dp
 cp
 (v  T ( ) p )
Dt
Dt
T
Dt
giving the transport equation for temperature
DT
cp

Dt
v Dp
T ( ) p
T Dt
this term is zero for
incompressible liquid and
reduces to Dp/Dt for ideal gas.
  q  :  
Q( R )
it is necessary to include
also reaction and phase
changes enthalpies (and
electric heat as previously)
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).
MHMT9
Fourier Kirchhoff equation
Diffusive (molecular) heat flux is proportional to the gradient of temperature
according to the Fourier’s law

q    T
where  is thermal conductivity of material.
Fourier Kirchhoff equation for temperature field reads like this
DT
cp
  (  T )   :   Q ( R )
Dt
As soon as thermal conductivity is constant the FK equation is
DT
cp
  2T   :   Q ( R )
Dt
MHMT9
Thermal conductivity 
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
a
Memorize some typical values:
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
c p
Thermal conductivity of nonmetals and gases increases with temperature (by
about 10% at heating by 100K), at liquids and metals  usually decreases.
MHMT9
Conduction - stationary
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)
 2T  2T  2T
0 T 
 2  2
2
x
y
z
2
cylinder
sphere
 2T 1  T
0 2 
(r )
x r r r
1 
T
0  2 (r 2 )
r r
r
The same equation written in
cylindrical and spherical
coordinate system (assuming
axial symmetry)
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.
MHMT9
1D conduction (plate)
Steady transversal temperature profile in a plate is described by FK equation
which reduces to
 2T
0 T 
x 2
2
with a general solution (linear temperature profile)
Tf1
T ( x)  c1 x  c2
Differential equations of the second order require two
boundary conditions (one BC in each point on boundary)
 Tw1
1. BC of the first kind (prescribed temperatures)
Tw2
x
T ( x)  (Tw 2  Tw1 )  Tw1
h
these boundary conditions with fixed values
are called Dirichlet boundary conditions
2. BC of the second kind (prescribed flux q0 in one point)
h
x
T ( x)  
q0

x  Tw 2 
q0 h

the boundary conditions of the second kind is
called Neumann’s boundary condition
3. BC of the third kind (prescribed heat transfer coefficient)
 (T f 1  Tw2 )
 hT f 1  Tw2
T ( x)  
x
 h
 h
the boundary condition of the third kind is called
Newton’s or Robin’s boundary condition
MHMT9
1D conduction (sphere and cylinder)
Steady radial temperature profile in a cylinder and sphere (for fixed
temperatures T1 T2 at inner and outer surface)
R1
R2
cylinder
T
r
 c1 ,
r
r2
T
 c1 ,
r
Sphere (bubble)
T=c1 ln r  c2
T=-
c1
 c2
r
T2  T1
T1 ln R2  T2 ln R1
c1 
c2 
ln R2 / R1
ln R2 / R1
c1 
R1 R2
T R  T1 R1
(T2  T`1 ) c2  2 2
R2  R1
R2  R1
MHMT9
1D conduction (cylinder-heating power)
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)
L
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
T
 c1 ,
T=c1 ln r  c2 ,
r
T0 =c1 ln R0  c2
r
Q
R0,T0
Positive value Q>0 represents
a line heat source, while
negative value heat sink.
Q=-2 L r
T
 2 L c1
r
Q
R0
T (r )  T0 
ln
2L
r
MHMT9
2D conduction (superposition)
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.
MHMT9
2D conduction (superposition)
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
y
of temperatures emitted by source Q and sink -Q
T(x,y)
rQ
rS
x
Source Q
Sink -Q
h
h
R0
R0
r
Q
Q
T ( x, y )  T0 
(ln
 ln )  T0 
ln S
2L
rQ
rS
2L rQ
Lines characterised by constant values k=rS/rQ are
isotherms.
rS2 ( x  h) 2  y 2
2
k 
2
Q
r

( x  h) 2  y 2
This equation describes a circle of radius R and with center at position m
2kh
R 2 ,
k 1
k 2 1
m 2 h
k 1
y
Resulting temperature field describes
T2
for example the following cases:
(see also the next slide)
y
T1
R
m
x
T2
T1
x
MHMT9
Thermal resistance
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
T T
R [K/W] thermal resistance
Q
T1
T2
1
2
T
RT
In this way it is possible to
express thermal resistance of
windows, walls, heat transfer
surfaces …
T1 T2
T2
S2
R2
S
Q
S1
R1
h
T1 T2
R1
T1
L
L
h
h1 h 2
RT 
1 h1 h2
(  )
S 1 2
Serial
RT 
h
S11  S 2 2
Parallel
RT 
ln R2 / R1
2L
Tube wall
RT 
ln 2h / R
2L
Pipe burried under surface
MHMT9
EXAM
Energy transport
MHMT9
What is important (at least for exam)
Kinetic energy
D 1 2
 ( u )   ( pu )  p u   ( u )   :    f u
Dt 2
dissipation of
reversible
work of
work done by
work done by
kinetic energy
pressure forces
acting at surface
expansion
viscous forces
mechanical
energy
to heat
external
forces
Conservation of all forms of energy

D
1
(uE  u 2 )   q   ( pu )   ( u )  Q ( g )
Dt
2
Internal energy (by subtraction kinetic energy, see also 1st law duE=dq-dw)
DuE

  q  p u   :   Q ( g )
Dt
Fourier Kirchhoff
cp
DT
  2T   :   Q ( R )
Dt
MHMT9
What is important (at least for exam)
Steady state heat conduction (cartesian, cylindrical, spherical coordinates)
 2T  2T  2T
0 T  2  2  2
x
y
z
2
 2T 1  T
0 2 
(r )
x
r r r
1  2 T
0  2 (r
)
r r
r
1D temperature profiles (cartesian, cylindrical, spherical coordinates)
T=c1 ln r  c2
T=-
c1
 c2
r
MHMT9
What is important (at least for exam)
Thermal resistance
Q
T1  T2
RT
Serially connected plates
1 h1 h2
RT  (  )
S 1 2
Cylinder
RT 
ln R2 / R1
2L