heat and mass transfer © prof. dr.-ing. h. bockhorn, karlsruhe 2012 heat and mass transfer 2....
TRANSCRIPT
Heat and Mass Transfer
© P
rof.
Dr.
-Ing
. H. B
ockh
orn,
Kar
lsru
he 2
012
Heat and Mass Transfer
2. Molecular transport processes: Heat conduction and mass diffusion 2.1 Stationary heat conduction
2.1.1 Fourier´s law of stationary heat conduction and heat conductivity
The general law of diffusive fluxes of properties
can be written in case of transport of energy (heat conduction):
Equation (2.1.1-2) is Fourier‘s law of stationary (one-dimensional) heat conduction. The heat flux is the product of heat conductivity, area and temperature gradient. The heat flux vector is anti-parallel to the temperature gradient (also vector). This is noted by the negative sign in equation (2.1.1-2). Heat, therefore, always is transported in the direction of lower temperatures (this is also a consequence of the second law of thermodynamics).
2.1.1/1 – 01.2012
)11.1.2(s
ofunit grad=diff
ATJ
)21.1.2()()( dx
dTT
dx
dTAT QQ jJ or
The heat conductivity [W/m K] according to Fourier‘s law is the heat flux density jQ [W/m2] by heat conduction per unit of negative temperature gradient dT/dx [K/m]. The heat conductivity is a material property.
2.1.2 Mechanisms of heat conduction in gases, liquids and solids
Gases. We consider a gas with a temperature gradient in one direction, see figure 2.1.2-1. The gas molecules move by thermal motion from an isotherm with a certain temperature to an other isotherm with higher (lower) temperature. Hopping between the isotherms the gas mole-cules carry their internal energy and transfer the energy by collisions to other molecules. Thereby, „hotter“ regions are cooled and „cooler“ regions are heated. If the temperature gradient is kept at constant, a heat flux is established from high temperature to low temperature.
Heat and Mass Transfer
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. H. B
ockh
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Heat and Mass Transfer
Fig. 2.1.2-1: Simple model for explication of heat conduction in gases.
x
T
Gaseous molecules accumulate their internal energy in form of kinetic energy (translation), rotational energy and oscillation energy (the latter forms only in multi-atom molecules, e.g. H2O, N2, CO2….). Rotational energy and oscillation energy are quantized.
Heat conductivity of gases. According to the simple model of heat conduction, the heat conductivity of ideal gases can be desribed with the kinetic theory of gases:
In equation (2.1.2-1) is the mean free pass of the molecules and the mean molecular velocity of the molecules.
Then:
2.1.2/1 – 01.2012
)12.1.2(3
1 VV cc
_
v
v
)22.1.2(8
M
RT
_
vp
RT
Nd A
22
1
)32.1.2(1
3
22
VA
cMRT
Nd
According to equation (2.1.2-3) the heat conductivity increases with increasing temperature (for atoms T0,5) and with increasing molar mass. According to the kinetic theory of gases, there is no pressure dependence of the heat conductivity for ideal gases.
Table 2.1.2-1 contains the heat conductivities of some gases. From the table a slight pressure dependence of the heat conductivity can be seen (in contrast to the simple kinetic gas theory).
Heat and Mass Transfer
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. H. B
ockh
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Heat and Mass Transfer
Table 2.1.2-1: Heat conductivity of some gases.
temperature T/oC -173 -73 25
substance heat conductivity /Wm-1K-1
hydrogen 0,06802 0,12626 0,17694
methane 0,01063 0,02185 0,03428
propane 0,01800
carbon dioxide 0,00950 0,01666
oxygen 0,00904 0,01833 0,02658
air 0,02640
air 0,03150 (10 MPa)
water vapor 0,01940 (0,0023 MPa)
2.1.2/2 – 01.2012
Table 2.1.2-2: Heat conductivity of some liquids.
Heat and Mass Transfer
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. H. B
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Heat and Mass Transfer
Heat conductivity of liquids. The mechanism of heat conduction in liquids is similar to that of heat conduction in gases, however, the interactions of molecules play an important role (hydrogen bridge bonds, van der Waals interactions, other long range and short range interactions).
Table 2.1.2-2 gives an overview about the heat conductivities of some liquids. The heat conductivity of most liquids is one order of magnitude higher than the heat conductivity of gases.
Water and other inorganic fluids (ammonia) exhibit a much higher heat conductivity than organic fluids; molten salts show even higher heat conductivities.
The heat conductivity of liquids increases generally with increasing temperature.
2.1.2/3 – 01.2012
Simple model for heat conduction in solids. In solids molecules (lattice building blocks) cannot move freely. Tey are fixed at their lattice places, see figure 2.1.2-2.
Solids store thermal energy in form of oscillations of the atoms about their rest positions in the lattice. In solids the transport of energy occurs via the oscillations of neigh-boring molecules. All molecules in the solid form a system of coupled oscillators (same mechanism as the transport of sound in solids).
Fig. 2.1.2-2: Simple model for explication of heat conduction in solids.
Heat and Mass Transfer
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Heat and Mass Transfer
The energy quanta released or accepted from this system of coupled oscillators are called phonons. Therefore, heat conduction in solids is accomplished by transport of phonons. In addition, in metals and electrical conductors the transport of phonons is superimposed by the transport of electrons in the conductivity bond. Therefore, heat conductivity in solids is caused by the transport of phonons and the thermal movement of conductivity electrons. The latter effect increases the heat conductivity.
The heat conductivity of metals is two to three orders of magnitudes higher than that of gases. Metals show a much higher thermal conductivity compared with other solids.
Heat conductivities of metals correspond to electrical conductivities (Fe < Al < Cu< Ag); Wiedemann-Franz-Lorenz law. contribution of conductivity electrons to heat conduction.
(: electrical conductivity in (Wm)-1).
2.1.2/4 – 01.2012
)42.1.2()/(1045,2 28
KVT
Figure 2.1.2-3 exhibits the temperature dependency of the heat conductivity of some solids. Table 2.1.2-3 gives some values for the heat conductivity of solids.
For metals and other electrical conductors the heat conductivity is a complicated function of temperature.
Difference between electrical conductors and semi-conductors and insolaters in the temperature dependency of the heat conduction.
Often the heat conductivity is non isotropic (graphite, wood).
Detailed information about heat conductivities of different materials, methods of estimation and methods of calculation can be found in VDI-Wärmeatlas.
Heat and Mass Transfer
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Heat and Mass Transfer
Stoff Temperatur T/oC
Wärmeleitfähigkeit /Wm-1K-1
Gusseisen 20 42-63
Normalstähle 20 46-50
Stahl 18Cr/8Ni 20 17
500 21
Kupfer 18 384
100 380
Zinn 0 63,9
100 59,8
Quarzglas 20 1,4
Fensterglas 20 0,76
Stahlbeton 20 1,5
Ziegelstein 20 0,8-0,9
Gummi 20 0,13-0,23
Polystyrol 20 0,17
Holz (trocken) 20 0,14-0,16
Asbestisolierung 20 0,16-0,24
Glas- und Steinwolle 20 0,03-0,45
Styropor 20 0,03-0,04
Table 2.1.2-3: Heat conductivities of some solids.
2.1.2/5 – 01.2012
Fig. 2.1.2-3: Temperature dependency of the heat conductivity of some solids.
10-1
100
101
102
103
104
105
0 200 400 600 800 1000
AluminiumEisenSilberConstantan
PyrexGraphit(senkrecht)Graphit(parallel)
Wär
mel
eitf
ähig
keit
/W
m-1K
-1
Temperatur T/K
Heat and Mass Transfer
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Heat and Mass Transfer
Setting x = s1 and T(s1) = T2 and resolving for JQ gives:
For 1 varying with temperature, no linear temperature profile!
2.1.3 Stationary heat conduction in planar and non-planar geometries, heat transfer coefficient
Stationary heat conduction through planar geometries. We consider a planar plate with homogeneous material properties, see figure 2.1.3-1. Boundary conditions at the left and right surface are kept at constant temperatures. Heat conductivity is 1 .
Using Fourier‘s law we obtain:
Assuming 1 being constant separation of variables leads to:
Integration within boundaries 0 and x results in:
For this problem we obtain a linear decrease of temperature within the plate.
)43.1.2(1
211
s
TTAQ J
JQ
Fig. 2.1.3-1:Stationary heat conduction in planar geometries.
T1
)13.1.2()(1 dx
dTATQ J
)23.1.2(1
dxA
dT Q
J
2.1.3/1 – 01.2012
)33.1.2(1
1
xA
TT Q
J
A
s1T
x
1
T2
Heat flux 0, if 0,Heat flux 0, if s ∞,Heat flux 0, if T 0.
1 increasing with tempe-rature
1 decreasing with tempe-rature
Heat and Mass Transfer
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Heat and Mass Transfer
Stationary heat conduction in spherical shells. We consider a spherical shell which is treated as one-dimensional problem (coordinate radius r). The temperatures at the inner and outer surface are constant in time and surface, see figure 2.1.3-2. The material has the heat conductivity 1.
According to Fourier‘s law we write:
In this case the area increases with radius. Therefore, according to Fourier‘s law we expect a steeper decrease of the temperature. Again the total heat flux is constant. Separation of variables in equation (2.1.3-4) gives:
Integration within the boundaries Ri and r gives:
The temperature in the sperical shell varies hyperbolicly with the radius r (at constant heat conductivity, compare planar problem: linear).
Fig. 2.1.3-2:Stationary heat conduction in non-planar geometries (sperical shells).
)43.1.2(4)( 211
dr
dTr
dr
dTrAQ J
)53.1.2(4 2
1
r
drdT Q
J
)63.1.2(11
4)(
1
rRTrT
i
Qi
J
T
r
Ti
Ta
Ri RaJQ
2.1.3/2 – 01.2012
Setting in equation (2.1.3-6) r = Ra and T(Ra) = Ta and resolving for the heat flux JQ we obtain:
With this we obtain the same form for the heat conduction equation as in the planar geometry.
)73.1.2()(
)(4
)(11
4
1
1
ia
aiia
ai
ai
Q
RR
TTRR
TT
RR
J
Heat and Mass Transfer
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Heat and Mass Transfer
At otherwise constant conditions (T, s, ) the heat flux in spherical geometries is smaller compared to planar geometries.
For Ra >> Ri in equation (2.1.3-6) 1/Ra may be neglected against1/Ri. Using this we obtain:
Heat flux --> 0, if -->Heat flux --> 0, if T --> 0,Heat flux --> minimum heat flux, if s --> ∞.
)83.1.2()(41 aiiQ TTRJ
Even at high wall thicknesses (strong isolation) a minimum residual heat flux remains in spherical shell geometries according to equation (2.1.3-8).
2.1.3/3 – 01.2012
Heat transfer coefficient. We consider the problem analogous to heat conduction in a spherical shell, figure 2.1.3-2. The cavity is now filled with material and the shell thickness is extended to infinity. The surface temperature of the sphere (filled cavity) is constant and the temperature far away from the surface is also kept at constant, see figure 2.1.3-3.
Fig. 2.1.3-3: Introduction of the heat transfer coefficient.
T
r
Ti
Ri
R --> ∞ T = Ta
JQ
The shell material (which is now the surrounding of the sphere) has the heat conductivity 1. This scenario describes the heat transfer from a sphere into surroundings at rest.
Heat and Mass Transfer
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Dr.
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. H. B
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Heat and Mass Transfer
Fig. 2.1.3-4: Introduction of the heat transfer coefficient.
For this problem a heat transfer coefficient is defined according to:
or:
With and and equa-tion (2.1.3-8) we obtain
T
r
Ti
Ri
R --> ∞ T = Ta)103.1.2(
gesbez
Q
TA
J
JQ
With the help of the heat transfer coefficient given in equation (2.1.3-9) any heat transfer problem can be now written as:
)143.1.2( TAQ J
2.1.3/4 – 01.2012
)93.1.2(
difference re temperatudrivingarea
fluxheat
)( aiges TTT 24 ibez RA
)113.1.2()(4
)(4 12
1
iaii
aii
gesbez
Q
RTTR
TTR
TA
J
Generally, the heat transfer coefficient is „non-dimensionalized“ with a characteristic length L and the heat conductivity :
This number is called Nusselt number. For the problem of heat transfer from a sphere into a surrounding at rest we obtain:
)123.1.2(
NuL
)133.1.2(21
1
1
K
iKK
DRD
Nu
Heat and Mass Transfer
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Heat and Mass Transfer
or
2.1.4 Stationary heat conduction in layered materials, over all heat transfer coefficient
Stationary heat conduction in layered materials. We consider a planar body consisting of layers of different materials. The over-all thickness is small compared with the lateral dimensions, so that the problem can be treated as a one-dimensional heat conduction problem. At the left and right wall constant temperatures are applied, seee figure 2.1.4-1. The different layers have the heat conductivities 1 ,2 and 2.
Applying Fourier‘s law we obtain:
and
and
Resolving for the temperature differences and adding results in:
)44.1.2(
)( 33/23/22/12/113
3
2
2
1
1
TTTTTTsss
AQ
J
JQ
Fig. 2.1.4-1:Stationary heat conduction in planar layered materials.
JQT1T1/2
)14.1.2(1
2/111
s
TTAQ J
)24.1.2(2
3/22/12
s
TTAQ J
)54.1.2()(1
31
3
3
2
2
1
1
TTAsssQ
J
2.1.4/1 – 01/2012
T3
T2/3
A
s1T
x1
s2
2 3
s3
JQ
)34.1.2(3
33/23
s
TTAQ J
Heat and Mass Transfer
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Heat and Mass Transfer
From equations (2.1.4-1) to (2.1.4-3) we find that the temperature differences in the single layers are inversely proportional to the heat conductivities per unit lenght (insolation problem).
Over-all heat transfer coefficient. For the stationary heat conduction problem through a layered material we define an over-all heat transfer coefficient k similar to the heat transfer coefficient in section 2.1.3:
)74.1.2(
difference re temperatudrivingarea
fluxheat k
)84.1.2()( 1
1
2/111
sTTAk Q J
2.1.4/2 – 01.2012
)64.1.2(/
/
/
/
22
33
2/22/1
33/2
22
11
2/11
3/22/1
s
s
TT
TT
s
s
TT
TT
JQ
Fig. 2.1.4-1:Stationary heat conduction in planar layered materials.
JQT1T1/2
T3
T2/3
A
s1T
x1
s2
2 3
s3
JQ
The heat transfer coefficients according to equations (2.1.4-8) to (2.1.4-10) are heat conductivities normalized to the thickness of the layers (specific conductivities). Then the reciprocal of the heat transfer coefficients are specific resistances (analogy to conduction of electricity).
)94.1.2()( 2
2
3/22/12
sTTAk Q J
)104.1.2()( 3
3
22/33
sTTAk Q J
Heat and Mass Transfer
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Heat and Mass Transfer
Using equations (2.1.4-8) to (2.1.4-10) in equation (2.1.4-5) yields:
or
With
For heat conduction through „heat conduction resistances“ in series the single resistances add to the total resistance! (reciprocal specific conductivities, see figure 3.1.4-2).
For heat conduction through „heat conduction resistances“ in parallel the single specific conductivities add to the total conductivity! (reciprocal specific restistances, see figure 3.1.4-3).
2.1.4/3 – 01.2012
)114.1.2()(111
131
321
TTA
kkk
QJ
)124.1.2()()(11
3131 TTAkTTA
k
QJ
)134.1.2(1111
321221
RRRRkkkk
Fig. 2.1.4-2: „Heat conduction resistances“ in series.
Fig. 2.1.4-3: „Heat conduction resistances“ in parallel.
)144.1.2()()( 31321
321
TTAkkk
QQQQ JJJJ
)154.1.2(1
)111
()(321
321 RRRR
kkkk
JQ
33
1
kR
22
1
kR
11
1
kR
JQ 1
33
1
kR
22
1
kR
JQ 2
JQ 3
11
1
kR
Heat and Mass Transfer
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Heat and Mass Transfer
Heat transfer to the ambient (U: circumference):
Heat balance for stationary conditions:
)55.1.2( übQxxQxQ JJJ
Problem: Temperature distribution in the slab, heat loss by the slab (thermal bridge, cold bridge).
For the calculation of the temperature profile in the slab we consider a heat balance for a differential volume of the slab. Within the slab we have heat conduction, at the surface we have heat transfer to the ambient. (Simplification: temperature constant over the radius (1-d problem).
Heat balance:
Heat flux in = heat flux out (2.1.5-1)
Heat flux into the balance volume:
Heat flux out of the balance volume:
Fig. 2.1.5-1: Stationary heat conduction in a cylindrical slab.
T
x
Ta
2.1.5 Examples of stationary heat conduction in engineering
Heat conduction in slabs with heat transfer into the ambient. We consider a cylindrical slab with the length L, mounted perpendicular to a surface. The surface is kept at temperature T0 and the ambient temperature is Ta. The material has the heat conductivity , see fig. 2.1.5-1 („thermal or cold bridge“).
T0
JQ
x x+x
JQx+xJQx
JQüb
)35.1.2()(
xxQ dx
dTAxx J
)25.1.2()(
xQ dx
dTAx J
)45.1.2()( aaussenübQ TTxUJ
L
2.1.5/1 – 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
Differentiating equation (2.1.5-9) twice by gives:
Equation (2.1.5-9) is a solution of (2.1.5-8) only if
Heat flux at x+x is obtained by a Taylor series expansion of the heat flux at x (linearisation):
For the heat balance we obtain:
Transformation with = (T-Ta)/(T0-Ta) and = x/L gives:
In equation (2.1.5-8) m2 = 4L(aussen/)(L/D). The ratio (aussenR/) can be understood as ratio of the heat transfer resistance slab/wall to wall/ambient. This ratio is called Biot number, Bi. Then m2 = 8 (L/D)2 Bi (Bi must be small for 1-d problem).
Solution of the ordinary differential equation of second order (2.1.5-8):
Fig. 2.1.5-1: Stationary heat conduction in a cylindrical slab.
)65.1.2()(
)()(
2
2
xdx
TdAx
xdx
dxxx
Q
QQQ
J
JJJ
)95.1.2( eC
)75.1.2()(02
2
aaussen TTxUx
dx
TdA
)85.1.2(22
2
2
mA
LU
d
d aussen
)105.1.2(222
2
eC
d
d
)115.1.2(8with
2122
BiD
Lm
mmm or
T
x
TaT0
JQ
x x+x
JQx+xJQx
JQüb
L
2.1.5/2 – 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
The solution then is:
General solutions: linear combination of particular solutions:
Integrations constants A and B form boundary conditions.
Boundary condition 1: For = 0 T =T0, therefore = 1. This gives:
Boundary condition 2: Finite long slab, heat transfer to the ambient at the front end side negligible. Then d/d = 0, for = 1. This gives:
Considering equation (2.1.5-13) we have:
)125.1.2( mm eBeA
)145.1.2(0
m
mmm
e
eABemBemA
)135.1.2(1 BA
)165.1.2(]cosh[
)]1(cosh[
m
m
)155.1.2(
mm
m
mm
m
ee
eB
ee
eA
Fig. 2.1.5-2: Solution according to equation (2.1.5-16) for stationary heat conduction in a cylindrical slab (m = 5).
2.1.5/3 – 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
Heat loss via the „cold bridge“ to the ambient as heat transfer problem: The total heat flux from the slab is supplied by heat conduction from the wall to the slab (see fig. 2.1.5-1). Therefore
The heat supplied by heat conduction into the slab is transferred to the ambient. Therefore, JQ = JQüb. Solution for the temperature distribution for the stationary problem at x 0:
With this we obtain from equation (2.1.5-17):
On the other hand, the heat transfer problem can be formulated based on the fundamental heat transfer equation (2.1.3-14):
Comparing equations (2.1.5-19) and (2.1.5-20) we obtain:
We learn from this example that heat transfer from the slab to the ambient and heat conduction into the slab are coupled processes that affect each other!
)175.1.2()(
0
0
0
d
d
L
TTA
dx
dTA a
xQJ
)185.1.2(therefore0
0
md
de m
)195.1.2()( 0
aübQQ TTL
mAJJ
)205.1.2()( 0 aübQ TTAJ
)215.1.2(82or
BiD
L
DLmNu
L
L
m aussenL
2.1.5/4 – 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
Exact temperature measurement. We mount a temperature sensor (thermo couple) into the slab to measure the temperature Ta. How close is the temperature at the front end to the ambient temperature Ta? In this case Ta > T0.
For the front end of the slab (= 1) we obtain:
For the measuring error to be smaller than 0,5% we have:
Transformation gives:
Finally:
Example: = 100 W/m2 K, = 60 W/m K, U/A = 1/s, s = 1 mm.
Fig. 2.1.5-3: For the demonstration of the thermometry problem.
T
x
Ta
)255.1.2(005,0
2 mm ee
)265.1.2(66400ln
1 mmm
L0,15m
JQ
)225.1.2(
]cosh[
)]1(cosh[
0
m
m
TT
TT
a
a
TL)235.1.2(]cosh[
1
0
mTT
TT
a
La
)245.1.2(005,02
]cosh[
1
0
mma
La
eemTT
TT
Then
For the prevailing conditions the slab has to have a length of minimum 15 cm!
LLLA
Um aussen
401600
T0
2.1.5/5 – 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
Summary of section 2.1
•The different mechanisms of heat conduction have been qualitatively discussed.
•The basics of heat conduction in fluids and solids have been elaborated.
•The material property „heat conductivity“ has been discussed.
•The heat transfer coefficient has been introduced.
•The over all heat transfer coefficient has been introduced.
•Stationary heat conduction for different geometries and some problems from process engineering has been elucidated.
2.1.5/6 – 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
2.2 Non stationary heat conduction
2.2.1 Fourier‘s law of non stationary heat conduction and thermal conductivity
Up to now only stationary heat conduction problems have been discussed. For this problems temperatures and heat fluxes are invariant against time. In process technology most problems are non stationary, so that non stationary heat conduction has to be discussed in the following.
Basics of non stationary heat conduction; Fourier‘s law of non stationary heat conduction. Again we consider a cylindrical slab with homogeneous material properties (, cp, ) which is mounted at a wall. (Simplification: temperature constant over the radius (1-d problem). The wall temperature is initially Ta and is increased at t = 0 to T0. The temperature distribution and heat fluxes are now dependent on distance and time!
For determinationof the temperature distribution in the slab we consider the heat balance for a differential volume of the slab. Within the slab we have heat conduction and accumulation, at the surface we have heat transfer to the ambient.
Heat balance:
accumulation of heat = heat flux in –
heat flux out (2.2.1-1)
Fig. 2.2.1-1:For derivation of Fourier‘s law for non stationary heat conduction.
2.2.1/1 – 01.2012
T
x
Ta
x x+x
JQ(x+x)JQ(x)JQüb
JQ
T0
JQacc
Heat and Mass Transfer
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Heat and Mass Transfer
Heat flux into the balance volume by heat conduction:
Heat flux out of the balance volume by heat conduction:
Heat transfer to the ambient (U: circumference):
Accumulation of heat:
Fig. 2.2.1-1:For derivation of Fourier‘s law for non stationary heat conduction.
T
x
Ta
x x+x
JQ(x+x)JQ(x)JQüb
)31.2.2(),(
),(
xxQ dx
txdTAtxx J
)21.2.2(),(
),(
xQ dx
txdTAtx J
)41.2.2()),((),( aaussenübQ TtxTxUtx J
Heat balance for non stationary conditions:
)61.2.2( übQxxQxQaccQ JJJJ
)51.2.2(),(
dt
txdTVcpaccQ J
JQ
T0
JQacc
2.2.1/2 – 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
Heat balance results in:
Temperature gradient is expanded into a Taylor series (linearization of the temperature gradient):
Using this in the heat balance gives:
Transformation results in:
For aussen --> 0, no heat transfer to the ambient, we obtain Fourier‘s law for non stationary heat conduction:
)71.2.2()(
aaussenxxx
p TTxUx
TA
x
TA
t
TVc
)71.3.3(.......
xx
T
xx
T
x
T
xxx
)81.2.2()(2
2
aaussenp TTxUx
x
TA
t
TVc
)91.2.2()(2
2
a
p
aussen
p
TTA
U
cx
T
ct
T
)101.2.2(2
2
2
2
x
Ta
x
T
ct
T
p
Stoff Temperaturleitzahl a in 10-6 m2 s-1
Luft (200C, 0,1 MPa) 21,4
Wasser 0,143
Org. Flüssigkeiten 0,09- 0,1
Stein, Keramik 0,5- 1,5
Glas 0,6 – 0,8
Metalle 0,5 - 100
The property a = / cp [m2/s] is called thermal conductivity (temperaure conductivity). It is a material property. Some values are given in table 2.2.1-1.
2.2.1/3– 01.2012
Table 2.2.1-1:Thermal conductivities of some materials.
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Heat and Mass Transfer
Initial and boundary conditions:
Normalization of temperature = (T-Ta)/(T0-Ta) gives:
Initial and boundary conditions :
Approach for solution: partial differential equation is transferred into an ordinary differential equation by use of a new variable which connects time and distance. For transformation the Fourier number is introduced:
The Fourier number has the meaning of a normalized time: a/x2: „Intrusion time“ for temperature. New variable:
)22.2.2(0,0,0 0 LxTTxTTt a forfor
)72.2.2(2
x
taFo
)12.2.2()(2
2
a
p
aussen TTA
U
cx
Ta
t
T
)32.2.2(,0,0 0 LxTTxTTt a forfor
)82.2.2(2
11
2
1
ta
x
Fo
T
x
Ta
JQ
T0
2.2.2 Examples of non stationary heat conduction in engineering
Solutions of Fourier‘s differential equation for non stationary heat conduction. Geometry: cylindrical slab with L/D >> 1, then TL = Ta, for the time being no heat transfer to the ambient.
)42.2.2(2
2
x
at
)52.2.2(00,01,0 Lxxt forfor
)62.2.2(0,01,0 Lxxt forfor
2.2.2/1– 01.2012
Fig. 2.2.2-1: Non stationary heat conduction in a cylindrical slab.
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Heat and Mass Transfer
Transformation rules:
Considering the transformation rules we obtain from the heat conduction equation:
And after transformation:
Finally we obtain by further transformation a system of coupled ordinary differential equations of first order:
2
2
xa
t
ta
x
2
1
Integration of the first equation after separation of variables considering the boundary condition Y = Y0 for = 0:
Integration of the second equation within the boundaries = 0 and and 1 und , respectively, yields:
)102.2.2(2
1
atxx
)112.2.2(2
12
2
2
2
atx
T
x
Ta
JQ
T0
)122.2.2(42
12
2
at
a
t
)132.2.2(22
2
)142.2.2(and2 Yd
dY
Y
)152.2.2(orln2
02
0
eYYY
Y
)162.2.2()1(00
2
de
d
d
)92.2.2(2
11
2
1
2 2/3
d
d
tta
x
tt
2.2.2/2– 01.2012
Fig. 2.2.2-1: Non stationary heat conduction in a cylindrical slab.
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Heat and Mass Transfer
The integral in equation (2.2.2-16) is given by 1/2/2.erf(), where erf() is the error function.
The remaining task is the determination of the wall gradient in equation (2.2.2-16) (d/d)0. The error function erf() converges assymptoticly at large versus 1. Therefore 1/2/2.erf() converges against 0,8862, compare figure 2.2.2-2. From equation (2.2.2-16) we then obtain for large ( = (T-Ta)/(T0-Ta)=0):
Or:
The general solution of the non stationary heat conduction equation for the cylindrical slab with the above initial and boundary conditions then is:
)172.2.2(5,2for2
10
d
d
)182.2.2(2
0
d
d
Fig. 2.2.2-2: Error function erf().
)192.2.2()()exp(2
10
2
erfcd
Fig. 2.2.2-3: Temperature distribution according to equation (2.2.2-19). 2.2.2/3– 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
Solution:
0n
Solutions of Fourier‘s differential equation for non stationary heat conduction. Geometry: cylindrical slab with L/D >> 1, then TL = Ta, now with heat transfer to the ambient!
Initial and boundary conditions identical to the case without heat transfer to the ambient. Normalization of temperature as bevor with = (T-Ta)/(T0-Ta) gives:
with n=(aussen/cpr)(4/D). Initial and boundary conditions:
If the solution for n = 0 is
then the solution for is (Dankwerts, Crank):
)202.2.2()(2
2
ap
aussen TTA
U
cx
Ta
t
T
T
x
Ta
JQ
T0
Fig. 2.2.2-4: Non stationary heat conduc-tion in a cylindrical slab with heat trans-fer to the ambient.
JQÜbertragung
)212.2.2(2
2
nx
at
)192.2.2()2
1(
0
at
xerfc
TT
TT
a
a
)242.2.2()exp()exp(00
t
a
a ntdtntnTT
TT
)222.2.2(00,01,0 Lxxt forfor )232.2.2(0,01,0 Lxxt forfor
)252.2.2(2
exp2
1
2exp
2
1
ntat
xerfc
a
nx
ntat
xerfc
a
nx
2.2.2/4– 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
Fig. 2.2.2-5: Non stationary heat conduction in a cylindrical slab (a = 2.10-5 m2 s-1, n = 10-3 s-1).
Heat loss via the „cold bridge“ to the ambient for non stationary conditions as heat transfer problem: The total heat flux from the slab is supplied by heat conduction from the wall to the slab (see fig. 2.2.2-5). From equation (2.2.2-18) we obtain (for the case without heat transfer to the ambient):
On the other hand, the heat transfer problem can be formulated with the help of the fundamental heat transfer equation:
Comparing equation (2.2.2-26) and (2.2.2-27) we obtain
In this case the heat transfer coefficient for the heat loss via the „cold bridge“ is time dependent!
)262.2.2()(2
2
)(
2
)(
00
0
0
0
aa
a
xQ
TTta
A
at
TTA
d
d
at
TTA
dx
dTA
J
T
x
Ta
JQ
T0
JQÜbertragung
2.2.2/5– 01.2012
)272.2.2()( 0 aQ TTAJ
)282.2.2(1
t
c
tap
Heat and Mass Transfer
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Heat and Mass Transfer
Averaging the heat transfer coefficient for the non stationary problem over time gives:
The averaged heat transfer coefficient is exactly double of the instantaneous heat transfer coefficient at time t.
For the case including heat transfer to the ambient (coupled inner and outer heat transfer) we obtain:
The heat transfer coefficient the is given by:
For t 0 error function 0 and exp(-nt) 1, so that
)292.2.2(2)(1
0
t
cdtt
tp
t
m
Fig. 2.2.2-6: Non stationary heat conduction in a cylindrical slab (a = 2.10-5 m2 s-1, n = 10-3 s-1); heat transfer coefficients.
)302.2.2(
)exp(1
)()( 0
0
ntat
nterfa
nTTA
dx
dTA
a
xQ
J
)312.2.2()exp(1
)(
nt
atnterf
a
n
2.2.2/6– 01.2012
)322.2.2(1
t
c
tap
For t large error function 1 and exp(-nt) 0, so that
)332.2.2(82
or4
BiD
L
DLNu
L
Da
n
aussenL
aussen
Heat and Mass Transfer
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Heat and Mass Transfer
Summary section 2.2
• Fundamentals of non-stationary heat conduction have been deduced.
• A method for the solution of the partial differential equation for non-stationary heat conduction has been discussed.
• Definition of the heat transfer coefficient has been extended to non-stationary heat conduction problems.
2.2.2/7– 01.2012
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Heat and Mass Transfer
2.3 Stationary diffusion
2.3.1 Fick´s law of stationary diffusion and diffusion coefficient
The general law of diffusive fluxes of properties
can be written in case of transport of chemical species (diffusion):
Equation (2.3.1-2) is Fick‘s law of stationary diffuison. The mole flux of species k is the product of diffusion coefficient, area and concentration gradient. The mole flux vector is anti-parallel to the concentration gradient, negative sign in equation (2.1.1-2). The species k, therefore, always is transported in the direction of lower concentrations.
The diffusion coefficient D [m2/s] according to Ficks‘s law is the mole flux density jnk [mole/m2.s] by diffusion per unit of negative concentration gradient dck/dx [mole/m4]. The diffusion coefficient D is a material property.
2.3.1/1 – 01.2012
)11.3.2(s
ofunit grad=diff
ATJ
)21.3.2( dx
dcD
dx
dcAD k
knkk
knk jJ or
2.3.2 Mechanisms of diffusion
Simple model for diffusion in a fluid (gas, liquid). Fluid particles (molecules in a gas) move driven by thermal molecular motion with a temperature dependent velocity. During this motion the molecules strike each other. Every strike leads to a change of the direction of the motion, so that the molecules in the average at constant pressure and constant temperature move homogeneously in all directions.
Fig. 2.3.2-1:Simple model for the explanation of diffusion.
x
cA > cB cA < cB
Diffusionflux of A Diffusionflux of B
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Heat and Mass Transfer
If in a closed system a concentration gradient for the single molecules exists, compare figure 2.3.2-1, on the time average more molecules from the region of higher concentration will move into the region of lower concentration. By this a diffusion flux is established into the direction of lower concentrations.
According to that simple model, the diffusion coefficient should be dependent on the mean molcular velocity and the mean free path of the molecules and from that on temperature and pressure.
Diffusion coefficients of gases. According to the above simple model for the diffusion of gases the diffusion coefficient of ideal gases can be calculated with the help of the kinetic gas theory:
In equation (2.3.2-1) is the mean free path of the molecules and v the mean molecular velocity (compare heat conductivity).
)12.3.2(3
1
_
vAAD
)32.3.2(8
M
RT
_
v
)22.3.2(2
12
p
RT
Nd A
)42.3.2(1
3
22/3
2
π
RT
pNdMD
A
AA
_
2.3.2/2 – 01.2012
According to equation (2.3.2-4), kinetic gas theory, the diffusion coefficients increase with decreasing molar mass of the molecules. Increasing temperatures lead to a strong increase of the diffusion coefficients (T1,5). According to the kinetic gas theory we have a pressure dependence according to p-1.
Diffusion coefficients of liquids. The diffusion coefficients of most liquids are about five orders of magnitude lower than those of gases.
The mechanism of diffusion in liquids is similar to the mechnism of diffusion in gases. However, the interactions of the molecules in the fluid (van der Waals interactions, hydrogen bridge bonds, long distance interactions) play an important role. Thereby, the diffusion coefficients of most species in liquids are dependent on concentration, compare table 2.3.2-2.
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Heat and Mass Transfer
Table 2.3.2-2: Diffusion coefficients of some fluid mixtures.
MixtureTemperature
T/oCXA DAB/m2.s-1*109
Ethanole-Water
25 0,050 1,130
0,275 0,410
0,50 0,900
0,70 1,400
0,950 2,200
Water- n-Butanole
30 0,131 1,240
0,222 0,920
0,358 0,560
0,454 0,437
0,524 0,267
2.3.2/3 – 01.2012
Table 2.3.2-1:Binary diffusion coefficients for some gas mixtures.
Temperature T/oC 0 15 25
Component Diffusion coefficient DAB/m2s-1
CO2-N2O 9,6.10-6
CO2-CO 1,39.10-5 CO2-N2 1,44.10-5
1,58.10-5
1,65.10-5
Ar-O2 2,0.10-5
H2-SF6 4,2.10-5
H2-CH4 7,26.10-5
Diffusion coefficients of solids. The mechanism of diffusion in solids is much more complicated (diffusion of empty places in the lattice, etc.).
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Heat and Mass Transfer
2.4 Non-stationary diffusion
2.4.1 Fick‘s law for non-stationary diffusion
Analogously to non stationary heat conduction we can derive the differential equation for non stationary diffusion with the help of a mass balance for the scenario given in figure 2.4.1-1. We focus again on a planar diffusion problem (e.g. solution of a solid in water), which can be treated 1-dimensionally. In the fluid initially a constant concentration cA0 prevails, compare figure 2.4.1-1. At time t = 0 the concentration is sudddenly increased to cAS (immersion of the solid into water). An equation of the evolution of the concentration cA with time is obtained from a mass balance for species A in a differential balance volume:
Accumulation of moles of A =
inflow of moles of A –
outflow of moles of A (2.4.1-1)
In the same way as for non stationary heat conduction we obtain finally
2.4.1/1 – 01.2012
Fig. 2.4.1-1: Mass balance for derivation of Fick‘s law for non-stationary diffusion.
cA
x
cAS
DA
JnA(x)
x
JnA(x+x)
x+x
SpeichnAJ
)21.4.2(2
2
x
cD
t
c AA
A
Equation (2.4.1-2) is Fick‘s law for non-stationary diffusion and is (formal) identical to the equation for non-stationary heat conduction!
The solutions of this equation then are identical to the solutions of the equation for non-stationary heat conduction (if initial and boundary conditions are identical).
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Heat and Mass Transfer
As can be seen from equations (2.1.1.-2), (2.2.2-1), (2.3.1-2) and (2.4.1-2) heat transport by (stationary and non stationary) conduction and mass transport by (stationary and non stationary) diffusion are closely related with vast analogies:
Also, the transport coefficients (heat conductivity, diffusion coefficients) are closely related, e.g. from kinetic theory of gases we obtain for ideal gases:
This gives then:
Due to this similarity, heat transfer problems and mass transfer problems can be often treated similarly!
general heat transfer equation !
general mass transfer equation !
2.5/1 – 01.2012
dx
dcAD
dx
dTAT k
knkQ JJ diffusion / conductionheat stationary )(
2
2
2
2
diffusion / conductionheat stationarynon x
cD
t
c
x
Ta
t
T AA
A
)15.2(1Sc1Pr
Da
tcoefficiendiffusion 3
1ty conductiviheat
3
1 viscosity
3
1
___
vvv AAV Dc
)25.2()( TAübQ J
L
LNu
2.5 Similarity of heat and mass transfer
)35.2()( kübkn cAJD
L
LSh
Heat and Mass Transfer
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Heat and Mass Transfer
Question: concentration profile in the pore of the catalyst, molar flux into the catalyst.
We perform a mass balance for the species A over a differential balance volume. Within the pore we have diffusion of the species A and the species A reacts at the wall of the pore into products according to a chemical reaction of first order. (Simplification: concentration only dependent on x, 1-dimensional problem.) Mass balance:
Diffusion flux of A into the volume:
Diffusion flux of A out of the volume:
Fig. 2.6-1: Stationary diffusion into a round pore of a catalyst.
c
x
c
2.6 Examples for stationary diffusion problems from process engineering
Diffusion into a porous catalyst with simultaneous chemical reaction. In a pellet of a catalysts we have round pores of the length L, into which from the surface of the pellet a chemical component diffuses and reacts at the inner surface of the pore. The pores can be treated for simplicity as cylinders. At the outer surface the concentration is c0, compare figure 2.6-1. The diffusion coefficient of the species is DA.
c0
DA
Jn
x x+x
Jnx+xJnx
JnReaktion
)36.2()(
xxAn dx
dcADxxJ
)26.2()(
xAn dx
dcADxJ
)46.2(Reaktion xUcknJ
Reaction of A in the volume by heterogeneous chemical reaction (U: pore circumference):
L
)16.2(reaction chemicalby change
A of moles of outflowA of moles of inflow
2.6/1 – 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
Differentiating with respect to gives:
Equation (2.6-9) is therefore only a solution of equation (2.6-8) if:
Mass balance for stationary conditions:
The molar flux at the position x+x is replaced by a Taylor series of the molar flux at the position x:
With this we have for the mass balance:
Transformation using = c/c0 and = x/L gives:
In equation (2.6-8) m2 = (k.U.L2)/(DA.A) = L2(2k/DA
.R) = 2. is called the Thiele-Modulus. The Thiele modulus can be interpreted as ratio of reatcion rate to diffusion rate.
Solution of the ordinary differential equation of second order for : e-Ansatz:
)66.2()(
)()(
2
2
xdx
cdADx
xdx
dxxx
An
nnn
J
JJJ
)96.2( eC
)76.2(02
2
xUckx
dx
cdADA
)86.2(22
2
2
mAD
LUk
d
d
A
)106.2(222
2
eC
d
d
)116.2(2
with
2122
RD
kLm
mmm
A
or
)56.2(Reaktion nxxnxn JJJ
c
x
c
c0
DA
Jn
x x+x
Jnx+xJnx
JnReaktion
L
Fig. 2.6-1: Stationary diffusion into a round pore of a catalyst.
2.6/2 – 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
General solution: linear combination of the particular solutions:
The integration constants A and B are calculated from the boundary conditions.
Boundary condition 1: For = 0 is c =c0, therefore = 1. This gives:
Boundary condition 2: The end of the pore is not permeable for material. Then d/d = 0, for = 1. Then it follows:
With equation (2.6-13) we obtain:
The solution then is:
)126.2( mm eBeA
)146.2(0
m
mmm
e
eABemBemA
)136.2(1 BA
)166.2(]cosh[
)]1(cosh[
m
m
)156.2(
mm
m
mm
m
ee
eB
ee
eA
Fig. 2.6-2: Solution of equation (2.6-8) for stationary diffusion in a pore of a catalyst.
With increasing Thiele modulus, that is with increasing reaction rate compared to the diffusion rate, the concentration decreases steeper and the concentration gradient at the outer surface increases.
2.6/3 – 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
Diffusion into the pores as mass transfer problem: The mass of the component A which is consumed by chemical reaction within the pore of the catalyst has to be transportet through the pore entrance by diffusion. Therefore
Concentration profile for the stationary problem:
This gives:
On the other hand, every mass transfer problem – in this case the mass transfer from the outer surface of the catalyst into the inner of the pore – can be written as (compare equation 2.5-3):
Comparing equation (2.6-19) and (2.6-20) we obtain (with coo = 0 for m>5):
The mass transfer through the pore entrance of the catalyst and the chemical reaction at the inner surface are coupled ! The faster the chemical reaction (the larger the Thiele modulus) the larger is the mass transfer coeffcient ! (compare stationary heat conduction in a slab ?)
)176.2(0
0
0
d
d
L
cAD
dx
dcAD A
xAgesnJ
)196.2(0
cL
mADAgesnJ
)206.2()( 0 ccAgesn J
)216.2(2
or
RD
kLm
D
LSh
L
mD
AAL
A
)186.2(5for)tanh(,]cosh[
)]1(cosh[
0
mmmmd
d
m
m
2.6/4– 01.2012
Heat and Mass Transfer
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Heat and Mass Transfer
Summary sections 2.3 to 2.6
• The different mechanisms of diffusion have been discussed.
• Stationary diffusion has been terated.
• Basics of diffusion in fluids have been elaborated.• The property „diffusion coefficient“ has been discussed.
• Non stationary diffusion has been introduced• Similarity between heat transport and mass transport has been identified.
• Simple problems for diffusion from process engineering have been discussed.
2.6/5– 01.2012