Momentum Heat Mass Transfer

1
Momentum Heat Mass Transfer
MHMT10
Heat transfer-conduction
Multidimensional heat conduction problems. Fins
and heat conduction with internal sources or
sinks. Unsteady heat conduction in solids.
Penetration theory.
Rudolf Žitný, Ústav procesní a zpracovatelské
techniky CVUT FS 2010
2
Heat transfer - conduction
MHMT10
Thermal resistance of fluid (thermal boundary
layer) can be expressed in terms of heat transfer
coefficients ? added to the thermal resistances
of solid layers.
For example resulting RT of serial resistances of
fluid and two concentric pipes can be expressed
as (see previous lecture)
Example (critical thickness of insulation) Let
us assume that only the thickness of insulation
(outer diameter D2) can be changed. Then the
thermal resistance (and effectiveness of
insulation) depends only upon D2, see graph
calculated for D10.02 m, Dm0.021, ?140 W/(m.K)
(steel), ?20.1 (insulation), ?11000, ?25
(natural convection) . Up to a critical D2 the
thermal resistance DECREASES with the increasing
thickness of insulation!
3
Heat transfer - conduction
MHMT10
Thermal resistance of a composite tube with
circular or spiral fins attached to outer tube
(fins can be also an integral part of the outer
tube).
Resulting thermal resistance is calculated
according to almost the same expression as in the
previous case
with the only but very important difference
Instead of the outer surface of plain tube ?D2L
is used the overall outer surface of fins S2L.
Such a modification assumes that the ?2 on the
surface of fins is the same as on the surface of
tube and first of all that the thermal resistance
of fin itself is negligible (simple speaking it
is assumed that the fin is made of material
having infinite value of thermal conductivity ?).
4
Heat transfer - conduction
MHMT10
The assumption of perfectly conductive fin is
unacceptable and in reality the thermal
resistance of fin must be respected by
multiplying the surface S2 by fins efficiency
?fin which depends upon thermal conductivity of
fin, its geometry (thickness and height) and also
upon the heat transfer coefficient ?2.
Efficiency of a thin rectangular fin can be
derived easily by solution of temperature profile
along the height of fin
This is FK equation with a source term,
representing heat transfer from both sides of
surface (2dx) to the control volume (bdx)
with boundary conditions at the heel of fin
(T(x0)Tw) and at top of fin dT/dx0 (there is
no heat flux at xH)
Efficiency of fin is calculated from temperature
gradient at the heel of fin (the gradient
determines heat flux at the heel)
5
Conduction - nonstationary
MHMT10
6
Conduction - nonstationary
MHMT10
Temperature distribution in unsteady case
generally depends upon time t and coordinates
x,y,z. Sometimes, when the temperature
distribution is almost homogeneous inside the
whole body, the partial differential Fourier
Kirchhoff equation reduces to an ordinary
differential equation. This simplification is
correct if the thermal resistance of solid is
much less than the thermal resistance of fluid,
more specifically if Biot number is small enough
here D is a characteristic diameter of a solid
object and ?s is thermal conductivity of solid
Fourier Kirchhoff equation can be integrated over
the whole volume of solid
which reduces to ordinary dif. equation as soon
as Ts depends only on time
7
Conduction - nonstationary
MHMT10
As soon as the Biot number is large (Bigt0.1,
therefore if the solid body is too big, for
example semi-infinite space) it is necessary to
solve the parabolic partial differential Fourier
Kirchhoff equation. For the case that the solid
body is homogeneous (constant thermal
conductivity, density and specific heat capacity)
and without internal heat sources the FK equation
reduces to with the boundary conditions of
the same kind as in the steady state case and
with initial conditions (temperature distribution
at time t0). This solution T(t,x,y,z) can be
expressed for simple geometries in an analytical
form (heating brick, plate, cylinder, sphere) or
numerically in case of more complicated
geometries.
The coefficient of temperature diffusivity
a?/?cp is the ratio of temperature conductivity
and thermal inertia
8
Conduction - nonstationary
MHMT10
Start up flow of viscous liquid in halfspace
(solved in lecture 4) was described by equation
which is identical with the Fourier Kirchhoff
equation for one dimensional temperature
distribution in halfspace and with the step
change of surface temperature as a boundary
condition
T(t0,x)T0 T(t,x0)Tw
Exactly the same solution as for the start up
flow (complementary error function erfc) holds
for dimensionless temperature ?
9
Conduction - nonstationary
MHMT10
Erfc function describes temperature response to a
unit step at surface (jump from zero to a
constant value 1). The case with prescribed time
course of temperature at surface Tw(t) can be
solved by using the superposition principle and
the response can be expressed as a convolution
integral.
t
Temperature at a distance x is the sum of
responses to short pulses Tw(?)d?
Time course Tw(t) can be substituted by short
pulses
The function E(t,?,x)E(t-?,x) is the impulse
function (response at a distance x to a
temperature pulse of infinitely short duration
but unit area Dirac delta function). The
impulse response can be derived from derivative
of the erfc function
10
Penetration theory
MHMT10
Still too complicated? Your pocket calculator is
not equipped with the erf-function? Use the
acceptable approximation by linear temperature
profile, (exactly the same procedure as with the
start up flow in a half-space)
Integrate Fourier equations (up to this step it
is accurate)
Approximate temperature profile by line
Result is ODE for thickness ? as a function of
time
Using the exact temperature profile predicted by
erf-function, the penetration depth slightly
differs ??(?at)
11
Penetration theory
MHMT10
???at penetration depth. Extremely simple and
important result, it gives us prediction of how
far the temperature change penetrates at the time
t. This estimate enables prediction of thermal
and momentum boundary layers thickness etc. The
same formula can be used for calculation of
penetration depth in diffusion, replacing
temperature diffusivity a by the diffusion
coefficient DA .
Wire Cu ?0.11 m ?398 W/m/K ?8930 kg/m3 Cp386
J/kg/K
12
Penetration theory and ?
MHMT10
The penetration theory can be applied also in the
case that the semi-infinite space is in contact
with fluid and surface temperature depends upon
temperature of fluid and the heat transfer
coefficient
Derive the result as a homework
13
PLATE - finite depth
MHMT10
In case of a finite thickness plate the
penetration theory can be used only for short
times (small Fourier number lt 0.1)
Let us define Fourier number and Biot number in
terms of half thickness of plate H/2
Long times (large Fourier) and finite Biot.. The
most complicated case/see next slide
Integral method
Fourier method
Penetration theory
14
PLATE - finite depth Fourier method
MHMT10
Using dimensionless temperature ?, distance ?,
time ? (Fourier number) and dimensionless heat
transfer coefficient Bi (Biot number)
the Fourier Kirchhoff equation, boundary and
initial conditions
are transformed to
Fourier method is based upon superposition of
solutions satisfying differential equation
and boundary conditions
15
PLATE - finite depth
MHMT10
Spatial component Gi(?) follows from
The function cos(??) automatically satisfies the
boundary condition at ?0 for arbitrary ?. The
boundary condition at wall is satisfied only for
yet undetermined values ?, roots of
transcendental equation
and this equation must be solved numerically,
giving infinite series of roots ?1, ?2,
These eigenvalues ?i determine also the temporary
components Fi
Final temperature distribution is the infinite
series of these elementary solutions
16
PLATE - finite depth
MHMT10
The coefficients ci are determined by the initial
condition
To solve the coefficients ci from this identity
(which should be satisfied for arbitrary ?) it is
convenient to utilise orthogonality of functions
Gi(?) and Gj(?) that follows from original
ordinary differential equations and boundary
conditions
for i?j
therefore
giving final temperature profile
17
PLATE/CYLINDER/SPHERE
MHMT10
Unsteady temperature profiles inside a sphere and
infinitely long cylinders can be obtained in
almost the same way, giving temperature profiles
in form of infinite series
where
and only eigenvalues ?I have to be calculated
numerically. This form of analytical solution is
especially suitable for description of
temperature field at longer times, because
exponential terms quickly decay and only few
terms in the series are necessary. For shorter
times the penetration theory can be applied
effectively
This analytical solution is presented in the book
Carslaw H.S., Yeager J.C. Conduction of Heat in
Solids. Oxford Sci.Publ. 2nd Edition, 2004
18
3D brick,finite cylinder
MHMT10
It is fantastic that an unsteady temperature
field in the finite 2D or 3D bodies can be
obtained in the form of PRODUCT of 1D solutions
expressed in terms of dimensionless temperatures
?x ?y ?z !!!
For example the temperature distribution in an
infinitely long rod with rectangular cross
section hx x hy is calculated as
Proof
and you see that the FK equation is satisfied if
?x ?y are solutions of 1D problem.
19
3D excercise
MHMT10
Calculate temperature in the center of a cube
Calculate temperature in the corner using erfc
solution
y
x
20
EXAM
MHMT10
Fins and Unsteady heat conduction
21
What is important (at least for exam)
MHMT10
Thermal resistance of finned tube
Efficiency of a planar fin
22
What is important (at least for exam)
MHMT10
Unsteady heat conduction
Transient heating of a semi-infinite space
Simplified solution by penetration depth
Temperature response to variable surface
temperature
23
What is important (at least for exam)
MHMT10
Find solution of equation for Dirichlet boundary
condition at ?1 and Neumann boundary condition
at ?0
Why is it necessary to use dimensionless
temperatures in the product solution of 2D and
3D problems?