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Thermal-Fluids I
Chapter 18 – Transient heat conduction
Dr. Primal Fernando
[email protected]
Ph: (850) 410-6323
1
Transient heat conduction
• In general, The temperature of a body varies with time as well as
position.
 In rectangular co-ordinates this variation is expressed as T(x,y,z,t)
 x,y,z → variations in x,y,z directions
 t → variation with time
• The studies in this chapter is focused on
 Lumped system analysis
 Transient heat conduction in large plane walls, long cylinders and
spheres with spatial effects
 Transient heat conduction in semi-infinite solids
 Transient heat conduction in multi-dimensional systems
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BROAD OBJECTIVE: INVESTIGATE THE PROBLEM
OF
HOW DO
SPHERES COMING
OUT OF A OVEN
COOL?
3
Consider …
• An engineer, a psychologist, and a
physicist were asked to make
recommendations to improve the
productivity of an under-producing
dairy farm …
• Engineer: more technology
• Psychologist: improve environment
• Physicist …
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“Consider a spherical cow …”
T(t)
• Great engineers and physicists are able to appropriately simplify
problems to extract the physics!
5
Lumped system
• A lumped system is one in which the dependent variables of interest
are a function of time alone. In general, this will mean solving a set of
ordinary differential equations (ODEs)
• A distributed system is one in which all dependent variables are
functions of time and one or more spatial variables. In this case, we will
be solving partial differential equations (PDEs)
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Lumped system
• Consider a small hot copper ball coming out from an oven.
– Temperature change with time.
– Temperature does not change much with position at any given time.
– Lumped system analysis are applicable to this system.
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Lumped system
• Consider a large roast in an oven.
– Temperature distribution not even.
– Temperature does change much with position at any given time.
– Lumped system analysis are not applicable to this system.
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Consider a body of arbitrary shape of mass m, volume V, surface area As,
density ρ, and specific heat Cp initially at a uniform temperature of Ti.
At time t=0, the body is placed into a medium at temperature T∞
Heat transfer take place between body and its environment
Temperature of the body change with the time
and the temperature of the body at a given time
T=T(t)
Heat transfer into the body at any given time
T=T(t)
Q = hA s [T∞ − T (t )]
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Heat transfer into the body at temperature T
Q = hA s (T∞ − T )
Heat transfer into the


 body during a time period dt

 The increase in the energy 
 = 


 of the body during time dt 
hA s (T∞ − T )dt = mc p ∆T
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hAs (T∞ − T )dt = mc p dT
m = ρV
hAs
dT
dt
=−
(T − T∞ )
ρVc p
∫
T (t )
Ti
t
dT
=
(T − T∞ )
∫
hAs
dt
−
ρVc p
ln(T − T∞ ) T
T (t )
i
hAs t
t0
=−
ρVc p
0
hAs
T (t ) − T∞
ln
=−
t
Ti − T∞
ρVc p
T (t ) − T∞
=e
Ti − T∞
−
hAs
t
ρVc p
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T (t ) − T∞
=e
Ti − T∞
−
hAs
t
ρVc p
T (t ) − T∞
= e −bt
Ti − T∞
hAs
where b =
ρVc p
→ units
1
s
Time constant
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Criteria for lumped system analysis
T (t ) − T∞
= e −bt
Ti − T∞
hAs
b=
ρVc p
→ units
1
s
V
Characteristic length
Lc =
As
Biot number Bi Bi = hLc
k
Bi =
Lc / k
Conduction resistance within the body
=
1 / h Convection resistance at the surface of the body
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Bi =
Lc / k
Conduction resistance within the body
=
1 / h Convection resistance at the surface of the body
Small Bi number indicate low conduction resistance, and
therefore small thermal gradient within the body
Lumped system is exact when
Bi = 0
Generally accepted lumped system analysis when,
Bi ≤ 0.1
If Bi < 0.1, there is a ± 5% error or less in estimating
temperature throughout body as a single-valued function
of time T(t)
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Remember 1st Major Assumption
Temperature is uniform throughout sphere.
- Temperature gradients are small inside sphere.
- Resistance to conduction within solid much less than
resistance to convection across fluid boundary layer.
Lc / k
Conduction resistance within the body
Bi =
=
1 / h Convection resistance at the surface of the body
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Remember 2nd Major Assumption
Heat transfer coefficient is
assumed not to be a function of ∆T.
Rate of heat energy passing through sphere
Q = - h As (Ts - T∞)
(W) = (W/[m2-Ko])(m2)(Ko)
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Problem: Steel balls 12 mm in diameter are annealed by heating to 1150 K
and then slowly cooling to 400 K in an air environment for which T∞=325
K and h=20 W/m2K. Assuming the properties of the steel to be k=40 W/mK,
ρ=7800 kg/m3, and cp=600 J/kgK, estimate the time required for the cooling
process.
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Solution
Biot number Bi
Characteristic length
Bi =
hLc
k
4 3
 πr 
V 3
r 6

−3
−3
=
=
Lc
10
2
10
=
=
×
=
×
As
4π r 2
3 3
(
(
hLc 20 × 2 × 10 −3
Bi =
=
k
40
)
)
 W  
  m 2 K (m )
  = 0.001 < 0.1

  W  
  mK  
Therefore, temperature of the steel balls remain approximately uniform:
lumped system analysis applicable
19
T (t ) − T∞
= e −bt
Ti − T∞
hAs
h
=
b=
ρVc p ρLc c p
Ti − T∞
1
t = ln
b T (t ) − T∞
t=
ρLc c p
h
Ti − T∞
ln
T (t ) − T∞
  kg   J  
 
 3 (m )

−3
ρLc c p
Ti − T∞
7800 × 2 × 10 × 600  1150 − 325    m   kgK  
ln
ln
t=
=


h
T (t ) − T∞
20
 W 
 400 − 325  
 2 


m
K




= 1122 s = 18.704 min
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Bi number provide-Measure of temperature drop in solid relative to
temperature difference between the surface and the fluid
Bi =
Lc / k
Conduction resistance within the body
=
1 / h Convection resistance at the surface of the body
Steady state system
 Ts ,1 − Ts , 2 



Ts ,1 − Ts , 2
Q


Bi =
=
 Ts , 2 − T∞  Ts , 2 − T∞



 Q 
21
Transient heat conduction in large plane walls,
long cylinders and spheres with spatial effects
In this section variation of temperature with time and position in one
dimensional problems such as those associated with large plane wall, long
cylinder and sphere.
A distributed system is one in which all dependent variables are functions
of time and one or more spatial variables. In this case, we will be solving
partial differential equations (PDEs)
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Transient heat conduction in large plane walls,
long cylinders and spheres with spatial effects
T∞ < Ti at t = 0
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large plane walls with spatial effects
Transient temperature distribution T(x,t) in
a wall results in a partial differential
equation, which can be solved using
advanced mathematical techniques. The
solution however, normally involves infinite
series , which are inconvenient and timeconsuming to evaluate. Therefore, there is
a clear motivation to present the solution in
tabular or graphical form.
Solution involves so many parameters
such as x, L, t, k, α, h, Ti and T∞. In order
to reduce the number of parameters, it is
defined dimensionless quantities
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Dimensionless parameters
Dimensionless temperature θ ( x, t ) =
T ( x, t ) − T∞
Ti − T∞
Dimensionless distance from the center
x
X =
L
ro for cylinder and
sphere (not V/A)
hL
Dimensionless heat transfer coefficient Bi =
k
Dimensionless time τ
=
αt
L2
(Biot number)
(Fourier number)
Nondimensionalization enables us to present temperature data in terms
of X, Bi and τ
The above defined dimensionless quantities can be used for cylinder or
sphere by replacing the variable x by r and L by ro.
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One dimensional transient heat conduction problem: For above
geometries, solutions involve in finite series, which are difficult to deal
with.
Solutions using one dimensional approximation
Plane wall:
Cylinder:
Sphere:
θ ( x, t ) wall
T ( x, t ) − T∞
−λ12τ
=
= A1e cos(λ1 x / L), τ > 0.2
Ti −T ∞
θ (r , t ) cyl
T (r , t ) − T∞
−λ12τ
=
= A1e J 0 (λ1 r / r0 ), τ > 0.2
Ti −T ∞
θ (r , t ) sph
T (r , t ) − T∞
−λ12τ sin(λ1 r / r0 )
, τ > 0.2
=
= A1e
Ti −T ∞
λ1 r / r0
A1 and λ1 are functions of Bi and their values are listed in Table 18-1
J0 is the Zeroth order Bessel function and values are listed in Table 18-2
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27
Temperature of the center of the plane wall,
cylinder and sphere
Plane wall:
Cylinder:
Sphere:
θ (0, t ) wall
T (0, t ) − T∞
− λ12τ
=
= A1e
Ti −T ∞
θ (0, t ) cyl
T (0, t ) − T∞
− λ12τ
=
= A1e
Ti −T ∞
θ (0, t ) sph
T (0, t ) − T∞
− λ12τ
=
= A1e
Ti −T ∞
Once Bi number is known, these relations can be use to
determine the temperature anywhere in the medium
(interpolations may be required to determine intermediate
values) .
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Heisler charts – M.P. Heisler 1947
• There are 3 charts associated with each geometry
– Chart 1: Determine the temperature of the of the center of the
geometry (T0) at a given time t.
– Chart 2: Determine the temperature of another location (T) in terms
of (T0) center temperature.
– Chart 3: Determine the total amount of heat transfer up to the time.
Note: these plots are valid to τ>0.2
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Heisler charts - Large plane walls (18-13)
(chart 1)
Fourier number
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Heisler charts - Large plane walls (18-13)
(chart 2)
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Heisler charts - Large plane walls (18-13)
(chart 3)
Note:
Total mount of heat transfer during
whole period
Q is amount of heat transfer
Qmax=mcp(T∞ - Ti) at finite time period t
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Heisler charts – Long cylinder (18-14)
Heisler charts - Sphere (18-15)
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Transient heat conduction in semi-infinite solids
• Semi-infinite solid is an idealized body that has an single plane surface
and extends to infinity in all other directions
– The earth
– A thick wall
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Transient heat conduction in semi-infinite solidsGraphical representation
Nondimensionalize
d temperature
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Transient heat conduction in multidimensional
systems
• The presented charts can be used to determine the temperature
distribution and heat transfer in one dimensional heat conduction
problems associated with, large plane wall, a long cylinder, a sphere
and a semi infinite medium.
• Using a superposition approach call product solution, these charts can
also be used to construct solutions for two dimensional transient heat
conduction problems encountered in geometries such as short cylinder, a
long rectangular bar, or semi-infinite cylinder or plate and even three
dimensional problems associated with geometries such as a rectangular
prism or semi-infinite rectangular bar…………
⇒ provided that: all surfaces of the solid are subjected to convection to
the same fluid at a T∞ with heat transfer coefficient h and no heat
generation in the body.
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Product solution short cylinder
Short cylinder
 T (r, x,t) −T

 T −T
∞
i
∞



short
cylinder
=
 T (x,t) −T

 T −T
∞
i
∞



plane
wall
 T (r,t) −T

 T −T
∞
i
∞



inf inite
cylinder
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Product solution –
rectangular profile
 T (r, x,t) −T∞ 


 Ti −T∞ 
rec tan gular
bar
 T (x,t) −T∞ 
=

 T −T 
 i ∞ 
plane
wall
 T ( y,t) −T∞ 


 Ti −T∞ 
plane
wall
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Look the examples in the book,
Friday.. work on problems, will give the home
work, probably a quiz
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