transient heat conduction with spatial variations · 10 5 m2=s;k= 215 w=mk;cp= 0:9 kj=kgk;ˆ= 2700...
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![Page 1: Transient heat conduction with spatial variations · 10 5 m2=s;k= 215 W=mK;cp= 0:9 kJ=kgK;ˆ= 2700 kg=m3 2.A long steel cylinder 12 cm in diameter and initially at 20 C is placed](https://reader034.vdocument.in/reader034/viewer/2022051910/6000512328ffde71c66e22fd/html5/thumbnails/1.jpg)
Cummins CollegeTransient heat conduction with spatial variations
Transient heat conduction problems for small bodies can be solves using lumped heat
capacity analysis. Lumped system analysis assumes a uniform temperature distribution
throughout the body, which is the case only when the thermal resistance of the body to heat
conduction (the conduction resistance) is zero. Thus, lumped system analysis is exact when
Bi = 0 and approximate when Bi > 0. The smaller the Biot number, the more accurate the
lumped system analysis. It is generally accepted that lumped system analysis is applicable if
Bi ≤ 0.1 Thus, when Bi ≤ 0.1 , the variation of temperature with location within the body
is slight and can reasonably be approximated as being uniform. Relatively small bodies of
highly conductive materials approximate this behavior.
In general, however, the temperature within a body changes from point to point as well as
with time. When geometries are large and spacial variations cannot be ignored lumped heat
capacity analysis approach for the solution cannot be used. These are the typical situation
where
Bi ≥ 0.1
Consider a plane wall of thickness 2L, initially at a uniform temperature Ti, At time t
= 0, is placed in a large medium that is at a constant temperature T∞ and kept in that
medium for t > 0. Heat transfer takes place between these bodies and their environments
by convection with a uniform and constant heat transfer coefficient h. The variation of the
temperature profile with time in the plane wall is illustrated in fig. 1
The above problem can be expresses in form of differential equations as
∂2T
∂x2=
1
α
∂T
∂t
1
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Figure 1: Transient temperature profiles in a plane wall exposed to convection
Boundary Conditions
∂T (0, t)
∂x= 0 and− k∂T (L, t)
∂x= h[T (L, t)− T∞]
The analytical solution obtained above for one-dimensional transient heat conduction in
a plane wall involves infinite series and implicit equations, which are difficult to evaluate.
Therefore it is required to present the solutions in tabular or graphical form using simple
relations. one-term approximation solutions of above equation are presented in graphical
form, known as the transient temperature charts.
The transient temperature charts for a large plane wall, long cylinder, and sphere were
presented by M. P. Heisler in 1947 and are called Heisler charts. They were supplemented
in 1961 with transient heat transfer charts by H. Grober. There are three charts associated
with each geometry: the first chart is to determine the temperature T0 at the center of the
geometry at a given time t. The second chart is to determine the temperature at other
locations at the same time in terms of T0. The third chart is to determine the total amount
of heat transfer up to the time t.
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The first chart for the plane wall is plotted using 3 different variables. Plotted along the
vertical axis of the chart is dimensionless temperature at the midplane. Fourier Number is
plotted along the horizontal axis.The curves within the graph are a selection of values for
the inverse of the Biot Number. The charts for the cylinder and sphere are also plotted on
the similar lines.
The values for the Biot number and Fourier number, as used in the Heisler charts are
evaluated on the basis of characteristics parameter, which is, semithickness for plate, and
surface radius for cylinders and spheres
Although Heisler-Grober Charts are a faster and simpler alternative to the exact solu-
tions of these problems, there are some limitations. First, the body must be at uniform
temperature initially. Additionally, the temperature of the surroundings and the convective
heat transfer coefficient must remain constant and uniform. Also, there must be no heat
generation from the body itself.
Exercise
1. A slab 10 cm thick at initially at 500 ◦C is immersed in liquid at 100 ◦C resulting in
heat transfer coefficient of 1200 W/m2K. Determine the temperature at the centerline
and at the surface 1 min after immersion. Properties for the slab are; α = 8.4 ×
10−5 m2/s, k = 215 W/mK, cp = 0.9 kJ/kgK, ρ = 2700 kg/m3
2. A long steel cylinder 12 cm in diameter and initially at 20 ◦C is placed into a furnace
at 820 ◦C with local heat transfer coefficient of 140 W/m2K. Calculate the time
required for the axis temperature to reach 800 ◦C. Also calculate the corresponding
temperature at radius of 5.4 cm. α = 6.11× 10−6 m2/s, k = 21 W/mK
3. A metallic sphere of radius 10 mm is initially at uniform temperature of 335 ◦C. It is
then quenched in water bath at 20 ◦C with h = 6000 W/m2K until the center of sphere
cools to 50 ◦C. Compute the time required for cooling of the sphere. Properties of
sphere are; α = 6.66× 10−6 m2/s, k = 20 W/mK, cp = 1000 kJ/kgK, ρ = 3000 kg/m3
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Fig
ure
2:T
emp
erat
ure
dis
trib
uti
onin
pla
ne
wal
lat
Mid
-Pla
ne
4
![Page 5: Transient heat conduction with spatial variations · 10 5 m2=s;k= 215 W=mK;cp= 0:9 kJ=kgK;ˆ= 2700 kg=m3 2.A long steel cylinder 12 cm in diameter and initially at 20 C is placed](https://reader034.vdocument.in/reader034/viewer/2022051910/6000512328ffde71c66e22fd/html5/thumbnails/5.jpg)
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Figure 3: Offset temperature distribution for wall
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Fig
ure
4:T
emp
erat
ure
dis
trib
uti
onin
acy
linder
atce
nte
rline
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![Page 7: Transient heat conduction with spatial variations · 10 5 m2=s;k= 215 W=mK;cp= 0:9 kJ=kgK;ˆ= 2700 kg=m3 2.A long steel cylinder 12 cm in diameter and initially at 20 C is placed](https://reader034.vdocument.in/reader034/viewer/2022051910/6000512328ffde71c66e22fd/html5/thumbnails/7.jpg)
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Figure 5: Offset temperature distribution for cylinder
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Fig
ure
6:T
emp
erat
ure
dis
trib
uti
onin
spher
eat
cente
r
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![Page 9: Transient heat conduction with spatial variations · 10 5 m2=s;k= 215 W=mK;cp= 0:9 kJ=kgK;ˆ= 2700 kg=m3 2.A long steel cylinder 12 cm in diameter and initially at 20 C is placed](https://reader034.vdocument.in/reader034/viewer/2022051910/6000512328ffde71c66e22fd/html5/thumbnails/9.jpg)
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Figure 7: Offset temperature distribution for sphere
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