Parag Chaware College · 2016. 9. 27. · Parag Chaware Department of Mechanical Engineering...

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Transient heat conduction with spatial variations

Parag Chaware

Department of Mechanical EngineeringCummins College of Engineering, Pune

Transient heat conduction with spatial variations Parag Chaware 1 / 13

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Limitations of Lumped capacity Analysis

Transient heat conduction problems for small bodies can be solvesusing lumped heat capacity analysis.

It is generally accepted that lumped system analysis is applicable ifBi ≤ 0.1The variation of temperature with location within the body is slightand can reasonably be approximated as being uniform.

Small bodies of highly conductive materials approximate this behavior.

When geometries are large and spacial variations cannot be ignoredlumped heat capacity analysis approach for the solution cannot beused.


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Effect of spatial variations

∂2T

∂x2=

1

α

∂T

∂t

Boundary Conditions

∂T (0, t)

∂x= 0 and − k ∂T (L, t)

∂x= h[T (L, t) − T∞]

The analytical solution involves infinite seriesand implicit equations, which are difficult toevaluate.

One-term approximation solutions of aboveequation are presented in graphical form,known as the transient temperature charts.


CumminsCollegeThe transient temperature charts for a large plane wall, long cylinder,

and sphere were presented by M. P. Heisler in 1947 and are calledHeisler charts.

Are supplemented in 1961 by H. Grober with three charts associatedwith each geometry:

The first chart is to determine the temperature T0 at the center ofthe geometry at a given time t.

The second chart is to determine the temperature at other locationsat the same time in terms of T0.

The third chart is to determine the total amount of heat transfer upto the time t.


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Figure: Temperature distribution in plane wall at Mid-Plane


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Figure: Offset temperature distribution for wall


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Figure: Temperature distribution in a cylinder at centerline


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Figure: Offset temperature distribution for cylinder


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Figure: Temperature distribution in sphere at center


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Figure: Offset temperature distribution for sphere


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A slab 10 cm thick at initially at 500 ◦C is immersed in liquid at 100 ◦Cresulting in heat transfer coefficient of 1200 W K/m2. Determine thetemperature at the centerline and at the surface 1 min after immersion.Properties for the slab are;α = 8.4 × 10−5m2/s, k = 215W K/m, cp = 0.9kJ K/kg, ρ = 2700kg/m3

Bi =hL

k=

1200 × 0.05215

= 0.28

Fo =αt

L2= 2.016

From chart 1Bi = 3.57

θcθi

=Tc − T∞Ti − T∞

= 0.63

Tc = 100 + (500 − 100) × 0.63 = 352◦C


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A long steel cylinder 12 cm in diameter and initially at 20 ◦C is placed intoa furnace at 820 ◦C with local heat transfer coefficient of 140 W K/m2.Calculate the time required for the axis temperature to reach 800 ◦C. Alsocalculate the corresponding temperature at radius of 5.4 cm.α = 6.11 × 10−6m2/s, k = 21W K/m

Bi =hL

k=

140 × 0.0321

= 0.2

Select the curve for 1/Bi = khro = 2.5

θcθi

=Tc − T∞Ti − T∞

=800 − 82020 − 820

= 0.0256

We get Fourier Number as ‘5’

Fo = 5 =6.11 × 10−6t

r2o= 2946S

Intermediate radius r 5.4 cm = 0.054 m


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From chart 1/2Bi = 2.5, So from chart we get

Tr − T∞T0 − T∞

= 0.85 = Tr = 820 + (800 − 820) × 0.85 = 803◦C


Parag Chaware College · 2016. 9. 27. · Parag Chaware Department of Mechanical Engineering...

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