Conduction
Through A Cylindrical Body:
For objects with cylindrical dimensions, the relevant equations are
provided in Table 5.1.
Example
5.2: A 2 " steel pipe carries
steam at 300 °F. It is lagged with 0.5 " of rock
wool. The surrounding air is at 70 °F. What will be the heat loss per linear
foot? The following data are
applicable.
Inside film
coefficient, hi = 1500 Btu/(hr·ft2·°F);
Outside film
coefficient, ho = 2.23 Btu/(hr·ft2·°F);
Thermal
conductivity of steel, k1 = 28 Btu/(hr·ft·°F);
Thermal
conductivity of rock wool, k2 = 0.033 Btu/(hr·ft·°F)
Solution:
Steam
temperature at the center of the pipe, t1, is 300 °F. The bulk temperature of the air, t5,
is 70 °F. Surface temperatures (t2,
and t4) are not known.
The temperature at the steel-wool interface (t3) is
also unknown. The inside diameter of a
2" pipe having standard thickness (represented by Schedule 40) is
2.067". Outside diameter of the
pipe can be found to be 2.374".
Now we can find the appropriate radii.
These are
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The
insulation is 0.5" thick, so r3 = 0.0989 + 0.5/12 =
0.1406 ft. Thickness of the cylindrical
sections can be found as follows
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This
gives
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The
length of the pipe is 1 foot. If R1
and R4 are equivalent resistances for film coefficients, then
their values can be found as
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However,
R2 and R3 are conductive resistances. Their values can be found as
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These
values are found to be R2 = 7.985´10-4
(hr·°F)/Btu, and R3
= 1.695 (hr·°F)/Btu. And the overall resistance is given as:
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The
heat flow can be found by dividing overall temperature difference by the
resistance.
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Using
basic concept, driving force is a product of flow and resistance.
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Δt1
= 0.13 °F, Δt2
= 0.08 °F, Δt3
= 176.83 °F, Δt4
= 52.96 °F
And
interface temperatures are t2 = 299.87 °F, t3
= 299.79 °F, t4
= 122.96 °F
The
bulk temperature can be verified to be 122.56 - 52.96 = 70 °F.