Thickness of thermal insulating material required to protect a lead sphere 18 inches in diameter

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Technical Note (National Research Council of Canada. Division of Building Research), 1960-03-01

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Thickness of thermal insulating material required to protect a lead sphere 18 inches in diameter

Shorter, G. W.; McGuire, J. H.

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DIVISION OF BUILDING RESEARCH

NATIONAL RESEARCH COUNCIL OF CANADA

'fE C1HI N II CAlL

NOT FOR PUBLICATION

PREPARED BY G.W. Shorter and CHECKED BY

J.R. McGuire

No.

307

NOTlE

FOR INTERNAL USE

APPROVED BY NBH DATE March 1960 PREPARED FOR _{C. Garrett, NRC Division of Applied Physics}

SUBJECT Thickness of Thermal Insulating Material

Required to Protect a Lead Sphere 18 inches in Diameter

4It

The DBR Fire Research Section received a request from Mr. Co

Garrett, Division of Applied Physics, NRC, to assist in evaluating shipping containers for radioactive materials. The National Research Council has been designated as the responsible organization to carry out such evalua-tion by the Board of Transport Commissioners. It was agreed at a meeting with Messrs. Garrett and Henry oj;'i'the Division of Applied Physics that the Fire Research Section would ｳｴｵ､ｾｴｨ･ question of the protection required in order that the container would remain intact and undamaged during a fire.

This Note sets forth a recommendation for such protection and the considerations leading to the recommendation. Certain approximations have been made in dealing with this problem, the first being that a lead sphere 18 inches in diameter be considered.

2

-1. Time required to melt lead

Lead sphere - 18 in. diameter, P = 11.3 gm/cc Mass

=

P v

=

4/3 P 7i r 3

= 4/3 x 11.3 x '71 x (9 x 2.54)3

=

5.64 x 105 gm If s is taken as 0.03:

the thermal capacity (c) of the sphere is c

=

s P v = 1.7 x 104 cal;oO

the surface area = 4 ,.." r 2

=

4 x 7r (9 x 2.54.)2

=

6.5 x 103 sq em the heat transfer from 1000°0 source to a surface at ambient temperature is approximately 4 cal/cm2/sec

d d t Q ₌ 0 de dt Therefore, de

4

x

6.5

x 10

3

= 1.5 °O/sec

dt

= 1.7 x 104

until the temperature rises appreciably compared with 1000°0 (i.e. say 200°0).

Therefore the melting point of lead (327.3°0) will be reached in about 300 seconds or 5 minutes.

Assume a latent heat of 6 cal/gm.

Heat required

=

6 x 5.6 x 105

=

3.4 x 106 cal.

Therefore the time reqUired at, say 3 cal/cm2/sec flux, = 3.4 x 106

3 x

6.5

x

10

3 • 170 seconds ;:::;::; 3 minutes

It is thus evident that the lead container will melt in something under 10 minutes if exposed to a fire unless adequate thermal insulation is provided.

3

-2. Thickness of Thermal Insulation ReSuired to Protect an IS-in. Diameter Lead Sphere

The solution of the problem, providing assumptions made are shown to be valid, is ,6 = eo (1 - e ｾＩＬ where R is the thermal resistance of the thickness of protecting material, and 0 is the thermal capacity of the lead sphere.

If eo is taken as 100000 and e as 32700,

thelL.2

=

0.327 and tables give t

=

0.4.

eo

ｾ

If t

=

3 hours then RO

=

3 x 34600,

=

2.7 x 104 seconds

o.

o

=

volume x P x s = 4 ｾ r 3 p s 3 and R = length = dr area x K "T'4..;.;f'(---r....'2...-_x-K Therefore RO = dr x 4:tr' r 3 p s 4 f ( r2 K x 3 = r d r P s 3K

dr

Note: f's refers to lead and K refers to insUlating material.

The above calculation rules that the ttutt value ＨｾＩ of the material

to be chosen must be less than

r P .

= 3RO (9

x

3 x2.54) 11.32.7 x 104x 0.03 = = 9.6 x 10-

5

cal/cm2;oO/seco

o.

71 Btu/hr/ft2

;OF

P

= 1:1.3 gm/cc s = 0.03 RO = 2.7 x 104 sec

.,

.

4

-1herefore

u

=

K

t

₌

_0.71 and therefore L = K x 12 =

0.71

17K inches

=

thickness of insulation, where the uni ts of K are Btu/ft;hr;OF

In choosing a sUitable material it should be ensured that the "K" value specified relates to a high temperature, e.g. l800oF.

Discussion of Assumptions

1. Neglect of Thermal Capacity of Insulant

If a substantial weight of insulation is used compared with the total weight of the lead then the thermal capacity of the insulant will contribute to keeping down the temperature of the lead. This effect may be approximately taken into account by adding to the thermal

capacity of the lead one half of the thermal capacity of the insulant.

e

2. _{Uniformity of Temperature of the Lead}

The calculation is based on the assumption that there are no

appreciable temperature"gradients within the lead. The dimensions of the sphere and the thermal properties of the lead indicate,

:. . "

however, that this assumption constitutes an appreciable departure from the truth. The extent of the departure suggests that the

calculated thickness of the insulation be increased by apprOXimately 20 per cent.

Summary

1. Lacking any specified requirements concerning the severity and

duration of a fire in which shipping containers might become involved it has been assumed that they might have to resist the temperature to be encountered in a severe building fire. A temperature of 10000C and a duration of 3 hours have been used in the calcuiations.

....

5

-2. It is evident that the present container (drawings of which were supplied by Mr. Henry of Applied Physics) would allow the lead to melt in under 10 minutes when subjected to the above conditions.

3. A thickness of insulation is recommended based on the uK" value of the insulating material (the ｦｾＢ value having been obtained at a high temperature, e.g. 1800PF). This thickness to be equal to 17K

inches (where the units of K are Btu/ft/hrj>F) with consideration being given when applicable to the ｴｨ･ｲｭ｡ｾ capacity of the insulating material. In addition due to the temperature gradient which will occur within the lead shield an arbitrary increase in thickness of 20 per cent should be applied to the calculated thickness to ensure that the lead will not melt on the exterior surface.

4.

The insulation may be incorporated directly with the container, provided that it will remain in place during a fire or be provided in a separate outer shell. In either case it would appear that there would be some design problems. However, if protection is to be_,

afforded to the lead shield, insulation must be prOVided.

e"

5.

There would be a choice of sUitable insulants that might be in the form of loose fill or castable units depending on how they are to be used, (e.g. in a separate enclosure or incorporated with present

container). TYpical insulants might be those used in the manufacture of safes such as a mixture of vermiculite, cement, and a binder or the possibility of using material ｦｯｲｭｩｾＮｦｩｲ･ﾷ｢ｲｩ｣ｫｳ might be considered.