Re: Thermal issues for Co-60

[William V Lipton <liptonw@DTEENERGY.COM>]

The formula to use is:



decay heat (watts) =  K*A*E



where:



K is a conversion factor = 5.9278 E-3 [(watt)(dis)]/[((MeV)(Ci)]



A = activity, in Curies



E = MeV/disintegration



For Co-60, E = 2.60 MeV/dis



This assumes that all of the photon energy is absorbed, which is conservative.



(I'm looking at the back of my AAA card, which says that if you want the answer

for  TBq, you have to figure it out, yourself.)



The opinions expressed are strictly mine.

It's not about dose, it's about trust.

Curies forever.



Bill Lipton

liptonw@dteenergy.com





mark.ramsay@ionactive.co.uk wrote:



> Dear All

>

> I'm sure I should work this out, my excuse - too tired :)

>

> Take a 10000 TBq Co-60 source (lets say its a point source). Are there any

> useful rules of thumb for working out the thermal (heat output) of that source

> ?

>

> cheers

>

> Mark

>

> Forum : http://ionactiveweb.adflix.com/IonActive_Forum/index.php

>

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