Click Here to
Order this information in Print

Click Here to
Order this information on CD-ROM

Click Here to
Download this information in PDF Format

 

Click here to make tpub.com your Home Page

Page Title: Effects of Temperature on Cross Section
Back | Up | Next

Click here for a printable version

Google


Web
www.tpub.com

Home

   
Information Categories
.... Administration
Advancement
Aerographer
Automotive
Aviation
Combat
Construction
Diving
Draftsman
Engineering
Electronics
Food and Cooking
Math
Medical
Music
Nuclear Fundamentals
Photography
Religion
USMC
   
Products
  Educational CD-ROM's
Printed Manuals
Downloadable Books
   
Back
Calculation of Macroscopic Cross Section and Mean Free Path
Up
Nuclear Physics and Reactor Theory Volume 1 of 2
Next
Neutron Flux

) ) ) ) DOE-HDBK-1019/1-93 NUCLEAR CROSS SECTIONS AND NEUTRON FLUX Reactor Theory (Neutron Characteristics) NP-02 Page 14 Rev. 0 Effects of Temperature on Cross Section As discussed, the microscopic absorption cross section varies significantly as neutron energy varies.  The microscopic cross sections provided on most charts and tables are measured for a standard neutron velocity of 2200 meters/second, which corresponds to an ambient temperature of 68   F.  Therefore, if our material is at a higher temperature, the absorption cross section will be lower than the value for 68   F, and any cross sections which involve absorption (for example, ) , )   , )    ) must be corrected for the existing temperature. a c f The following formula is used to correct microscopic cross sections for temperature.  Although the example illustrates absorption cross section, the same formula may be used to correct capture and fission cross sections. where: ) =   microscopic cross section corrected for temperature ) =   microscopic cross section at reference temperature (68   F or 20   C) o T =   reference temperature (68   F) in degrees Rankine (   R) or Kelvin (   K) o T =   temperature for which corrected value is being calculated NOTE:   When using this formula, all temperatures must be converted to   R or   K.   R =     F  +  460   K =     C  +  273 Example: What is the value of )    for uranium-235 for thermal neutrons at 500   F?  Uranium-235 f has a )    of 583 barns at 68   F. f Solution:

Privacy Statement - Press Release - Copyright Information. - Contact Us - Support Integrated Publishing