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Half-value layers
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Half-value layers

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Author Topic: Half-value layers  (Read 12920 times)
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opus_two
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« on: Jun 15, 2005, 09:38 »

I looking for half-value, quarter-value, and tenth-value layers numbers for CS-137

For lead and Tungsten shelding

Thanks
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« Reply #1 on: Jun 16, 2005, 05:21 »

http://www.nukeworker.com/study/hp/rct/pdf/core_1-11.pdf
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« Reply #2 on: Jun 16, 2005, 09:30 »

The numbers we use here at work are in HVL and we deal pretty exclusively with Cs-137 and some Co-60.  Our "cheat sheets" state HVL for lead is 0.25 in. and 0.16" for tungsten.  Using the given equations from Rennhack's link you can solve for QVL and TVL.  For QVL I got 0.5" for lead and 0.32" for Tungsten.  For TVL I got 1.25" for lead and 0.8" for tungsten.  Let me know if the math is wrong or if my assumptions on HVL were rounded since that could lead to a discrepancy. 
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opus_two
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« Reply #3 on: Jun 16, 2005, 04:22 »

Thanks for the info!!

Its a start.. It is hard to find tables on the internet that give hvl, etc for various elements and energies
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« Reply #4 on: Jun 17, 2005, 10:09 »

Actually for Cs 137 the TVL for lead is .82 inches (2.1 cm) and fo Co 60 the TVL for lead is 1.57 in (4.0cm)
« Last Edit: Jun 17, 2005, 10:36 by Rad Gal » Logged

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« Reply #5 on: Jun 17, 2005, 11:53 »

Lead (11.35 g/cm )

Photon Energy(keV)    500  HVL (cm) 0.38
Photon Energy(keV)  1000  HVL (cm) 0.86
Photon Energy(keV)  1500  HVL (cm) 1.2
Photon Energy(keV)  2000  HVL (cm) 1.3
Photon Energy(keV)  3000  HVL (cm) 1.5
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« Reply #6 on: Jun 17, 2005, 12:09 »

Hey my math works if you are willing to forget Log rules in math.  Like since 2^n=4.....then n=2


Therefore, 2^n=10.....n=5 since before 2*2=4 then 2*5=10.  It has to be.

But hey lets try it the right way and 2^n=10....n=3.32.., then from a HVL of 0.25in you get 0.83in. approximately for a TVL of lead for Cs-137.
TVL for Tungsten becomes 0.53in fir Cs-137.

Luckily for me I dont have to admit to error due to my disclaimer in my post Smiley Smiley Smiley

Reminds me of a story.  NASA buillds a rocket (I believe to orbit some distant planet or something like that)and they have one set of enginneers calculate one set of numbers and give said numbers to another group of engineers.  But, since none of them had attended Naval Nuclear Power School and missed points on a test for their mistake, their math made them "Unitless" men in their work.  So they missed their mark and found out one group used english units the other thought they were in metric.  They probably are still out there looking for their "unit"....of measurement of course....and a new job.
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opus_two
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« Reply #7 on: Jun 17, 2005, 12:54 »

Thanks

I am rebuilding attentuators on one of the irradiator that uses two CS137 sources and three attenuators.  The HVL atteuator has the lead insert beat up in it so I want to replace it with a tungsten insert. While I was replacing it we decided to replace all three attenuators instead. The hvl, qvl and tvl.

So I am chasing down tungten values.

I located some alloy values for tungsten from Wolfmet.com at a density of 19.3 vs Lead at 11.3

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« Reply #8 on: Jun 17, 2005, 01:04 »

Reminds me of a story.  NASA buillds a rocket (I believe to orbit some distant planet or something like that)and they have one set of enginneers calculate one set of numbers and give said numbers to another group of engineers.  But, since none of them had attended Naval Nuclear Power School and missed points on a test for their mistake, their math made them "Unitless" men in their work.  So they missed their mark and found out one group used english units the other thought they were in metric.  They probably are still out there looking for their "unit"....of measurement of course....and a new job.

That was a Mars lander. Calculations in meters, measurements in feet, results in 'excessive landing speed at incorrect angle' otherwise known as crash. Oops... looks like NASA needs a little three-part-communications training (as well as attention to detail, peer checking, STAR, etc). Guess they got the STAR -- as in 'S**t That Ain't Right!
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