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- since 1996 -

IMG ULTRASUONI SRL
Ultrasonic transducers for industrial applications (NDT) and medical (Doppler effect). Probes, Ultrasonic instruments, Special systems for NDT.

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Technical Discussions
gowrishankar
R & D, scientist
ISRO, India, Joined Nov 2008, 34

gowrishankar

R & D, scientist
ISRO,
India,
Joined Nov 2008
34
08:00 Nov-05-2004
eddy current testing of aluminum welds

Can we test 4.5mm aluminium(aa 2219) weld by eddy current ?testing, if so please give the optimum probe details.

My second question is Radioactive sources like Iridium and Cobalt etc will be decaying continuously( based on half life time). But at last what will be formed from Ir-192 and what is the final product from Co-60 etc..
THANK YOU.
GOWRISHANKAR



    
 
 
John
John
09:18 Nov-05-2004
Re: eddy current testing of aluminum welds
----------- Start Original Message -----------
: Can we test 4.5mm aluminium(aa 2219) weld by eddy current ?testing, if so please give the optimum probe details.
: My second question is Radioactive sources like Iridium and Cobalt etc will be decaying continuously( based on half life time). But at last what will be formed from Ir-192 and what is the final product from Co-60 etc..
: THANK YOU.
: GOWRISHANKAR
------------ End Original Message ------------

Answer to your second question is Radioactive waste.



    
 
 
S.V.Swamy
Engineering, - Material Testing Inspection & Quality Control
Retired from Nuclear Fuel Complex , India, Joined Feb 2001, 787

S.V.Swamy

Engineering, - Material Testing Inspection & Quality Control
Retired from Nuclear Fuel Complex ,
India,
Joined Feb 2001
787
06:10 Nov-06-2004
Re: eddy current testing of aluminum welds
----------- Start Original Message -----------
: : Can we test 4.5mm aluminium(aa 2219) weld by eddy current ?testing, if so please give the optimum probe details.
: : My second question is Radioactive sources like Iridium and Cobalt etc will be decaying continuously( based on half life time). But at last what will be formed from Ir-192 and what is the final product from Co-60 etc..
: : THANK YOU.
: : GOWRISHANKAR
: Answer to your second question is Radioactive waste.
------------ End Original Message ------------

Dear John,

Sorry to differ with you but if the daughter product of a radioactive element is radioactive, that too will decay with a specific half-life. The ultimate end product must necessarily be non-radioactive.

I will post the radioactive decay chain of Ir-192 and Co-60 in the next few days.

Swamy


    
 
 
John
John
04:47 Nov-07-2004
Re: eddy current testing of aluminum welds
----------- Start Original Message -----------
: : : Can we test 4.5mm aluminium(aa 2219) weld by eddy current ?testing, if so please give the optimum probe details.
: : : My second question is Radioactive sources like Iridium and Cobalt etc will be decaying continuously( based on half life time). But at last what will be formed from Ir-192 and what is the final product from Co-60 etc..
: : : THANK YOU.
: : : GOWRISHANKAR
: : Answer to your second question is Radioactive waste.
: Dear John,
: Sorry to differ with you but if the daughter product of a radioactive element is radioactive, that too will decay with a specific half-life. The ultimate end product must necessarily be non-radioactive.
: I will post the radioactive decay chain of Ir-192 and Co-60 in the next few days.
: Swamy
------------ End Original Message ------------

Of course it will, but it will be disposed of as radioactive waste. It will never be considered non radioactive. The half life will never reach true zero. Although unmeasurable, you can still divide by 2.
Yes, I agree, it is an isotope of basic element, and is fighting to get back to its normal state, but will never be non radioactive. I disagree with you.


    
 
 
Ed Ginzel
R & D, -
Materials Research Institute, Canada, Joined Nov 1998, 1252

Ed Ginzel

R & D, -
Materials Research Institute,
Canada,
Joined Nov 1998
1252
08:04 Nov-07-2004
Re: eddy current testing of aluminum welds
John:
Swamy's reply is quite correct and indicates the correct academic understanding of the issue. The fact is, any sample of radioactive mateiral is a defined mass, i.e. a specific quantity of atoms. Therefore you cannot just keep on dividing by 2 as you suggest. Unlike the ideal quantity of "length" there is a quantised end to a mass as it relates to radioisotopes. After the last atom in the mass decays to its stable state the process of "radioactive decay" is completed.
The von Weizsacker equation predicts the binding energy and it is usual to plot this as a mass parabola (plotting nuclear binding energy versus Z) as calculated for the lowest energy level of a specific isotope (a dual parabola occurs for mass numbers being even). We see, for example, there are several isotopes of Cobalt but only Co59 is stable. All others move to lower masses (more stable binding energy states) by decay processes.

More problematic in the "concern" for decay is the decay constant, as the final few atoms in the mass move to more stable binding energies. The decay constant is a statistical value used for large numbers of atoms decaying and describes the "probability" of an atom disintegrating in a unit time. Since it is a probablistic value, the actual time involved for the final few atoms is not so simply defined by a "specific time" because the assumptions made for large population statistics breaks down and the final disintigrations in the original mass may occur well outside the time "estimate" assumed by the decay constant for the "half-life" of a large mass.

It might be informative for you to read "Nuclear and Radiochemistry" by Friedlander, Kennedy, Miller (John Wiley). I have noted in libraries that newer editions than my Second edition are now available.

Ed
----------- Start Original Message -----------
: : : : Can we test 4.5mm aluminium(aa 2219) weld by eddy current ?testing, if so please give the optimum probe details.
: : : : My second question is Radioactive sources like Iridium and Cobalt etc will be decaying continuously( based on half life time). But at last what will be formed from Ir-192 and what is the final product from Co-60 etc..
: : : : THANK YOU.
: : : : GOWRISHANKAR
: : : Answer to your second question is Radioactive waste.
: : Dear John,
: : Sorry to differ with you but if the daughter product of a radioactive element is radioactive, that too will decay with a specific half-life. The ultimate end product must necessarily be non-radioactive.
: : I will post the radioactive decay chain of Ir-192 and Co-60 in the next few days.
: : Swamy
: Of course it will, but it will be disposed of as radioactive waste. It will never be considered non radioactive. The half life will never reach true zero. Although unmeasurable, you can still divide by 2.
: Yes, I agree, it is an isotope of basic element, and is fighting to get back to its normal state, but will never be non radioactive. I disagree with you.
------------ End Original Message ------------




    
 
 
John
John
05:32 Nov-08-2004
Re: eddy current testing of aluminum welds
----------- Start Original Message -----------
: John:
: Swamy's reply is quite correct and indicates the correct academic understanding of the issue. The fact is, any sample of radioactive mateiral is a defined mass, i.e. a specific quantity of atoms. Therefore you cannot just keep on dividing by 2 as you suggest. Unlike the ideal quantity of "length" there is a quantised end to a mass as it relates to radioisotopes. After the last atom in the mass decays to its stable state the process of "radioactive decay" is completed.
: The von Weizsacker equation predicts the binding energy and it is usual to plot this as a mass parabola (plotting nuclear binding energy versus Z) as calculated for the lowest energy level of a specific isotope (a dual parabola occurs for mass numbers being even). We see, for example, there are several isotopes of Cobalt but only Co59 is stable. All others move to lower masses (more stable binding energy states) by decay processes.
: More problematic in the "concern" for decay is the decay constant, as the final few atoms in the mass move to more stable binding energies. The decay constant is a statistical value used for large numbers of atoms decaying and describes the "probability" of an atom disintegrating in a unit time. Since it is a probablistic value, the actual time involved for the final few atoms is not so simply defined by a "specific time" because the assumptions made for large population statistics breaks down and the final disintigrations in the original mass may occur well outside the time "estimate" assumed by the decay constant for the "half-life" of a large mass.
: It might be informative for you to read "Nuclear and Radiochemistry" by Friedlander, Kennedy, Miller (John Wiley). I have noted in libraries that newer editions than my Second edition are now available.
: Ed

Yes Theoretically. By the time a 100 curie source of Iridium decays to its original state, we will be long gone. And guess what! It will still be stored as radioactive waste. Of course theoretically an isotope of Ir192 will decay to the basic element Iridium.
I guess to answer the mans question, Ir and Co60 will decay to Ir and Co59.
: : : : : Can we test 4.5mm aluminium(aa 2219) weld by eddy current ?testing, if so please give the optimum probe details.
: : : : : My second question is Radioactive sources like Iridium and Cobalt etc will be decaying continuously( based on half life time). But at last what will be formed from Ir-192 and what is the final product from Co-60 etc..
: : : : : THANK YOU.
: : : : : GOWRISHANKAR
: : : : Answer to your second question is Radioactive waste.
: : : Dear John,
: : : Sorry to differ with you but if the daughter product of a radioactive element is radioactive, that too will decay with a specific half-life. The ultimate end product must necessarily be non-radioactive.
: : : I will post the radioactive decay chain of Ir-192 and Co-60 in the next few days.
: : : Swamy
: : Of course it will, but it will be disposed of as radioactive waste. It will never be considered non radioactive. The half life will never reach true zero. Although unmeasurable, you can still divide by 2.
: : Yes, I agree, it is an isotope of basic element, and is fighting to get back to its normal state, but will never be non radioactive. I disagree with you.
------------ End Original Message ------------




    
 
 

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