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Saturday, February 1, 2014

Principles Of Nuclear Technology Course

Q1 . Activity of one gm of the cover up of Turin i .e . R is cypher belowR (3 counts per bit 100 300 counts per min 5 counts /s 5 BqActivity of one gm of the bed sheet of Turin i .e . Ro is calculated belowRo 512 pCi (512 10-12 3 .7 1010 Bq 18 .94 BqR and Ro is related by the next equation ?tWhere ? is decline unvaried of C14 and given by 0 .693 /t1 /2t1 /2 of C14 is 5760 yearsand t is feel of the Shroud of TurinTherefore , life of the Shroud of Turin forget be The calculation shows that the sample is 11072 yrs elder . This but means that the shroud is not realQ2 . allow us move into there were N atoms of Pu238 in the RHU after 8 yearsThen , regularize of insubordination will be R NWhere ? is the decay constant of Pu238 and is given as (0 .693 /t1 /2where t1 /2 of Pu238 is 87 .7 years 87 .7 365 24 3600 s 2 .77 109 sNow , annihilation of each(prenominal) Pu238 atom gives a ? particle having energyE 5 .45 MeV 5 .45 1 .6 10-13 8 .72 10-13 which is absorbed in the RHUTherefore , power due to insubordination of Pu238 will be For this power to be 1 .5W 1 .5 Now let us try to calculate mathematical function of Pu238 at the time of launch let us comeback that as NoNo and N is related by the succeeding(a) equationhj-JLZ\ATZrtzjf Noe- ?t 7 .33 1021Therefore , the initial mass of Pu238 will be Q3 . Let Do be the battery-acid rate without the access and D is the dit rate after the Pb entry of ponderousness t has been installed . Then , oppressiveness t has to be calculated such that D and Do is related by the following equationD Doe- ?tWhere ? is the bilinear attenuation coefficient of Pb for 2 MeV ?-rays and t is ponderousness of Pb penetrationLinear attenuation coefficient of Pb for 2 MeV ?-rays is 51 .8 m-1Therefore , thickness of the Pb door for 100 times attenuation of the 2 MeV ? dose rate will be Area of the door A 8 sq ft 8 30! .48 30 .48 cm2 7432 .2432 cm2Volume of the door A t 7432 .2432 8 .892 cm3 66087 .51 cm3Mass of the start door will bem ?V 11 .35 66087 .51 gm (11 .35 66087 .51 /453 .6 lb 1653 .65 lbTherefore cost of the lead door will beC 0 .80 1653 .65 1323 ReferenceGlassstone S and Sesonske A . nuclear Reactor Engineering Volume 1 Chapman dorm fashion Inc . New York...If you want to get a full essay, redact it on our website: OrderCustomPaper.com

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