RADIATION CALCULATIONS QUESTION ANSWERS NDT CSWIP

Keywords article for this are RADIATION CALCULATIONS QUESTION ANSWERS NDT CSWIP. WELDING INSPECTION COURSE. WELDING COURSE. COMPLETE FREE COURSE WELDING INSPECTOR. 

RADIATION CALCULATIONS QUESTION ANSWERS NDT CSWIP

ISOTOPE

HALF LIFE

OUTPUT

HVL Fe

HVL Pb
COBALT 60 5.3 yrs 13 mSv/Ci @ 1m 22 12.5
IRIDIUM 192 74 days 4.8 mSv/Ci @ 1m 13 4.8
YTTERBIUM 169 32 days 1.2 mSv/Ci @ 1m 3 0.2

 

  1. Calculate the dose rate at one metre from an unshielded 75 curie Iridium 192 source.

    75 x 4.8 = 360 mSv/h

  2. Calculate the dose rate at 75 metres from an unshielded 75 curie Iridium 192 source.

    (360 x 12)/752 = 0.064 mSv/h = 64 mSv/h

  3. How many whole half value thicknesses of lead shielding are needed to reduce the dose rate at 75 metres from a 75 curie Iridium 192 source to below 7.5 mSv/h?

    1 HVL, 64 —–32: 2 HVL, 32 ——- 16: 3 HVL, 16 ——- 8: 4 HVL, 8 ——- 4

     therefore 4 HVL = 4 x 4.8 = 19.2 mm

  4. Calculate the distance at which the dose rate due to an unshielded 75 curie Iridium 192 source falls to 7.5 mSv/h.

    square root of (360 x 1000)/7.5 = 219.09 m

  5. Calculate the dose rate at one metre from an unshielded 55 curie Cobalt 60 source.

    55 x 13 = 715 mSv/h

 

  1. Calculate the dose rate at 55 metres from an unshielded 55 curie Cobalt 60 source.

    (715 X 12)/552 = 0.236 mSv/h = 236 mSv/h

  2. How many whole half value thicknesses of steel shielding are needed to reduce the dose rate at 55 metres from a 55 curie Cobalt 60 source to below 7.5 mSv/h?

    1 HVL, 236 ——– 118: 2 HVL, 118 ———- 59: 3 HVL, 69 ———–5:
    4 HVL, 29.5 ——— 14.75, 5 HVL, 14.75 ———- 7.375

     therefore 5 HVL = 5 x 22 = 110 mm

  3. Calculate the distance at which the dose rate due to an unshielded 55 curie Cobalt 60 source falls to 7.5 mSv/h.

    square root of (715 x 1000)/7.5 = 308.76 m

  4. Calculate the dose rate at one metre from an unshielded 20 curie Ytterbium 169 source.

    20 x 1.2 = 24 mSv/h

  5. Calculate the dose rate at 5 metres from an unshielded 20 curie Ytterbium 169 source.

    24/52 = 96 mSv/h = 960 mSv/h

  6. How many whole half value thicknesses of lead shielding are needed to reduce the dose rate at 5 metres from a 20 curie Ytterbium 169 source to 7.5 mSv/h?

    1 HVL, 960 ——– 480: 2 HVL, 480 ——– 240: 3 HVL, 240 ——— 120:
    4 HVL, 120 ——- 60, 5 HVL, 60 ———- 30: 6 HVL, 30 ——— 15: 7 HVL, 15 ———5

    therefore 7 HVL = 7 x 0.2 = 1.4 mm

  7. Calculate the distance at which the dose rate due to an unshielded 20 curie Ytterbium 169 source falls to 7.5 mSv/h.

    square root of (24 x 1000)/7.5 = 56.57 m

Keywords article for this are RADIATION CALCULATIONS QUESTION ANSWERS NDT CSWIP. WELDING INSPECTION COURSE. WELDING COURSE. COMPLETE FREE COURSE WELDING INSPECTOR. 

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