Introduction to Health Physics: Fourth Edition - Ruang Baca FMIPA UB

560 CHAPTER 10
Table 10-7 lists the first TVL, TVL1, and the equilibrium TVL, TVLe, for concrete,
steel, and lead for various primary-beam quantum energies. The TVL1 differs from
TVLe in order to account for changes in the spectral distribution of the radiation as
it penetrates the shielding barrier. In Table 10-7, we find TVL1 to be 37-cm concrete,
and TVLe to be 33-cm concrete. The required thickness of concrete is
t barrier = TVL1 + (n − 1)TVLe
t barrier = 37 cm + (5.0 − 1) × 33 cm = 169 cm(66.5in.)
(10.31)
Now we must see whether this barrier thickness meets the TADR limit of 20 × 10−6 Sv
in 1 week (2 mrems in 1 week). The maximum IDR at the dose point, with the
calculated transmission factor of 7.2 × 10−6 is calculated as follows:
600
IDR =
Sv
h × 9.92 × 10−6 × (1m) 2
(6.3m) 2
−4 Sv
= 1.5 × 10
h .
When we substitute this value into Eq. (10.28), we have
IDR
RW =
Sv Gy
× Wpri × Upri
h wk
Gy
˙D0
=
h
−4 Sv Gy
1.5 × 10 × 800 × 0.5
h wk
600 Gy
−6 Sv Sv
= 100 × 10 = 100 μ
wk wk .
h
This TADR is five times greater than the required value of 20 μSv/wk. Increasing
the thickness of the concrete barrier by 3 HVLs (1 TVL = 3.3 HVL) would decrease
the TADR to 12.5 μSv/wk, which is reasonably conservative and would allow for an
increased utilization of the facility. If we choose to do this, then the barrier thickness
would be
33 cm
t barrier = 165 cm + 3 HVL ×
TVL
= 195 cm.
3.3 HVL
TVL
Since the primary barrier is much thicker than the other walls, which shield the
secondary (scatter and leakage) radiation, the width of the primary barrier is usually
restricted to one that is functionally effective. Good design practice specifies that the
width of the primary barrier must be greater by at least 30.5 cm (1 ft.) on each side
of the maximum-sized projection of the primary beam on the barrier (Fig. 10-21).
That is, its width is 61 cm (2 ft.) greater than the projection of the primary beam
at its greatest possible size. The widest possible beam projection occurs when the
collimator is rotated by 45◦ . Since the diagonal of a square is equal to 1.414 × length
of the side, and the length of the projected side is given by
s1 cm
d1
= s2 cm
,
d2