food irradiation technology irradiation technolo… · for 1-side e-beam irradiation normalizing...
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FOOD IRRADIATION TECHNOLOGY
BAEN-625- Advances in Food Engineering
Food irradiation
KE produced by accelerators system is limited by regulation to
10 MeV for direct electron irradiation5 or 7.5 MeV for indirect irradiation using X-rays
E-beam Typical Installation
Key elementsAccelerator systemScanning systemMaterial handling systemOthers
Vacuum & cooling subsystemsShieldingSafety system
X-ray facility
Accelerator system
Bean scanning system
Scan hornScan magnet
Material handling
Key concepts and parameters
Dose uniformity and utilization efficiency for e-beamsDose uniformity and utilization efficiency for X-raysDose and dose rate estimationThroughput estimates for electrons and X-rays
Dose uniformity and utilization efficiency for e-beams
When energetic electrons pass through matter they lose energy via Coulomb interaction with atomic and molecular electrons and nucleiRadiation shower –primary electrons produce
Secondary electrons, tertiary, so on
Energy deposition profile
10 MeV electron normally incident onto the surface of water absorberAbsorbed dose at a particular depth can be calculated as
tIExD Aab ×= ")(
Current density [amp/cm2]
Example
I” = 10-6 amp/cm2 incident at surfacet = 1 s
kgkJDCeV
JsA
Cscm
Ag
cmeV
scm
Ag
cmeV
scm
Ag
cmMeV
tIEkGyD Aab
/85.1106022.1
11
106022.11
1185.1
1101085.1
11085.1
][
19
19
2
2
26
26
26
2
"
=×
××
××
×××=
×××=
××−
=
=
−
−
−
−
Depth-dose profile
For water [1 g/cm3]Can be extended to other absorbers with different densities
For 0.5 g/cm3 the MSP would have the same max (2.5 MeV-cm2/g], but it would occur at 5.5 cm
]/[ 2cmkgxDepth ×= ρ
Electron energy deposition is not constantPosition in product will receive the minimum dose and another the maximumDose uniformity ratio
Dmin and Dmax
minmax / DDDUR =
At 2.75 cm (maximum Eab)Dmax/Dmin = 1.35 =2.5/1.85
It remains constant up to A = 3.8 g/cm2
Beyond this depth, the minimum dose decreases and the ratio increases
overdose
unabsorbed
A
Dose uniformity ratio
Dose-depth curve for water-10MeV
]/[5.65.2])75.2(27.0exp[48.2
5.2025.084.1
2
7.2
gcmMeVMSPddMSP
ddMSP
−
<<−−=
<<+=
DUR –water – 10 MeV
EDUR /84.2=
DUR –water – 10 MeV
EDUR /84.2=
Useful energythat is absorbedby the product
Maximum efficiency
Minimum dose:
A = 1.85 MeV-cm2/gOptimum depth:
B = 3.75 g/cm2Max.Eff.
=A*B = 7 MeVFor 10 MeV electrons
Maximum efficiency is 70%
overdose
unabsorbed
A
B
Useful measure of the electron penetration power
Depth max. throughput occur
20]MeV[40][g/cmdepth Optimum 2 .- E . =
The entrance (surface) dose is 100%. The various ranges are identified as:
rmax = the depth at which the maximum dose occurs, ropt = the depth at which the dose equals the entrance dose, r50 is the depth at which the dose equals half of the maximum dose r33 is the depth at which the dose equals a third of the maximum dose.
10 MeV electrons in water
For the example, Emean is calculated to be 10.6 MeV for r50 = 4.53 cm (of water).
A typical dose distribution along the scan direction for an electron irradiator
For 1-side e-beam irradiation
Normalizing the surface dose to 100%,
the maximum dose Dmax of 130% occurs at a depth rmax = 2.8 cm, the entrance dose equals the exit dose at ropt = 4.0 cm.
For a process load of thickness between 2.8 and 4.0 cm,
the DUR is constant with a value of about 1.3.
If the process can allow a uniformity ratio of 2,
the maximum useful thickness of the process load is r50 = 4.5 cm the exit dose equals half of the maximum dose.
For 1-side e-beam irradiation
If a uniformity ratio of 3 is acceptable, the
maximum allowable thickness can be as much as r33 = 4.8 cmthe exit dose equals a third of the maximum dose
A steep increase in the DUR is observed as soon as the thickness exceeds the optimal range ropt. It approaches infinity when the maximum range of the electrons (about 6.5 cm for 10 MeV) is exceeded;
any product behind that range remains untreated.
Two sided irradiation can overcome this restriction and extend the processable thickness
Depth–dose distributions for 10 MeV electrons for two sided irradiation
Each curve refers to the dose distribution for one sided irradiation; dashed curve, irradiation from top side; solid curve, irradiation from bottom side.
2-sided e-beam irradiation
The position of Dmax will be somewhere midway between the two outside planes, along two lines parallel to the motion of the product and about halfway between the top and bottom surfaces. Dmin will be found along lines parallel to the direction of motion of the product either through the side edges at the top and bottom of the process load or in the midplane at the side edges.
Depth–dose distributions for 10 MeV electrons in varying thicknesses (widths) of water
0.00
0.50
1.00
1.50
2.00
2.50
3.00
0.00 2.00 4.00 6.00
Nor
mal
ized
Dos
e
Depth [cm]
top beam bottom beam double beamThickness = 6 cmDUR = 2.7
Depth–dose distributions for 10 MeV electrons in varying thicknesses (widths) of water
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
0.00 2.00 4.00 6.00 8.00
Nor
mili
zed
Dos
e
Depth [cm]
top beam bottom beam double beamThickness = 8.5 cmDUR = 1.35
Depth–dose distributions for 10 MeV electrons in varying thicknesses (widths) of water
0.00
0.20
0.40
0.60
0.80
1.00
1.20
1.40
1.60
0.00 5.00 10.00
Nor
mili
zed
Dos
e
Depth [cm]
top beam bottom beam double beamThickness = 10 cmDUR = 5.4
Single-sided IrradiationMin-Max & Energy Efficiency
Mono-energetic beamsNormally incident 10 MeV electronsThe integral under the curve gives a total absorbed energy of 9.61 MeV/electrons
46 keV is deposited in a 3-mil exit window
Goal: a minimum dose to be delivered to all portions of product
Too much dose in any portion of product: energy wasted and quality deterioration
dmax = maximum thickness @ minimum required doseOptimum depth thickness in water
Utilization efficiency – Single-sided
)](),0(min[
])75.2(27.0exp[48.2
5.20,25.084.1]/[
min
minmax
max
7.2
2
rababab
ab
u
ab
dEEEE
Ed
dd
d
ddgcmMeVE
=
=
=
−−=
<<+=−
ρ
η
2.0][4.0][ −= MeVEcmdopt
dopt = at the max throughput efficiency
DUR and Efficiency
r
r
dddEMeVdddEMeVd
ddEffEDUR
>=≤≤=
==
)(/100)0(/10
//84.2
max
max
max
Example of calculation – 10 MeV in water
Thick [cm]Eab
[MeV-cm2/g] max/min d/dmax dmax0.00 1.84 1.00 0.00 5.430.50 1.97 1.07 0.10 5.091.00 2.09 1.14 0.21 4.781.50 2.22 1.20 0.33 4.512.00 2.34 1.27 0.47 4.272.50 2.47 1.34 0.62 4.062.80 2.50 1.36 0.70 4.003.00 2.46 1.36 0.74 4.063.20 2.40 1.36 0.77 4.163.50 2.19 1.36 0.77 4.573.78 1.85 1.36 0.70 5.403.80 1.82 1.36 0.69 5.494.00 1.51 1.65 0.61 6.604.50 0.73 3.43 0.33 13.705.00 0.22 11.24 0.11 44.95
C5 = 10 Mev/C2C4 = C1/C5
C3 = C2/Eabminuntil Eabmaxthen Eabmax/Eabminthen Eabmax/C2
C1 = column 1C2 = ….
Optimum depth thickness in water
T = product thicknessx = depth into product measured from either surfaceEab(x) = single-sided energy deposition
Utilization efficiency – double sided
)](),0(min[
2
)()(
min
minmax
max
rababab
ab
u
ababt
dEEEE
Ed
dT
xTExEE
=
=
=
−+=
ρ
η
4.0][9.0][ −= MeVEcmdopt
dopt = at the max throughput efficiency
The energy deposition for X-rays decrease exponentially with timeFor mono-energetic X-ray beam of intensity I, the decrease in intensity in passing through a material of thickness s is:
For X-rays
soeII μ−=
Linear Attenuation coefficientIntensity at the surface
Single sided X-ray
X-ray Utilization efficiency
)exp()( dd mmu ρμρμη −=
d
s
Conveyor direction
X-rays flux Fo
Mass absorption coefficient
It is never used because the max:min ratio is very poor ~ 2.72Single-sided X-ray treatment is used when the product is rotated for a second pass or the product makes a single pass through two X-ray beams
Double-sided X-ray
d
Conveyor direction
X-rays flux Fo X-rays flux Fo
Dose distribution for double sided X-ray
[ ]
2/
2/minmax
2/min
max
)(
]1[5.0/
2)2/(
)1(
)(
du
dd
do
do
sdso
de
eeDD
eFdsD
eFD
eeFsD
μρ
μρμρ
μρ
μρ
μρμρ
μρη
μ
μ
μ
−
−
−
−
−−−
=
+=
==
+=
+=
Dose & Dose rate estimation
w
v
Electron beams
Electron beams
Wb = e-beam width [cm], A[cm2], t[s], w[cm], v[cm/s], IA [amp]
Dose & Dose rate estimation
)/(108.1]/[
)/(108.1 surface @][
//
)/()()(/
/)()(
6
6
bAs
As
bp
Aabs
Aabs
vWIskGyD
vwIkGyD
WDvtDD
vwIdEdDvwtA
AtIdEdD
×=
×==
==
===
&
&
Example
Scan with, w =100 cm; conveyor speed v = 10 cm/sCalculate the front surface dose delivered by a10MeV and 1mA beam
kgkJDCeV
JsA
Cscm
Ag
cmeV
scm
Ag
cmeV
scmcm
Ag
cmMeV
wwIdEkGyD Aab
/85.1106022.1
11
106022.11
1185.1
1101085.1
)/1000/(1085.1
/)(][
19
19
2
2
26
26
22
32
=×
××
××
×××=
×××=
×−
=
=
−
−
−
−
X-ray systemsIt more difficult to estimate because angular dependence of X-ray produced in the converter targetThe total electron beam energy (Qbeam) incident on the converter is multiplied by the X-ray efficiency
Dose & Dose rate estimation
Dose & Dose rate estimation
)/(7.2)]/([08.0
)/(08.0/
08.008.008.0 MeV, 5
60/)(
c
vwPDvwPFD
vwPAQF
PtQQFor
MeVE
s
mm
rayxo
beamrayx
c
===
==
===
=
−
−
μμ
ηη Power e-beam
incident ontothe converter
Throughput Estimates
][ dose minimum
[kg/s]product oft throughpumassMeV 7.5at rays-Xfor 0.045
MeV 5at rays-Xfor 0.03 electronsfor 0.45
efficiency throughput
)(/)()/(
min
p
min
kGyD
M
kGyDkWPskgM
t
tp
=
=≈≈≈=
=
η
η
Throughput rates versus e-beam power
Technology choices
Product characteristics
• irradiation goal• Safety• disinfestations
• fresh/frozen• radiation response• density • packaging• throughput peak/average
Process requirements Technology selection
• e-beam/X-ray• Kinetic energy• System power• material handling• single-/double-sided• absorbers?• refrigeration• ozone removal• shielding
• min. required dose• Max:Min ratio • mass thickness Max/variation• power Max/average• mass thickness• local regulations
Yearly costs
Costs estimates for food irradiation facilities[$K]Based on 2000 hours of production per year
15 kW 150 kWFixed costs
capital costsaccelerator 1176 2176installation 235 435shielding 353 653materla hdling 250 250building 2800 2800engineering 340 370total capital costs 5302 6962
investiment 619 813labor 300 300maintainance 71 121Total fixed costs 991 1235
Variable costselectricity 19 192labor 140 140maintanance 71 121Total variable costs 231 453
TOTAL YEARLY COSTS $1221 $1688
Cost element
Accelerator –LINAC [$K] =103log(P[kW)])
Installation [$K] = 20% accelerator
Shielding [$K] = 30% accelerator
Material handling system [$K] = 250
Building [$K] = 2800
Engineering [$K] = 10%( shield+ mat.hand+build)
Total Capital [$K], TC = 3350+1660*log(P[kW])
Investment expense [$K], I = 0.117*TC
Labor [$K], L = 300
Maintenance [$K], M = 5% (accelerator + mat.hand.)
Total Fixed Costs [$K], TFC = I + L + M
Electricity [$K], E = 6.4x10-4*P[kW]*U [hr] ; U=prod.hr/year
Labor [$K], L2 = 0.07*U [hr]
Maintenance [$K], M2 = 5% (accelerator + mat.hand.)
Total variable costs [$K], TV = E + L2 + M2
Total Yearly Costs [$K] = TFC + TV