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