FOOD IRRADIATION TECHNOLOGY. BAEN-625- Advances in Food Engineering

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Oscar French
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1 FOOD IRRADIATION TECHNOLOGY BAEN-625- Advances in Food Engineering

2 Food irradiation KE produced by accelerators system is limited by regulation to 10 MeV for direct electron irradiation 5 or 7.5 MeV for indirect irradiation using X-rays

3 E-beam Typical Installation Key elements Accelerator system Scanning system Material handling system Others Vacuum & cooling subsystems Shielding Safety system

4 X-ray facility

5 Accelerator system

6 Bean scanning system Scan magnet Scan horn

7 Material handling

8 Key concepts and parameters Dose uniformity and utilization efficiency for e- beams Dose uniformity and utilization efficiency for X-rays Dose and dose rate estimation Throughput estimates for electrons and X-rays

9 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 nuclei Radiation shower primary electrons produce Secondary electrons, tertiary, so on

10 Energy deposition profile 10 MeV electron normally incident onto the surface of water absorber Absorbed dose at a particular depth can be calculated as D( x) = " Eab I A t Current density [amp/cm 2 ]

11 Example I = 10-6 amp/cm 2 incident at surface t = 1 s D[ kgy] = 1.85MeV cm = g = = 1.85eV E ab ev cm g I " A D = 1.85kJ / kg 6 2 t 2 cm g A cm A cm 2 A cm 1s 2 1s 1C s 1A s 1eV 19 J C

12 Depth-dose profile For water [1 g/cm 3 ] Can be extended to other absorbers with different densities 2 Depth = ρ x [ kg / cm ] For 0.5 g/cm 3 the MSP would have the same max (2.5 MeV-cm 2 /g], but it would occur at 5.5 cm

13 D min and D max Electron energy deposition is not constant Position in product will receive the minimum dose and another the maximum Dose uniformity ratio DUR = D max / D min

14 Dose uniformity ratio At 2.75 cm (maximum E ab ) D max /D min = 1.35 =2.5/1.85 It remains constant up to A = 3.8 g/cm 2 Beyond this depth, the minimum dose decreases and the ratio increases overdose A unabsorbed

15 Dose-depth curve for water-10mev MSP = d 0 < d < 2.5 MSP = 2.48exp[ 0.27( d 2.75) MSP[ MeV cm 2 / g] 2.7 ] 2.5 < d < 6.5

16 DUR water 10 MeV DUR = 2.84 / E

17 DUR water 10 MeV DUR = 2.84 / E Useful energy that is absorbed by the product

18 Maximum efficiency Minimum dose: A = 1.85 MeV-cm 2 /g Optimum depth: B = 3.75 g/cm2 overdose Max.Eff. =A*B = 7 MeV For 10 MeV electrons Maximum efficiency is 70% A B unabsorbed

19 Depth max. throughput occur Useful measure of the electron penetration power 2 Optimum depth [g/cm ] = 0. 4 E [MeV]-0. 2

20 10 MeV electrons in water The entrance (surface) dose is 100%. The various ranges are identified as: r max = the depth at which the maximum dose occurs, r opt = the depth at which the dose equals the entrance dose, r 50 is the depth at which the dose equals half of the maximum dose r 33 is the depth at which the dose equals a third of the maximum dose. For the example, E mean is calculated to be 10.6 MeV for r 50 = 4.53 cm (of water).

21 A typical dose distribution along the scan direction for an electron irradiator

22 For 1-side e-beam irradiation Normalizing the surface dose to 100%, the maximum dose D max of 130% occurs at a depth r max = 2.8 cm, the entrance dose equals the exit dose at r opt = 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 r 50 = 4.5 cm the exit dose equals half of the maximum dose.

23 For 1-side e-beam irradiation If a uniformity ratio of 3 is acceptable, the maximum allowable thickness can be as much as r 33 = 4.8 cm the 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 r opt. 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

24 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.

25 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.

26 Depth dose distributions for 10 MeV electrons in varying thicknesses (widths) of water Thickness = 6 cm DUR = top beam bottom beam double beam Normalized Dose Depth [cm]

27 Depth dose distributions for 10 MeV electrons in varying thicknesses (widths) of water Thickness = 8.5 cm DUR = 1.35 Normilized Dose top beam bottom beam double beam Depth [cm]

28 Depth dose distributions for 10 MeV electrons in varying thicknesses (widths) of water Thickness = 10 cm DUR = 5.4 Normilized Dose top beam bottom beam double beam Depth [cm]

29 Single-sided Irradiation Min-Max & Energy Efficiency Mono-energetic beams Normally incident 10 MeV electrons The 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

Utilization efficiency Single-sided d max = maximum thickness @ minimum required dose Optimum depth thickness in water d opt [ cm] = 0.4E[ MeV ] 0.

30 Utilization efficiency Single-sided d max = maximum minimum required dose Optimum depth thickness in water d opt [ cm] = 0.4E[ MeV ] 0.2 d opt = at the max throughput efficiency E ab [ MeV / g] = d, 2.7 = 2.48exp[ 0.27( d 2.75) ] d ηu = dmax E dmax = ρeabmin E = min[ E (0), E ( d )] abmin cm 2 ab ab r 0 < d < 2.5

31 DUR and Efficiency DUR = 2.84 / E Eff d d max max = d / d max = 10MeV / E(0) = 10MeV / E( d) 0 d d > d r d r

32 Example of calculation 10 MeV in water Thick [cm] Eab [MeV-cm 2 /g] max/min d/dmax dmax C1 = column 1 C2 =. C5 = 10 Mev/C2 C4 = C1/C5 C3 = C2/E abmin until E abmax then E abmax /E abmin then E abmax /C2

33 Utilization efficiency double sided Optimum depth thickness in water d opt [ cm] = 0.9E[ MeV ] 0.4 d opt = at the max throughput efficiency T = product thickness x = depth into product measured from either surface E ab (x) = single-sided energy deposition E η d E u abt = max = abmin E T d = ab max 2E ρe ( x) + abmin = min[ E E( T ab x) (0), E ab ( d r )]

34 For X-rays The energy deposition for X-rays decrease exponentially with time For mono-energetic X-ray beam of intensity I, the decrease in intensity in passing through a material of thickness s is: I = I o e μs Intensity at the surface Linear Attenuation coefficient

35 X-ray Utilization efficiency Single sided X-ray η u = ( μ ρd)exp( μ ρd m m ) Conveyor direction Mass absorption coefficient X-rays flux F o s d

36 Double-sided X-ray It is never used because the max:min ratio is very poor ~ 2.72 Single-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 Conveyor direction X-rays flux F o X-rays flux F o d

37 Dose distribution for double sided X- ray [ ] 2 / 2 / min max 2 / min max ) ( ] 0.5[1 / 2 2) / ( ) (1 ) ( d u d d d o d o s d s o de e e D D F e d s D e F D e e F s D μρ μρ μρ μρ μρ μρ μρ μρ η μ μ μ = + = = = + = + =

38 Dose & Dose rate estimation Electron beams w v

39 Dose & Dose rate estimation Electron beams D( d) = A / t D( d) = E D& = D / t D D& s s = E vw p abs abs [ kgy] = ( d) I ( d) I /( surface = [ kgy / s] = /( vw W b = e-beam width [cm], A[cm 2 ], t[s], w[cm], v[cm/s], I A [amp] A A t / = Dv / W b A 6 I A b 6 ) I A /( vw)

40 Example Scan with, w =100 cm; conveyor speed v = 10 cm/s Calculate the front surface dose delivered by a 10MeV and 1mA beam D[ kgy] = E 1.85MeV = g = ab cm ev ( d) I cm g 2 cm A = 1.85eV g cm D = 1.85kJ / kg 6 2 A / ww A cm 2 A cm /(1000cm 2 1s / s) 1C s 1A s 1eV 2 19 J C

41 Dose & Dose rate estimation X-ray systems It more difficult to estimate because angular dependence of X-ray produced in the converter target The total electron beam energy (Q beam ) incident on the converter is multiplied by the X-ray efficiency

42 Dose & Dose rate estimation ηc = E( MeV ) / 60 For 5 MeV, η = 0.08 Q F o D = D x ray s = Q = m = 2.7P c 0.08Q x ray μ F = beam / A = 0.08P 0.08[ μ P /( vw)] /( vw) = 0.08Pt m /( vw) Power e-beam incident onto the converter

43 Throughput Estimates M M D p p min ( kg / s) = η P( kw t ) / min = mass throughput of product [kg/s] D = minimum dose [ kgy] ( kgy) ηt = throughput efficiency 0.45 for electrons 0.03 for X - rays at 5 MeV for X - rays at 7.5 MeV

44 Throughput rates versus e-beam power

45 Technology choices Product characteristics Process requirements Technology selection irradiation goal Safety disinfestations fresh/frozen radiation response density packaging throughput peak/average min. required dose Max:Min ratio mass thickness Max/variation power Max/average mass thickness local regulations e-beam/x-ray Kinetic energy System power material handling single-/doublesided absorbers? refrigeration ozone removal shielding

46 Yearly costs Costs estimates for food irradiation facilities [$K] Based on 2000 hours of production per year 15 kw 150 kw Fixed costs capital costs accelerator installation shielding materla hdling building engineering total capital costs investiment labor maintainance Total fixed costs Variable costs electricity labor maintanance Total variable costs TOTAL YEARLY COSTS $1221 $1688

47 Cost element Accelerator LINAC [$K] =10 3 log(p[kw)]) Installation [$K] = 20% accelerator Shielding [$K] = 30% accelerator Material handling system [$K] = 250 Building [$K] = 2800 Engineering [$K] = 10%( shield+ mat.hand+build) Total Capital [$K], TC = *log(P[kW]) Investment expense [$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

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