Modified atmosphere packaging (MAP) of fruits requires careful selection of pack

Question

Modified atmosphere packaging (MAP) of fruits requires careful selection of packaging materials in order to achieve optimal equilibrium headspace O2 and CO2 concentrations, which are critical to prevent undesirable anaerobic respiration. Consider a tropical fruit (0.5 kg) packaged in a LDPE bag that has a total surface area of 0.03 m^2. At 20 C, the product has a respiration rate that can be described by R = 0.92 x [O2], where R is respiration rate of the fruit in mL/kg.h and [O2] is concentration of oxygen in %.

(i) Using the Arrhenius equation, calculate O2 and CO2 permeability coefficients for the LDPE film at 20 C using the data below:

   permeability(mL.mil/m2.h.atm), Temperature (C), Activation energy (kJ/mol)

Oxygen: 110, 10, 30.2

CO2: 366, 10, 31.1

(ii) If the desirable equilibrium O2 and CO2 concentrations in the headspace are 2% and 6%, respectively, determine the thickness of the LDPE film suitable for the modified atmosphere package. Assume that the respiration quotient (RQ) of the fruit is 1 (i.e., the rate of O2 consumption is equal to the rate of CO2 generation)?

(iii) Using the thickness value obtained from part (ii), determine the equilibrium O2 and CO2 concentrations in the package headspace.

Explanation / Answer

Ln(Pi) = Ea/RT x Ln(A) where A is permeability coefficient

where Pi is PO2 or PCO2, PO2 and PCO2 are measured O2 and CO2 permeability coefficients (mmol·cm-1 per cm 2 per hour per kPa), respectively ; Ea is the energy of activation of O2 or CO2 permeation (kJ·mol-1); and R is the gas constant (0.0083144 kJ/mol per °K).

permeability of O2 is mmol/m2.h.atm

Ea/RT for O2 is 30.2 kJ/mol /0.0083144 kJ/mol K x 293 K = 12.39

Ea/RT for CO2 is 31.1 kJ/mol /0.0083144 kJ/mol K x 293 K = 12.766

Ln(PO2) = 12.39 x lnA

ln(110)/12.39 = lnA

A = e(0.3793)

A for O2=1.46

Ln(PCO2) = 12.76 x lnA

ln(366)/12.766 = lnA

A = e(0.4623)

A for CO2=1.587

2% oxygen will mean a partial pressure of 15.2 mmHg which is 2.026kPa

we have the formula

RRO2 = {(PO2.A/x) x [(O2)atm-(O2)pkg]}/W; RRO2=1, W = 0.5 A= 3 cm2

where RRO2 and RRCO2 are the rates of O2 uptake and CO2 production (mmol·kg-1·h-1), respectively; PO2 and PCO2 are measured O2 and CO2 permeability coefficients (mmol·cm-1 per cm 2 per hour per kPa), respectively, for our LDPE at the storage temperature; A is film area (cm2 ); x is film thickness (cm); (O2 ),,m and (O2 )pkg are atmospheric and package partial pressures of 02 (kPa), respectively; (CO2 )pkg and (CO2 )atm are the package and atmospheric CO2 partial pressures (kPa), respectively, and W is fruit weight (kg)

1 = {((0.03 x 110)/x) x (0.21-0.02 atm)}/0.5

0.5= ((3.3/x) x 0.19)

3.3/x = 0.5/19.18

x=3.3/2.64

x = 1.254 mm

O2 and CO2 concentrations in the head space are 0.02 atm and 0.06 atm

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