Packaging Design for Moisture-Sensitive Foods PDF

1 J. M. Krochta FST 131 PACKAGE DESIG N F OR M OIST UR E-SENSI TIVE F OODS The Food-Pa cka ge System Water content is central to definition, stability and quality of food categories. In foods packag...

1 J. M. Krochta FST 131 PACKAGE DESIG N F OR M OIST UR E-SENSI TIVE F OODS The Food-Pa cka ge System Water content is central to definition, stability and quality of food categories. In foods packaged in polymer films, moisture: permeates through the packaging film, diffuses through the package free volume air space, and sorbs on, or desorbs from, and diffuses within the food. H2O in Package Film pW1 > pW2 PW Food A pW1 pW2 WS m = m{aW,T} H2O out pW1 < pW2 Permeation. The rate of moisture permeation depends on: 1) nature of the film (PW) 2) film thickness (X) 3) film area (A) 4) difference between outside and inside partial pressures (pW1 - pW2) 5) temperature (T) Food Moisture Sorption. The rate of moisture sorption depends on: 1) nature of the food (m = m{aW,T}) 2) food weight (WS) 3) water vapor partial pressure inside package (aW= pW/pWo) 4) temperature (m = m{aw,T})  2 J. M. Krochta FST 131 Food-Pa ckage Materia l Ba lan ce for W ater Vapor We make certain assumptions: 1) The package head space is very small. 2) The package head space offers little resistance to mass transfer, resulting in uniform moisture in the head space at any given time. 3) The resistance to moisture diffusion in the food is negligible, resulting in uniform moisture in the food at any given time. 4) The food moisture content (m) is in equilibrium with the package environment, and follows some simple relationship with the package water vapor partial pressure (pW2). 5) Temperature and relative humidity (water vapor partial pressure, pW1) of environment outside package are constant. Recall the general material balance equation: Input + Generation = Output + Accumulation (1) The resulting equation for moisture transport into a food-package system (Taoukis et al., 1988) is dm WS = PW A (pW1 - pW2) (2) dt X m is the moisture per unit dry weight of food (g water/kg dry solids) pW1 is the partial pressure of water vapor in the outside air (atm) pW2 is the partial pressure of water vapor inside the package (atm) PW is the permeability of the packaging film to water (g/m2.atm.d/m) WS is the weight of dry solids of the food in the package (kg dry solids) A is the package area (m2) X is the film thickness (m)  3 J. M. Krochta FST 131 Materia l Ba lan ce Solution We must now relate m with pW. Moisture Isotherm. The linear isotherm is as follows: m = b aW + c = b (pW/pWo) + c (3) b and c are constants specific to a given food at a given temp. pWo is the vapor pressure of pure water at the same temp. Using equation (3) for our system: pW2 = (pWo/b) m - pWoc/b (4) Also: pW1 = (pWo/b) me - pWoc/b (5) me is the moisture content of the food product which would be at equilibrium with the atmosphere outside the package. Therefore: pW1 - pW2 = (pWo/b)(me - m) (6) Substituting equation (6) into the material balance equation (2): dm WS = PW A pWo (me - m) (7) dt b X Rearranging equation (7): dm = PW A pWo dt (8) (me - m) b WS X The integrated result is ln (me - mi) = PW A pWo t (9) (me - m) b WS X  Show the equation necessary for solution of the problem. List values for the components of the equation that you will use to solve the problem, always being sure to use units along with any value. These should be the values and units you will use with the equation. Use the equation with your selected values (including units) to arrive at an answer, and place a rectangle around your answer. Make sure your units cancel out appropriately to arrive at an MOISTURE SENSITIVE answer with FOOD the correct units. >>> You have developed a new intermediate-moisture fruit bar. This product becomes unacceptable when it gains too much moisture, because then it becomes vulnerable to microbial growth. It also becomes unacceptable when it loses too much moisture, because then it becomes too hard to chew. You are most concerned about the latter for distribution of this product in a part of the country where it is likely to be stored at approximately 25°C and 20% relative humidity (RH). The data that you have collected on the product includes the following: The fruit bar moisture content immediately after production and just before packaging is 0.20 g water/g dry solids. Sensory testing has revealed that the fruit bar has significantly lower acceptability after it reaches 0.10 g water/g dry solids. Thus, this moisture content is selected to represent the limit of the product's shelf life. In establishing the 25°C moisture isotherm for the fruit bar (moisture content (dry basis) vs water activity), you find that the relevant part of the isotherm curve can be represented by a straight line with slope 0.12 g water/g dry solids per unit aw and that the moisture content of the fruit bar at 20% RH is 0.05 g water/g dry solids. The pouch required to package four fruit bars of 125 g (dry solids) each (total weight is thus 500 g) will have a total surface area of 0.15 m2. The vapor pressure of pure water at 25°C is 0.0313 atm (3.17 kPa). 1) Calculate t he va lue o f P W / X in ( g H 2 0) / (m 2. d a y. kP a) req u ire d t o a ch ie ve a s he lf life of 12 m on ths. 2) P W va lue s o f L DPE , PET & P VC a re 0.0 3 1, 0.168 & 0.6 17 ( g. mm )/ (m 2. da y. k Pa ), resp ect ive ly. Ass um ing that the pa ck a ge film t hick nes s is limite d t o 0.0 1-0.1 m m, dete rm in e w het he r L DPE , P ET o r P VC com es closes t to g ivin g a 1 2 mo. s helf life. >> H o me wo rk is due at be ginn ing o f cla ss o n Fr id a y, No ve m be r 13.  1 J. M. Krochta FST 131 PACKAGE DESIG N F OR OXYG EN- SENSITI VE F OODS The Food- Package System In foods packaged in polymer films, oxygen: 1) permeates (sorbs, diffuses, desorbs) through the packaging film, 2) diffuses through the package free volume air space, 3) sorbs on and diffuses within the food, and 4) reacts with the food Package Film PO pO1 pO2 pO2 A Food X O2 W pT rrO Vu QOs Film Oxygen Permeation. The rate of oxygen permeation depends on: 1) nature of the film (PO), 2) film thickness (X), 3) film area (A), 4) difference between external and internal oxygen partial pressures (pO1 - pO2) 5) temperature (PO = PO{T}) Food Oxidation. Rate of oxidative reactions depends on: 1) nature of the food (affects rrO and QOs) 2) food weight (W) 3) oxygen partial pressure inside package (pO2) 4) temperature (rrO = rrO{T}) 5) food water activity (aw)  2 J. M. Krochta FST 131 Unsteady-State Material Balance We make certain assumptions: 1) The permeation rate across the film can be described by the equation describing steady-state mass transport across a film. 2) The package headspace offers little resistance to mass transfer. Thus, it is assumed to have uniform concentration at any time. 3) The resistance to oxygen diffusion in the food is negligible. Thus, it is assumed to have uniform concentration at any time. 4) The food is in equilibrium with the package environment. Thus, the oxygen reaction can be expressed in terms of the package headspace pO2. 5) Constant temperature and constant food moisture content. ACCUMULATION = INPUT - OUTPUT + GENERATION For oxygen transport into a food-package system with oxygen reaction: dVOu = PO A (pO1 - pO2) - dVOr (1) dt X dt VOu = yO Vu = (pO2/pT) Vu (2) dpO2 Vu = PO A (pO1 - pO2) - rrO W (3) dt pT X VOu is the volume of oxygen in unfilled package volume (headspace) (cm3) Vu is unfilled package volume (headspace) (cm3) VOr is the volume of oxygen sorbed and reacted (cm3) pO1 is the partial pressure of oxygen in the external air (atm) pO2 is the partial pressure of oxygen in the package inside air (atm) pT is total pressure in the package (atm) PO is the film oxygen permeability (cm3/cm2.atm.s/cm) A is the package area (cm2) X is the film thickness (cm) W is the weight of the food placed in the package (g), rrO is the oxygen reaction rate (cm3/s.g) QOs is quantity of oxygen food can react with before unacceptable (cm3/g)  3 J. M. Krochta FST 131 Unsteady-State Material Balance Sol ution Often the oxidation kinetic expression can be approximated by a simple expression: rrO = b pO2 (4) where b is a constant for a specific food at a specific temperature. Substituting eq'n (4) into eq'n (3) and then rearranging: dpO2 = P' dt (5) pO1 - b' pO2 where P' = PO A pT/X Vu (6) and b' = 1 + W b X/PO A (7) Therefore: ln (pO1 - b' pO2o) = b' P' t (8) (pO1 - b' pO2) or pO2 = b' pO2o - pO1 + pO1 (9) b' exp(b'P't) b' NOTE: pO2(t = ∞) = pO1 =. pO1. (10) b' 1 + W b X/PO A NOTE: Eq'n (10) shows that the package-food system eventually reaches a steady-state which is different from the package interior reaching pO1. NOTE: Eq'n (10) could have been obtained by assuming steady state at eq'n (3). For achievement of a low pO(t = ∞): 1) PO and A should be as small as possible. 2) W and L should be as large as possible. NOTE: End of shelf-life (acceptable quality) can again be related to rrO. VOs/W = QOs = ∫rrO dt = ∫b pO2 dt (11) where Qos is the maximum amount of oxygen per unit food weight with which the food can react before it reaches end of shelf life.  4 J. M. Krochta FST 131 Simplified Steady-State Model Equation (3) can be simplified by assuming that: 1) No O2 accumulates in the package. Instead, it reacts with the food as soon as it enters the package. 2) The food-package system reaches steady-state quickly. 3) After reacting with a certain amount of O2, the food is unacceptable. (See Table 12-3.) This amount is QOs (cm3/g). And, the time it takes to reach QOs is ts (s). Recall: dpO2 Vu = PO A (pO1 - pO2) - rrO W (12) dt pT X Recognize that: rrO = QOs (13) ts At steady-state and pO2 = 0: PO A pO1 = QOs W (14) X ts Thus: ts = QOs W X (15) PO A pO1 NOTE: This approach gives a conservative value of ts, because it assumes the most rapid permeation possible (pO2 = 0).  List values for the components of the equation which you will use to solve the problem, always being sure to use units along with any value. These should be the values and units you will use with the equation. Use the equation with your selected values (including units) to arrive at an answer, and place a rectangle around your answer. Make sure your units cancel out appropriately to arrive at an answer with the correct units. O2 SENSITIVE FOOD >>> You are considering replacing the glass jar your company has used for its peanut butter with a plastic jar to eliminate the problem of jar breakage. You wish to use a plastic material that is recyclable, so you decide to evaluate PET and HDPE jars. The critical design issue is oxygen permeability through the container surface from the atmosphere to the peanut butter. The maximum quantity of oxygen that can be absorbed by the peanut butter before an unacceptable level of oxidative rancidity occurs is 100 ppm. The following information is essential to your evaluation: The jar contains 1 pound (454 g) of peanut butter. The jar surface area is 500 cm2. The closure of the jar can be considered impermeable, because it contains an aluminum foil layer that is well-sealed against the jar sealing surface. Thickness of the jar is 0.05 cm over the entire surface. The surrounding atmosphere is 21% oxygen. Assume that the jar headspace upon sealing is at 0% oxygen and remains at this value, because all oxygen that permeates into the jar is immediately absorbed and reacted by the peanut butter. The density of oxygen is 1.43 x 10-3 g/mL at 30°C. Assume a design temperature of 30°C, for which the 02 permeabilities of the PET and HDPE are 0.22 x 10-11 mL 0.cm/cm2.sec.cm Hg and 10.6 x 10-11 mL 0.cm/cm2.sec.cm Hg, 2 2 respectively. Calculate t he s he lf life o f t he p ean ut b utte r fo r a P ET j ar a nd for a H DPE j ar. >> Homework is due at beginning of class on Friday, November 20.  1 J. M. Krochta FST 131 PACKAGE DESIG N F OR M OIST UR E-SENSI TIVE F OODS The Food-Pa cka ge System Water content is central to definition, stability and quality of food categories. In foods packaged in polymer films, moisture: permeates through the packaging film, diffuses through the package free volume air space, and sorbs on, or desorbs from, and diffuses within the food. H2O in Package Film pW1 > pW2 PW Food A pW1 pW2 WS m = m{aW,T} H2O out pW1 < pW2 Permeation. The rate of moisture permeation depends on: 1) nature of the film (PW) 2) film thickness (X) 3) film area (A) 4) difference between outside and inside partial pressures (pW1 - pW2) 5) temperature (T) Food Moisture Sorption. The rate of moisture sorption depends on: 1) nature of the food (m = m{aW,T}) 2) food weight (WS) 3) water vapor partial pressure inside package (aW= pW/pWo) 4) temperature (m = m{aw,T})  2 J. M. Krochta FST 131 Food-Pa ckage Materia l Ba lan ce for W ater Vapor We make certain assumptions: 1) The package head space is very small. 2) The package head space offers little resistance to mass transfer, resulting in uniform moisture in the head space at any given time. 3) The resistance to moisture diffusion in the food is negligible, resulting in uniform moisture in the food at any given time. 4) The food moisture content (m) is in equilibrium with the package environment, and follows some simple relationship with the package water vapor partial pressure (pW2). 5) Temperature and relative humidity (water vapor partial pressure, pW1) of environment outside package are constant. Recall the general material balance equation: Input + Generation = Output + Accumulation (1) The resulting equation for moisture transport into a food-package system (Taoukis et al., 1988) is dm WS = PW A (pW1 - pW2) (2) dt X m is the moisture per unit dry weight of food (g water/kg dry solids) pW1 is the partial pressure of water vapor in the outside air (atm) pW2 is the partial pressure of water vapor inside the package (atm) PW is the permeability of the packaging film to water (g/m2.atm.d/m) WS is the weight of dry solids of the food in the package (kg dry solids) A is the package area (m2) X is the film thickness (m)  3 J. M. Krochta FST 131 Materia l Ba lan ce Solution We must now relate m with pW. Moisture Isotherm. The linear isotherm is as follows: m = b aW + c = b (pW/pWo) + c (3) b and c are constants specific to a given food at a given temp. pWo is the vapor pressure of pure water at the same temp. Using equation (3) for our system: pW2 = (pWo/b) m - pWoc/b (4) Also: pW1 = (pWo/b) me - pWoc/b (5) me is the moisture content of the food product which would be at equilibrium with the atmosphere outside the package. Therefore: pW1 - pW2 = (pWo/b)(me - m) (6) Substituting equation (6) into the material balance equation (2): dm WS = PW A pWo (me - m) (7) dt b X Rearranging equation (7): dm = PW A pWo dt (8) (me - m) b WS X The integrated result is ln (me - mi) = PW A pWo t (9) (me - m) b WS X  Show the equation necessary for solution of the problem. List values for the components of the equation that you will use to solve the problem, always being sure to use units along with any value. These should be the values and units you will use with the equation. Use the equation with your selected values (including units) to arrive at an answer, and place a rectangle around your answer. Make sure your units cancel out appropriately to arrive at an MOISTURE SENSITIVE answer with FOOD the correct units. >>> You have developed a new intermediate-moisture fruit bar. This product becomes unacceptable when it gains too much moisture, because then it becomes vulnerable to microbial growth. It also becomes unacceptable when it loses too much moisture, because then it becomes too hard to chew. You are most concerned about the latter for distribution of this product in a part of the country where it is likely to be stored at approximately 25°C and 20% relative humidity (RH). The data that you have collected on the product includes the following: The fruit bar moisture content immediately after production and just before packaging is 0.20 g water/g dry solids. Sensory testing has revealed that the fruit bar has significantly lower acceptability after it reaches 0.10 g water/g dry solids. Thus, this moisture content is selected to represent the limit of the product's shelf life. In establishing the 25°C moisture isotherm for the fruit bar (moisture content (dry basis) vs water activity), you find that the relevant part of the isotherm curve can be represented by a straight line with slope 0.12 g water/g dry solids per unit aw and that the moisture content of the fruit bar at 20% RH is 0.05 g water/g dry solids. The pouch required to package four fruit bars of 125 g (dry solids) each (total weight is thus 500 g) will have a total surface area of 0.15 m2. The vapor pressure of pure water at 25°C is 0.0313 atm (3.17 kPa). 1) Calculate t he va lue o f P W / X in ( g H 2 0) / (m 2. d a y. kP a) req u ire d t o a ch ie ve a s he lf life of 12 m on ths. 2) P W va lue s o f L DPE , PET & P VC a re 0.0 3 1, 0.168 & 0.6 17 ( g. mm )/ (m 2. da y. k Pa ), resp ect ive ly. Ass um ing that the pa ck a ge film t hick nes s is limite d t o 0.0 1-0.1 m m, dete rm in e w het he r L DPE , P ET o r P VC com es closes t to g ivin g a 1 2 mo. s helf life. >> H o me wo rk is due at be ginn ing o f cla ss o n Fr id a y, No ve m be r 13.  1 J. M. Krochta FST 131 PACKAGE DESIG N F OR OXYG EN- SENSITI VE F OODS The Food- Package System In foods packaged in polymer films, oxygen: 1) permeates (sorbs, diffuses, desorbs) through the packaging film, 2) diffuses through the package free volume air space, 3) sorbs on and diffuses within the food, and 4) reacts with the food Package Film PO pO1 pO2 pO2 A Food X O2 W pT rrO Vu QOs Film Oxygen Permeation. The rate of oxygen permeation depends on: 1) nature of the film (PO), 2) film thickness (X), 3) film area (A), 4) difference between external and internal oxygen partial pressures (pO1 - pO2) 5) temperature (PO = PO{T}) Food Oxidation. Rate of oxidative reactions depends on: 1) nature of the food (affects rrO and QOs) 2) food weight (W) 3) oxygen partial pressure inside package (pO2) 4) temperature (rrO = rrO{T}) 5) food water activity (aw)  2 J. M. Krochta FST 131 Unsteady-State Material Balance We make certain assumptions: 1) The permeation rate across the film can be described by the equation describing steady-state mass transport across a film. 2) The package headspace offers little resistance to mass transfer. Thus, it is assumed to have uniform concentration at any time. 3) The resistance to oxygen diffusion in the food is negligible. Thus, it is assumed to have uniform concentration at any time. 4) The food is in equilibrium with the package environment. Thus, the oxygen reaction can be expressed in terms of the package headspace pO2. 5) Constant temperature and constant food moisture content. ACCUMULATION = INPUT - OUTPUT + GENERATION For oxygen transport into a food-package system with oxygen reaction: dVOu = PO A (pO1 - pO2) - dVOr (1) dt X dt VOu = yO Vu = (pO2/pT) Vu (2) dpO2 Vu = PO A (pO1 - pO2) - rrO W (3) dt pT X VOu is the volume of oxygen in unfilled package volume (headspace) (cm3) Vu is unfilled package volume (headspace) (cm3) VOr is the volume of oxygen sorbed and reacted (cm3) pO1 is the partial pressure of oxygen in the external air (atm) pO2 is the partial pressure of oxygen in the package inside air (atm) pT is total pressure in the package (atm) PO is the film oxygen permeability (cm3/cm2.atm.s/cm) A is the package area (cm2) X is the film thickness (cm) W is the weight of the food placed in the package (g), rrO is the oxygen reaction rate (cm3/s.g) QOs is quantity of oxygen food can react with before unacceptable (cm3/g)  3 J. M. Krochta FST 131 Unsteady-State Material Balance Sol ution Often the oxidation kinetic expression can be approximated by a simple expression: rrO = b pO2 (4) where b is a constant for a specific food at a specific temperature. Substituting eq'n (4) into eq'n (3) and then rearranging: dpO2 = P' dt (5) pO1 - b' pO2 where P' = PO A pT/X Vu (6) and b' = 1 + W b X/PO A (7) Therefore: ln (pO1 - b' pO2o) = b' P' t (8) (pO1 - b' pO2) or pO2 = b' pO2o - pO1 + pO1 (9) b' exp(b'P't) b' NOTE: pO2(t = ∞) = pO1 =. pO1. (10) b' 1 + W b X/PO A NOTE: Eq'n (10) shows that the package-food system eventually reaches a steady-state which is different from the package interior reaching pO1. NOTE: Eq'n (10) could have been obtained by assuming steady state at eq'n (3). For achievement of a low pO(t = ∞): 1) PO and A should be as small as possible. 2) W and L should be as large as possible. NOTE: End of shelf-life (acceptable quality) can again be related to rrO. VOs/W = QOs = ∫rrO dt = ∫b pO2 dt (11) where Qos is the maximum amount of oxygen per unit food weight with which the food can react before it reaches end of shelf life.  4 J. M. Krochta FST 131 Simplified Steady-State Model Equation (3) can be simplified by assuming that: 1) No O2 accumulates in the package. Instead, it reacts with the food as soon as it enters the package. 2) The food-package system reaches steady-state quickly. 3) After reacting with a certain amount of O2, the food is unacceptable. (See Table 12-3.) This amount is QOs (cm3/g). And, the time it takes to reach QOs is ts (s). Recall: dpO2 Vu = PO A (pO1 - pO2) - rrO W (12) dt pT X Recognize that: rrO = QOs (13) ts At steady-state and pO2 = 0: PO A pO1 = QOs W (14) X ts Thus: ts = QOs W X (15) PO A pO1 NOTE: This approach gives a conservative value of ts, because it assumes the most rapid permeation possible (pO2 = 0).  List values for the components of the equation which you will use to solve the problem, always being sure to use units along with any value. These should be the values and units you will use with the equation. Use the equation with your selected values (including units) to arrive at an answer, and place a rectangle around your answer. Make sure your units cancel out appropriately to arrive at an answer with the correct units. O2 SENSITIVE FOOD >>> You are considering replacing the glass jar your company has used for its peanut butter with a plastic jar to eliminate the problem of jar breakage. You wish to use a plastic material that is recyclable, so you decide to evaluate PET and HDPE jars. The critical design issue is oxygen permeability through the container surface from the atmosphere to the peanut butter. The maximum quantity of oxygen that can be absorbed by the peanut butter before an unacceptable level of oxidative rancidity occurs is 100 ppm. The following information is essential to your evaluation: The jar contains 1 pound (454 g) of peanut butter. The jar surface area is 500 cm2. The closure of the jar can be considered impermeable, because it contains an aluminum foil layer that is well-sealed against the jar sealing surface. Thickness of the jar is 0.05 cm over the entire surface. The surrounding atmosphere is 21% oxygen. Assume that the jar headspace upon sealing is at 0% oxygen and remains at this value, because all oxygen that permeates into the jar is immediately absorbed and reacted by the peanut butter. The density of oxygen is 1.43 x 10-3 g/mL at 30°C. Assume a design temperature of 30°C, for which the 02 permeabilities of the PET and HDPE are 0.22 x 10-11 mL 0.cm/cm2.sec.cm Hg and 10.6 x 10-11 mL 0.cm/cm2.sec.cm Hg, 2 2 respectively. Calculate t he s he lf life o f t he p ean ut b utte r fo r a P ET j ar a nd for a H DPE j ar. >> Homework is due at beginning of class on Friday, November 20.  1 J. M. Krochta FST 131 PACKAGE DESIG N F OR M OIST UR E-SENSI TIVE F OODS The Food-Pa cka ge System Water content is central to definition, stability and quality of food categories. In foods packaged in polymer films, moisture: permeates through the packaging film, diffuses through the package free volume air space, and sorbs on, or desorbs from, and diffuses within the food. H2O in Package Film pW1 > pW2 PW Food A pW1 pW2 WS m = m{aW,T} H2O out pW1 < pW2 Permeation. The rate of moisture permeation depends on: 1) nature of the film (PW) 2) film thickness (X) 3) film area (A) 4) difference between outside and inside partial pressures (pW1 - pW2) 5) temperature (T) Food Moisture Sorption. The rate of moisture sorption depends on: 1) nature of the food (m = m{aW,T}) 2) food weight (WS) 3) water vapor partial pressure inside package (aW= pW/pWo) 4) temperature (m = m{aw,T})  2 J. M. Krochta FST 131 Food-Pa ckage Materia l Ba lan ce for W ater Vapor We make certain assumptions: 1) The package head space is very small. 2) The package head space offers little resistance to mass transfer, resulting in uniform moisture in the head space at any given time. 3) The resistance to moisture diffusion in the food is negligible, resulting in uniform moisture in the food at any given time. 4) The food moisture content (m) is in equilibrium with the package environment, and follows some simple relationship with the package water vapor partial pressure (pW2). 5) Temperature and relative humidity (water vapor partial pressure, pW1) of environment outside package are constant. Recall the general material balance equation: Input + Generation = Output + Accumulation (1) The resulting equation for moisture transport into a food-package system (Taoukis et al., 1988) is dm WS = PW A (pW1 - pW2) (2) dt X m is the moisture per unit dry weight of food (g water/kg dry solids) pW1 is the partial pressure of water vapor in the outside air (atm) pW2 is the partial pressure of water vapor inside the package (atm) PW is the permeability of the packaging film to water (g/m2.atm.d/m) WS is the weight of dry solids of the food in the package (kg dry solids) A is the package area (m2) X is the film thickness (m)  3 J. M. Krochta FST 131 Materia l Ba lan ce Solution We must now relate m with pW. Moisture Isotherm. The linear isotherm is as follows: m = b aW + c = b (pW/pWo) + c (3) b and c are constants specific to a given food at a given temp. pWo is the vapor pressure of pure water at the same temp. Using equation (3) for our system: pW2 = (pWo/b) m - pWoc/b (4) Also: pW1 = (pWo/b) me - pWoc/b (5) me is the moisture content of the food product which would be at equilibrium with the atmosphere outside the package. Therefore: pW1 - pW2 = (pWo/b)(me - m) (6) Substituting equation (6) into the material balance equation (2): dm WS = PW A pWo (me - m) (7) dt b X Rearranging equation (7): dm = PW A pWo dt (8) (me - m) b WS X The integrated result is ln (me - mi) = PW A pWo t (9) (me - m) b WS X  Show the equation necessary for solution of the problem. List values for the components of the equation that you will use to solve the problem, always being sure to use units along with any value. These should be the values and units you will use with the equation. Use the equation with your selected values (including units) to arrive at an answer, and place a rectangle around your answer. Make sure your units cancel out appropriately to arrive at an MOISTURE SENSITIVE answer with FOOD the correct units. >>> You have developed a new intermediate-moisture fruit bar. This product becomes unacceptable when it gains too much moisture, because then it becomes vulnerable to microbial growth. It also becomes unacceptable when it loses too much moisture, because then it becomes too hard to chew. You are most concerned about the latter for distribution of this product in a part of the country where it is likely to be stored at approximately 25°C and 20% relative humidity (RH). The data that you have collected on the product includes the following: The fruit bar moisture content immediately after production and just before packaging is 0.20 g water/g dry solids. Sensory testing has revealed that the fruit bar has significantly lower acceptability after it reaches 0.10 g water/g dry solids. Thus, this moisture content is selected to represent the limit of the product's shelf life. In establishing the 25°C moisture isotherm for the fruit bar (moisture content (dry basis) vs water activity), you find that the relevant part of the isotherm curve can be represented by a straight line with slope 0.12 g water/g dry solids per unit aw and that the moisture content of the fruit bar at 20% RH is 0.05 g water/g dry solids. The pouch required to package four fruit bars of 125 g (dry solids) each (total weight is thus 500 g) will have a total surface area of 0.15 m2. The vapor pressure of pure water at 25°C is 0.0313 atm (3.17 kPa). 1) Calculate t he va lue o f P W / X in ( g H 2 0) / (m 2. d a y. kP a) req u ire d t o a ch ie ve a s he lf life of 12 m on ths. 2) P W va lue s o f L DPE , PET & P VC a re 0.0 3 1, 0.168 & 0.6 17 ( g. mm )/ (m 2. da y. k Pa ), resp ect ive ly. Ass um ing that the pa ck a ge film t hick nes s is limite d t o 0.0 1-0.1 m m, dete rm in e w het he r L DPE , P ET o r P VC com es closes t to g ivin g a 1 2 mo. s helf life. >> H o me wo rk is due at be ginn ing o f cla ss o n Fr id a y, No ve m be r 13.  1 J. M. Krochta FST 131 PACKAGE DESIG N F OR OXYG EN- SENSITI VE F OODS The Food- Package System In foods packaged in polymer films, oxygen: 1) permeates (sorbs, diffuses, desorbs) through the packaging film, 2) diffuses through the package free volume air space, 3) sorbs on and diffuses within the food, and 4) reacts with the food Package Film PO pO1 pO2 pO2 A Food X O2 W pT rrO Vu QOs Film Oxygen Permeation. The rate of oxygen permeation depends on: 1) nature of the film (PO), 2) film thickness (X), 3) film area (A), 4) difference between external and internal oxygen partial pressures (pO1 - pO2) 5) temperature (PO = PO{T}) Food Oxidation. Rate of oxidative reactions depends on: 1) nature of the food (affects rrO and QOs) 2) food weight (W) 3) oxygen partial pressure inside package (pO2) 4) temperature (rrO = rrO{T}) 5) food water activity (aw)  2 J. M. Krochta FST 131 Unsteady-State Material Balance We make certain assumptions: 1) The permeation rate across the film can be described by the equation describing steady-state mass transport across a film. 2) The package headspace offers little resistance to mass transfer. Thus, it is assumed to have uniform concentration at any time. 3) The resistance to oxygen diffusion in the food is negligible. Thus, it is assumed to have uniform concentration at any time. 4) The food is in equilibrium with the package environment. Thus, the oxygen reaction can be expressed in terms of the package headspace pO2. 5) Constant temperature and constant food moisture content. ACCUMULATION = INPUT - OUTPUT + GENERATION For oxygen transport into a food-package system with oxygen reaction: dVOu = PO A (pO1 - pO2) - dVOr (1) dt X dt VOu = yO Vu = (pO2/pT) Vu (2) dpO2 Vu = PO A (pO1 - pO2) - rrO W (3) dt pT X VOu is the volume of oxygen in unfilled package volume (headspace) (cm3) Vu is unfilled package volume (headspace) (cm3) VOr is the volume of oxygen sorbed and reacted (cm3) pO1 is the partial pressure of oxygen in the external air (atm) pO2 is the partial pressure of oxygen in the package inside air (atm) pT is total pressure in the package (atm) PO is the film oxygen permeability (cm3/cm2.atm.s/cm) A is the package area (cm2) X is the film thickness (cm) W is the weight of the food placed in the package (g), rrO is the oxygen reaction rate (cm3/s.g) QOs is quantity of oxygen food can react with before unacceptable (cm3/g)  3 J. M. Krochta FST 131 Unsteady-State Material Balance Sol ution Often the oxidation kinetic expression can be approximated by a simple expression: rrO = b pO2 (4) where b is a constant for a specific food at a specific temperature. Substituting eq'n (4) into eq'n (3) and then rearranging: dpO2 = P' dt (5) pO1 - b' pO2 where P' = PO A pT/X Vu (6) and b' = 1 + W b X/PO A (7) Therefore: ln (pO1 - b' pO2o) = b' P' t (8) (pO1 - b' pO2) or pO2 = b' pO2o - pO1 + pO1 (9) b' exp(b'P't) b' NOTE: pO2(t = ∞) = pO1 =. pO1. (10) b' 1 + W b X/PO A NOTE: Eq'n (10) shows that the package-food system eventually reaches a steady-state which is different from the package interior reaching pO1. NOTE: Eq'n (10) could have been obtained by assuming steady state at eq'n (3). For achievement of a low pO(t = ∞): 1) PO and A should be as small as possible. 2) W and L should be as large as possible. NOTE: End of shelf-life (acceptable quality) can again be related to rrO. VOs/W = QOs = ∫rrO dt = ∫b pO2 dt (11) where Qos is the maximum amount of oxygen per unit food weight with which the food can react before it reaches end of shelf life.  4 J. M. Krochta FST 131 Simplified Steady-State Model Equation (3) can be simplified by assuming that: 1) No O2 accumulates in the package. Instead, it reacts with the food as soon as it enters the package. 2) The food-package system reaches steady-state quickly. 3) After reacting with a certain amount of O2, the food is unacceptable. (See Table 12-3.) This amount is QOs (cm3/g). And, the time it takes to reach QOs is ts (s). Recall: dpO2 Vu = PO A (pO1 - pO2) - rrO W (12) dt pT X Recognize that: rrO = QOs (13) ts At steady-state and pO2 = 0: PO A pO1 = QOs W (14) X ts Thus: ts = QOs W X (15) PO A pO1 NOTE: This approach gives a conservative value of ts, because it assumes the most rapid permeation possible (pO2 = 0).  List values for the components of the equation which you will use to solve the problem, always being sure to use units along with any value. These should be the values and units you will use with the equation. Use the equation with your selected values (including units) to arrive at an answer, and place a rectangle around your answer. Make sure your units cancel out appropriately to arrive at an answer with the correct units. O2 SENSITIVE FOOD >>> You are considering replacing the glass jar your company has used for its peanut butter with a plastic jar to eliminate the problem of jar breakage. You wish to use a plastic material that is recyclable, so you decide to evaluate PET and HDPE jars. The critical design issue is oxygen permeability through the container surface from the atmosphere to the peanut butter. The maximum quantity of oxygen that can be absorbed by the peanut butter before an unacceptable level of oxidative rancidity occurs is 100 ppm. The following information is essential to your evaluation: The jar contains 1 pound (454 g) of peanut butter. The jar surface area is 500 cm2. The closure of the jar can be considered impermeable, because it contains an aluminum foil layer that is well-sealed against the jar sealing surface. Thickness of the jar is 0.05 cm over the entire surface. The surrounding atmosphere is 21% oxygen. Assume that the jar headspace upon sealing is at 0% oxygen and remains at this value, because all oxygen that permeates into the jar is immediately absorbed and reacted by the peanut butter. The density of oxygen is 1.43 x 10-3 g/mL at 30°C. Assume a design temperature of 30°C, for which the 02 permeabilities of the PET and HDPE are 0.22 x 10-11 mL 0.cm/cm2.sec.cm Hg and 10.6 x 10-11 mL 0.cm/cm2.sec.cm Hg, 2 2 respectively. Calculate t he s he lf life o f t he p ean ut b utte r fo r a P ET j ar a nd for a H DPE j ar. >> Homework is due at beginning of class on Friday, November 20.