Packaging Material Analysis for Food Preservation

Packaging Material Analysis for Food PreservationTroullinos YannisPackaging paperIn this practical class, diverse films for food package were examined as far as their physical properties and their ability to preserve grapes, cheese, meat and potatoes. Appropriate measurements and tests we done on detail time separations.ResultsTable 1. Results for the rapid tests for the identification of packaging materials* Materials were identified using the packaging materials identification chart for films in the practical booklet.Table 2. Results of the mechanic and physical properties of the packaging materialDiscussionCalculations and Questions1. Calculate the tensile power of the three packaging materials tested.Table 3. Physical properties of different packaging materials* More than 25% difference from the meanBy using N = 100 g, Force mean values for each of the materials can be comprise. Also, Area = Width (m) x Gauge (m) = X m2. Tensile strength = Force (N) / Area (m2) so for the above materials we haveCellu regress 340 DMS Tensile strength = 25.6 / 2.125 x 10-6 = 12.0 x 106 N/m2 = 12.0 x 106 Pa = 12.0 MPa, as 1 N/m2 = 1 Pa, piece of music 1 MPa = 1,000,000 PaPolypropylene Tensile strength = 108.7 / 2.5 x 10-6 = 43.5 MPaPolyethylene Tensile strength = 12.5 / 0.75 x 10-6 = 16.7 MPa2.Define tensile strength and discuss what factors will affect the tensile strength of the packaging materialTensile strength is the maximum load that a material can support without fracture when being stretched, divided by the original cross-sectional argona of the material. Generally, as tensile strength increases, the tougher the material is considered (Hui, 2008). Factors affecting the tensile strength are (Yam, 2010 Fel low-spiriteds, 2009)Plasticiser levels (increased values give less tensile strength and more than elasticityDegree of crystallinity (crystal structure)Density of the material (increasing tautness gives more tensile strength)Manufacturing process (orientatio n, treatment, coatings)TemperaturePhysical properties of the material (branching, side groups, chain length, molecular weight)Duration of the time that the force is applied3.Compare your tensile strength results to those found in literature.According to Goodfellow Cambridge Ltd. tensile strength for regenerated cellulose is 50 MPa, which, as mentioned, is affected by a lot of different factors. In our experiment, tensile strength of the cellulose employ is a lot lower (12MPa).Paine (1990) gives values of 30 MPa for polypropylene, while in this experiment a value of 43.5 MPa was calculated.Finally, polyethylene gave an experimental value of 16.7 MPa, while Goodfellow Cambridge Ltd. reports 5-25 MPa for low density polyethylene (LDPE) and 15-40 MPa for high density polyethylene (HDPE). In this experiment it is unknown which exactly was the type of PE used, as there are many different types in market.As explained, duration of the force applied affects the tensile strength, so differen t testing machines give different results. There are numerous more factors as noted in question 2, which greatly affect the measurements and results. Thus, comparing values to literature cannot give objective judgement of the experiment.4.Calculate the moisture vaporisation infection rate (g m-2 day-1) for each of the films testedTable 4. Results of the water vaporisation permeableness testCircle area = r2 = 0.005 m2 (r = 40mm = 0.04m)Number of Days = 4, as Day 1 is the day we started the storageCellulose 340 DMS1st measurements congeries moisture gained = Weight of Day 5 Weight of Day 1 = 84.8 83.9 = 0.9 gMoisture gained per day = center moisture gained (g) / Nr Days = 0.9/4 = 0.225 g day-1Water evaporation permeability per 24h = Moisture gained per day / Circle Area = 0.225 / 0.005 = 45 g/m2 24h (1)2nd measurementsTotal moisture gained = 87.6 87.1 = 0.5 gMoisture gained per day = 0.5 / 4 = 0.125 g day-1Water vapour permeability per 24h = 0.125 / 0.005 = 25 g/m2 24h (2)Me an value of water vapour permeability per 24h = (1) + (2) / 2 = 35 g/m2 24hPolypropylene1st measurementsTotal moisture gained = 86.0 85.9 = 0.1 gMoisture gained per day = 0.1/4 = 0.025 g day-1Water vapour permeability per 24h = 0.025 / 0.005 = 5 g/m2 24h2nd measurementsTotal moisture gained = 87.1 87.1 = 0.0 gMoisture gained per day = 0.0 / 4 = 0 g day-1Water vapour permeability per 24h = 0 g/m2 24hMean value of water vapour permeability per 24h = 2.5 g/m2 24hPolyethylene1st measurementsTotal moisture gained = 84.6 84.5 = 0.1 gMoisture gained per day = 0.1/4 = 0.025 g day-1Water vapour permeability per 24h = 0.025 / 0.005 = 5 g/m2 24h2nd measurementsTotal moisture gained = 84.6 84.5 = 0.1 gMoisture gained per day = 0.1/4 = 0.025 g day-1Water vapour permeability per 24h = 0.025 / 0.005 = 5 g/m2 24hMean value of water vapour permeability per 24h = 5 g/m2 24h5.Discuss the results of the water vapour permeability test.Water vapour permeability is a measure for breathing roomability or for a textiles ability to canalize moisture. The results show that PP and PE have comparatively low water permeability, while cellulose has a lot more. These values agree with literature (Brennan and Grandison, 2012), which states that PP has lower permeability than PE. Cellulose is similarly stated as a low barrier of water vapour permeability. These results show that using cellulose to pack food sensitive to humidity such as powders is not considered wise.6.Discuss the results of the packaging and storage of fresh fruit experiment. Explain what is do the observed changes in the fruit and how the different packaging/storage conditions cultivate the ledge life of the fruit.Table 5. Fresh fruit (grapes) 3 days interval observationsFirstly, the tissues of fruits are alive after harvest and they only die through natural senescence, rotting or when they are consumed, cooked or similarly processed. All these tissues breath, a phenomenon called respiration with obvious relations t o maintenance of the quality and prolonging the shelf life of the product. Specifically, grapes do not respire very intensively and this is the reason they chafe harvested when they are ripe. Reducing respiration can extend the shelf life but stalling it will make tissues senesce and die. Cooling temperatures can also lower undesirable effects on fruits (Jongen, 2002).As far as grapes concerned, mould is primarly because of the fungus Botrytis cinerea.Browning spotted is a chemical process caused by specific enzymes changing the tissues comment to brown, while shrinkage is caused by increased respiration (tissues eventually lose water as shown in the weight measurements causing them to lose volume). Sweating is caused once again because of the respiration in packages where go down on permeability is low or very low.In the above experiments, it is shown that when using MS and fondness seal, grapes got sweaty in day 2 and 3, while in the same packaging with 2 holes, sudor was onl y slight. This makes sense as the 2 holes allowed the argument transfer between package and the environment, lowering the humidity because of the respiration in the package.In PE and heat seal, sweating was even more obvious as PE has lower gas permeability than MS.Finally, in the open tray, sweating was absent but mould started to show at day 3, as it partially did in the package with 2 holes. This was caused by a microorganism, probably fungus since grapes have low pH. another(prenominal) change which was spotted in the open tray was the soft, modify and oxidised appearance of the grapes because of the large amounts of respiration. manner temperatures and total contact with the environment allowed this level of respiration, lowering shelf life dramatically.7.What changes would you make to the packaging/storage conditions to extend the shelf life of the grapes?The most eventful change to the storage conditions would be to lower the storage temperature, as it would significantly reduce respiration. The package should not have holes, as they allow environmental air to get in allowing microorganisms to grow faster.8.Discuss the results of the packaging and storage of cheese experiment. Explain what is causing the observed changes in the cheese and how the different packaging/storage conditions influence the shelf life of the cheese.Table 6. Cheese 3 days interval observationsBrowning of cheese is significant in high storage temperatures (37C), less in medium (20C) and absent in low temperatures of 5C. Light causes the formation of lipid peroxides in medium temperatures, while compounds such as riboflavin are affected by light unrelated to storage temperature (Kristensen et al., 2001).Cheese tend to produce free oil when they melt and sweats during storage in relatively high temperatures because of the high humidity of it. When in open air sweating is more and drying out occurs (Wang and Sun, 2004).From the above, it becomes more obvious in ours experiments w hy cheese dried out during storage in open tray and why this drying out is more than in aluminium foil (which was not folded enough to keep air from contacting cheese). Another way to see the above is the greater loss of weight in open tray rather in aluminium foil. On the other hand, in twain MS and cryovac packages no drying out was noted, as can be seen from the differences in initial and final weight (0.1g).Relatively high storage temperatures (about 25C) caused the oiling and sweating of the cheese.9.What changes would you make to the packaging/storage conditions to extend the shelf life of the cheese?The storage temperature should be as low as about 5C (refrigerator) in dark and should be kept either in MS or cryovac packaging. Ideally, a modified atmosphere packaging should be used (Khoshgozaran et al., 2012), extending shelf life even more than the usual packages.10.Discuss the results of the packaging and storage of fresh meat experiment. Explain what is causing the observ ed changes in the meat and how the different packaging/storage conditions influence the shelf life of the meat.Table 7. Fresh meat 4 days intervals observationsFilmDayWeight (g)ChangesIn colourclouding upoverMoisteningInternal and external appearance of packageStoragetemp.Type ofspoilageGeneral appearanceof productStoragehumidity different change and/or notesPPAndHeat seal121.04C75%221.0slight green slight4Ccolour changes75%321.0slight green slight4Ccolour changes75%421.0slight green and browning